WIND ENERGY SYSTEMS - Cd3wd

Chapter 5—Electrical Network 5–27
Figure 15: Single-phase transformer connected to two resistive loads.
The first step is to get all impedance values computed on the same base. Any choice of base will
work, but minimum effort will be exerted if we choose the transformer base as the reference base. This
yields V base = 240 V and S base = 10 kVA. The per unit values of the electric heater resistances would
be unity on their nameplate ratings. The per unit values referred to the transformer rating would be
R 1,pu =(1)
( ) 2 230 10
240 1.5 =6.12
R 2,pu =(1)
( ) 2 220 10
240 1 =8.40
The equivalent impedance of these heaters in parallel would be
The per unit current is then
I pu =
The load voltage magnitude is
V oc,pu
Z pu + R pu
=
R pu = 6.12(8.40)
6.12 + 8.40 =3.54
1
=0.282/ − 0.48o
0.005 + j0.03 + 3.54
|V 1 | = I pu R pu V base =0.282(3.54)(240) = 239.6 V
The voltage V 1 has decreased only 0.4 V from the open circuit value for a current of 28.2 percent
of rated. This indicates the voltage varies very little with load changes, which is quite desirable for
transformer outputs.
5 THE INDUCTION MACHINE
A large fraction of all electrical power is consumed by induction motors. For power inputs of
less than 5 kW, these may be either single-phase or three-phase, while the larger machines are
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001