MD Singh Power Electronics Solution Manual to Chapter 15

C

HAPTER

15

15.1 (a) Zin = (R1 + jx1) + Zf where Zf =

(

2 2

)

2/ ( 2 R j x jxm R s j x xm + + + =

(

) (

)

(

)

(

)

0.95 / 0.04 0.95 81.4 0.95 / 0.04 1.63 81.4 j j j + + + =

(

)

(

)

1933.25 77.33 23.75 83.03 j j -+ = 21.27 + j 7.02 \ Zin = (0.15 + j1.63) + 21.27 + j 7.02 = 21.42 + j8.65 = 23.10 – 22°. Stator current I1 = 2300 / 3 23.10 Iph in V Z = = 57.49 A.

Input power Pi = 3 Viph . I1 . cosQ

= 3 2300 57.49

3

¥ ¥

= 229023.88 W

Pi = 229 kW.

(b) haft power P = Pgross – (Pfw + Pcore + Pstray) (i)

Air gap power, pg = 3 I21 Rf = 3¥57.492¥21.27

= 210898.43 W. Pgross = Pg(1 – S) = 210898.43(1 – 0.04) = 202462.49 W Pstray = 1 100¥ 3521.7 = 35.22 W from (i), \ P = 202462.48 – (8300 + 35.22) P0 = 194.12 kW Now, Wr = Ws (1 – S) Also, S = s s W W W

-\ 0.04 = s 1100 s W W -\ Ws = 120 rad/s. Also, wr = 120 (1 – 0.04) = 115.2 rad/s. Torque, T = 210898.43 120 g s P W = = 1757 NM. (c) Efficiency = 210898.43₃ 229.02 10¥ = 92%.

(a) Stator current I1 = 1ph

in V Z where, Zin = (R1 + jx1) + Zf where Zf =

(

) (

)

(

)

2 2 2 2 / / R s j s j xm R s j x xm -+ + At starting s = 1 \ Zf =

(

) (

)

(

)

0.95 1.63 81.4 77.33 132.682 0.95 1.63 81.4 0.95 83.03 j j j j j + -= + + + = 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 = 2300 / 3 3.41 = 389.6 A. (b) Input power = 3 V1 I1 cos q1

= 3 2300 389.6cos 71.84¥ ¥

= 484 kW. (c) Starting developed torque (Td)

At starting, Ws = W = 1100 rpm. ` using equations (15.33), Tstart =

(

)

(

)

(

)

2 2 2 3 2300 / 3 0.95 1100 2 0.15 0.95 1.63 1.63 60 p ¥ È ˘ ¥ _Î + + + _˚ = 3684.79 Nm 15.3 From Eq.(15.29), Smax. T =

(

)

2 2 2 1 1 2 R R + x +x

= ( )2

₍

₎

2 0.95 0.15 + 1.63 1.63+ = 0.291 From Eq. (15.30), Tmax =

(

)

( )

₍

₎

2 2 2 3 2300 / 3 2 120 0.15 0.15 1.63 1.63 ¥ È ˘ ¥ Î + + + ˚ = 6457.28 N-m.

15.4 In a constant power zone, at a constant rotor frequency, T a

( )

1 ₁ 2

F . or, 2 2 2 1 F F = 1 2 T T , or F2 = F1 1 2 T T \ F2 = 2000 50 1500 = 57.74 Hz.

15.5 From equation (15.34), k = f/frated =

12 60 = 0.2

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

(

)

(

)

( ) ( )

( )

( ) 2 2 max 2 max ₂ . 0.2 0.024 0.024 0.24 1 _{0.024 0.024} 0.24 0.2 0.2 T k T k = ± + = = ± + For motoring,

(

)

(

)

max max 0.2 1 T k T k = = = 0.68 For breaking,

(

)

(

)

max max 0.2 1 T k T k = = = 1.46.

(2) Substituting S = 1 in Eq. (15.51), gives

Starting torque. Thus,

T =

(

)

(

)

2 2 2 2 1 2 1 / 3 _rated s V R k w R R x x k È ˘ Í ˙ + Í _{+ +} ˙ Í ˙ Î ˚ (i)

From equation (i), the ratio of starting torques for k = 0.2 and k = 1 is

(

)

(

)

0.2 1 s s T k T k = = =

₍

₎

( ) ( ) ( ) 2 2 2 2 0.024 / 0.2 0.048 / 0.2 0.24 0.024 0.048 0.24 + + = 2.6.

