blob: 62d9e231e4276d9502e1335dccac8fa477201dd2 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART II : TRANSMISSION AND DISTRIBUTION
// CHAPTER 7: UNDERGROUND CABLES
// EXAMPLE : 7.14 :
// Page number 219-220
clear ; clc ; close ; // Clear the work space and console
// Given data
c_s = 0.90 // Capacitance b/w all conductors(µF)
C_0 = 0.4 // Capacitance b/w two conductor(µF)
V = 11.0*10**3 // Line Voltage(V)
f = 50.0 // Frequency(Hz)
// Calculations
C_s = c_s/3.0 // Capacitance measured(µF)
C_c = (C_0-C_s)/2.0 // Capacitance(µF)
C_a = 3.0/2*(C_c+(1/3.0)*C_s) // Capacitance b/w any two conductors(µF)
C_b = 2.0*C_c+(2.0/3)*C_s // Capacitance b/w any two bounded conductors and the third conductor(µF)
C_o = 3.0*C_c+C_s // Capacitance to neutral(µF)
I_c = 2.0*%pi*f*C_o*V/3**0.5*10**-6 // Charging current(A)
// Results
disp("PART II - EXAMPLE : 7.14 : SOLUTION :-")
printf("\nCase(a): Capacitance between any two conductors = %.3f µF", C_a)
printf("\nCase(b): Capacitance between any two bounded conductors and the third conductor = %.1f µF", C_b)
printf("\nCase(c): Capacitance to neutral, C_0 = %.2f µF", C_o)
printf("\n Charging current taken by cable, I_c = %.3f A \n", I_c)
printf("\nNOTE: ERROR: Calculation mistakes in textbook answer")
|
