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1. Transformers

Electrical Machines

(Solutions for Volume‐ 1 Class Room Practice Questions)

01. Ans: (b) Sol: Given data: 400/200 V 50 Hz B (^) max = 1.2 T 800V, 50 Hz linear dimension all double

2

N N^11

12 ^2

N N^21

B (^) max2 =? l 2 = 2 l 1 and b2 = 2b (^1) A 1 = l 1 b1 A 2 = 4A 1

2 B AN f

2 B AN f E

E

max 1 11

max 2 12 11

12 1

2  

11

12 1

max 2 1 N

N A

4 A

1. 2

B 400

(^800)   

B (^) max2 = 2 1. 2 4

2  1. 2   T

02. Ans: (c)

Sol: Given data:  = b = 2

(^40) c.m

Anet = 0.9

2 2

40 

 

= 7.2 10 -2^ m^2

TURN

EMF = 4.44  1 7.2 10 -250 = 16 V

03. Ans: (d)

Sol: Induced emf E 2 = dt

Mdi

(Where, dt

di is slope of the waveform)

= 800 V

3

3 



As the slope is uniform, the induced voltage is a square waveform.  Peak voltage = 800 V  Note: As given transformer is a 1: transformer, the induced voltage on both primary and secondary is same.

04. Ans: (a) Sol: i(t) = 10 sin (100 t) A Induced emf on secondary E 2 = M dt

di

E 2 =

(^400)  10 -3 (^)  10  100  cos(100t)

= 400 cos (100t) E 2 = 400 sin (100t + 2

When S is closed, the same induced voltage appears across the Resistive load  Peak voltage across A & B = 400V

05. Ans: (a) Sol: E 1 = – dt N d 1

 (^) (where E 1 =^ ^ e^ pq)

1

E 200 0.

  ^ ^ 

e (^) pq = 30 V (Between 0 & 0.06)

1

E 200 0.

  ^  

e (^) pq =  90 V (Between 0.1 & 0.12)

: 4 : Electrical Machines

08. Ans: (c) Sol: ZT = (0.18+j0.24) and Z (^) L = (4+j3)

line T L

I 480 0 480 0

Z Z 0.3 53.13 5 36.

  ^   

= 90.76–37.77A

Voltage at the load, Vload = (90.76–37.77)  (536.86) = 453.8 –0.91 V And power loss in tr.line = (I (^) line )^2 0. = (90.76) 2  0. = 1482 W

09. Ans: (b) Sol: 200V, 60Hz, Wh 1 = 250W, Wh 2 =? We1 = 90W We2 =?

2

2 1

1 f

V f

V (^) 

1. 6

2

1

1. 6

1

2 h 1

h 2 f

f V

V

W

W

 

1. 6 0. 6 h 2 50

W 

Wh2 = 348. When f

V (^) ratio is not constant

We  v^2 2 1

2 e 1

e 2 V

V W

W 

  

90 119. 02 W

W^2302

e 2   

Wi = Wh2 + We2 = 467.81 W

10. Ans: (a) Sol: V 1 = 440 V ; f 1 = 50Hz ; Wi = 2500 W

V 2 = 220 V ; f 2 = 25Hz ; Wi = 850 W

1

1 2

2 f

V f

V (^)  = Constant

Wi = Af + Bf^2 2500 = A  50 + B  502 ….……… (1) 850 = A  25 + B  25 2 …..………(2) By solving (1) & (2) A = 18 ; B = 0. We = Bf^2 = 0.64  502 = 1600 W Wh = Af = 18  50 = 900 W

11. Ans: (b) Sol: Given data: 2 W Wi h 1 ^ ;^2 W Wi e 1 

1. 6 1

2 h 1

h 2 V

V W

W 

  

h 1

1. 6 1 h 2 1 W V W 0.^9 V  

  

Wh2 = 0.844 Wh1 = 0.422 Wi 2 1

2 e 1

e 2 V

V W

W 

  

We2 = 0.81 We1 = 2 0.81 Wi We2 = 0.40 Wi Wi2 = Wh2 + We2 = 0.422 Wi + 0.40 Wi Wi2 = 0.822 Wi Reduction in iron loss is = 1 – 0. = 0.  0. i.e., 17.3% reduction

12. Ans: (a) Sol: At 50 Hz; Given, Pcu = 1.6% , Ph = 0.9%, Pe = 0.6%

: 5 : Postal Coaching Solutions We know that, Ph  f -0.

