Induction Motors: Important Solved Examples Problems

 

SOLVED EXAMPLES

 

EXAMPLE 1

A 3ϕ - induction motor has 6 poles. If supply frequency is 50 Hz. Calculate the speed of rotating stator field.

Given Data: P=6, f = 50 Hz

Solution

Speed of rotating stator field is,

N_S = 120f / p

N_S = (120 × 50) / 6

 

EXAMPLE 2

A 4 pole 3ϕ induction motor is connected to 50 Hz ac supply. If it is running at 1400 rpm. Find the slip.

Given Data: P=4, f = 50 Hz, N = 1400 rpm

To find: Synchronous speed and slip.

Solution

Synchronous speed

N_S = 120f / P

= (120 × 50) / 4

N_S = 1500 rpm

% Slip = (N_S‒N)/N_S × 100

= (1500‒1400)/1500 × 100

= 6.6%

 

EXAMPLE 3

A 4 pole 3ϕ induction motor is connected 50 Hz frequency supply. Calculate (i) Synchronous speed, (ii) The speed of the rotor when the slip is 0.05, (iii) The frequency of the rotor when the slip is 0.04, (iv) The frequency of the rotor at standstill.

Given Data: P=4, f= 50 Hz

To find: (i) Synchronous speed (ii) Speed of the rotor, (iii) Frequency of the rotor current, (iv) Frequency of the rotor current at standstill.

Solution

(i) Synchronous speed

Ns = 120f / P

= 120×50 / 4

N_S = 1500 rpm

(ii) Speed of the rotor

N=N_S(1‒s)

Here  s = 0.05

N = 1500 (1‒0.05) = 1425 rpm

N=1425 rpm

(iii) Frequency of the rotor currents,

ƒ' = sf

Here, s = 0.04

f '(0.04) × (50)

f ' = 2 r.p.s or 2 Hz

= 2×60 rpm

 ƒ ' = 120 rpm

(iv) Frequency of the rotor current at standstill.

Here s = 1

:. f ' = sf

f ' = 1 × 50

f ' = 50 Hz

 

EXAMPLE 4

An 18 kW, 6‒pole, 50 Hz, 3ϕ slipping induction motor runs at 900 rpm on full load with a rotor current per phase of 35 A. Allowing 1 kW for frictional and windage losses, find the resistance per phase of 3‒phase rotor winding.

Given Data: Motor output 18 kW, p=6, f = 50 Hz, N=900 rpm, I₂=35 A, Mechanical losses = 1 kW

To find: Resistance per phase of 3‒phase rotor winding

Solution

Mechanical power developed by rotor

 P_m = Motor output + Mechanical losses

  = 18+1

 P_m = 19 kW

Synchronous speed, N_S = 120f / P

N_S = (120×50) / 6

N_S = 1000 rpm

Slip, s = (N_S‒N) / N_S

= (1000‒900) / 1000

s = 0.01

Rotor Cu loss = [ s / (1‒s) ] × P_m

= [ 0.01 / (1‒0.01) ] × 19

= 0.19 kW

= 190 W

3‒phase Rotor Cu loss = 3I₂²R₂

= 190 W

R₂ = 190 / 3I₂²

R₂ = 190 / 3(35)²

R₂ = 0.052Ω/phase

 

EXAMPLE 5

The input power to a 6‒pole, 3‒phase, 50 Hz induction motor is 40 kW, the speed is 950 rpm. The stator losses are 1 kW and the Mechanical losses are 1.5 kW. Find (i) the rotor cu loss and (ii) the efficiency of the motor.

