Chapter 1 part II.ppt

Sinusoidal Steady-State
Power Calculation
1
Chapter 1 – Part
II

Introduction
• Power is the most important quantity in electrical utilities,
electronic, and communication system because such
system involve transmission of power from one point to
another.
• Most of electrical appliances has a power rating to
indicate the maximum power required.
• In AC, the high voltage power is allowed to transmit from
the power generating plant to the consumer.
2
Sinusoidal Steady-State Power
Calculation

3
Calculation
Instantaneous and Average Power
• The instantaneous power p(t) absorbed by an element is the product
of the instantaneous voltage v(t) across the element and the
instantaneous current i(t) through it.
• Lets
)
(
)
(
)
( t
i
t
v
t
p 
)
cos(
)
(
)
cos(
)
(
i
m
v
m
t
I
t
i
t
V
t
v









4
Calculation
• Therefore,
• Apply trigonometric identity
• Hence,
 
)
cos(
)
cos(
2
1
cos
cos B
A
B
A
B
A 



)
cos(
)
cos(
)
(
)
(
)
(
i
v
m
m t
t
I
V
t
i
t
v
t
p



 



   
   
)
2
cos(
2
1
)
cos(
2
1
)
cos(
2
1
)
cos(
2
1
)
(
i
v
m
m
i
v
m
m
i
v
m
m
i
v
m
m
t
I
V
I
V
t
t
I
V
t
t
I
V
t
p



























5
Calculation
• The instantaneous power change with time, therefore difficult to
measure.
• The average Power is more convenient to measure.
• In wattmeter, the instruments are responds to average power.
• The average power is
















T
i
v
m
m
T
i
v
m
m
T
i
v
m
m
T
i
v
m
m
T
dt
t
T
I
V
dt
T
I
V
dt
t
I
V
T
dt
I
V
T
dt
t
p
T
P
0
0
0
0
0
)
2
cos(
1
2
1
1
)
cos(
2
1
)
2
cos(
2
1
1
)
cos(
2
1
1
)
(
1











6
• Therefore
• Means that p(t) is time-varying and P does not depend on time.
• From
• Notice that,
Calculation
)
cos(
2
1
i
v
m
mI
V
P 
 

)
cos(
)
(
)
cos(
)
(
i
m
v
m
t
I
t
i
t
V
t
v








To phasor form i
m
v
m
I
I
V
V






 
)
sin(
)
cos(
2
1
2
1
2
1 *
i
v
i
v
m
m
i
v
m
m
j
I
V
I
V
VI














7
Calculation
• The real part of the expression is average power
• If (Purely resistive)
• If (Purely reactive)
)
cos(
2
1
]
Re[
2
1 *
i
v
m
m I
V
VI
P 
 


i
v 
 
*
2
2
2
2
1
2
1
2
1
I
I
I
where
R
I
R
I
I
V
P m
m
m






90


 i
v 

0
90
cos
2
1

 
m
m I
V
P

• Example 1
Find instantaneous power and average power
8
Calculation
A
t
t
i
V
t
t
v
)
10
377
cos(
10
)
(
)
45
377
cos(
120
)
(






 
)
cos(
)
cos(
2
1
cos
cos B
A
B
A
B
A 




9
Calculation
• Example 2
Calculate the average power absorbed by impedance Z
when the voltage across it


 70
30 j
Z
V
V 
0
12


10
Calculation
Average and Reactive Power

11
Calculation
Average and Reactive Power
• What is the corresponding relation in phasor domain?

12
Calculation
The RMS Value and Power Calculations
• The effective value, Ieff of a periodic current i(t) is the equivalent dc
current that delivers the same average power to a resistor as the
periodic current.

13
Calculation
The RMS Value and Power Calculations

14
Calculation
Complex Power

15
Calculation
Complex Power

16
Calculation
Complex Power

17
Calculation
Complex Power
Q is measure of the energy exchange between the source and the
reactive part of the load.
Reactive power represents the lossless interchange between the
and the source.

18
Calculation
Complex Power

19
Calculation
Complex Power
Lagging pf means current lags
voltage
Leading pf means current leads
voltage

20
Calculation
Power Calculations

21
Calculation
Power Calculations

22
Calculation
Power Calculations

23
Calculation
Power Calculations

24
Calculation
Power Calculations

25
Calculation
Power Calculations

26
Calculation
Power Calculations
• Example
Given the load voltage and current
Find:
a) S
b) P and Q
c) pf and ZL

27
Calculation
Power Calculations
• Example (Solution)

28
Calculation
Power Calculations
• Conservation of AC Power
The complex power supply by the
source

29
Calculation
Power Calculations
• Conservation of AC Power
The complex power supply by the
source
Similarly
** The complex (Real or Reactive) power of the source equals
the respective sum of the complex (Real or Reactive) powers
of the individual load.

30
Calculation
Power Calculations
• Most industrial loads are inductive, and operated at low lagging
power factor

31
Calculation
Power Calculations
• This will sacrificing some generator real power output capability to
provide the reactive power.
• Also from the users’ view point, to reduce energy cost due to the
penalty of low pf. It is worth to increase the pf.
• Solution:
» for inductive loads,
» add parallel capacitor,
» without influencing the average power, P.

32
Calculation
Power Calculations
• This will sacrificing some generator real power output capability to
provide the reactive power.
• Also from the users’ view point, to reduce energy cost due to the
penalty of low pf. It is worth to increase the pf.
• Solution:
» for inductive loads,
» add parallel capacitor,
» without influencing the average power, P.

33
Calculation
Power Calculations
Original load,
After power factor correction,
Then,

34
Calculation
Power Factor Correction
2
2
1
2
2
2
2
1
2
1
2
2
2
2
2
1
1
1
1
1
1
1
)
tan
(tan
,
)
tan
(tan
tan
cos
,
tan
sin
cos
rms
rms
c
rms
c
rms
c
c
V
P
V
Q
C
Finally
CV
X
V
Q
P
Q
Q
Q
P
Q
Therefore
S
P
P
P
Since
P
S
Q
S
P





























35
Calculation
Power Factor Correction
• Example 1
Voltage source of
The electrical load absorbs 4 kW at a lagging power factor 0.8. Find C
to raise the power factor to 0.95.
• Example 2
Fine the value of parallel capacitance needed to correct a load of 140
kVAR at 0.85 lagging pf to unity pf. The source is 110 V, 60 Hz line.

Chapter 1 part II.ppt

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Chapter 1 part II.ppt