MD Singh Power Electronics Solution Manual to Chapter 15

CHAPTER

15

15.1 (a)

Zin = (R1 + jx1) + Zf

where

Zf =

( R2 + j x2 ) jxm R2 / s + j ( x2 + xm

=

(0.95 / 0.04 + j 0.95) ( j 81.4) (0.95 / 0.04) + j (1.63 + 81.4)

=

(1933.25 j -77.33) = 21.27 + j 7.02 ( 23.75 + j 83.03)

Zin = (0.15 + j1.63) + 21.27 + j 7.02

\

= 21.42 + j8.65 = 23.10 – 22°. I1 =

Stator current Input power

VIph Zin

=

2300 / 3 = 57.49 A. 23.10

Pi = 3 Viph . I1 . cosQ = 3¥

2300 3

¥ 57.49

= 229023.88 W Pi = 229 kW. (b)

haft power

P = Pgross – (Pfw + Pcore + Pstray) 2

2

Air gap power, pg = 3 I 1 Rf = 3 ¥ 57.49 ¥ 21.27 = 210898.43 W. Pgross = Pg(1 – S) = 210898.43(1 – 0.04) = 202462.49 W Pstray =

1 ¥ 3521.7 = 35.22 W 100

from (i), P = 202462.48 – (8300 + 35.22)

\

P0 = 194.12 kW Now, Also,

Wr = Ws (1 – S) S =

Ws - W Ws

(i)

Solution Manual 79

0.04 =

\ Also,

wr = 120 (1 – 0.04) = 115.2 rad/s.

Torque, (c) Efficiency =

(a)

Ws - 1100 \ Ws = 120 rad/s. Ws

T = 210898.43 229.02 ¥ 103

Stator current

I1 =

Pg Ws

=

210898.43 = 1757 NM. 120

= 92%. V1ph Zin

Zin = (R1 + jx1) + Zf

where,

Zf =

where At starting

( R2 / s ) ( - j s2 ) j xm R2 / s + j ( x2 + xm )

s =1 Zf =

\

=

(0.95 + j 1.63) ( j 81.4) 77.33 j - 132.682 = 0.95 + j (1.63 + 81.4) 0.95 + j 83.03 153.57 – 149.77∞ = 1.85 – 60.43∞ 83.04 – 89.34∞

Zin = (0.15 + j1.63) + 0.913 + j1.61 = 3.41 –71.84° I1 = (b)

Input power =

2300 / 3 = 389.6 A. 3.41

3 V1 I1 cos q1 =

3 ¥ 2300 ¥ 389.6 cos 71.84

= 484 kW. (c)

Starting developed torque (Td) At starting, Ws = W = 1100 rpm.

`

using equations (15.33), 2

Tstart =

3 ¥ ( 2300 / 3 ) 0.95 1100 ¥ 2 p ÈÎ(0.15 + 0.95)2 + (1.63 + 1.63)2 ˘˚ 60

= 3684.79 Nm 15.3 From Eq.(15.29), Smax. T =

R2 R12 + ( x1 + x2 )2

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0.95

= (0.15

)2

+ (1.63 + 1.63)2

= 0.291

From Eq. (15.30), Tmax =

3 ¥ ( 2300 / 3 )

2

2 ¥ 120 ÈÎ0.15 + (0.15)2 + (1.63 + 1.63)2 ˘˚

= 6457.28 N-m. 15.4 In a constant power zone, at a constant rotor frequency, T a F2 2

or,

F12

1

T1 T , or F2 = F1 1 T2 T2 2000 = 57.74 Hz. 1500

F2 = 50

\

12 = 0.2 60 From equation (15.52), the ratio of breakdown torques for k = 0.2 and k = 1.

15.5 From equation (15.34), (1)

=

( 1 F )2 .

k = f/frated =

Tmax . ( k = 0.2) 0.024 ± (0.024)2 + (0.24)2 = Tmax ( k = 1) 0.024 0.024 2 ( ± + 0.24)2 0.2 0.2

( )

(2)

For motoring,

Tmax ( k = 0.2) = 0.68 Tmax ( k = 1)

For breaking,

Tmax ( k = 0.2) = 1.46. Tmax ( k = 1)

Substituting S = 1 in Eq. (15.51), gives Starting torque. Thus,

˘ V 2 rated R2 / k 3 È Í ˙ 2 ws Í R1 + R2 2 ˙ + ( x1 + x ) ÍÎ ˙˚ k From equation (i), the ratio of starting torques for k = 0.2 and k = 1 is T =

