Transcript Slide 1

AC POWER CALCULATION
Instantaneous, average and reactive power
Apparent Power and Power Factor
Complex Power
SEE 1023 Circuit Theory
Dr. Nik Rumzi Nik Idris
1
Instantaneous, Average and Reactive Power
i(t)
+
v(t)

Passive, linear
network
Instantaneous power absorbed by the network is,
p =v(t).i(t)
Let v(t) = Vm cos (t + v) and i(t) = Imcos(t + i)
Which can be written as
v(t) = Vm cos (t + v  i) and i(t) = Imcos(t)
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v(t) = Vm cos (t + v  i) and i(t) = Imcos(t)
p = Vm cos(t + v – i ) . Im cos(t)
Example when v  i = 45o
45o
2
v
1
i
0
-1
-2
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
positive p = power
transferred from source
to network
2
1.5
Instantaneous
Power (p)
1
negative p = power
transferred from network
to source
0.5
0
-0.5
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
3
v(t) = Vm cos (t + v  i) and i(t) = Imcos(t)
p = Vm cos(t + v – i ) . Im cos(t)
Using trigonometry functions, it can be shown that:
VmIm
VmIm
VmIm
cos( v  i ) 
cos(v  i ) cos 2t 
sin( v  i ) sin 2t
p=
2
2
2
Which can be written as
p = P + Pcos(2t)  Qsin(2t)
P
VmIm
cos( v  i )
2
VmIm
Q
sin(  v  i )
2
= AVERAGE POWER (watt)
= REACTIVE POWER (var)
4
VmIm
VmIm
VmIm
cos( v  i ) 
cos(v  i ) cos 2t 
sin( v  i ) sin 2t
p=
2
2
2
5
p=
VmIm
VI
VI
cos( v  i )  m m cos(v  i ) cos 2t  m m sin( v  i ) sin 2t
2
2
2
2
1
0
-1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Example for
v-i = 45o
1.5
1
0.5
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
1
0.5
0
-0.5
-1
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p=
VmIm
VI
VI
P cos(2t)
Q sin(2t)
cos( v  i )  p =mPm+cos(
v  i )cos
2t  m m sin( v  i ) sin 2t
2
2
2
2
1
0
-1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
1.5
P = average power
1
0.5
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
1
0.5
Q = reactive power
0
-0.5
-1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
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p = P + P cos(2t)  Q sin(2t)
P = AVERAGE POWER
• Useful power – also known as ACTIVE POWER
• Converted to other useful form of energy – heat, light, sound, etc
• Power charged by TNB
Q = REACTIVE POWER
• Power that is being transferred back and forth between load and source
• Associated with L or C – energy storage element – no losses
• Is not charged by TNB
• Inductive load: Q positive, Capacitive load: Q negative
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Power for a resistor
Voltage and current are in phase,
p=
p=
(v  i )  0
VmIm
VI
VI
cos( v  i )  m m cos(v  i ) cos 2t  m m sin( v  i ) sin 2t
2
2
2
VmIm
VI
cos 0  m m cos 0 cos 2t
2
2
p=
VmIm
(1  cos 2t )
2
2
1
0
-1
-2
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
2.5
P = average power =
2
1.5
VmIm
2
1
0.5
Q = reactive power = 0
0
-0.5
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
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Power for an inductor
(v  i )  90o
Voltage leads current by 90o,
p=
VmIm
VI
VI
cos( v  i )  m m cos(v  i ) cos 2t  m m sin( v  i ) sin 2t
2
2
2
p= 
VmIm
sin( 90O ) sin 2t
2
VmIm
sin 2t
p= 
2
2
v
1
i
0
-1
-2
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
2
P = average power = 0
1
0
-1
-2
Q = reactive power =
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
VmIm
2
10
Power for a capacitor
(v  i )  90o
Voltage lags current by 90o,
p=
p =
VmIm
VI
VI
cos( v  i )  m m cos(v  i ) cos 2t  m m sin( v  i ) sin 2t
2
2
2
VmIm
sin( 90O ) sin 2t
2
p=
VmIm
sin 2t
2
2
v
1
i
0
-1
-2
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
1.5
P = average power = 0
1
0.5
0
Q = reactive power = 
-0.5
-1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
VmIm
2
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Apparent Power and Power Factor
Consider v(t) = Vm cos (t + v) and i(t) = Imcos(t + i)
We have seen,
P
VmIm
cos( v  i )
2
Vm Im

