A balanced three-phase load requires 480 kW at a lagging power factor of 0.8. The load is fed from a line having an impedance of 0.005 + j0.025 Ω/Φ . The line voltage at the terminals of the load is 600 V.
a) Construct a single-phase equivalent circuit of the system.
b) Calculate the magnitude of the line current.
c) Calculate the magnitude of the line voltage at the sending end of the line.
d) Calculate the power factor at the sending end of the line.
a) Figure 11.17 shows the single-phase equivalent circuit. We arbitrarily selected the line-to-neutral voltage at the load as the reference
b) The line current \pmb{I}^{*}_{aA} is given by
\left(\frac{600}{\sqrt{3}}\right)\pmb{I}^{*}_{aA}= \left(160 + j120\right )10^{3},
or
\pmb{I}^{*}_{aA}= 577.35 \angle 36.87^{\circ} A.Therefore, \pmb{I}_{aA}= 577.35 \angle -36.87^{\circ} A. The magnitude of the line current is the magnitude of \pmb{I}_{aA} :
\pmb{I}_{L}= 577.35 A.We obtain an alternative solution for I_{L} from the expression
P_{T}=\sqrt{3}V_{L}I_{L}\cos \theta_{p} = \sqrt{3}(600)I_{L}(0.8)= 480,000 W;
I_{L}=\frac{480,000}{\sqrt{3}(600)(0.8)} =\frac{1000}{\sqrt{3}}= 577.35 A.
c) To calculate the magnitude of the line voltage at the sending end, we first calculate \pmb{V}_{an} . From Fig. 11.17,
\pmb{V}_{an}=\pmb{V}_{AN}+\pmb{Z}_{l}\pmb{I}_{aA} =\frac{600}{\sqrt{3}}+\left(0.005 + j0.025\right)+\left(577.35 \angle -36.87^{\circ}\right) = 357.51 \angle 1.57^{\circ}V.Thus
V_{L}=\sqrt{3}\left|\pmb{V}_{an}\right|= 619.23 V.
d) The power factor at the sending end of the line is the cosine of the phase angle between \pmb{V}_{an} and \pmb{I}_{aA}:
pf = \cos\left [1.57^{\circ}- (-36.87^{\circ})\right] = \cos 38.44^{\circ}= 0.783 lagging.
An alternative method for calculating the power factor is to first calculate the complex power at the sending end of the line:
S_{\phi} = \left(160 + j120\right)10^{3} + \left(577.35\right)^{2} \left(0.005 + j0.025\right)= 161.67 + j128.33 kVA
= 206.41 \angle 38.44^{\circ} kVA.
The power factor is
pf = \cos 38.44^{\circ}= 0.783 lagging.
Finally, if we calculate the total complex power at the sending end, after first calculating the magnitude of the line current, we may use this value to calculate V_{L}. That is,
\sqrt{3}V_{L}I_{L}= 3(206.41)\times 10^{3} V_{L}=\frac{3(206.41)\times 10^{3} }{\sqrt{3}\left(577.35\right)}= 619.23 V.