(

)

(i)

Ts ( k = 0.2) 0.024 / 0.2 = Ts ( k = 1) (0.048 / 0.2)2 + (0.24)2 0.024 (0.048)2 + (0.24)2 = 2.6. The starting motor current is given by Ims =

Vrated

(

R1 + R2 k

(ii)

2

) +(

x1 + x2 )

Solution Manual 81

The ratio of starting current for k = 0.2 and 1 is I ms ( k = 0.2) = I ms ( k = 1)

(0.048)2 + (0.24)2

(

0.048 2 ( ) + 0.24)2 0.2

= 0.72. The preceding ratios of starting torques and starting rotor currents show that the constant (v/F) control provides a high starting torque with a reduced motor current. k = 30/60 = 0.5 From Eq. (15.49), for rated torque and f = 60, Trated =

=

3 È V 2 rated (0.024 / 0.04) ˘ Í ˙ ws Í (0.024 + 0.024 ) 2 + (0.24)2 ˙ Î ˚ 0.04 3V 2 rated (1.34) ws

(a)

and for 30 Hz from equation (15.51), 0.024 ˆ È ˘ V 2 rated ÁÊ ˜ Í ˙ Ë 0.5 S ¯ 3 Trated = Í ˙ 2 ws Í Ê 0.024 0.024 ˆ 2 ( ) ˙ ÍÎ ÁË 0.5 + 0.5 S ˜¯ + 0.24 ˙˚ Equating equations (a) and (b), we get 0.024 0.5 S 2 Ê 0.024 + 0.024 ˆ (0.24)2 ÁË 0.5 ˜ 0.5 S ¯

or or

= 1.34

26.04 s2 – 0.98 S + 1 = 0 S =

14.62 ±

(14 . 62)2 - 4 ¥ 26.04 2 ¥ 26.04

= 0.089 or 0.43.

\

The slip on the stable part of the speed torque curve will be 0.089. Now synchronous speed for 30 Hz = 900 rpm. Motor speed = 900 (1 – 0.089) = 820 rpm.

15.7 From the rated conditions of operations, Ns =

120 f 120 ¥ 50 = 1000 rpm. = p 6

Ws =

1000 ¥ 2p = 104.7 rad/s. 1000

(b)

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Power Electronics

Slip, Rotor impedance

S =

1000 - 960 = 0.04 1000

Zr =

R2 + jx2 = 5 + j1.5 s

= 5.22 – 16.7° W Stator impedance

Zs = 0.4 + j1.5 = 1.55 – 75° W

Machine impedance

ZQ = Zs +

Z2 . Zm Z2 + Zm

I1 =

400 / 3 = 38.5 A. 6

I2 =

Zm 30 ¥ 38.5 Is = Zm + Z2 31.89

= 36.22 A E = I2|Z2| = 36.22 ¥ 5.22 = 189 V. Rated torque

=

3 I 2 2 . R2 3 ( 0.2 . 36.22)2 = Ws S 104 . 7 0.04

= 188 Nm. (a)

25 = 0.5 50 Substitute values in equation (15.41), we get At 25 HZ,

k =

(189)2 ¥ 0.2 / 0.55 188 3 = . 2 104 . 7 (0.2)2 / (0.5 S )2 + 1.52

S = 0.0374.

\ From Eq. (15.39)

wr = kWs( 1 – S)

or

N = kNs(1 – S) = 0.5 ¥ 1000(1 – 0.0374) = 481.3 rpm.

At 25 Hz1

E = 0.5 ¥ 189 = 94.5 V. R2 0.2 + j k. x2 = + j 0.75 S 0.0374

Z2 =

= 5.4 – 8° W. Taking E as a reference vector, I2 =

E 94.5 = 17.5 – – 8°. = Z 2 5.4 – - 8∞

Im =

E 94.5 = = 6.30– - 90∞ A J k . xm j 0.5 ¥ 30

I1 = I2 + Im = 17.5 – – 8° + 6.30 – – 90°

Solution Manual 83

I1 = 19.85 A.

