MD Singh Power Electronics Solution Manual to Chapter 14

CHAPTER

14

14.1 Back emf at 800 rpm, Eb1 = E – Ia . Ra = 230 – (7 ¥ 0.2) = 228.6 V Rated speed Now,

W1 =

800 ¥ 2p = 83.78 rad/s. 60

ka.f1 ¥ 83.78 = Eb1 = 228.6 Kaf1 = 2.73

\

(i)

(i) Back emf at 500 rpm is Eb2 500 . = 800 Eb1 Eb2 =

\

500 ¥ 228.6 = 142.9 V 800

Motor terminal voltage = Eb2 + Ia Ra = 142.9 + (7 ¥ 0.2) = 144.3 V (ii)

(ii)

Let us take the new flux f2 = Kf1 Since

Eb = Ka . f . w : Eb3 = Ka . f2 ¥

1100 ¥ 2p 60

= Ka . Kf1 ¥ 115.19 Substituting from Eq. (i), gives Eb3 = K · ¥ 2.73 ¥ 11.19 = 314.47 K.

(iii)

T = Ka . f . Ia, We have

(iii) Since Ka . f1. Ia1 = Ka . f2 . Ia2 or

Ia2 = =

f1 Ia . Ia1 or Ia2 = 1 f2 K 7 K

E = Eb3 + Ia2 . Ra

\ or

230 = Eb3 + 0.2 Ia2

Substituting (iii), and (iv) in five gives

(v)

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Power Electronics

230 = 314.47 K +

7 . ¥ 0.2 k

314.47 k2 – 230 k + 1.4 = 0 feasible value of k is 0.71. Hence, the flux must be reduced to 0.71 of its rated value. 14.2 From example 14.1,

Eb1 = 228. 6 V,

Rated motor torque = T1=

\

W1 = 83.78 rad/s

Eb1 ¥ Ia1 228.6 ¥ 7 = w1 83.78

= 19.10 Nm. Now,

T = Ka . f . Ia.

19.10 = Ka . f ¥ 7 and and 800 = Ka . f Ia2 Where IA2 is the current under regenerative breaking. From equations (i) and (ii), 800 ¥ 7 Ia2 = = 293.19 A 19.10

\

Back emf,

Eb2 = E + Ia2 . Ra = 230 + 293.19 ¥ 0.2 = 288.64 V

New speed,

W2 =

Eb2 288.64 ¥ W1 = ¥ 83.78 Eb1 228.6

= 105.78 rad/s = 1010 rpm. 14.3 (i) For single phase semiconverter controlled d.c. drive, we can write for field circuit ds, Ef = Efmax = Now,

Ifmax =

Em (1 + cosa) p 2 Em 2 2 ¥ 210 = = 189 V. p p E f max Rf

=

189 = 1.26 A 150

(ii) Now, we have the relation Ia = Also,

T 80 = 60.18 A = Ia. If 1.055 ¥ 1.26

2p ˆ Ê ¥ 1.26 Eb = Ka . W . If = 1.055 ¥ Ë960 ¥ 60 ¯ = 133.64 V.

Now,

Ea = Eb + Ia . Ra = 133.64 + 60.18 ¥ 0.25 = 148.69 V

(i) (ii)

Solution Manual 71

Armature Voltage is given by Ea = 148.69 =

Em . (1+ cosa) p 2 ¥ 210 (1 + cos a) \ a = 55°. p

(iii) Output power, P0 = Eq. By neglecting losses, we can write

Ia = 148.69 ¥ 60.18 = 8948.16 W.

Pa = P0 = 8948.16 Rms input current

Irms

p -aˆ = Ia Ê Ë p ¯ = 60.18 Ê Ë

1

2

p - 55 ˆ p ¯

1

2

= 50.15 Now, input voltage-ampere rating is E.I. = E. Irms = 210 ¥ 50.15 = 1053.5 Input power-factor assuming negligible harmonics is given by PF = 14.4 (a) (1) kaf = 0.45 V/rpm Now, (2)

Ermature voltage,

=

Pa 8948.16 = = 0.85 EI 10531.5 0.45 ¥ 60 = 4.3 V-s/rad. 2p

T = kaf . Ia = 4.3 ¥ 35 = 150.4 N-m. Ea = =

2 Em cos a. p 2 2 ¥ 480 cos 60 = 216.08 V. p

Eb = Ea – Ia Ra = 216.08 = (35 ¥ 0.15) = 210.83 V.

