Department of Technical Vocational Education

DEPARTMENT OF TECHNICAL VOCATIONAL EDUCATION

B.Tech (First year)

EP – 03041

ELECTROMECHANICS [First Semester]

Questions and Answers

ELECTRICAL POWER ENGINEERING 2

CHAPTER 1 MAGNETIC CIRCUIT AND MAGNETIC MATERIALS

Q 1. (a) A coil of 300 turns is wound uniformly on a ring of non-magnetic material. There has a mean circumference of 40cm and uniform cross sectional area of 4 cm2 . If the current in the coil is 5A, calculate (i) the magnetic field strength (ii) the flux density and (iii) the total magnetic flux in the ring. (b) An iron ring of mean diameter 10cm is uniformly wound with 2000turns of wire. When a current of 0.25A is passed through the coil of flux density of 0.4T is set up in iron. Find (i) the magnetizing force and (ii) the relative permeability of the iron under this conditions.

–7 Ans: (a) N = 300 turns, r = 1, 0 = 4  10 A = 4 cm2 = 4  10–4 m2 I = 5 A, l = 40 cm = 40  10–2 m (i) H = ? F = HL NI = HL 300 5 H = NI/L = = 3750 At/m 40 102 (ii) B = ? B = H

–7 -3 = 0rH = 4  10  1  3750 = 4.7124  10 T (iii)  = ?  = BA = 4.7124  10-3  4  10–4 = 1.885  10–6 Wb (b) D = 10 cm = 10  10–2 m, N = 2000 t, I = 0.25 A, B = 0.4 T (i) H = ?

–2 Lc = D =   10  10 = 0.3142 m

NI = HLc 2000 0.25 H = NI/L = = 1591.343 At/m c 0.3142

(ii) r = ? 3

H = B/ = B/0r 0.4 r = B/H0 = = 200.0259 1591.343 4 107

Q 2. A closed magnetic circuit of cast steel contains a 6cm long path of cross sectional area 1 cm2 and a 2cm path of cross-sectional area 0.5cm2. A coil of 200 turn is wound around the 6cm length of the circuit and a current of 0.4A flows. Determine the flux densely in the 2cm path; if the relative permeability of cast steel is 750.

Ans: L1 = 6 cm = 0.06 m, L2 = 2 cm = 0.02 m

2 -4 2 A1 = 0.0001 m , A2 = 0.5  10 m

N = 200 turns, i = 0.4 A, r = 750, B2 = ?

L 0.06 R  1   636620 1 7 4 At/wb 0rA1 4 10  75010 L 0.02 R  2   424413 2 7 4 At/wb 0rA2 4 10  750 0.510 NI 200 0.4     75.4 wb R1  R 2 636620  424413

 75.4106 B    1.508 T  2 4 A2 0.510

Q 3. A section through a magnetic circuit of uniform cross-sectional area 2cm2 is shown in figure (1). The cast steel core has a mean length of 25cm. The air gap is 1mm wide and the coil has 500 turns. The relative permeability of the cast steel is 750. Determine the current in the coil to produce a flux density of 0.8T in the air gap, assuming that all the flux passes through both part of the magnetic circuit. 4

2 -4 2 -2 2 -3 Ans: A = 2 cm = 2  10 m , Lc = 25 cm = 25  10 m , g = 1 mm = 1  10 m

N = 500 turns, r = 750, B = 0.8 T, i = ? B 0.8 Hc = = 7 = 848.83 At/m 0r 410  750 B 0.8 Hg = = 7 = 636619.22 At/m 0 410

F = Ni = HcLc + Hg g H L H g 848.83 25102 636619.22 1103 i = c c  g =  = 1.69765 A N N 500 500

Q 4. An iron ring has external radius of 11cm and an internal radius of 9cm and a cross- sectional area of 4cm2. For air gap 5mm in width is cut in the ring. When the current flowing in coil is 6amp, the total flux is 0.48mwb. The relative permeability of the iron under this condition is 800, leakage coefficient is 1.2. Find the numbers of turns wound on the coil.

2 -2 R0 = 11 cm = 11  10 m, Ri = 9 cm = 9  10 m

2 -4 2 -3 Ac = Ag = 4 cm = 4  10 m , g = 5 mm = 5  10 m, i = 6 A, c = 0.48 mWb

r = 800, Leakage coefficient, L.F = 1.2, N = ? R  R (9 11) 102 radius, r = 0 i = = 0.1 m 2 2

Lc = 2r – g = 2  0.1 – 0.005 = 0.6233 m 5

3 c 0.4810 Bc = = 4 = 1.2 T Ac 410

Bc 1.2 Hc = = 7 = 1193.662 At/m 0r 410 800

Total flux in magnet c L.F = = Useful flux in airgap g

3 c 0.4810 -4 Useful flux, g = = = 4  10 Wb L.F 1.2

4 g 410 Bg = = 4 = 1 T Ag 410

Bg 1 Hg = = 7 = 795774.715 At/m 0r 410 1

F = HL = Ni = HcLc + Hgg 3 3 HcLc  Hgg 1193.662 0.6233  795.77410  510 N = = i 6 = 787.1465 turns  788 turns

Q 5. A soft iron ring, 10in mean diameter and circular cross-section 2in, is wound with a magnetizing coil. Four amperes flowing in the coil, produce a flux of 2.5 mwb in the

airgap which is 0.1in wide. Taking r to be 1000 at this flux density and allowing for a leakage coefficient of 1.2, find the number of turns on the coil.

