Download Solved_Problems_to_Chapter_11

CHAPTER
11
Additional Problems
Solved Problems
11.1 A single phase voltage controller has input voltage of 230 V, 50 Hz and a load of R = 15W. For 6
cycles on and 4 cycles off, determine.
(a) rms output voltage
(b) input pF and
(c) average and rms thyristor currents.
Sol. (a) For n on-cycles and m off-cycles,
Eor = Es .
(b) Input
n
= 230
n+m
n
=
n+m
pF =
6
= 178.16 V.
6+4
6
= 0.78 lag
6+4
Also power delivered to load = Ior2 . R
= Vor2/R. =
178.157 2
15
= 2116 W.
Input
VA = 230 ¥
\ Input
PF =
230 6
= 2731.74 VA.
15
2116
= 0.78 lag
2731.74
(c) Peak thyristor current, Im =
230 6
= 21.68 A.
15
Average value of thyristor current
ITA =
=
F
h
◊ m
n+m p
0.6 ¥ 21.68
= 4.14 A
p
Solution Manual 2
Im
\ RMS value of thyristor current, ITe =
=
n
n+m
2
21.681 ¥ 0.6
= 8.397 A
2
11.2 A three-phase, three-wire bidirectional controller supplies a star-connected resistance load of
R = 10 W per phase. Determine the output voltage and power consumed by the load for
following cases:
(a)
a = 30° (b) a = 75° (c) a = 120°
Assume
Es = 230 V
Sol. (a) From Eq. (11.41),
Eo =
È1
6 ◊ ( 230) Í
ÎÍ p
1/ 2
Ê p p / 6 sin (p /3) ˆ ˘
ÁË 6 - 4 +
˜¯ ˙
8
˚˙
= 225 V.
Power consumed,
P=
3 Eo 2
R
2
=
(b) for
3 ¥ ( 225)
= 15.19 kW.
10
a = 75° (= 5p/12)
Eo =
È1
6 ( 230) Í
ÍÎ p
1/ 2
Ê p 3 sin (150)
3 cos (150) ˆ ˘
+
Á 12 +
˜˙
16
10
Ë
¯ ˙˚
= 162.65 V.
2
P=
(c) for
3 (162.65)
= 7.93 KW.
10
Ê 2p ˆ
a = 120° Á = ˜
Ë 3¯
Eo =
È1
6 ( 230) Í
ÍÎ p
1/ 2
Ê 5p 2p /3 sin ( 240)
3 cos ( 240) ˆ ˘
+
Á 24 - 4 + 16
˜˙
16
Ë
¯ ˙˚
= 47.83 V
P= 3¥
(47.83)2
10
= 686.3 W.
11.3 In an on-off control circuit using single-phase 230 V, 50 Hz supply, the on-time is 10 cycles and
off-time is 4 cycles. Determine the RMS value of the output voltage.
Sol. Output voltage is equal to the supply voltage during on-time and zero during off-time.
\ RMS value of the output voltage is given by
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Power Electronics
1/ 2
Ï È( 230)2 ¥ 10 + 0 ¥ 4 ˘ ¸
Ô
˚Ô
Vrms = Ì Î
˝
14
Ô
Ô
Ó
˛
= 230 ¥ 0.84 = 193 V.
11.4 The circuit in Fig. E 11.4 is used for controlling furnance temperature. If the pedestal voltage
VP = OV. Calculate the firing angle a and rms load-voltage. If the pedestal voltage is 3 V,
calculate the change in a.
+
+
15 V
R3
R1
R4
VP
330 sin 314 t
R2
furnace
(a)
n = 0.8
D
G
C
0.1mF
Pedestal voltage
(b)
K
Fig. E 11.4
Sol. (i) When pedestal voltage, VP = O. With this Fig. E 11.4 (b) simply work as UJT relaxation
oscillator. Time taken by the capacitor to charge from O to nVBB is ‘t1’ which corresponds to the
firing angle a.
\ Peak voltage on capacitor, VP = nVBB + VD.
= (0.8 ¥ 15) + 0.7 = 12.7 Volts.
Now,
(
VP = VBB 1 - e-t1/ R 3c
3
\
12.7 = 15 È1 - e -t1 / 3 ¥ 10
Î
\
t1 = 562.6 msec.
)
¥ 0.1 ¥ 10-6
˘
˚
One half cycle of input a.c. voltage corresponds to 1800 and the duration of one half cycle is 10
msec.
\ If
10 msec fi 180°,
then
0.562 ms fi 2∞
0.562
¥ 180∞
10
\
a=
\
a = 10.116∞
Now, rms value of load voltage is
ÏÔ 1
ELrms = Ì
ÔÓ p
1/ 2
180∞
ÚE
10
m
2
¸Ô
sin wt dwt ˝
Ô˛
2
Solution Manual 4
ÏÔ 1
= Ì
p
ÓÔ
1/ 2
180∞
Ú (330 sin wt )
10
ÏÔ (330)2
= Ì
◊
p
ÓÔ
2
¸Ô
d wt ˝
˛Ô
1/ 2
180∞
Ú
10
¸Ô
1 - cos 2 wt
d wt ˝
2
˛Ô
1/ 2
180
ÏÔ (330)2 ÈÊ
˘ ¸Ô
pˆ
= Ì
◊ ÍÁ p - ˜ - 1/ 2 (sin 2 wt )˙ ˝
18 ¯
˚10 Ô˛
ÔÓ 2p ÎË
= 232.95 Volts.
(b) With the pedestal voltage VP = 3 V., the initial voltage on capacitor is 3 V
\ Therefore, time t1. is given by
nVBB + VD = 3 + Es. ÈÎ1 - e-t1 / R3c ˘˚
\
\
So if
\
-3
12.7 – 3 = 15 È1 - e-t1 / 0.3 ¥ 10 ˘
Î
˚
t1 = 312.1 msec.
10 msec fi 180°,
a=
312.1 msec fi?
312 ¥ 10-6
¥ 180°
10 ¥ 10-3
\ a = 5.6°.
11.5 An a.c. full-wave voltage controller operating in integral cycle mode feeds a resistive load of
10W from a single phase a.c. voltage source related at 230 V, 50 Hz. The thyristor switch is ON
for 25 cycles followed by 75 cycles of extinction period. Determine:
(a) rms value of load voltage and load current
(b) input power factor and (i) average and rms value of thyristor current.
Sol. (a) Duty ratio,
d=
Ton
n.
=
Ton + Toff m + n.
In a integral cycle power controller, control is achieved by allowing a number of cycles of
supply voltage to be applied to the load and with a number of cycles of supply voltage
withdrawn from the load. This type of control is known as burst-firing.
d=
25
=¼
25 + 75
Rms value of load voltage, is given by
Eorms = E .
d
= 230 1/ 4 = 115 V.
and
Iorms =
(b)
pF =
E orm 115
=
= 11.5 A.
R
10
d = 1/ 4 = 0.5
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Power Electronics
(c) Average value of thyristor voltage
Eoavg =
=
=
1
.
2p f
Ú
p
0
. 2 ◊ E ◊ sin wt d wt .
2E
2 2 E.d .
2 E.d
p
- cos wt ]o =
=
[
p
2p / d
2p
2 ¥ 230 ¥ 0.25
= 25.88 V.
p
Hence, average value of thyristor current
ITav =
E oavg
R
=
Rms value of thyristor current (IRMS)
=
E o rms
2 ◊R
=
Io rms
2
=
11.5
2
= 8.14 A
25.88
= 2.6 A.
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