An industrial load consumes 88 kW at a pf of 0.707 lagging from a 480-V rms line. The transmission line resistance from the power company’s transformer to the plant is 0.08 Ω.

Let us determine the power that must be supplied by the power company (a) under present conditions and (b) if the pf is somehow changed to 0.90 lagging. (It is economically advantageous to have a power factor as close to one as possible.)

Question

An industrial load consumes 88 kW at a pf of 0.707 lagging from a 480-V rms line. The transmission line resistance from the power company's transformer to the plant is 0.08 Ω.

Let us determine the power that must be supplied by the power company (a) under present conditions and (b) if the pf is somehow changed to 0.90 lagging. (It is economically advantageous to have a power factor as close to one as possible.)

Accepted Answer

a. The equivalent circuit for these conditions is shown in Fig. 9.11. Using Eq. (9.27) ( [latex]\mathrm{pf}=\frac{P}{V_{\mathrm{rms}} I_{\mathrm{rms}}}=\cos \left( heta_{v}- heta_{i} ight)[/latex] ), we obtain the magnitude of the rms current into the plant:

[latex]\begin{aligned}I_{\mathrm{rms}} &=\frac{P_{L}}{(\mathrm{pf})\left(V_{\mathrm{rms}} ight)} \\&=\frac{(88)\left(10^{3} ight)}{(0.707)(480)} \\&=259.3 \mathrm{Arms}\end{aligned}[/latex]

The power that must be supplied by the power company is

[latex]\begin{aligned}P_{S} &=P_{L}+(0.08) I_{\mathrm{rms}}^{2} \\&=88,000+(0.08)(259.3)^{2} \\&=93.38 \mathrm{~kW}\end{aligned}[/latex]

b. Suppose now that the pf is somehow changed to 0.90 lagging but the voltage remains constant at 480 V. The rms load current for this condition is

[latex]\begin{aligned}I_{\mathrm{rms}} &=\frac{P_{L}}{(\mathrm{pf})\left(V_{\mathrm{rms}} ight)} \\&=\frac{(88)\left(10^{3} ight)}{(0.90)(480)} \\&=203.7 \mathrm{~A} \mathrm{rms}\end{aligned}[/latex]

Under these conditions the power company must generate

[latex]\begin{aligned}P_{S} &=P_{L}+(0.08) I_{\mathrm{rms}}^{2} \\&=88,000+(0.08)(203.7)^{2} \\&=91.32 \mathrm{~kW}\end{aligned}[/latex]

Note carefully the difference between the two cases. A simple change in the pf of the load from 0.707 lagging to 0.90 lagging has had an interesting effect. Note that in the first case the power company must generate 93.38 kW in order to supply the plant with 88 kW of power because the low power factor means that the line losses will be high- 5.38 kW. However, in the second case the power company need only generate 91.32 kW in order to supply the plant with its required power, and the corresponding line losses are only 3.32 kW.

[ hint ]

Technique
1. Given [latex]P_{L}, \mathrm{pf}[/latex], and [latex]V_{\mathrm{rms}}[/latex], determine [latex]I_{\mathrm{rms}}[/latex].
2. Then [latex]P_{s}=P_{L}+I_{ ext {rms }}^{2} R_{ ext {line }}[/latex], where [latex]R_{ ext {line }}[/latex] is the line resistance.

Question 9.10: An industrial load consumes 88 kW at a pf of 0.707 lagging f...

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