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authorprashantsinalkar2017-10-10 12:27:19 +0530
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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 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS
+
+// EXAMPLE : 1.12 :
+// Page number 473-474
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA_G = 625.0 // Generator rating(kVA)
+V_G = 2.4 // Voltage rating of generator(kV)
+X_st_G = 8.0/100 // Sub-transient reactance of generator
+rating_M = 250.0 // Motor rating(HP)
+V_M = 2.4 // Voltage rating of motor(kV)
+n = 90.0/100 // Efficiency of motor
+X_st_M = 20.0/100 // Sub-transient reactance of motor
+
+// Calculations
+kVA_base = 625.0 // Base kVA
+input_M = rating_M*0.746/n // Each motor input(kVA)
+X_st_m_pu = X_st_M*kVA_base/input_M // Sub-transient reactance of motor(p.u)
+I_base = kVA_base/(3**0.5*V_M) // Base current(A)
+Z_th = %i*X_st_m_pu/3*X_st_G/(X_st_m_pu/3+X_st_G) // Thevenin impedance(p.u)
+I_st = 1.0/Z_th // Initial symmetrical current at F(p.u)
+I_st_g = I_st*(X_st_m_pu/3/(X_st_m_pu/3+X_st_G)) // Fault current rating of generator breaker(p.u)
+I_st_m = (I_st-I_st_g)/3 // Fault current rating of each motor breaker(p.u)
+
+// Results
+disp("PART III - EXAMPLE : 1.12 : SOLUTION :-")
+printf("\nSub-transient fault current at F = %.2fj p.u", imag(I_st))
+printf("\nFault current rating of generator breaker = %.1fj p.u", imag(I_st_g))
+printf("\nFault current rating of each motor breaker = %.2fj p.u", imag(I_st_m))