Lecture 23 EE743

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Lecture 23 - EE743
3-Phase Induction Machines Performance calculation
Professor Ali Keyhani
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Equivalent circuit
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Power calculations
Pc core losses
Power crossing the air gap
Rotor losses
Developed power
Electromagnetic power
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Power calculations
Rotational losses
To load
Pi
P0
Rotational losses
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Power calculations
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Power calculations
Recall
assume
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Power calculations

• Also note

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Induction generator

• For induction generators
• ?mgt?sync that is mechanical power supplied by the
  prime mover
• ?m and ?sync are rotating in the same direction
• generator action Electrical power is supplied to
  the network via stator terminals
• Also note for this case (?mgt?sync)

N
?sync
S
?sync
N
Tm
?m

• that is the slip s is negative. Therefore
  negative slip means generator action.

Text
S
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Special cases

• Stand still condition (?m0)that is s1
  implies stand still
• s0 implies ?m?sync that is the induction
  machine is running at synch speed. Induction
  machine can not run at ?sync ?m since frsfs gt
  fr0 (No power transmission)

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Example 1

• A six-pole induction motor is supplied by a
  synchronous generator having four poles and
  running at 1500 rpm. If the speed of the
  induction motor is 750 rpm, what is the rotor
  frequency

nm1500 rpm
nm750 rpm
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Example 1
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Example 2

• The input to the rotor circuit of a four-pole
  60Hz induction motor, running at 1000 rpm is 3kW.
  What is the rotor copper losses?

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Example 2
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Example 3

• The stator current of a 400 V, 3-phase
  WYE-connected, four-pole, 60Hz induction motor
  running at s0.06 is 60A at P.F.0.866. The
  stator copper losses is 2700 W, the total iron
  and rotational losses are 3600 W. Calculate the
  efficiency of the motor.

400 V
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Example 3