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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 II : TRANSMISSION AND DISTRIBUTION

+// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES

+

+// EXAMPLE : 11.4 :

+// Page number 332-333

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

+

+// Given data

+X = 10.0               // Reactance(ohm)

+I_a = 220.0            // Armature current(A)

+PF = 1.0               // Unity power factor

+V = 11000.0            // Phase voltage(V)

+emf_raised = 0.2       // EMF rasied by 20%

+

+// Calculations

+I_X = I_a*X                        // Reactance drop(V)

+E_0 = (V**2+I_X**2)**0.5           // EMF(V)

+E_00 = (1+emf_raised)*E_0          // New value of induced emf(V)

+U = ((E_00**2-I_X**2)**0.5-V)/X    // Current(A)

+I_1 = (I_a**2+U**2)**0.5           // Current(A)

+PF_1 = I_a/I_1                     // Lagging power factor

+I_X_2 = (E_00**2+V**2)**0.5        // Reactance drop(V)

+I_2 = I_X_2/X                      // Current corresponding to this drop(A)

+PF_2 = E_00/I_X_2                  // Leading power factor

+P_max = V*I_2*PF_2/1000            // Maximum power output(kW)

+

+// Results

+disp("PART II - EXAMPLE : 11.4 : SOLUTION :-")

+printf("\nNew value of machine current = %.1f A", I_1)

+printf("\nNew vaue of power factor, p.f = %.4f (lagging)", PF_1)

+printf("\nPower output at which alternator break from synchronism = %.f kW", P_max)

+printf("\nCurrent corresponding to maximum load = %.f A", I_2)

+printf("\nPower factor corresponding to maximum load = %.4f (leading) \n", PF_2)

+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")