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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 : 1.11 :

+// Page number 472-473

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+X_d_st_G = 0.15      // Sub-transient reactance of generator(p.u)

+X_d_st_M = 0.45      // Sub-transient reactance of motor(p.u)

+X = 0.10             // Leakage reactance of transformer(p.u)

+V = 0.9              // Terminal voltage of the generator(p.u)

+I_G = 1.0            // Output current of the generator(p.u)

+PF = 0.8             // Power factor of the load

+

+// Calculations

+sin_phi = (1-PF**2)**0.5

+I = I_G*(PF+%i*sin_phi)            // Load current(p.u)

+E_st_G = V+%i*I*X_d_st_G           // Sub-transient voltage of the generator(p.u)

+E_st_M = V-%i*I*X_d_st_M           // Sub-transient voltage of the motor(p.u)

+I_st_g = E_st_G/(%i*(X_d_st_G+X))  // Sub-transient current in the generator at fault(p.u)

+I_st_m = E_st_M/(%i*(X_d_st_M-X))  // Sub-transient current in the motor at fault(p.u)

+

+// Results

+disp("PART III - EXAMPLE : 1.11 : SOLUTION :-")

+printf("\nCase(a): Sub-transient current in the fault in generator = %.3f∠%.3f° p.u", abs(I_st_g),phasemag(I_st_g))

+printf("\nCase(b): Sub-transient current in the fault in motor = %.3f∠%.2f° p.u \n", abs(I_st_m),180+phasemag(I_st_m))

+printf("\nNOTE: ERROR: Sub-transient reactance of motor is 0.45 p.u & not 0.35 p.u as mentioned in textbook statement")