BEF 23803 - Lecture 7 - Complex Power Calculation.ppt

Lecture 7
Single-Phase Complex
Power Calculations

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

i. is excited by a sinusoidal input, v(t) = Vmcos(), and
ii. the terminal current has reached its steady state value,
i(t) = Imcos(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:

The complex power delivered to the circuit is defined to be


 0
m
I
I


 0
m
V
V
and
*
2
1
I
V
S 

where is the complex conjugate of the current . Therefore,
*
I I









 0
2
1
0
0
2
1
m
m
m
m I
V
I
V
S


 0
m
I
I


 0
m
V
V
The input current and input voltage in phasor form are
VA

The complex power in rectangular form is
This can be written as
   



 0
sin
2
1
0
cos
2
1
m
m
m
m I
V
j
I
V
S
jQ
P
S 

  rms
rms
m
m I
V
I
V
P 

 0
cos
2
1
  0
0
sin
2
1


 m
mI
V
Q
where
is the average, real or active
power [W]
is the reactive power [VAR]
This last result tells us that a resistance does not consume
reactive power.

The power triangle for a purely resistive load is shown below.
Re
Im
S = P ; Q = 0
S
P

1. The COMPLEX Power S contains all the information
pertaining to the power absorbed by a given load.
2. The REAL Power is the only useful power delivered to the load.
3. The REACTIVE Power represents 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

Worked Example
Compute the power absorbed by R
Solution
  V
o
s
S t
v
P
V 15
20
)
( 








 
12
10
120
100 3
j
j
L
j

Solution
Phasor domain circuit:
R
V
L
V

12
j
V
15
20 o
S
V 


I


 12
25 j
Z
I
R
VR 

12
25
15
20
j
Z
V
I
o
S





Compute circuit current, .
I
A
64
.
40
721
.
0 o



Compute voltage drop .
R
V
V
64
.
40
03
.
18
64
.
40
721
.
0
25







 o

Complex power delivered to R is
*
2
1
I
V
S R
R 

W
5
.
6
64
.
40
721
.
0
64
.
40
03
.
18
2
1






 o
o
Solution
R
V
I
R = 25 Ω
Since SR has no reactive component, the 25 Ω resistance
absorbs only active power and no reactive power.

Alternative Solution
12
25
15
20
j
Z
V
I
o
S





I
A
64
.
40
721
.
0 o



Complex power consumed by circuit
is
R
V
L
V

12
j
V
15
20 o
S
V 


I
Z
*
2
1
rms
s I
V
S 

VA
64
.
25
21
.
7
64
.
40
721
.
0
15
20
2
1












R
V
L
V

12
j
V
15
20 o
S
V 


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 Ω resistance. Therefore, active power
absorbed by the 25 Ω resistance is
 
 
W
5
.
6
64
.
25
21
.
7
Re
Re




 S
P

 



 90
cos
sin
)
( t
I
t
I
t
i m
m 

Let
Complex Power Absorbed by a Purely Inductive Circuit
i(t)
v(t)
Linear
circuit
Assume a linear inductive circuit.
From Lecture 6, we found for an inductor,
 
t
V
t
LI
t
I
dt
d
L
dt
di
L
t
v
m
m
m




cos
cos
sin
)
(




where
2
m
m LI
V 





 90
m
I
I


 0
m
V
V
and
*
2
1
I
V
S 

where is the complex conjugate of the current .
*
I
In phasor form, we have
I



 90
m
I
I


 0
m
V
V
Complex Power Absorbed by a Purely Inductive Circuit

Therefore, complex power consumed by the circuit is
Thus, average, real or active power consumed by the purely
inductive circuit is
   
o
m
m
o
m
m I
V
j
I
V 90
sin
2
1
90
cos
2
1


  0
90
cos
2
1

 o
m
mI
V
P
and reactive power consumed by the inductive circuit is
Power Absorbed by a Purely Inductive Circuit








 90
2
1
90
0
2
1
m
m
m
m I
V
I
V
S
  rms
rms
m
m
o
m
m I
V
I
V
I
V
Q 


2
1
90
sin
2
1

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

Worked Example
Compute the power absorbed by L.
Solution
  V
o
s
S t
v
P
V 15
20
)
( 