Ans: D = 10 in = 10  2.54  102 m = 0.254 m

2 2 -4 -3 A = 2 in = 2  2.54  10 m = 1.29  10 m, i = 4 A, g = 2.5 mWb, 6

-2 g = 0.1 in = 0.1  2.54 = 0.254  10 m, r = 1000, L.F = 1.2, N = ? Total length = D =   0.254 = 0.798 m

-2 Lc = 0.798 – 0.254  10 = 0.79546 m

3 g 2.510 Bg = = 3 = 1.938 T Ag 1.29 10 B g 1.938 6 Hg = = 7 = 1.5422  10 At/m 0r 410 1

-3 -3 c = L.F  g = 1.2  2.5  10 = 3  10 Wb

3 c 310 Bc = = 3 = 2.326 T Ac 1.2910

Bc 2.326 Hc = = 7 = 1851 At/m 0r 410 1000

F = HL = Ni = HcLc + Hgg H L  H g 1851 0.79546 1.5422106  0.254102 N = c c g = i 4 = 1347.396 turns  1348 turns

Q 6. A composite iron ring is made up partly of wrought iron and partly of cast iron. Wrought iron has a mean magnetic length of 25cm with a cross sectional are a 7.5cm 2. The cast iron has a mean length of 40cm with a cross-sectional area of 24cm2 and a radial air gap of length 5mm in it. Calculate the current in the coil of 800turns, wound uniformly on the ring necessary to produce a flux of 1mwb in the air gap allowed a leakage coefficient of 1.2.

H(AT/m) 1000 2000 3000 7

B(wb/m2) (WI) 1.35 1.5 1.6 B(wb/m2) (CI) 0.25 0.5 0.62 -2 -2 Ans: Lw = 25 cm = 25  10 m, Lc = 40 cm = 40  10 m,

2 -4 2 2 -4 2 Ac = 24 cm = 24  10 m , Aw = 7.5 cm = 7.5  10 m

-3 -3 g = 5 mm = 5  10 m , N = 800 turns, g = 1  10 Wb L.F = 1.2, i = ?

3 g 110 Bg = = 4 = 0.4167 T Ag 2410 B g 0.4167 3 Hg = = 7 = 331.573  10 At/m 0 410

c = L.F  g = 1.2  0.001 = 0.0012 Wb

c 0.0012 Bc = = 4 = 0.5 T Ac 2410

From Table, Hc = 2000 At/m

w 0.0012 Bw = = 4 = 1.6 T Aw 7.510

From Table, Hw = 3000 At/m

F = HL = Ni = HcLc + Hgg + HwLw H L  H g  H L i = c c g w w N 2000 40102  3000 25102  331.573103  5103 = 800 = 4.01 A

Q 7. A magnetic circuit made of wrought iron is arranged as in figure. The center limb is wound with 500turns and has a cross sectional area of 6cm2 and each of the limbs has a 8

cross sectional area of 4cm2. Calculate the amp-turns required to produce a flux of 0.9mwb in the center limb, assuming the magnetic leakage to be negligible. B(wb/m2) 1.125 1.5 H(AT/m) 500 2000

2 -4 2 2 -4 2 Ans: Ac = 6 cm = 6  10 m , A0 = Ag = 4 cm = 4  10 m

-3 N = 500 turns, c = 0.9  10 Wb, L0 = 20 cm, Lc = 10 cm g = 1  10-3 m, i = ?

3 c 0.910 For center limb, Bc = = 4 = 1.5 T Ac 610

From Table, Hc = 2000 At/m

3 c 0.910 -3 0 = = = 0.45  10 Wb 2 2

-3 0 0.45 10 For outer limb, B0 = = 4 = 1.125 T A0 410

From Table, H0 = 500 At/m

3 c 0.910 -3 For air gap, g = = = 0.45  10 Wb 2 2

-3 g 0.45 10 Bg = = 4 = 1.125 T Ag 410 B 1.125 g 6 Hg = = 7 = 0.895246  10 At/m 0 4 10

F = Ni = HcLc + Hgg + H0L0 500i = 2000  10  10-2 + 500  20  10-2 + 0.895  106  1  10-3 i = 2.39 A (OR) MMF = 1195 At Q 8. For the magnetic circuit as shown in figure assume that the iron portion is infinitely

permeable. If all three coils are excited simultaneously such that I1 = 8A, I2 = 4A, with the 9

directions of current as shown. What is the flux density in gap g3 = ? g1 = g2 = 2mm, g3 =

2 4mm, A3 = 2A1 = 2A2 = 10cm , N1 = 60turns, N2 = 80turns, N3 = 120turns.

Ans: I1 = 8 A, I2 = 6 A, I3 =4 A

-3 -3 g1 = g2 = 2 mm = 2  10 m, g3 = 4 mm = 4  10 m

2 -4 2 -4 2 A3 = 2A1 = 2A2 = 10 c m = 10  10 m , A1 = A2 = 5  10 m

N1 = 60 turns, N2 = 80 turns, N3 = 120 turns

g 3 1 210 6 Rg1 = Rg2 = = 7 4 = 3.1831  10 At/Wb 0A1 410  510 g 3 3 410 6 Rg3 = = 7 4 = 3.1831  10 At/Wb 0A3 410 1010

By super position, F2 only, 80 6 F2 N2I2 6 6 T = = = 6  3.183110  3.183110  RT Rg1  (Rg2 // Rg3) 3.183110  6 6  3.183110  3.183110  10

= 1.0053  10-4 Wb

6 Rg2 3.183110  4 -5 3 = T  = 1.005310  6 = 5.0265  10 Wb Rg2  Rg3 2 3.183110

F3 only, 120 4 F N I 3 3 3  6 6  T = 3 = = = 3.183110  3.183110 6 RT (Rg1 // Rg2 )  Rg3  6 6   3.183110 3.183110  3.183110  = 1.0053  10-4 Wb

F1 only, 608 F1 N1I1 6 6 T = = =  3.183110  3.183110  6 RT (Rg1 // Rg3)  Rg2  6 6   3.183110 3.183110  3.183110  = 1.0053  10-4 Wb

