BEF 23803 - Lecture 7 - Complex Power Calculation.ppt

Complex Power Calculation

1. Lecture 7
Single-Phase Complex
Power Calculations

2. Learning Outcomes:
After completing this study unit you will be able to:
1.Calculate complex power, apparent power and power factor
2.Apply the principle of conservation of complex power
3.Do power factor correction

3. i.is excited by a sinusoidal input, v(t) = V
mcos(), and
ii.the terminal current hasreached its steady state value,
i(t) = I
mcos(t)
i(t)
v(t)
Linear
network
The time domain circuit
Complex Power Absorbed by a Resistive Circuit
Consider a two-terminal, linear and purely resistive network,
as shown in the figure below. We assume that the circuit to be
analysed:

4. The complex powerdelivered to the circuit is defined to be 0
mII  0
mVV
and*
2
1
IVS 
where is the complex conjugate of the current . Therefore,*
I I  0
2
1
00
2
1
mmmm IVIVS 0
mII  0
mVV
The input current and input voltage in phasor form are
Complex Power Absorbed by a Resistive Circuit
VA

5. The complex powerin rectangular form is
This can be written as  0sin
2
1
0cos
2
1
mmmm IVjIVS jQPS 
rmsrmsmm IVIVP  0cos
2
1 00sin
2
1

mmIVQ
where
is the average, real or active
power [W]
is the reactive power [VAR]
Complex Power Absorbed by a Resistive Circuit
This last result tells us that a resistance does not consume
reactive power.

6. The power triangle for a purely resistive load is shown below.
Re
Im
S = P; Q = 0
S
P
Complex Power Absorbed by a Resistive Circuit

7. 1.The COMPLEX PowerScontains all the information
pertaining to the power absorbed by a given load.
2.The REAL Poweris the only useful power delivered to the load.
3.The REACTIVE Powerrepresents the energy exchange
between the source and reactive part of the load. It is being
transferred back and forth between the load and the source
Notes
Complex Power Absorbed by a Resistive Circuit

8. Worked Example
Compute the power absorbed by R
Solution V
o
sS tvPV 1520)(  

1210120100
3
jjLj

9. Solution
Phasor domain circuit:RV LV  12j V 1520
o
SV  I  1225jZ IRV
R  1225
1520
jZ
V
I
o
S



Compute circuit current, .I A 64.40721.0
o

Compute voltage drop .
RV V 64.4003.18
64.40721.025


o

10. Complex powerdelivered to R is*
2
1
IVS
RR  W5.6
64.40721.064.4003.18
2
1


oo
Solution
RV I
R = 25 Ω
Since S
Rhas no reactive component, the 25 Ωresistance
absorbs only active power and no reactive power.

11. Alternative Solution1225
1520
jZ
V
I
o
S



Compute circuit current, .I A 64.40721.0
o

Complex power consumed by circuit
is RV LV  12j V 1520
o
SV  I
Z*
2
1
rmssIVS VA 64.2521.7
64.40721.01520
2
1



12. Alternative SolutionRV LV  12j V 1520
o
SV  I
Z
Since there is only one resistance present in the circuit, the
active power absorbed by the circuit is also the active power
absorbed by the 25 Ωresistance. Therefore, active power
absorbed by the 25 Ωresistance is 
 
W5.6
64.2521.7Re
Re


SP

13.   90cossin)( tItIti
mm  Let
Complex Power Absorbed by a Purely Inductive Circuit
i(t)
v(t)
Linear
circuit
The time domain circuit
Assume a linear inductive circuit.
From Lecture 6, we found for an inductor, 
tVtLI
tI
dt
d
L
dt
di
Ltv
mm
m


coscos
sin
)(



where2
mmLIV

14. The complex powerdelivered to the circuit is defined to be  90
mII  0
mVV
and*
2
1
IVS 
where is the complex conjugate of the current . *
I
In phasor form, we haveI  90
mII  0
mVV
Complex Power Absorbed by a Purely Inductive Circuit

15. Therefore, complex power consumed by the circuit is
Thus, average, real or active power consumed by the purely
inductive circuit is 
o
mm
o
mm IVjIV 90sin
2
1
90cos
2
1
 090cos
2
1

o
mmIVP
and reactive power consumed by the inductive circuit is
Power Absorbed by a Purely Inductive Circuit 90
2
1
900
2
1
mmmm IVIVS 
rmsrmsmm
o
mm IVIVIVQ 
2
1
90sin
2
1

