Power system fundamentals - Wind Power Integration: Connection and System Operational Aspects - page 42

In power systems work we are more likely to work to MVA and kV bases:

MVA b
kV b

Assuming that we use the equivalent single-phase approach, the impedance
base Z b may be obtained as follows:

Z b ¼ V b = I b ¼ð kV b 10 3

Þ=ðð MVA b 10 6 Þ= kV b 10 3 Þ¼ kV b = MVA b

Therefore

Z MVA b
kV b

Z pu ¼ ð 2 : 18 Þ

Also

S
MVA b ¼

P
MVA b þ j

Q
MVA b ¼ P pu þ j Q pu

S pu ¼

Note that the voltage, current and impedance bases are real, while the corre-
sponding per unit quantities are complex.

2.4.9.1 Networks with multiple voltage levels
Practical power networks contain several voltage levels. For example, wind
generators often have a nominal line voltage of 690 V. These feed a distribution
network of 11 or 33 kV (in the United Kingdom) through a transformer. The dis-
tribution network is in turn connected, at a substation, to a transmission network of
perhaps 275 kV. The per unit system provides a convenient means of analysing
such multi-voltage networks, because the entire network may be considered to be a
single voltage network with a nominal voltage of 1.0 pu. The actual voltages may
be obtained easily from the calculated per unit values and the base values at the
nodes of interest.
If the voltage base is changed in proportion to the nominal turns ratios of
transformers, an equivalent transformer-less network with a nominal voltage level
of 1.0 pu is created.

Example
The simple network in Figure 2.22 could be used to model a length of 132 kV line
of impedance j 100 W /phase supplying a 10 W load through a 132/11 kV step-down
transformer and a line of impedance j 0.50 W .
Suppose we wish to determine the impedance between A and B, perhaps to
assess the load on a generator connected between A and B. The conventional
approach is to refer the impedances to the high or low voltage side. Suppose we
refer the impedances to the 132 kV level. The impedances on the low voltage side
must be modified to dissipate the same power. Taking the step-down ratio as n , the
impedances referred to the high-voltage side will now experience a current 1/ n th