ETO COC NUMERICAL QUESTIONS AND ANSWERS - SOLVED I

SOLVED  NUMERICAL QUESTIONS IN ASKED IN ETO COC WRITTEN EXAM - PART I
Let me remind the basic electrical formula

BASIC FORMULAS FOR 3 PHASE AC MOTOR

SLIP

S = (Ns - N) / Ns    or    fr / fs

Where

Ns is synchronous speed

N rotor speed

fr is Rotor frequency = x/60

fs is supply frequency

STATOR INPUT P1

P1 = P2 + PF + PCS (Rotor input + stator losses)

Where

P2 is Rotor input

PF is stator core loss

PCS is stator copper loss

ROTOR INPUT P2

Rotor input P2 = Pm / (1-S)

Where

Pm is Rotor gross output

Or

P2 = P1 – Stator loss

ROTOR GROSS OUTPUT Pm

Pm = Pout + Pw

Where

Pout is motor or rotor output

Pw is windage and friction loss

Or

Pm = P2 – Rotor loss

Where

P2 is Rotor input

Or

Pm = Tg x ωr

Where

Tg is gross Torque

ωr is Rotor radian speed

Tg = Tsh + Torque loss

Where Tsh is shaft torque

ωr = (2Nr) / 60   or 2nfL ; here nfL = NfL / 60 and Nr = Ns (1-S)

Where Nr is rotor speed or actual speed

NET MECHANICAL OUTPUT OR SHAFT OUTPUT OR MOTOR OUTPUT OR ROTOR OUTPUT

Pout = Pm – Pw   or   Pout = Tsh x ωr

EFFICIENCY

ղ % = (Pout / P1) X 100

ROTOR CU LOSS

Rotor Cu Loss = S x P2

Where

S is SLIP

P2 is Rotor input

FREQUENCY OF ROTOR CURRENT

fr = Sf x F

Where

Sf is Slip at frequency

F is supply frequency

ETO COC CHAPTERWISE SOLVED NUMERICALS

CHAPTER 1: 3 PHASE AC MOTOR

1.1) Question: The shaft output of a three-phase 60- Hz induction motor is 80 KW. The friction and windage losses are 920 W, the stator core loss is 4300 W and the stator copper loss is 2690 W. The rotor current and rotor resistance referred to stator are respectively 110 A and 0.15 Ω. If the slip is 3.8%, what is the percent efficiency? (Apr, Aug 2016)

Solution:

Shaft output (Pout) = 80 KW

Windage and Friction losses (Pw) = 920W 

Stator core loss (p.f.) = 4300 W

Stator copper loss (Pcs) = 2690W 

Slip (s) = 3.8% = 0.038

Rotor Gross output (Pm) = Pout + windage and friction losses

= 80 KW + 920 W 

= 80.92KW

Rotor gross output (Pm)/ Rotor Input (P2) = 1–s

80.92/P2= 1-0.038

= 0.962 P2

= 80.92/0.962 = 84.11 KW

We know that;

Stator input (P1) = rotor input (P2) + stator core loss (Pf) + stator cu loss (Pcs)

= 84.11 KW +4300 W + 2690 W 

= 91.1 KW

% Efficiency (% η) = (rotor output / stator input) x 100 

= (Pout/P1) x 100

= (80/91.1) x 100

Answer: % of Efficiency = 87.81%

1.2) Question: The power input to a 500 V, 50 Hz, 6 pole 3 phase squirrel cage induction motor running at 975 rpm is 40 KW. The stator losses are 1 KW and the friction and windage losses are 2 KW.

Calculate

(i) Slip

(ii) Rotor copper loss

(iii) Mechanical power developed

(iv) The efficiency.

(MAR 2015, JUN 2015 & JUN 2016)

Solution:

(i) Rotor Speed (N) = 975 RPM

Synchronous speed (Ns) = 120 f /P

= 120 X 50/6 = 1000 RPM

Slip (s) = (Ns-N)/Ns

= (1000-975)/ 1000 

= 0.025

(ii) Stator Input (P1) = 40 KW,

Stator Output or Rotor Input (P2) = P1-Stator loss

= 40 -1 = 39KW

Rotor copper loss = s x P2 

= 0.025 x 39 

= 0.975KW

(iii) Gross mechanical output (Pm) = P2 – Rotor Cu loss = 39 - 0.975 = 38.025KW

Net mechanical output or Shaft output (Pout) =Pm - friction and winding loss

= (38.025-2.000) KW 

= 36.025KW

(iv) Efficiency % (% η) = (Pout x 100)/ P1 

= (36.025 x 100) /40 

= 90.06 %

Note: Assume that the core loss is included in friction and windage loss and the total loss under this head is 2 kW.