The starting motor current is given by

Ims =

(

)

(

)

2 1 2 1 2 rated V R R x x k + + + (ii)

The ratio of starting current for k = 0.2 and 1 is

(

)

(

)

( ) ( ) ( ) 2 2 2 2 0.2 0.048 0.24 1 0.048 ( ) 0.24 0.2 ms ms I k I k = + = = + = 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 =

(

)

( ) 2 2 2 0.024 / 0.04 3 0.024 (0.024 ) 0.24 0.04 rated s V w È ˘ Í ˙ Í + + ˙ Î ˚ = 2 3 _rated s V w (1.34) (a)

and for 30 Hz from equation (15.51),

Trated = ( ) 2 2 2 0.024 3 0.5 0.024 0.024 0.24 0.5 0.5 rated s V S w S È Ê ˆ ˘ Á ˜ Í _{Ë ¯} ˙ Í ˙ ÍÊ ₊ ˆ ₊ ˙ Á ˜ Í_{Ë ¯} ˙ Î ˚ (b)

Equating equations (a) and (b), we get

( ) 2 2 0.024 0.5 0.024 0.024 0.24 0.5 0.5 S S Ê ₊ ˆ Á ˜ Ë ¯ = 1.34 or 26.04 s2 – 0.98 S + 1 = 0 or S =

(

)

2 14.62 14 . 62 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 = 120 120 50 6 f p ¥ = = 1000 rpm. Ws = 1000 2 1000 p ¥ _{= 104.7 rad/s.}

Slip, S = 1000 960 1000 - _{= 0.04} Rotor impedance Zr = 2 2 R jx s + = 5 + j1.5 = 5.22 – 16.7° W Stator impedance Zs = 0.4 + j1.5 = 1.55 – 75° W Machine impedance ZQ = Zs + 2 2

.

_m m Z Z Z

Z

+ I1 = 400 / 3 6 = 38.5 A. I2 = 2 30 38.5 31.89 m s m Z I Z +Z = ¥ = 36.22 A E = I2|Z2| = 36.22 ¥ 5.22 = 189 V. Rated torque = ( ) 2 2 2 2 3 . 3 0.2 . 36.22 104 . 7 0.04 s I R W S = = 188 Nm. (a) At 25 HZ, k = 25 50 = 0.5 Substitute values in equation (15.41), we get

188 2 = ( ) ( )

₍

₎

2 2 2 2 3 189 0.2 / 0.55 . 104 . 7 0.2 / 0.5S 1.5 ¥ + \ 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. Z2 = 2 R S + j k. x2 = 0.2 0.0374 + j 0.75 = 5.4 – 8° W.

Taking E as a reference vector,

I2 = 2 94.5 5.4 8 E Z = – - ∞= 17.5 – – 8°. Im = 94.5 6.30 90 . 0.5 30 E A J k xm= j ¥ = – - ∞ I1 = I2 + Im = 17.5 – – 8° + 6.30 – – 90°

\ 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,

Nss = 0.5 ¥ 40 = 20 rpm. At 25 Hz, Ns = 25

50 ¥ 1000

= 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 S = 2 20 500 s s N N = = 0.04 Z2 = 2 . 2 R jk x S = 5 + j0.75 = 5.06 – 8.5° W

Taking E as a reference vector, I2 = 2 94.5 18.7 8.5 5.06 8.5 E A Z = – ∞= – - ∞ . Im = m E jkx = 94.5 6.30 90 15 A j = – - ∞ . I1 = I2+Im =18.7– -8.5∞ +6.30– -90∞ Hence I1 = 20.6 A.

(c) 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 50 ¥ 189 = 143.64 V. S = – 40/760 = – 0.0526 2 Z = R2 jk x₂ 38.8 j1.14 s = - + = 3.97 – 163.3° 1 W

I2 = 2 143.64 36.2 163.3 3.97 163.3 E A Z = – - = – - ∞ m

I remains the same as the foregoing

I1 = 36.2 – – 163.3° + 6.30 – – 90°

= – 34.62 – j16.88 = 38.52 – – 154°.

V = E+Z I_{5 1}= 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.

