Lab Manual of Ac Machines

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SR.N

OLAB WORK

PAG

E

01 Working and construction of Transformer 2

02 Study of step and step down Transformer 7

03 Short circuit test of a Transformer to find out its iron losses 10

04 Open circuit test of a Transformer to find out its iron losses 13

05 Efficiency of single phase step down Transformer 16

06 Voltage regulation of single phase step down Transformer 19

07 Working principle and construction of induction motor 21

08 Production of flux and polarity checking of dc motor 25

09 Run the 3 phase induction motor on no load 27

10 Relation b/w synchronous speed & poles of induction motor 29

11 Relation b/w synchronous speed & frequency of induction motor 31

12 Start the induction motor through star delta switch 33

List of Experiments of Lab

Performed in “Advance Machines and Drives”

Department Of Electrical Engineering, UCE&T, Bahawalpur

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The Islamia University of BahawalpurUniversity College of Engineering & Technology

Advance Machines & Drives 5rd Semester (3rd Year)(LAB EXPERIMENT NO. 01)

To demonstrate the constructional parts and working principle of A transformer

PERFORMANCE OBJECTIVES:-Upon unsuccessful of this completion of this experiment, the students will be able to:

Understand the basic working principle of transformer Identify the constructional parts of Transformer

EQUIPMENT: Transformer

DISCUSSION:

Working Principal of a TransformerTransformer is a device that transfers electrical energy from one circuit to another by electromagnetic induction (transformer action). The electrical energy is always transferred without a change in frequency, but may involve changes in magnitude of voltage and current. The physical basis of transformer is mutual induction between two circuits. Linked by a common magnetic flux. The two coils posses’ high mutual inductance. If one coil is connected to a source of alternating voltage, an alternating flux is set up in a laminated core, most of which is linked with the other coil in which it produces mutually induced emf. According to Faraday’s law of electromagnetic Induction:

e= N dødt

Where:

e =induced voltage (electromotive force, emf) (V)

N = number of series-connected turns

dødt

=rate of change of flux through window (Wb/s)

If the, second coil circuit is closed & current floes in it and so electric energy is transferred (entirely magnetically) from the first coil to the second coil. The 1st coil in which electric energy is fed from the A.C. supply mains; is called primary winding and the other from which energy is drawn out, is called secondary winding. In brief a transformer is a device that:

Transfer electric power from one circuit to another.

It does so without a change of frequency.

It accomplishes this by Electro-magnetic induction.Where the two electric circuits are mutual inductive influence of each other.

Types and Construction of Transformers

Principle purpose of a transformer is to convert ac power at one voltage level to ac power of the frequency at another voltage level. Power transformers are given a variety of different names, depending on their use in power system. A transformer connected to the output of a generator and used


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to step up the voltage is called “unit transformer”. Transformer connected at the other end of the transmission line, which step down the voltage at the distribution level is called “substation transformer”. The transformer that takes the distribution voltage and step down the voltage at which the power is actually used (110, 208, 220 V, etc.) is called “distribution transformer” in addition to the various power transformers, two special-purpose transformer are used in the power system.A. Potential Transformer (PT). For Voltage sampling.

B. current Transformer (CT). For Current sampling.

Power transformers are constructed on one of two types of cores:

Core – typeShell - type

Core Type Transformer

This type of transformer consists of a simple rectangular laminated piece of steel with the transformer winding wrapped around two sides of the rectangle. Figure 1(a) shows the core type transformer.

Shell Type Transformer

It consists of a three – lagged laminated core with the windings wrapped one on top of the other with the low – voltage-winding innermost. It serves two purposes:

Simplifies the problem of insulating the high – voltage winding from the core.

Less leakage flux than if the two windings wrapped by distance on the core.

CONSTRUCTIONAL PARTS OF THE TRANSFORMER

CORE

The main purpose of the core offer low reluctance path for the flux. The cores are made of lamination of silicon steel to minimize the eddy current and hysteresis loss. Laminations are insulated from each other by means of varnish, impregnated paper or enamel.

