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~ IS A ~ + V ~ S ~ IL
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The current through the load is VS VS 0 IL = = 2 (R j L) R + j L R + 2 L2 VS R VS L j 2 2 L2 + R + 2 L2 The current through the capacitor is = R2 VS 0 IC = = j VS C 1/j C
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The source current to be measured is IS = IL + IC = VS R VS L + j VS C 2 R 2 + 2 L2 R + 2 L2
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The magnitude of the source current is IS = VS R 2 + 2 L2 R
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VS L 2 + 2 L2 R
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We know that when the load is a pure resistance, the current and voltage are in phase, the power factor is 1, and all the power delivered by the source is dissipated by the load as real power This corresponds to equating the imaginary part of the expression for the source current to zero, or, equivalently, to the following expression: R2 VS L = VS C + 2 L2
in the expression for IS Thus, the magnitude of the source current is actually a minimum when the power factor is unity! It is therefore possible to tune a load to a unity pf by observingthe readout of the ammeter while changing the value of capacitor and selecting the capacitor value that corresponds to the lowest source current value
EXAMPLE 710 Power Factor Correction
Problem
A capacitor is used to correct the power factor of the load of Figure 734 Determine the reactive power when the capacitor is not in the circuit, and compute the required value of capacitance for perfect pf correction
~ IC
~ IL
Solution
Known Quantities: Source voltage; load power and power factor Find:
~ VS + ~
100 kW pf = 07
1 Q when the capacitor is not in the circuit 2 Value of capacitor required for power factor correction resulting in pf = 1 pf = 07 lagging Schematics, Diagrams, Circuits, and Given Data: VS = 480 (0); P = 105 W;
Assumptions: Use rms values for all phasor quantities in the problem Analysis:
1 With reference to the power triangle of Figure 711, we can compute the reactive power of the load from knowledge of the real power and of the power factor, as
7
AC Power
shown below: |S| = P P 105 = = = 1429 105 VA cos( ) pf 07
Since the power factor is lagging, we know that the reactive power is positive (see Table 72), and we can calculate Q as shown below: Q = |S| sin( )
= arccos(pf ) = 0795
Q = 1429 10 sin(0795) = 102 kVAR 2 To compute the reactance needed for the power factor correction, we observe that we need to contribute a negative reactive power equal to 102 kVAR This requires a negative reactance, and therefore a capacitor with QC = 102 kVAR The reactance of such a capacitor is given by the expression: XC = and, since C= we have C= 1 1 = 1,175 F = XC 377 2258 1 XC |VL |2 (480)2 = = 2258 QC 102 105
Comments: Note that it is not necessary to know the load impedance to perform power
factor correction; it is suf cient to know the apparent power and the power factor
Focus on Computer-Aided Tools: A le containing the computer-generated solution to
this problem may be found in the CD-ROM that accompanies this book
EXAMPLE 711 Power Factor Correction
Problem
A second load is added to the circuit of Figure 734, as shown in Figure 735 Determine the required value of capacitance for perfect pf correction after the second load is added Draw the phasor diagram showing the relationship between the two load currents and the capacitor current
~ IS IC ~ + V ~ S ~ IL ~ I1 100 kW pf = 07 ~ I2 50 kW pf = 095
Solution
Known Quantities: Source voltage; load power and power factor
Part I
Circuits
Find:
1 Power factor correction capacitor 2 Phasor diagram pf1 = 07 lagging; P2 = 5 104 W; pf2 = 095 leading Schematics, Diagrams, Circuits, and Given Data: VS = 480 (0) V; P1 = 105 W;
Assumptions: Use rms values for all phasor quantities in the problem Analysis:
1 We rst compute the two load currents, using the relationships given in equations 728 and 729: 1 P = |VS ||I | cos( 1 ); 1 |I | = I1 = P1 |VS | cos( 1 ) ;
P1 105 (arccos(07)) arccos(pf 1 ) = S |pf 1 480 07 |V = 298 (0795) A P2 5 104 (arccos(095)) arccos(pf 2 ) = S |pf 2 480 095 |V = 360 ( 0318) A
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