WIND ENERGY SYSTEMS - Cd3wd

Chapter 5—Electrical Network 5–26
E pu = V pu + I pu (jX s,pu )=1+0.693/33.69 o (1.38)/90 o
= 1+0.956/123.69 o =1− 0.530 + j0.796
= 0.470 + j0.796 = 0.924/59.46 o
The per unit generator inductance per phase is
The actual generator inductance per phase is
L pu = X s,pu
= 1.38 =1.38
ω pu 1
The per unit load capacitance per phase is
L = L pu L base =1.38(14.7) = 20.3 mH
C pu =
The actual load capacitance per phase is
1
= 1
X C,pu ω pu 0.8(1) =1.25
C = C pu C base =1.25(480) = 600 µF
The base of any device such as an electrical generator, motor, or transformer is always
understood to be the nameplate rating of the device. The per unit impedance is usually
available from the manufacturer.
Sometimes the base values need to be changed to a common base when several devices are
connected together. Solving an electrical circuit requires either the actual impedances or the
per unit impedances referred to a common base. The per unit impedance on the old base can
be converted to the per unit impedance for the new base by
Example
Z pu,new = Z pu,old
(
Vbase,old
V base,new
) 2
Sbase,new
S base,old
(46)
A single-phase distribution transformer secondary is rated at 60 Hz, 10 kVA, and 240 V. The open
circuit voltage V oc is 240 V. The per unit series impedance of the transformer is Z = 0.005 + j0.03.
Two electric heaters, one rated 1500 W and 230 V, and the other rated at 1000 W and 220 V, are
connected to the transformer. Find the per unit transformer current I pu and the magnitude of the
actual load voltage V 1 , as shown in Fig. 15.
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001