Transcript AC Machines 2 - UniMAP Portal

Syafruddin Hasan
Motor
input
in
stator
Stator
copper
and
iron
losses
Rotor
input
Rotor
copper
losse
Mech
Power
developed
or
Gross
Torque
Windage
and
Friction
losses
Rotor
output
or
BHP
Power Flow Diagram
Air gap
Input power
Psup
Air gap
power Pag
Developed power
Pdv = 3 Irot2 Rrot (1-s)/s
Output power
Pout
Ventilation and
loss friction losses
Stator Iron loss Rotor Copper
3 Irot2 Rrot
3 Vsta2 / Rc
Stator Copper loss
3 Ista2 Rsta
Figure Motor energy balance flow diagram.
Induction Motors

Figure shows the energy balance in a motor.

The supply power is:

Psup  Re(S sup )  Re 3 Vsup I
*
sta


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.
Induction Motors

The electrically developed power can be
computed from the power dissipated in the second
term of rotor resistance:
Pdv  3 I rot_t

2
1 s 

 Rrot _ t

s 

The subtraction of the mechanical ventilation and
friction losses (Pmloss) from the developed power
gives the mechanical output power
Pout  Pdv  Pmloss
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
Induction Motors

The motor efficiency:
Pout

Psup

Motor torque (shaft load torque):
T
Pout
m
Equivalent Circuit
X sta =  sy L sta
V sup
Ista
Stator
R sta
Irot_t
X rot_m =  rot L rot
R rot
Irot
Rc
Xm
V sta
V rot = s V rot_s
Rotor
Figure Single-phase equivalent circuit of a threephase induction motor.
Induction Motors
Xsta
Vsup
Rsta
Ista
Stator
Rc
Irot_t
Xm
Vsta
Xrot
Vrot_s
Rrot/s
Irot
Rotor
Figure Modified equivalent circuit of a three-phase induction
motor.
The rotor impedance is transferred to the stator side. This
eliminates the transformer
Induction Motors
Xsta
Vsup
Rsta
Ista
Xrot_t
Rc
Xm
Vsta
Rrot_t/s
Irot_t
Stator
Rotor
Air gap
Figure Simplified equivalent circuit of a three-phase
induction motor.
Induction Motors

The last modification of the equivalent circuit is
the separation of the rotor resistance into two
parts:
Rrot _ t
s

 Rrot _ t

1 s 

R
s
rot _ t
The obtained resistance represents the outgoing
mechanical power
1  s  R
s
rot _ t
Blocked-Rotor test (cont…)
Vblocked_ln 
Pblocked_A 
Vblocked
3
Pblocked
3
Vblocked_ln  21.939 V
Re 
P
Pblocked_A
2 160 W
blocked_A
I
blocked
The stator resistance was measured directly
Rrot_t  Re  Rsta
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
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
Q U I Z 04-10-2005

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
Speed Control of IM


ns = 120 f / p
by changing the electrical frequency (f)
by changing the number of poles (p)
(1) The methode of consequent poles
(2) Multiple stator windings



by changing the line voltage:
n proportional to V2
by changing the rotor resistance
Solid-state IM drives
Motor Protection
Over Current Protection
 Over Load Protection

IM Ratings
1.
2.
3.
4.
5.
6.
7.
8.
9.
Output power
Voltage
Current
Power Factor
Speed
Nominal efficiency
NEMA design Class
Starting Code
Service Factor (SF)