Chapter_03 ESP Motor

Electric Submersible Pumps Mohamed Dewidar 2013

Chapter 3

1

Submersible Motor

Table of Content

Section Content Page

1 General 3

2 Motor construction 3

21 Stator

22 Rotor

23 Rotor bearing

24 Motor thrust bearing

25 Pothead

3 Electromagnetism 8

31 Magnetic field

32 Magnetic flux and flux density

33 Magnetic field due to current in a solenoid

34 Changing polarity

35 Induced voltage

36 Electromagnetic attraction

4 Start coil arrangement 15

5 Power supply 15

51 Start

52 Time 1

53 Time 2

54 360 degree rotation

6 Mathematical analysis of rotating magnetic

field due to 3 phase current 19

7 Slip 24

8 Rotor current frequency 24

9 Magneto-motive force and magnetic field

Strength 25

10 Force in current carrying conductor in

magnetic field 26

11 Torque on a current carrying coil in

magnetic field 27

12 Theory of operation 28

13 Motor configurations 30

14 Motor current 32

15 Motor rating 32

16 Motor protection 35

17 Application of ESP motor 37


Chapter 3

2

18 Fundamentals of electricity 39

19 Equivalent circuit of induction motor 63

191 Effective circuit of induction motor at

Standstill

192 Effective circuit of induction motor under

Operating conditions (rotor is shorted)

193 Power relations

20 Determination of motor parameters 70

21 NEMA standard for squirrel cage IM 75

22 Torque of squirrel cage IM 77


Chapter 3

3

Submersible Motor

31 General

Motor is an electric machine which converts electric

energy into mechanical energy

Three phase induction motors are the most frequently

encountered in industry They are simple rugged low priced

and easy to maintain They run at essentially constant speed

from zero to full load The speed is frequency dependent

however variable speed electronic drives are being used more

and more to control the speed of the motors

ESP motor Classified as 3 phases squirrel cage 2 pole

induction Alternating current motor

The position of the motor in ESP integrity is just below the

protector (seal)

32 Motor Construction

The induction motor is a three phase squirrel cage two

pole induction design consists of

1 Stator which supports windings which receive energy from

the mains circuit

2 Rotor which carries windings in which working current is induced

3 Shaft which transfers the mechanical energy to the pump 4 Bearings 5 Housing 6 Insulated Magnet Wire 7 Thrust bearing

Fig (31) Most of the motor construction


Chapter 3

4

321 Stator

The stator is the stationary electrical part of the motor

The stator core of a National Electrical Manufacturers

Association (NEMA) motor is made up of several hundred thin

laminations

Fig (32) stator laminations

Stator laminations are stacked together forming a hollow

cylinder (fig 32) Coils of insulated wire are inserted

into slots of the stator core (fig 33)

Fig (33) stator core

Electromagnetism is the principle behind motor operation

Each grouping of coils together with the steel core it

surrounds form an electromagnet The stator windings are

connected directly to the power source

The stator winding consists of three individual windings

which overlap one another and are offset by an electrical

angle of 120deg (fig 34) When it is connected to the power

supply the incoming current will first magnetize the

stator This magnetizing current generates a rotary field

which turns with synchronous speed Ns


Chapter 3

5

Fig (34) stator windings

When the alternating current passes through a coil group a

magnetic field of fixed shape and sinusoidally varying

amplitude will result A magnetic pole is formed at the

center of this coil group The internal stator winding

connections determine the number of poles the voltage

applied to individual windings and the direction of

rotation In a three phase induction motor rotating

magnetic field is obtained by three separate single phases

with currents that differ in phase by 120 degrees

Three phases reach their maximum and minimum in a rapid

succession sequence As currents change the effect is to

rotate the magnetic fields The magnetic field rotates

continuously at a constant speed determined by the line

frequency and number of poles

The laminations wound with three very big loops of wire one

for each phase When current is flowing through a phase

magnetic flux is induced as shown in fig (35)

Fig (35) induced magnetic flux due to current flow

Because of this configuration the inside of the stator

holds a strong magnetic field

The strength of the field will depend on the amount of

current flowing through the wire loop (ie the phase

winding)


Chapter 3

6

The more copper that is in the stator the more the winding

losses are reduced making the motor more efficient

The winding is two pole because two magnetic poles are

created (one North and one South) Motors can be wound

differently to create more than two poles such as a four

pole motor

Remember that the direction of the magnetic field in the

stator depends on the direction of current flowing in the

wire

With AC or Alternating Current the direction of current

flow is changing 60 times every second for 60 Hz power (or

50 times per second for 50 Hz power)

322 Rotor

The rotor is the rotating part of the electromagnetic

circuit

The most common type of rotor is the ldquosquirrel cagerdquo rotor

Fig (36) squirrel cage rotor

The rotor consists of a stack of steel laminations

The squirrel cage rotor consists of copper or aluminum bars

accommodated in slots of rotor core (fig 37)

Fig (37) rotor core


Chapter 3

7

At each end the bard are connected to heavy conducting end

rings which serve the purpose of short circuit bars In the

absence of the end rings emfs would be induced in the rotor

bars but no current would flow through them and no torque

would be produced

The wound rotor has a 3 phase winding placed in the slots of

the rotor core The rotor is wound for the same number of

poles as the stator

The terminals of the rotor winding are brought out of three

slip rings mounted on the machine shaft

During running condition the slip ring are short circuited

so as to close the rotor circuit

The air gap or more accurately the clearance between the

stator and rotor should be as small as possible in order to

the primary and secondary leakage fluxes to minimum

323 Rotor bearing

Rotor Bearings are one of the most vital parts of the

motor The Bearing Material is Babbitt-lined steel and

machined after processing There are fluid holes to insure oil

circulation and wide angle oil grooves on the OD to distribute

lubrication evenly over the entire length of the bearing

surface

Fig (38) rotor bearing

The bearing sleeve is a bronze material for the sleeve

construction of the bearing This part is keyed to the shaft

and the hole on the sleeve is aligned with the hole on the

shaft to insure proper cooling and lubrication


The motor thrust bearing is installed at the top of the

rotor string It is designed to hold the weight of the entire

rotor string


Chapter 3

8

Fig (39) motor thrust bearing

Currently three types of motor thrust bearings are used

Babbitt

Glacier

KMC Bronze pads The thrust bearing limits on the system will indicate the type

of load required for the selected bearing material

325 Pothead

Pothead is the place where the motor lead extension

cable is connected to the motor three phase windings (fig

310) There are two types of pothead they are

1 Tape in type where tape wrapped around individual connector leads inside motor

2 Plug in type where mating block mounted in motor

Fig (310) motor pothead

33 Electromagnetism

331 Magnetic field

When an electric current is passed through a conductor

a magnetic field is set up around the conductor The direction


Chapter 3

9

of the magnetic field can be found by using right hand rule or

the right hand screw The right hand rule states ldquoGrasp the

wire in the right hand with the thumb pointing in the

direction of the current The fingers will curl around the

wire in the direction of the magnetic fieldrdquo

The right hand screw is explained in this way as a wood screw

is turned clockwise it progresses into the wood The

horizontal direction of screw is analogous to the direction of

current in a conductor The circular motion of the screw shows

the direction of the magnetic flux around the conductor (fig

311)

Fig (311) Magnetic field around the conductor carrying current

In fig 312 (a) the dot inside the circle is the standard

symbol used to show that the direction of current flow is out

of the page Then by right hand rule the magnetic field is

counterclockwise In fig 312 (b) the cross is the standard

symbol used to show that the current is entering the page The

magnetic field is clockwise The strength of the magnetic

field is proportional to the current ie if the current is

doubled the magnetic field will be doubled

(a) Current coming out of the page (b) Current entering the page

Fig (312) Magnetic field surrounding the conductor


Chapter 3

10

Since a current carrying conductor has a magnetic field around

it when two current carrying conductors are brought close

together there will be interactive between the fields When

the currents in the two conductors are in opposite direction

the fields are as shown in fig 313 (a) and the force of

repulsion is experienced

When the currents are in the same direction the field are as

shown in fig 313 (b) and a fore of attractive is experienced

Fig (313) force between parallel current carrying conductors

Consider a single turn coil carrying current As shown in

figure 35 the hole of the magnetic flux generated by electric

current passed through the centre of the coil Therefore the

coil acts like a little magnet and has a magnetic field with

identifiable N and S poles The coil may also have more than

one turn The flux generated by each of the individual turns

of the coil tends to link up and pass out of one end of coil

and back into the other end Such an arrangement is known as

solenoid (figure 36) and has a magnetic field pattern very

similar to that of bar magnet The right hand rule for

determining the direction of flux from solenoid states ldquoGrasp

the solenoid in the right hand such that the fingers point in

the direction of current flow in the coil The thumb will

point towards N pole of fieldrdquo

As discussed above a current flow in the conductors produces

a magnetic field The converse is also possible that is a

magnetic field can produce a current in a conductor This is

known as the phenomenon of electromagnetic induction (this

will be discussed later)


Chapter 3

11


The total lines of force in a magnetic field are called

magnetic flux Flux density is the flux per unit area of cross

section

Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the

SI unit of flux density and represents Wbm2 Flux density is

also known as magnetic induction From its definition

AB

Where B is the flux density in teslas φ is the total flux in

webers and A is the cross sectional area in m2


When an electric current passed through a solenoid the

resultant magnetic flux is very similar to that of a bar

magnet The magnetic flux lines make complete circuit inside

and outside the coil each line is a closed path The side at

the flux emerges is the North Pole the other side where the

magnetic flux reenters is the South Pole

The strength of the magnetic field in the DC electromagnet can

be increased by increasing the number of turns andor current

in the coil The greater the number of turns the stronger the

magnetic field will be See fig (314) and (315)