1. 115 50

f

f P

P 0.^60.^6

1

2 h

h 2

P 0.^009

 h 2   Eddy current loss = constant, (since Pe V^2 ) and given total losses remains some. Ph 1 Pcu 1 Pe 1 Ph 2 Pcu 2 Pe 2

3. 1 % 0. 806 %Pcu 2  0. 6 % P (^) cu 2  1. 694 % Pcu 2 is directly proportional to I 2 2

2

1 cu

cu I

I

P

P

2

1 

I 2  1. 028 I 1

Output kVA = VI 2 = 1.028 VI (^1)

13. Ans: (d) Sol: Given data: 20 kVA, 3300/220V, 50Hz No load at rated voltage i,e W 0 = 160Watt cos 0 = 0. % R = 1% %X = 3% Input power = output Power + Total loss of power %.R = %FL cu loss = 100 VArating

FL culoss

FL cu loss = %R  VA rating = 0.01 20,000 = 200 Watt I (^) F2 = 90. 9 A 220

E

VArating 2

Iload = 85 A 220 0. 8

14. 96 k   At 90.9A  Cu loss = 200 W

85A  Cu loss =?  Cu loss at

85A= 200 174. 8 Watt

90. 9

Total loss when 14.96 kW o/p = Iron l oss + cu l oss at 85A = 160+174. = 334.8 W Input power = 14.96 kW+334.8W = 15294.8W

14. Ans: (a) Sol: Given data: At 50Hz: 16 V, 30 A, 0.2 lag At 25 Hz , 16 V, Isc =? and p.f =?

I

Z V

Z ^16 

R = Z cos R = 0.533  0. R 1 = 0.106  X 1 = Zsin = 0.533 0.979= 0.522  Reactance at f = 25 Hz

50

X

X

1

X 2 = 0.2611 

Z  R^2 X^2

 ( 0. 106 )^2 ( 0. 2611 )^2

Z = 0.281

56. 78 A 56. 65 A

Z

I V  

p.f = 0. 376

1. 2817

Z

cos R  (^) sc    lag

Z X

R

: 7 : Postal Coaching Solutions Wi + 0.25 WCu = 0.0102 ……… (2)

By solving equation (1) & (2) Wi = 6.8  10 –3^ ; WCu = 0.

1. 75 1 1 6. 8 10 ( 0. 75 ) 0. 0136

2. 75 1 1 3 / (^4)     (^3)  (^2)       = 98.1%

19. Ans: (a) Sol: Percentage of load at which maximum

efficiency possible is = Cu

i W

W

6. 8  10 ^3 

max (^)      3 

 = 98.1 %

20. Ans: (d) Sol: Given data: 10 kVA,2500/250 V OC: 250V, 0.8A, 50W SC: 60V, 3A, 45W Iron losses = 50 W = WI 4 A 2500

I 10000

( HV)^  (Rated current) Copper loss at 3A = 45W Copper loss at 4A =?

 45 80 W 9

4 ^2    

kVA at (^) max kVAFL culoss   Iron^ loss

10 kVA 7. 9 kVA 80

^50  

21. Ans: (c) Sol: 100 7. 9 08 10 ( 2 50 )

3

3 0 max. 8 pf^

22. Ans: (c) Sol: Given data: 1000/ 200 V, R 1 = 0.25  ; R 2 = 0.014 , Iron loss = 240W 2 1 R 02  R 1 R = K

2 R 1 + R 2

200 ^2  

I (^) 2 max = R 02

Ironloss

100A

^240 

23. Ans: (c) Sol: Given data: Max.  = 98 %, at 15 kVA, full load kVA = 20, UPF for 12 hours

15 k 1 2 W i

1. 98 15 k^0.^1  

Wi = 153.06W

allday

output in kWh output kwh losses

kW = kVA cos kW = 20  1 = 20 kW kWh output = 2012 = 240 kWh Wi = 153.06  24 = 3.673 kWh WCu  S^2