Answer:

(i) Slip = 0.025

(ii) Rotor copper loss = 0.975KW

(iii) Mechanical power developed = 36.025KW

(iv) The efficiency = 90.06 %

1.3) Question: A 400V, 4-pole, 50 Hz, 3-phase, 10 Hp, star connected induction motor has a no-load slip of 1% and full load slip of 4%. Find the following:

(i) Syn. speed (ii) no-load speed (iii) full-load speed.

(iv) Frequency of rotor current at full-load (v) full-load torque. (JUL 2016, OCT 2017)

Given Data:

VL = 400 volts

P = 4 nos

50 Hz

Po = 10 HP = 735.5 x 10 = 7355-watt (For British unit 1 (B) HP = 746 W)

Solution:

(i) Synchronous speed Ns = 120 f / p 

= 120 x 50 / 4 

= 1500 rpm

(ii) No load speed at slip s = 0.01

No = Ns (1 – s) 

= 1500 (1- 0.01) 

= 1485 rpm

(iii) Full load speed at slip sf = 0.04

Nfl = Ns (1-sf) 

= 1500 (1-0.04) 

= 1440 rpm

(iv) Frequency of rotor current (fr) = sf. f = 0.04 x 50 = 2.0 Hz

(v) Full load torque at shaft

TSh = 9.55 Po / Nfl 

= 9.55 x 7355 /1440 

= 48.78 Nm

Answer:

(i) Syn. Speed = 1500 rpm

(ii) No-load speed = 1485 rpm

(iii) full-load speed = 1440 rpm

(iv) Frequency of rotor current at full-load = 2.0 Hz

(v) full-load torque 48.78 Nm

1.4) Question: A 3 phase Induction Motor which is wound for 4 pole, when running full load develops a useful torque of 100 Nm; also rotor emf is observed to make 120 cycles/ min. It is known that the torque lost on account of friction and core loss is 7 Nm. Calculate the shaft Power output, Rotor cu loss, Motor Input and Efficiency. (SEP & OCT 2016, FEB 2018)

Given Data:

Shaft Torque Tsh = 100Nm &

Rotor frequency fr= 120 / 60 = 2 Hz

Solution:

Assume supply frequency is 50 Hz, so, fs = 50Hz 

Slip s = fr/fs = 2/ 50 

=0.04

Synchronous speed Ns = 120f/p 

= 120x50/4 

= 1500 rpm 

Rotor speed Nr = (1-s) Ns 

= (1 – 0.04) x 1500 

= 1440 rpm 

Rotor’s radian speed ωr = 2 π X1440 / 60

= 150.7 rad/s

Shaft power output Pout = Tsh x ωr 

= 100 x 150.7 

= 15.07 KW.

Gross Torque Tg = Tsh + Torque loss 

= (100+7) 

= 107Nm 

Also, Tg = Pm/ωr

or, Rotor gross output Pm = Tg x ωr 

= 107 x 150.7 

= 16.12kW

Rotor Input P2 = Pm/(1-s) 

= 16.12/(1-0.04) 

= 16.12/0.96 

= 16.79kW

So, Rotor copper loss = s x P2 

= 0.04 x 16.79 

= 0.67 kW

Motor Power input Pin = P2 + Stator Cu Loss 

= 17.49 kW. (Assume Stator Cu Loss = 0.7 KW)

% Motor efficiency (η) = (Shaft Power Output/ Motor Power Input) x 100

= (Pout/Pin) x 100

= (15.07 / 17.49) x 100 = 86.16 %

Answer:

Shaft Power output = 15.07 KW

Rotor Cu loss = 0.67 kW

Motor input = 17.49 kW

Efficiency = 86.16 %

1.5) Question: The output of an induction motor running at 4% slip is 36.775kW and the friction and windage losses are 1500W. Find the rotor copper loss and motor efficiency if stator losses are 3 kW. (10 Marks) (Jan, Apr 2019)

Solution:

Shaft output (Pout) = 36.775 KW

Windage and Friction losses (Pw) = 1500W = 1.5 KW Stator loss = 3 KW

Slip (s) = 4% = 0.04

Rotor Gross output (Pm) = Pout + windage and friction losses

= 36.775 KW + 1.5 KW = 38.275KW

Rotor gross output (Pm)/ Rotor Input (P2) = 1–s

38.275/P2= 1-0.04=0.96

P2= 38.275/0.96 = 39.87 KW

Rotor copper loss = s x P2   

= 0.04 X 39.87     = 1.595KW

(Alternate Method: Rotor Copper Loss = Rotor Input (p2) – Rotor Gross Output (Pm))

We know that;

Stator input (P1) = rotor input (P2) + stator loss

= 39.87 KW + 3KW = 42.87 KW

% Efficiency (% η) = (rotor output / stator input) x 100 = (Pout/P1) x 100

= (36.775/42.87) x 100

= 85.78%

Answer:

Rotor copper loss = 1.595KW

Motor efficiency = 85.78% (when stator losses are 3 kW)

1.6) Question: If a 6-Pole motor is supplied at 60Hz and runs with a slip of 5%, what is the actual Rotor Speed? (4 marks) (Aug, Sep 2018, Mar 2019)

Given Data:

Given That f = 60 Hz

Slip% = 5%

No. of poles P = 6

Solution:

Synchronous speed

Ns = 120 f/P 

= 120 x 60/6 

= 1200RPM 

Let Actual speed is N,

Then, Slip% = {(Ns – N)/ Ns} *100

5 = {(1200 – N)/ 1200} * 100

6000/100 = 1200 – N

N = 1200 – 60 = 1140 RPM

Answer:

Actual Rotor speed = 1140 RPM

1.7) Question: A 3-phase induction motor is wound for four poles and is supplied from a 50Hz system. Calculate:

(i) The synchronous speed,

(ii) The speed of the rotor when the slip is 4%.

(iii) The rotor frequency when the speed of the rotor is 600r/min. (10 Marks) (Jul 2019)

Solution:

(i) Synchronous speed Ns = 120 f/P = 120x50/4 =1500rpm

(ii) Rotor Speed N = Ns (1-s) = 1500 (1 – 0.04) = 1500x0.96 = 1440rpm

(iii) If Rotor Speed N = 600rpm

Then slip s = (Ns – N)/Ns = (1500 -600)/ 1500 = 0.6 So, Rotor current frequency f’ = s.f = 0.6x50 = 30 Hz

Answer:

(i) Synchronous speed Ns =1500rpm

(ii) Rotor Speed N = 1440rpm

(iii) The rotor frequency when the speed of the rotor is 600r/min = 30 Hz

CHAPTER 2: DC MACHINES (MOTOR AND GENERATOR)

2.1) Question: A shunt generator has an induced emf of 254 V. When the generator is loaded, the terminal voltage is 240 V. Neglecting armature reaction, find the load current if the armature resistance is 0.04 ohm and the field circuit resistance is 24 ohms. (Apr, Aug 2016)

   

Given Data:

Eg = 254 V

V = 240 V

Ra = 0.04 Ω

Rsh = 24 Ω

Solution:

Ish = V / Rsh = 240/24 = 10A.

Substituting the values in the below expression,

Eg = V + Ia Ra

Eg = V + (IL + Ish) Ra

254 = 240 + (IL + 10) 0.04

Answer: LOAD CURRENT = 340A

2.2) Question: A 220 V Shunt Motor takes 5 A at No load. Armature resistance is 0.25 Ohm and Field Resistance is 200 ohm.

Calculate

(i) Efficiency when taking Full Load current of 50A

(ii) % Changes in Speed between No load to Full Load (MODEL PAPER 2015)

Solution:

(i)    

No Load Io = 5 A,

Ra = 0.25 ohm

Full Load IL = 50 A,

Rsh = 200 ohm

Field current Ish = 220/200 = 1.1 A

No Load Ia = 5-1.1=3.9 A

Full load Ia = 50-1.1 = 48.9 A

No Load Power I/P = 220x5 = 1100 W

Full load Power I/P = 220x 50 = 11000W

No Load Armature Loss = Ia² Ra = 3.9 x 3.9 x 0.25 =3.8 W

Constant Loss = N/L Power – N/L Ia² Ra loss = 1100-3.8 = 1096.2 W F/L

Armature Cu Loss = Ia2 Ra = 48.9 x 48.9 x 0.25 = 597.8 W

Total F/L Losses = Constant Loss + F/L Armature Cu Loss = 1096.2 + 597.8 = 1694 W F/L Power Output = 11000-1694 = 9306W