Fig (314) Magnetic field in coils of different number of

turns


Chapter 3

12

Fig (315)

Magnetic field in coils of different currents

The magnetic flux density in the interior of a solenoid

carrying an electric current depends on the current intensity

passing through the coil (I) and number of turns per unit

length (n) ie B is proportionally change with I an n

InB

Where μ is the permeability of the core material The equation

can be written as follows

Where N is the number of turns of a solenoid and l is its

length


The magnetic field of an electromagnet has the same

characteristics as a natural magnet including a north and

South Pole However when the direction of current flow

through the electromagnet changes the polarity of the

electromagnet changes The polarity of an electromagnet

connected to an AC source will change at the same frequency as

the frequency of the AC source This can be demonstrated in

the following illustration (fig 37) At Time 1 current flow

is at zero There is no magnetic field produced around the

electromagnet At Time 2 current is flowing in a positive

direction A magnetic field builds up around the

electromagnet The electromagnet assumes a polarity with the

South Pole on the top and the North Pole on the bottom At

Time 3 current flow is at its peak positive value The

strength of the electromagnetic field is at its greatest

value At Time 4 current flow decreases and the magnetic field


Chapter 3

13

begins to collapse until Time 5 when current flow and

magnetic field are at zero Current immediately begins to

increase in the opposite direction At Time 6 current is

increasing in a negative direction The polarity of the

electromagnetic field has changed The north pole is now on

top and the south pole is on the bottom The negative half of

the cycle continues through Times 7 and 8 returning to zero

at Time 9 This process will repeat 50 times a second with a

50 Hz AC power supply (fig 316)

Fig (316)

335 Induced voltage

A conductor moving through a magnetic field will have a

voltage induced into it This electrical principle is used in

the operation of AC induction motors In the following

illustration an electromagnet is connected to an AC power

source Another electromagnet is placed above it The second

electromagnet is in a separate circuit There is no physical

connection between the two circuits Voltage and current are

zero in both circuits at Time 1 At Time 2 voltage and current

are increasing in the bottom circuit A magnetic field builds

up in the bottom electromagnet Lines of flux from the

magnetic field building up in the bottom electromagnet cut

across the top electromagnet A voltage is induced in the top

electromagnet and current flows through it At Time 3 current

flow has reached its peak Maximum current is flowing in both

circuits The magnetic field around the coil continues to


Chapter 3

14

build up and collapse as the alternating current continues to

increase and decrease As the magnetic field moves through

space moving out from the coil as it builds up and back

towards the coil as it collapses lines of flux cut across the

top coil As current flows in the top electromagnet it creates

its own magnetic field (fig 317)

Fig (317) magnetic field increases as the current increases


The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field

in the bottom electromagnet Since opposite poles attract the

top electromagnet will follow the bottom electromagnet when it

is moved (fig 318)

Fig (318)


Chapter 3

15

34 Start coil arrangement

The following schematic (fig 319) illustrates the

relationship of the coils In this example six coils are used

two coils for each of the three phases The coils operate in

pairs The coils are wrapped around the soft iron core

material of the stator These coils are referred to as motor

windings Each motor winding becomes a separate electromagnet

The coils are wound in such a way that when current flows in

them one coil is a north pole and its pair is a south pole

For example if A1 were a north pole then A2 would be a south

pole When current reverses direction the polarity of the

poles would also reverse

Fig (319)

35 Power supply

The stator is connected to a 3-phase AC power supply In

the following illustration phase A is connected to phase A of

the power supply Phase B and C would also be connected to

phases B and C of the power supply respectively

Fig (320)

Phase windings (A B and C) are placed 120deg apart In this

example a second set of three-phase windings is installed


Chapter 3

16

The number of poles is determined by how many times a phase

winding appears In this example each phase winding appears

two times This is a two-pole stator If each phase winding

appeared four times it would be a four-pole stator

Fig (321) 2 poles stator winding

When AC voltage is applied to the stator current flows

through the windings The magnetic field developed in a phase

winding depends on the direction of current flow through that

winding The following chart is used here for explanation

only It will be used in the next few illustrations to

demonstrate how a rotating magnetic field is developed It

assumes that a positive current flow in the A1 B1 and C1

windings result in a north pole

Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start

It is easier to visualize a magnetic field if a start

time is picked when no current is flowing through one phase

In the following illustration for example a start time has

been selected during which phase A has no current flow phase

B has current flow in a negative direction and phase C has

current flow in a positive direction Based on the above

chart B1 and C2 are south poles and B2 and C1 are north

poles Magnetic lines of flux leave the B2 North Pole and

enter the nearest South Pole C2 Magnetic lines of flux also


Chapter 3

17

leave the C1 North Pole and enter the nearest South Pole B1

A magnetic field results as indicated by the arrow fig

(322)

Fig (322) start

352 Time 1

If the field is evaluated at 60deg intervals from the

starting point at Time 1 it can be seen that the field will

rotate 60deg At Time 1 phase C has no current flow phase A has

current flow in a positive direction and phase B has current

flow in a negative direction Following the same logic as used

for the starting point windings A1 and B2 are north poles and

windings A2 and B1 are south poles fig (323)

Fig (323) time 1


Chapter 3

18

353 Time 2

At Time 2 the magnetic field has rotated 60deg Phase B

has no current flow Although current is decreasing in phase A

it is still flowing in a positive direction Phase C is now

flowing in a negative direction At start it was flowing in a

positive direction Current flow has changed directions in the

phase C windings and the magnetic poles have reversed

polarity

Fig (324) time 2


At the end of six such time intervals the magnetic

field will have rotated one full revolution or 360deg This

process will repeat 60 times a second on a 60 Hz power supply


Chapter 3

19

Fig (325) time 1

36 Mathematical analysis of rotating magnetic field due to 3 phase current

When a 3-phase winding is energized from a 3-phase supply

a rotating magnetic field is produced This field is such that

its poles do no remain in a fixed position on the stator but

go on shifting their positions around the stator

For this reason it is called a rotating magnetic field It

will be shown that magnitude of this rotating field is

constant and is equal to 15 fm where fm is the maximum flux

due to any phase

To see how rotating field is produced consider a 2-pole 3

phase winding as shown in fig 326(i) The three phases A B

and C are energized from a 3-phase source and currents in

these phases are indicated as IA IB and IC Referring to Fig

326 (ii) the fluxes produced by these currents are given by

Here φm is the maximum flux due to any phase We shall now

prove that this 3-phase supply produces a rotating field of

constant magnitude equal to 15 φm


Chapter 3

20

(i) (ii)

Fig (326)

At start fig 326 (ii) and fig 327 (i) the current in phase

A is zero and currents in phases B and C are equal and

opposite The currents are flowing outward in the top

conductors and inward in the bottom conductors This

establishes a resultant flux towards right The magnitude of

the resultant flux is constant and is equal to 15 φm as

proved under

At start ωt = 0deg Therefore the three fluxes are given by

(i) (ii)

Fig (327)


Chapter 3

21

The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that

Fig (328)

At time 1 fig (15 (ii)) ωt = 60deg Therefore the three

fluxes are given by

The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that

Fig (329)


Chapter 3

22

At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by

The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that

It follows from the above discussion that a 3-phase supply

produces a rotating field of constant value (= 15 φm where φm

is the maximum flux due to any phase)

Fig (330)

We shall now use another useful method to find the magnitude

and speed of the resultant flux due to three-phase currents

The three-phase sinusoidal currents produce fluxes φ1 φ2 and

φ3 which vary sinusoidally The resultant flux at any instant

will be the vector sum of all the three at that instant

The fluxes are represented by three variable magnitude

vectors fig (331)

In fig (331) the individual flux directions are fixed but

their magnitudes vary sinusoidally as does the current that

produces them To find the magnitude of the resultant flux

resolve each flux into horizontal and vertical components and

then find their vector sum


Chapter 3

23

Fig (331)

The resultant flux is given by

Thus the resultant flux has constant magnitude (= 15 φm) and

does not change with time The angular displacement of φR relative to the OX axis is

So

Thus the resultant magnetic field rotates at constant angular

velocity ω(= 2πf) radsec For a P-pole machine the rotation

speed (ωs) is


Chapter 3

24

Thus the resultant flux due to three-phase currents is of

constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a

synchronous speed of 120fP rpm

For example for a 2-pole 50 Hz 3-phase induction motor N

= 120x502 = 3000 rpm It means that flux rotates around the

stator at a speed of 3000 rpm

37 Slip

We have seen above that rotor rapidly accelerates in the

direction of rotating field In practice the rotor can never

reach the speed of stator flux If it did there would be no

relative speed between the stator field and rotor conductors

no induced rotor currents and therefore no torque to drive

the rotor The friction and windage would immediately cause

the rotor to slow down Hence the rotor speed (N) is always

less than the suitor field speed (Ns) This difference in speed

depends upon load on the motor

The difference between the synchronous speed Ns of the rotating

stator field and the actual rotor speed N is called slip It

is usually expressed as a percentage of synchronous speed

ie

The quantity Ns-N is sometimes called slip speed

When the rotor is stationary (ie N = 0) slip s = 1 or 100

In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a

constant-speed motor

38 Rotor current frequency

The frequency of a voltage or current induced due to the

relative speed between a vending and a magnetic field is given

by the general formula

Where

n = Relative speed between magnetic field and the winding

P = Number of poles


Chapter 3

25

For a rotor speed N the relative speed between the rotating

flux and the rotor is Ns-N Consequently the rotor current

frequency fr is given by

ie Rotor current frequency = slip x Supply frequency

When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply

frequency (fr = sf = 1 fr = f)