153. 06 15

W^202

Cu 2  

: 8 : Electrical Machines WCu2 = 272. Transformer is ON load for 0 to 12 hrs. So, WCu2 = 272. 106  12 = 3.265 kWh

3 3 3

3 all day 240 10 3. 673 10 3. 265 10

240 10         %all day = 97.19%  97.2%

24. Ans: () Sol:* Given Iron loss = 1.25 kW, cos = 0. Find equivalent resistance R 01 on H.V side k = 231 0. 11000

R 01 = 8.51 +^ 0.0038 2

k

Full load current on H.V side =

= 9.09 A

Full load Cu loss = (9.09) 2  17. = 1.415 kW Efficiency = 100 0.85^100 100 0.85 1.415 1.

25. Ans: (c) Sol: Given data: 1100/400 V, 500 kVA, max = 98% 80% of full load UPF % Z = 4.5% PF  max V.R = %Z

%R

For min. secondary 10%

0.98 = 0.8 500 10 2IronLoss

3

3   

Iron loss = 4081.63 W  Cu loss at 80 % of FL = 4081.

(.8) 2 Cu loss of FL = 4081. FL cu loss = 6377. 54 W %R = % FL cu loss = VARating

FLculoss

^3 

PF  max. VR= 0. 283 lag

4. 5

%Z

%R  

26. Ans: (b) Sol: Terminal voltage =? % X %Z^2 %R^2  ( 4. 5 )^2 ( 1. 27 )^2 = 4.317% %VR = %R cos 2 +%Xsin 2 = (1.27  0.283) + (4.317  0.959) % VR = 4.49% = 0 .0449 Pu Total voltage drop on secondary side = PU VR E 2 = 0.0449  400 = 18V V 2 = E 2 Voltage drop = 400 18 = 382V 27. Ans: (a) Sol: R 02 = R 1 ^ + R 2 X 02 = X 1 ^ + X 2 R 1 ^ = K (^2) R 1  (Resistance referred to secondary side)

3. 4 10

R^12

X 1  k^2 X 1 = (0.01  7.2)

: 10 : Electrical Machines

32. Ans: () Sol:*

The equivalent circuit refer to L.V side is

48. 91 A

I^90103

Where V 1 = voltage applied across the transformer. V 1 = V 2 +I 2 (0.12  cos  + 0.5  sin) =2300+48.91[0.120.8+0.50.6) = 2300+19. V 1 = 2319.36V % Regulation= 100 2400

33. Ans: 96.7%

Sol: copper losses = I (1.18^22 0.12)

= (48.91)^2 1. = 3109.8 W

%  = 100 90 10 3109. 8

3

3   

34. Ans: 218. Sol:

Equivalent circuit refer to H.V side is

Z^1 L  4275.625.

Transformer impedance = R 01 + jX 01 = 310.4875. 1 2

I 7967

= 1.78–28.15A

V 1   I 2 ^ ZL

t Now V 7600.6^230 8000

^ 

35. Ans: 4.9% Sol: Voltage regulation = 100 E

E V

2

2  t

^230 ^218.^52 

j150

0.12 (^) I (^2)

40 

V (^) s

j0.5

P =90kW 2300V 0.8pf

14/2.4kV

V (^) s

1.18 4.408

0.12 j0.5 (^) I (^2)

2300V

V (^1)

7967V

80  j300

350k (^) 70k ZL = 3.2+j1.5

15kVA and 8000/230V

V (^) r

R (^01) X (^01)

Z^1 L 3871. 4 j 1814. 7

80 

7967V 350k

j300

Vt

I 2

: 11 : Postal Coaching Solutions

C (^2) B 2

A (^2) A

C B

b (^2)

a 2

a c (^2) b

b

c a c

36. Ans: () Sol:* Given data, f = 60 Hz, 30 kVA, 4000 V/120 V, Zpu = 0.0324 pu, I 0 = 0.0046 pu, W 0 = 100 W, Wcu = 180 W P 0 = 20 kW & cos = 0.8lag

Load current I 2 =

= 208.33 A

Rated load current =

 = 250 A

The copper losses for 208.33 A is 208.33 2 180 250

  = 124.99 watt

Efficiency =

3 3

The equivalent circuit wrt primary is

Primary rated current

I (^) P =

30 10^3

 = 7.5 A

Given cu losses = 180 W  R 1 = (^2) P

I

Given, Zpu = 0.