2 2
=
Vrms  Irms
S  Vrms  Irms
Is known as the APPARENT POWER
VA
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Apparent Power and Power Factor
We can now write,
P  S cos(v  i )
The term cos(v  i ) is known as the POWER FACTOR
POWER FACTOR  pf 
P
 cos( v  i )
S
For inductive load, (v  i) is positive  current lags voltage  lagging pf
S  Vrms  Irms
For capacitive load, (v  i) is negative  current leads voltage  leading pf
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Apparent Power and Power Factor
POWER FACTOR  pf 
P
 cos( v  i )
S
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Apparent Power and Power Factor
POWER FACTOR  pf 
P
 cos( v  i )
S
Irms = 5- 40o
Vrms =
+
25010o
Source
+

VL
Load

Power factor of the load = cos (10-(-40)) = cos (50o) = 0.6428
(lagging)
Apparent power, S = 1250 VA
Active power absorbed by the load is 250(5) cos (50o)= 1250(0.6428) = 803.5 watt
Reactive power absorbed by load is 250(5) sin (50o)= 1250(0.6428) = 957.56 var
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Complex Power
Defined as:
S
Where,
If we let
V  Vmv
V I *
2
(VA)
and
I  Imi
V
Vrm s  m v  Vrmsv
2
S  Vrm s  I * rm s
and

Irm s 
I*  Im  i
Im
i  Irms i
2
(VA)
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Complex Power
S
Where,
S

V I *
2
(VA)
1
Vm v  Im  i
2
1
VmIm( v  i )
2
 VrmsIrms(v  i )
 VrmsIrms cos(v  i )  jVrmsIrms sin( v  i )
 P  jQ
P  Re( S)  VrmsIrmscos( v - i )
Q  Im( S)  VrmsIrmssin( v - i )
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Complex Power
S  P  jQ
The complex power contains all information about the load
o
Irms = 540
We have seen before:
Apparent power, S = 1250 VA
Vrms =
+
25010o
Source
+

VL
Active power, P = 803.5 watt
Load
Reactive
power, Q = 957.56 var
 With complex power,
S = 25010o (5-40o) VA
S = 1250 50o VA
S
50o
803.5 watt
957.56 var
S = (803.5 + j957.56) VA
|S| = S = Apparent power
= 1250 VA
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Complex Power
Other useful forms of complex power
S  Vrm s  I * rm s
We know that Vrm s  ZIrm s

S  ZIrms  I *rms
S  Z Ir m s
2
S  Irm s (R  jX )
2
S  ( Irm s R  j Irm s X)
2
P
2
Q
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Complex Power
Other useful forms of complex power
S  Vrm s  I * rm s
We know that

Irm s 
Vrm s
Z
 Vrm s 
S  Vrm s  

 Z 
For a pure resistive element,
For a pure reactive element,
P
Q
*

Vrm s
S
Vrm s
2
Z
2
R
Vrm s
2
X
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Conservation of AC Power
Complex, real, and reactive powers of the sources equal the respective
sums of the complex, real and reactive powers of the individual loads
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Conservation of AC Power
Complex, real, and reactive powers of the sources equal the respective
sums of the complex, real and reactive powers of the individual loads
Ss = Ps +jQs = (P1 + P2 + P3) + j (Q1 + Q2 + Q3)
But
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Maximum Average Power Transfer
Max power transfer in DC circuit can be applied to AC circuit analysis
ZTh
I
+
VTh
AC linear
+
circuit

V
ZL

What is the value of ZL so that maximum average power is transferred to it?
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Maximum Average Power Transfer
ZTh
I
+
VTh
+

V
ZL

What is the value of ZL so that maximum average power is transferred to it?
24
Maximum Average Power Transfer
What is the value of ZL so that maximum average power is transferred to it?
ZTh
I
ZTh= RTh + jXTh
+
VTh
+

V
ZL
ZL= RL + jXL

1 2
P  I RL
2
P max when
P
P
 0 and
0
RL
XL
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Maximum Average Power Transfer
What is the value of ZL so that maximum average power is transferred to it?
ZTh
I
I
+
VTh
+

V
1 2
P  I RL
2
ZL
P

P max when
(R Th
VTh
 jX Th )  (RL  jX L )
VT h
(RTh
2
RL
 RL )2  ( XL  XTh )2 2
P
P
 0 and
0
RL
XL
XL = XTh , RL= RTh
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