6 Rg1 3.18310    -4 -5 3 = T = – 1.0053  10  6 = – 5.0265  10 Wb Rg1  Rg3 2 3.18310

-5 -4 -5 3 = 3 +3 +3 = 5.0265  10 + 1.0053  10 – 5.0265  10 = 100.53 Wb

3 100.53 B3 = = 4 = 0.10053 T A3 1010 ************************************************* 11

CHAPTER 1 THE IDEAL TRANSFORMER

Q 1. A coil having 90 turns is connected to a 120 V, 60 Hz source. If the effective value of the magnetizing current is 4 A, calculate the following: a. The peak value of flux b. The peak value of the mmf c. The inductive reactance of the coil d. The inductance of the coil Ans: N = 90 turns I = 4 A

Eg = 120 V Ip = 2  4

Eg (a)   max 4.44 fN

120  = 0.005 = 5 mWb 4.446090

(b) The peak value of mmf = NIP = 90 × 2  4 = 509.1 A-turns

(c) Xm = Eg/I = 120/4 = 30 

(d) L = Xm/2f = 30/(2 × 60) = 79.6 mH

Q 2. A not quite ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 120 V, 60 Hz source. The coupling between the primary and secondary is perfect, but the magnetizing current is 4 A. Calculate: a. The effective voltage across the secondary terminals b. The peak voltage across the secondary terminals c. The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V

Ans: N1 = 90 turns N2 = 2250 turns 1/a = N2/N1 = 2250/90 = 25

(a) E2 = ?

E2/E1 = N2/N1

E2 = 3000V (b) The peak secondary voltage;

E2(peak) = 2  E2 = 2 3000  4242 V 12

(c) When e1 = 37 V

e2 = 25 × 37 = 925 V

Q 3. An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A at a power factor of 80 percent lagging (Fig 1.6a). Calculate: a. The effective value of the primary current b. The instantaneous current in the primary when the instantaneous current in the secondary is 100 mA c. The peak flux linked by the secondary winding d. Draw the phasor diagram

Figure 1.6 (a) See Example 1-4 (b) Phasor relationships

Ans: N1 = 90 turns N2 = 2250 turns 1/a = N2/N1 = 2250/90 = 25

(a) I2 = 2 A, I1 = ?

I1/I2 = 1/a

I1 = 25 × 2 = 50 A

(b) I2,instantaneous = 100 mA, I1,instantaneous = ?

I1,instantaneous = 25 I2,instantaneous = 25 × 0.1= 2.5 A

(c) max = Eg/(4.44 f N1) = 200/(4.44 × 50 × 90) = 0.01 = 10 mWb (d) To draw phasor diagram,

E2 = 25 × E1 = 25 × 200 = 500 V power factor = 80% power factor = cos  cos  = 0.8  = cos-1 0.8  = 36.9 13

Q 4. The ideal transformer in figure 2 has 500 turns on the primary and 300 turns on the

secondary. The source produces a voltage Eg of 600V, and the load Z is a resistance of 12Ω. Calculate the following:

(a) the voltage E2

(b) the current I1

(c) the current I2 (d) the power delivered to the primary [W] (e) the power output from the secondary [W]

Figure 2. For Question No.4

Ans: N1 = 500 turns N2 = 300 turns 1/a = N2/N1 = 300/500 = 0.6

(a) E1 = 600 V, E2 = ?

E2/E1 = N2/N1

E2 =600 × 0.6 = 360 V (b) Z = 12 

I2= E2/Z = 360/12 = 30 A

(c) I1= I2 × N2/N1 = 30 × 0.6 = 18 A

(d) P1 = ?

P1 = E1 I1 = 600× 18 = 10800 W

(e) P1 = ?

P2 = E2 I2 = 360 × 30 = 10800 W

Q 5. A coil with an air core has a resistance of 14.7Ω. When it is connected to a 42V, 60Hz ac source, it draws a current of 1.24A. Calculate the following: (a) the impedance of the coil (b) the reactance of the coil, and its inductance 14

(c) the phase angle between the applied voltage (42V) and the current 1.24A. Ans: R = 14.7 , E = 42 V, f = 60 Hz, I = 1.24 A (a) Z = E/I \= 42/1.24 = 33.87 

2 2 2 2 (b) XL = Z  R = (33.87)  (14.7) = 30.5 

XL = 2fL

L = XL/2f = 30.5/(2 × 60) = 0.08094 H (c)  = tan-1 X/R = tan-1 3.5/14.7 = 64.27

Q 6. Calculate voltage E and current I in the circuit of Fig(3a), knowing that ideal transformer T has a primary to secondary turns ratio of 1:100.

Figure 3. (a) For Question No 6. (b) Equivalent circuit of Fig(3a). Ans:

2 2 2 2 ZT = R  (XL XC ) = (40)  (3 2) = 5 

I1 = Eg/Z = 10/5 =2 A

 E = I1R = 2 × 4 = 8 A But ratio is 1 : 100, a = 1/100

 E2 = E1/a = 8 × 100 = 80 V

Q 7. A 40F, 600V paper capacitor is available, but we need one having a rating of about 300F. It is proposed to use a transformer to modify the 40F so that it appears as 300F. The following transformer ratios are available: 120V/330V; 60V/450V; 480V/150V. Which transformer is the most appropriate and what is the reflected value of the 40F capacitance? To which side of the transformer should the 40F capacitance? To which side of the transformer should the 40F capacitor be connected? Ans: available paper capacitor = 40 F need paper capacitor = 300 F 15

1 1 XC1 = = = 66.314  2fC1 260 40 1 1 XC2 = = = 8.8419  2fC2 260300 2 Z1  N1     Z2  N2  But Z  L & Z  1/C { Z = R + j2πfL – j 1/2πfC } 2 C2  N1  X C1 1       7.49997  C1  N2  X C2 0.1333

 N1  1      N2  0.365 N E 1  1 N2 E2

E1 120 For transformer (1)  = = 0.364 E2 330

E1 60 For transformer (2)  = = 0.133 E2 450

E1 480 For transformer (3)  = = 3.2 E2 150  Transformer (1) is most appropriate  The 40 F capacitor should be connected to low voltage side (at 120V terminal). At that case, the capacitor seen by high voltage side is 2 2  330   330  C2  C1     40     302.5µF   120  120 

************************************* 16

CHAPTER 2 PRACTICAL TRANSFORMERS

Q 1. A large transformer operating at no-load draws an exciting current I0 of 5 A when the primary is connected to a 120 V, 60 Hz source (Fig 2.2a). From a wattmeter test it is known that the iron losses are equal to 180 W.