Given Data: P=6, f = 50 Hz Stator Input power 40 kW, N=950 rpm

To find: Rotor Cu loss and (ii) Efficiency of the motor

Solution

Mechanical losses = 1 kW

Synchronous Speed, N_S = 120f / P

N_S = (120×50) / 6

N_S = 1000 rpm

Slip, s = (N_S‒N) / N_S

= (1000‒950) / 1000

s = 0.05

Stator output = Stator input ‒ Stator losses

 = 40 ‒1 = 39 kW

Rotor input = Stator output = 39 kW

Rotor Cu loss = s × Rotor input

= 0.05 × 39

Rotor Cu loss = 1.95 kW

Motor output power = Rotor input ‒ Rotor cu loss ‒ Mechanical losses

= 39 ‒ 1.95 ‒ 1.5

= 35.55 kW

Motor efficiency, η = [ Motor output power / stator input power ] × 100

= (35.55/40) × 100

 η = 88.8%

 

EXAMPLE 6

A 400 V, 50 Hz, 6 pole delta connected 3ϕ induction motor consumes 75 kW with a line current of 70 A and running at 975 rpm at a slip 2%. If stator iron loss is 2 kW and windage and friction loss is 1.2 kW and resistance between two stator terminals is 0.32Ω Calculate (i) Power supplied to the rotor P_r (i) rotor cu loss, (iii) output power of the motor, P_out (iv) motor efficiency and (v) shaft torque developed.

Given Data: V_L=400 V, f = 50 Hz, p=6

To find: (i) Rotor Cu loss (ii) Power supplied to load, P_out (iii) Motor efficiency (iv) Shaft torque

Solution

Stator input power p₁ = 75 kW

 N = 975 rpm

 I_L = 70 A

 s = 2%

Stator iron loss = 2 kW

Friction and windage loss = 1.2 kW

Stator resistance perphase, R =  0.32/2 = 0.16Ω

(i) Find power supplied to the rotor

Stator Cu loss = 31²R

= 3 × (70)² × 0.16

= 2.35 kW

Total stator loss = 2.35 + stator iron loss

= 2.35+2

= 4.35 kW

Rotor power, P_r = Stator input power ‒ total stator losses

= 70‒4.35

Rotor power = 65.65 kW

(i) Rotor Cu loss = sP_r

= 0.02 × 65.65

Rotor Cu loss = 1.313 kW

(ii) Power supplied to load, P_out = P_r ‒ Rotor Cu loss ‒ mechanical losses

= 65.65 ‒ 1.313 ‒ 1.2

P_out = 63.137 kW

(iii) Motor efficiency, η = [ P_out/ P_in ] × 100

= 63.137/75 × 100

 η = 84.2%

(iv) Shaft torque, T_sh = 9.55 × P_out/N

= (9.55×63137) / 975

T_sh = 618.41 N‒m

 

EXAMPLE 7

A 6‒pole, 50 Hz, 3ϕ induction motor runs at 960 rpm. When the torque on the shaft 200 N‒m. If the stator losses are 1.5 kW and mechanical losses are 0.5 kW, find (i) Rotor cu loss and (ii) Motor efficiency.

Given Data: p=6, f = 50 Hz, N=960 rpm, T_sh = 200 N‒m, Stator loss = 1.5 kW, Mechanical loss = 0.5 kW

To find: (i) Rotor Cu loss, (ii) Efficiency the motor, (iii) Efficiency the motor

Solution

Synchronous Speed, N_S = 120f / P

= (120×50) / 6

Ns = 1000 rpm

Slip, s = (N_S‒N) / N_S

= (1000‒960) / 1000

s = 0.04

(i) Rotor Cu loss

Rotor output, P_out = 2πNT_sh / 60

= 2π×960×200 / 60

= 20096 kW

Gross rotor output, P_m = 20096 + 500 = 20596 W

(ii) Rotor Cu loss

Gross rotor output = s / 1‒s

Rotor Cu loss = [ 0.04 / 1‒0.04 ] × 20596

= 858 kW

(iii) Efficiency the motor

Rotor input = Stator output = 20596 + 858 = 21454 kW

Stator input, P_i = 21454 + 1500 = 22954 W

Efficiency of the motor, η = ( P_out / P_i ) × 100

= (20096/22954) × 100

 η = 87.5%

 

EXAMPLE 8

A 3 phase, 4 pole, 50 Hz induction motor is running at 1440 rpm. Determine the slip speed and slip.