(a) Im =

m

E

X = 189/30 = 6.3 A

Substituting values in Equation (15.66),

We get, (6.30)2 ₌

(

)

( )

(

)

( ) ( ) 2 2 2 2 2 0.2 / 1.5 60 0.2 / 31.5 s s È + ˘ ¥ Í ˙ + Î ˚ \ S = 0.067. Slip speed in rpm = S.N rpm = 0.067 ¥ 1000 = 67 rpm.

(b) (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 =

(

)

2 2 2 2 2 . / 3 / rated s E R s W R s x È ˘ = Í ˙ ¥ Î ˚ 139 = ( )

(

)

(

)

( ) 2 2 2 3 189 0.2 / 104.7 0.2 / 1.5 s s È ¥ ˘ Í ˙ + Î ˚ or, ( ) 2 2 0.2 1.472 1.5 s s Ê ˆ _{+ =}Ê ˆ Ë ¯ Ë ¯ \ 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

1000 ¥ = 26.42 Hz.

S = 28.4 500 ss s N N = = 0.0568 K.S.= 0.528 ¥ 0.0568 = 0.03

Substituting the known values in equation (15.68),

We get Im = 6.3 ( )

₍

₎

( ) ( ) 1/ 2 2 2 2 2 0.2 / 0.04 30 1.5 0.2 / 0.04 1.5 È + + ˘ Í ˙ + Î ˚ = 38.5 A.

(c) Since, at a constant flux, speed-torque curves for different frequencies are straight lines, slip-speed

Nss =

139 40

188¥ = 29.56 rpm.

Hence, Synchronous speed

Ns = N + Nss = 500 + 29.56 = 529.56 rpm. Frequency = 529.56 50 1000 Ê _{ˆ ¥} Ë ¯ = 26.48 Hz S = 29.56 529.56 ss s N N = = 0.056 K = 26.48 50 = 0.53, S.K = 0.03. Substituting the known values in equation (15.68), we get

Im = ( )

₍

₎

( ) ( ) 2 2 2 2 0.2 / 0.03 30 1.5 6.30 297 0.2 / 0.03 1.5 A È + + ˘ =_{Í ˙}= + Î ˚ 15.9

(a) The fundamental Rms live voltage of a six-step inverter is given by E1 =

6

p . Edc (i)

where, Edc = 3/p . Em cosa (ii)

Em is the peak of ac source live voltage.

\ E1 = 3 6₂ p Em . cosa cosa = 2 1 3 6 E Em p (iii) Here, E1 = 460 V, Em = 460 2 V. \ cos a = 2 ₄₆₀ 460 2 3 6 p ¥ ¥

\ 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 6 ¥ – 1180 = 20 rpm

Hence, synchronous speed at 600 rpm.

Ns = N + Nss = 600 + 20 = 620 rpm. Ns = 120 f/p = 120 60 6 ¥ = 1200 rpm. \ Inverter frequency = 620 60 1200 Ê _{ˆ ¥} Ë ¯ = 31 Hz.

(ii) Back emf at the rated operation Erated = I2 =

(

)

1 2 2 2 2/ R s x È + ˘ Î ˚ where I2 =

(

)

(

)

1 2 2 1 2 1 2 / 3 / E R +R s + x +x Also, S = 1200 1180 1200 = 0.017 \ I2 =

(

)

(

)

2 2 460 / 3 0.07 0.19 0.75 0.67 0.017 + + + = 58.5 A. Erated = ( ) 1 2 2 2 0.07 58.5 0.67 0.017 È_{Ê ˆ +} ˘ Í_{Ë ¯} ˙ Î ˚ = 244 V Now, torque at constant flux is given by

T = ( ) ( ) 2 2 2 2 2 2 2 / . 3 . / rated s E R K S W R KS x È ◊ ˘ Í ˙ + Î ˚ (iv) Ws and x2 in this equation are for rated frequency.

where Ws = 1200 2 60 ¥ p = 125.66 rad/s T = 3/Ws . I22 Rs/S and, T = 3 (58.50)2 0.07 125.66 0.017 Ê ˆ Ë ¯

= 336.42 Nm. Equation (iv) becomes

336.42 2 = ( )

( )

( )

( ) 2 2 2 0.07 244 3 _. 125.66 0.07 0.67 K S KS È _¥ ˘ Í ˙ Í ˙ Í ₊ ˙ Í ˙ Î ˚ Now, k =

(

)

(

) (

)