Figure 1. For Question No.1. Calculate

a. The reactive power absorbed by the core

b. The value of Rm and Xm

c. The value of If, Im, and I0

Ans: I0 = 5 A, E1 = 120 V, f = 60 Hz, Pin = 180 W

(a) Qm = ?

Sm = E1I0 = 120 × 5 = 600 VA

(b) Xm = ?, Rm = ?

2 2 E1 120 Rm =   80  Pm 180

2 2 E1 120 Xm =   25.2  Qm 572

(c) If, Im, I0 = ?

E1 120 If =   1.5A R m 80

E1 120 Im =   4.8A Xm 252

2 2 2 2 I0 = If  Im  1.5  4.8  5A

Q 2. A single-phase transformer rated at 3000 kVA, 69 kV/4.16 kV, 60 Hz has a total internal

impedance Zp of 127 , referred to the primary side. 17

Calculate a. The rated primary and secondary currents b. The voltage regulation from no-load to full-load for a 2000 kW resistive load, knowing that the primary supply voltage is fixed at 69 kV. c. The primary and secondary currents if the secondary is accidentally short-circuited. Ans:

S = 3000 kVA, f = 60 Hz, Ep = 69 kV, Es = 4.16 kV, ZP = 127  a = 69/4.16 = 16.58

(a) rated Is = S/Es = 3000k/4.16k = 721.2 A

rated Ip = Is/a = 721.2/16.58 = 43.5 A

(b) PL = 2000 kW, V.R = ?

E2 (4.16k)2 R = s = = 8.65  PL 2000k a2 R = (16.58)2 (8.65) = 2380  S = 3000 kVA > 500 kVA

 Rp neglect

Xp = Zp = 127 

2 2 2 2 2 Z = Zp  a R = 127  2380 = 2383 

Ip = Ep/Z = 69 k/ 2383 = 28.95 A

2 aEs = (a R) Ip = 238028.95 = 68902 V

Es(F.L) = 4154 V

Es(N.L) = 4160 V

ES(N.L)  ES(F.L) 4160  4154 %V.R = 100 = 100 = 0.14% ES(F.L) 4154

(c) Ip = Ep/Xp = 69 k/127 = 543 A

Is = aIp = 16.58543 = 9006 A

Q 3. The following information is given for the transformer circuit of Figure 3.

R1 = 18Ω Ep = 14.4kV (nominal)

R2 = 0.005Ω Es = 240V (nominal)

Xf1 = 40Ω Xf2 = 0.01Ω 18

Figure 3. For Question No 3 If the transformer has a nominal rating of 75kVA, calculate the following: (a) The transformer impedance [Ω] referred to the primary side (b) The percent impedance of the transformer (c) The impedance referred to the secondary side (d) The percent impedance referred to the secondary side (e) The total copper losses at full load (f) The percent resistance and percent reactance of the transformer

Ans: R1 = 18 , R2 = 0.005 , Xf1 = 40 , Ep = 14.4 kV, Es = 240 V, Xf2 = 0.01 , S = 75 kVA

a = Ep/Es = 14.4k/240 = 60

2 2 Rp = R1 + a R2 = 18 + (60 0.005) = 36 

2 2 Xp = Xf1 + a Xf2 = 40 + (60 0.01) = 76 

2 2 2 2 (a) Zp = R p  Xp = 36  76 = 84 

(b) Ip = S/Ep = 75 k/14.4 k = 5.2 A

Base impedance, Znp = Ep/Ip = 14.4k/5.2 = 2764 

p.u Z = Zp/Znp = 84/2769 = 0.03 %Z = 0.03100% = 3 %

2 2 (c) Rs = (R1/a ) + R2 = (18/60 ) + 0.005 = 0.01 

2 2 Xs = (Xf1/a ) + Xf2 = (40/60 ) + 0.01 = 0.02 

2 2 2 2 Zs = Rs  Xs = 0.01  0.02  0.022 

Is = aIp = 605.2 = 312 A

(d) Zns = Es/Is = 240/312 = 0.77 

p.u Z = Zs/Zns = 0.022/0.77 = 0.028 % Z = 0.028100 % = 2.8 %

2 2 (e) Total copper loss at F.L = IpR p = 5.2 36 = 976.44 W

(f) p.u X = Xp/Znp = 76/2769 = 0.02754 %X = 0.0275 100 % = 2.75 % 19

p.u R = Rp/Znp= 36/2769 =0.013 %R = 0.013100 % =1.3%

Ans: S = 500 kVA, 69 kV/4.16 kV, f = 60 Hz,

Esc = 2600 V, Isc = 4 A, Psc = 2400 W, Es = 4160 V, I0 = 2 A, Pm = 5000 W

Zp = Esc/Isc = 2600/4 = 650 

2 2 R;p = Psc/ Isc = 2400/4 = 150 

2 2 2 2 Xp = Zp  R p = 650 150 = 632 

Sm = EsI0 = 4160 2 = 8320 VA

2 2 2 2 Qm = Sm  Pm = 8320  5000 = 6650 VAR L.V side H.V side

2 2 Es 2 2  69  (a) Rm = = 4160 /5000 = 3461  Rm  a R m     3461=952.2 k Pm  4.16 