Given Data: P=4, f = 50 Hz, N = 1440 rpm

To find: (a) Slip speed, (b) Slip

Solution

(a) Slip speed = N_S ‒ N = [ 120f / P ] ‒ 1440

= [ 20(50) / 4 ] ‒ 1440 = 1500‒1440

 Slip speed = 60 rpm

(b) Slip, s = (N_S‒N) / N_S

= (1500‒1440) / 1500

 s = 0.04 or s = 4%

 

EXAMPLE 9

A 6 pole, 3 phase, 60 Hz induction motor runs at 4% slip at a certain load. Determine the synchronous speed, rotor speed. frequency of the rotor currents, speed of the rotor rotating field with respect to the stator, the speed of the rotor rotating field with respect to the stator magnetic field.

Given Data: P=6, f = 60 Hz, s = 0.04 or 4%

To find: (a) Synchronous speed, (b) Rotor speed, (c) Frequency of the rotor currents, (d) Speed of the rotor field with respect to stator, Speed of rotor field with respect to the stator magnetic field

Solution

(a) Synchronous speed, N_S ̧= 120f / P

= 120(60) / 6

N_S = 1200 rpm

(b) Rotor speed, N = N_S(1 − s) = 1200 (1‒0.04)

N = 1152 rpm

(c) Frequency of the rotor currents f ' = sf = (0.04) (60)

 f ' = 2.4 Hz

(d) Speed of the rotor field with respect to stator

= N_r + (N_S − N_r)

N_S = 1200 rpm

(e) Speed of rotor field with respect to the stator magnetic field

The two fields are stationary with respect to each other.

 

EXAMPLE 10

A 373 kW, 3ϕ, 440 V, 50 Hz, induction motor has a speed of 950 rpm. On full load, the motor has 6 poles. Calculate slip. How many complete alternations will the rotor voltage make per minute.

Given Data:

 P = 373 kW, V = 440 V, f = 50 Hz, N = 950 rpm, P=6

To find: (a) Slip, (b) Alternations of rotor volt/min.

Solution

(a) Slip: N_S = 120f / P

= 120(50) / 6 = 1000 rpm

Slip, s = (N_S‒N) / N_S

= (1000‒950) / 950

 s = 0.05 or s = 5%

(b) Alternations of rotor volt/min f ' = sf

= 0.05 × 50 = 2.5 Hz (or) cycle/sec

 f ' = 150 cycle/minute

∴ 150 complete alternations will be made by the rotor voltage per minute.

 

EXAMPLE 11

A 4 pole, 50 Hz, 3 phase induction motor has a rotor resistance of 0.024 Ω per phase and standstill reactance of 0.6 Ω per phase. Determine the speed at which the maximum torque is developed.

Given Data: P=4, f = 50 Hz, R₂ = 0.024 Ω/phase, X₂ = 0.6Ω/phase

To find: Speed at which max torque is developed

Solution

Slip at T_max, S_mT = R₂/X₂

= 0.024 / 0.6

S_mT = 0.04

N_S = 120ƒ / P

= 120(50) / 4

N_S = 1500 rpm

N_mT = N_S(1‒ S_mT)

= 1500 (1 ‒0.04)

N_mT = 1440 rpm

 

EXAMPLE 12

A 400 V, 60 Hz, 6 pole 3 phase induction motor runs at a speed of 1140 rpm when connected to a 440 V line. Calculate the speed if voltage is increased to 550 V.

Given Data: V=400 V, f = 60 Hz, P = 6, N₁ = 1140 rpm, V₁ = 440 V, V₂ = 550 V

To find: (a) Speed for V₂

Solution

N_S = 120ƒ / P = 120(60) / 6

= 1200 rpm

s₁ = (N_S‒N₁) / N_S

= (1200‒1140) / 1200

= 0.05 or 5%

s₂ = s₁ (T₂/T₁) (R₂/R₂) (V₁/V₂)²

s₂ = s₁ (V₁/V₂)²

[as other parameters are remaining the same]

s₂ = 0.05 (440/550)²

s₂ = 0.032

N₂ = N_S (1‒s₂)

= 1200 (1‒0.032)

N₂ = 1161.6 rpm

 

EXAMPLE 13

A 6 pole, 50 Hz, 3ϕ induction motor has a full load speed of 950 rpm. and has rotor copper loss of 5 kW. Calculate rotor input.

