1. Example 10.6 Calculating Power in Parallel Loads
Load 1 absorbs an average power of 8 kW at a leading power factor of 0.8.
Load 2 absorbs 20 kVA at a lagging power factor of 0.6.
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2. Determine the power factor of the two loads in parallel.
The total complex power absorbed by the two loads is
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3. Power Triangle for Load 1
8 kW at a leading power factor of 0.8.
cos-1(0.8) leading = °
S1 = 8,000 –j6,000 VA
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4. Power Triangle for Load 2
20 kVA at a lagging power factor of 0.6.
cos-1(0.6) lagging = 53.13°
S2 = 12,000 +j16,000 VA
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5. The resulting Power Triangle
S = S1 + S2 = 8,000 – j6,000 = 12,000 +j16,000
S = 20,000 +j10,000 VA
The resulting Power Triangle
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6. (net reactive power is positive)
The power factor of the two loads in parallel is cos(26.57°) = lagging
(net reactive power is positive)
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7. Determine the apparent power required to supply the loads, the magnitude of the current IS, and the average power loss in the transmission line.
The apparent power which must be supplied to these loads is kVA as shown in the power triangle below.
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8. The magnitude of the current that supplies the required 22.36 kVA is
The average power lost in the line is due to the current flowing through the line resistance.
Note: the total power supplied is equal to 20,400 W, even though the loads require only 20,000 W.
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9. Given that the frequency of the source is 60 Hz, compute the value of capacitor that would correct the power factor to 1 if placed in parallel with the two loads.
As seen from the power triangle, the capacitor would have to supply 10 kVAR of magnetizing reactive power.
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10. Calculation of the capacitance from the required reactance
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11. Look at the resulting power triangle
is the sum of the power triangles for loads 1 and 2.
(b) is the power triangle for the µF capacitor at 60 Hz.
(c) is the “corrected” power triangle (for a power factor equal to 1).
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12. Once the power factor has been corrected,
The total power supplied is now
20, = 20,320 W
The line loss has been reduced by correcting the power factor.
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