 Z 1 = (kV)^2

MVA

X 1 = Z^21  R 12 = 17.28^2 3.2^2

Load current wrt primary is

2 2

I I 120

 = 6.24 A

Necessary primary voltage VS = V 2   I 2   R cos 1   X sin 1  = 4000 + 6.24[3.2  0.8 + 16.98 0.6] = 4079.5 V

37. Ans: (b) Sol:

 The Possible Connection is Yd

38. Ans: (a)

Sol: R = (^)   

0. 012 0.^42

X =  

0. 05 0.^42

V

I P

3

3 (^2)   

0.019 j0. 0.4 10 3 V

E 2

250/36.

VS

20 kW 0.8 pf

4 k/120V

R 01 X 01

Z (^1) 30 kVA

4000V

: 13 : Postal Coaching Solutions

E 0  = (^) VRs E 120 2

 VRs E 0 E 120 2

= 3 E 30

44. Ans: (d) Sol: The flux linkages in phase ‘b’ and ‘c’

windings is 2

 (^). Therefore induce voltage

is also becomes half

KVL:

V 0  + V^0 E

 E 3 V 0

45. Ans: (b) Sol:

I (^) Y2 is  120  lagging w.r.t I (from 3 system)  I (^) Y2 = I 120 

And (^) I     I 120   180  = I   60 

46. Ans: (a) Sol: I (^) rated = I (^) base = 1. Vrated = Vbase = 1. Under short circuit, I (^) scz (^) e1 = Vsc Since I (^) sc = I (^) rated ; 1z (^) e1 = (0.03)(1) Or z (^) e1 = 0. Short circuit pf = cossc = 0.25, sinsc = 0. In complex notation, z (^) e 1 = 0.03(0.25 +j0.968) = (0.0075 + j0.029) pu Similarly z (^) e 2 = 0.04(0.3 + j0.953) = 0.012 + j0.0381 pu (a) When using pu system, the values of z (^) e and z (^) e2 should be referred to the common base kVA. Here the common base kVA may be 200 kVA. 500 kVA or any other suitable base kVA. Choosing 500 kVA base arbitrarily, we get z (^) e 1  200500 ( 0. 0075 j 0. 029 ) = 0.01875 +j0. = 0.07575. z (^) e 2  500500 ( 0. 012 j 0. 0381 ) = 0.0472.54 S = 0. 8

(^560) = 700 kVA

 S = 700cos^1 0. = 70036.9 From Eq. e 1 e 2 1 e^2 z z S S z  

V 0 





• (^) +

 V/2 0 

• E 

V/2 0 

I V 0 

V 120 

IY

I

3- balanced V 120  load

: 14 : Electrical Machines

= (70036.9) (^) o

o

1. 114 74. 74

2. 04 72. 54 



= 46036.1 kVA S 2 = (460 )(cos36.1o^ ) at pf cos36.1o^ lag = 372 kW at pf of 0.808 lag (Check. Total power = 190 + 372 = 562 kW, almost equal to 560 kW)

47. Ans: (d)

Sol: Current shared by transformer 1 = 200245

= 1.225 pu Transformer 1 is, therefore, overloaded by 22.5%, i.e., 45 kVA Current shared by transformer 2 = 500460 = 0.92 pu Transformer 2 is, therefore, under loaded by 8%, i.e. 40 kVA. Voltage regulation, from Eq. (1.40), is given by rcos 2 + x sin 2 For transformer 1, the voltage regulation at 1.225 pu current is = 1.225 (r cos 2 + x cos 2 ) = 1.225 (0.0075  0.76 + 0.0290  0.631) = 1.225(0.024119) = 0. Or 2

2 2 E

E  V = 0.