As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip

S and hence rotor current frequency decreases

39 Magneto-motive force and magnetic field strength

An emf causes a current to flow in an electric circuit

Similarly a magneto motive force (mmf) symbol F produces a

magnetic flux in a magnetic circuit The mmf of a coil is the

product of current in the coil and the number of turns of the

coil and has the unit of ampere turns (AT)

The magnetic flux which can be set up in a magnetic circuit

depends on the mmf and the length of the magnetic circuit If

the length is large the mmf has to act over a long distance

and the resulting magnetic flux is small The magnetic field

strength H is defined as the mmf per unit length of magnetic

circuit ie

Where I is the current in amperes and N is the number of

turns and l is the length of the magnetic circuit in meter

Example

The total flux of an electro magnet is 4x10-4 Wb

a If the cross sectional area of the core is 1 cm2 find the flux density in the core

b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is

20 cm find the mmf and the magnetic field strength

Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT

Field strength 2501020

502

m

AT

l

NIH ATm

310 Force in current carrying conductor in magnetic field

Figure 332 (a) shows a current carrying conductor (the

current entering the page) laying in magnetic field flux

density B The current in the conductor sets up a flux in a

clockwise direction around the conductor When the external

field is in the vertically downward direction the field of

the conductor assists the external field on the right hand

side of the conductor The effect of this is to produce a

force that pushes the conductor to the lift If the direction

of the current is reversed as shown in figure 36 (b) the flux

around the conductor is in counterclockwise direction and the

resulting force pushes the conductor to the right In both

cases maximum force is generated if the conductor is at right

angle to the direction of the magnetic flux The force is

always in a direction perpendicular to both the conductor and

the field

The magnitude of the force F is given by

Where B is the flux density in telsas I is the current in

amperes and l is the length of the conductor in meters

A force of one Newton is exerted on a 1 meter long conductor

carrying a current of 1 ampere and situated at right angle to

a magnetic field having a flux density of 1 tesla

(a) Conductor current entering the page (b) Conductor current coming out of the page

Fig (332) Force on a current carrying conductor in a magnetic

field

+ Force

Flux set up

by current

in on

conductor Flux set up

by current

in on

conductor

Force


Chapter 3

27

If the conductor is not perpendicular to magnetic field but

inclined at angle Ө to the magnetic field the force is given

by

311 Torque on a current carrying coil in magnetic field

Figure 333 shows a current carrying coil placed in

magnetic field From the discussion in section 35 it follows

that a downward force is exerted on the left hand conductor

and an upward force is exerted on the right hand conductor

The force on each conductor is given by the equation 33 The

total force is given by

Fig (333) Force on a coil carrying current in a magnetic

field

If the coil has N turns the force is

Since the force is acting at a radius r meters the torque on

the coil is

The above simple arrangement is the basic part of electric

measuring instrument The operation of an electric motor is

also based on this principle

Example

A 30 cm long conductor is carrying a current of 10 A and

situated at a right angle to a magnetic field having flux

density of 08 T Calculate the force on the conductor

F = 08 x 10 x 30 x 10-2 = 24 N

Example

A 200 turn coil having an axial length of 3 cm and radius of 1

cm is pivoted in magnetic field having a flux density of 08

S

N

I

2r

Flux


Chapter 3

28

T The coil carries a current of 05 A Calculate the torque

acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm

Figure 334 is the same as figure 333 but with other

direction of the magnetic flux

Fig (334) Force on a coil in a magnetic field

If the loop is in line with the magnetic field the secondary

magnetic field will be perpendicular to the main field This

will cause two equal and opposite forces (a torque) on the

loop causing it to rotate until the forces balance (fig 335)

The forces will reach a steady state and hold the magnet in

place as long as current is applied

To cause rotation the field must rotate This is accomplished

with the alternating current where the field is rotated

This is accomplished with the alternating current going

through the windings in the stator of the induction motor


312 Theory of operation

An induction motor consists of a stator and rotor The

stator carries a 3 phase winding in its slots and is connected

to a 3 phase supply The rotor carries a cage winding and it

is free to rotate within the stator


Chapter 3

29

The 3 phase currents flowing in the stator winding produce a

rotating field The rotor winding cuts the rotating field and

an emf is induced in the rotor winding When the rotor is at

rest the frequency of this emf is the same as the supply

frequency If the rotor circuit is closed a current flows in

the rotor winding The rotor current produces an mmf which

rotates and is directed in opposition to stator mmf The

interaction of the stator and the rotor fields produces a

torque which causes the rotor to rotate in the direction of

the rotor field

If the rotor shaft is not loaded the machine has only to

rotate itself against the mechanical losses and the rotor

speed is very close to the synchronous speed However the

rotor speed cannot become equal to the synchronous speed

because if it does so the emf induced in the rotor winding

would become zero and there will be no torque Hence the rotor

speed remains slightly less than the synchronous speed

If the motor shaft is loaded the rotor will slow down and the

relative speed of rotor with respect to the stator rotating

field will increase The emf induced in the rotor winding will

increase and this will produce more rotor current which will

increase the electromagnetic torque produced by the motor

Conditions of equilibrium are attained when the rotor speed

has adjusted to a new value so that the electromagnetic torque

is sufficient to balance the mechanical or load torque applied

to the shaft The speed of the motor when running under full

load conditions is somewhat less than the no load speed

The speed of rotation of the field mmf is called synchronous

speed is related to the frequency and number of poles by the

expression

np

f s

2

Where

f is the frequency in Hz

p is the number of poles and

ns is the synchronous speed in revolution per second

An alternative form of above expression is

p

fN s

120

Where

Ns is the synchronous speed in revolution per minute (rpm)


Chapter 3

30

313 Motor configurations

Motors come in single sections (head and base) as well as

tandem configurations

The tandems can include the UT (upper tandem - head but no

base) the CT (center tandem - no head or base) or the LT

(lower tandem - base but no head)

An upper tandem motor can be used as a single section if it is

completed on the bottom with either a Universal Motor Base

(UMB) or Downhole Monitoring System (Sensor)

If additional horsepower is required over what can be achieved

in one piece a CT or LT motor can be added

Submersible electric motors can be designed in tandem

configuration to create the desired Horsepower required for

each application

So based on the above there are four different motor

sections they are

1 Single section

Where the motor has head and base a certain horsepower we

cannot increase and could not attach any equipment below

2 Upper Tandem motor (UT)

UT motor has head no base (open circuit) and the

horsepower can be increased by adding another central tandem

(CT) or lower tandem (LT) The circuit must be closed either

by Universal Motor Base (UMB) or sensor

3 Central Tandem (CT)

CT motor has no head no base and cannot use alone UT must

be attached on top of it Another central tandems or LT can

be attached on the bottom of it

4 Lower Tandem (LT)

Where the motor has a base no head and cannot use alone

it must be attached with upper tandem (as there is no head

attached)


Chapter 3

31

Single UT CT

Fig (336) motor configurations

Notes

1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a

particular Hp if we have a 1000 V 50 A motor a 2000 V

motor would be 25 amps and a 500V motor would be 100 amps

In other words KVA is constant

2 When putting more than one motor together in tandem

combinations always keep the sections the same Hp and

voltage For example a 300 Hp 540 motor should be made of

two 150 Hp motors

3 With two motors we double the Hp (add the two Hps

together) We also double the voltage but the amperage

remains the same With three motors we triple the Hp and

voltage but the amperage still does not change

For example a 140 Hp 1299 V 695 A UT motor coupled to a

140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V

695 A motor

4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not

try to put 3500 volts on a 3 kV cable Surface controllers

transformers wellhead feedthru mandrels etc will all have

voltage limits we need to be concerned with

5 For any given Hp there will be several voltages and

amperages available why have more than one voltage


Chapter 3

32

The answer is not in the motor but in the power cable Lower

voltage means higher current and this results in higher

voltage lost in the power able

So even though the motor efficiency does not change the

overall system efficiency will decrease with higher

amperage

If the amperage is too high the motor may not even be able

to start as we will see when we discuss power cable chapter

6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating

314 Motor current

Induction motor current consists of reactive (magnetizing)

and real (torque) components

The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor

close to 100

The magnetizing current would be purely inductive except

that the winding has some small resistance and it lags the

voltage by nearly 90deg

The magnetizing current has a very low power factor close to zero

The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total

current is approximately the same for all loads

The torque current increases as the load increases

At full load the torque current is higher than the

magnetizing current

For a typical motor the power factor of the resulting

current is between 85 and 90

As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting

current has a lower power factor

The smaller the load the lower the load current and the

lower the power factor Low power factor at low loading

occurs because the magnetizing remains approximately the same

at no load as at full load

315 Motor rating

If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break

down the various sizes into several voltages and amperages as


Chapter 3

33

the table below

KMH 562 SERIES MOTORS

HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348

Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are

bottom-hole temperature and fluid flow rate past the motor in

this industry

The motor will put out exactly as much horsepower as the pump wants no more and no less

Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo

In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes

We in fact use this information in the form of an amp chart

to see how the motor performs at downhole

We can very easily anticipate this relationship by simply

looking at the equation for motor horsepower


Chapter 3

34

Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear

relationship

In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary

the greater will be the change in amperage

One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the

motor winding which is not good for efficiency

There will be a practical limit to how far this can continue

We can look at the laminations to understand the basics of

this concept As we increase current on a motor we increase

the flux density induced in the laminations

Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the

motor

Fig (337)

If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as

shows in fig 338

Fig (338)


Chapter 3

35

If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is

called SATURATION

Any more horsepower beyond this point will severely overheat

our motor

Another practical consideration on rating a motor is the

speed We know that the motor will slow down with load If

the motor speed is too low we will lose pump performance

so we must set the Hp at a point where the speed is

acceptable

One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must

be dissipated by the fluid flowing past the outside of the

motor

Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at

different rates since not all the materials are the same

Even if the motor were all one material expansion would vary

since the internal temperature changes within the motor

itself

The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with

(overheating) tolerances might be exceeded and we could have

bearing failures or other damage

316 Motor protection

In this discussion we will address proper protection for ESP motors operating down-hole