2 Es 2 Xm = = 4160 /6650 = 2600  Qm

2 2  69  Xm  a Xm     2602  716 k  4.16 

(b) SL = 250 kVA, cos  = 0.8 (lag),  = ?

a = Ep/Es = 69k/4.16k = 16.6 20

I2 = SL/Es = 250k/4.16k = 60 A

I1 = I2/a = 60/16.6 = 3.62 A

2 2 Pcopper = I1 R p  3.62 150  1960 W

Piron = Pm = 5000 W

P0 = SL cos  = 250k  0.8 = 200 kW

Pi = P0 + Pcopper + Piron = 200 k + 1960 + 5000 = 207 kW

 = (Pout/Pin) 100% = (200k/207k) 100% = 0.966 100% = 96.6 %

Q 5. During a short-circuit test on a 10MVA, 66kV/7.2kV transformer (see Fig 5), the following results were obtained:

Eg = 2640V Isc = 72A Psc = 9.85kW

Figure 5. For question 5. Calculate the following: (a) The total resistance and the total leakage reactance referred to the 66kV primary side (b) The nominal impedance of the transformer referred to the primary side (c) The percent impedance of the transformer (d) If the iron losses at rated voltage are 35kW, calculate the full-load efficiency of the transformer if the power factor of the load is 85percent.

Ans: Eg = Esc = 2640 V, Isc = 72v A, Psc = 9.85 kW, 66kV/7.2kV, S = 10 MVA[

(b) Zp = Esc/Isc = 2640/72 = 36.67 

2 2 (a) Rp = Psc/ Isc = 9.852 k/72 = 1.9 

2 2 2 2 Xp = Zp  R p = 36.67 1.9 = 36.6 

(c) Ep = 66kV, Ip = S/Ep = 10 M/66 k = 152 A

Znp = Ep/Ip = 66k/152 = 434 

p.u Z = Zp/Znp = 36.67/434 = 0.084 %Z = 0.084100% = 8.4 %

p.uX = Xp/Znp = 36.6/434 = 0.08 21

%X = 0.08100% = 8 %

p.u R = Rp/Znp = 1.9/434 = 0.004 %R = 0.004100% = 0.4 %

(d) Piron = 35 kW, cos  = 85 % = 0.85,  = ?

2 2 Pcu = IpR p  152 1.9 = 43900 W

Po = S cos  = 10 M 0.85 = 8500 kW

Pi = Pout + Pcu + Piron = 8500k + 43900 + 35k = 8600 kW

F.L  = (Pout/Pin) 100% = (8500k/8600k)100% = 99.1 %

Q 6. The 500 kVA, 69kV/4160V, 60Hz transformer shown in Fig 6 has are resistance Rp of 150

 and a leakage reactance Xp of 632 . Using the perunit method, calculate: a. the voltage regulation when the load varies between zero and 250 kVA at a lagging power factor of 80 b. the actual voltage across the 250 kVA load

c. the actual line current I1.

Figure 6. For question 6.

Ans: Base P, Pb = 500 kVA, Rp = 150 , Xp = 632 , cos  = 0.8, S = 250 kVA

Base V, Eb = 69kV

Ib = Pb/Eb = 500k/69k = 7.25 A

Zb = Eb/Ib = 69k/7.25 = 9517 

Rp(p.u) = Rp/Zb = 150/9517 = 0.0158 p.u

Xp(p.u) = Xp/Zb = 632/9517 = 0.0664 p.u

E12(p.u) = 69k/69k = 1 p.u

S(p.u) = S/Pb = 250/500 = 0.5 p.u

P(p.u) = S(p.u)cos  = 0.50.8 =0.4 p.u

2 2 2 2 Q(p.u) = Sp.u  Pp.u = 0.5  0.4 = 0.3 p.u 22

2 2 Ep.u 1 RL(p.u) =   2.5 p.u Pp.u 0.4

2 2 Ep.u 1 XL(p.u) =   3.333 p.u Qp 0.3 2.5 j3.333 Z = R //X =  1.6  j1.2 34(p.u) L L 2.5  j3.333

Z12(p.u) = Rp + jXp + Z34(p.u) = 0.0158 + j0.0664 + 1.6 + j1.2 = 1.616 + j1.266

I1(p.u) = E12(p.u)/ Z12(p.u) = 0.4872-38.07

E34(p.u)= I1(p.u)  Z34(p.u) = 0.4872-38.07(1.6 + j1.2) = 0.9744-1.2 = E34(F.L)

E34(N.L) = 1

E34(N.L)  E34(F.L) 1 0.9744 p.u voltage regulation =   0.0263 E34(F.L) 0.9744 (a) % V.R = 0.0263100% = 2.63 %

(b) E34 = E34(p.u)  EB = 0.794469k = 67.23 kV 4160 E = E  = 4053 V 56 34 69k

(c) I1 = I1(p.u) IB = 0.4872 7.25 = 3.53 A

Q 7. A 100 kVA transformer is connected in parallel with an existing 250kVA transformer to supply a load of 330kVA. The transformers are rated 7200V/240V, but the 100 kVA unit has an impedance of 4 percent while the 250kVA transformer has an impedance of 6 percent (Fig 7). Calculate a. The nominal primary current of each transformer b. The impedance of the load referred to the primary side c. The impedance of each transformer referred to the primary side d. The actual primary current in each transformer.

