Question 23.7: Each transmission line of the three-wire, three-phase system...
Each transmission line of the three-wire, three-phase system in Fig. 23.26 has an impedance of 15 Ω+j 20 Ω. The system delivers a total power of 160 kW at 12,000 V to a balanced three-phase load with a lagging power factor of 0.86.
a. Determine the magnitude of the line voltage E_{A B} of the generator.
b. Find the power factor of the total load applied to the generator.
c. What is the efficiency of the system?
\text { a. } \quad V_{\phi}( load )=\frac{V_{L}}{\sqrt{3}}=\frac{12,000 V }{1.73}=6936.42 V.
P_{T}( load )=3 V_{\phi} I_{\phi} \cos \theta.
and
I_{\phi}=\frac{P_{T}}{3 V_{\phi} \cos \theta}=\frac{160,000 W }{3(6936.42 V )(0.86)}.
= 8.94 A.
\text { Since } \theta=\cos ^{-1} 0.86=30.68^{\circ}, \text { assigning } V _{\phi} \text { an angle of } 0^{\circ} \text { or } V _{\phi}=V_{\phi} \angle 0^{\circ} \text {, a lagging power factor results in }
I _{\phi}=8.94 A \angle-30.68^{\circ}.
For each phase, the system will appear as shown in Fig. 23.27, where
E _{A N}- I _{\phi} Z _{\text {line }}- V _{\phi}=0.
or
E _{A N}= I _{\phi} Z _{\text {line }}+ V _{\phi}.
=\left(8.94 A \angle-30.68^{\circ}\right)\left(25 \Omega \angle 53.13^{\circ}\right)+6936.42 V \angle 0^{\circ}.
=223.5 V \angle 22.45^{\circ}+6936.42 V \angle 0^{\circ}.
=206.56 V +j 85.35 V +6936.42 V.
=7142.98 V + j 85.35 V.
=7143.5 V \angle 0.68^{\circ}.
Then E_{A B}=\sqrt{3} E_{\phi g}=(1.73)(7143.5 V ).
= 12,358.26 V.
\text { b. } P_{T}=P_{\text {load }}+P_{\text {lines }}.
=160 kW +3\left(I_{L}\right)^{2} R_{\text {line }}.
=160 kW +3(8.94 A )^{2} 15 \Omega.
=160,000 W +3596.55 W.
= 163,596.55 W.
and
P_{T}=\sqrt{3} V_{L} I_{L} \cos \theta_{T}.
or \cos \theta_{T}=\frac{P_{T}}{\sqrt{3} V_{L} I_{L}}=\frac{163,596.55 W }{(1.73)(12,358.26 V )(8.94 A )}.
and F_{p}= 0 . 8 5 6 <0.86 \text { of load }.
\text { c. } \eta=\frac{P_{o}}{P_{i}}=\frac{P_{o}}{P_{o}+P_{\text {losses }}}=\frac{160 kW }{160 kW +3596.55 W }=0.978.
= 97.8%.
