MD Singh Power Electronics Solution Manual to Chapter 10

CHAPTER

10

10.1 (i) Using Eq. (10.3), we can write 480 = (3/p) ¥ Emph ◊ sin(p/3) Emph = 580 V/ph. (ii) maximum value of cycloconverter load = 72 2 i.e. peak current = 101.8 A Thyristor Rms current = Irms = PIV=

72 2 3

= 58.79 A

3 Ephcmax =

3 ¥ 580 ¥ 2 = 1420.70 V

Rms value for one-third period conduction Irms = Input power per phase =

(72) 2 / 3 = 41.57 A

1 ¥ load power 3 1 = ¥ 480 ¥ 72 ¥ 0.85 3 = 9792 W

Power factor

=

Power 9792 = Erms .I rms 580 ¥ 41.57

= 0.41 Load power

= 480 ¥ 72 ¥ 0.85 = 29.4 kW

10.2 (i) Per phase input voltage to transformer = 415 V per phase input voltage to converter = Eph =

415 = 207.5 V 2

Voltage reduction factor, r = cos (180 – 170) = cos 10∞

54

Power Electronics

= 0.9848 For 3-phase pulse device, \

m =3

RMS value of fundamental voltage 3 Eor = cos10∞ ÈÍ207.5 Ê ˆ ·sin p / 3˘˙ Ëp¯ Î ˚ = 169 V (ii) RMS output current =

169 2

3 +2

2

- tan -1

2 3

= 46.87 -33.690 A (iii) Output power = I2 or R = (46.94)2 ¥ 3 = 6590.39 W 10.3 (i) Per phase input voltage to converter = 207.5 V Live voltage input to bridge converter = 207.5 3 = 359.40 V Voltage reduction factor, For 6-phase device,

r = cos 10. m =6

Rms value of output voltage is 6 È Êp ˆ˘ Eor = cos 10 Í207 ◊ 5 3 Ê ˆ ◊ sin Ë ¯˙ Ë ¯ Î p 6 ˚ .

= 337.99 V (ii) Rms output current

=

337.99 32 + 22

– tan – 1

2 3

= 93.74 – 33 ◊ 69∞ . (iii) Rms output power

= (93.74)2 . ¥ 3 = 26361.56 W.

10.4 For a three-phase, six pulse cycloconverter, input power is given by Pi = 3.E. Imphase = 3.E.I

cos q 2

cos f∞

\

50 ¥ 103 = 3 ¥ 415 ¥ I ¥

\

I @ 142 A .

1 2

¥ 0.5 ¥ 0.8