MD Singh Power Electronics Solution Manual to Chapter 11

CHAPTER

11

11.1 (i)

The waveforms of VT1 and VT2 in Fig 11.3 (b) reveals that the greatest forward or reverse p voltage would appear across either of the thyristor when a ≥ . 2 Magnitude of these voltages, Em =

2 Es 2 .230 = 325.27 V

= (ii)

RMS value of the output voltage from Eq. (11.3) is given by 1 sin 2 a ˘1/ 2 Eorms = Es ÈÍ p – a + ˙˚ Îp 2

(

(a)

)

a = 120∞ / , Eorms = 135.10V

for

max forward or reverse voltage = 135.10 ¥ 2 = 191 V.

\ (b)

for

a = 60∞ , Eorms = 188.616 V

\

maximum forward or reverse voltage

= 266.74 V.

(iii) From Eq. (11.8) 2 ¥ 230 È sin (5 + 1) p / 4 sin (5 – 1) p / 4 ˘ – ÍÎ p (5 + 1) (5 – 1) ˙˚

A5=

= – 17.29 From (11.9). 2 ¥ 230 È cos (6) ◊ p 4 – 1 cos ( 4) p 4 - 1 ˘ – ÍÎ ˙˚ 6 4 p = 34.48 From Eq. (11.10), peak amplitude of 5th harmons B5 =

output voltage,

E5 =

A52 + B52 =

= 38.57 Phase

fn = tan

–1

= tan

-1

B5 A5 34.48 - 17.29

= - 63.44∞

( - 17.29)2 + (34.48)

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Power Electronics

11.2 (a)

For controlling the load, the minimum value of firing angle a = load phase angle,

f = tan -1

WL R

= tan -1 4 3 = 53.13∞. Maximum possible value of a is 180∞ .

\

Firing angle control range is 53.13∞ £ a £ 180∞ .

(b)

At a = f = 53.13∞ , the maximum value of rms load current occurs. But, at this value of firing angle, the power circuit of ac voltage controller behaves as if load is directly connected to ac source. Therefore, maximum value of rms load current is given by 230

I0 = (c)

(d)

R 2 + (WL )2

=

230 32 + 42

Maximum power

= I 0 2 R = 462 ¥ 3 = 6348 W .

Power factor

=

46 A.

I 0 2 R 462 ¥ 3 = = 0.6. E5 I 0 230

when a = f, average thyristor current is maximum and conduction angle ° = p. from Fig. 11.5 (c),we can write ITavg =

\

1 2p

a +p

Úa

Em sin (wt - f ) d (wt ) z

Em 2 ¥ 230 = = 20.707 A . p . z. p ¥ 32 + 42 Similarly, maximum value of thyristor current is ITM

1 = ÈÍ Î 2p

= (e)

Maximum value of

a +p

Úa

{

Em 2 ¥ 230 = = 32.527 A. 2Z 2¥5

dio occurs when dt

a = f. From eq. (11.25), dio w. Em cos (cot - f ) = 0 . dt z Its value is maximum, when cos(w t – f) = 1.

\ÊË

}

Em sin (w t - a ) d wt ˘˙ z ˚

dio ˆ 2. 230. 2p . 50 = = 2.0437 ¥ 104 A / Sec. ¯ dt max 5

1

2

Solution Manual 57

(f)

For a = 0, Fig 11.6 shows that conduction angle r is 180° For a = 120° and 53.13°, Fig 11.6 gives a conduction angle of about 95°.

11.3 Given: R = 12 W, L = 24 mH, Em = Z =

2 ¥ 240 = 339.41 V ., f = 50 Hz. R 2 + (WL )2 = (12)2 + ( 2p ¥ 50 ¥ 24 ¥ 10-3 )

2

= 14.17 W

f = tan -1

( )= WL R

Ê 2p ¥ 50 ¥ 24 ¥ 10-1 ˆ tan -1 ÁË ˜¯ 12

= 32.13∞ = 0.56 rad . (a)

With zero firing delay, normal steady state theory applies. Load impedance,

Z = 14.17 32.13∞W

Load current

=

Em sin (wt - f ) z

=

339.41 sin ( 2p 50 t - 0.56) 14.17

= 23.95 sin ( 2 p 50 t - 0.56) A. Load power

=

240 ¥ 23.95 ¥ cos (32.13) 2

= 3441 W. Note that a train of firing pulses will be required as the thyristor will not turn-on until after 32.13°. (b)

