6.14 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending end is 220 kV. The parameters of the line are R = 0.2 /mi, X = 0.8 N/mi = and Y 5.3 μS/mi. Find the sending-end current when there is no load on the line. Solution: Z = Y = yl (0.2+0.8) x 250 = 206.1/75.96° 250 x 5.3 x 10-6 = 1.325 × 10-³/90° √√ZY = √206.1 × 1.325 × 10−3³ / 165.96° = 0.5226/82.98° = 0.0639+0.5187 206.175.96° Zc = √Z/Y = = 394-7.02° N 1.325 x 10-3/90° By Eq. (6.39) for IR = 0, sinh nh Is == (Vs/Zc) coshyl 0.5187 rad = 29.72° Bl cal €31 == €-al-JBL = coshy! = 0.9258+0.5285 0.8147j0.4651 1 (0.9258+0.8147 +0.5285j0.4651) = 0.8709/2.086° sinh === Is = 394-7.02° 10.9258 -0.8147 + j (0.5285 +0.4651)] = 0.4999/83.61° 220,000/√3 0.4999/83.61° = 185.0/88.54° A 0.8709/2.086° GIVEN THE SOLUTION ABOVE I DID EVERYTHING RIGHT EXCEPT WHEN FINDING THE SENDING-END CURRENT. I USED THIS FORMULA BELOW: Is = Ir*cosh(yl)+(Vr/Zc)* senh(yl) Is=0+ 127017|0° 394.447-7,02° Is = 161.006|90.65° *0,583.68° ALSO, THE BOOK IN QUESTION "POWER SYSTEM ANALYSIS” BY WILLIAM D. STEVENSON GIVES THOSE TWO FORMULAS. BUT SHOULDN'T THEY GIVE THE SAME RESULT? Vs IR = 15 cosh y - sinh yl Z c (6.39) V R Is IR cosh y/ + sinh yl (6.36) Z PAGE 208
6.14 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending end is 220 kV. The parameters of the line are R = 0.2 /mi, X = 0.8 N/mi = and Y 5.3 μS/mi. Find the sending-end current when there is no load on the line. Solution: Z = Y = yl (0.2+0.8) x 250 = 206.1/75.96° 250 x 5.3 x 10-6 = 1.325 × 10-³/90° √√ZY = √206.1 × 1.325 × 10−3³ / 165.96° = 0.5226/82.98° = 0.0639+0.5187 206.175.96° Zc = √Z/Y = = 394-7.02° N 1.325 x 10-3/90° By Eq. (6.39) for IR = 0, sinh nh Is == (Vs/Zc) coshyl 0.5187 rad = 29.72° Bl cal €31 == €-al-JBL = coshy! = 0.9258+0.5285 0.8147j0.4651 1 (0.9258+0.8147 +0.5285j0.4651) = 0.8709/2.086° sinh === Is = 394-7.02° 10.9258 -0.8147 + j (0.5285 +0.4651)] = 0.4999/83.61° 220,000/√3 0.4999/83.61° = 185.0/88.54° A 0.8709/2.086° GIVEN THE SOLUTION ABOVE I DID EVERYTHING RIGHT EXCEPT WHEN FINDING THE SENDING-END CURRENT. I USED THIS FORMULA BELOW: Is = Ir*cosh(yl)+(Vr/Zc)* senh(yl) Is=0+ 127017|0° 394.447-7,02° Is = 161.006|90.65° *0,583.68° ALSO, THE BOOK IN QUESTION "POWER SYSTEM ANALYSIS” BY WILLIAM D. STEVENSON GIVES THOSE TWO FORMULAS. BUT SHOULDN'T THEY GIVE THE SAME RESULT? Vs IR = 15 cosh y - sinh yl Z c (6.39) V R Is IR cosh y/ + sinh yl (6.36) Z PAGE 208
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter4: Transmission Line Parameters
Section: Chapter Questions
Problem 4.32MCQ
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Question
Given this solution below explain why answer was incorrect. I couldn't find anything that explains it.
![6.14 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending
end is 220 kV. The parameters of the line are R = 0.2 /mi, X = 0.8 N/mi
=
and Y 5.3 μS/mi. Find the sending-end current when there is no load on the
line.
