Electrical Power Systems

84 Electrical Power Systems
| V | | |
VR = nl - Vrated
| E| -|
Vrated|
´ 100 =
´ 100 ...(4.17)
| V |
| V |
rated
4.4 POWER ANGLE CHARACTERISTICS
rom ig. 4.3, the three phase complex power at the generator terminal is
Also
S 3 f = 3VI a *
|| E d -| V|
0º
Ia =
| | b
Z s
rom eqns. (4.18) and (4.19), we get
2
| E|| V|
| V|
S3 f = 3 3
| Zs|
| Zs|
\ Three phase real and reactive power are:
rated
...(4.18)
...(4.19)
b - d - b
...(4.20)
2
| E|| V|
| V|
P3 f = 3 cos( ) 3
| Zs|
Zs
2
| E|| V|
| V|
Q3 f = 3 sin( ) 3
| Zs|
Zs
If r a is neglected, then Z s = jx s and b = 90º,
Therefore,
| | cos
b - d - b
...(4.21)
| | sin
b - d - b
...(4.22)
P3 f = 3 | || | E V
sin d ...(4.23)
Xs V
Q3 f = 3 | | (| E|cos | V|)
X s
d- ...(4.24)
Example 4.1: A 50 MVA, 13.8 kV, three-phase, 50 Hz synchronous generator has a synchronous
reactance of 3 W/phase and r a » 0. One generator is delivering rated power at a 0.85 pf lagging
at the rated terminal voltage to an infinite bus.
(a) Compute the excitation voltage per phase E and the power angle d.
(b) With the excitation held constant at the value found in (a), the driving torque is reduced
until the generator is delivering 22 MW. Determine the armature current and the power
factor.
(c) If the generator is operating at the excitation voltage of part (a), what is the maximum
power the machine can deliver before losing synchronism?
Solution:
(a) cos q = 0.85 \q = 31.8º
S3f = 50 318 . º = ( 42. 5 + j26. 34)
MVA