Or V 2 = (0.970454)(400) = 388.182 V

48. And: (c) Sol: Here I (^) Z (^) e  (^) f 1  360 V, IZe (^) f 2  400 V

and I^ Z (^) e f^  3  480 V Transformer 1 is loaded first to its rated capacity, because I (^) z (^) ef (^)  1 has lowest magnitude. Thus the greatest load that can be put on these transformers without overloading any one of them is, I   kVA II^  kVA  ^ II   kVA 3 ..... z f 3 2 Z f^1 Z f 2 z f 3 1 Z f^1 e

E e e   e   

   

400 360 400 360 400 400 480       1060 kVA The total load operates at unity p.f. and it is nearly true to say that transformer 1 is also operating at unity p.f.

49. Ans: (c) Sol: Secondary rated current 60. 6 Amp

6. 6

^400 

Since transformer 1 is fully loaded, its secondary carries the rated current of 60.6 A. For transformer 1,  

  0. 825 

60. 6

r 3025 e (^22) Full-load voltage drop for transformer 1, E 2 V 2 I 2 re 2 cos 2 I 2 xe 2 sin 2 = (60.6) (0.825) (1) + 0 = 50 V  Secondary terminal voltage V 2 = 6600 50 = 6550 V

50. Ans: (a) Sol: Voltage rating of two winding transformer = 600 / 120V, 15 KVA voltage rating of auto

: 16 : Electrical Machines

500V

3

500 440V

743.6 A

I.M

kVA transformed = (1K) kVAAT and kVA conducted = 210– = 200 kVA.

55. Ans: (d) Sol:

Current through 480 V winding is

1000 A 480

I^480103

kVA rating of auto transformer = 8400  1000 = 8.4 MVA For two winding transformer

= 480 10 W

3

3  

W = 10.79 kW Efficiency = 100

8. 4 10 1 10. 79 10

9. 4 10 1 6 3

6     

 

= 99.87%

56. Ans: (a) Sol:

I^6100.^745103

 743. 69 A

By equation

3 I^1

I 1 = 845.11 A

I 1 – I 2 = 100 A

57. Ans: (a) Sol:

The voltage per turn = 100

400 = 4V

For 80 turns = 80  4  1320 V For 60 turns = 60  4 240 V

I (^) d =^320 5.33 A 60

I (^) c=^240 12 A 20

VA rating fo 20 load is 240 I (^) c  240  12 = 2880VA VA rating for 60  load is 320 Id  320  5. 33 = 1705.6 VA Primary current I 1 = Total load VA 400 = 400

I 1 = 11.464 A

8000 V

400

8000 8400V

1000

240 V 60 80

400v50Hz 100 D

20Ω

60Ω

I 1 A Id Ic

B

C 320V

: 17 : Postal Coaching Solutions For resistive load power factor is at unity.

58. Ans: (c) Sol:

Load current = 4 –45 + 1 0 = 4.75 –36. mmf = 400  4.75  –36..55 + 200 0 = 1900 –36.55 + 200 = 1726.3 – j 1131. Total secondary mmf = 2064.07–33. Primary current =^2064 20.64 A 100

59. Ans: (b) Sol:

Sec. mmf = 2000 0 + 20 2 (500)– = 2000 0 + 10000 2 – = 1000 [20 + 10 2 –45] = 1000[2+10–j 10] = 1000[12 – j 10]

mmf = 15620.4 –39. Primary current = 15620.4^ 39. 400

= 39 A at 0.76 lag

60. Ans: (b) Sol: From power balance V 1 I 1 cos 1 = V 2 I 2 cos 2 + V 3 I 3 cos 3 10 : 2 : 1

5

1 N

N 1

10

1 N

N 1

(^3) 

cos 2 = 0.8   2 = 36. cos 3 = 0.71   3 = 44.

