Example 10.6Calculating 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.

1ECE 201 Circuit Theory 1

ECE 201 Circuit Theory 1 2

Determine the power factor of the two loads in parallel.

The total complex power absorbed by the two loads is

21

*2

*1

*21

*

)250()250(

))(250(

)250(

SSSIIS

IIS

IS S

Power Triangle for Load 1

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8 kW at a leading power factor of 0.8.

cos-1(0.8) leading = -36.87°

S1 = 8,000 –j6,000 VA

Power Triangle for Load 2

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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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S = S1 + S2 = 8,000 – j6,000 = 12,000 +j16,000S = 20,000 +j10,000 VA

The resulting Power Triangle

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AIAjIAjI

jI

S

S

S

S

57.2644.8940804080250

000,10000,20

*

*

The power factor of the two loads in parallel is cos(26.57°) = 0.8944 lagging

(net reactive power is positive)

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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 22.36 kVA as shown in the power triangle below.

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The magnitude of the current that supplies the required 22.36 kVA is

AjIS 44.894080

The average power lost in the line is due to the current flowing through the line resistance.

WP

RIP

line

Sline

400)05.0()44.89( 2

2

Note: the total power supplied is equal to 20,400 W, even though the loads require only 20,000 W.

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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.

25.6000,10)250( 2

2

C

effC

XQ

VX

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Calculation of the capacitance from the required reactance

FCX

C

CX

C

C

4.424)25.6)(60)(2(

11

1

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Look at the resulting power triangle

(a) is the sum of the power triangles for loads 1 and 2.

(b) is the power triangle for the 424.4 µF capacitor at 60 Hz.

(c) is the “corrected” power triangle (for a power factor equal to 1).

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Once the power factor has been corrected,

WRIP

AI

kVAPS

Sline

S

320)05.0()80(

80250000,20

20

22

The total power supplied is now20,000 + 320 = 20,320 W

The line loss has been reduced by correcting the power factor.