16. Power triangle for a purely inductive circuit
Re
Im
S = jQ; P = 0
SjQ
Power Absorbed by a Purely Inductive Circuit

17. Worked Example
Compute the power absorbed by L.
Solution V
o
sS tvPV 1520)(  

1210120100
3
jjLj

18. Solution
Phasor domain circuit:RV LV  12j V 1520
o
SV  I  1225jZ 1225
1520
jZ
V
I
o
S



Compute circuit current, .I A 64.40721.0
o

Peak voltage drop across L isV 36.498.65
64.40721.09012



o
LLL IXV

19. Solution*
2
1
IVS
LL 
Complex powerdelivered to L isVA 12.3
46.40721.036.4965.8
2
1

j
o


Hence, reactive power consumed by L is VAR 12.3SImQ  L LS
j12 Ω

20. Alternative Solution VAR 12.312721.0
2
1
Im
2
2
2









 Z
I
Q
m 1225
1520
jZ
V
I
o
S



Compute circuit current, .I A 64.40721.0
o

Therefore,

21.   90cossin)( tVtVtv
mm  Complex Power Absorbed by a Purely Capacitive Circuit
i(t)
v(t)
Linear
circuit
The time domain circuit
Assume a linear and purely capacitive circuit.
Then, current flowing into capacitor is  
tItCV
tV
dt
d
C
dt
dv
Cti
mm
m


coscos
sin
)(



where2
mmCVI

22. The complex powerdelivered to the circuit is defined to be
In phasor form,0
mII  90
mVV
and*
2
1
IVS 
where is the complex conjugate of the current . Therefore,*
I I  90
2
1
090
2
1
mmmm IVIVS 0
mII  90
mVV
Complex Power Absorbed by a Purely Capacitive Circuit

23. Therefore, complex power consumed by the circuit is
Thus, average, real or active power consumed by circuit is 
o
mm
o
mm IVjIVS 90sin
2
1
90cos
2
1
  090cos
2
1
Re 
o
mmIVSP  
rmsrmsmm
o
mm IVIVIVSQ 
2
1
90sin
2
1
Im
and reactive power consumed by the circuit is
Complex Power Absorbed by a Purely Capacitive Circuit

24. Re
Im
S = -jQ; P = 0
Power triangle for a purely capacitive circuit
S -jQ
Complex Power Absorbed by a Purely Capacitive Circuit

25. The (driving point) impedanceof a two-terminal circuit can
be expressed as
In rectangular form  
IV
Im
Vm
Z
I
V
I
V
Z 





    
IV
m
m
IV
m
m
I
V
j
I
V
Z   sincos
orjXRZ  
IV
m
m
I
V
R cos  
IV
m
m
I
V
X sin
where
andm
m
I
V
Z where
Complex Power Consumed by an Impedance
Linear
networkI V

26. The complex powercan also be expressed in terms of
the impedanceZ, as shown below.   
IVmmIVmm IVjIVS   sin
2
1
cos
2
1    
IVm
m
m
IVm
m
m
I
I
V
jI
I
V
 













 sin
2
1
cos
2
1 22  Z
I
jZ
I
mm
Im
2
Re
2
22
















Comparing the above equation with , we hence obtainjQPS Z
I
P
m
Re
2
2






 Z
I
Q
m
Im
2
2









and
Complex Power Consumed by an Impedance

27. 1.Resistive-inductive load
Re
Im S = P + jQ
S
jQ
P
Power Triangle

28. 2.Resistive-capacitive load
Re
Im
S = P -jQ
S
P
-jQ
Summary: Power Triangle

29. 3.Resistive –capacitive -inductive load
Re
Im
S = P + j(Q
L–Q
C)
S
j(Q
L-Q
C)
Summary: Power Triangle

30. Worked Example100cos1000 V
=100 0
svt

Calculate the active and reactive powers supplied by the
voltage source to the load.
Given:

31. Solution
1. Find load current.( ) 7.07 45
1
s
R j L j
C



    

V
I
2. Use Ohm’s law to get the element voltage phasors.( ) ( ) 70.7 45
( ) ( ) 141.4 45
( ) ( ) 70.7 135
R
L
C
R
jL
j
C

  


    
   

    
VI
VI
VI

32. 3. Compute complex power of each element.*
2
353.5 45 VA
s
V
  
VI
S
Complex power suppliedby the source
For the resistor*
2
250 0 VA
R
R
  
VI
S
Complex power absorbedby the resistor
For the inductor*
2
500 90 VA
L
L
  
VI
S
Complex power deliveredto the inductor
Solution

33. For the capacitor*
2
250 90 VA
C
C
   
VI
S
Complex power deliveredto the
capacitor250 0 500 90 250 90
353.5 45
R L C
V
           
   
S S S
S
The total power absorbed by all elements (except source)*
0
2
kk
all
elements

VI
For all elements
Solution

34. Active power supplied to the resistor is2
Re( )
2
mI
P




Z 2
250 W 0
2
m
R L C
I
P R P P

   


Solution
Active power supplied to the inductor and capacitor is
Active power supplied by the voltage source is
*
Re Re
2
Re(353.5 45 ) 250 W
s
VV
P




   
VI
S

35. Calculate the total active and reactive powers supplied by the source
to the resistors.
Exercise

36. This result shows that for a parallel-
connected circuit the complex, real and
reactive power of the sources equal the
respective sum of the complex, real and
reactive power of the individual loads.
Consider the parallel connected circuit shown below.
Conservation of Complex Power*

