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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 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES
// EXAMPLE : 9.4 :
// Page number 237-238
clear ; clc ; close ; // Clear the work space and console
// Given data
V_a = 1.0 // Voltage(p.u)
V_b = 1.0*exp(%i*-36.87*%pi/180) // Voltage(p.u)
V_c = 1.0 // Voltage(p.u)
Z_1 = complex(0,1) // Reactance(p.u)
Z_2 = complex(0,1) // Reactance(p.u)
Z_3 = complex(0,1) // Reactance(p.u)
Z_13 = complex(0,0.4) // Reactance(p.u)
Z_23 = complex(0,0.4) // Reactance(p.u)
Z_14 = complex(0,0.2) // Reactance(p.u)
Z_24 = complex(0,0.2) // Reactance(p.u)
Z_34 = complex(0,0.2) // Reactance(p.u)
Z_12 = complex(0,0) // Reactance(p.u)
// Calculations
I_1 = V_a/Z_1 // Current injection vector(p.u)
I_2 = V_b/Z_2 // Current injection vector(p.u)
I_3 = V_c/Z_3 // Current injection vector(p.u)
I_4 = 0.0 // Current injection vector(p.u)
y1 = 1.0/Z_1 // Admittance(p.u)
y2 = 1.0/Z_2 // Admittance(p.u)
y3 = 1.0/Z_3 // Admittance(p.u)
y13 = 1.0/Z_13 // Admittance(p.u)
y23 = 1.0/Z_23 // Admittance(p.u)
y14 = 1.0/Z_14 // Admittance(p.u)
y24 = 1.0/Z_24 // Admittance(p.u)
y34 = 1.0/Z_34 // Admittance(p.u)
y12 = 0.0 // Admittance(p.u)
Y_11 = y1+y13+y14 // Equivalent admittance(p.u)
Y_12 = y12 // Equivalent admittance(p.u)
Y_13 = -y13 // Equivalent admittance(p.u)
Y_14 = -y14 // Equivalent admittance(p.u)
Y_21 = Y_12 // Equivalent admittance(p.u)
Y_22 = y2+y23+y24 // Equivalent admittance(p.u)
Y_23 = -y23 // Equivalent admittance(p.u)
Y_24 = -y24 // Equivalent admittance(p.u)
Y_31 = Y_13 // Equivalent admittance(p.u)
Y_32 = Y_23 // Equivalent admittance(p.u)
Y_33 = y3+y13+y23+y34 // Equivalent admittance(p.u)
Y_34 = -y34 // Equivalent admittance(p.u)
Y_41 = Y_14 // Equivalent admittance(p.u)
Y_42 = Y_24 // Equivalent admittance(p.u)
Y_43 = Y_34 // Equivalent admittance(p.u)
Y_44 = y14+y24+y34 // Equivalent admittance(p.u)
Y_bus = [[Y_11, Y_12, Y_13, Y_14],
[Y_21, Y_22, Y_23, Y_24],
[Y_31, Y_32, Y_33, Y_34],
[Y_41, Y_42, Y_43, Y_44]] // Bus admittance matrix
K = Y_bus([1,2],1:2)
L = Y_bus([1,2],3:4)
M = Y_bus([3,4],3:4)
N = Y_bus([3,4],1:2)
inv_M = inv([M(1,1:2);M(2,1:2)]) // Multiplication of marix [L][M^-1][N]
Y_bus_new = K-L*inv_M*N // New bus admittance matrix
// Results
disp("PART II - EXAMPLE : 9.4 : SOLUTION :-")
printf("\n[Y_bus]_new = \n"); disp(Y_bus_new)
printf("\nNOTE: ERROR: Mistake in representing the sign in final answer in textbook")
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