Sheet #4
Example 5.1: A 3-phase star connected alternator is rated at 100 kVA. On a shortcircuit a field current of 50 amp gives the full load current. The e.m.f. generated on
open circuit with the same field current is 1575 V/phase. Calculate the voltage
regulation at (a) 0.8 power factor lagging, and (b) 0.8 power factor leading. Assume
armature resistance is 1.5W.
Solution:
Let the rated terminal voltage of the alternator
V = 1575 volts per phase
1000 X 103
 Full load current
I
 211.64 Amp
3 X 1575
Synchronous impedance
O.C voltage
for same field excitation
S .C.current
1575
Zs 
 7.442 Ω
211.64
Zs 
Or
X s  (Z s2  Ra2  (7.442)2  (1.5)2  7.289
Ω
(a) At lagging power factor, the no. load voltage E f is given by the equation
E f  (Vt cos  I a Ra )2  (Vt sin   I a X s )2
 (1575 X 0.8  211.64 X 1.5) 2  (1575 X 0.6  211.64 X 7.289) 2
= 2945.63
Volts
2945.63  1575
X 100 = 87.02%
 % Regulation =
1575
(b) At leading power factor, the no. load voltage E f is given by the equation
E f  (Vt cos  I a Ra )2  (Vt sin   I a X s )2
 (1575 X 0.8  211.64 X 1.5) 2  (1575 X 0.6  211.64 X 7.289) 2
= 1686.878
Volts
1686.878  1575
X 100 = 7.1%
 % Regulation =
1575
Example 5.2: 20 MVA, 13kV, 3-phase, Y-connected alternator has an armature
resistance of 0.2 Ohms per phase and a synchronous armature reactance of 1.2 ohms
per phase. Find the full load generated voltage per phase at:
(a) Unity power factor
(b) a power factor of 0.7 lagging
(c) a power factor of 0.95 leading
Then calculate then voltage regulation for the 3 power factor conditions.
Solution:
13000
= 7505.8
volts
3
MVA X 1000000 20 X 1000000
= 888.25 Amp.
Ia 

3V
3 X 7505.8
Vt 
I a Ra  888.25 X 0.2 = 177.65
volts
I a X s  888.25 X 1.2 = 1065.9
volts
(a) at unity power factor
E f  Vt  I a Z s  (7505.8  177.65)  j1065.9
 77577.90
 % Regulation =
volts
7757  7505.8
X 100 = 3.35%
7505.8
(b) at 0.7 lagging power factor
E f  (Vt cos  I a Ra )2  (Vt sin   I a X s )2
 (7505.8 X 0.7 177.65) 2  (7505.8 X 0.714 1065.9) 2
 8413.36
volts
8413.36  7505.8
X 100 = 12%
 % Regulation =
7505.8
(c) at 0.95 leading power factor
E f  (Vt cos  I a Ra )2  (Vt sin   I a X s )2
 (7505.8 X 0.95 177.65) 2  (7505.8 X 0.312 1065.9) 2
 7419
volts
7419  7505.8
X 100 = -1.15%
 % Regulation =
7505.8
3. A 1000 kVA, 3300-V, 3-phase, Y-connected alternator delivers full-load current at
rated voltage at 0.80 p.f. Lagging. The resistance and synchronous reactance of the
machine per phase are 0.5 Ω and 5 Ω respectively. Estimate the terminal voltage for
the same excitation and same load current at 0.80 p.f. leading.