Electrical Power Systems

100 Electrical Power Systems
and also V pu = Z pu I pu ...(5.8)
The power consumed by the load at its rated voltage can also be expressed by per-unit
impedance. The three-phase complex load power can be given as:
*
Sload(3f) = 3 Vphase IL
Here S load(3f) = three-phase complex load power
V phase = phase voltage
The phase load current can be given as:
I L * = complex conjugate of per-phase load current IL .
IL = Vphase
where Z L is load impedance per phase.
Substituting I L from eqn. (5.10) in eqn. (5.9), we get,
Z
L
S load(3f) = 3. V phase
\ Sload(3f) = 3|
Vphase|
Z
\ ZL = 3|
Vphase|
*
S
L *
HG
2
load( 3f)
2
V
I
phase
ZL
KJ *
Also, load impedance in per-unit can be given as
Zpu = ZL
ZB
Substituting Z L from eqn. (5.11) and Z B from eqn. (5.6) into eqn. (5.12), we obtain
Zpu = 3
2
Vphase
*
Sload
( 3f)
Now |VL–L| = 3 | Vphase| ...(5.9)
...(5.10)
...(5.11)
...(5.12)
| | (MVA) B
× ...(5.13)
2
(KV)
B
\ 3|V phase | 2 = |V LL | 2 ...(5.14)
Using eqns. (5.13) and (5.14), we get
| VL-L|
(MVA)
Zpu = ×
2 *
( KV)
S
| Vpu|
\ Zpu =
*
S
2
B
2
load (pu)
B
load(
3f)
...(5.15)