1. EET 306
ELECTRIC MACHINES
Syafruddin Hasan

2. Power Flow Diagram of an IM

3. Figure Motor energy balance flow diagram.
Power Flow Diagram
Figure Motor energy balance flow diagram.

4. Induction Motors Figure shows the energy balance in a motor.
The supply power is:
The power transferred through the air gap by the magnetic coupling is the input power (Psup) minus the stator copper loss and the magnetizing (stator iron) loss.
The electrically developed power (Pdv) is the difference between the air gap power (Pag) and rotor copper loss.

5. Induction Motors
The electrically developed power can be computed from the power dissipated in the second term of rotor resistance:
The subtraction of the mechanical ventilation and friction losses (Pmloss) from the developed power gives the mechanical output power

6. Power and Torque in an IM
From Fig.7-12 Input current : I1 = VΦ / Zeq Pcu Stator: Pscl = 3 I12R1 Core Losses : Pcore = 3 E12 Gc Air-gap power : PAG = Pin – Pscl – Pcore = 3I22R2/s Pcu Rotor: PRcl = 3 IR2RR = 3 I22R2 Note PAG : Pconv : PRCL = 1 : (1-s) : S Example 7-3

7. Induction Motors The motor efficiency:
Motor torque (shaft load torque):

8. Equivalent Circuit
Figure Single-phase equivalent circuit of a three- phase induction motor.

9. Figure Modified equivalent circuit of a three-phase induction motor.
Induction Motors
Figure Modified equivalent circuit of a three-phase induction motor.
The rotor impedance is transferred to the stator side. This eliminates the transformer

10. Induction Motors
Figure Simplified equivalent circuit of a three-phase induction motor.

11. Induction Motors
The last modification of the equivalent circuit is the separation of the rotor resistance into two parts:
The obtained resistance represents the outgoing mechanical power

12. Blocked-Rotor test (cont…)
The stator resistance was measured directly

13. IM Design Classes
NEMA & IEC : Class A: normal Tst, normal Ist, normal S Class B: normal Tst, low Ist, low S Class C: high Tst, low Ist, low S Class D: high Tst, low Ist, high S Class E: low Tst, normal Ist, low S Class F: low Tst, low Ist, normal S

14. Starting Induction Motors
Problems : - High Starting current
- Low starting torque
Determining of Istart
Read the rated voltage, hp and code letter from name plate.
Starting apparent power :
Sstart = (rated hp)(code letter factor)
Istart = Sstart / √3 VT
Example 7-7

15. Q U I Z
1. Why the circuit equivalent of an induction motor can be approached by equivalent circuit of a transformer and draw it
2. The output of a 3 phase induction motor is 9 kW. Rotor copper losses is 0.5 kW.
Motor runs at 5 % of slip. Stator loss is 0.75 kW
a). Calculate the mechanical power that converter by this
motor
b). Determine the input power
c). Calculate the efficiency

16. Speed Control of IM ns = 120 f / p
by changing the electrical frequency (f)
by changing the number of poles (p)
The methode of consequent poles
Multiple stator windings
by changing the line voltage:
n proportional to V2
by changing the rotor resistance
Solid-state IM drives

17. Motor Protection
Over Current Protection
Over Load Protection

18. IM Ratings Output power Voltage Current Power Factor Speed
Nominal efficiency
NEMA design Class
Starting Code
Service Factor (SF)