WINDINGS

The windings are placed on the core. The winding is done with the insulated copper conductors. The winding which is connected to supply, is called Primary Winding and to which load is connected is called Secondary Winding. According to construction of windings are classified as under:

Cylindrical type: In this type of winding the length of the coil is equal to the length of the limb. The coils of high tension or low tension or say primary and secondary are wound keeping low tension near the cores.Sandwich type: In this type of winding the primary and secondary are placed one over the other alternately. These are suitably employed for the shell type transformer.

TERMINALS AND BUSHING

The leads of both windings are connected to the terminal so that the supply can be taken and connected to. Simple porcelain bushings are used to 20kV. Oil filled bushing are used to 33kV lines. For 132kV and above oil impregnated paper condenser bushing is used.

TANK


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The tank is used to accumulate the windings, core etc. in it. The tank of small transformer is made with iron sheets having the provision of ventilation and for connections to the load and supply.

CONSERVATIORIt is a small tank mounted over the top of the main tank. Conservator sometimes called as the expansion tank. A level indicator is fitted to check the level and color of the transformer oil. The main tank is completely filled with the transformer oil and the conservator partially.

With the increase and decrease of current (load) the heat produced is also increased and decreased; as a result the expansion and contraction of the oil takes place; so the conservator is not filled completely to facilitate the expansion etc. an instrument is also attached to indicate the temperature of the oil.

BREATHER

It is fitted between the air space of conservator and the outside vent. When oil in transformer expands, air is driven out. When oil cools, outside air enters through the breather. The incoming air is taken through Silica Gel contained in the breather, so that moisture of air is arrested and oil is prevented from getting contaminated with moisture of air. Dry silica gel is of pink color. It turns blue as it absorbs moisture. Drying can be regenerating the wet silica gel.

BUCHHOLZ RELAY

It is used for the protection of the oil filled transformer from developing faults. This relay is connected in the tank between pipe and conservator. At minor fault (over loading) in the transformer tank Buchholz relay gives alarm, but during major fault (short circuit) Buchholz relay closes trip circuit to cut off the supply. Buchholz relays are not provided for transformers below 500 kVA. (This is for economic consideration).

TAP CHANGER

A tap changer is provided with transformer for adjusting secondary voltage occasional adjustment in secondary voltage is made by off circuit changes. Daily or short time adjustments are made by on load tap changer. The tap changers are installed within the transformer tank.

TRANSFORMER OIL

Transformer oil (dielectric oil) is used as insulation and cooling medium in power transformer and instrument transformer. The fresh dielectric has pale clear yellow color. Dielectric oil should never contain suspended particles, water-soluble acids and bases. Moisture in the oil lowers the dielectric strength, thereby causing internal flashover. Viscosity indicates fluidity. Oil with low viscosity has more fluidity and gives better cooling.


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Cutaway view of large 3- phase oil – cooled power transfer

REVIEW QUESTION:

Q#1 what is the principle purpose of transformer?

______________________________________________________________________________________________________________________________________________________________________

Q#2 what is the function of Buchholz relay in the oil – immersed transformer?

______________________________________________________________________________________________________________________________________________________________________

Q#3 Write the name of special-purpose transformer.

______________________________________________________________________________________________________________________________________________________________________


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Q#4 what is the function of transformer oil?

________________________________________________________________________________________________________________________________________________________

Q#5 how eddies current and hysteresis losses are reduced in the transformer core?

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Q#6 why silica gel is used in breather?

______________________________________________________________________________________________________________________________________________________________________

Q#7 what is the function of taps?