Figure 7. for question 7 23

Ans: S1 = 250 kVA, S2 = 100 kVA, SL = 330 kVA, Ep = 7200 V, Es = 240 V

(a) Ip1 = 250k/7200 = 34.7 A

Ip2 = 100k/7200 = 13.9 A

2 2 Ep 7200 (b) ZL =   157  SL 330k

IL = SL/Ep = 330k/7200 = 46 A

(c) Znp1 = Ep/Ip (Ip = Sn1/Ep)

2 2 Ep 7200 Znp1 =   207  Sn1 250k

Zp1 = Znp1 Zp1(p.u) = 2070.06 = 12.4 

2 2 Ep 7200 Znp2 =   518  Sn2 100k

Zp2 = Znp2 Zp2(p.u) = 5180.04 = 20.7  Z p2 20.7 46  28.8 Zp1  Zp2 (d) I1 = IL = 12.4  20.7 A

I2 = IL – I1 = 46 – 28.8 = 17.2 A

Q 8. A 50kVA, 2400:240V 60Hz distribution transformer has the leakage impedance of 0.72 + j0.92 Ω in the high-voltage winding and 0.007 + j0.009 Ω in the low-voltage winding. At rated voltage and frequency, the admittance of shunt branch accounting for the exciting current is (0.324 – j2.24) x 10-2 mho when viewed from the low-voltage side. It is used to step down the voltage at the load end of a feeder whose impedance is 0.30 + j1.60Ω. The

voltage Vs at the sending end of the feeder is 2400V. Find the voltage at the secondary terminals of the transformer when the rated load connected to its secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops in the transformer and feeder caused by the exciting current. Ans: 24

2 Zeq = ( 0.3 + j1.6 ) + ( 0.72 + j0.92 ) +( 0.007 + j0.009 )(2400/ 240) = 1.72 + j3.42 Ω θ = cos-1 0.8 = 36.87

I2FL = 50 k/240 = 208.33 -36.87 A

I1FL = I2FL  240/2400 = 20.833-36.87 A

-2400 + I1FLZeq + E1 = 0

E1 = 2400 - I1FLZeq = 2400- 20.833-36.87 (1.72 + j3.42 ) = 2328.85-0.87 V

E2 = E1  240/2400 = 232.885-0.87 V

 V2 = 232.885-0.87 V

Q 9. A single-phase load is supplied through a 35kV feeder whose impedance is 115 + j380Ω and a 35kV:2400V transformer whose equivalent impedance is 0.26 + j1.21Ω referred to its low-voltage side. The load is 180kW at 0.87 leading power factor and 2320V. (a) Compute the voltage at the high-voltage terminals of the transformer. (b) Compute the voltage at the sending end of the feeder. (c) Compute the power and reactive power input at the sending end of the feeder. Ans: 25

(a) V1 =?

I2FL = 180k/(2320  0.87 ) = 89.1795 29.54 A

I1FL = I2FL  2400/35 k = 6.115 29.54 A

2 Zeq1 = (0.26 + j1.21 ) ( 35 k/ 2400) = 55.295 + j257.335 Ω

E2 = VL = 23200 V

E1 = E2  35 k/ 2400 = 33833.330 V

V1 = E1 + I1 Zeq1 =33833.33 + (6.11529.54)(55.295+j257.335) = 33387.012.64 V

(b) Vs = ?

Vs =V1 + I1FL Zeq = 33387.012.64 + 6.115 29.54 (115+j380) = 330496.8

(c) P = Vs I1FL cos  = 33049 6.115 cos 22.74 = 186.37 kW

Q = Vs I1FLsin  = 33049 6.115 sin 22.74 = 78.15 kVAR

Q 10. A 3.46kV:2400V transformer has a series leakage reactance of 38.3Ω as referred to the high-voltage side. A 20kW load (unity power factor) is connected to the low-voltage side, and the voltage is measured to be 2380V. Calculate the corresponding voltage and power factor as measured at the high-voltage terminals. Ans:

I2FL = 20 k/( 2380 1 ) = 8.403  0 A

I1FL = I2FL  2400/3.46 k = 5.83 0 A 26

E2 = VL = 2380 V

E1 = E2  3.46 k/2400 = 3431.17 V

V1 = E1 + I1FL X1 = 3431.17 + 5.83 ( j38.3 ) = 3438.43 3.72 V P.F = cos  = cos 3.72 = 0.9979

Q 11. The resistances and leakage reactance in ohms of a 25kVA, 60Hz, 2400:240V distribution transformer are

R1 = 0.650 R2 = 0.00650

Xf1 = 8.40 Xf2 = 0.0840 where subscript 1 denotes the 2400V winding and subscript 2 denotes the 240V winding. Each quantity is referred to its own side of the transformer. (a) Draw the equivalent circuit referred to (i) the high- and (ii) the low-voltage sides. Label the impedances numerically. (b) Consider the transformer to deliver its rated kilovoltamperes at 0.8 power factor lagging to a load on the low-voltage side with 240V across the load. Find the high- tension terminal voltage. Ans:

(a) The equivalent circuit referred to high-voltage side:

Zeq1  Z1  Z2

2 Zeq = ( 0.65 + j8.4 ) + ( 0.0065 + j0.084 )( 2400/240 ) = 1.3 + j16.8  27

The equivalent circuit referred to low-voltage side:

Zeq2  Z1  Z2

2 Zeq = ( 0.0065 + j0.084 ) + ( 0.65 + j8.4 ) ( 240/2400 ) =0.013 + j0.168  (b)

I2FL = 25 k/240 = 104.17 -36.87 A

I1FL = I2FL  240/2400 = 10.417 -36.87 A

V1 = E1 + I1FL Zeq1 = 2400 + 10.417 -36.87 ( 1.3 + j16.8) = 2519.3 -3 V

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CHAPTER 3 SPECIAL TRANSFORMERS

Q 1. (a) Define CT and PT.

(b) The autotransformer in Fig 3.3 has an 80 percent tap and the supply voltage E1 is 300V. If a 3.6kW load is connected across the secondary calculate: a. The secondary voltage and current b. The currents that flow in the winding c. The relative size of the conductors on windings BC and CA

Figure 1. For Question No1(b). Ans: (a) Current Transformers (CT)  Current transformers are high-precision transformers in which the ratio of primary to secondary current is a known constant, which changes very little with the burden.

