Transcript Complex Power

Complex Power
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Designate as S
Magnitude of S = Apparent Power
Units are Volt-Amperes
The complex sum of Real Power (P) and
Reactive Power (Q)
S  P  jQ
S  P Q
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Graphical Interpretation of S
Power Triangle
  v i
S  P  jQ
S  P Q
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  v i
Q
tan  
P
Vm I m
sin(  v   i )
Q
 2
P Vm I m cos(   )
v
i
2
Q
 tan( v   i )
P
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Summary
  v i
S  P  jQ
Q

P
  tan 1 
S  P2  Q2
S  ApparentPower
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Summary -- 2
Apparent Power – the volt-amp capacity required to supply P
Q = Power exchanged between
source and load
P = Useful Output from the device being powered
(Watts)
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Summary -- 3
If the power factor angle is = 0,
Apparent Power = Average Power (and Reactive Power = 0).
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Example 10.4
Calculating Complex Power
• An electrical load operates at 240 V rms.
The load absorbs an average power of
8 kW at a lagging power factor of 0.8.
• Calculate the complex power of the load.
• Calculate the impedance of the load.
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The power factor is “Lagging”, so we have an
inductive load. The Reactive Power, Q, is
Positive.
P  S cos 
Q  S sin 
P  S cos 
Q  S sin 
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  cos (0.8)  36.87
sin   sin( 36.87)  0.6
1
P
8kW
S 

 10kVA
cos
0.8
Q  S sin   10kVA(0.6)  6kVARS
S  P  jQ  (8  j 6)kVA
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Vrms Veff
Z

I rms I eff
P  Veff I eff cos( v   i )
P  (240) I eff (0.8)  8000W
I eff
8000W

 41.67 A
(240)(0.8)
Z 
Veff
I eff
240

 5.76
41.67
Z  cos 1 (0.8)  36.87
Z  5.7636.87  4.608  j 3.456
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