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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 18: POWER DISTRIBUTION SYSTEMS
// EXAMPLE : 18.7 :
// Page number 441-442
clear ; clc ; close ; // Clear the work space and console
// Given data
r_out = 0.05 // Resistance of each outer per 100 metre length(ohm)
r_neutral = 0.10 // Resistance of each neutral per 100 metre length(ohm)
V_A = 200.0 // Potential at point A(V)
V_B = 200.0 // Potential at point B(V)
l_AC = 100.0 // Length between A & C(m)
l_CD = 150.0 // Length between C & D(m)
l_DB = 200.0 // Length between D & B(m)
l_AF = 200.0 // Length between A & F(m)
l_FE = 100.0 // Length between F & E(m)
l_EB = 150.0 // Length between E & B(m)
I_C = 20.0 // Current at point C(A)
I_D = 30.0 // Current at point D(A)
I_F = 60.0 // Current at point F(A)
I_E = 40.0 // Current at point E(A)
// Calculations
x = poly(0,"x") // Current in positive outer alone(A)
equ_1 = r_out*(l_DB*(I_D-x))-r_out*(l_AC*(I_C+x)+l_CD*x)
x_sol = roots(equ_1) // Current in positive outer alone(A)
y = poly(0,"y") // Current in negative outer alone(A)
equ_2 = r_out*((I_E-y)*l_FE+(I_E+I_F-y)*l_AF)-r_out*(l_EB*y)
y_sol = roots(equ_2) // Current in negative outer alone(A)
I_pos_out = I_C+x_sol // Current entering positive outer(A)
I_neg_out = I_E+I_F-y_sol // Current returning via negative outer(A)
I_middle = I_neg_out-I_pos_out // Current in the middle wire towards G(A)
r_CD = r_out*l_CD/100.0 // Resistance between C & D(ohm)
r_D = r_out*l_DB/100.0 // Resistance between D & B(ohm)
r_IH = r_neutral*l_FE*0.5/100.0 // Resistance between I & H(ohm)
r_IJ = r_neutral*l_FE*0.5/100.0 // Resistance between I & J(ohm)
r_GH = r_neutral*l_AF*0.5/100.0 // Resistance between G & H(ohm)
r_AF = r_out*l_AF/100.0 // Resistance between A & F(ohm)
I_CD = x_sol // Current flowing into D from C(A)
I_out_D = I_D-x_sol // Current flowing into D from outer side(A)
I_GH = I_C+I_middle // Current flowing into H from G(A)
I_IH = I_F-I_GH // Current flowing into H from I(A)
I_BJ = I_E-(I_D-I_IH) // Current flowing into J from B(A)
I_FE = y_sol-I_E // Current flowing into E from F(A)
I_IJ = I_D-I_IH // Current flowing into J from I(A)
V_C = V_A-(I_pos_out*r_out-I_middle*r_neutral) // Potential at load point C(A)
V_D = V_C-(I_CD*r_CD+I_IH*r_IH-I_GH*r_GH) // Potential at load point D(A)
V_F = V_A-(I_middle*r_neutral+I_GH*r_neutral+I_neg_out*r_AF) // Potential at load point F(A)
V_E = V_F-(-I_IH*r_IH+I_IJ*r_IJ-I_FE*r_out) // Potential at load point E(A)
// Results
disp("PART II - EXAMPLE : 18.7 : SOLUTION :-")
printf("\nPotential difference at load point C = %.3f V", V_C)
printf("\nPotential difference at load point D = %.3f V", V_D)
printf("\nPotential difference at load point E = %.3f V", V_E)
printf("\nPotential difference at load point F = %.3f V", V_F)
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