Taking t = 0 at the instant of firing (a - 90∞) then

(

i = 23.95sin 2p .50 t +

)

(

)

12 ˆ p p t - 0.56 - 23.95 sin - 0.56 - ÊË 2 2 0.024 ¯

= 23.95 sin(2p.50.t + 1.01) – 20.29 e500t A. Current will cease when i = 0 at t = 7.33 ms/, that is, at an angle of 221.9° on the voltage wave. Mean load power =

1 0.01

0.00733

Ú

(

240 2 sin 2p 50 t

0

= 1370 w (c)

Similarly, for mean load power

a = 120°, = 177.2 W.

)

p 500t ÎÈ23.95sin ( 2p.50 t + 1.01) - 20.29e ˚˘ dt 2

58

Power Electronics

(

)

1 sin 2a ˘ 2p - a + 11.4 From Eq. (11.36), rms output voltage is given by E0 = Es ÈÍ Î 2p 2 ˚˙ (i)

1

2

a = 0° 1 E0 = Es . ÈÍ ( 2p - 0 + 0)˘˙ Î 2p ˚

1

2

E0 = E s If R is the resistance of heater element (load) and the maximum power or rating of heater, p = 1 kW. Therefore, R =

\

E 2 (230)2 = = 52.9W P 1 ¥ 103

Output power or power dissipated by the heater element at a = 0°, Po =

E0 2 Es 2 2302 = = = 1kW . k k 52.9

a = 180∞

(ii)

(

)

È 1 2p - p + sin 2 p ˘ Eo = Es Í Î 2p 2 ˙˚

1

2

1

= Es [0.5] 2 = 0.707 Es Po =

Eo 2 (0.707 Es )2 (0.707 ¥ 230)2 = = @ 500w R R 52.9

a = 70∞

(iii)

(

)

È 1 2p - 70∞ + sin 140∞ ˘ Eo = Es Í Î 2p ˚˙ 2

1

2

= 0.926 Es Po = 11.5 (a) Rms live current=

(0.926 Es )2 R

=

(0.926 ¥ 230)2

24000 = 33.39 A. 3 ¥ 415

Hence, required triac rms current rating = 33.39 A. (i)

For 16 kW, 16 (p - a + 12 sin 2 a ) = p 24

Solving the above equation, we get

a = 74.49°

\ (ii)

For 8 kW,

52.9

856.71w

Solution Manual 59

8 (p - a + 1 2 sin 2 a ) = p 24 Solving the above equation we get a = 105.29°. (b)

If thyristors were used, they would control for only one half-cycle of the sine wave, have their required rms current rating =

33.39 2

= 23.61 A. The voltage requirement would be identical to that for the triac. 11.6 Given e1 = 110 V,

e2 = 110 V,

Ep = 220 V,

E6 = 170 V,

R = 8 W,

a = 72°. (a)

Rms current of thyristor T1 & T2 can be obtained from eq. (11.38), as

\

I1 =

(

)

110 + 110 È 1 sin 144∞ ˘ p - 72∞ + Í Î ˚˙ 2 p 2 ¥8

1

2

= 16.19 A. (b)

Rms current of thyristors T3 and T4 can be obtained from (Eq. 11.33), as I2 =

(

)

110 È 1 sin 144∞ ˘ 72∞ Í Î ˚˙ 2 p 2 ¥8

1

2

= 5.382 A. (c)

The Rms current of second (top) secondary winding is Iw2 =

2 I1

= 22.90 A Rms current of first (lower) secondary winding, which is the total Rms current of thyristors T1, T2, T3 & T4 is 2 2 Iw1 = ÈÎ( 2 I1 ) + ( 2 . I 2 ) ˘˚

1

2

2 2 = ÈÎ( 2 ¥ 16.91) + ( 2 ¥ 5.382) ˘˚

= [(571.896) + (57.931)]

1

2

= 25.095 A. The volt-ampere rating of primary or secondary, EA = e1 . Iw1 + e2 . Iw2 = 110 ¥ 25.095 + 110 ¥ 22.90 = 52.79.45 load power,

po =

Eo 2 R

1

2

60

Power Electronics

=

\

The power-factor,

pf = =

(170)2 = 3612.5 8

po EA 3612.5 5279.45

= 0.684 (lagging)