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// 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 : 8.6 :
// Page number 626
clear ; clc ; close ; // Clear the work space and console
// Given data
MVA = 50.0 // Alternator rating(MVA)
kV = 11.0 // Alternator voltage(kV)
X = 2.0 // Synchronous reactance per phase(ohm)
R = 0.7 // Resistance per phase(ohm)
R_n = 5.0 // Resistance through which alternator is earthed(ohm)
ofb = 25.0 // Out-of-balance current(%)
// Calculations
I_fl = MVA*1000/(3**0.5*kV) // Full load current(A)
I_ofb = ofb/100*I_fl // Out-of-balance current(A)
x = R_n/((kV*1000/(3**0.5*100*I_ofb))-(R/100)) // Portion of winding unprotected(%)
// Results
disp("PART III - EXAMPLE : 8.6 : SOLUTION :-")
printf("\nPortion of winding unprotected, x = %.f percent", x)
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