Transcript Average Power

CHAPTER 3: AC
POWER ANALYSIS
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Instantaneous & Average Power
Max. Average Power Transfer
RMS Value & Apparent Power
Complex Power & Power Factor
Correction
Instantaneous Power
Instantaneous power (in watts) : the power at
any instant of time
p( t )  v( t )  i ( t )
Where:
v( t )  Vm cos(t  v )
i(t )  I m cos(t  i )
p( t ) 
Vm I m
cos( v  i )  cos( 2t   v  i )
2

Instantaneous Power
The instantaneous power p(t) entering a circuit
Average Power
Average Power (in watts) : the average of
the instantaneous power over one period
Vm I m
P
cos( v  i )
2
Vm I m
P
cos 
2
Where:
  v  i
Average Power
P  Veff Ieff cos 
Vrms or Veff 
Vm
2
I rms or I eff
Im

2
Example 1
Given that
v(t )  120 cos(377t  450 )V
i( t )  10 cos(377 t  100 )A
Find the instantaneous power and
average power absorbed by the passive
linear network.
Reference : Alexander, Sadiku Chapter 11 - page 461
Exercise 1
Calculate the instantaneous power and average
power if
v( t )  80 cos(10t  20 )V
0
i( t )  15 sin( 10t  600 )A
p( t )  385.7  600 cos( 20t  100 ) W
P  385.7W
Circuit Elements
(a) Resistors:
In purely resistive circuit, v and i are in phase.
θv = θi. Therefore θ = 0.
2
Veff
Vm I m
Vm I m
0
P
cos 0 
 Veff I eff 
 I eff
2
2
R
2 R
The average power is only dissipated in a purely resistive
circuit. For a purely inductive and capacitive, the average
power is zero.
Circuit Elements
(b) Inductors:
In purely inductive circuit, v leads by 90o,
therefore θ = 90o
Vm I m
P
cos(90 o )  0
2
(c) Capacitors:
In purely capacitive circuit, I leads by 90o,
therefore θ = - 90o
Vm I m
P
cos( 90 o )  0
2
Example 2
Find the average power supplies by the source
and the average power absorbed by the resistor.
Reference : Alexander, Sadiku Chapter 11 - page 462
Exercise 2
Calculate the average power absorbed by the resistor and
inductor. Find the average power supplies by the voltage source.
PR  9.6W
PL  0W
P  9.6W
Maximum Average Power
Transfer
Finding the maximum average power transfer:
a) circuit with a load
b) the Thevenin equivalent
Maximum Average Power
Transfer
In rectangular form, Thevenin impedance and Load impedance:
ZTh  R Th  jX Th
ZL  R L  jX L
For maximum average power transfer, the load impedance ZL must
be equal to the complex conjugate of the Thevenin impedance ZTh
ZL  R L  jX L  R Th  jX Th  Z*Th
ZL  Z*Th
Maximum Average Power
Transfer
Pmax 
VTh
2
8R Th
In a situation in which the load is purely real or purely
resistive load (XL=0), the load impedance (or resistance RL)
is equal to the magnitude of the Thevenin impedance.
RL  R
2
Th
X
2
Th
 ZTh
Example 3
Determine the load impedance ZL that maximizes the average
power drawn from the circuit of figure below. Calculate the
maximum average power.
Reference : Alexander, Sadiku Chapter 11 - page 466
Exercise 3
Find the load impedance ZL that absorbs the maximum average power
for the circuit of figure below. Calculate the maximum average power.
ZL  Z*Th  3.415  j0.7317
Pmax  1.429W
Exercise 4
In Figure below, the resistor RL is adjusted until it absorbs the
maximum average power. Calculate RL and the maximum
average power absorbed by it.
R L  30
Pmax  9.883W
Effective or RMS Value
The effective value of a periodic current is the dc
current that delivers the same average power to a
resistor as the periodic current.
Finding the effective current:
a) ac circuit
b) dc circuit
Effective or RMS Value
For any perodic function x(t) in general, the rms value is
given by:
X rms
1 T 2

x dt

T 0
The effective value of a periodic signals is its root mean
square (rms) value.
Effective or RMS Value
I rms
1 T 2
2

