Question 16.25: In Figure P16-25 the voltage across the two loads is |VL| = ......
In Figure P16-25 the voltage across the two loads is \left|\mathbf{V}_{\mathrm{L}}\right|= = 4.8 kV (rms). The load Z_1 draws an average power of 12 kW and a lagging power factor of 0.75 lagging. The load Z_2 draws an apparent power of 15 kVA and a lagging power factor of 0.8 lagging. The line has an impedance of Z_W = 10 + j54 Ω per wire. Find the apparent power produced by the source and the rms value of the source voltage.
Use MATLAB to perform the calculations.
Script File
clear all
MagVL = 4.8e3;
pf1 = 0.75;
PZ1 = 12e3;
pf2 = 0.8;
PZ2 = 15e3*pf2;
ZW = 10+54j;
% Find the complex power for each load
S1 = PZ1/pf1*(pf1+j*sqrt(1-pf1^2));
S2 = PZ2/pf2*(pf2+j*sqrt(1-pf2^2));
% Find the total complex power at the load
SL = S1+S2;
% Find the magnitude of the current
MagIL = abs(SL)/MagVL;
% Find the complex power in the line
SW = 2*MagIL^2*ZW;
% Find the total complex power
Ss = SL+SW;
% Find the apparent source power
MagSs = abs(Ss)
% Find the rms value of the source voltage
MagVs = MagSs/MagIL
MagSs =
34.5912e+003
MagVs =
5.3603e+003
The apparent power produced by the source is 34.5912 kVA.
The rms value of the source voltage is 5.3603 kV.