Electrical Power Systems

280 Electrical Power Systems
Eqn. (11.20), can then be rewritten as:
\
G
G
machine
HG
system
H
system
p f
HG
I
H
machine
p f
I KJ
2
b g machine
system
...(11.21)
d d
2
dt
= P P G
i – e .
G
d d
KJ dt
.
2
2 =(Pi – Pe ) pu on system base ...(11.22)
where Hsystem = Gmachine
. H
G
machine
system KJ
HG
I
= machine inertia constant in system base.
11.4 MACHINES SWINGING IN UNISON (COHERENTLY)
Let us consider the swing equations of two machines on a common system base, i.e.,
H
1
p f
H
2
d
d
dt
...(11.23)
2
d 1
= P
2 i1 – Pe1 ...(11.24)
2
d 2
= P
2 i2 – Pe2 ...(11.25)
p f dt
Since the machines rotor swing in unison,
d1 = d2 = d ...(11.26)
Adding eqns. (11.24) and (11.25) and substituting
d1 = d2 = d, we get
H
eq
p f
2
d d
2
dt
= Pi – Pe Where P i = P i1 + P i2
P e = P e1 + P e2
H eq = H 1 + H 2
Equivalent inertia H eq can be expressed as:
HG
I
Heq = G1,machine
. H1, machine +
Gsystem
KJ G2,machine
H2, machine
Gsystem
KJ
HG
I
...(11.27)
...(11.27)
...(11.28)
Example 11.1: A 60 Hz, 4 pole turbogenerator rated 100 MVA, 13.8 KV has an inertia constant
of 10 MJ/MVA.