blob: a61e5ba94c23ab751d88f2e23db08b6702ebaa4b (
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
|
// 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 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS
// EXAMPLE : 1.7 :
// Page number 471-472
clear ; clc ; close ; // Clear the work space and console
// Given data
MVA_G1 = 50.0 // Generator rating(MVA)
kV_G1 = 15.0 // Voltage rating of generator(kV)
X_G1 = 0.2 // Reactance of generator(p.u)
MVA_G2 = 25.0 // Generator rating(MVA)
kV_G2 = 15.0 // Voltage rating of generator(kV)
X_G2 = 0.2 // Reactance of generator(p.u)
kV_T = 66.0 // Voltage rating of transformer(kV)
X_T = 0.1 // Reactance of transformer(p.u)
kV_fault = 66.0 // Voltage at fault occurence(kV)
kv_base = 69.0 // Base voltage(kV)
MVA_base = 100.0 // Base MVA
// Calculations
X_d_G1 = X_G1*MVA_base/MVA_G1 // Sub-transient reactance referred to 100 MVA(p.u)
E_G1 = kV_fault/kv_base // Voltage(p.u)
X_d_G2 = X_G2*MVA_base/MVA_G2 // Sub-transient reactance referred to 100 MVA(p.u)
E_G2 = kV_fault/kv_base // Voltage(p.u)
X_net = X_d_G1*X_d_G2/(X_d_G1+X_d_G2) // Net sub-transient reactance(p.u)
E_g = (E_G1+E_G2)/2 // Net voltage(p.u). NOTE: Not sure how this comes
I_fault = E_g/(%i*(X_net+X_T)) // Sub-transient fault current(p.u)
// Results
disp("PART III - EXAMPLE : 1.7 : SOLUTION :-")
printf("\nSub-transient fault current = %.3fj p.u \n", imag(I_fault))
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
|