% Efficiency at 50 A F/L current = (9306/11000) x 100 = 84.6 %

(ii)    % Changes in Speed between No load to Full Load

Ebo = V - IaRa

= 220 – 3.9 x 0.25

= 220- 0.975 = 219.025V 

Eb = V - IaRa

= 220 – 48.9 x 0.25

= 220 – 12.225

= 207.775V

We know that;

Nfl / No = Eb / Ebo = 207.775 / 219.025  if ϕ = ϕ

% Changes in Speed between No load to Full Load = [(No–Nfl )/  ] x 100

= [(219.025–207.775) / 207.775] x 100

= 5.41%

Answer:

(i) The Efficiency is 84.6 % when taking Full Load current of 50A

(ii) % Changes in Speed between No load to Full Load = 5.41%

2.3) Question: Two DC Shunt Generators are connected in parallel and supply a load to DC shunt motor taking a current of 150 A. One generator emf of 250 V and has an armature resistance of 0.2 Ohm and the other has 255 V with an armature resistance of 0.3 ohm. What is the motor output voltage and power output of each generator? (Neglect field current) (JULY 2016 & SEP 2015)

Solution:

Let Motor Output voltage = V

Load Current IL = 150 A

IL = Ia1+Ia2 = 150 A

So, Ia2 = 150- Ia1 ----------- (1)

Now,

In Gen 1,

V = E1- Ia1 Ra1

= 250 – 0.2 Ia1 --------------- (2)

And,

In Gen 2,

V= E2- Ia2 Ra2

= 255- 0.3(150-Ia1)

= 255-45 + 0.3 Ia1

= 210 + 0.3 Ia1 ---------------- (3)

From eq. (2) & (3)

V = 250 – 0.2 Ia1 = 210 + 0.3 Ia1

0.5 Ia1 = 40

Ia1 = 40/0.5

= 80 A

So, Ia2 = 150- Ia1

= 150 - 80

= 70 A

Now Motor output voltage V = E1 – Ia1 Ra1 ---------------- (2)

= 250 - 0.2 x 80

= 234 V

Power Output of First Generator = E1 x Ia1 = 250x 80 = 20 KW

Power Output of Second Generator = E2 x Ia2 = 255x 70 = 17.85 KW

Answer:

The motor output voltage = 234 V

Power Output of First Generator = 20 KW

Power Output of Second Generator = 17.85 KW

2.4) Question: Two 220V dc generators each having linear external characteristics, operated in parallel. One machine has a terminal voltage of 270V on no-load and 220V at a load current 35A, while the other has a voltage of 280V   at no-load and 220V at 50A. Calculate the output current of each machine and the bus bar voltage when the total load is 60A. What is the kW output of each machine under this condition? (AUG 2018, Nov 2018)

Solution:

Generator 1

No-load voltage = 270V

Full-load voltage = 220V

Full-load current = 35A

Voltage drop for 35A = 270V - 220V = 50V

Therefore Voltage drop/Amp of output current 50V / 35A = 1.429V/A

So, R1 = 50V/35A

= 1.43 Ohm

Generator 2

No-load voltage = 280V

Full-load voltage = 220V

Full-load current = 50A

Voltage drop for 50A = 280V - 220V = 60V

Therefore Voltage drop/Amp of output current 60V / 50A = 1.2V/A

So, R2 = 60V/50A

= 1.2 Ohm

Let V = bus-bar voltage

I1 = current output of generator 1

I2 = current output of generator 2

Then V = 270 – 1.43 I1------------- for Gen 1

& V = 280 – 1.2 I2 ------------------for Gen 2

Also, I1 + I2 = 60A

Ia2 = 60A - Ia1

To operate in parallel both generator terminal Voltages must be equal under the 60 Amp total load current. Therefore;

270V - 1.43 I1 = 280V - 1.2 I2 or

270 - 1.43 I1 = 280 - 1.2(60 - I1)

2.63 I1 = 62

So, I1 = 23.57 Amps

&, I2 = 60 - 23.57

= 36.43 Amps

Now bus Voltage V = 270 – 1.43 I1 = 236.29 Volts

Generator 1 output power = V * I1 = 236.29 *23.57 = 5.57 kW

Generator 2 output power = V * I2 = 236.29 * 36.43 = 8.60 kW

Answer:

Generator 1 output power = 5.57 kW

Generator 2 output power = 8.60 kW

Solved by The senior Electro Technical Officer Mr. Mobin ETO

Next Part> Chapter 3, 4, 5 and 6