are
Generator G1: 10 MVA, 12 percent reactance
Generator G:: 5 MvA, 8 percent reactance
Transformer: 15 MVA, 6 percent reactance
Transmission line: (4 + j60) Q, 230 kV
where the percent reactances are computed on the basis of the individual component ratings.
Express the reactances and the impedance in percent with 15 MVA as the base value.
Equation (2.7) gives, for generator G1.
Percent reactance = l2(15/10) = 18 percent
For generator G,
Percent reactance = 8(15/5) =24 percent
For the transformer,
Percent reactance = 6(15/15) = 6 percent
And for the transmission line, from (2.2) and (2.7),
2.14 Draw an impedance diagram for the system shown in Fig. 2-4(a), expressing all values as
per-unit values.
2.1s Draw an impedance diagram for the system shown in Fig. 2-5(a) expressing all values as
percent values.
Let us arbitrarily choose 10 MvA to be the base MVA.
Then, for generator G1 Percent impedance = 10(10/10) = 10 percent
For generator G2,
Percent impedance = 8(10/5) =16 percent
For the transformer,
Percent impedance = 6(10/15) = 4 percent
And for the transmission line,
6
Percent impedance = (4 + j40)
(10x10 )
3
(66x10 )
2
x100 = (0.918 + j9. 18) percent
These values produce Fig. 2.5(b).
2.16 Draw a per-unit reactance diayram for the system shown in Fig. 2-6(a).
We arbitrarily choose 20 MvA and 66 kV as base values. The per-unit reactance diagram is that
shown as Fig. 2-6(b), where, for G., X,,. = JO. is Pu because its Percent reactance is 15 percent
with the same kVA base. Also for G2 and G3,
X pu=(20/10)(j0.1) =jO.2pu
For T1 and T2,
X pu= (20/30) (jO.15)=jO.1 Pu
For T3,
20.008 = jO.64 .u
Xpu=
20
j0.08
2.5
and for the line,
Xpu= Xline
base K V A20
2
(base K V A) 1000
=j60
20,000
2
(66) (1000)
=j0.276pu kVA = jr) 20,000 = 10.276 Pu