\

Speed

(3)

\

N =

Eb 210.75 = 468.6 rpm = ka f 0.45

I = 35 A. Supply volt ampere,

EI = 480 ¥ 35 = 16800 VA Ps = Ea. Ia = 216.08 ¥ 35 = 7562.8 W.

\

PF =

7562.8 = 0.45 16800

\

Supply power factor

(b)

(1) Back emf at the time of polarity reversely Eb = 210.83 V Now,

Ea = Eb + Ia . Ra = – 210.83 + (35 ¥ 0.15)

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Power Electronics

= – 205.58 V Also, (2)

Ea =

2 2 ¥ 480 . cos a \ a = 118.40 p

Power from d.c. machine is Pg = 210.83. ¥ 35 = 7379.05

Power loss in the armature resistance is PR = Ia2 Ra = (35)2 ¥ 0.15 = 183.75 W. Power fed back to ac supply is Ps = 7379.05 – 183.75 = 7195.3 W Ps = Ea Ia = 205.58 ¥ 35 = 7196 W.

Also,

Em = 230 ¥ 2 = 325.27 V

14.5 Now,

Eb = Ea – Ia Ra = 220 – 12 ¥ 2.5 = 190 V 1500 ¥ 2p = 157.07 rad/s. 60 Eb 190 = = 1.21 Kaf = w 157.07 Ea = Eb + Ia Ra. W =

Now, or (a)

2 Em cos a = Ia Ra + Eb p At rated torque, Ia = 12 A. Back emf at 1000 rpm =Eb1 =

1000 ¥ 190 1500

= 126.67 V. Now, from Eq. (i), 2 ¥ 325.27 cosa = (12 ¥ 2.5) + 126.67 p

a = 40.44°.

\ (b)

For the speed of 1500 rpm, Eb = – 190 V

\

From (i),

2 ¥ 325.27 cosa = (12 ¥ 2.5) – 190 p

a = 140.56°.

\ (c)

From (i), For

a = 150° & Ia = 12 A.

2 ¥ 325.27 cos 150 = 12 ¥ 2.5 + Eb p

\ Since

Eb = – 209.33 V kaf = –1.21

(i)

Solution Manual 73

Now,

W =

Eb - 209.33 = 173 rad/s. = ka f - 1.21

= 1652.03 rpm. 14.6 (a) Semiconductor controlled d.c. drive. (1)

N = 1200 rpm =

1200 ¥ 2p = 125.66 rad/sec 60

From Eq. (14.41),

È ( 2 ¥ 230 / p ) (1 + cos30) - (0.08 ¥ 125.66) ˘ T = 0.02 Í ˙ 0.3 + (0.02 ¥ 125.66) Î ˚ = 84.96 N-M. (2)

From Eq. (14.40), 1

Ê84.96 ˆ 2 = 65.18 A Ia = Ë 0.02 ¯ (3)

Motor terminal voltage is given by Ea =

2 ¥ 230 (1 + cos.30) = 193.20 V. p

Input power is given by Ps = Ea . Ia = 193.20 ¥ 65.18 = 12592.79 W. Input volt-ampere = 230 ¥ 65.18 ¥ ( 5 6 )

1

2

= 13685.213 VA. Supply power factor (b) (i)

PF =

12592.78 = 0.92 13685.213

full-converter controlled d.c. series motor.

From Eq. (14.41),

È ( 2 2 ¥ 230 / p ) . cos 30 - 0.08 ¥ 125.66 ˘ T = 0.02 Í ˙ 0.3 + (0.02 ¥ 125.66) Î ˚ = 72.58 N-m. (2)

From Eq. (14.40), 1

Ê72.58 ˆ 2 = 60.24 A Ia = Ë 0.02 ¯ (3)

Now,

Ea =

2 2 ¥ 230 cos 30 = 179.33 V. p

Ps = 179.33 ¥ 60.24 = 10802.84 W

\ Input volt-ampere

= 230 ¥ 60.24 = 13855.2 VA

2

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Power Electronics

PF =

10802.84 = 0.78. 13855.2

14.7 Given: Eb1 = 405 V, N1 = 960 rpm, N2 = 750 rpm. Eb2 Eb1

Now,

=

N2 750 = 316.41 V \ Eb2 = 405 ¥ 960 N1

Eb1 = E – Ia Ra. \ 405 = 440 – Ia Ra\ Ia Ra = 35 V But,

Ia =

100 ¥ 1000 = 227.27 A 440

Ra = 0.154 W.

