Page 310 - Solutions Manual to accompany Electric Machinery Fundamentals

P  33.6 MW 8.6 MW   42.2 MW
The cylindric al rotor term is 33.6 MW, and the reluctance term is 8.6 MW, so the reluctance torque
accounts for about 20% of the power in this generator.
C-2. A 14-pole Y-connected three-phase water-turbine-driven generator is rated at 120 MVA, 13.2 kV, 0.8 PF
lagging, and 60 Hz. Its direct-axis reactance is 0.62  and its quadrature- axis reactance is 0.40 . All
rotational losses may be neglected.
(a) What internal generated voltage would be required for this generator to operate at the rated
conditions?

(b) What is the voltage regulation of this generator at the rated conditions?

(c) Sketch the p ower-versus-torque-angle curve for this generator. At what angle is the power of the
generator maximum?
(d) How does the maximum power out of this generator compare to the maximum power available if it
were of cylindrical rotor construction?
SOLUTION

(a) At rated conditions, the line and phase current in this generator is
P 120 MVA
I A  I L    5249 A at an angle of –36.87
L 3  3V 13 2 kV  .

 E A   V R A A  I jX q I
A

  E  7621 0   0  j . 0   40 5249  36 . 87 A 
A
  E  9038 10 7 . V 
A
Therefore the torque angle  is 10.7. The direct-axis current is
,
I d  I A  sin        90 
 5  249 sin  A . 57   47   79  3 .
 I
d

I d  3874  79 3 . A 
u
The quad at re-axis current is
r
 

I q  I A  cos   

 I
57
 5  249 cos  A 47 .  10  7 .
q
I q  3541  10 7 . A 
,
h
Therefore t e internal generated voltage of t he machine is
E  V  R I  jX I  jX I

d
d
q
A
A
A
q
E A  7621 0   0   j . 0  62 3874   79  3 .  j  40.0 3541 10   7 .
E A  9890 10 7 . V 
(b) The voltage regulation of this generator is

V nl  V fl  100 %  9890  7621  100 %  29 8 . %
V fl 7621
(c) The power supplied by this machine is given by the equation
304