Question

A three-phase, 318.75-kVA, 2300-V alternator has an armature resistance of
0.35 ­/phase and a synchronous reactance of 1.2 ­/phase. Determine the no-load
line-to-line generated voltage and the voltage regulation at
(a) Full-load kVA, 0.8 power factor lagging, and rated voltage.
(b) Full-load kVA, 0.6 power factor leading, and rated voltage

Answer 1

3 S= 38-75 kVA etnge (+知 2 vo lood line to ive dl Vol 24406 Vol -1409.1%-1329. 9 32.9 Vo

1230412 270-412-132チ 327.9 E-4.321

Answer 2

For 318:75 kVA, 0:6 power factor leading, S = 3187506 ¡53:13± VA.
Ia = S¤
3V ¤
Á
=
3187506 53:13±
(3)(1327:9
= 806 53:13± A
Nextq
The two-winding transformer rated currents are:
ILV =
9000
90
= 100 A
38 CONTENTS
IHV =
9000
120
= 75 A
The auto transformer connection is as shown in Figure 23.
90V
120V
100A
75A
175A
+
¡
+
¡
120V
²
²
FIGURE 23
Auto transformer connection for Problem 3.7.
(a) With windings carrying rated currents, the auto transformer rating is
S = (120)(175)(10¡3) = 21 kVA
(b) Since the windings are subjected to the same rated voltages and currents as
the two-winding transformer, the auto transformer copper loss and the core loss at
the rated values are the same as the two-winding transformer. Therefore, the auto
transformer efficiency at 0:8 power factor is
´ =
(21000)(0:8)
(21000)(0:8) + 200 + 500
£ 100 = 96 %

Answer 3

A three-phase, 318.75-kVA, 2300-V alternator has an armature resistance of 0.35

Answer 4

Peh Vi 8o A

#2300 : 1310.2.42 12300小07 2453 8