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3 Phase Induction Motor

Solved Problems on 3 φ Induction Motor - Electrical Engineering 

(By Prof Sunil Kumar, Electrical Engineering, IEC College, Greater Noida, Uttar Pradesh) 

Ques1: A 3 φ 4 pole 50 hz induction motor runs at 1460 r.p.m. find its %age slip.
Solution
N s = 120f/p = 120*50/4 = 1500r.p.m.
Running speed of motor = n= 1460r.p.m.
Slip S=( N s–N)/ N s*100 =(1500-1460) x 100 / 1500 = 2.667% 

Ques2: A 12 pole 3 φ alternator driver at speed of 500 r.p.m. supplies power to an 8 pole 3 φ induction motor. If the slip of motor is 0.03p.u, calculate the speed.
Solution
Frequency of supply from alternator, f=PN/120
=12*500/120 = 50hz

where P= no of poles on alternatev
N=alternator speed is r.p.m.
Synchronous speed of 3 φ induction motor
N=120f/Pm
=120*50/8 = 750 r.p.m.
Speed of 3 φ induction motor N=Ns (1-s)
=750(1-0.03) = 727.5 r.p.m.

 

Ques3: A motor generates set used for providing variable frequency ac supply consists of a 3-φ synchronous and 24 pole 3 φ synchronous generator. The motor generate set is fed from 25hz, 3 φ ac supply. A 6 pole 3 φ induction motor is electrically connected to the terminals of the synchronous generator and runs at a slip of 5%. Find
  i) the frequency of generated voltage of synchronous generator
  ii) the speed at which induction motor is running
Solution
Speed of motor generator set
Ns=(120*f1(supply freq))/(no of pole on syn motor)
=120*25/10 = 300 r.p.m.

(1) frequency of generated voltage
fz=speed of motor gen set voltage *no of poles on syn gen/120
= 300*24/120 = 60hz
(2) Speed of induction motor , Nm=Ns(1-s)
=120 fz /Pm(1-s) = 120*60/6(1-0.05) = 1140r.p.m.  

 

Ques4: A 3-φ 4 pole induction motor is supplied from 3φ 50Hz ac supply. Find
(1) synchronous speed
(2) rotor speed when slip is 4%
(3) the rotor frequency when runs at 600r.p.m.
Sulution
1) Ns =120f/p
=120*50/4 = 1500 r.p.m.

2) speed when slip is 4% or .04
N=Ns (1-s)
=1500(1-0.04) = 1440 r.p.m.
3) slip when motor runs at 600 r.p.m.
S’=(Ns –N)/Ns
=(1500-600)/1500 = 0.6

Rotor frequency f’ = S’f = 0.6*50 = 30Hz. 

 

Ques5: A 12 pole 3-φ alternator is coupled to an engine running at 500r.p.m. If supplied a 3φ induction motor having full speed of 1440r.p.m.
Find the %age slip, frequency of rotor current and no of poles of rotor.
Ans
Frequency of supply from alternator f=Pa*Na/120
=12*500/120 = 50Hz

Full load speed Nf =1440 r.p.m.
The no of poles (nearest to and higher than full load speed of motor =1440) should be in even nos.
P=120f/n = 120*50/1440 = 4
Ns = 120f/Pm = 120*50/4 = 1500 r.p.m.
% Slip s = (Ns-N)/Ns x 100 =(1500-1440) x 100 / 1500 = 4%

Rotor frequency f’ = sf = 0.04*50 = 2Hz

No A poles of the motor = 4 

 

Ques6: The rotor of 3φ induction motor rotates at 900r.p.m. when states is connected to 3φ supply .find the rotor frequency.
Solution Nr =980 r.p.m., f=50Hz, Ns=120f/p
When P=2, Ns=3000r.p.m.,P=4, Ns=1500

P=6, Ns=1000, P=8, Ns=750r.p.m.

As we know that synchronous speed is slightly greater than rotor speed.