Motor controllers can provide very simple protection to very sophisticated protection

Simple controllers will look at overload and under-load

conditions only

More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power

factor kw back-spin and many others

In either type of controller overload and under-load

protection is of primary importance and it is critical that

these both be set correctly in order to properly protect the

motor from damage

Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb

While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some

particular examples where the standard rule-of-thumb settings


Chapter 3

36

may not give adequate protection

Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V

355 Amp motor Should we set the OL at 115 of 355A and

the UL at 80 of 355A

Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the

motor should draw only 32 amps rather than the 355 of the

nameplate This should be the basis for our calculations and

settings

If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively

Would this protect this unit

At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps

or 22 amps Since our UL is set at 256 the motor should

trip on UL if the well is shut-in

Note that a shut-in pump is NOT a no-load condition for the

motor What amperage would you expect for this motor if the

pump shaft were broken at the intake (a true no-load

condition)

For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good

indication of a broken shaft However even if the current

reads higher than this it still could be a broken shaft so

do not rule that out on this basis alone

Lets take another case Say we have a 150 stage DN1750

producing a fluid of 086 gravity on a 456 series 50 Hp

885V 355A motor Should we set the OL at 115 of 355A and


This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the

motor should draw only 34 amps rather than the 355 amp

nameplate rating This should be the basis for our

calculations



Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of

NP amps or 28 amps Since our UL is set at 272 the motor

may not trip on UL if the well is shut-in In this case we

should set the UL higher

The point is that these standard rules-of-thumb are not

always perfect Every application should be considered


Chapter 3

37

independently to ensure that the settings selected are

adequate to properly protect the down-hole equipment

317 Application of ESP Motors

Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of

Centrilift and Reda motors

With all these choices which motor should we use for a given application

The process to select the best motor for the application will depend on the economic compromises of the user but in

general after defining the customers objectives and the pump

horsepower load for the application we can resume the

process of selection of the motor as an iterative process

which includes

Motor Series

Motor Type

Motor configuration Voltage and Amperage

Actual motor performance amp Operating Temperature and

compare against max temperature


Chapter 3

38

All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water

Lower flow rates or higher oil cut can lower the effective Hp rating

After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the

HP requirement of the pump

Now we should look at Volts and Amps

For any given Hp there will be several voltages and amperages available

For any particular horsepower the product of the volts and amps will be essentially constant

For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100

amps In other words KVA is constant High voltage motors

(single motors) are no more or less efficient than low

voltage motors so why have more than one voltage

The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher

voltage lost in the power able So even though the motor

efficiency does not change the overall system efficiency

will decrease with higher amperage

If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable

This explains why the various voltages but why such odd

voltages Surface motors for example are rated at 460V

4160V 2300V etc

These motors are made to standard voltages So why do the

motor voltages turn out to be such strange numbers


Chapter 3

39

In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a

very long length of power cable which surface motors do not

If we have a 460 V surface supply we would probably only

want about 430V down-hole (for a low Hp motor) to give us the

necessary 460 V at the surface including the cable loss

So in determining motor voltages we are really limited by surface equipment

Motor control panels come in certain voltage ranges such as

600V 1000V 1500V 2400V etc

Motor voltages are selected assuming a length of cable such

that the total voltage (motor plus cable loss) will fall just

below one of the panel ratings

Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating

Higher voltage motors require smaller gauge wire and very low

Hp motors simply cannot be wound at very high voltages

because the wire would be too small to work with

As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can

include the UT (upper tandem head but no base) the CT

(center tandem no head or base) or the LT (lower tandem

base but no head) An upper tandem motor can be used as a

single section if it is completed on the bottom with either a

Universal Motor Base (UMB) or Downhole Monitoring System

(Sensor)

If additional horsepower is required over what can be

achieved in one piece a CT or LT motor can be added

318 Fundamentals of electricity

This section is not an attempt to present a course in

electricity but is intended as a review of the terms and

basic formulas associated with ESP applications

Electricity

Since the electrons are normally distributed evenly

throughout a substance a force called electromotive force

(emf) is required to detach them from the atoms and make them

flow in a definite direction This force is also often called

ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this

electromotive force is the ldquovoltrdquo

The higher the voltage the greater the number of electrons

which will be caused to flow

It has been found experimentally that the charge on a single

electron is 1602e-19 coulomb Hence when a current of 1


Chapter 3

40

ampere (or 1 coulombsecond) flows in a conductor the number

of electrons passing any given point must be such that

1602x10-19

x no of electronssecond = 1 coulombsecond

So no of electronssecond = 624e18

Ie when the current in a circuit is 1 ampere electrons are

passing any given point of the circuit at the rate of 624e18

per second

When a potential or voltage of sufficient strength is applied

to a substance it causes the flow of electrons This flow of

electrons is called an electric current The rate of this flow

of current is measured in amperes An ampere is the rate of

flow of electric current represented by movement of a unit

quantity of electricity (coulomb) per second

Every substance is a conductor of electricity but it flow

very easily through some materials such as copper aluminum

iron and called electric conductors Wire and cables are the

common forms of conductors

Materials such as rubber glass certain plastics fibers dry

paper and air allow almost no electricity to pass through

them Such materials are called non-conductors insulators or

dielectrics When an insulator is continuous as for instance

around a wire it is commonly called insulation

The property of any material to oppose the flow of electricity

through it is called impedance The unit of the measurement of

this impedance or opposition to the flow of current is the

ldquoohm(Ω)rdquo Even the best conductors have some impedance poor

conductors have much impedance insulators (dielectrics) have

very high impedance The unit for the measurement of very low

impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth

of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and

is equal to one million ohm

An element of impedance called resistance in a conductor

varies directly as its length and inversely as its are (see

item 569)

Resistance may be compared to the friction encountered by a

flow of water through a pipe A straight pipe smooth inside

conducts water with little loss of pressure If the pipe is

rough inside and has many bends the loss of pressure and the

rate of flow will be greatly reduced Similarly a good

conductor allows electricity to flow with small loss of

voltage a poor conductor offers a large resistance and so

causes a corresponding large drop in voltage The energy used

in overcoming resistance is converted into heat


Chapter 3

41

The voltage required to make a current flow depends upon the

impedance of the circuit A voltage of one volt will make one

ampere flow through an impedance of one ohm

Z

EI or

I

EZ or ZIE

Where

I = Current in amperes

E = Voltage in volts

Z = Impedance in ohms

This formula is known as ldquoohmrsquos lowrdquo

Many alternating current circuits contain coils that produced

magnetic effects These magnetic effects in turn react upon

the current They retard the current and cause it to lag

behind the voltage It means that the voltage has reaches its

maximum and started to fall some time before the current

reaches a maximum Some current will be flowing in the circuit

at the instant when the voltage is zero This magnetic reaction

is called ldquoinductancerdquo or ldquoself inductancerdquo

Another kind of influence on an alternating current is caused

by the presence in the circuit of alternate plates of

conducting material separated by insulation This commonly

referred to as ldquocapacitorrdquo and its effect on the current is

to cause it to lead ahead of the voltage This reaction is

called ldquocapacitancerdquo It tends to counteract the inductance in

a circuit and is useful in overcoming the inductive lag in the

current inherent in most alternating current motors

Therefore in an alternating current circuit there is

resistance inductance and reactance affecting the current

The combination of any two or all three of these effects is

referred to as ldquoimpedancerdquo of the circuit

Power

Power is defined as the rate of doing work Electric power

is measured in ldquohorsepowerrdquo One horsepower equals 746 watts

One watt is rather small unit of power consequently when

speaking of power required by motors the term ldquokilowattrdquo is

used one kilowatt being thousand watts To obtain the power

delivered to an alternating current apparatus you can not

merely multiply effective amperes by effective volts If the

circuit contains inductance the apparatus circuits always

contain it the product of the effective current and effective

voltage will be greater than the real power This ldquoapparent

powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit

1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated

to ldquoKVArdquo

In alternating current power system the voltage and current

follow an approximate sine wave They build up from zero to a


Chapter 3

42

maximum in one direction then diminish to zero build up again

to a maximum but in the opposite direction and again diminish

to zero thus completing one cycle or two alternations and 360

electrical degree

The power factor is said to be 10 or unity if the voltage and

current reach their respective maximum values simultaneously

However as discussed previously in most alternating current

systems the voltage reaches its maximum value in a given

direction before the current attains its maximum value then

the current is said to lag behind the voltage This lag may

measure in degree The actual current drawn by apparatus of

this class may be considered as having two components one

known as the magnetizing current or that current which must

overcome the choking effect produced by the characteristics of

the apparatus and which lags 90 electric degree behind the

voltage The value of this lagging current is zero when the

voltage has reached its maximum value This lagging or

magnetizing current is called ldquoreactive currentrdquo

The other component is known as ldquoreal currentrdquo and it is in

phase with the voltage This real current and voltage reach

maximum values simultaneously

The actual line current is therefore the vector sum of the

reactive and real currents furthermore it is the current

that would be registered if an ammeter is connected in the

circuit Since there one component lagging 90 electric degree

or at right angles to the voltage the resultant or actual

line current of which this component is a part must

consequently be out of phase with the voltage and lag behind

it The degree or amount that it lags depends upon the

magnitude of this reactive current component and is a measure

of power factor

Resistance

Consider a circuit having resistance R ohms connected

across the terminals of an alternator A as in the following

figure and suppose the alternating voltage to be represented

by the sine wave as in the next figure


Chapter 3

43

If the value of the voltage at any instant is v volts the

value of the current at that instant is given by Rvi

amperes When the voltage is zero the current is also zero

and since the current is proportional to the voltage the wave

from the current is exactly the same as that of voltage Also

the two quantities are in phase with each other that they

pass through their zero value at the same instant and attain

their maximum values in a given direction at the same instant

If Vm and Im are the maximum values of the voltage and the

current respectively it follows that RVI mm

If the instantaneous value of the applied voltage is

represented by

sinVv m

Then the instantaneous value of current in resistive circuit

is

sinR

Vi m

Vector representing the voltage and the current in resistive

circuit is as follows

Inductive Reactance

Let us consider the effect of alternating current flowing

through a coil having an inductance of L henrys and negligible

resistance as the following figure

Current

Voltage

0

Vm

Im i v

-

+

Time

I V

V and I curve for a resistive Circuit


Chapter 3

44

Suppose the instantaneous value of the current through

inductance L henrys to be represented by

Where

t is the time in second

f is number of cycles per second

Suppose the current to increase by di ampere in dt second

then

Instantaneous value of induced emf is

Since f represents the number of cyclessecond the duration

of 1 cycle = 1f second consequently

Hence the wave of the induced emf is represented by the curve

in the figure below lagging the current by a quarter of cycle

(90O)