1 1 1 12 2 10 V 1 I 3 cos 3 VI cos^1 5 V Icos ^1    I 1 cos 1 = 9–36.86 + 5–44. = 13.969 –39.6o I 1 = 14A p.f = cos(39.6) = 0.77 lag

61. Ans: (a) Sol: Given R 1 = 1.6, L 1 = 21mH, R 2 = 1.44m, f = 60Hz, L 2 = 19H, Rc = 160k, L (^) m = 450 H, P = 20 kW,V 2 = 120V and cos = 0.85lag. X 1 = 2fL 1 = 2 60  21  10 -3^ = 7.91  X 2 =2fL 2 = 2 60  19  10 -6^ = 9.55 m The equivalent circuit is,

(^2000) –  0

200

C

1  0

-^4000 ^0

A+ 4 –45 + 1 (^0 4) –45 1000 –45 1  0 

6 kW

1000  0 100

400 6000  0

100

20  0

20 2 –45 (^20) – 1000  0 20  0  8000  0 400

20 –

500 1200  0

Vs

1.6 j7.91

I (^) line

160k 450H

HV

1.44m j9.55m

I (^) L

120  0  20kW, 0.85pf

LV 4000/

: 19 : Postal Coaching Solutions  N = 1609 rpm

02. Ans: 6.9 (Range: 6 to 7) Sol: Given data: Vt = 250 V,  = constant Ra = 0.1  P 1 = 100 kW and P 2 = 150 kW Case (i): P 1 = Vt I (^) a 100 k = 250  I (^) a  I (^) a1 = 0.4  103 A Eg1 = Vt + I (^) a1  Ra = 250 + 400  0. = 290 V Case (ii): P 2 = Vt I (^) a 150  103 = 250  I (^) a  I (^) a2 = 600 A Eg2 = Vt + I (^) a2 Ra = 250 + 600  0. = 310 V From emf equation of generator, Eg  N

 290

E

E

N

N

g 1

g 2 1

% Increase in speed = 100 N

N N

1

N

N

1

03. Ans: (a) Sol: Given data: Load current = 250 A Generator (A): 50 kW, 500 V, % drop = 6% Generator (B): 100 kW, 500 V, % drop = 4%

The no-load voltage of generator (A) = (^)  

 ^ 

= 530 V

Generator (B) = (^)  

 ^ 

= 520 V

6 x 50 k

P 1  

 P 1 =  6 x 6

4 x 100 k

P 2  

 P 2 =  4 x 4

Total load power, 250  500 =    4 x 4 6 x^10010 6

50  103   ^3 

 125 =  4 x 4 ( 6 x)^100 6

 5 = ( 4 x) 3

( 6 x) 

530

6% 4%

520

4

4 x 6 x

x P 2 500 P (^1) 100kW (^) 50kW

: 20 : Electrical Machines

x = 4

Load shared by generator (A), P 1 = (^)  

= 43.75 kW  Current I = 500

43. 75 = 87.5 A

Load shared by generator (B),

P 1 = (^)  

= 81.25 kW  Current I = 500

81. 25 = 162.5 A

04. Ans: (d) Sol: Terminal voltage = 500 + x% of 500

= 500 + 4

(^3) % of 500

= 503.75 V

05. Ans: (b)

Sol: KC

r r KCT

V

a

a s a e

m t ^ 

Speed is directly proportional to applied voltage.

06. Ans: 100  Sol: Given data:

Vt = 200 V, Rf = 100  and f

f 1 0. 5 I

I

N 0 = 1000 rpm and N 1 =1500 rpm Re =?

We know that   speed(N)

0

1 1

0 N

N

1

Field current I (^) f0 = 100

R

V

f

t (^)  = 2A

f

f 1 0. 5 I

I

f 0

f 1 f 1

f 0 1

0 1 0. 5 I

1 0. 5 I

I

I

1 0. 5 I

I

(^2) f 1 f 1 1.5I (^) f1 = 1 + 0.5I (^) f  I (^) f1 = 1 A Field current I (^) f  R f

f

f e f 1

f 0 R

R R

I

I  

 Rf + Re = 2 Rf  Re = 100 

07. Ans: 32. 95 Nm Sol: Given data: 500 V, 60 hp, 600 rpm Ra = 0.2  and Rsh = 250 

Losses = (^)  

(^1 1) output power

= 4973.33 watt Input power =

1. 9

efficiency

Output power 