Ns=1000 r.p.m. P=6

Fr=Sf=(Ns-N)/Ns*f=Sf = (1000-980) x 50 / 1000 

 

Ques7: A 3 φ 50Hz induction motor has a full load speed of 960 r.p.m
(a) find slip
(b) No of poles
(c) Frequency of rotor induced e.m.f
(d) Speed of rotor field w.r.t. rotor structure
(e) Speed of rotor field w.r.t. Stator structure
(f) Speed of rotor field w.r.t. stator field
Solution:
Given f = 50 Hz(supply frequency)
N = 960r.p.m
The no. of pole will be 6 only(because at P=6, Ns = 1000 which is nearer nad greater then 960 r.p.m.)

(a) Slip, S = (Ns-N)/Ns * 100 = (1000 – 960) / 1000 * 100 = 4%
(b) No of poles = 6
(c) Frequency of rotor induced emf = fr = SF = .04 * 50 = 2Hz
(d) Speed of rotor field w.r.t rotor structure = 120fr/p = 120*2/6 = 40 r.p.m.
(e) Speed of rotor field w.r.t. stator structure os actually the speed of stator filed w.r.t stator structure, Ns = 1000r.p.m
(f) Speed of rotor field w.r.t stator field is zero

 

Ques8: A 3 φ, 400V wound rotor has delta connected stator winding and star connected rotor winding. The stator has 48 turns/phase while rotor has 24 turns per phase. Find the stand still or open circuited voltage across the slip rings
Solution
Stator e.m.f/phase E1 = 400V
Statur turns/phase N1 = 48
Rotor turns/phase N2 = 24
K= N2/N1 = 24/48 = 1/2

Rotor e.m.f/phase = KE1 = 1/2 * 400 = 200V
Voltage between slip rings = Rotor line voltage = √ 3 x 200 = 346 volt

Ques9: A 6 pole 3 φ 50Hz induction motor is running at full load with a slip of 4%. The rotor is star connected and its resistance and stand still reactance are 0.25 ohm and 1.5 ohm per phase. The e.m.f between slip ring is 100V. Find the rotor current per phase and p.f, assuming the slip rings are short circuited.
Solution
Rotor e.m.f./phase at stand still E2 = 100√3 = 57.7V
Rotor e.m.f./phase at full load = sE2 = 0.04 * 57.7 = 2.31 V
Rotor reactance/phase at full Load = SX2 = .04 * 1.5 = .06 ohm
Rotor impedance/phase at full load = √ ((0.25)2 + (0.06)2) = .257 ohm
Full load Rotor current/phase = 2.31/0.257 = 9A
Rotor P.f = 0.25/0.257 = 0.97 lag

 

Quest10: A 50 Hz, 8 pole induction motor has full load slip of 4%. The rotor resistance and stand still reactance are 0.01 ohm and 0.1 ohm per phase respectively. Find:
i) The speed at which maximum torque occurs
ii) The ratio of maximum torque to full load torque
Solution:
Synchronous speed Ns = 120f/P = 120*50/8 = 750r.p.m.
Slip at which maximum torque occurs = R2/X2 = 0.01/0.1 = 0.1
Rotor speed at maximum torque = (1-0.1) Ns = (1- 0.1) 750 = 675 r.p.m. Tm/Tf = (a2 + s2)/2as
Where s = Full load slip = 0.04

a = R2/X2 = 0.01/0.1 = 0.1
Tm/Tf = ((0.1)2 + (0.04)2)/(2*0.1*0.04) = 1.45

 

Ques 11: An 8 pole 3 φ, 50 Hz induction motor has rotor resistance of 0.025 ohm/phase and rotor standstill reactance of 0.1ohm/phase. At what speed is the torque maximum? What proportion of maximum torque is the starting torque?
Solution
Ns = 120f/P = 120*50/8 = 750 r.p.m.
R2 = SX2 ------------ for maximum torque
S = R2/X2 = 0.025/0.1 = 0.25
Corresponding speed N = (1-s)Ns = (1 – 0.25)750 = 562.5 r.p.m.