Since the resistance of the circuit is assumed negligible the

whole of the applied voltage is absorbed in neutralizing the

induce emf

So instantaneous value of the applied voltage is

v L

i

Circuit with inductance only

A Coil


Chapter 3

45

Comparison of expression (1) and (3) shows that the applied

voltage leads the current by a quarter of cycle (90O) Also

from (3) it follows that that maximum value Vm of the applied

voltage is

So that

The inductive reactance is expressed in ohms and is

represented by XL

Hence

Capacitance reactive

The property of a capacitor to store an electric charge

when its plates are at different potentials is referred to

capacitance as the following figure

The unit of capacitance is termed the farad (F) Farad is

defined as the capacitance of capacitor which required a

potential difference (pd) of 1 volt to maintain a charge of

one coulomb on that capacitor


through a capacitor having a capacitance of C farad and

negligible resistance


Chapter 3

46

Circuit with capacitance only

Suppose the instantaneous value of the voltage applied to the

capacitance to be represented by

If the applied voltage increases by dv volt in dt second as in

the following figure

Instantaneous value of current flow through capacitor is

Comparison of (4)and (5) shows that current leads applied

voltage by a quarter of cycle (90O)and the current and voltage

can be represented vectorially as the following figure


Chapter 3

47

From expression (5) it follows that the maximum value Im of the

current is

So

The capacitive reactance is expressed in ohms and is

represented by the symbol Xc

Resistance Inductance and Capacitance in series

An actual circuit may have resistance and inductance or

resistance and capacitance or resistance conductance and

capacitance in series Hence if we consider the general case

of R L and C in series we can adapt the results to the

other two cases by merely omitting the capacitive or the

inductance from the expression for general case

In following fig OB vector represents L (2πfLI) leading the

current by 900 and OC vector represents C (I2πfC) lagging the

current by 900

Since OD = OB-OC OB being assumed greater than OC and supply

voltage is the vectorial sum of OA and OD namely OE

I

v

90 0


Chapter 3

48

Vector diagram for above figure

From this expression it is seen that

Resultant reactance = inductive reactance -capacitive

reactance

Φ = phase difference between the current and the supply

voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49

If the inductive reactance is greater than the capacitive

reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying

that the current leads the supply voltage by angle Φ

If a circuit consists of a coil having a resistance R ohms and

inductance L henrys such a circuit can be considered as a

resistance and inductive in series and

And the phase angle in which the current lags the supply

voltage is given by

Example 1

A coil having a resistance of 12 Ω and inductance of 01 H is

connected across a 100 V 50 cs supply Calculate

a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply

voltage

Solution

(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω

Impedance = Z = radicR2+XL

2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2

A metal filament lamp rated at 750 watt 100 v is to be

connected in series with a capacitance across a 230 v 60 cs

supply Calculate

a) The capacitance required

b) The phase angle between the current and the supply voltage

Solution

From vector diagram below

(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts

Rated current of lamp = 750 w 100 v = 75 A

75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad

= 96 microfarad (μF)

i


Chapter 3

50

(b) If φ = phase angle between current I and supply V

Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3

A resistance of 12 Ω an inductance of 015 H and capacitance

of 100 μF are connected in series across 100 V 50 cs supply

Calculate

a) The impedance

b) The current

c) The voltage across R L and C

d) The phase difference between current and supply voltage

Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A

(c) Voltage across R = VR = 12x515 = 618 V

Voltage across L = VL = 471x515 = 2425 V

Voltage across C = VC = 3185x515 = 164 V

(d) Phase difference between current and supply voltage =

φ = cos-1(VRV)= cos

-1(618100) = 51

0 48rsquo

Or φ = tan-1 (VL-VCVR ) = tan

-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo

See vector diagram below


Chapter 3

51

Power in series RL circuit

From the above analysis it seen that the voltage applied to

a series RL circuit the voltage leads the current by angle φ

It is equally valid to say that the current lags the voltage

by an angle φ If a voltage Emsinωt is applied to a series RL

circuit the current is Imsin(ωt-φ) The instantaneous power

is

p = e x i

= Emsin ωt x Imsin(ωt-φ)

= tIEIE mmmm 2cos

2cos

2

The second term of the right side has an average value of

zero

p = coscos2

EIIE mm -----------(11)

Where E and I are the rms (Root Mean Square) values of

voltage and current and φ is the phase angle between the

voltage and current

The following figure is the plot of instantaneous voltage

current and power The instantaneous power is positive during

the time interval when both e and i are simultaneously

positive or negative During the interval when one of the two

quantities out of e and i is positive and the other is

negative instantaneous power is negative The positive area

is more than the negative area and therefore the average power

is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360

i=ImSin(ωt-φ)e=EmSinωt

p=EmSinωtImSin(ωt - φ) c o s φ

0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit

Active power reactive power and power factor (single phase)

The average power in the circuit ie EIcosφ is the actual

power supplied by the source to the circuit This is known as

active power of the circuit The active power is measured in

watts The bigger units of active power are KW (kilowatt=103

watts) and MW (megawatt=106 watts) The product of voltage and

current ie EI called apparent power and is measured in

volt-ampere (VA)

The ratio of active power to apparent power equals cosφ This

term cosφ is called power factor of the circuit It is the

factor by which the apparent power (EI) must be multiplied to

give the active power The power factor for purely resistive

circuit is 1 Therefore the apparent power and active power

are equal for purely resistive circuit A circuit may be


Chapter 3

53

characterized as having leading or lagging power factor A

leading power factor means that the current leads the voltage

This occurs in the capacitive circuits An inductive circuit

is described as having lagging power factor since the current

lags the voltage The active power in ac circuit can also be

written as I2R

Example

A 10 ohm resistor and 20 mH (mille Henry) inductor are

connected in series across a 230 volts 50 cs supply Find

the circuit impedance current voltage across resistor

apparent power active power reactive power and power

factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes

Voltage across resistor = RI = 10x1949 = 1499 volts

Voltage across inductor = IXL = 1224 volts

Apparent power = EI = 230x 1949 = 44827 VA

Power factor = cos(321) = 0847 lagging

Active power = EIcosφ = 230x1949x0847 = 37968 watts

Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w

Practical importance of power factor (PF)

If an alternator is rated at a given say 2000 A at a

voltage of 400 v it means that these are the highest current

and voltage values of the machine can give without the

temperature exceeding a safe value Consequently the rating of

the alternator is given as 400x20001000=800 KVA The phase

difference between the voltage and the current depends upon

the nature of the load and not upon the generator Thus if the

power factor of the load is unity the 800 KVA are also 800

kW and the engine driving the generator has to be capable of

developing this power together with the losses in the

generator But if the pf of the load is say 05 the power

is only 400 kW so that the engine is only developing about

one half of the power which it is capable through the

alternator is supplying its rated output 800 KVA

Similarly the conductors connecting the alternator to the load

have to be capable of carrying 2000 A without excessive

temperature rise Consequently they can transmit 800 kW if the

power factor is unity but only 400 kW at 05 pf for the

same rise of temperature


Chapter 3

54

It is therefore evident that the higher the pf of the load

the greater is the power that can be generated by a given

alternator and transmitted by a given conductor

The matter may be put another way by saying that for a given

power the lower the pf the larger must be the size of the

alternator to generate that power and the greater must be the

cross sectional area of the conductor to transmit it in other

words the greater is the cost of generation and transmission

of the electric energy This is the reason why supply

authorities do all they can to improve the pf for their

loads (by ex installing capacitors)

Three phase alternating current

Three phase alternating current is the best suites for long

distance transmission because it may be easily generated at

low to moderately high voltages and can then have the voltage

raised to very high values suitable for efficient

transmission and then the voltage can be reduced to a value

suitable for general use by means of stationary device known

as a transformer The higher the voltage the smaller the wire

required to carry a given amount of power The following

figures are the curve and diagram representing three phase

alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55

Since the angle between ENR and EYN is 600

EYNR = 2ENRcos300 = radic3ENR

Line voltage = 173 x phase voltage

In general

If VL = pd between any two line conductors

= line voltage

And VP = pd between a line conductor and neutral point

= phase voltage

And if IL and IP = line and phase current respectively

then for a star connection system

VL = 173 VP

IL = IP

For delta connection

IL = 173 IP

Power in three phase system with balanced load

If Ip = value of the current in each phase

Vp = value of pd across each phase

Power per phase = IpVp x power factor

And total power = 3IpVp x power factor

If IL and VL be the value of the line current and voltage

respectively then for star connection system

VP = VL173 and IP = IL so

Total power in watts = 173 x ILVL x power factor

For delta connection system

120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)

