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Induction motor

Construction

• Stator + Rotor

•

Stator circuit

has

three

sets

of

coils

which

are

separated by 120o and are excited by a three‐phase power supply.

• The rotor circuit is also composed of three‐phase windings that are shorted internally (within the rotor structure) or externally (through slip rings and brushes)

• Rotor has two types: squirrel cage rotor and

wound rotor


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Rotating

magnetic

field

•

Stator is

supply

by

a balanced

three

phase

system

Flux

direction


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Flux

equation

•

At time

t1

• At time t2

•

At

time

t3

Some

important

equation

• Synchronous speed

• Ns is the synchronous speed

• F is the frequency

• PP is the number of pairs of pole

• P is the number of poles

• Slip ‐ S is slip. S is between 0 and 1

‐ Ns

is

the

synchronous

speed

‐ N is the mechanical speed‐


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Equivalent

circuit

of

induction

motor

‐ R1: stator resistance

‐ X1:

stator

reactance

‐ Rm and Xm: represent

the core of the motor

‐ E1: is the voltage of the

source minus the voltage

drop on the stator

‐ N1:N2 is the equivalent

number of turns of stator

comparing to

that

of

rotor

‐ R2: rotor resistance

‐ X2: rotor reactance

‐ S: slip

Equivalent

circuit

of

induction

motor


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More

equivalent

circuit

of

induction

motor

Power

flow


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Power

flow

Example

• A 50 hp, 60 Hz, three‐phase, Y‐connected induction motor operates at full load at a speed of 1764 rpm. The rotational losses of the motor are

950W,

the

stator

copper

losses

are

1.6

kW,

and

the

iron

losses

are

1.2 kW. Compare the motor efficiency?


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Torque

characteristics

More

regions

of

torque

characteristics

• Large slip region

• Starting torque


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More

regions

of

torque

characteristics

•

Small slip

region

• Slip at maximum torque

Example

• A 50 hp, 440 V, 60 Hz, three‐phase, four‐pole induction motor develops a maximum torque of 250% at slip of 10%. Ignore the stator resistance

and

rotational

losses.

Calculate

the

following

• Speed of the motor at full load

• Copper losses of the rotor

• Starting torque of the motor


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Starting

procedure

Starting

procedure


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Example

•

An induction

motor

has

a stator

resistance

of

3 Ohm,

and

the

rotor

resistance referred to the stator is 2 Ohm. The equivalent inductive reactance Xeq=10 Ohm. Calculate the change in the starting torque if the voltage is reduced by 10%. Also, compute the resistance that should be added to the rotor circuit to achieve the maximum torque at starting.

Speed

control

of

induction

motor

• Armature or rotor resistance

• Armature or rotor inductance

• Magnitude of terminal voltage

• Frequency of terminal voltage

• Voltage/Frequency control

• Rotor voltage injection

• Slip energy recovery


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Controlling

speed

by

using

rotor

resistance

•

In steady

state

condition,

the

motor

operates

near

the

synchronous

speed. Hence, small slip region.

Controlling speed by using rotor resistance

• Consequences of change in rotor resistance

• The synchronous speed does not change

• The maximum torque does not change

• The slip at maximum torque change

• With the increase in rotor resistance, the starting torque decreases

• Inconveniences

• Small range of speed variation (i.e speed change from position 1 to position 2 at torque T1)


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Example

•

A three

phase,

Y connected,

30

hp (rated

output),

480

V,

six

pole,

60

Hz, slip ring induction motor has a stator resistance R1=0.5 Ohm and a rotor resistance referred to stator R’2=0.5 Ohm. The rotational losses are 500 W and the core losses are 600 W. Assume that the change in the rotational losses due to the change in speed is minor. The motor load is a constant‐torque type.

• At full load torque, calculate the speed of the motor?

• Calculate the added resistance to the rotor circuit needed to reduce the speed by 20%?

•

Calculate

the

motor

efficiency

without

and

with

the

added

resistance?

• If the cost of energy is 0.05 USD/kWh, compute the annual cost of operating the motor continuously with the added resistance. Assume that the motor operates 100 hours a week.

Controlling speed using inductance

• Adding inductance to the motor windings is an unrealistic option for the following reasons:

• The physical size of the inductance required t make a sizable change in speed is likely to be larger than the motor itself.

• Variable inductance requires expensive and elaborate design

• The insertion of inductance reduce the starting torque

• The insertion of inductance consumes reactive power that further lowers the already low power factor of induction motor


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Controlling speed by adjusting the stator voltage

Controlling speed by adjusting the stator voltage

• The torque of the motor is proportional to the square of its stator voltage

• Synchronous speed does not change

• Decreasing stator voltage will decrease also the starting torque

• Slip at maximum torque does not move


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Example

•

For the

motor

given

in

the

above

example,

assume

that

the

load

torque is constant and equal to 120 Nm. Ignore the rotational losses and calculate the motor speed at full voltage. Repeat the computation if the voltage is reduced by 20%.

Controlling speed by adjusting the supply frequency

• Synchronous speed


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Controlling speed by adjusting the supply frequency

•

Decreasing the

frequency

increases

in

return

the

starting

current

Effects

of

excessively

high

frequency

• An increase in the no‐load speed

•

A decrease

in

the

maximum

torque


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Effects

of

excessively

high

frequency

•

A decrease

in

starting

torque

• An increase in speed at the maximum torque

• A decrease in the starting current

Example

• A 480 V, two pole, 60 Hz, Y connected induction motor has an inductive reactance of 4 Ohm and a stator resistance of 0.2 Ohm. The rotor

resistance

referred

to

the

stator

is

0.3

Ohm.

The

motor

is

driving

a constant‐torque load of 60 Nm at a speed of 3500 rpm. Assume that this torque includes the rotational losses.

• Compute the maximum frequency of the supply voltage that would not result in stalling the motor.

• Calculate the motor current at 60 Hz, and at the maximum frequency.

• Calculate the power delivered to the load at 60 Hz, and at the maximum frequency.


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Effect of excessively low frequency

•

Reducing the

supply

frequency

reduces

the

speed

of

the

motor.

However, frequency reduction may results in an increase in motor current. At very low frequencies, the equivalent reactance of the motor Xeq is very low. Since Xeq is the limiting parameter for motor current at starting, its large reduction could lead to an excessive current beyond the ratings of the machine.

Example

• For the motor described in the above example, compute the motor speed and starting current if the frequency is decreased to 50 Hz.


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Voltage/Frequency

control

Voltage/Frequency

control


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Example

•

Repeat the

above

example

with

constant

v/f

control?

Q&A