CHAPTER 14

14.1 Back emf at 800 rpm,

Eb1 = E Ia . Ra = 230 (7 0.2)

= 228.6 V

Rated speed W1 = 800 260

p = 83.78 rad/s.

Now, ka.f1 83.78 = Eb1 = 228.6

\ Kaf1 = 2.73 (i)

(i) Back emf at 500 rpm is

21

EbEb

= 500800

.

\ Eb2 = 500 228.6

800 = 142.9 V

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 2

60p

= Ka . Kf1 115.19

Substituting from Eq. (i), gives

Eb3 = K 2.73 11.19 = 314.47 K. (iii)

(iii) Since T = Ka . f . Ia, We have

Ka . f1. Ia1 = Ka . f2 . Ia2

or Ia2 = 12

ff

. Ia1 or Ia2 = 1IaK

= 7K

\ E = Eb3 + Ia2 . Ra

or 230 = Eb3 + 0.2 Ia2 (v)

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

70 Power Electronics

230 = 314.47 K + 7 .k

0.2

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, W1 = 83.78 rad/s

\ Rated motor torque = T1= 1 11

228.6 783.78

Eb Iaw =

= 19.10 Nm.

Now, T = Ka . f . Ia. \ 19.10 = Ka . f 7 and (i) and 800 = Ka . f Ia2 (ii) Where IA2 is the current under regenerative breaking. From equations (i) and (ii),

Ia2 = 800 719.10

= 293.19 A

Back emf, Eb2 = E + Ia2 . Ra = 230 + 293.19 0.2

= 288.64 V

New speed, W2 = 2 11

288.64 83.78228.6

Eb WEb

=

= 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 = Emp

(1 + cosa)

Efmax = 2 2 2 210Emp p

= = 189 V.

Now, Ifmax = max 189

150f

f

ER

= = 1.26 A

(ii) Now, we have the relation

Ia = 80. 1.055 1.26T

Ia If=

= 60.18 A

Also, Eb = Ka . W . If = 1.055 2960 1.2660p

= 133.64 V.

Now, Ea = Eb + Ia . Ra

= 133.64 + 60.18 0.25

= 148.69 V

Solution Manual 71

Armature Voltage is given by

Ea = Emp

. (1+ cosa)

148.69 = 2 210p (1 + cos a) \ a = 55.

(iii) Output power, P0 = Eq. Ia = 148.69 60.18 = 8948.16 W. By neglecting losses, we can write

Pa = P0 = 8948.16

Rms input current Irms = Ia1

2p ap-

= 60.181

255pp-

= 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 = 8948.1610531.5

PaEI

= = 0.85

14.4 (a) (1) kaf = 0.45 V/rpm = 0.45 602p

= 4.3 V-s/rad.

Now, T = kaf . Ia = 4.3 35 = 150.4 N-m.

(2) Ermature voltage, Ea = 2 Emp

cos a.

= 2 2 480p cos 60 = 216.08 V.

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

\ Speed N = 210.750.45

Ebka f

= = 468.6 rpm

(3) I = 35 A.

\ Supply volt ampere, EI = 480 35 = 16800 VA

\ Ps = Ea. Ia = 216.08 35 = 7562.8 W.

\ Supply power factor PF = 7562.8 0.4516800

=

(b) (1) Back emf at the time of polarity reversely

Eb = 210.83 V

Now, Ea = Eb + Ia . Ra = 210.83 + (35 0.15)

72 Power Electronics

= 205.58 V

Also, Ea = 2 2 480p . cos a \ a = 118.40

(2) 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

Also, Ps = Ea Ia = 205.58 35 = 7196 W.

14.5 Em = 230 2 = 325.27 V

Now, Eb = Ea Ia Ra = 220 12 2.5 = 190 V

W = 1500 260

p = 157.07 rad/s.

Kaf = Eb 190157.07w

= = 1.21

Now, Ea = Eb + Ia Ra.

or 2 Emp

cos a = Ia Ra + Eb (i)

(a) At rated torque, Ia = 12 A.

Back emf at 1000 rpm =Eb1 = 1000 1901500

= 126.67 V.

Now, from Eq. (i),

2 325.27p

cosa = (12 2.5) + 126.67

\ a = 40.44.

(b) For the speed of 1500 rpm, Eb = 190 V

\ From (i), 2 325.27p

cosa = (12 2.5) 190

\ a = 140.56.

(c) From (i), For a = 150 & Ia = 12 A.

2 325.27p

cos 150 = 12 2.5 + Eb

\ Eb = 209.33 V

Since kaf = 1.21

Solution Manual 73

Now, W = 209.331.21

Ebka f

-=-

= 173 rad/s.

= 1652.03 rpm.

14.6 (a) Semiconductor controlled d.c. drive.

