MD Singh Power Electronics Solution Manual to Chapter 16

CHAPTER

16

16.1 Given: Load = 800 w, PF = 0.8, e = 85%. Edc = 48 V, Back-up time = 30 min. (i) KVA rating of inverter

\ Total rms

PF =

Active Power Total Rms Power

0.8 =

800 Total Rms Power

power = 1000 VA

\ VA rating of inverter = 1 KVA (ii) Voltage of rectifier =

KVA rating ¥ UPF Efficiency =

1 KVA ¥ 0.8 = 1 kW 0.850

(iii) Battery kW rating = Voltage ¥ current 1 kW = 48 ¥ Idc Idc = 20.83 A.

\

A-H rating = 20.83 ¥ 1/2 = 10.5 AH. 16.2 (i)

Battery

(kW) = =

\ (ii)

50 ¥ 0.8 = 44.44 kW. 0.90

Battery needs to supply 44.44 kW power. No. of cells = =

(iii)

Load KVA ¥ PF Efficiency

No. of jars =

Minimum allowable voltage battery Final voltage per cell. 210 = 120. 1.75 120 No. of cells = 120 = 6 No. of cells per jar

94

Power Electronics

(iv) Cell size

(kW) =

44.44 = 120. 120

= 0.370 kW. Two 75-AH cell battery is selected which has 0.426 kW discharge rate at 15 min back-up time which is greater than required 0.370 KW. 16.3 E0 = 80 V, Es = 110 V, f = 16 kHz. E0 = Es d /1 - d

(i)

80 d = 0.73 \ d = 0.42 = (1 - d ) 110

\

T =

1

f

= 62.5 usec

P0 = E0 . I0

\

\ I0 =

120 = 1.5 A. 80

For a single transistor flyback smps, the minimum value of the inductance is given by Lmin = (Es . d)2.

T 2 P0

= (110 ¥ 0.42)2 ◊

62.5 ¥ 10-6 2 ¥ 120

= 0.559 mH. 16.4 We have d = 25 cm = 0.25 cm N = 20 turns I = 200 A F = 450 ¥ 103 Hz.

r = 6.8 ¥ 10–8 W - m. Depth of heating is given by

d = 503 assuming mr = 1 \

604 ¥ 10-8 r m. ◊ = 503 ur. f 1 ¥ 450 ¥ 103

d = 190 ¥ 10–6 meters d = 0.19 mm.

Now, heat per unit area of the surface is given by H = pa = 2p

Ê NI ˆ ÁË ˜¯ l

2

. r . mr ◊ F 107

2 Ê 20 ¥ 200 ˆ . = 2p Ë 0.25 ¯

(6.8 ¥ 10-8 ) ¥ 1 ¥ 450 ¥ 103 107

= 6.28 ¥ 256 ¥ 106 ¥ (55.31 ¥ 10–6) = 88.92 kW/m2 16.5 Area A = (20 ¥ 15) cm2 = 300 cm2 = 0.0300 m2 b = distance between electrodes = 1 cm = 0.01 m.

Solution Manual 95

Er = 3, cosq = 0.4, f = 20 ¥ 106 Hz. V = 750 V. C =

Œ0 . Œr A 8.854 ¥ 10-12 ¥ 3 ¥ 0.0300 = = 79.69 ¥ 10–12 F. b 0.01

cos q = 0.4 \ q = 66.42, d = 900 – 66.42 = 23.58 \ tan d = 0.44 Also,

P = 2p f.c. V 2 tan d = 2p ¥ 20 ¥ 106 ¥ 79.69 ¥ 10–12 ¥ 750 ¥ 750 ¥ 0.44 = 2478.5 Watts.