blob: a2975e3420e5c0040dc25ae3697e7456ed3e16ee (
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
|
// 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 : 7.3 :
// Page number 212
clear ; clc ; close ; // Clear the work space and console
// Given data
V = 66.0*10**3 // Line Voltage(V)
l = 1.0 // Length of cable(km)
d = 15.0 // Core diameter(cm)
D = 60.0 // Sheath diameter(cm)
e_r = 3.6 // Relative permittivity
f = 50.0 // Frequency(Hz)
// Calculations
C = e_r/(18.0*log(D/d))*l // Capacitance(µF)
I_ch = V/3**0.5*2*%pi*f*C*10**-6 // Charging current(A)
// Results
disp("PART II - EXAMPLE : 7.3 : SOLUTION :-")
printf("\nCapacitance of single-core cable, C = %.3f µF", C)
printf("\nCharging current of single-core cable = %.2f A", I_ch)
|