(1) N = 1200 rpm = 1200 260

p = 125.66 rad/sec

From Eq. (14.41),

T = ( ) ( ) ( )

( )

22 230 / 1 cos30 0.08 125.660.02

0.3 0.02 125.66p + -

+

= 84.96 N-M.

(2) From Eq. (14.40),

Ia = 1

284.960.02

= 65.18 A

(3) Motor terminal voltage is given by

Ea = ( )2 230 1 cos.30p + = 193.20 V.

Input power is given by

Ps = Ea . Ia = 193.20 65.18 = 12592.79 W.

Input volt-ampere = ( )1256230 65.18 = 13685.213 VA.

Supply power factor PF = 12592.78 0.9213685.213

=

(b) full-converter controlled d.c. series motor.

(i) From Eq. (14.41),

T = ( )

( )2 2 230 / . cos 30 0.08 125.660.02

0.3 0.02 125.66p -

+

= 72.58 N-m.

(2) From Eq. (14.40),

Ia = 1

272.580.02

= 60.24 A

(3) Now, Ea = 2 2 230 cos 30p = 179.33 V.

\ Ps = 179.33 60.24 = 10802.84 W

Input volt-ampere = 230 60.24 = 13855.2 VA

74 Power Electronics

PF = 10802.8413855.2

= 0.78.

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

Now, 21

EbEb

= 21

NN

\ 2750405960

Eb = = 316.41 V

Eb1 = E Ia Ra. \ 405 = 440 Ia Ra\ Ia Ra = 35 V

But, Ia = 100 1000440 = 227.27 A

\ 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 = 3 3p

Em . cosa.

335.25 = 3 3 23p

. Eac . cosa

where Eac is rms value of a.c. voltage

335.25 = 3 2 415 cosap

\ a = 53.26

14.8 (a)

Ef = 2 cosEm ap

Take a = 0

= 2 2 410p

cos 0 = 369.13 V

If = 369.13150

EfRf

= = 2.46 A.

Now, T = Ia . ka . If

= 30 1.2 2.46

= 88.56 N-m.

(b) For a = 45, Ea = 2 2 410p

cos 45

= 261 V.

Now, Eb = 261 30 0.3 = 252 V

Solution Manual 75

\ W = 252. 1.2 2.46

Ebka If

=

=

= 85.36 rad/s = 815.22 rpm

(c) Ps = Ea . Ia = 30 261 = 7830 W EA = 410 30 = 12300

Pf = 783012300

= 0.63

14.9 (a) Phase voltage Ep = 2103

= 121.24 V

\ Em = 2 121.24 = 171.46 V

Field controlled voltage Ef = 3 3 Emp

cos a

For maximum field current, a = 0.

\ Ef = 3 3 171.46p

= 283.5 gv.

If = 283.59250

EfRf

= = 1.134 A.

Ia = 746 25 746Volts 380

HP = = 49 A.

T = Ia . ka . If = 49 1.2 1.134

= 66.68 N-M

Eb = ka . If . W = 1.2 1.134 1800 260p

= 256.50 V.

Ea = Eb + Ia . Ra = 256.50 + 49 0.15

= 263.85 V

\ Eq = 263.85 = 3 3 171.46p

cos a

a = 21.50

(b) Ea = 3 3p

Em cos a

for a = 21.50, Ea = 263.86 V

Eb = Ea Ia . Ra.

= 263.86 10 65.18 0.15100

= 262.88 V.

76 Power Electronics

Now, N = Ea Ia RaKf-

\ f = 263.86 49 0.151.2 1800

= 0.118.

No-load speed is given by

NNL = 262.88

1.2 0.118Eb

ka f=

= 1856.5 rpm.

(c) Speed regulation = 1856.5 1800 1001800

-

= 3.13%.

14.10 Ea = 3

310 10. . 22022 10

Ton EdcT

-

-=

= 100 V.

Now, back-emf, 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 = 2203 3

EL = = 127 V.

Ea = 3 6p

Ep cos a.

= ( )3 6 127 cos 22.5p

= 274.45 V.

ia = Tkt

= 702.1

= 33.33 A.

Average speed is given by

N = . .a aEa i Rka f-

= 274.45 33.33 0.780.209-

= 1188 rpm.

a = 180 22.5 = 157.5.

(b) Ea = ( )3 6 127 cos 157.5p

Solution Manual 77

= 274.45 V.

Eb = Ea + Ia . Ra

= 274.45 + 33.33 0.78

= 300.45V.

Speed N = 300.450.209

Ebka f

= = 1437 rpm.

(c) Ea = ( )3 6 127 cos 67.5p

= 113.68 V.

Speed, N = Ea ia Raka f-

720 = 113.68 0.780.209

ia-

\ 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 . 0.780.209

ia+

\ ia = 47.17 A

T = 2.1 471

= 99 N-m.