500 0.42 1 1200 1 1 r s w w -S = -S = -S \ 336.42 2 =  ( ) ( ) ( ) ( ) 2 2 2 0.07 1 244 3 0.42 125.66 0.07 1 0.67 0.42 S s S s -È _¥ ˘ Í ˙ Í ˙ ÍÈ _{¥ -} ˘ ₊ ˙ ÍÍ ˙ ˙ 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 = ( ) 2 2 244 0.07 0.67 0.425 0.012 Ê _{ˆ +} Á ˜ Ë ¥ ¯ = 17.76 A Machine fundamental phase voltage,

V1 =

(

)

1 2 2 2 2 2 2 1 1 2 R I R K x x s È_{Ê ˆ} ˘ + + + Í_{Ë ¯} ˙ Î ˚ = ( )

₍

₎

1 2 2 2 2 0.07 17.76 0.19 0.425 0.75 0.67 0.012 È_{Ê ˆ} ˘ + + + Í_{Ë ¯} ˙ Î ˚ V1 = 107.51 V

Now, taking V1 as reference vector,

I2 = I2

(

)

(

)

1 1 2 1 2 tan / k x x R R s - + -+ = 17.76 ( ) ( ) 1 0.425 1.42 tan 6.02 = 17.76 – – 88.10° A. m I = 1 ₉₀ . _m E K X – - ∞ = 107.51 90 12.65 90 0.425 20¥ – - ∞ = – - ∞

Now, I = ₁ I₂ + I_m =17.76– -88.10∞ +12.65– -90∞

1

I = 30.40 A.

Rms harmonic current is given by

(

)

1 2 1 4 1 2 3, 5, 11, 13 1 I_h h V K x x h • = È ˘ = _{Í ˙} + _Î

Â

_˚

Neglecting harmonics higher than 13, gives

h I =

(

)

(

)

1 1 2 0.046 0.046 107.51 0.425 0.75 0.67 V K x x ¥ = + ¥ + = 8.195 A. \ Rms input current =

(

)

12 1 h I +I = _È(_30.40)2 (_8.195)2_˘12 + Î ˚ = 31.49.

Ex 15.10. At the rated operations,

Ns = 120 120 60 6 f P ¥ = = 1200 rpm = 125.66 rad/s Rated slip = 1200 1176 1200 = 0.02. Rotor Impedance, Z2 = 0.145 0.02 + j 0.5 = 7.27 – 3.95°W Machine Impedance = 2 1 2 m m Z Z Z Z Z + + = 0.19 + j 0.75 +

(

) (

)

(

)

7.25 0.5 15.3 7.25 15.8 j j j + ◊ + = 5.81 + j3.81 = 0.95 – 33.26°W 1 I = 460 / 3 6.95–33.26= 38.21 – – 33.26° A 2 I = ₁ 2 15.3 90 17.38 65.35 Zm I Zm Z – = + – (38.21 – – 33.26°) = 33.64 – – 8.61 A. m I =

(

) (

)

(

)

2 1 2 7.27 3.95 38.21 33.26 17.38 65.35 Z I Z Zm – – - ∞ = + – = 15.98 – – 95.61° Torque = 22

(

2

)

3 / s I R s W

= 3 (33.64)2 0.145

125.66 0.02

Ê ˆ

◊ _{Ë ¯}

T = 195.87 Nm.

(a) For the three phase current source six-step inverter, fundamental Rms current is given by I1 = 6 Id. p ◊ \ 1 . 6 I Id=p = 38.21 6 p ¥ = 49 A. Also, rms stator current is given by

Irms =

(

23

)

Id= 23. 49 40¥ = A.

(b) Slip speed at rated torque and frequency,

Nss = s . Ns = 0.02 ¥ 1200 = 24 rpm.

\ At motor speed of 600 rpm, synchronous speed Ns = 600 + 24 = 624 rpm.

Inverter frequency = 624 60

1200¥ = 31.2 Hz.

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.

(c) 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 2 2 2 2 2 / 3 / rated s E R s W _{R s x} È ˘ Í ˙ + Î ˚ 195.87 2 = ( )

₍

₎

(

)

( ) 2 2 3 244.49 0.145/ 125.66 0.145/ 0.5 s s È ¥ ˘ Í ˙ + Î ˚

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 = 30

60 = 0.5

\ Synchronous speed, Ns = 0.5 ¥ 1200 = 600 rpm.

Hence, motor speed = 600 – 12 = 588 rpm.