\ (b)

Slip speed in rpm at the rated torque and frequency Nss = S . Ns = 0.04 ¥ 1000 = 40 rpm. Since the speed torque curve is a straight line, slip speed at half the rated torque, 25 ¥ 1000 Nss = 0.5 ¥ 40 = 20 rpm. At 25 Hz, Ns = 50 = 500 rpm. Since the slip speed remains constant for a given torque, Motor speed N = Ns – Nss = 500 – 20 = 480 rpm. For a constant flux the E/F ratio must be constant. Hence, at 25 Hz, E = 0.5 ¥ 189 = 94.5 V N s 2 20 = 0.04 = N s 500

S =

R2 jk .x2 = 5 + j0.75 S

Z2 =

= 5.06 – 8.5° W Taking E as a reference vector, I2 =

E 94.5 = = 18.7 – - 8.5∞ A . Z 2 5.06 – 8.5∞ E 94.5 = = 6.30– - 90∞ A . jkxm j 15

Im =

I1 = I 2 + I m = 18.7 – - 8.5∞ + 6.30 – - 90∞ Hence (c)

I1 = 20.6 A.

At the rated breaking torque, the slip speed will be the negative of the slip speed at rated motoring torque. Hence, slip speed = Ns3 = – 40 rpm. Synchronous speed = N + Ns3 = 800 – 40 = 760 rpm. Frequency = (760/1000) ¥ 50 = 38 Hz. k = 38/50 = 0.76. At 38 Hz,

E =

38 ¥ 189 = 143.64 V. 50

S = – 40/760 = – 0.0526 Z2 =

R2 jk x2 = - 38.8 + j 1.14 s

= 3.97 – 163.3° 1 W Taking E as a reference vector,

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Power Electronics

I2 =

E 143.64 = = 36.2 – - 163.3∞ A Z 2 3.97 – - 163.3

I m remains the same as the foregoing I1 = 36.2 – – 163.3° + 6.30 – – 90° = – 34.62 – j16.88 = 38.52 – – 154°.

\

V

= E + Z5 I1 = 143.64 + (0.4 + j0.76 ¥ 1.5) ¥ (38.52 – – 154°)

V

= 156 – – 17.3 V

Since phase angle between V and I is more than 90°, hence power flows from motor to the source. 15.8 From example 15.7, for 50 Hz operations Ns = 1000 rpm, Ws = 104.7 rad/s Rated torque = 188 N-m. Slip speed at rated torque = 40 rpm. E at rated conditions = 189 V. Im =

(a)

E = 189/30 = 6.3 A Xm

Substituting values in Equation (15.66), We get,

S = 0.067.

\

(b)

È (0.2 / s )2 + (1.5)2 ˘ 2 (6.30)2 = Í ˙ ¥ (60) 2 2 Î (0.2 / s ) + (31.5) ˚

Slip speed in rpm = S.N rpm = 0.067 ¥ 1000 = 67 rpm. (i) Since the flux is constant for a given torque, the slip speed will also be constant for all frequencies. Thus, the slip speed can be evaluated from the rated frequency operation. Now,

T =

139 =

3 104.7

È (189)2 ¥ (0.2 / s ) ˘ Í 2 2˙ Î (0.2 / s ) + (1.5) ˚

2 Ê0.2 ˆ + (1.5)2 = Ê1.472 ˆ Ë s ¯ Ë s ¯

or,

\

2 . R2 / s ˘ 3 È Erated =Í ˙ Ws Î ( R2 / s )2 ¥ x2 ˚

S = 0.0284 Slip-speed, Nss = 0.0284 ¥ 1000 = 28.4 rpm. Now for operation at 500 rpm Synchronous speed N = N + Nss = 500 + 28.4 = 528.4 rpm. Frequency =

528.4 ¥ 50 = 26.42 Hz. 1000

K = 26.42/50 = 0.528.

Solution Manual 85

S =

N ss 28.4 = 0.0568 = Ns 500

K.S.= 0.528 ¥ 0.0568 = 0.03 Substituting the known values in equation (15.68), We get 1/ 2

È (0.2 / 0.04)2 + (30 + 1.5)2 ˘ Im = 6.3 Í ˙ 2 2 Î (0.2 / 0.04) + (1.5) ˚ = 38.5 A. (c)

Since, at a constant flux, speed-torque curves for different frequencies are straight lines, slipspeed 139 Nss = ¥ 40 = 29.56 rpm. 188 Hence, Synchronous speed Ns = N + Nss = 500 + 29.56 = 529.56 rpm.

