(A) What Is the Synchronous Speed of the Rotor in Rad/Sec ?

Module PE2

Problem 1 A three-phase, 4 pole, 60 Hz induction motor has a line-to-line voltage applied across its terminals equal to 1214 volts. The parameter values for this motor are: R1=0.3 ohms, X1=1 ohm, Rc=large, Xm=30 ohms, R’2=0.3 ohms, X’2=1 ohm. The slip corresponding to maximum torque is 0.1507. The current flowing in the rotor circuit is |I’2|=279 amperes.

(a) What is the synchronous speed of the rotor in rad/sec ? (b) What is the difference in speed between rotor speed at maximum torque and synchronous rotor speed ? (c) Compute the Thevenin equivalent voltage and impedance “looking left” from the rotor circuit. (d) Compute the maximum torque of this motor.

Solution to problem 1 2  f  2  2 (a) s     377   188.5radians / sec 1  p  4

s m (b) s   s m  s s  0.2626188.5 49.5radians / sec s

Z th  Z a Z b  Z a  0.3  j1, Z b 0.3  j1 j30 Z th   0.28027  j0.9705 (c) 0.3  j31 1214  j30  Vth      678.23  j6.564  678.30.554V  3  0.3  j31

' 2 ' R2 2 0.3 (d) Td  3 I 2   3 279V    1415.6ntu  m s  s 0.2626188.5rad / sec

Problem 2 A three-phase, 4 pole, 60 Hz induction motor has a line-to-line voltage applied across its terminals equal to

1214 volts. The parameter values for this motor are: R1=0.3 , X1=1 , Rc=, Xm=30 , R’2=0.3 , X’2=1 . The slip corresponding to maximum torque is 0.1507. You will recall that the expression for torque is

' 2 ' 3 I2  R2 TD  s s

(a) Find the Thevenin equivalent voltage and impedance seen “looking left” from the rotor circuit. (b) Compute the speed of this motor, in mechanical rad/sec, under the maximum torque condition. (c) Compute the maximum torque of this motor. Solution to problem 2 0.3  j1 j30 (a) Z   0.2809  j0.9705 th 0.3 j31 1214 j30 Vth    678.23  j6.5635  678.260.554V 3 0.3  j31 2 (b)  s   377  188.5 p

 m   s 1 s  188.51 0.0.1507  160.093rad / sec (c) Method 1 : V 678.260.554 I '  th   225.55  40.39 2 R ' 0.3 Z  2  jX ' 0.2809  j0.9705   j1 th s 2 0.1507 0.3 T  3 225.552   1611.77N  m Dmax 188.5 0.1507 Method 2 : ' R2 0.3 ST    0.1507 2 ' 2 2 2 Rth  X th  X 2  0.2809  0.9705 1 3678.262 Tmax   1611.74N  m 2188.5 0.2808  0.28092  0.9705 12   

Problem 3 A three-phase, 4 pole, 60 Hz induction motor operates at a slip of s=0.08. The mechanical power developed at the shaft is PD = 5 kW. (a) Compute the mechanical speed of this motor. (b) Compute the power flowing across the air gap.

Solution to problem 3  2   2  s   377   377 188.5rad / sec (a)  p   4 

m  1.08188.5rad / sec 173.42rad / sec 1656.9rpm

' 2 ' R2  1 s ' 2 PD  3 I 2 ' R2 1 s s ' 3 I 2 2 ' R2 PD s (b) ; Pg  3 I 2   '  1 s s P 2 R P 5kW g 3 I ' 2 P  D   5.435kW 2 s g 1 s 1 .08 Problem 4 A 10 hp, 440 volt (line-to-line), three-phase induction motor has synchronous speed of 1800 and runs at 1700 rpm at the full load power output of 10 hp. The stator copper loss is 200 W and the rotational loss is

400 W. Assume the core losses are zero, i.e., Pc = 0. Determine

(a) Power developed, Pm (otherwise known as PD) (b) Slip (c) Input power to the rotor, Pg (d) Rotor copper losses, Prc (e) Total power input, Pin (f) The magnitude of the current drawn by the motor, assuming the motor power factor is 0.80

Solution to problem 4 (a) Pm  Pout  Protationalloss  7460  400  7860W 1800 1700 (b) s   0.0556 1800 1 (c) P  P  8322.7W g 1 s m s (d) P  P  P  s  P   P  462.7W rc g m g 1 s m

(e) Pin  Pg  Psc  8522.7W (f) P  3V  I  cos P 8522.7 I    13.98A 3V cos 3  440 0.8

Problem 5

A three phase, 4 pole, 60 Hz induction motor operates with the parameter values: R1=0 , X1=3 , Rc=,

Xm=30 , R’2=1 , X’2=3 .

(a) Compute the synchronous rotor speed of this motor, ws , in rad/sec.

(b) Compute the slip corresponding to a rotor speed of wm =160 rad/sec ' (c) Compute the mechanical torque developed at the rotor shaft when I 2  80 amperes and the speed is

wm =160 rad/sec.

Solution to problem 5 2  f  2  2 (a)  s     377     188.5rad / sec 1  p 4

 s  m 188.5 160 (b) s    0.1512  s 188.5

' 2 R  1 s 2 1 1 .1512 P  3 I ' 2  3 80  107.78kW 2 s .1512 (c) P 107.78kW T    673.7ntu  m  m 160rad / sec Problem 6 A three-phase, 6 pole, 60 Hz, wye-connected induction motor has a line to line voltage applied across its terminals of 220 volts. The parameter values for this motor are: R1 = 0.3 ohms, X1 = 0.5 ohms, Rc = large, Xm = 15 ohms, R’2 = 0.15 ohms, X’2 = 0.2 ohms. For a slip of s = 0.02, compute:

(a) The synchronous speed ws and the rotor speed wr in rad/sec. (b) The Thevenin equivalent voltage and impedance “looking left” from the rotor circuit. (c) The current magnitude in the rotor part of the equivalent circuit. (d) The power delivered to the rotor across the airgap. (e) The power and torque developed at the shaft.

Solution to problem 6

(a) ws=(2/p) (377)=(2/6)(377)=125.67 rad/sec (=1200 rpm), wr=(1-s) ws=(1-0.02)125.67=123.16 rad/sec (b) Za=R1+jX1=0.3+j0.5, Zb=(Rc//jXm)=jXm=j15 Zth=(ZaZb)/(Za+Zb)=(0.3+j0.5)(j15)/(0.3+j0.5+j15)=0.2809+j0.4893

Vth=V1(Zb)/(Za+Zb)=(220/Ö3)(j15)/(0.3+j0.5+j15)= 122.9 Ð 1.1 °@ 122.9 (c) I’2=Vth/(Zth+(R’2/s)+jX’2)=122.9/(0.2809+j0.4893+(0.15/0.02)+j0.2)=122.9/(7.781+j.6893)

=15.6721-j1.3884=15.734Ð0.5° è |I’2| =15.734 amperes 2 2 (d) Pag=3|I’2| R’2/s=3(15.734) (0.15/0.02)=5570 watts 2 2 (e) Pd=3|I’2| R’2(1-s)/s=3(15.734) *0.15(1-.02)/0.02=5459 watts

Td=Pd/wr=5459/123.16=44.32 ntn-m