\ S = 12 600 ss s N N = = 0.02 \ I2 =

(

)

(

2 2

)

2 2 . ( / ) . rated K E R S + K x

\ = ( )2

₍

₎

2 0.5 244.46 0.145 / 0.02 0.5 0.5 ¥ + ¥ = 16.81 A. From Eq. (15.62), we have

I22 = 2 2 2 2 2 1 m m I I x X -+ or (16.81)2 = ( ) 2 2 1 15.98 2 0.5 1 15.3 I -¥ + or I1 = 23.59 A.

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

Id = 1 6 I p ¥ = 23.59 6 p ¥ = 30.26 A. Rms stator current, Irms = 2

3 ¥30.26 = 24.71 A. 15.11. Ns =120 120 60 6 f p ¥ = = 1200 rpm. V = 460/ 3 = 265.58 V Ws = 1200 2 60 p ¥ = 125.66 rad/s. Full-load slip = 1200 1164 1200 = 0.03. Without rotor resistance control,

T =

(

)

(

)

(

)

2 2 2 2 1 2 1 2 / 3 / s V R s W R R S x x È ◊ ˘ ◊ Í ˙ + + + Î ˚ = ( ) ( )

(

)

(

)

2 2 2 3 265.58 . 0.6 / 0.03 125.66 0.6 0.4 1.8 1.8 0.03 È ˘ Í ˙ Í _{+ + +} ˙ Í ˙ Î ˚ = 78.48 Nm. (a) Also, we have

1 Rm S =

(

) (

)

1 2 2 2 1 1 1 k 1 2 R R x x È _{+ + +} ˘ Î ˚

When the breakdown torque occur at standstill,

(

1

)

2

₍

1

₎

2

₍

₎

2 1 1 2 Rm = R +Rk + x +x or (Rm1)2 = R1K2 + 2RK1 R1 + R12 + (x1 + x2)2 (i) Also, Rk1 = 0.0966 Rm1 (ii) \ (Rm1)2 – 9.33 ¥ 10–3 (Rm1)2 – 0.0579 Rm2 – 0.09 – 10.24 = 0 \ Rm1 = 3.26 W Rk1 = 0.315 W Also, R1 = R1 – R21 = 3.26 – 0.6 = 2.66. Re* = ( ) 2 2 1 2.66 2.66 2.5 aT = = 0.426 W Also, for a = 0, R = 2Re* – Rd = 2 ¥ 0.426 = 0.02 = 0.832 W

(b) With rotor resistance control,

T =

(

)

(

)

(

)

2 1 2 ₂ 1 1 1 1 2 / 3 / m s _{k m} V R s W _{R R R s x x} È ˘ Í ˙ + + + + Í ˙ Î ˚ Where s = 1200 960 1200 = 0.2. \ 1.5 ¥ 78.48 = ( )

(

)

( ) 2 1 2 1 2 1 265.58 / 0.2 3 125.66 0.4 0.0966 3.6 0.2 m m m R R R È ¥ ˘ Í ˙ Ê ˆ Í ˙ + + + Á ˜ ÍË ¯ ˙ Î ˚ which gives Rm1 = 2.171 W Hence Re* = ( ) 1 2 2 2 2 1 2.171 0.6 2.5 m T R R a - -= = 0.2513 W Also, (1 – a) = 2 R *e 2 0.2513( ) 0.02 0.832 Rd R - -= \ a = 0.42 (c) 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.

1.5 ¥ 78.48 = ( ) ( )

(

)

( ) 2 2 2 3 265.58 0.6 / 125.66 1.7025 0.4 0.165 3.6 s s È ◊ ˘ ◊ Í ˙ Í _{+ + +} ˙ Í ˙ Î ˚ which gives , s = 0.15 \ N = Ns(1 – s) = 1200 (1 – 15) = 1020 rpm. 15.12 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, wb = 2 376.99 6 ¥ _{= 125.66 rad/s.} Now, ratio K = Eb/Wb = 265.58 125.66 = 2.11 At 720 rpm, TL = (398) ¥ 2 720 1200 Ê ˆ Ë ¯ = 143.28 Nm. 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 =

(

)

10803.31 3 159.09¥ = 22.64 A (c) E f = 159.09 – 22.64 ¥ (1 + j0) (j2.5) = 169.1 – – 19.52°

(d) torque angle (d) = – 19.52°. from Eq. (15.121),

(e) Tp = 3 159.09 169.1

2.5 75.4

¥ ¥