Page 99 - Electric Machinery Fundamentals

TRANSFORMERS 75

and the line losses are

Ploss = (llincr R 1ine
~ (90S A)' (0 1S!l) ~ 14S4 W
(b) Figure 2-6b shows the power system with the transformers. To analyze this sys-
tem, it is necessary to convert it to a common voltage level. This is done in two
steps:

1. Eliminale transformer T2 by refelTing the load over to the transmission line's
voltage leveL
2. Eliminate transfonner Tl by referring the transmission line's elements and
the equivalent load at the transmission line's voltage over to the source side.

The value of the load's impedance when reflected to the transmission system's
voltage is
(
2
Ziood = a Zload
~ C?)'(4!i + j3 fi)
~ 400 fi + j300 fi
The total impedance at the transmission line level is now

~ 400.1 S + j300.24fi = 500.3 L36.S8° fi
This equivalent circuit is shown in Figure 2-7a. The total impedance at (he transmission
line level (Zlinc + 2 1 00<1) is now reflected across TI (0 the source's voltage level:
2
Z~q = a Zeq
= a2CZjinc + Z load)
~ (/0)'(0.1S fi + jO.24 ii + 400 fi + j300 fi)

= (O.OOIS fi + j O.OO24 fi + 4 ii + j3 fi)
= 5.003 L36.88° fi

Notice that Z'1'oad = 4 + )3 nand Ziinc = 0.00 18 + )0.0024 n. The resulting equivalent cir-
cuit is shown in Figure 2-7b. The generator's Cl.1lTent is
_ 4S0LOoV _ °
Ia - 5.003 L36.S80 fi - 95.94L-36.SS A

Knowing the current la. we can now work back and find I line and I 1oad ' Working back
through T , we get
1

_ Np ! I
Iline - N SI G

= 1~(95.94L-36.88° A) ~ 9.594 L-36.8So A