Solution:
Z
=
Y
=
yl
(0.2+0.8) x 250 = 206.1/75.96°
250 x 5.3 x 10-6 = 1.325 × 10-³/90°
√√ZY = √206.1 × 1.325 × 10−3³ / 165.96° = 0.5226/82.98°
= 0.0639+0.5187
206.175.96°
Zc =
√Z/Y
=
= 394-7.02° N
1.325 x 10-3/90°
By Eq. (6.39) for IR = 0,
sinh nh
Is ==
(Vs/Zc)
coshyl
0.5187 rad = 29.72°
Bl
cal €31 ==
€-al-JBL =
coshy! =
0.9258+0.5285
0.8147j0.4651
1
(0.9258+0.8147 +0.5285j0.4651) = 0.8709/2.086°
sinh
===
Is
= 394-7.02°
10.9258 -0.8147 + j (0.5285 +0.4651)] = 0.4999/83.61°
220,000/√3 0.4999/83.61°
= 185.0/88.54° A
0.8709/2.086°
GIVEN THE SOLUTION ABOVE I DID EVERYTHING RIGHT EXCEPT WHEN FINDING THE SENDING-END CURRENT.
I USED THIS FORMULA BELOW:
Is = Ir*cosh(yl)+(Vr/Zc)* senh(yl)
Is=0+
127017|0°
394.447-7,02°
Is = 161.006|90.65°
*0,583.68°
ALSO, THE BOOK IN QUESTION "POWER SYSTEM ANALYSIS” BY WILLIAM D. STEVENSON GIVES THOSE
TWO FORMULAS. BUT SHOULDN'T THEY GIVE THE SAME RESULT?
Vs
IR = 15 cosh y
-
sinh yl
Z c
(6.39)
V
R
Is
IR cosh y/ +
sinh yl
(6.36)
Z
PAGE 208](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cdc93ce-e2c1-450b-a108-334f1039b730%2F06ef57dc-56b1-4aac-98c5-c19166874bcb%2Fbf104e_processed.png&w=3840&q=75)
Transcribed Image Text:6.14 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending
end is 220 kV. The parameters of the line are R = 0.2 /mi, X = 0.8 N/mi
=
and Y 5.3 μS/mi. Find the sending-end current when there is no load on the
line.
Solution:
Z
=
Y
=
yl
(0.2+0.8) x 250 = 206.1/75.96°
250 x 5.3 x 10-6 = 1.325 × 10-³/90°
√√ZY = √206.1 × 1.325 × 10−3³ / 165.96° = 0.5226/82.98°
= 0.0639+0.5187
206.175.96°
Zc =
√Z/Y
=
= 394-7.02° N
1.325 x 10-3/90°
By Eq. (6.39) for IR = 0,
sinh nh
Is ==
(Vs/Zc)
coshyl
0.5187 rad = 29.72°
Bl
cal €31 ==
€-al-JBL =
coshy! =
0.9258+0.5285
0.8147j0.4651
1
(0.9258+0.8147 +0.5285j0.4651) = 0.8709/2.086°
sinh
===
Is
= 394-7.02°
10.9258 -0.8147 + j (0.5285 +0.4651)] = 0.4999/83.61°
220,000/√3 0.4999/83.61°
= 185.0/88.54° A
0.8709/2.086°
GIVEN THE SOLUTION ABOVE I DID EVERYTHING RIGHT EXCEPT WHEN FINDING THE SENDING-END CURRENT.
I USED THIS FORMULA BELOW:
Is = Ir*cosh(yl)+(Vr/Zc)* senh(yl)
Is=0+
127017|0°
394.447-7,02°
Is = 161.006|90.65°
*0,583.68°
ALSO, THE BOOK IN QUESTION "POWER SYSTEM ANALYSIS” BY WILLIAM D. STEVENSON GIVES THOSE
TWO FORMULAS. BUT SHOULDN'T THEY GIVE THE SAME RESULT?
Vs
IR = 15 cosh y
-
sinh yl
Z c
(6.39)
V
R
Is
IR cosh y/ +
sinh yl
(6.36)
Z
PAGE 208
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