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Write a MATLAB function named [ABCD] = abcdm(z, y, Lngt) to evaluate and return

ID: 1925939 • Letter: W

Question

Write a MATLAB function named [ABCD] = abcdm(z, y, Lngt) to evaluate
and return the ABCD transmission matrix for a medium-length transmission line
where z is the per phase series impedance per unit length, y is the shunt admittance
per unit length, and Lngt is the line length. Then, write a program that uses the
above function and computes the receiving end quantities, voltage regulation, and
transmission efficiency when sending end quantities are specified. The program
should prompt the user to enter the following quantities:
The sending end line-to-line voltage magnitude in kV
The sending end voltage phase angle in degrees
The three-phase sending end real power in MW
The three-phase sending end reactive power in Mvar
Use your program to obtain the solution for the following case:
A three-phase transmission line has a per phase series impedance of z = 0:03 +
j0:4 ­ per Km and a per phase shunt admittance of y = j4:0 £ 10¡6 siemens per
Km. The line is 125 Km long. Obtain the ABCD transmission matrix. Determine
the receiving end quantities, voltage regulation, and the transmission efficiency
when the line is sending 407 MW, 7.833 Mvar at 350 kV.

Explanation / Answer

z=input('Line series impedance per phase per unit length z='); y=input('Line shunt admittance per phase per unit length y='); Lngt = input('Transmission line length = '); ABCD = abcdm(z, y,Lngt) VL_s=input('Sending end line-to-line voltage magnitude in kV='); AngV_s=input('Sending end voltage phase angle in degree = '); P_s =input('Three-phase sending end real power in MW '); Q_s =input('Three-phase sending end reactive power in Mvar '); S_s = P_s + j*Q_s; % MVA AngV_srd = AngV_s*pi/180; % Radian V_s = VL_s/sqrt(3)*(cos(AngV_srd) + j*sin(AngV_srd)); %kV I_s = conj(S_s)/(3*conj(V_s)); % kA IL_s= abs(I_s)*1000; AngI_srd = angle(I_s); AngI_s = AngI_srd*180/pi; VI_r = inv(ABCD)*[V_s; I_s]; V_r = VI_r(1); VL_r = sqrt(3)*abs(V_r); AngV_rrd = angle(V_r); AngV_r = AngV_rrd*180/pi; I_r = VI_r(2); IL_r = abs(I_r)*1000; AngI_rd = angle(I_r); AngI_r = AngI_rd*180/pi; S_r = 3*V_r*conj(I_r); P_r = real(S_r); Q_r = imag(S_r); A = abs(ABCD(1,1)); Reg = (VL_s/A - VL_r)/VL_r*100; Eff = P_r/P_s*100; fprintf('Sending end line-to-line voltage =%g KV ', VL_s) fprintf('Sending end voltage phase angle =%g Degree ',AngV_s) fprintf('Sending end real power = %g MW ', P_s) fprintf('Sending end reactive Power = %g Mvar ', Q_s) fprintf('Sending end current = %g A ', IL_s) fprintf('Sending end current phase angle=%g Degree ',AngI_s) fprintf('receiving end line-to-line voltage = %g KV ',VL_r) fprintf('receiving end voltage phase angle=%g Degree ',AngV_r) fprintf('receiving end real power = %g MW ', P_r) fprintf('receiving end reactive Power = %g Mvar ', Q_r) fprintf('receiving end current = %g A ', IL_r) fprintf('receiving end current phase angle=%g Degree ',AngI_r) fprintf('Voltage regulation = %g percent ',Reg) fprintf('Transmission efficiency = %g percent ',Eff) typing ch5p7 at the MATLAB prompt result in Line series impedance per phase per unit length z = 0.03+j*0.4 Line shunt admittance per phase per unit length y = j*4.0e-6 Transmission line length = 125 ABCD = 0.9875+ 0.0009i 3.7500+50.0000i 0.0000+ 0.0005i 0.9875+ 0.0009i Sending end line-to-line voltage magnitude in kV = 350 Sending end voltage phase angle in degree = 0 Three-phase sending end real power in MW 407 Three-phase sending end reactive power in Mvar 7.883 Sending end line-to-line voltage = 350 kV Sending end voltage phase angle = 0 Degree Sending end real power = 407 MW Sending end reactive Power = 7.883 Mvar Sending end current = 671.502 A Sending end current phase angle = -1.1096 Degree Receiving end line-to-line voltage = 345.003 kV Receiving end voltage phase angle = -9.63278 Degree Receiving end real power = 401.884 MW Receiving end reactive Power = 0.0475969 Mvar Receiving end current = 672.539 A Receiving end current phase angle = -9.63957 Degree Voltage regulation = 2.73265 percent Transmission efficiency = 98.7429 percent

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