______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Q#8 why tap changer is connected to HT winding?

______________________________________________________________________________________________________________________________________________________________________

Q 10 Write names against the number of the dc machine as shown in the table

No Name of the component 71 82 93 10


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4 115 126 13



To Study the Step-up and Step-down Transformer



EQUIPMENT: Transformer

DISCUSSION:A transformer is a static piece of equipment used either for raising or lowering the voltage of an a.c. supply with a corresponding decrease or increase in current. It essentially consists of two windings, the primary and secondary, wound on a common laminated magnetic core as shown in Fig. (2.1). the winding connected to the a.c. source is called primary winding (or primary) and the one connected to load is called secondary winding (or secondary). The alternating voltage V1 whose magnitude is to be changed is applied to the primary. Depending upon the number of turns of the primary (N1) and secondary (N2), an alternating e.m.f. E2 is induced in the secondary. This induced e.m.f. E2 in the secondary causes a secondary current I2. Consequently, terminal voltage V2 will appear across the load. If V2 > V1, it is called a step up-transformer. On the other hand, if V2 < V1, it is called a step-down transformer.


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Fig:- 2.01An ideal transformer is one that has(i) no winding resistance(ii) no leakage flux i.e., the same flux links both the windings(iii) no iron losses (i.e., eddy current and hysteresis losses) in the coreAlthough ideal transformer cannot be physically realized, yet its study provides a very powerful tool in the analysis of a practical transformer. In fact, practical transformers have properties that approach very close to an ideal transformer.E.M.F. Equation of a TransformerConsider that an alternating voltage V1 of frequency f is applied to the primary as shown in Fig.(2.01).

Above equations are calculated from the e.m.f. E1 induced in primary and rms values of E

In an ideal transformer, E1 = V1 and E2 =V2.

Voltage Transformation Ratio (K):From the above equations of induced e.m.f.,

The constant K is called voltage transformation ratio. Thus if K = 5 (i.e. N2/N1 = 5), then E2 = 5 E1.

For an ideal transformer:(i) E1 = V1 and E2 = V2 as there is no voltage drop in the windings.

(ii) there are no losses. Therefore, volt-amperes input to the primary are equal to the output volt- mperes i.e.

V1 I1 = V2 I2

Hence, currents are in the inverse ratio of voltage transformation ratio. This simply means that if we raise the voltage, there is a corresponding decrease of current.

A practical transformer differs from the ideal transformer in many respects. The practical transformer has (i) iron losses (ii) winding resistances and (iii) magnetic leakage, giving rise to leakage reactances.


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Step-Down TransformerBecause the same magnetic flux lines cut both coils of a transformer, the induced EMF in the secondary winding, is proportional to the number of turns on both the primary and secondary windings.

If the number of turns on the secondary winding is less than the number of turns on the primary winding, then the secondary output voltage will be less than the primary input voltage. This type of transformer is called a step-down transformer and is illustrated in Figure 02.

Applied AlternatingCurrent Supply

Primary W inding Secondary W inding

Having less turns thanthe Primary W inding

Step-Down Transformer

Fig:- 2.02

Step-Up Transformer

If the number of turns on the secondary winding of a transformer is greater than the number of turns on the primary winding, then the secondary output voltage will be greater than the primary input voltage. This type of transformer is called a step-up transformer and is illustrated inFigure 4.

Secondary W indingPrimary W inding

Applied AlternatingCurrent Supply

Having more turns thanthe Primary W inding

Step-Up Transformer

Fig:- 2.03


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Power Rating of Transformers

The power rating of a transformer may be calculated by multiplying the secondary AC voltage by the full load secondary AC current.

Rating = Secondary Voltage x Secondary CurrentRating = V x A ( U2 x I2 )Rating = VA

A rating quoted in VA will apply to small transformers. The rating of larger transformers will be quoted in kVA or MVA


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To perform an open circuit test on a single phase transformer to find out the iron losses.



EQUIPMENT: Single Phase Transformer Voltmeter Ammeter Wattmeter Connecting leads

DISCUSSION:

Circuit Diagram:

Theory:

This test is performed to find out the no load losses or iron losses (also called core losses) and no load current Io

which is helpful in determining Xo and Ro.