\

Now, terminal voltage of dc motor at 960 rpm and 75% rated torque, = Eb2 + Ia . Ra = 316.41 + (0.5 ¥ 227.27 ¥ 0.154) = 335.25 V Neglecting voltage drop in converter circuit, Ea = 335.25 =

3 3 Em . cosa. p 3 3 2 . Eac . cosa ¥ p 3

where Eac is rms value of a.c. voltage 335.25 =

3 2 ¥ 415 cos a p

a = 53.26

\ 14.8 (a)

Ef = = If = Now,

2 Em cos a Take a = 0 p 2 ¥ 2 ¥ 410 cos 0 = 369.13 V p Ef 369.13 = 2.46 A. = Rf 150

T = Ia . ka . If = 30 ¥ 1.2 ¥ 2.46 = 88.56 N-m.

(b)

For

a = 45°,

Ea =

2 ¥ 2 ¥ 410 cos 45 p

= 261 V. Now,

Eb = 261 – 30 ¥ 0.3 = 252 V

Solution Manual 75

W =

\

Eb 252 = = ka.If 1.2 ¥ 2.46

= 85.36 rad/s = 815.22 rpm (c)

Ps = Ea . Ia = 30 ¥ 261 = 7830 W EA = 410 ¥ 30 = 12300 Pf =

14.9 (a) Phase voltage

Ep =

7830 = 0.63 12300 210 3

2 ¥ 121.24 = 171.46 V

Em =

\ Field controlled voltage For maximum field current,

\

Ef =

= 121.24 V

3 3 Em cos a p

a = 0. Ef =

3 3 ¥ 171.46 = 283.5 gv. p

If =

Ef 283.59 = 1.134 A. = Rf 250

Ia =

HP ¥ 746 25 ¥ 746 = 49 A. = Volts 380

T = Ia . ka . If = 49 ¥ 1.2 ¥ 1.134 = 66.68 N-M Eb = ka . If . W = 1.2 ¥ 1.134 ¥ 1800 ¥ = 256.50 V. Ea = Eb + Ia . Ra = 256.50 + 49 ¥ 0.15 = 263.85 V 3 3 ¥ 171.46 cos a p

Eq = 263.85 =

\

a = 21.50° 3 3 Em cos a p

(b)

Ea =

for

a = 21.50°, Ea = 263.86 V Eb = Ea – Ia . Ra. = 263.86 –

10 ¥ 65.18 ¥ 0.15 100

= 262.88 V.

2p 60

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Power Electronics

Now,

N =

Ea - Ia Ra Kf

\

f =

263.86 ¥ 49 ¥ 0.15 = 0.118. 1.2 ¥ 1800

No-load speed is given by NNL = (c)

Speed regulation

14.10 Now, back-emf,

Eb 262.88 = 1856.5 rpm. = ka f 1.2 ¥ 0.118

1856.5 - 1800 ¥ 100 1800 = 3.13%. Ton 10 ¥ 10-3 ¥ 220 = 100 V. . Edc. = Ea = T 22 ¥ 10-3 =

Eb = K. N. = 0.495 ¥ 146.60 = 72.57 V.

Now,

Ea = Eb – Ia . Ra. 100 = 72.57 – 3Ia Ia = 9.2 A.

\ 14.11 (a)

Ep =

Ea = =

EL 3

=

220 3

= 127 V.

3 6 Ep cos a. p 3 6 ¥ 127 cos (22.5) p

= 274.45 V. ia =

70 T = = 33.33 A. kt 2.1

Average speed is given by N = =

Ea - ia . Ra . ka f 274.45 - 33.33 ¥ 0.78 0.209

= 1188 rpm.

a = 180° – 22.5° = 157.5°. (b)

Ea =

3 6 ◊ ¥ 127 ¥ cos (157.5∞) p

Solution Manual 77

= – 274.45 V. Eb = – Ea + Ia . Ra = 274.45 + 33.33 ¥ 0.78 = 300.45V. Speed

(c)

N =

Ea =

Eb 300.45 = 1437 rpm. = ka f 0.209 3 6 ¥ 127 ¥ cos (67.5) p

= 113.68 V. Speed,

N = 720 =

Ea - ia Ra ka f 113.68 - ia ¥ 0.78 0.209

\

ia = – 47.18 A.

\

T = – 2.1 ¥ 47.18 = – 99 Nm (d)

\ \ \

a = 180 – 67.5 = 112.5°. Ea = – 113.68 V 720 =

113.68 + ia. 0.78 0.209

ia = 47.17 A T = 2.1 ¥ 471 = 99 N-m.