1. Explain what is meant by the synchronous speed of a three-phase induction mot
ID: 2990171 • Letter: 1
Question
1. Explain what is meant by the synchronous speed of a three-phase induction motor.
2. What two conditions determine the synchronous speed of a three-phase induction
motor?
6. A six-pole, three-phase, 60-Hz induction motor has a full-load speed of 1140 r/
min. Determine
a. the synchronous speed.
b. the revolutions slip.
c. the percent slip.
d. the rotor frequency.
10. A three-phase, 60-Hz, four-pole, 220-V, squirrel-cage induction motor takes 52
A per terminal at full load. The power factor is 0.85 lag, and the efficiency is 88%.
The slip is 3.0%. At the rated load, determine
a. the speed in r/min.
b. the horsepower output of the motor.
c. the total losses.
11. A three-phase, 60-Hz, six-pole, 220-V, squirrel-cage induction motor has a full-
load output of 15 hp. The full-load efficiency is 87%, and the power factor is 0.88
lag. The windings of the motor are connected in delta. At the rated load, determine
a. the line current.
b. the phase winding voltage.
Explanation / Answer
Answer-1
Synchronous Speed- It is the speed by which magnetic field rotates in space with respect to the stator. The induction motor try to get the synchronous speed but fails becauuse at synch speed the developed torque becomes zero.
synchronous speed Ns= 120*f/P
f= supply frequency, P= no. of poles in the motor.
Answer-2
two conditions determinining the synchronous speed -
sincesynchronous speed Ns= 120*f/ P
f= supply frequency, P= no. of poles in the motor.
So the conditions are 1. supply frequency and 2. no. of poles in the motor
Answer 6.
given poles P=6 , frequency f=60 , N=1140
a. the synchronous speed Ns= 120*f/ P= 120*60/6= 1200 RPM
b. the revolutions slip s= 1- (N/ Ns) = 1- 1140/1200= 0.05
c. the percent slip. s= 0.05*100=5 %
d. the rotor frequency fr= s*f= 0.05*60=3 Hz
Answer 10.
Given f=60, P=4 , s=3%=0.03, efficiency n= 88%=0.88, power factor pf=0.85
line voltage VL= 220 V line current IL= 52 A
a. the speed in r/min. N= Ns (1-s)= (120*60/4) *(1-0.03)=1800*0.97=1746 RPM
b. the horsepower output of the motor.
Output= input *efficiency
input= root(3) VL*IL *pf= 1.73*220*52*0.85= 16842.5 W
output= input*efficiency= 16842.5 * 0.88=14821.366 W.= 14.8 kW
c. the total losses.= input- output= 16842.5-14821.36= 2021.13 W= 2.021 kW
Answer= 11.
given VL =220, pf=0.88
a. the line current.
given Output= 15hp= 15*0.746 KW= 11.19KW
full load input= output/ efficiency = 11.19/0.87= 12.86 kW
since input= root(3) VL*IL *pf
so 12.86*1000= 1.73*220*0.88*IL
Solving line current IL = 38.35 A
b. the phase winding voltage
since winding is connected in delta so phase winding voltage= input voltage= 220 V
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