In this test, the rated voltage is applied to the primary (usually low voltage winding) while the

secondary is left open circuited. The applied primary voltage V1 is measured by the voltmeter, the no load

current Io by ammeter and no load input power Wo by wattmeter as shown in circuit diagram.

As the normal rated voltage is applied to the primary, therefore normal iron losses will occur in the

transformer core. Hence wattmeter will record the iron losses and small copper losses in the primary. Since no

load current Io is very small (usually 2 to 10% of rated current), Cu losses in the primary under no load

condition are negligible as compared with iron losses. Hence, wattmeter reading practically gives the iron losses

in the transformer.


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Procedure:

1. Make connections according to the given circuit.

2. Connect primary of transformer with rated ac voltage supply.

3. Note down readings of instruments connected and calculate different parameters.

Observations:

Sr.No Observations Calculations

Primary Side P.f=

P0/V1I0

R0=

V1/I0Cosθ

Xm=

V1/I0Sinθ

Voltage

V1

Current

I0

Power P0 Ohms Ohms

Volts Amps Watts

1

2

3

Results and conclusion:

The open circuit test on a transformer gives the parameters Ro and Xm and gives the

core losses.


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Review Questions:

Q1: Calculate the no-load power Po (core loss) using the following relation

P0= I2WR0 ____________________________ W

and compare it with the value of Po in Table

Q2: Does transformer draw any current when its secondary is open?

Ans:

Q3: If any current flow when secondary is open then why it flow?

Ans:

Q4: Why h.v side of transformer is open?

Ans:

Q5: Why are iron losses Constant at all loads in a transformer?

Ans:

Q6: What effects are produced by change in voltage?


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Ans:



To perform a short circuit test on a single phase transformer to find out the copper losses.



EQUIPMENT: Single Phase Transformer

Voltmeter

Ammeter

Wattmeter

Connecting leads

DISCUSSION:Circuit Diagram:

Theory:

In this test, the secondary (usually low voltage winding) is short circuited by a thick conductor and an ammeter, and variable low voltage is applied to the primary as shown in the circuit diagram. The low input voltage is gradually raised till a level (VSC) where full load current ISC flows in the primary. Then current in the secondary also has full load value. Under such conditions, the copper loss in the winding is the same as that on full load.


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There is no output from the transformer under short circuit conditions. Therefore input power is all loss and this loss is almost entirely copper loss. It is because iron loss in the core is negligibly small since the voltage is very small. Hence, wattmeter will practically register the full load copper losses in the transformer windings.

Procedure:


2. Connect primary of transformer with variable ac voltage supply.

3. Note down transformer rated current from name plate data and keep on increasing voltage until you get rated current read by Ammeter connected.

4. Once you get rated current at any specific voltage level, note down readings of instruments connected and calculate different parameters.

Observations:

Sr.N

o

Observations Calculations

Primary Side Secondar

y side

P.f=

P

sc/VscIs

Zsc=

Vsc/I0

R01=

Z

sc.Cosө

X01=

Zsc.Sin

ө

Voltag

e

Vsc

Curren

t Isc

Powe

r Psc

Current Ohm

s

Ohms

Volts Amps Watts Amps

1

2

3

Results and conclusion:

The short circuit test on a transformer gives the parameters R01 and X01 and gives the

copper losses.


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Review Questions:

Q1: Why L.V side of transformer is short?

Ans:

Q2: Cu-losses are constant or variable?

Ans:

Q3: If variable then how it can be varied?

Ans:

Q4: Is cu-loss affected by power factor?

Ans:


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Q5: If cu-loss is affected by power factor then why?



To find out the efficiency of a single -phase step down transformer.





Theory:


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A step-down transformer transforms the high voltage at primary side to a lower voltage at the secondary side. It works on the principle of mutual induction i.e. the transformer secondary winding has an induced emf due to the change in voltage across the primary winding. The efficiency of a transformer at a particular load and p.f is defined as the ratio between output power and input power.