(voltage to neutral)


Chapter 3

56

IP = IL173 and VP = VL so


Example

A three phase motor operating off 400 v system is developing

25 bhp at an efficiency of 087 and a power factor of 082

Calculate (a) the line current and (b) the phase current if

the windings are delta-connected

(a) wattsinpowerinput

wattsinpoweroutputEfficiency

FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A

(b) For delta-connected windings

Aline

currentphase 821731

837

731

current

Complex Numbers

The mathematics used in Electrical Engineering to add

together resistances currents or DC voltages uses what are

called real numbers But real numbers are not the only kind

of numbers we need to use especially when dealing with

frequency dependent sinusoidal sources and vectors As well as

using normal or real numbers Complex Numbers were introduced

to allow complex equations to be solved with numbers that are

the square roots of negative numbers radic-1

In electrical engineering this type of number is called an

imaginary number and to distinguish an imaginary number from

a real number the letter j known commonly in electrical

engineering as the j-operator The letter j is used in front

of a number to signify its imaginary number operation

Examples of imaginary numbers are j3 j12 j100 etc Then

a complex number consists of two distinct but very much

related parts a Real Number plus an Imaginary Number

Complex Numbers represent points in a two dimensional complex

or s-plane that are referenced to two distinct axes The

horizontal axis is called the real axis while the vertical

axis is called the imaginary axis

The rules and laws used in mathematics for the addition or

subtraction of imaginary numbers are the same as for real

numbers j2 + j4 = j6 etc The only difference is in

multiplication because two imaginary numbers multiplied

together becomes a positive real number as two negatives make


Chapter 3

57

a positive Real numbers can also be thought of as a complex

number but with a zero imaginary part labeled j0

The j-operator has a value exactly equal to radic-1 so successive

multiplication of j (j x j ) will result in j having the

following values of -1 -j and +1 As the j-operator is

commonly used to indicate the anticlockwise rotation of a

vector each successive multiplication or power of j j2

j3 etc will force the vector to rotate through an angle of

90o anticlockwise as shown below Likewise if the

multiplication of the vector results in a -j operator then the

phase shift will be -90o ie a clockwise rotation

Vector Rotation of the j-operator

So by multiplying an imaginary number by j2 will rotate the

vector by 180o anticlockwise multiplying by j

3 rotates

it 270o and by j

4 rotates it 360

o or back to its original

position Multiplication by j10 or by j

30 will cause the vector

to rotate anticlockwise by the appropriate amount In each

successive rotation the magnitude of the vector always

remains the same There are different ways in Electrical

Engineering to represent complex numbers either graphically or

mathematically One such way that uses the cosine and sine

rule is called the Cartesian or Rectangular Form

Complex Numbers using the Rectangular Form

Complex number is represented by a real part and an imaginary

part that takes the general form of

Where

Z is the Complex Number representing the Vector

x is the Real part or the Active component

y is the Imaginary part or the Reactive component

j is defined by radic-1

In the rectangular form a complex number can be represented

as a point on a two-dimensional plane called the complex or s-

plane So for example Z = 6 + j4 represents a single point

whose coordinates represent 6 on the horizontal real axis and

4 on the vertical imaginary axis as shown


Chapter 3

58

Complex Numbers using the Complex or s-plane

But as both the real and imaginary parts of a complex number

in the rectangular form can be either a positive number or a

negative number then both the real and imaginary axis must

also extend in both the positive and negative directions This

then produces a complex plane with four quadrants called

an Argand Diagram as shown below

Four Quadrant Argand Diagram

On the Argand diagram the horizontal axis represents all

positive real numbers to the right of the vertical imaginary


Chapter 3

59

axis and all negative real numbers to the left of the vertical

imaginary axis All positive imaginary numbers are represented

above the horizontal axis while all the negative imaginary

numbers are below the horizontal real axis This then produces

a two dimensional complex plane with four distinct quadrants

labeled QI QII QIII and QIV The Argand diagram can also

be used to represent a rotating phasor as a point in the

complex plane whose radius is given by the magnitude of the

phasor will draw a full circle around it for

every 2πω seconds

Complex Numbers can also have zero real or imaginary parts

such as Z = 6 + j0 or Z = 0 + j4 In this case the points

are plotted directly onto the real or imaginary axis Also

the angle of a complex number can be calculated using simple

trigonometry to calculate the angles of right-angled

triangles or measured anti-clockwise around the Argand

diagram starting from the positive real axis

Then angles between 0 and 90o will be in the first quadrant

( I ) angles ( θ ) between 90 and 180o in the second quadrant

( II ) The third quadrant ( III ) includes angles between 180

and 270o while the fourth and final quadrant ( IV ) which

completes the full circle includes the angles between 270 and

360o and so on In all the four quadrants the relevant angles

can be found from tan-1(imaginary componentreal component)

Addition and Subtraction of Complex Numbers

The addition or subtraction of complex numbers can be done

either mathematically or graphically in rectangular form For

addition the real parts are firstly added together to form

the real part of the sum and then the imaginary parts to form

the imaginary part of the sum and this process is as follows

using two complex numbers A and B as examples

Example 1

Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine

the sum and difference of the two vectors in both rectangular

(a + jb) form and graphically as an Argand Diagram

Addition

Subtraction


Chapter 3

60

Graphical Addition and Subtraction

Multiplication and Division of Complex Numbers

The multiplication of complex numbers in the rectangular form

follows more or less the same rules as for normal algebra

along with some additional rules for the successive

multiplication of the j-operator where j2 = -1 So for

example multiplying together our two vectors from above

of A = 4 + j1 and B = 2 + j3 will give us the following

result

Mathematically the division of complex numbers in rectangular

form is a little more difficult to perform as it requires the

use of the denominators conjugate function to convert the

denominator of the equation into a real number This is called

rationalizing Then the division of complex numbers is best

carried out using Polar Form which we will look at later

However as an example in rectangular form lets find the value

of vector A divided by vector B

Multiply top and bottom by conjugate (2-j3)

The Complex Conjugate

The Complex Conjugate or simply Conjugate of a complex number

is found by reversing the algebraic sign of the complex

numbers imaginary number only while keeping the algebraic sign


Chapter 3

61

of the real number the same and to identify the complex

conjugate of z the symbol z is used For example the

conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate

of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram

for a complex conjugate have the same horizontal position on

the real axis as the original complex number but opposite

vertical positions Thus complex conjugates can be thought of

as a reflection of a complex number The following example

shows a complex number 6 + j4 and its conjugate in the

complex plane

Conjugate Complex Numbers

The sum of a complex number and its complex conjugate will

always be a real number as we have seen above Then the

addition of a complex number and its conjugate gives the

result as a real number or active component only while their

subtraction gives an imaginary number or reactive component

only The conjugate of a complex number is an important

element used in Electrical Engineering to determine the

apparent power of an AC circuit using rectangular form

Complex Numbers using Polar Form

Unlike rectangular form which plots points in the complex

plane the Polar Form of a complex number is written in terms

of its magnitude and angle Thus a polar form vector is

presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector

and θ is its angle or argument of A which can be either

positive or negative The magnitude and angle of the point

still remains the same as for the rectangular form above this

time in polar form the location of the point is represented in

a triangular form as shown below


Chapter 3

62

Polar Form Representation of a Complex Number

As the polar representation of a point is based around the

triangular form we can use simple geometry of the triangle

and especially trigonometry and Pythagorass Theorem on

triangles to find both the magnitude and the angle of the

complex number As we remember from school trigonometry deals

with the relationship between the sides and the angles of

triangles so we can describe the relationships between the

sides as

Using trigonometry again the angle θ of A is given as

follows

Then in Polar form the length of A and its angle represents

the complex number instead of a point Also in polar form the

conjugate of the complex number has the same magnitude or

modulus it is the sign of the angle that changes so for

example the conjugate of 6 ang30o would be 6 angndash30o

Converting between Rectangular Form and Polar Form

In the rectangular form we can express a vector in terms of

its rectangular coordinates with the horizontal axis being

its real axis and the vertical axis being its imaginary axis

or j-component In polar form these real and imaginary axes

are simply represented by A angθ Then using our example

above the relationship between rectangular form and polar

form can be defined as

Converting Polar Form into Rectangular Form ( PrarrR )


Chapter 3

63

We can also convert back from rectangular form to polar form

as follows

Converting Rectangular Form into Polar Form ( RrarrP )

Polar Form Multiplication and Division

Rectangular form is best for adding and subtracting complex

numbers as we saw above but polar form is often better for

multiplying and dividing To multiply together two vectors in

polar form we must first multiply together the two modulus or

magnitudes and then add together their angles

Multiplication in Polar Form

Multiplying together 6 ang30o and 8 angndash 45o in polar form gives

us

Division in Polar Form

Likewise to divide together two vectors in polar form we

must divide the two modulus and then subtract their angles as

shown

319 Equivalent circuit of induction motor

To analyze the operating and performance characteristics

of an induction motor an Equivalent Circuit can be drawn We

will consider a 3ndashphase Y connected machine the Equivalent

Circuit for the stator is as shown below


Chapter 3

64

V1 = Stator Terminal Voltage

I1 = Stator Current

R1 = Stator Effective Resistance

X1 = Stator Leakage Reactance

Z1 = Stator Impedance (R1 + jX1)

Io = Exciting Current (this is comprised of the core loss

component = Ic and a magnetizing current = Im)

The rotor winding can be represented as

Rotor Circuit

I2 = Rotor Current

R2 = Rotor winding Resistance

X2 = Rotor Leakage Reactance

Z2 = Rotor Impedance (R1 + jX1)

