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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 : 10.12 :
// Page number 303-304
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 generator(MVA)
H = 9.0 // Inertia constant(kWsec/MVA)
P_m = 26800.0 // Rotational loss(hp)
P_e = 16000.0 // Electric power developed(kW)
// Calculations
GH = G*H // Energy stored in rotor at synchronous speed(MJ)
P_m_kW = P_m*0.746 // Rotational loss(kW)
P_a = P_m_kW-P_e // Acceleration power(kW)
P_a1 = P_a/1000.0 // Acceleration power(MW)
M = GH/(180*f) // Angular momentum
acceleration = P_a1/M // Acceleration(°/sec^2)
acceleration_1 = acceleration*%pi/180.0 // Acceleration(rad/sec^2)
// Results
disp("PART II - EXAMPLE : 10.12 : SOLUTION :-")
printf("\nKinetic energy stored in the rotor at synchronous speed, GH = %.f MJ", GH)
printf("\nAcceleration = %.f°/sec^2 = %.2f rad/sec^2 \n", acceleration,acceleration_1)
printf("\nNOTE: ERROR: H = 9 kW-sec/MVA, not 9 kW-sec/kVA as mentioned in the textbook statement")
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