η= (VSIS / VPIP) × 100 %

Procedure:



3. Increase the load on the secondary side in steps.

4. Note down readings of instruments connected and calculate different parameters

Observations:

Sr.No Resistance

(Ohms)

Vp (Volts) Ip (Amps) Vs (Volts) Is (Amps)

1

2

CALCULATIONS:

For First Load:

1. η= ( VsIs / VpIp ) × 100%=_______________________________

For Second Load:

1. η= ( VsIs / VpIp ) × 100%=_______________________________

RESULT

The efficiency have been calculated and increase in both parameters is observed with increases in load.


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Review Questions:

Q1: Draw the relation between I2 and V2 at

a) unity power factor (resistive load)b) Lagging power factor (Inductive load)c) Leading power factor (capacitive load)

Q2: If the transformer is loaded by purely capacitive load, what is the expected efficiency?

Ans:

Q3: List the different losses which take place in a transformer?

Ans:

Q3: What are various causes of voltage drop in a transformer?

Ans:


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Q4: In above question, under what conditions will only one of the components of voltage drop be present?

Ans:

Q5: In this experiment, even after the load on secondary is thrown off, wattmeter connected to

primary does read some power. Where is this power consumed?

Ans:

Q6: Draw the phasor diagram of transformer at no-load with

a) P.f 0.5 lagging

b) P.f 0.5 leading



To find out the voltage regulation of a single -phase step down transformer.






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Theory:

A step-down transformer transforms the high voltage at primary side to a lower voltage at the secondary side. It works on the principle of mutual induction i.e. the transformer secondary winding has an induced emf due to the change in voltage across the primary winding. When we increase load at the secondary terminals of a transformer, current drawn by transformer will increase. This increase in current will cause will increase in load dependant losses, Cu loss and leakage magnetic loss, hence causes decrease in output voltage. The change in secondary voltage from no load to full load with respect to no load voltage or with respect to full load voltage is called voltage regulation.

(Voltage Regulation) VR = [(VSN – VSL) / V SN] × 100 %

Procedure:



3. Switch on primary supply and read the no load secondary voltage.

4. Increase the load on the secondary side in steps.

5. Note down readings of instruments connected and calculate different parameters

Observations:

No load secondary voltage VSN = _______ Volts

Sr.No Vp (Volts) Ip (Amps) Vs (Volts) Is (Amps)

1

2

CALCULATIONS:


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For First Load:

1. VR = [(VSN – VSL) / VSN] × 100% =________________________________For Second Load:

1. VR = [(VSN – VSL) / VSN] × 100% =________________________________RESULT

The voltage regulation have been calculated and increase in both parameters is observed with increases in load.

Review Questions:

Q1: Is it possible to get a -ve value for the voltage regulation? Give reason.

Ans:

Q2: What is the best value of voltage regulation? Is it possible to get it practically?

Ans:

Q3: Calculate the power factor which yields the voltage regulation best value.

Ans:



To demonstrate the constructional parts and working principle of A 3-phase Induction Motor


Understand the basic working principle of 3-phase induction motor Identify the constructional parts of a induction motor

EQUIPMENT: Induction motor

DISCUSSION:


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Induction motor:

Induction motors are most common type of electrical motors. It is widely used because of its simple construction, economical cost and ruggedness.

As the name suggests, the motor works on principle of electromagnetic induction. In a way it can be called as rotating transformer because of the close similarity in principle of operation.

A schematic diagram of an induction motor is shown above. Pairs of electromagnetic poles are housed in a casing called stator. The e electromagnetic poles are wound with conductors to produce magnetism which is called stator windings. A rotating part called rotor is placed in the annular gap of the stator by suitable mountings in such a way that the rotor can rotate freely. The rotor may be either wire wound or may simply be bars of metal. In the latter case, the rotor is called squirrel cage type.