ii) Ts/Tm = 2a/(a2+1) = 0.47 where a = R2/X2 = 0.025/0.1 = 0.25

Ques12: A 500 V, 3 φ, 50 Hz induction motor develops an output of 15 KW at 950 r.p.m. If the input p.f. is 0.86 lagging, Mechanical losses are 7.30 W and stator losses 1500W, Find
i) the slip
ii) the rotor Cu loss
iii) the motor input
iv) the line current
Solution:
VL = 500V, motor output Pr = 15KW
N = 950 r.p.m. P.f. = cos Ø = 0.86lags
Mech. Loss = 730 W
Stator loss = 1500 W
Ns = 120f/P = 120 * 50/6 = 1000r.p.m.
i) S = (Ns-N)/Ns * 100 = (1000 – 960)/1000 *100 = 0.05*100 = 5%
ii) Rotor output = Motor output + Mechanical output = 15 + .730 watt = 15.73 KWatt
There fore (Rotor Cu loss)/(Rotor output) = s/(s-1)
Or Rotor Cu loss = 15.73 * (0.05)/(1-0.05) = 827.89 watt

Power flow diagram for finding the motor input

Motor input = 15kw + 730 + 1500 + 827.89 = 18.058KW
Line Current = √3V2I2Cosφ
I2 = 24.25A 

 

Ques13: A 6 pole 3φ induction motor develops 30hp including 2 hp mechanical losses at a speed of 950 r.p.m. on 550V, 50Hz Mains. The P.F. is 0.88 lagging.
Find:
1) Slip
2) Rotor Cu loss
3) Total input if stator losses are 2kw
4) η
5) Line current

Solution
Ns = 120f/P = 120 * 50/6 = 1000 r.p.m.
1) S = (Ns – N)/Ns = (1000 – 950)/1000 = 0.05
Rotor output Pmech = 30hp = 30 * 735.5 = 22065 watt
Power input to rotor = Pmech/(1-S) = 22065/(1-0.05) = 23,226
2) Rotor Cu loss = s * rotor input = 0.05 * 23226 = 1161 Watt
3) Total input = Power input to rotor + stator losses = 23226 + 2000 = 25226 Watt

Motor output = Rotor output – Mech loss = 30 – 2 = 28 HP = 28 * 735.5 = 20594 Watt
4) η = (Motor output)/(Motor input) * 100 = 81.64%
5) IL = (Motor Input)/( √3 * 550 * 0.88) = 30A

 

Ques14: A 4 pole 50 Hz 3 φ induction motor running at full load, develops a torque of 160N-m, when rotor makes 120 complete cycles per minute, find what power output
Solution
Supply frequency f = 50Hz
Rotor e.m.f. frequency = f = 120/60 = 2Hz
Slip S = f’/f = 2/50 = 0.04
Ns = 120f/p = 120 *50/4 = 1500 r.p.m.
Shaft power output = Tsh * 2πN/160 = 160 * 2 π * 1440/60 = 24127W

Ques15: The power input to a 500V 50Hz, 6 pole, 3 φ squirrel case inductor motor running at 975 r.p.m. is 40kw. The stator losses are 1 kw and friction and windage losses are 2kw. Find:
1) Slip
2) Rotor Cu loss
3) Brake hp
Solution:
i) Ns = 120f/P = 120*50/6 = 1000 r.p.m.
S= ( Ns – N)/Ns = (1000 – 975)/1000 = 0.025
Power input to station P1 = 40Kw
Stator output power = P1 – stator losses = 40 -1 = 39kw
Power input to rotor P2 = Stator output power = 39 KW
ii) Rotor Cu loss = sP2 = 0.025 * 39 = 0.975KW
Pmech = P2 – Pcu = 39 – 0.975 = 38.025
iii) Motor output = Pmech – friction and windage loss = 38.025 – 2 = 36.025KW