blob: 404de3c5595879856d13c3403139bf47c53761bd (
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
35
36
37
38
39
40
41
42
43
44
|
// 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 2: FAULT LIMITING REACTORS
// EXAMPLE : 2.5 :
// Page number 481-482
clear ; clc ; close ; // Clear the work space and console
// Given data
kVA_base = 10000.0 // Base kVA
V = 6.6*10**3 // Voltage of generator(V)
X_A = 7.5 // Reactance of generator A(%)
X_B = 7.5 // Reactance of generator B(%)
X_C = 10.0 // Reactance of generator C(%)
X_D = 10.0 // Reactance of generator D(%)
X_E = 8.0 // Reactance of reactor E(%)
X_F = 8.0 // Reactance of reactor F(%)
X_G = 6.5 // Reactance of reactor G(%)
X_H = 6.5 // Reactance of reactor H(%)
// Calculations
Z_1 = X_B*X_C/(X_H+X_B+X_C) // Impedance(%). Fig E2.7
Z_2 = X_H*X_C/(X_H+X_B+X_C) // Impedance(%). Fig E2.7
Z_3 = X_B*X_H/(X_H+X_B+X_C) // Impedance(%). Fig E2.7
Z_4 = Z_2+X_F // Impedance(%). Fig E2.8 & Fig 2.9
Z_5 = Z_3+X_E // Impedance(%). Fig E2.8 & Fig 2.9
Z_6 = X_D*Z_1/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10
Z_7 = X_D*Z_4/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10
Z_8 = Z_1*Z_4/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10
Z_9 = Z_7+X_G // Impedance(%). Fig E2.11 & Fig 2.12
Z_10 = Z_8+Z_5 // Impedance(%). Fig E2.11 & Fig 2.12
Z_11 = Z_9*Z_10/(Z_9+Z_10) // Impedance(%). Fig 2.12 & Fig 2.13
Z_12 = Z_6+Z_11 // Impedance(%). Fig 2.13
Z_eq = X_A*Z_12/(X_A+Z_12) // Final Impedance(%). Fig 2.13 & Fig 2.14
MVA_SC = kVA_base*100/(Z_eq*1000) // Instantaneous symmetrical short-circuit MVA for a fault at X(MVA)
// Results
disp("PART III - EXAMPLE : 2.5 : SOLUTION :-")
printf("\nInstantaneous symmetrical short-circuit MVA for a fault at X = %.f MVA \n", MVA_SC)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook")
|