Construction:

The stator consists of wound 'poles' that carry the supply current to induce a magnetic field that penetrates the rotor. In a very simple motor, there would be a single projecting piece of the stator (a salient pole) for each pole, with windings around it; in fact, to optimize the distribution of the magnetic field, the windings are distributed in many slots located around the stator, but the magnetic field still has the same number of north-south alternations. The number of 'poles' can vary between motor types but the poles are always in pairs (i.e. 2, 4, 6, etc.).

Induction motors are most commonly built to run on single-phase or three-phase power, but two-phase motors also exist. Single-phase power is more widely available in residential buildings, but cannot produce a rotating field in the motor (the field merely oscillates back and forth), so single-phase induction motors must incorporate some kind of starting mechanism to produce a rotating field. Three-phase motors have three salient poles per pole number. This allows the motor to produce a rotating field, allowing the motor to start with no extra equipment and run more efficiently than a similar single-phase motor.There are three types of rotor:

a) Squirrel-cage rotor:The most common rotor is a squirrel-cage rotor. It is made up of bars of either solid copper

(most common) or aluminum that span the length of the rotor, and those solid copper or aluminium strips can be shorted or connected by a ring or sometimes not, i.e. the rotor can be closed or semi closed type. The rotor bars in squirrel-cage induction motors are not straight, but have some skew to reduce noise and harmonics.b) Slip ring rotor:

A slip ring rotor replaces the bars of the squirrel-cage rotor with windings that are connected to slip rings. When these slip rings are shorted, the rotor behaves similarly to a squirrel-


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cage rotor; they can also be connected to resistors to produce a high-resistance rotor circuit, which can be beneficial in startingc) Solid core rotor:

A rotor can be made from solid mild steel. The induced current causes the rotation.

Principle of operation:

A 3-phase power supply provides a rotating magnetic field in an induction motor.

The basic difference between an induction motor and a synchronous AC motor with a permanent magnet rotor is that in the latter the rotating magnetic field of the stator will impose an electromagnetic torque on the magnetic field of the rotor causing it to move (about a shaft) and a steady rotation of the rotor is produced. It is called synchronous because at steady state the speed of the rotor is the same as the speed of the rotating magnetic field in the stator.

By way of contrast, the induction motor does not have any permanent magnets on the rotor; instead, a current is induced in the rotor. To achieve this, stator windings are arranged around the rotor so that when energized with a polyphase supply they create a rotating magnetic field pattern which sweeps past the rotor. This changing magnetic field pattern induces current in the rotor conductors. These currents interact with the rotating magnetic field created by the stator and in effect cause a rotational motion on the rotor.

However, for these currents to be induced the speed of the physical rotor must be less than the speed of the rotating magnetic field in the stator (the synchronous frequency ns) or else the magnetic field will not be moving relative to the rotor conductors and no currents will be induced. If by some chance this happens, the rotor typically slows slightly until a current is re-induced and then the rotor continues as before. This difference between the speed of the rotor and speed of the rotating magnetic field in the stator is called slip. It is unit less and is the ratio between the relative speeds of the magnetic field as seen by the rotor (the slip speed) to the speed of the rotating stator field. Due to this, an induction motor is sometimes referred to as an asynchronous machine.

Synchronous speed:

To understand the behavior of induction motors, it is useful to understand their distinction from a synchronous motor. A synchronous motor always runs at a synchronous speed- a shaft rotation frequency that is an integer fraction of the supply frequency. The synchronous speed of an induction motor is the same fraction of the supply.

It can be shown that the synchronous speed of a motor is determined by the following formula:


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Where ns is the (synchronous) speed of the rotor (in rpm), f is the frequency of the AC supply (in Hz) and p is the number of magnetic poles per phase.

Slip:

The slip is a ratio relative to the synchronous speed and is calculated using:

Where

s is the slip, usually between 0 and 1

nr = rotor rotation speed (rpm)

ns = synchronous rotation speed (rpm)

Typical torque curve as a function of slip:

Procedure:

Firstly I identified different parts of the motor. They were as follows: Stator, Bearings, Frame, connection ports etc.

