blob: b736b34c051ce127bff0a875139c733e2e3656c7 (
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
34
|
// 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 18: POWER DISTRIBUTION SYSTEMS
// EXAMPLE : 18.2 :
// Page number 437-438
clear ; clc ; close ; // Clear the work space and console
// Given data
I = 600.0 // Constant current drawn(A)
D = 8.0 // Distance b/w two sub-stations(km)
V_A = 575.0 // Potential at point A(V)
V_B = 590.0 // Potential at point B(V)
R = 0.04 // Track resistance(ohm/km)
// Calculations
x = poly(0,'x') // x(km)
I_A = ((-V_B+R*I*D+V_A)-(R*I)*x)/(D*R) // Simplifying
V_P = V_A-I_A*R*x // Potential at P in terms of x(V)
dVP_dx = derivat(V_P) // dV_P/dx
x_sol = roots(dVP_dx) // Value of x(km)
I_A_1 = ((-V_B+R*I*D+V_A)-(R*I)*x_sol)/(D*R) // Current drawn from end A(A)
I_B = I-I_A_1 // Current drawn from end B(A)
// Results
disp("PART II - EXAMPLE : 18.2 : SOLUTION :-")
printf("\nPoint of minimum potential along the track, x = %.2f km", x_sol)
printf("\nCurrent supplied by station A, I_A = %.f A", I_A_1)
printf("\nCurrent supplied by station B, I_B = %.f A \n", I_B)
printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")
|