 
12
10
120
100 3
j
j
L
j

Solution
R
V
L
V

12
j
V
15
20 o
S
V 


I


 12
25 j
Z
12
25
15
20
j
Z
V
I
o
S





I
A
64
.
40
721
.
0 o



Peak voltage drop across L is
V
36
.
49
8.65
64
.
40
721
.
0
90
12










o
L
L
L I
X
V

Solution
*
2
1
I
V
S L
L 

Complex power delivered to L is
VA
12
.
3
46
.
40
721
.
0
36
.
49
65
.
8
2
1
j
o







Hence, reactive power consumed by L is
  VAR
12
.
3
S
Im
Q 
 L
L
S
j12 Ω

  VAR
12
.
3
12
721
.
0
2
1
Im
2
2
2












 Z
I
Q m
12
25
15
20
j
Z
V
I
o
S





I
A
64
.
40
721
.
0 o



Therefore,

 



 90
cos
sin
)
( t
V
t
V
t
v m
m 

Complex Power Absorbed by a Purely Capacitive Circuit
i(t)
v(t)
Linear
circuit
Assume a linear and purely capacitive circuit.
Then, current flowing into capacitor is
 
t
I
t
CV
t
V
dt
d
C
dt
dv
C
t
i
m
m
m




cos
cos
sin
)
(




where
2
m
m CV
I 


In phasor form,


 0
m
I
I



 90
m
V
V
and
*
2
1
I
V
S 

where is the complex conjugate of the current . Therefore,
*
I I











 90
2
1
0
90
2
1
m
m
m
m I
V
I
V
S


 0
m
I
I



 90
m
V
V

Therefore, complex power consumed by the circuit is
Thus, average, real or active power consumed by circuit is
   
o
m
m
o
m
m I
V
j
I
V
S 90
sin
2
1
90
cos
2
1




    0
90
cos
2
1
Re 


 o
m
mI
V
S
P
    rms
rms
m
m
o
m
m I
V
I
V
I
V
S
Q 






2
1
90
sin
2
1
Im
and reactive power consumed by the circuit is

Re
Im
S = - jQ ; P = 0
Power triangle for a purely capacitive circuit
S - jQ

The (driving point) impedance of a two-terminal circuit can
be expressed as
In rectangular form
 
I
V
I
m
V
m
Z
I
V
I
V
Z 










   
I
V
m
m
I
V
m
m
I
V
j
I
V
Z 


 


 sin
cos
or
jX
R
Z 

 
I
V
m
m
I
V
R 
 
 cos  
I
V
m
m
I
V
X 
 
 sin
where
and
m
m
I
V
Z 
where
Complex Power Consumed by an Impedance
Linear
network
I
V

The complex power can also be expressed in terms of
the impedance Z, as shown below.
   
I
V
m
m
I
V
m
m I
V
j
I
V
S 


 


 sin
2
1
cos
2
1
       
I
V
m
m
m
I
V
m
m
m
I
I
V
j
I
I
V



 














 sin
2
1
cos
2
1 2
2
   
Z
I
j
Z
I m
m
Im
2
Re
2
2
2
















Comparing the above equation with , we hence obtain
jQ
P
S 

 
Z
I
P m
Re
2
2






  
Z
I
Q m
Im
2
2









and
Complex Power Consumed by an Impedance

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

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

3. Resistive – capacitive - inductive load
Re
Im
S = P + j(QL – QC)
S
j(QL- QC)
Summary: Power Triangle

Worked Example
100cos1000 V
=100 0
s
v t

 
Calculate the active and reactive powers supplied by the
voltage source to the load.
Given:

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
j L
j
C
 
  
 

    
   

    
V I
V I
V I

3. Compute complex power of each element.
*
2
353.5 45 VA
s
V 
  
V I
S
Complex power supplied by the source
For the resistor
*
2
250 0 VA
R
R 
  
V I
S
Complex power absorbed by the resistor
For the inductor
*
2
500 90 VA
L
L 
  
V I
S
Complex power delivered to the inductor
Solution

For the capacitor
*
2
250 90 VA
C
C 
   
V I
S
Complex power delivered to the
capacitor
250 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
k k
all
elements


V I
For all elements
Solution

Active power supplied to the resistor is
2
Re( )
2
m
I
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
V V
P
 
   
 
   
V I
S

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

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
I
V
S 

 
*
2
*
1
2
1
I
I
V 


*
*
2
1
2
1
2
1
I
V
I
V 



2
1 S
S 

S

The same result is obtained for series-connected circuit, as
shown below.
*
2
1
I
V
S 

