Module T1
Solution to problem 1
Two ways to work it.
1.
Find
pq
P
and
'
pq
P
then
'
pq
pq
loss
P
P
P

=
where
'
pq
P
=
qp
P

2.
Find
pq
I
and then
R
I
P
pq
loss
2
3
=
We will do #2.
(
29
(
29
MW
MW
P
j
jX
R
V
V
I
loss
q
q
p
p
pq
3
61
.
0
183
.
0
30
45
3
26
.
27
45
150
30
0
06
.
65
0
.
6
51
.
63
10
2
3
×
=
=
=
°
∠
=
+
°
∠

°
∠
×
=
+
∠

∠
=
θ
#1:
666
.
2
=
pq
P
605
.
2
'
=
pq
P
061
.
0
/
=
φ
loss
P
Solution to problem 2
(
29
(
29
(
29
[
]
[
]
MW
B
V
V
P
q
p
q
p
pq
885
.
2
10472
.
0
0067
.
0
10
06
.
65
10
51
.
63
3
3
=
⋅
⋅
×
⋅
×
=

⋅
⋅
⋅
=
(
29
(
29
[
]
[
]
MVAR
V
V
B
V
Q
q
p
p
pq
6566
.
0
10
06
.
65
51
.
63
0067
.
0
10
51
.
63
3
3

=
×

⋅
⋅
×
=

⋅
⋅
=
(
29
(
29
MVAR
B
V
Q
c
p
sp
4034
.
0
0001
.
0
10
51
.
63
2
2
3
2

=
⋅
×

=
⋅

=
So,
MVAR
Q
Q
Q
sp
pq
06
.
1
4034
.
0
6566
.
0

=


=
+
=
MVAR
Q
MW
P
18
.
3
06
.
1
3
,
655
.
8
885
.
2
3

=
×
=
=
×
=
⇒
p q
p q
Q
P
,
Q
P
,
s p
Q
P
Q
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentSolution to problem 3
80
.
19
98
.
1
1
j
jX
R
jB
G

=
+
=

(a)
P
pq
=
(
29
(
29
q
p
q
p
q
p
q
p
p
B
V
V
G
V
V
G
V
φ

+


sin
cos
2
=
)
10
sin(
)
80
.
19
)(
95
.
0
(
15
.
1
(
)
10
cos(
)
98
.
1
)(
95
.
0
)(
15
.
1
(
)
98
.
1
(
15
.
1
2
+

=
4.2445
So real power into line = 100(4.245) =
424.5 MW
P'
pq
=P
pq
=
)]
sin(
)
cos(
[
2
p
q
p
q
p
q
p
q
q
B
V
V
G
V
V
G
V

+



=
)]
10
sin(
)
8
.
19
)(
95
.
0
)(
15
.
1
(
)
10
cos(
)
98
.
1
)(
15
.
1
)(
95
.
0
(
)
98
.
1
(
95
[.
2

+



=
4.0996
So real power out of the line = 100(4.0996) =
409.96 MW
(b) P
loss
= P
pq
P'
pq
=424.5409.96=
14.49 MW
(c)
F = (0.03 / kwhr)(14.49MW)(1000kW/MW) =
$434.70 / hr
(d)
5
.
424
96
.
409
=
=
in
out
P
P
γ
=
96.6 %
Solution to problem 4
(a) What are the typical voltage levels for bulk transmission lines?
138, 101, 230, 345, 500, 765
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 mccalley
 Electric power transmission, Electric power, Transmission line, Impedance matching, Ppq

Click to edit the document details