A three-phase 60-hz transmission line has its conductors arranged in a triangular formation so that two of the distances between conductors are 20ft and the third is 40 ft. The conductors are ACSR Rook. Determine the capacitance to neutral in microfarads per mile and the capacitive reactance to neutral in ohm-miles. If the line is 115 mi long, find the capacitance to neutral and capacitive reactance of the line.

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A three-phase 60-hz transmission line has its conductors arranged in a triangular formation so that two of the distances between conductors are 20ft and the third is 40 ft. The conductors are ACSR Rook. Determine the capacitance to neutral in microfarads per mile and the capacitive reactance to neutral in ohm-miles. If the line is 115 mi long, find the capacitance to neutral and capacitive reactance of the line.

Note: In the below i have added similer type of math and its ans.In the above math, there is just changes of values and Main Change is ACSR ROOK which is given to then question and in the below question which is solved it was ask ACSR Parakeet. So notice that.

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AND DESIGN;FIFTH EDITION, SI;J. DUNCAN GLOVER"

Now solve the math according to the given way solve the math.

"A three-phase 60-hz transmission line has its conductors arranged in a triangular formation so that two of the distances between conductors are 25 ft and the third is 42 ft. The conductors are ACSR Parakeet. Determine the capacitance to neutral in microfarads per mile and the capacitive reactance to neutral in ohm-miles. If the line is 150 mi long, find the capacitance to neutral and capacitive reactance of the line."--- This math solves are given in the picture.

 

 

Freom the ta ble A.3:
The outer diameten of parraKeet is: 0.914 in.
oe, 0.038ft.
So,
the oquilitrial spacing is: Dea =2/25x25X42:
%3D
ニ 29.719ft.
The capacitane to neutral:
27 X 8.85X112
%3D
27 xK
: Free spaee
permitivity , Ko=
8.85X10-12
'1mi = 1609 m
Cn=
InDa)
129.72x12
In0.914/2
27メ8.85×10
6.6597
8.25 x10-12 F/m
= 8.35 x (612x10 x 1609 uf/mi
= 0.01 3435 uf/mi. (Am:)
.'. Cm
A Capacitive Reactanee to eutral:
106
Xe =
2万×60×0.01344
= 197360.11 2.mi
ニ o.1973×10°2.mi。
(Am?)
A If the line is 150 mi long, thom:
long, thon:
Cn= (150x 0.013435)= 2.015 uFAm:
Transcribed Image Text:Freom the ta ble A.3: The outer diameten of parraKeet is: 0.914 in. oe, 0.038ft. So, the oquilitrial spacing is: Dea =2/25x25X42: %3D ニ 29.719ft. The capacitane to neutral: 27 X 8.85X112 %3D 27 xK : Free spaee permitivity , Ko= 8.85X10-12 '1mi = 1609 m Cn= InDa) 129.72x12 In0.914/2 27メ8.85×10 6.6597 8.25 x10-12 F/m = 8.35 x (612x10 x 1609 uf/mi = 0.01 3435 uf/mi. (Am:) .'. Cm A Capacitive Reactanee to eutral: 106 Xe = 2万×60×0.01344 = 197360.11 2.mi ニ o.1973×10°2.mi。 (Am?) A If the line is 150 mi long, thom: long, thon: Cn= (150x 0.013435)= 2.015 uFAm:
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