  *
*
2
*
1
2
1
I
V
V 


*
2
*
1
2
1
2
1
I
V
I
V 



2
1 S
S 
 S
This can be written as
0


 2
1 S
S
S or 0


elements
all
i
S

Mathematically, we can write
*
0
2
k k
all
elements


V I
Thus, for either the parallel circuit or the series circuit, we
have shown that the sum of complex power absorbed by all
the elements of the circuit is zero.
Now, the complex power is conserved implies that both average
power and reactive power are conserved. That is,
0 and 0
k k
all all
elements elements
P Q
 
 

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

Solution
Complex power consumed by load 1,
 VA
0
100
1 j
jQ
P
S 




 VA
700
200
2 j
jQ
P
S 




 VA
1500
300
3 j
jQ
P
S 





1 2 1 2 1 2
( ) ( )
S P jQ S S P P j Q Q
       
Total complex power consumed by the loads,
VA
)
1500
700
(
)
300
200
100
( 



 j
VA
o
j 13
.
53
1000
800
600 




Solution

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

Solution
2
2
60
240 I


A
2
60
240
2 

I
Solving for I2, we obtain
Let be the current through the 60-Ω resistor. Now ,
therefore
2
I R
I
P 2
2


(rms)
A
o
j
I 0
2
0
2
2 



Let be the reference phasor. Therefore, we can write
2
I
    (rms)
A
40
120
0
2
20
60 j
j
V 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
.
2
2
.
3
10
30
1 j
j
V
I o
Next, KCL gives us the current equation
A
.
j
.
j
I
I
I 4
2
2
5
2
4
.
2
2
.
3
2
1 






Solution

    V
40
120
48
104
20 j
j
V
I
V o 






V
o
j 45
.
21
67
.
240
88
224 



For the 20-Ω resistor,
V
o
j
I
V 8
.
4
5
.
114
48
204
20 





   A
4
.
2
2
.
5
4
2
2
5
*
*
j
.
j
.
I 



 
4
.
2
2
.
5
8
.
4
5
.
114
*
3 j
I
V
S o






A
654 o
S 97
.
19
3 


Therefore,
Solution

For the (30 - j10) Ω impedance,
V
o
o j
V 8
.
24
5
.
126
40
120 




A
o
j
I 87
.
36
4
4
.
2
2
.
3
1 



VA
o
o
o I
V
S 87
.
36
4
43
.
18
5
.
126
*
1
1 






  VA
-j
S o
160
480
44
.
18
506
1 



Therefore,
Solution

VA
43
.
18
253
0
2
43
.
18
5
.
126
*
2
2 






 o
o
o I
V
S
VA
.
.
.
.
S o
o
T 8
24
727
5
45
21
67
240 




Therefore,
For the (60 + j20) Ω impedance, A
o
I 0
2
2 

  A
V
j
S 80
240
2 

The overall complex power supplied by the source is
  VA
j
ST 80
1736 

giving
Solution

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.

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

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.

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
IS

Solution
600 V Load
P = 120 kW
Q = 160 kVAr
IS
L
S
Let be the reference phasor.
Complex power supplied by the generator to the load is
000
,
160
000
,
120
0
600 *
*
j
I
I
V
S
S
o
S
S
L






S
V
Therefore,
(rms)
A
3
.
51
33
.
333
0
600
000
,
160
000
,
120
* o
o
S
j
I 




and
(rms)
A
3
.
51
33
.
333 o
S
I 



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
IS
Reactive
load
To obtain unity power factor, we need ensure that
has no imaginary part. Now, since
000
,
160
000
,
120 j
SL 

Let be the complex power
consumed by the compensating load.
Then, complex power supplied by the
generator to the two loads is
L
C
G S
S
S 

C
S
G
S
Solution

Therefore, we require
000
,
160
j
SC 

so that
kW
120
000
,
160
000
,
160
000
,
120






j
j
S
S
S L
C
G
Therefore, supply current at unity power factor is
S
o
o
S
G
S I
V
S
I 




 A
0
200
0
600
kW
120
*
Solution

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 = VSIS = 600 x 200 = 120 kVA
Solution

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

BEF 23803 - Lecture 7 - Complex Power Calculation.ppt

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BEF 23803 - Lecture 7 - Complex Power Calculation.ppt