December 2012
Beginner
528 pages
17h 31m
English
Content preview from Introduction to Electrical Power and Power Electronics
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66 Introduction to Electrical Power and Power Electronics
Example 3.3
A 30-MVA, 11-kV, 60-Hz, three-phase, Y-connected synchronous generator has
X
s
= 2 Ω/ph and negligible R
a
. Determine its power angle when delivering rated
power at unity power factor.
Solution:
On a per-phase basis, V
T
= 11 kV ÷ √3 = 6.351 kV and 1/3 × 30,000 kVA = 6.351
kV × I
a
.
Therefore, armature current I
a
= 1575 A. From the phasor diagram shown below
at unity pf, we get
δ
V
T
I
a
jI
a
X
s
E
f
E
P
f
Vk
Va
nd
=+×= =
=
6351 1575 2 7089 7 089
63
22
() .
.
max
551 7 089
2
22 51
×
=
.
.MW.
The power delivered by the generator P = 1/3 × 30 MVA × 1.0 pf = 10 MW/phase.
The power angle is then derived from Eq
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