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Circuits
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Table 72 Important facts related to complex power
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Resistive load Ohm's law ~ ~ VL = ZL IL Capacitive load ~ ~ VL = ZL IL Inductive load ~ ~ VL = ZL IL
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Complex impedance
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ZL = RL
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ZL = RL jXL
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ZL = RL + jXL
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Phase angle Im Complex plane sketch
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u= u Im
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u< u Im
u> u
=0 ~ ~ I V
~ I ~ V Re
Explanation Power factor Reactive power
The current is in phase with the voltage Unity 0
The current leads the voltage Leading, < 1 Negative
The current lags the voltage Lagging, < 1 Positive
EXAMPLE 77 Complex Power and Power Triangle
Problem
Find the reactive and real power for the load of Figure 716 Draw the associated power triangle
R ~ VS
+ ~
jXL jXC
Complex load
Solution
Known Quantities: Source voltage; load impedance Find: S = Pav + j Q for the complex load
j XL = j 9
Schematics, Diagrams, Circuits, and Given Data: VS = 60 (0) V; R = 3 ; j XC = j 5
7
AC Power
Assumptions: Use rms values for all phasor quantities in the problem Analysis: First, we compute the load current:
60 (0) VL 60 (0) = = 12 ( 0644) A = IL = ZL 3 + j9 j5 5 (0644) Next, we compute the complex power, as de ned in equation 728: L S = VL I = 60 (0) 12 (0644) = 720 (0644) = 432 + j 576 W Therefore Pav = 432 W Q = 576 VAR
Im QL S Q
If we observe that the total reactive power must be the sum of the reactive powers in each of the elements, we can write Q = QC + QL , and compute each of the two quantities as follows: QC = |IL |2 XC = (144)( 5) = 720 VAR QL = |IL |2 XL = (144)(9) = 1,296 VAR
P Re
and Q = QL + QC = 576 VAR
Comments: The power triangle corresponding to this circuit is drawn in Figure 717
QC Note: S = PR + jQC + jQL
The vector diagram shows how the complex power, S, results from the vector addition of the three components, P , QC , and QL
The distinction between leading and lagging power factors made in Table 72 is important, because it corresponds to opposite signs of the reactive power: Q is positive if the load is inductive ( > 0) and the power factor is lagging; Q is negative if the load is capacitive and the power factor is leading ( < 0) It is therefore possible to improve the power factor of a load according to a procedure called power factor correction that is, by placing a suitable reactance in parallel with the load so that the reactive power component generated by the additional reactance is of opposite sign to the original load reactive power Most often the need is to improve the power factor of an inductive load, because many common industrial loads consist of electric motors, which are predominantly inductive loads This improvement may be accomplished by placing a capacitance in parallel with the load The following example illustrates a typical power factor correction for an industrial load
EXAMPLE 78 Power Factor Correction
Problem
Calculate the complex power for the circuit of Figure 718 and correct the power factor to unity by connecting a parallel reactance to the load
Part I
Circuits
Solution
Known Quantities: Source voltage; load impedance Find:
~ IS + R
+ ~ ~ VS
1 S = Pav + j Q for the complex load 2 Value of parallel reactance required for power factor correction resulting in pf = 1 j XL = j 867 Schematics, Diagrams, Circuits, and Given Data: VS = 117 (0) V; RL = 50
Assumptions: Use rms values for all phasor quantities in the problem Analysis:
~ VL jXL
1 First, we compute the load impedance: ZL = R + j XL = 50 + j 867 = 100 (1047) Next, we compute the load current: VL 117 (0) 117 (0) = = = 117 ( 1047) A IL = ZL 50 + j 866 100 (1047) and the complex power, as de ned in equation 728: L S = VL I = 117 (0) 117 (1047) = 137 (1047) = 684 + j 1185 W Therefore Pav = 684 W Q = 1185 VAR
7V 13 S= A
The power triangle corresponding to this circuit is drawn in Figure 719 The vector diagram shows how the complex power, S, results from the vector addition of the two components, P and QL To eliminate the reactive power due to the inductance, we will need to add an equal and opposite reactive power component, QL , as described below 2 To compute the reactance needed for the power factor correction, we observe that we need to contribute a negative reactive power equal to 1185 VAR This requires a negative reactance, and therefore a capacitor with QC = 1185 VAR The reactance of such a capacitor is given by the expression: XC = and, since C= we have C= 1 1 = 231 F = XC 377 ( 115) 1 XC |VL |2 (117)2 = 115 = QC 1185
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