Ê529.56 ˆ ¥ 50 = 26.48 Hz Frequency = Ë 1000 ¯ S =

N ss 29.56 = 0.056 = N s 529.56

K =

26.48 = 0.53, S.K = 0.03. 50

Substituting the known values in equation (15.68), we get

È (0.2 / 0.03)2 + (30 + 1.5)2 ˘ Im = 6.30 = Í ˙ = 297 A 2 2 Î (0.2 / 0.03) + (1.5) ˚ 15.9 (a) The fundamental Rms live voltage of a six-step inverter is given by E1 =

6 . Edc p

(i)

where, Edc = 3/p . Em cosa

(ii)

Em is the peak of ac source live voltage. E1 =

Here,

\

3 6

Em . cosa p2 E p2 cosa = 1 Em 3 6

\

E1 = 460 V, Em = 460 cos a =

p 2 ¥ 460 460 2 ¥ 3 6

(iii) 2 V.

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Power Electronics

a = 18.25°.

\

(b) (i) For a given torque the motor operates at a fixed slip speed for all frequencies when the flux is maintained constant. \ Sleep speed in rpm at the rated torque, Nss = Ns – N =

120 ¥ 60 – 1180 6

= 20 rpm Hence, synchronous speed at 600 rpm. Ns = N + Nss = 600 + 20 = 620 rpm. Ns = 120 f/p =

\

120 ¥ 60 = 1200 rpm. 6

Ê 620 ˆ ¥ 60 = 31 Hz. Inverter frequency = Ë 1200 ¯ (ii) Back emf at the rated operation Erated =

I2 = ÎÈ( R2 / s )2 + x 2 ˚˘

where

I2 =

Also,

S = I2 =

\

1

2

E1 / 3

( R1 + R2 / s )2 + ( x1 + x2 )2 1200 - 1180 = 0.017 1200 460 / 3

(

0.19 +

0.07 0.017

2

) +(

0.75 + 0.67 )2

= 58.5 A. Erated

ÈÊ 0.07 ˆ2 ˘ + (0.67 )2 ˙ = 58.5 ÍË ¯ Î 0.017 ˚

1

2

= 244 V Now, torque at constant flux is given by T =

2 ◊ R2 / ( K .S ) ˘ 3 È Erated Í 2 ( )2 ˙ Ws . Î R2 / KS + x2 2 ˚

Ws and x2 in this equation are for rated frequency. where

Ws =

1200 ¥ 2p = 125.66 rad/s 60

T = 3/Ws . I22 Rs/S and,

T =

3 ( Ê 0.07 ˆ 58.50)2 Ë 125.66 0.017 ¯

(iv)

Solution Manual 87

= 336.42 Nm. Equation (iv) becomes

( )

È ( 244)2 ¥ 0.07 ˘ 336.42 3 Í K .S ˙ = Í ˙ 2 2 125.66 Í 0.07 + (0.67 )2 ˙ ÍÎ KS ˙˚

( )

Now,

\

k =

wr 500 0.42 = = ws (1 - S ) 1200 (1 - S ) (1 - S )

È ( 244)2 ¥ 0.07 (1 - S ) ˘ ˙ 336.42 3 Í 0.42 s = Í ˙ 2 125.66 Í È 0.07 2 2˙ ˘ ( ) ( ) Í Í 0.42 s ¥ 1 - S ˙˚ + 0.67 ˙˚ ÎÎ

which gives

S = 0.012

and

K = 0.425

Thus, frequency = 0.425 ¥ 60 = 25.5 Hz Substituting the known values in Eq. (15.38), yields, I2 =

244 2 0.07 Ê ˆ + (0.67 )2 ÁË 0.425 ¥ 0.012 ˜¯

= 17.76 A Machine fundamental phase voltage,

ÈÊ ˘ R ˆ2 V1 = I 2 ÍË R1 + 2 ¯ + K 2 ( x1 + x2 )2 ˙ Î ˚ s

1

2

ÈÊ ˘ 0.07 ˆ2 ( + 0.425)2 (0.75 + 0.67 )2 ˙ = 17.76 ÍË0.19 + ¯ Î ˚ 0.012 V1 = 107.51 V Now, taking V1 as reference vector, I2 = I2 - tan -1

k ( x1 + x2 ) ( R1 + R2 / s )

= 17.76 - tan -1

0.425 (1.42) (6.02)

= 17.76 – – 88.10° A. Im = =

E1 – - 90∞ K .X m 107.51 – - 90∞ = 12.65 – - 90∞ 0.425 ¥ 20

1

2

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Power Electronics

Now,

I1 = I 2 + Im = 17.76 – - 88.10∞ + 12.65 – - 90∞ I1 = 30.40 A.