Given Data: P= 6, f = 50 Hz, 3ϕ I.M, N=950 rpm, Rotor Copper loss = 5 kW.

To find: Rotor input = ?

Solution

Slip: N_S = 120f / P

= 120(50) / 6 = 1000 rpm

Slip, s = (N_S‒N) / N_S

= (1000‒950) / 950

 s = 0.05 or s = 5%

Rotor copper loss = s × rotor input

Rotor Input = Rotor copper loss / s

Rotor Input = 5×10³ / 0.05

Rotor Input = 100 kW

 

EXAMPLE 14

The power input to the rotor of a 3 phase, 50 Hz, 6 pole induction motor is 80 kW. The rotor emf makes 100 complete alternations per minute. Find, (a) Slip, (b) Motor speed, (c) Mechanical power developed, (d) Rotor copper loss per phase, (e) Rotor resistance per phase if rotor current is 65 A.

Given Data: f = 50 Hz, P=6, Rotor power i/p = 80 kW = P₂

To find: (a) Slip frequency of rotor emf, (b) Motor speed, (c) Mech power developed, (d) Rotor copper loss / phase, (e) Rotor resistance per phase

Solution

(a) Slip

Frequency of rotor emf, f ' = sf = 100/60 = 5/3

= 5/3 × 1/50

s = 1 / 30

s = 0.0333

(b) Motor speed, Ns = 120f / P = 120(50) / 6

N_S = 1000 rpm

N=N_S(1‒s) = 1000 (1‒0.033)

N=967 rpm

(c) Mech power developed, P_m = P₂ (1 − s)

= 80 × 10³(1 − 0.033)

P_m = 77.335 W

(d) Rotor copper loss/phase,

Rotor Cu loss / ϕ = sP₂

= (0.033)80 × 10³

Rotor Cu loss = 2,664 W

(e) For I₂=65 A, rotor resistance per phase

Per phase copper loss = I₂²R₂=2664

R₂ = 2644 / I₂²

R₂ = 2644 / 65²

R₂ = 0.631 Ω/phase

 

EXAMPLE 15

A 6 pole, 50 Hz three phase induction motor running on full load develops a useful torque of 160 N‒m. When the rotor emf makes 120 complete cycles per minute. Calculate the shaft power output. If the mechanical torque lost in friction and that for core loss is 10 Nm. Compute (a) copper loss in the rotor winding (b) the input power to the motor and (c) the efficiency. The total stator loss is given to be 800 W.

Given Data: P=6, f = 50 Hz, T_FL useful = 160 N‒m = T_st , Total stator loss = 800 W, T_lost = 10 Nm, Frequency of rotor emf, f ' = 120/60 = 2 Hz

To find: (a) Shaft power output, (b) Rotor copper loss, (c) Power input, (d) Efficiency

Solution

(a) Shaft power output

Synchronous speed, N_S =  120f / P = 120(50) / 6

N_S = 1000 rpm

 f ' = s_f

⇒ s = ƒ' / f = 2 / 50

Slip, s=0.04

Rotor speed N = N_S (1‒s)

= 1000 (1 ‒0.04)

N = 960 rpm

Shaft power output, P_out = T_st (2πΝ/60)

= 160 [2π(960) / 60]

P_out = 16076.8 W

(b) Rotor copper loss

Gross torque, T_g = T_st + T_lost

= 160+ 10

T_g = 170 Nm

T_g = 9.55 (P₂/N_S)

Rotor input, P₂ = TgNs / 9.55

= [(170)(1000)] / 9.55

P₂ = 17801.047 W

Rotor Cu loss = sP₂

= (0.04) (17801.047)

Rotor Cu loss = 712.04 W

(c) P₁, input power:

Input power P₁ = P₂ + stator loss

= 17801.047 + 800

P₁ = 18601.047 W

(d) Efficiency.

 η = P_out/P_i × 100

 = (16076.8/18601.047) × 100

 η = 86.431 %