E2 = Induced EMF in the rotor (generated by the air gap flux)

The EMF (E2) is equal to the stator terminal voltage less the

voltage drop caused by the stator leakage impedance

Note

Never use three-phase equivalent circuit Always use per- phase equivalent circuit

The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor

Induction machine equivalent circuit is composed of stator circuit and rotor circuit

3191 Effective circuit of induction motor at standstill

Standstill means rotor circuit is open At open

circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2

Refer to section 37 and 38


Chapter 3

65

Effective circuit at standstill

E2 = V1 - I1 (Z1)

Where E2 is rotor induced emf at standstill

E2 = V1 - I1 (R1 + jX1) = I1 Z1


operating conditions (rotor winding is shorted)

Now suppose induction-motor is loaded down As motorrsquos load

increases its slip rises because of which the rotor speed

falls

Greater relative motion produces a stronger rotor voltage E2

which in turn produces a larger rotor current I2

Then the rotor magnetic field φ also increases Since rotor

slip is larger rotor frequency rises (fr = s fs ) and rotor

reactance increases (ωLr) Therefore rotor current lags

further behind the rotor voltage

Accordingly the parameters will be changed to be as follows

The stator and rotor sides are in the figure below separated

by an air gap

I2 = Rotor current in running condition

It is important to note that as load on the motor changes the

motor speed changes Thus slip changes As slip changes the


Chapter 3

66

reactance of standstill X2 changes to be sX2 which is shown

variable

The rotor impedance can be represented by

Effective circuit of induction motor under operating

conditions

In the running condition the variable resistance R2s can be

rearranged as follows

So the variable rotor resistance R2s has two parts

1 Rotor resistance R2 itself which represents copper loss

2 R2(1 - s)s which represents load resistance RL So it is

electrical equivalent of mechanical load on the motor

Key Point Thus the mechanical load on the motor is

represented by the pure resistance of value R2(1 -s)s

The effective circuit of the induction motor under operating

condition shall be as following

It means that R2s = rotor copper resistance + load resistance


Chapter 3

67

3193 Power Relations

1 Input Power

2 Stator copper losses

3 Rotor copper losses

4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1

A 480-V 60 Hz 50-hp three phase induction motor is drawing

60A at 085 PF lagging The stator copper losses are 2 kW and

the rotor copper losses are 700 W The friction and windage

losses are 600 W the core losses are 1800 W and the stray

losses are negligible Find the following quantities

1 The air-gap power PAG

2 The power converted Pm

3 The output power Pout

4 The efficiency of the motor

Solution

Example 2

A 480V 60 Hz 6-pole three-phase delta-connected induction

motor has the following parameters

R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω

Rotational losses are 2450W The motor drives a mechanical

load at a speed of 1170 rpm Calculate the following

information

Synchronous speed in rpm

Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69

Note the core losses is so small as the IoltltltltltI1 so it is

neglected


Slip

Line Current

Phase current is given by

Note that the machine is delta connected so V1 = VLL = 480 V


Chapter 3

70

Therefore IL = radic3 times 466 = 806 A

Input Power

Air gap Power

Air gap power is the input power minus stator losses

Torque Developed

Output Power in Hp

Neglecting friction and windage losses so Pout = Pm

Efficiency

320 Determination of motor parameters

The motor parameters are determined from three tests

Stator DC resistance measurement Determines the stator

resistance value (R1)

o The motor is supplied by DC voltage between two terminals A and B at the figure below

o The dc voltage and current are measured

o If the stator is Y-connected the per phase stator

resistance is

o If the stator is Delta-connected the per phase stator

resistance is


Chapter 3

71

No load test

No-load test Provides the magnetizing reactance and core

resistance (Rc and Xm) In this course we will only find Xm

and ignore Rc

o The motor is allowed to spin freely

o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are

measured

o The only load on the motor is the friction and windage

losses so all Pm is consumed by mechanical losses

o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the

rotor branch of the equivalent circuit is very high

The equivalent circuit reduces tohellip

Combining Rc amp RF+W we gethelliphellip


Chapter 3

72

o At the no-load conditions the input power measured by

meters must equal the losses in the motor

o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large

o The input power equals

o The equivalent input impedance is thus approximately

o The value of the stator leakage reactance X1 can be

determined from the blocked rotor test The value of the

magnetizing reactance can then be determined

Blocked Rotor Test

o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the

resulting voltage current and power are measured

o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value

o The locked-rotor power factor can be found as

o The magnitude of the total impedance

Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the

test frequency respectively


Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example

The following test results are obtained from three phase 100

hp 460 v eight pole star connected induction motor design A

rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw

Average DC resistor between two stator terminals is 0152

Determine

1 The parameters of the equivalent cicuit

2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine

a The input current

b The input power

c Air gap power

d Rotor copper loss

e Mechanical power developed

f Output power

g Efficiency of the motor

3 The speed of the rotor field relative to stator structure and stator rotating field

Solution


Chapter 3

74

From No load test

From block rotor test

As rotor is design A then

Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss

Mechanical power developed

From no load test

321 NEMA standard for squirrel cage induction motor

Design A Motor

Hp range 05 ndash 500 hp

Starting current 6 to 10 times full-load current

Good running efficiency 87 - 89

Good power factor 87 - 89

Low rated slip 3 ndash5

Starting torque is about 150 of full load torque

Maximum torque is over 200 but less than 225 of full-load

torque

Typical applications constant speed applications where high

starting torque is not needed and high starting torque is

tolerated


Chapter 3

76

Design B motor

Hp range 05 to 500 hp

Higher reactance than the Design A motor obtained by means of

deep narrow rotor bars

The starting current is held to about 5 times the full-load

current

This motor allows full-voltage starting

The starting torque slip and efficiency are nearly the same

as for the Design A motor

Power factor and maximum torque are little lower than class A



tolerated

Unsuitable for applications where there is a high load peak

Design C motor


This type of motor has a double-layer or double squirrel-

cage winding

It combines high starting torque with low starting current

Two windings are applied to the rotor an outer winding having

high resistance and low reactance and an inner winding having

low resistance and high reactance

Operation is such that the reactance of both windings decrease

as rotor frequency decreases and speed increases

On starting a much larger induced currents flow in the outer

winding than in the inner winding because at low rotor speeds

the inner-winding reactance is quite high

As the rotor speed increases the reactance of the inner

winding drops and combined with the low inner-winding

resistance permits the major portion of the rotor current to

appear in the inner winding

Starting current about 5 times full load current

The starting torque is rather high (200 - 250)

Full-load torque is the same as that for both A and B designs

The maximum torque is lower than the starting torque maximum

torque (180-225)

Typical applications constant speed loads requiring fairly

high starting torque and lower starting currents


Chapter 3

77

Design D motor

Produces a very high starting torque-approximately 275 of

full-load torque

It has low starting current

High slip 7-16

Low efficiency

Torque changes with load

Typical applications used for high inertia loads

The above classification is for squirrel cage induction motor

322 Torque of squirrel cage induction motor

In order to perform useful work the induction motor must

be started from rest and both the motor and load accelerated

up to full speed Typically this is done by relying on the

high slip characteristics of the motor and enabling it to

provide the acceleration torque

Induction motors at rest appear just like a short circuited

transformer and if connected to the full supply voltage draw

a very high current known as the Locked Rotor Current They

also produce torque which is known as the Locked Rotor

Torque The Locked Rotor Torque (LRT) and the Locked Rotor

Current (LRC) are a function of the terminal voltage to the

motor and the motor design As the motor accelerates both

the torque and the current will tend to alter with rotor speed

if the voltage is maintained constant

The starting current of a motor with a fixed voltage will

drop very slowly as the motor accelerates and will only begin

to fall significantly when the motor has reached at least 80

full speed The actual curves for induction motors can vary

considerably between designs but the general trend is for a

high current until the motor has almost reached full speed

The LRC of a motor can range from 500 Full Load Current (FLC)

to as high as 1400 FLC Typically good motors fall in the

range of 550 to 750 FLC


Chapter 3

78

The starting torque of an induction motor starting with a

fixed voltage will drop a little to the minimum torque known

as the pull up torque as the motor accelerates and then rise

to a maximum torque known as the breakdown or pull out torque

at almost full speed and then drop to zero at synchronous

speed The curve of start torque against rotor speed is

dependent on the terminal voltage and the motorrotor design

The LRT of an induction motor can vary from as low as 60 Full

Load Torque (FLT) to as high as 350 FLT The pull-up torque

can be as low as 40 FLT and the breakdown torque can be as

high as 350 FLT Typical LRTs for medium to large motors are

in the order of 120 FLT to 280 FLT


Chapter 3

79

Figure above graph shows that starting torque known as locked

rotor torque (LRT) is higher than 100 of the full load torque

(FLT) the safe continuous torque rating

The locked rotor torque is about 175 of FLT for the example

motor graphed above Starting current known as locked rotor

current (LRC) is 500 of full load current (FLC) the safe

running current The current is high because this is analogous

to a shorted secondary on a transformer

As the rotor starts to rotate the torque may decrease a bit

for certain classes of motors to a value known as the pull up

torque This is the lowest value of torque ever encountered by

the starting motor As the rotor gains 80 of synchronous

speed torque increases from 175 up to 300 of the full load

torque This breakdown torque is due to the larger than normal

20 slip

The current has decreased only slightly at this point but

will decrease rapidly beyond this point As the rotor

accelerates to within a few percent of synchronous speed both

torque and current will decrease substantially Slip will be

only a few percent during normal operation For a running

motor any portion of the torque curve below 100 rated torque

is normal

The motor load determines the operating point on the torque

curve While the motor torque and current may exceed 100 for

a few seconds during starting continuous operation above 100

can damage the motor Any motor torque load above the

breakdown torque will stall the motor The torque slip and

current will approach zero for a ldquono mechanical torquerdquo load

condition This condition is analogous to an open secondary

transformer

There are several basic induction motor designs (Figure below)

showing considerable variation from the torque curve above

The different designs are optimized for starting and running

different types of loads The locked rotor torque (LRT) for

various motor designs and sizes ranges from 60 to 350 of

full load torque (FLT) Starting current or locked rotor

current (LRC) can range from 500 to 1400 of full load

current (FLC) This current draw can present a starting

problem for large induction motors


Chapter 3

80

Torque of NEMA designs


Chapter 3

2

18 Fundamentals of electricity 39

19 Equivalent circuit of induction motor 63

191 Effective circuit of induction motor at

Standstill


Operating conditions (rotor is shorted)

