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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 10: POWER SYSTEM STABILITY
// EXAMPLE : 10.21 :
// Page number 305-306
clear ; clc ; close ; // Clear the work space and console
// Given data
f = 50.0 // Frequency(Hz)
P = 4.0 // Number of poles
G = 20.0 // Rating of turbo-generator(MVA)
V = 13.2 // Voltage(kV)
H = 9.0 // Inertia constant(kW-sec/kVA)
P_s = 20.0 // Input power less rotational loss(MW)
P_e = 15.0 // Output power(MW)
// Calculations
KE = G*H // Kinetic energy stored(MJ)
M = G*H/(180*f) // Angular momentum(MJ-sec/elect.degree)
P_a = P_s-P_e // Accelerating power(MW)
alpha = P_a/M // Acceleration(elect.degree/sec^2)
alpha_deg = alpha/2.0 // Acceleration(degree/sec^2)
alpha_rpm = 60.0*alpha_deg/360 // Acceleration(rpm/sec)
// Results
disp("PART II - EXAMPLE : 10.21 : SOLUTION :-")
printf("\nCase(a): Kinetic energy stored by rotor at synchronous speed, GH = %.f MJ", KE)
printf("\nCase(b): Acceleration, α = %.f degree/sec^2", alpha_deg)
printf("\n Acceleration, α = %.2f rpm/sec", alpha_rpm)
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