0195136047.pdf

118 TIME-DEPENDENT CIRCUIT ANALYSIS

−

+

+

−

+

−

V 1 = 462 ∠0 ̊ V

I 1

VL =?

P 1 = 5 kW

Q 1 = kVAR G 1

−

+

G 2

S 1 = kVA

P 2 =?

Q 2 =?

(1.4 + j1.6) Ω (0.8 +^ j1.0) Ω

IL = I 1 + I 2

4

15

4

25

PL = 10 kW

QL = kVAR

SL = kVA

4

30

4

50

I 2

+

−

Load V 2 =?

Figure E3.1.7

(b) From Equation (3.1.37) applied to the load, 433. 8 IL( 0. 8 )= 10 × 103 ,orIL= 28 .8A.

As a phasor, withV ̄ 1 as reference,I ̄ 1 = 28. 8  −( 0 .78°+ 36 .9°)= 28. 8  − 37 .68° A.

The KCL equation yields:

I ̄ 2 =I ̄L−I ̄ 1 = 15. 22  − 38 .33°

The KVL equation yields

V ̄ 2 =V ̄L+( 15. 22 − 38 .38°)( 0. 8 +j 1. 0 )= 4. 52. 75  − 0 .22° V

The terminal voltage ofG 2 is thus 452.75 V (rms).

(c) The power triangle values forG 1 and the load are shown in Figure E3.1.7 so that the

student can check it out.

P 2 =V 2 I 2 cosφ 2 = 452. 75 × 15. 22 ×cos( 38 .38°− 0 .22°)

=5418 W,or 5.418 kW

(Note thatφ 2 is the angle between phasorsV ̄ 2 andI ̄ 2 .)

Q 2 =V 2 I 2 sinφ 2 = 452. 75 × 15 .22 sin( 38 .38°− 0 .22°)

=4258 VAR, or 4.258 kVAR

The conservation of power (real and reactive) is satisfied as follows:

PL+loss in feeder resistances= 10 +

( 13. 53 )^21. 4

1000

+

( 15. 22 )^20. 8

1000

∼= 10 .4kW

which is the real power delivered byG 1 andG 2 , which in turn is the same asP 1 +P 2 =

5 + 5. 4 = 10 .4kW.

QL+(I^2 XL)of feeders=

30

4

+

( 13. 53 )^21. 6

1000

+

( 15. 22 )^21. 0

1000

∼=8kVAR

which is the same as the reactive power delivered byG 1 andG 2 , which in turn is the

same as

Q 1 +Q 2 =

15

4

+ 4. 26 ∼=8kVAR