Rms harmonic current is given by • È V1 1˘ Ih = Â Í ˙ K ( x1 + x2 ) Îh = 3, 5, 11, 13 h 4 ˚

1

2

Neglecting harmonics higher than 13, gives 0.046 V1 0.046 ¥ 107.51 = 8.195 A. = K ( x1 + x2 ) 0.425 ¥ (0.75 + 0.67 )

Ih =

\

Rms input current = ( I1 + I h )

1

2

= ÈÎ(30.40)2 + (8.195)2 ˘˚

1

2

= 31.49. Ex 15.10. At the rated operations, Ns = Rated slip Rotor Impedance, Machine Impedance

= Z2 =

120 f 120 ¥ 60 = = 1200 rpm = 125.66 rad/s P 6 1200 - 1176 = 0.02. 1200 0.145 + j 0.5 = 7.27 – 3.95°W 0.02

= Z1 +

Z2 Zm Z2 + Zm

= 0.19 + j 0.75 +

(7.25 + j 0.5) ◊ ( j 15.3) (7.25 + j 15.8)

= 5.81 + j3.81 = 0.95 – 33.26°W I1 =

460 / 3 = 38.21 – – 33.26° A 6.95 – 33.26

I2 =

Zm 15.3 – 90 I1 = (38.21 – – 33.26°) Zm + Z 2 17.38 – 65.35

= 33.64 – – 8.61 A. Im =

Z2 (7.27 – 3.95) (38.21 – - 33.26∞) I1 = Z 2 + Zm (17.38 – 65.35)

= 15.98 – – 95.61° Torque

=

3 2 I 2 ( R2 / s ) Ws

Solution Manual 89

=

3 0.145 ˆ ◊ (33.64)2 ÊË 125.66 0.02 ¯

T = 195.87 Nm. (a)

For the three phase current source six-step inverter, fundamental Rms current is given by I1 = =

6 ◊ Id . p

\ Id =

p . I1 6

p ¥ 38.21 = 49 A. 6

Also, rms stator current is given by Irms = ( (b)

2

3

) Id =

2

3 . ¥ 49

= 40 A .

Slip speed at rated torque and frequency, Nss = s . Ns = 0.02 ¥ 1200 = 24 rpm.

\

(c)

At motor speed of 600 rpm, synchronous speed Ns = 600 + 24 = 624 rpm. 624 ¥ 60 = 31.2 Hz. Inverter frequency = 1200 When the motor is controlled at 9 constant flux, for a given torque, the stator current remains constant at all speeds. Since the stator current is constant, the dc link current also remains constant at 49 A. Slip-speed is constant at all frequencies as the flux is constant for a given torque. Slip speed for 30 Hz operation at half the rated torque can be determined from 60 Hz operation. For 60 Hz operation, Erated = Im . Xm = 15.98 ¥ 15.3 = 244.49 V Now,

T =

2 R2 / s ˘ 3 È Erated Í ˙ 2 Ws Î ( R2 / s ) + x2 2 ˚

195.87 3 È (244.49)2 ¥ (0.145/ s ) ˘ = Í ˙ 2 125.66 Î (0.145/ s ) + (0.5)2 ˚ After solving, it gives

\

S = 0.01 Slip speed, Nss = S.Ns = 0.01 ¥ 1200 = 12 rpm Now, consider the operation at 30 Hz, K =

\

Synchronous speed, Hence,

30 = 0.5 60

Ns = 0.5 ¥ 1200 = 600 rpm.

motor speed = 600 – 12 = 588 rpm.

\

S =

\

I2 =

N ss 12 = 0.02 = N s 600 K . Erated

(( R2 / S )2 + ( K . x2 )2 )

90

Power Electronics

\

0.5 ¥ 244.46

=

(0.145 / 0.02)2 + (0.5 ¥ 0.5)2

= 16.81 A.

From Eq. (15.62), we have I22 =

(16.81)2 =

or

I 22 - I m 2 2x 1+ 2 Xm I12 - (15.98)2 2 ¥ 0.5 1+ 15.3

I1 = 23.59 A.

or

Also, d.c. link current Id can be given by the formula, Id = =

p ¥ I1 6 p ¥ 23.59 6

= 30.26 A. Rms stator current, 15.11.

Irms = Ns =

2

3

¥ 30.26 = 24.71 A.