Voltage Transformers (VT) (OR) Potential Transformers (PT)  Voltage transformers are high-precision transformers in which the ratio of primary to secondary voltage is a known constant, which changes very little with the load.

(b) E1 = 300 V, P = 3.6 kW, 80% tap

(i) E2 = 80% E1 = 0.8 300 = 240 V

I2 = P/E2 = 3.6k/240 = 15 A

(ii) I1 = P/E1 = 3.6k/300= 12 A

the current in winding BC = I1 = 12 A

the current in winding CA = I2 -I1= 15 – 12 = 3 A (iii)  The current in winding CA is 4 times smaller than the current in winding BC.

(ICA= 1/4IBC) Size of conductors in winding CA = (1/4)size of conductors in winding BC 29

VBC = E1 – E2 = 300- 240 = 60 V

VCA = E2 = 240 V

VCA = 4 VBC number of turns in winding CA = 4 number of turns in winding BC

NCA = 4 NBC The two windings require the same amount of copper.

Q 2. (a) Why must we never open the secondary of a current transformer?. (b) A single-phase transformer has a rating of 100kVA, 7200/600V, 60Hz. If it is reconnected as an autotransformer having a ratio of 7800V/7200V, calculate the load it can carry. Ans: (a) - We must never open the secondary circuit of a current transformer while current is flowing in the primary circuit.

- If the secondary is accidentally opened, the primary current I1 continues to flow unchanged because the impedance of the primary is negligible compared to that of the electrical load. - The line current thus becomes the exciting current of the transformer.

I1 = 1000 k/7200 = 13.88 A

IH = I2 = 1000/600 = 166.67 A

IX = I1 + I2 = 180.55 A

New kVA rating = VH IH = 7800 166.67 = 1300 kVA

New kVA rating = VX IX = 7200 180.55 = 1300 kVA 30

Q 3. A120:480V, 10kVA transformer is to be used as an autotransformer to supply a 480V circuit from a 600V source. When it is tested as a two-winding transformer at rated load, unity power factor, its efficiency is 0.975. (a) Make a diagram of connections as an autotransformer. (b) Determine its kVA rating as an autotransformer. (c) Find its efficiency as an autotransformer at full load, with 0.85 power factor lagging. Ans: (a)

(b) IH = I1 =10 k/120 = 83.33 A

I2 = 10 k/480 = 20.83 A

IX = I1 + I2 = 104.167 A

kVA rating = VH IH = 600 83.33 = 50 kVA

= VX IX = 480 104.167 = 50 kVA (c) For U.P.F,  = 0.975; (2 winding transformer )

PO = 10 k1 = 10 kW

Pi = PO/ = 10 k/0.975 = 10.2564 kW

PLosses = Pi - PO = 0.2564 kW For autotransformer; P.F = 0.85 (Lag) { Same core and wires, i.e. same losses }

PO = 50 k0.85 = 42.5 kW

Pi = PO + PLosses = 42.7564 kW

 = PO/Pi =(42.5/42.7564) 100% = 99.4%

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CHAPTER 4 THREE-PHASE TRANSFORMERS

Q 1. Three single-phase step-up transformers rated at 40MVA, 13.2kV/80kV are connected in delta-wye on a 13.2 kV transmission line (Fig 4.5). If they feed a 90MVA load, calculate the following: a. The secondary line voltage b. The currents in the transformer windings c. The incoming and outgoing transmission line currents Ans:

(a) Es(line) = ?

Ep(ph) = Ep(line) = 13.2 kV ()

 Es(ph) = 80 kV (Y)

Es(line) = 3  80 k = 138 kV

(b) S = 90 MVA (3-phase), Ip = ?, Is = ?

Sph = S/3 = 90/3 = 30 MVA (per phase)

Ip = Sph/Ep(ph) = 30M/13.2k = 2272 A

Is = Sph/Es(ph) = 30M/80k = 375 A (c) the current in each coming line A,B,C is

Iin = 3 Ip = 3  2272 = 3932 A the current in each outgoing line 1,2,3, is

Iout = Is = 375 A

Q 2. Three single-phase transformers are connected in delta-delta to step down a line voltage of 138kV to 4160V to supply power to a manufacturing plant. The plant draws 21MW at a lagging power factor of 86 percent. Calculate 32

a. The apparent power drawn by the plant b. The apparent power furnished by the HV line c. The current in the HV lines d. The current in the LV lines e. The currents in the primary and secondary windings of each transformer f. The load carried by each transformer Ans: a. The apparent power drawn by the plant is S = P/cos  = 21/0.86 = 24.4 MVA b. The transformer bank itself absorbs a negligible amount of active and reactive power because the I2R loses and the reactive power associated with the mutual flux and the leakage fluxes are small. It follows that the apparent power furnished by the HV line is also 24.4 MVA. c. The current in each HV line is

I1 = S/(3 E) = (24.4 x 106)/(3 x 138000) = 102 A d. The current in the LV lines is

I2 = S/(3 E) = (24.4 x 106)/(3 x 4160) = 3386A e. Referring to Fig 4.2, the current in each primary winding is

Ip = 102/3 = 58.9A The current in each secondary winding is

Is = 3386/3 = 1955 A f. Because the plant load is balanced, each transformer carries one-third of the load, or 24.4/3 = 8.13 MVA. The individual transformer load can also be obtained by multiplying the primary voltage times the primary current:

S = EpIp = 138000 x 58.9 = 8.13 MVA 33

Q 3. Three single-phase transformers rated at 250kVA, 7200V/600V, 60Hz are connected in wye-delta on a 12470V, 3-phase line. If the load is 450kVA, calculate the currents in the incoming and outgoing transmission lines and also in the primary and secondary windings.

Ans: SLL = 250 kVA, 7200 V/600 V, S = 450 kVA, Iin= ?, Iout = ?

