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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)