EE 295  Homework 3. Solutions
March 3, 2010
1
Readings
1. Chapters 5, 6 and 7 of Bergen & Vittal
2
From the textbook (Bergen/Vittal)
2.1
Problem 5.5
In this problem the feeder is a long medium voltage line (24kV), with
Z
F
= 50+
j
400
feeding a step
down transformer, which in turn feeds a 2.4kV load. The transformer impedance is
Z
T
(
LV
)
= 0
.
2+
j
1
(on the load voltage side). The load consumes 200kW at 0.9 lagging power factor.
(a)
Find the sending end voltage
V
s
:
V
s
=
V
R
+
IZ
F
=
nV
L
+
I
(
Z
F
+
Z
T
(
HV
)
)
The problem is that we are given
Z
T
(
LV
)
not
Z
T
(
HV
)
.
Therefore it is neccessary to do a
conversion to get the transformer impedance reflected on the high voltage side, which we can
do from the equations on pp.130131 in the book:
Z
T
(
HV
)
=
Z
T
(
LV
)
1
n
2
= 20 +
j
100
Now we can calculate the current in the circuit (reflected to the primary side) and solve:
I
=
P
(
nV
L
)0
.
9
∠

cos

1
0
.
9
=
9
.
66
∠

25
.
8
◦
V
s
=
nV
L
+
I
(
Z
F
+
Z
T
(
HV
)
) = 26
.
0
.
3
k
∠
8
.
96
◦
(b)
Find
V
R
.
V
R
=
nV
L
+
IZ
T
(
HV
)
= 23
.
6
kV
∠
1
.
9
◦
(c)
Find the total transmission efficiency (including the transformer)
P
LOSS
=

I

2
R
= 9
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 gully
 Alternating Current, Impedance matching, zt

Click to edit the document details