1a. Problem 4.1 on page 117. Given a 138 kV three phase line iwth series impedance z=0.17+j0.79 ohms/mi and shunt admittance y=j5.4 uSiemens/mi, find the characteristic impedance Zc, the propagation constant, and attenuation constant and the phase constant. Given z
( 0.17
0.79j ) .
ohm
6 siemens 5.4j . 10 . mi
y
mi propagation constant
γ
z .y
γ = 2.21 . 10
4
γ = 1.373 . 10
2.077i . 10
3
1.291i . 10
7
mi
1
6
m
1
γ = 2.09 . 10
3
γ = 1.3 . 10
m
6
mi 1
1
arg( γ ) = 83.9 deg arg( γ ) = 83.9 deg
attenuation constant
α
α = 1.373 . 10
Re( γ )
7
m
α = 2.21 . 10
1
4
mi
1
phase constant
β
β = 1.291 . 10
Im( γ )
1
β = 2.077 . 10
40.92i Ω
Z c = 386.8 Ω
6
m
3
mi
1
Characteristic impedance
Zc
z y
Z c = 384.67
arg Z c = 6.1 deg
1b. Problem 4.5 on page 117. Given a transmission line described by a total series impedance Z=zl=20+j80 and a shunt admittance Y=yl=0.0005, find Zc, gamma*l,e^gamma*l, sinh(gamma*l), cosh(gamma*l)
State the given Z
( 20
80j ) . ohm
Y
4 5j . 10 . siemens
Characteristic impedance Z
Zc
Z c = 403.066
Y
49.62i Ω
Z c = 406.1 Ω
arg Z c = 7 deg
γl = 0.203
arg( γl ) = 83 deg
propagation constant * length Z .Y
γl γl
e = 1.004
γl = 0.025
0.202i
0.205i
sinh( γl ) = 0.024 cosh( γl ) = 0.98
e
0.2i 4.967i . 10
3
γl
= 1.025
arg e
γl
= 11.5 deg
sinh( γl ) = 0.202
arg( sinh( γl ) ) = 83.1 deg
cosh( γl ) = 0.98
arg( cosh( γl ) ) = 0.29 deg
1c. Problem 4.18 on page 118. Find the ABCD constants of the 150 mile line in Problem 4.8. The line is the same as in problem 4.1, but the length is now given as 150 miles. Recall z
( 0.17
0.79j ) .
ohm
6 siemens 5.4j . 10 . mi
y
mi propagation constant
γ
z .y
γ = 2.21 . 10
4
γ = 1.373 . 10
2.077i . 10
3
1.291i . 10
7
mi
1
6
m
1
γ = 2.09 . 10
3
γ = 1.3 . 10
m
6
mi 1
1
arg( γ ) = 83.9 deg arg( γ ) = 83.9 deg
attenuation constant α
α = 1.373 . 10
Re( γ )
7
m
α = 2.21 . 10
1
4
mi
1
phase constant β
Im( γ )
β = 1.291 . 10
1
β = 2.077 . 10
40.92i Ω
Z c = 386.8 Ω
6
m
3
mi
1
Characteristic impedance Zc
z y
Z c = 384.67
Length is 150 miles γl
γ .l
l
150 . mi
γl = 0.033
0.312i
arg Z c = 6.1 deg
ABCD Constants A
cosh( γl )
A = 0.952
0.01i
A = 0.952
arg( A ) = 0.6 deg
B
Z c . sinh( γl )
B = 24.69
116.7i Ω
B = 119.3 Ω
arg( B ) = 78.1 deg
C
1 . sinh( γl ) Zc
C = 2.762 . 10
6
7.971i . 10
4
S
C = 7.971 . 10
D
cosh( γl )
C = 2.762 . 10
6
7.971i . 10
4
S
D = 0.952
4
S
arg( C ) = 90.2 deg arg( D ) = 0.6 deg
2a. Problem 4.3 on page 117. Given a transmission line described by (4.10), we perform two tests and obtain the following results. Open circuit test: Zoc=(800 at -89 deg) ohms Short circuit test: Zsc=(200 at 77 deg) ohms Find the characteristic impedance Zc and gamma*l. Z oc
B D cosh( γl ) . Z c . sinh( γl ) 1 . sinh( γl ) . cosh( γl ) Zc
Z oc . Z sc Z c = 400 Ω
arg Z c = 6 deg
γl
0
1
Find( γl ) = 0.049
0.462i
Problem 2b. Problem 4.4 on page 117. Jack and Jill measure Zoc and Zsc as in problem 4.3. Jill says, "With this information, we can now calculate V1 and I1 in terms of V2 and I2 for any termination." Agree or disagree wiht Jill and give your reasons. Agree. Zc and gamma*l specify the general relation between (V1 and I1) and (V2 and I2). If they have Zoc and Zsc, they can find Zc and gamma * l, as we did in problem 4.3. Hence, we can find the relationship between V1 and I1 and any termination.
