Schaum's Outline of Electric Circuits by Mahmood Nahvi

Author:Mahmood Nahvi
Language: eng
Format: epub, pdf
Publisher: McGraw-Hill Education
Published: 2018-03-14T04:00:00+00:00

EXAMPLE 10.14 A load with P = 1000 kW and pf = 0.5 lagging is fed by a 5-kV source. A capacitor is added in parallel such that the power factor is improved to 0.8. Find the reduction in current drawn from the generator.

Before the improvement:

After the improvement:

Hence, for the same amount of real power, the current is reduced by (400 − 250)/400 = 0.375 or 37.5 percent.

EXAMPLE 10.15 A fourth load Q4 is added in parallel to the three parallel loads of Example 10.12 such that the total power factor becomes 0.8 lagging while the total power remains the same. Find Q4 and the resulting S. Discuss the effect on the current.

In Example 10.12 we found the total real and reactive powers to be P = P1 + P2 + P3 = 45 kW and Q = Q1 + Q2 + Q3 = 54.6 kvar, respectively. For compensation, we now add load Q4 (with P4 = 0) such that the new power factor is pf = cos θ = 0.8 lagging or θ = 36.87°.

Then, .

The results are summarized in Table 10-2. Addition of the compensating load Q4 reduces the reactive power from 54.6 kvar to 33.75 kvar and improves the power factor. This reduces the apparent power S from 70.75 kVA to 56.25 kVA. The current is proportionally reduced.

Table 10-2

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