2
1
IVS   
*
2
*
1
2
1
IIV  **
21
2
1

2
1
IVIV 
21SS S

37. The same result is obtained for series-connected circuit, as
shown below.
Conservation of Complex Power*

2
1
IVS   
**
2
*
1
2
1
IVV  *
2
*
1
2
1

2
1
IVIV 
21SS S
This can be written as0 
21SSS
or0
elements all
iS

38. Mathematically, we can write*
0
2
kk
all
elements

VI
Thus, for either the parallel circuit or the series circuit, we
have shown that the sum of complex powerabsorbed by all
the elements of the circuit is zero.
Now, the complex poweris conservedimplies that both average
power and reactive power are conserved. That is,0 and 0
kk
all all
elements elements
PQ
Conservation of Complex Power

39. Finding the total complex power supplied by the source to
the three loads.
Worked Example

40. Solution
Complex power consumed by load 1,  VA 0100
1 jjQPS 

Complex power consumed by load 2,  VA 700200
2 jjQPS 

Complex power consumed by load 3,  VA 1500300
3
jjQPS 


41. 1 2 1 2 1 2
( ) ( )S P jQ S S P P j Q Q        Total complex power consumed by the loads, VA)1500700()300200100(  j VA
o
j 13.531000800600 
Solution

42. The 60 resistor absorbs 240 Watt of average power.
Calculate Vand the complex power of each branch. What is
the total complex power?
Worked Example

43. Solution
Phasor domain circuit:2
260240 I A 2
60
240
2 I
Solving for I
2, we obtain
Let be the current through the 60-Ω resistor. Now ,
therefore2I
RIP
2
2


44. (rms) A
o
jI 0202
2  Let be the reference phasor. Therefore, we can write2I    (rms) A 40120022060 jjV
o
o 
Application of Ohm’s law to the right branch impedance gives us the
voltage drop
Application of Ohm’s law to the left branch impedance gives us the
branch current(rms) A 

 4.22.3
1030
1
j
j
V
I
o
Next, KCL gives us the current equation A .j. jIII 422524.22.3
21 
Solution

45.    V401204810420 jjVIV
o  V
o
j 45.2167.24088224  For the 20-Ω resistor, V
o
jIV 8.45.1144820420     A4.22.54225
**
j .j.I   4.22.58.45.114
*
3 jIVS
o
 A 654
o
S 97.19
3 
Therefore,
Solution

46. For the (30 -j10) Ω impedance, V
o
o jV 8.245.12640120  A
o
jI 87.3644.22.3
1  VA
oo
oIVS 87.36443.185.126
*
11    VA -jS
o
16048044.18506
1 
Therefore,
Solution

47. VA 43.182530243.185.126
*
22 
oo
oIVS VA....S
oo
T 8247275452167240  Therefore,
For the (60 + j20) Ω impedance, A
o
I 02
2  A VjS 80240
2 
The overall complex power supplied by the source is  VAjS
T 801736
giving
Solution

48. Exercise
Two loads are connected in parallel. Load 1 has 2 kW,
pf=0.75 leading and Load 2 has 4 kW, pf=0.95 lagging.
Calculate the pf of the two loads and the complex power
supplied by the source.

49. Power Factor Correction
To adjust the power factor by adding a compensating
impedance to the load.
Objective
The goal of power factor correction is to deliver maximum
power to the load using the lowest source current.
Goal

50. In the following worked example we will first
determine the current that the generator needs to
supply to load when its power factor is not
corrected to unity. Then, we will demonstrate the
advantage of correcting the power factor of the
load on the magnitude of the current that needs
to be supplied by the generator to the load.

51. Worked Example
For the circuit shown below, calculate
(i)the supply current,
(ii)the reactive power that needs to be supplied by a
capacitor bank to increase the power factor to unity.
(iii)the value of the supply current at unity power factor.
600 V
Load
P = 120 kW
Q = 160 kVAr
I
S

52. Solution
600 V
Load
P = 120 kW
Q = 160 kVAr
I
SLS
Let be the reference phasor.
Complex power supplied by the generator to the load is000,160000,120
0600
*
*
j
I
IVS
S
o
SSL


 SV
Therefore,(rms)A 3.5133.333
0600
000,160000,120

* o
oS
j
I 



and(rms)A 3.5133.333
o
SI 

53. To achieve unity power factor, we need to connect a
compensating reactive load in parallel with the original load
that cancels out the reactive power.
600 V
Load
P = 120 kW
Q = 160 kVAr
I
S
Reactive
load
To obtain unity power factor, we need ensure that
has no imaginary part. Now, since000,160000,120 jS
L 
Let be the complex power
consumed by the compensating load.
Then, complex power supplied by the
generator to the two loads isLCG SSS  CS GS
Solution

54. Therefore, we require000,160jS
C
so thatkW 120
000,160000,160000,120



jj
SSS
LCG
Therefore, supply current at unity power factor isS
o
o
S
G
S
I
V
S
I 

 A 0200
0600
kW 120
*
Solution

55. Thus, by correcting the load power factor, we have managed to
deliver the active power required by the load and at the same
time significantly reduced the supply current. This reduces the
size of the cable used to supply the load current and also the
required VA rating of the generator.
The VA rating of the generator for the power factor corrected
load is
S= V
SI
S= 600 x 200 = 120 kVA
Solution

56. Summary
In this study unit we have looked at
1.Complex power
2.Conservation of complex power
3.Power factor correction.