Problem 2c. Problem 4.6 on page 117. Suppose that the line in problem 4.5 is terminated in it characteristic impedance. Find the efficiency of transmission in this case. Given Z
( 20
j . 80 ) . ohm
Z
Zc
Y
Z c = 403.066
Y
4 j . 5 . 10 . siemens
49.62i Ω
γl
Y.Z
γl = 0.025
0.202i
Find the ABCD parameters. A
cosh( γl )
B
Z c . sinh( γl )
C
we are given V2=Zc*I2 Therefore, V 1 A.V 2
B .I 2
V 1 A.V 2
B.
V2 Zc
I 1 C .V 2
I 1 C .Z c .I 2
Substituting into an expression for power V 1.I 1
A
Efficiency is
η
Re V 2 . I 2 Re V 1 . I 1
D.I 2
B . V 2. C .Z c Zc
D .I 2
D.I 2
1 . sinh( γl ) Zc
D
cosh( γl )
1
η Re A
B . . C Zc
Zc
Solving
η = 0.952
D
Problem 3a. Problem 4.8 on page 117. The 138 kV three phase line described in problem 4.1 is 150 miles long and is delivering 15 MW at 132 kV at a 100% power factor. Find the sending end voltage and current, the power angle across the line, and thre transmission efficiency. Use the long line model. j
Given in problem 4.1 z
0.79j ) .
( 0.17
ohm
y
mi
6 siemens 5.4j . 10 . mi
l
1
150 . mi
propagation constant γl
z .y .l
γl = 0.033
0.312i
Characteristic impedance
z
Zc
Z c = 384.7
y
40.9i Ω
Receiving end 132 .
V2
kV
pf
15 . MW 3 . V 2 . pf
I2
3
1.0
MW
6 10 . watt
Find the ABCD parameters. cosh( γl )
A
B
Z c . sinh( γl )
C
1 . sinh( γl ) Zc
D
cosh( γl )
Transmission line equations V1 I1
A .V 2 C .V 2
B .I 2 D .I 2
Solving V 1 = 7.42 . 10 I 1 = 62.273
4
8.431i . 10 V 3
61.414i A
V 1 = 74.7 kV
arg V 1 = 6.48 deg arg I 1 = 44.6 deg
I 1 = 87.5 A
Efficiency and angle
η
Re V 2 . I 2 Re V 1 . I 1
η = 0.973
θ 12
arg V 1
arg V 2
θ 12 = 6.48 deg
Problem 3b . Problem 4.12 on page 118. Repeat problem 4.8using the nominal pi circuit. j
Given in problem 4.1 z
0.79j ) .
( 0.17
ohm
y
mi
6 siemens 5.4j . 10 . mi
150 . mi
l
Receiving end 132 .
V2
kV
pf
15 . MW 3 . V 2 . pf
I2
3
1.0
MW
6 10 . watt
Find series Z and shunt Y for pi model z . l Z = 25.5
Z
118.5i Ω
Y
y .l
C
Y. 1
Y = 8.1i . 10
4
S
Find ABCD parameters A
Z .Y
1
B
Z
Z .Y
2
D
1
4
Z .Y 2
Transmission line equations A .V 2
V1
C .V 2
I1
B .I 2 D .I 2
Solving V 1 = 7.423 . 10 I 1 = 62.141
4
8.562i . 10 V 3
V 1 = 74.7 kV
arg V 1 = 6.58 deg arg I 1 = 44.4 deg
I 1 = 87 A
60.927i A
Efficiency and angle
η
Re V 2 . I 2 Re V 1 . I 1
η = 0.974
θ 12
arg V 1
θ 12 = 6.58 deg
arg V 2
Problem 3c. Problem 4.13 on page 118. Repeat Problem 4.8 for a short line. Recall the given V2