193 Power relations

20 Determination of motor parameters 70

21 NEMA standard for squirrel cage IM 75

22 Torque of squirrel cage IM 77


Chapter 3

3

Submersible Motor

31 General












protector (seal)





the mains circuit





Chapter 3

4

321 Stator




laminations

















Chapter 3

5
























winding)


Chapter 3

6






pole motor



wire




322 Rotor


circuit






Fig (37) rotor core


Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

3

Submersible Motor

31 General












protector (seal)





the mains circuit





Chapter 3

4

321 Stator




laminations

















Chapter 3

5
























winding)


Chapter 3

6






pole motor



wire




322 Rotor


circuit






Fig (37) rotor core


Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

4

321 Stator




laminations

















Chapter 3

5
























winding)


Chapter 3

6






pole motor



wire




322 Rotor


circuit






Fig (37) rotor core


Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

5
























winding)


Chapter 3

6






pole motor



wire




322 Rotor


circuit






Fig (37) rotor core


Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

6






pole motor



wire




322 Rotor


circuit






Fig (37) rotor core


Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

7





would be produced



poles as the stator








323 Rotor bearing






surface









rotor string


Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

8



Babbitt

Glacier



325 Pothead







33 Electromagnetism

331 Magnetic field




Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

9











311)













Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

10






























Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

11




section




AB















turns


Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

12

Fig (315)






InB




length



















Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

13










Fig (316)

335 Induced voltage

















Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

14












is moved (fig 318)

Fig (318)


Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

15













Fig (319)

35 Power supply





Fig (320)




Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

16














Winding

Current Flow

Direction

Positive Negative

A1 North South

A2 South North

B1 North South

B2 South North

C1 North South

C2 South North

351 Start











Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

17



(322)

Fig (322) start

352 Time 1








Fig (323) time 1


Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

18

353 Time 2







polarity

Fig (324) time 2






Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

19

Fig (325) time 1









due to any phase










Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

20

(i) (ii)

Fig (326)







proved under


(i) (ii)

Fig (327)


Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

21


Fig (328)


fluxes are given by


Fig (329)


Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

22






Fig (330)







vectors fig (331)







Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

23

Fig (331)




So



speed (ωs) is


Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

24







37 Slip













ie









Where


P = Number of poles


Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

25




















circuit ie



Example





Flux density 4101

10424

4

m

Wb

AB

T


Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

26

mmf 50150 NIF AT


502

m

AT

l

NIH ATm
















the field









field

+ Force

Flux set up

by current

in on


by current

in on

conductor

Force


Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

27



by









field



the coil is




Example




F = 08 x 10 x 30 x 10-2 = 24 N

Example



S

N

I

2r

Flux


Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

28


acting on the coil

Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10

-2 = 48 x 10

-2 Nm





















Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

29










the rotor field




















expression

np

f s

2

Where





p

fN s

120

Where



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

30














each application


sections they are

1 Single section












4 Lower Tandem (LT)



attached)


Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

31

Single UT CT


Notes








two 150 Hp motors







695 A motor








Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

32






amperage




314 Motor current




close to 100









magnetizing current









315 Motor rating




Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

33

the table below


HP Volt Amp

60 HZ 50 HZ 60 HZ 50 HZ

38 32

43553 36353

87526 72926

131518 109618

57 48

43081 35881

87040 72540

131526 109626

76 63 86553 72153

136034 113334

95 79 84069 70069

133044 110844

114 95

86081 71781

130053 108353

233030 194230

133 111

83098 69298

134560 112160

220537 183837

152 127 134069 111769

232540 193840

171 143 129081 107581

239044 199244

190 158

118598 98898

143081 119281

241548 201348



this industry









Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

34


relationship










motor

Fig (337)


shows in fig 338

Fig (338)


Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

35


called SATURATION


our motor





acceptable



motor





itself








conditions only






motor from damage





Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

36








settings









condition)












calculations










Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

37











which includes

Motor Series

Motor Type





Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

38



















4160V 2300V etc




Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

39








600V 1000V 1500V 2400V etc














(Sensor)







Electricity












Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

40



1602x10-19





per second



























item 569)











Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

41




Z

EI or

I

EZ or ZIE

Where

























Power













to ldquoKVArdquo




Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

42




electrical degree



























of power factor

Resistance






Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

43












represented by

sinVv m


is

sinR

Vi m



Inductive Reactance




Current

Voltage

0

Vm

Im i v

-

+

Time

I V



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

44



Where




then






(90O)



induce emf


v L

i


A Coil


Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

45




voltage is

So that


represented by XL

Hence













Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

46











Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

47


current is

So












current by 900



I

v

90 0


Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

48




reactance


voltage

B

D

O

E

A I

C

[L]

[C]

φ


Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

49








voltage is given by

Example 1




voltage

Solution



2 = radic12

2+314

2 = 336 Ω

(b) Current = VZ = 100 336 = 2975 A

(c) Tan Φ = XLR = 314 12 = 2617

Φ = 690

Example 2



supply Calculate



Solution


(a) V2 = VR

2+VC

2

(230)2 = (100)

2 + VC

2

VC = 207 Volts


75 = 2 x 314 x 60 x C x 207

C = 96 x 10-6 farad


i


Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

50


Cos φ = VR V = 100 230 = 0435

Φ = 640 12rsquo

Example 3



Calculate

a) The impedance

b) The current



Solution

(a)

fCfLRZ

2

12

2

2

4198631147144

100501415932

10615050141593212

2

2

2z

(b) Current = VZ = 100 194 = 515 A






-1(618100) = 51

0 48rsquo


-1(2425-164)618) = 51

0 48rsquo

Or alternatively

φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51

0 48rsquo



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

51







is

p = e x i


= tIEIE mmmm 2cos

2cos

2


zero

p = coscos2

EIIE mm -----------(11)



voltage and current








is positive

L 471

3185

1525

12

R

C

φ

VL 2425

164

785

618

VR

VC


Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

52

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360



0 φ

+ ve area

-ve

+ ve area

-ve

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360


0

φ

+ ve area

-ve

+ ve area

-ve

p=EIcosφ

Power in RL circuit








volt-ampere (VA)








Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

53






written as I2R

Example





factor

XL = ωL = 2πfL = 2x324x50x20x10-3

= 628 Ω

Z = radicR2+XL

2 = radic10

2+(628)

2 = 118 Ω

Φ = tan-1(XLR) = 321

0

I = EZ = 230118 = 1949 amperes




























Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

54






















alternating current

0 30 60 90 120 150 180 210 240 270 300 330 360

phase 1

phase 2 phase 3


Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

55




In general


= line voltage


= phase voltage



VL = 173 VP

IL = IP


IL = 173 IP











120o

120o

30o

ENR

EBNY

ENY

ENB

EYN

EYNR

ERNB

EBN

Vector diagram

Line voltage (VL)

Between any line

conductors

phase voltage (VP)



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

56



Example







FxPVI

bhpx

LL 731

746

820400731

74625870

xxxI

x

L

IL = 378 A


Aline

currentphase 821731

837

731

current

Complex Numbers



























Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

57















3 rotates

it 270o and by j

4 rotates it 360













Where











Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

58












Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

59










every 2πω seconds






















Example 1




Addition

Subtraction


Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

60









result















Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

61










complex plane





















Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

62









sides as


follows
















Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

63


as follows










us




shown







Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

64


I1 = Stator Current







Rotor Circuit

I2 = Rotor Current







Note









Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

65


E2 = V1 - I1 (Z1)


E2 = V1 - I1 (R1 + jX1) = I1 Z1





falls









by an air gap





Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

66


variable



conditions













Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

67


1 Input Power



4 Air gap power

5 Mechanical power

6 Output power

7 Output torque

Power flow diagram


Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

68

Example 1










Solution

Example 2



R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω



information


Slip

Line Current

Input Power

Air gap Power

Torque Developed

Output Power in Hp

Efficiency

Solution


Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

69


neglected


Slip

Line Current




Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

70


Input Power

Air gap Power


Torque Developed

Output Power in Hp


Efficiency








resistance is


resistance is


Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

71

No load test



and ignore Rc



measured








Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

72









Blocked Rotor Test









Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

73

X1 and X2 as

function of

XLR

Rotor Design X1 X2

Design A 05

XLR

05

XLR

Design B 04

XLR

06

XLR

Design C 03

XLR

07

XLR

Design D 05

XLR

05

XLR

Example



rotor

No load test

460 v 60 Hz 40 A 42 Kw

Blocked rotor test

100 v 60 Hz 140 A 8 Kw


Determine



a The input current

b The input power

c Air gap power

d Rotor copper loss


f Output power



Solution


Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

74

From No load test



Equivalent circuit

Input impedance


Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

75

Input Power

Stator copper loss

Air gap power

Rotor copper loss


From no load test


Design A Motor








torque



tolerated


Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

76

Design B motor





current







tolerated


Design C motor



cage winding


















torque (180-225)




Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

77

Design D motor


full-load torque


High slip 7-16

Low efficiency





























Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

78














Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

79















20 slip







is normal








transformer











Chapter 3

80



Chapter 3

80


Documents

Chapter_03 ESP Motor