Page 415 - Power Electronic Control in Electrical Systems

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394 Examples, problems and exercises

From open-circuit test
20e j arc cos(0:5) 230 2
Y sh ˆ 66 66 ˆ 0:01148 j0:02276S
0:25 p 10 3

3
giving R c ˆ 87:11
and X m ˆ 43:94
.
From short-circuit test
230
0:02 p 
3 j arc cos(0:08) j85:411
Z se ˆ e ˆ 0:02213e ˆ 0:00177 ‡ j0:02206
120
giving R e ˆ 0:00177
and X e ˆ 0:002206
.
21. (i) Draw a circuit diagram and a phasor diagram showing the use of two
wattmeters to measure the total power in a three-phase three-wire AC load.
(ii) Derive an expression for the power factor of a balanced three-phase AC load
in terms of the wattmeter readings P 1 and P 2 .
(iii) An unbalanced delta-connected load on a three-phase 415-V 50-Hz supply
has the following impedances in each phase:

Z ab ˆ 15 ‡ j0
Z bc ˆ 1:2 ‡ j16:5
Z ca ˆ j18:2
:
Determine the three-line currents in magnitude and phase, and the readings
of two wattmeters connected such that P 1 measures I a and V ac , while P 2
measures I b and V bc . Take V ab as reference phasor.

(i)

Fig. 9.16

(ii) P 1 ˆ RefV ac I gˆ V L I L cos (30 f)

a
P 2 ˆ RefV bc I gˆ V L I L cos (30 ‡ f)

b
p
3V L I L cos f
P 1 ‡ P 2 ˆ V L I L 2 cos 30 cos ( f) ˆ
P 1 P 2 ˆ V L I L 2 sin 30 sin ( f) ˆ V L I L sin f

p
\ tan f ˆ 3(P 1 P 2 )/(P 1 ‡ P 2 ) .. . : then power factor ˆ cos f.
(iii) I ab ˆ 415/(15 ‡ j0) ˆ 27:6667 ‡ j0 A
I bc ˆ 415e j120 /(1:2 ‡ j16:5) ˆ 415e 120 /16:54e j85:84 ˆ 25:085e j205:84 A
I ca ˆ 415e j120 /( j18:2) ˆ 22:80e j210 ˆ 19:75 j11:40A