Two stores, as shown in Fig. 11.23, are located at a busy intersection. The stores are fed from a balanced three-phase 60-Hz source with a line voltage of 13.8 kV rms. The power line is constructed of a #4ACSR (aluminum cable steel reinforced) conductor that is rated at 170 A rms.

Question

Two stores, as shown in Fig. 11.23, are located at a busy intersection. The stores are fed from a balanced three-phase 60-Hz source with a line voltage of 13.8 kV rms. The power line is constructed of a #4ACSR (aluminum cable steel reinforced) conductor that is rated at 170 A rms.
A third store, shown in Fig. 11.23, wishes to locate at the intersection. Let us determine (1) if the #4ACSR conductor will permit the addition of this store, and (2) the value of the capacitors connected in wye that are required to change the overall power factor for all three stores to 0.92 lagging.

Accepted Answer

1. The complex power for each of the three loads is
[latex]S _{1}=700 \underline{/ 36.9^{\circ}} =560+j 420 kVA[/latex]
[latex]S _{2}=1000 \underline{/60^{\circ}} =500+j 866 kVA[/latex]
[latex]S _{3}=800 \underline{/ 25.8^{\circ}} =720+j 349 kVA[/latex]
Therefore, the total complex power is
[latex]S _{T}= S _{1}+ S _{2}+ S _{3}[/latex]
= 1780 + j1635
[latex]=2417 \underline{/ 42.57^{\circ}} kVA[/latex]
Since
[latex]S _{T}=\sqrt{3} V_{L} I_{L}[/latex]
the line current is
[latex]I_{L}=\frac{(2417)\left(10^{3} ight)}{\sqrt{3}(13.8)\left(10^{3} ight)}[/latex]
= 101.1 A rms
Since this value is well below the rated value of 170 A rms, the conductor is sized properly and we can safely add the third store.
2. The combined power factor for the three loads is found from the expression
[latex]\cos heta= pf =\frac{1780}{2417}=0.7365 ext { lagging }[/latex]
By adding capacitors we wish to change this power factor to 0.92 lagging. This new power factor corresponds to a [latex] heta_{ ext {new }}[/latex] of 23.07°. Therefore, the new complex power is
[latex] S _{ ext {new }}=1780+j 1780 an \left(23.07^{\circ} ight)[/latex]
= 1780 + j758.28 kVA
As illustrated in Fig. 9.17, the difference between [latex] S _{ ext {new }} ext { and } S _{T}[/latex] is that supplied by the purely reactive capacitor and, therefore,
[latex]S _{ ext {cap }}=j Q_{C}= S _{ ext {new }}- S _{T}[/latex]
or
[latex]j Q_{C}=j(758.28-1635)[/latex]
= −j876.72 kVA
Thus,
[latex]-j \omega C V_{ rms }^{2}=\frac{-j 876.72 k }{3}[/latex]
and
[latex]377\left(\frac{13.8 imes 10^{3}}{\sqrt{3}} ight)^{2} C=\frac{876.72}{3} imes 10^{3}[/latex]
Therefore,
C = 12.2 μF
Hence, three capacitors of this value connected in wye at the load will yield a total power factor of 0.92 lagging.

Question 11.12: Two stores, as shown in Fig. 11.23, are located at a busy in...

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