0195136047.pdf

11.4 TRANSFORMER PERFORMANCE 487

Solution

(a) The equivalent circuit of the transformer, referred to the high-voltage (primary) side,

neglecting the exciting current of the transformer, is shown below in Figure E11.4.1(a).

Note that the voltage at the load terminals referred to the high-voltage side is 240× 10 =

2400 V. Further, the load current corresponding to the rated (full) load condition is

50 × 103 / 2400 = 20 .8 A, referred to the high-voltage side. With a lagging power factor

of 0.8,

I ̄ 1 =I ̄ 2 /a= 20. 8  −cos−^10. 8 = 20. 8  − 36 .9°

Using KVL,

V ̄ 1 = 2400  0°+( 20. 8  − 36 .9°)( 1. 5 +j 2. 0 )= 2450  0 .34°

If there is no load, the load-terminal voltage will be 2450 V. Therefore, from Equation

(11.4.1b),

% voltage regulation=

V 1 −aV 2

aV 2

× 100 =

2450 − 2400

2400

× 100 = 2 .08%

V 1 aV 2 = 2400 ∠^0 ° V

I 1 = I 2 /a = 20.8∠−36.9° A

• 
  

−

+

−

Load

1.5 Ω

Req

j 2.0 Ω

jXeq

(a)

Figure E11.4.1

0.34° aV 2 = 2400 ∠^0 ° V

cos−^1 0.8 = 36.9°

I 1 =20.8∠−36.9° A

|I 1 Req| = 31.2 V

V 1 = 2450 ∠0.34°

|j I 1 Xeq| = 41.6 V

(b)

VS

• 
  

−

+

−

V 1

I 1 = I 2 /a = 20.8∠−36.9° A

+

−

aV 2 = 2400 ∠ 0 ° Load

0.5 Ω j 2.0 Ω Req = 1.5 Ω j Xev = j2.0 Ω

(c)

Feeder