Then connected the leads and checked the working of motor.

Conclusion:

Different parts of 3 phase induction motor were successfully identified.

Review question:

Q1) What do you mean by slip of a motor? Also draw typical torque curve as a function of slip?

Ans.: Difference between the speed of the rotor and speed of the rotating magnetic field in the stator is called slip.


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Q2) Label the following diagram:

Ans.:



Production of flux and to check the polarity of main poles of the dc motor.



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EQUIPMENT: DC motor

DC supply

Connecting wires

Compass needle

Iron rod.

THEORETICAL BACKGROUND:

Motor is an electrical device which converts the electrical energy into

mechanical energy. If the applied supply is dc and motor is operated on it then this type of motor is called dc

motor. When the voltage is applied across its stator, the applied voltage produces the flux in the stator, whose

direction depends upon the direction of the applied current. If the direction of the current is reversed the flux

direction will also reversed.

Fig: direction of current and flux


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Fig: DC Motor

Procedure:

Connect the dc motor to the dc supply.

Switch on the supply.

Take the compass needle and bring it within the stator core.

The compass needle will direct according to the flux direction produced by the applied dc supply.

Now change the direction of the dc supply and again bring the compass needle within the stator core,

this time the compass needle will erect in opposite direction to the first case.

Take the iron bar and bring it near the stator part. The electronically magnetized stator will attract the

iron bar. By increasing the amount of the applied supply, the stator will attract the iron bar more

strongly.


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To Run the 3-ø Induction Motor on no Load



EQUIPMENT: 3- ø induction motor AC supply Volt meter Watt meter Connection leads

THEORY:Assume that the induction rotor is already rotating at no load conditions, hence its rotating

speed is near to synchronous speed. The net magnetic field Bnet is produced by the magnetization current IM . The magnitude of IM and Bnet is directly proportional to voltage E1 . If E1 is constant, then Bnet is constant. In an actual machine, E1 varies as the load changes due to the stator impedances R1

and X1 which cause varying volt drops with varying loads. However, the volt drop at R1 and X1 is so small, that E1 is assumed to remain constant throughout.

At no-load, the rotor slip is very small, and so the relative motion between rotor and magnetic field is very small, and the rotor frequency is also very small. Since the relative motion is small, the voltage ER induced in the bars of the rotor is very small, and the resulting current flow IR is also very small. Since the rotor frequency is small, the reactance of the rotor is nearly zero, and the max rotor current IR is almost in phase with the rotor voltage ER . The rotor current produces a small magnetic field BR at an angle slightly greater than 90 degrees behind Bnet. The stator current must be quite large even at no-load since it must supply most of Bnet .

The induced torque which is keeping the rotor running, is given by:

PROCEDURE:As shown in figure connect the motor with ac supply source at no load.Connect the measuring

instruments like ammeter,volt meter and watt meter for calculation voltage and no load core and iron losses.Vary the voltage at a constant frequency with the induction motor uncoupled from its load. Measured the total power input, the voltage and the current .


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CIRCUIT DIAGRAM:

Motor connection for no loadCALCULATION:

The Core loss resistance can be found at rated voltage-

where

The magnetizing reactance can be found by:

Where Im = I NL sin

Reading.No Phase Voltage / Volts

PhaseCurrents / Amps

Watt per phase Speed / rpm

12


# 31



Check the relation between synchronous speed and poles of 3-Ф induction motor



EQUIPMENT: 3- Ф induction motor (2 poles) 3- Ф induction motor (4 poles) Ac supply Techo meter Connection lead Torque meter

THEORY:

To understand the behaviour of induction motors, it is useful to understand their distinction from a synchronous motor. A synchronous motor always runs at a synchronous speed- a shaft rotation frequency that is an integer fraction of the supply frequency. The synchronous speed of an induction motor is the same fraction of the supply.


where ns is the (synchronous) speed of the rotor (in rpm), f is the frequency of the AC supply (in Hz) and p is the number of magnetic poles per phase.

