blob: 561a59f5d5c80c1c598d4c6eaff30d22639537d4 (
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
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART III : SWITCHGEAR AND PROTECTION
// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS
// EXAMPLE : 8.8 :
// Page number 627
clear ; clc ; close ; // Clear the work space and console
// Given data
kV = 11.0 // Alternator voltage(kV)
P = 100.0 // Alternator maximum rating(MW)
PF = 0.8 // Power factor
X = 0.1 // Reactance of alternator(pu)
i = 500.0 // Current(A)
per = 10.0 // Windings unprotected(%)
// Calculations
I = P*1000/(3**0.5*kV*PF) // Rated current of alternator(A)
a = i/I // Relay setting
I_n = a*I*100/per // Current through neutral(A)
R = kV*1000/(3**0.5*I_n) // Magnitude of neutral earthing resistance(ohm)
// Results
disp("PART III - EXAMPLE : 8.8 : SOLUTION :-")
printf("\nMagnitude of neutral earthing resistance, R = %.2f ohm\n", R)
printf("\nNOTE: ERROR: Unit of resistance is not mentioned in textbook solution")
|