120 f 120 ¥ 60 = 1200 rpm. = p 6

V = 460/ 3 = 265.58 V Ws = Full-load slip =

1200 ¥ 2p = 125.66 rad/s. 60 1200 - 1164 = 0.03. 1200

Without rotor resistance control, T =

=

˘ V 2 ◊ ( R2 / s ) 3 È ◊Í 2 2˙ Ws Î ( R1 + R2 / S ) + ( x1 + x2 ) ˚ 3 È (265.58)2 . (0.6 / 0.03) ˘ Í ˙ 125.66 Í 0.6 2 2˙ 0.4 + + (1.8 + 1.8) ÍÎ ˙˚ 0.03

(

)

= 78.48 Nm. (a)

Also, we have Rm1 S

2 2 = ÈÎ( R1 + Rk1 ) + ( x1 + x12 ) ˘˚

1

2

Solution Manual 91

When the breakdown torque occur at standstill,

( Rm1 )2 = ( R1 + Rk1 )2 + ( x1 + x2 )2 or (Rm1)2 = R1K2 + 2RK1 R1 + R12 + (x1 + x2)2

(i)

Also, Rk1 = 0.0966 Rm1

(ii) (Rm1)2

\ Rm1

\ Also,

– 9.33 ¥ 10

–3

(Rm1)2

– 0.0579

Rm2

– 0.09 – 10.24 = 0

= 3.26 W

Rk

1

= 0.315 W

R

1

= R1 – R2 1 = 3.26 – 0.6 = 2.66.

R e* =

2.66 aT12

=

2.66 = 0.426 W (2.5)2

Also, for a = 0, R = 2Re* – Rd = 2 ¥ 0.426 = 0.02 = 0.832 W (b)

With rotor resistance control,

Where

T =

˘ V 2 ( Rm1 / s ) 3 È Í ˙ 2 2 1 1 Ws Í ( R + R + R / s ) + ( x + x ) ˙ Î 1 ˚ k m 1 2

s =

1200 - 960 = 0.2. 1200

1.5 ¥ 78.48 =

\

3 125.66

(265.58)2 ¥ ( Rm1 / 0.2) È ˘ Í ˙ 1 2 Rm ˆ ÍÊ ˙ 2 1 ÍÎ ÁË 0.4 + 0.0966 Rm + 0.2 ˜¯ + (3.6) ˙˚

which gives

Rm1 = 2.171 W

Hence

R e* =

Rm1 - R2 2 aT 12

=

2.171 - 0.6 (2.5)2

= 0.2513 W Also,

2 R e * - Rd 2 (0.2513) - 0.02 = R 0.832

a = 0.42

\ (c)

(1 – a) =

Now,

Re* = 0.5[0.02 + (1 – 0.6)0.832] = 6.1764 W Rm1 = R2 + aT12 . Re* = 0.6 + (2.5)2 ¥ 0.1764 = 1.7025 W. Rk1 = 0.0966 ¥ 1.7025 = 0.165 W.

Substituting all known values in torque equation,

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Power Electronics

1.5 ¥ 78.48 =

which gives ,

\

(

)

s = 0.15 N = Ns(1 – s) = 1200 (1 – 15) = 1020 rpm.

\ 15.12

(265.58)2 ◊ (0.6 / s ) È ˘ 3 ◊Í ˙ 2 125.66 Í 1.7025 0.4 + 0.165 + + (3.6)2 ˙ ÎÍ ˚˙ s

Pf = cosf = 1 \ f = 0 Eq(rated) = Eb = Es = 460 / 3 = 265.58 V Poles, P = 6, W = 2pf = 2p ¥ 60 = 376.99 rad/s Base speed, Now, ratio

wb =

2 ¥ 376.99 = 125.66 rad/s. 6

K = Eb/Wb =

265.58 = 2.11 125.66

At 720 rpm, 2

Ê 720 ˆ = 143.28 Nm. TL = (398) ¥ Ë 1200 ¯ Ws = Wm = 720 ¥ 2p/60 = 75.4 rad/s. P0 = (143.28) ¥ 75.4 = 10803.31 W. (a)

Eq = k.Ws = 2.11 ¥ 75.4 = 159.09 V

(b)

P0 = 3Eq Ia Pf = 10803.31

\

Ia =

(c) (d) (e)

10803.31 = 22.64 A (3 ¥ 159.09)

E f = 159.09 – 22.64 ¥ (1 + j0) (j2.5) = 169.1 – – 19.52° torque angle (d) = – 19.52°. from Eq. (15.121), 3 ¥ 159.09 ¥ 169.1 = 428.82 N-m Tp = 2.5 ¥ 75.4