For example, a 6 pole motor operating on 60 Hz power would have a speed of:

Note on the use of p - some texts refer to number of pole pairs per phase instead of number of poles per phase. For example a 6 pole motor, operating on 60 Hz power, would have 3 pole pairs. The equation of synchronous speed then becomes:

with P being the number of pole pairs per phase.

Procedure:


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For this two induction motors 1st of 2 poles and 2nd of 4 poles is required. First connect 2 pole induction motor with ac regulated supply source through a torque meter and measure the speed. Same procedure rpeat for 4 poles machine and compare their speeds.

Observation Table:

Sr.No Input Voltage / Volts

2 poles / RPM 4 poles / RPM

1 29.33 1437 960

Remarks:

So it proves from observation table that speed of rotor is inversly proportional to the no.of poles.


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Check the relation between synchronous speed and frequency of 3-Ф induction motor



EQUIPMENT: 3- Ф induction motor Ac supply Frequency variable drive Connection lead

THEORY:To understand the behaviour of induction motors, it is useful to understand their distinction from a synchronous motor. A synchronous motor always runs at a synchronous speed- a shaft rotation frequency that is an integer fraction of the supply frequency. The synchronous speed of an induction motor is the same fraction of the supply.


where ns is the (synchronous) speed of the rotor (in rpm), f is the frequency of the AC supply (in Hz) and p is the number of magnetic poles per phase.

For example, a 6 pole motor operating on 60 Hz power would have a speed of:

Note on the use of p - some texts refer to number of pole pairs per phase instead of number of poles per phase. For example a 6 pole motor, operating on 60 Hz power, would have 3 pole pairs. The equation of synchronous speed then becomes:

with P being the number of pole pairs per phase.

Procedure:


# 34

Connect the three phase induction motor with ac supply source through frequency variable drive.By changing the frequency of ac supply observe effect on motor’s rpm.

Observation Table:

Sr.No Input Voltage / Frequency

4 poles / RPM

1 25 6052 40 9683 50 1200

Remarks:

From observations it is clear that rpm of induction motor is directly proportional to frequency of ac supply source.


# 35



To study the Star-Delta starting method of 3-phase induction motors.



EQUIPMENT: 3-phase Induction motor 3-phase supply Star-Delta starter

THEORY:As we know, once a supply is connected to a three phase induction motor a rotating magnetic field will be set up in the stator, this will link and cut the rotor bars which in turn will induce rotor currents and create a rotor field which will interact with the stator field and produce rotation. Of course this means that the three phase induction motor is entirely capable of self starting. The need for a starter is to reduce heavy starting currents.Star Delta starter:This is the most common form of starter used for three phase induction motors. It achieves an effective reduction of starting current by initially connecting the stator windings in star configuration which effectively places any two phases in series across the supply. Starting in star not only has the effect of reducing the motor’s starting current but also the starting torque. Once up to a particular running speed a double throw switch changes the winding arrangements from star to delta whereupon full running torque is achieved. This is applicable to motors designed for delta connection in normal running conditions. Both ends of each phase of the stator winding are brought out and connected to a 3-phase change-over switch.For starting, the stator windings are connected in star and when the machine is running the switch is thrown quickly to the running position, thus connecting the motor in delta for normal operation.The phase voltage of the

motor in star connection is reduced to 1

√3 of the direct-on-line value in delta.

A disadvantage of this method is that the starting torque (which is proportional to the square of the applied voltage) is also reduced to 1/3 of its delta value.


# 36

PROCEDURE:1. To start the induction motor first set the switch on STAR for some time, in this condition its

current and torque are low.2. After some time (about 10 seconds) set the switch to DELTA position. Now motor will run at

its rated speed and take its rated current.

CONNECTION DIAGRAM:


# 37

QUESTIONS:1. What is the disadvantage of star-delta starter?

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Documents

Lab Manual of Ac Machines