Ep(LL) = 12470 V

Ep(phase) = Ep(LL)/ 3 = 12470/ 3 = 7199.56 V (Y)

Es(phase) = Ep(phase)  600/7200 = 7199.56  600/7200 = 599.96 V

Es(LL) = Es(ph) = 599.96 V()

Sph = S/3=450/3 = 150 KVA

Ip = Sph/Ep(phase) = 150k/7199.56 = 20.83 A

Iin = Ip = 20.83 A

Is = Sph/Es(phase) = 150k/599.96 = 250 A

Iout = 3 Is = 3  250 = 433 A

Q 4. In order to meet an emergency, three single-phase transformers rated 100kVA, 13.2kV/2.4kV are connected in wye-delta on a 3-phase 18kV line. (a) What is the maximum load that can be connected to the transformer bank? (b) What is the outgoing line voltage?

Ans: S = 100 kVA, 13.2kV/2.4 kV, Ep(LL) = 18 kV 34

(a) maximum load = S = 3  Sph = 3  100 = 300 kVA

(b) Ep(LL) = 18 kV

Ep(phase) = Ep(LL)/ 3 = 18k/ 3 = 10392.3 V

Es(phase) = Ep(phase)  2.4k/13.2k = 1889.5 V

Es(LL)= Es(phase)= 1889.5 V  the outgoing line voltage = 1889.5 V

Q 5. The 3-phase step-up transformer is rated 1300MVA, 24.5kV/345kV, 60Hz, impedance 11.5percent. It steps up the voltage of a generating station to power a 345kV line. (a) Determine the equivalent circuit of this transformer, per phase. (b) Calculate the voltage across the generator terminals when the HV side of the transformer delivers 810MVA at 370kV with a lagging power factor of 0.9. Ans: Assume Y-Y connection

a = 345/24.5 = 14.08 35

the base voltage is EB = 345/ 3 = 199.2 kV (secondary)

SB = S/3 = 1300/3 = 433.3 MVA this is a very large transformer. (R is neglected)  Z = X

Zr(pu) = j0.115 (a) the equivalent circuit per phase is

(b) S = 810 MVA, SL = S/3 = 810/3 = 270 MVA

EL = 370k/ 3 = 213.6 kV, cos  = 0.9 (lag), Eg = ?  = cos-1 0.9 = 25.84

SL(pu) = 270/433.33 = 0.6231

EL(pu) = 213.6/199.2 = 1.0723 0 (reference)

IL(pu) = SL(pu)/EL(pu) = 0.6231/1.0723 = 0.5811 -25.84

Es(pu) = EL(pu) + IL(pu)  Zr(pu) = 1.0723 0 + 0.5811 -25.84 (j0.115) = 1.103 3.12

Es = 1.103  345 kV = 380.5 3.12 kV

Ep(pu) = 1.103 3.12

Eg = Ep(pu)  EB(primary) = 1.103  24.5 kV = 27.02 kV (L-L)

Q 6. (a) Three 150kVA, 480V/4000V, 60Hz single-phase transformers are to be installed on a 4000V, 3-phase line. The exciting current has a value of 0.02pu. Calculate the line current when the transformers are operating at no-load. (b) The core loss in a 300 kVA 3-phase distribution transformer is estimated to be 0.003 pu. The copper losses are 0.0015 pu. If the transformer operates effectively at no-load 50 percent of the time, and the cost of electricity is 4.5 cent per kWh, calculate the cost of the no-load operation in the course of one year. 36

Ans: (a) SB = 3x150 kVA = 450kVA, 480V/4000, ELL = 4000 V, I0 = 0.02 pu

EB(primary) = ELL = 4000 V ()

IB(primary) = SB/EB(primary) = 450kVA/ 3 x 4000V= 64.95 A = 1 pu

at no load, I0 = 0.02 pu

Ip(line) = Ip(phase) = 0.02 pu = 0.02  64.95 = 1.3 A

(b) SB = 300 kVA = 1 pu,

Pcore = 0.003 pu  300k = 900 W

Pcu = 0.0015 pu  300k = 450 W the cost of electricity = 4.5 cent per kWh no-load time = 50% of the time = 0.5  8760Hr/yr = 4380 Hr/yr

Pno-load = Pcore = 900 W no-load energy = 900  4380 = 3942 kWh/yr The cost of no-load operation in one year = 4.5  3942 = 17739 cent = 177.39 $/yr

Q 7. A three-phase Y- transformer is rated 225 kV:24kV, 400 MVA and has a series reactance of 13.3  as referred to its high-voltage terminals. The transformer is supplying a load of 325 MVA, with 0.9 power factor lagging at a voltage of 24 kV (line to line) on its low-voltage side. It is high-voltage feeder whose impedance is 0.15 + j 1.9  connected to its high-voltage terminals. For these conditions, calculate (a) the voltage at the high-voltage terminals of the transformer and (b) the voltage at the sending end of the feeder. Ans:

The equivalent ckt per phase is SL = 325/3 = 108.33 MVA

-1  = cos 0.9 = 25.84

VL(phase) = 24 kV() 37

(a) I2 = SL/VL = (325/3) M/24k = 4513.89 -25.84 A

I1 = I2  24k/(225k/ 3 ) = 833.95 -25.84 A

E2 = VL = 24 kV

E1 = E2  (225/ 3 )k/24k = 24k  (225/ 3 )k/24k = (225/ 3 ) kV = 129.90 kV

V1 = E1 + I1  j13.3 = 129904 + (833.95-25.848)(j13.3) = 135107.64.237 kV

V1(LL) = 3  135107.6 = 234 kV

(b) Vs = V1 + I1 (0.15 + j1.9) = 135107.64.237 + 833.95 -25.84 (0.15 + j1.9) = 136016 4.788 V

Vs(line) = 3  136016 = 235.6 kV

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