I
cos
tdt
m

T 0
I rms
I 2m

T
Vrms
Vm

2

T
0
1
Im
(1  cos 2t )dt 
2
2
Effective or RMS Value
The average power can be written as:
P  Vrms I rms cos(v  i )
The average power absorbed by resistor R can be
written as:
2
Vrms
P
 I 2rms R
R
Example 4
Determine the rms value of the current waveform in
figure below. If the current is passed through a 2Ω
resistor, find the average power absorbed by the resistor.
i(t)
10
0
2
4
6
8
10
t
I rms  8.165A
-10
Reference : Alexander, Sadiku Chapter 11
- page 469
P  133.3W
Exercise 5
Find the rms value of the full wave rectified sine wave in figure
below. Calculate the average power dissipated in a 6Ω resistor.
v(t)
8
0

Vrms  5.657V
2
3
P  5.334W
t
Apparent Power
The apparent power (in VA) is the product of
the rms value of voltage and current.
P  Vrms I rms cos(v  i )
P  S cos(v  i )
S  Vrms I rms
S is known as the apparent power.
Power Factor
The power factor is the cosine of the phase difference
between voltage and current. It is also the cosine of
the angle of the load impedance.
Power Factor :
where
pf  cos(v  i )
v  i
is Power Factor Angle
 pf is lagging if the current lags voltage (inductive load)
 pf is leading if the current leads voltage (capacitive load)
For purely resistive circuit, pf=1. With inductors and
capacitors in the circuit, pf may reduced to less than 1.
Example 5
A series connected load draw
a current
i( t )  4 cos(100t  100 )A
0
when the applied voltage is v(t )  120 cos(100t  20 )V
Find the apparent power and the power factor of the
load. Determine the element values that form the series
connected load.
S  240V
pf  0.866(leading )
C  212.2uF
Reference : Alexander, Sadiku Chapter 11 - page 472
Exercise 6
Calculate the power factor of the circuit below as
seen by the source. What is the average power
supplies by the source?
pf  0.936(lagging )
P  118W
Complex Power
Complex power (in VA) is the product of the rms voltage
phasor and the complex conjugate of the rms current
phasor. As a complex quantity, its real part is real power P
and its imaginary part is reactive power Q.
Complex power :
1 *
S  P  jQ  VI  Vrms I rms  v  i
2
V I
*
rms rms
2
V
S  I 2rms Z  rms
Z*
VA
Complex Power
Apparent power :
S  S  Vrms Irms  P  Q
2
2
Real power :
Reactive power :
P  Re(S)  S cos(v  i )
Q  Im( S)  S sin( v  i )
VAR
Q = 0 for resistive loads (unity power factor)
Q < 0 for capacitive loads (leading power factor)
Q > 0 for inductive loads (lagging power factor)
W
VA
Complex Power
S

Q
P
Power triangle
IZI

X
R
Impedance triangle
P
pf  cos 
S
Complex Power
Power Triangle
Example 6
The voltage across a load is
v( t )  60 cos(t  100 )V
and the current through the element in the direction
of the voltage drop is
Find
i( t )  1.5 cos(t  500 )A
a) the complex and apparent powers
b) the real and reactive powers
c) the power factor and the load impedance
Reference : Alexander, Sadiku Chapter 11 - page 475
Example 7
A load Z draws 12kVA at a power factor of 0.856
lagging from a 120 Vrms sinusoidal source.
Calculate:
a) the average and reactive powers delivered to
the load
b) the peak current
c) the load impedance
Reference : Alexander, Sadiku Chapter 11 - page 476
Exercise 7
A sinusoidal source supplies 10kVAR reactive
0
power to load Z  250  75 
Determine:
a) the power factor
b) the apparent power delivered to the load
c) the peak voltage
Power Factor Correction
The process of increasing the power factor without altering
the voltage or current to the original load is known as power
factor correction.
Most loads are inductive. A load power factor is improved
(to make closer to unity, pf=1) by installing a capacitor in
parallel with the load.
a) Original inductive load
b) inductive load with improved
power factor
Power Factor Correction
Phasor diagram showing the effect of adding a
capacitor in parallel with the inductive load
Power Factor Correction
Power triangle illustrating power factor correction
Power Factor Correction
QC  Q1  Q 2  P(tan 1  tan 2 )
2
Vrms
2
QC 
 CVrms
XC
Value of required shunt capacitance :
QC
P(tan 1  tan 2 )
C

2
2
Vrms
Vrms
Example 8
When connected to a 120 V (rms), 60Hz power
line, a load absorbs 4kW at a lagging power
factor of 0.8. Find the value of capacitance
necessary to raise the pf to 0.95.
Reference : Alexander, Sadiku Chapter 11 - page 482
Exercise 8
Find the value of parallel capacitance needed to
correct a load of 140kVAR at 0.85 lagging pf to
unity pf. Assume that the load is supplied by a
110V (rms), 60Hz line.