Transcript ac power analysis_sdnt

AC POWER ANALYSIS
Tunku Muhammad Nizar Bin Tunku Mansur
Pegawai Latihan Vokasional
Pusat Pengajian Kejuruteraan Sistem Elektrik
Content
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Average Power
Maximum Average Power Transfer
Complex Power
Power Factor Correction
2
AVERAGE POWER
3
Average Power
Average Power, in watts (W), is the average
of instantaneous power over one period
1
P  Vm I m cos( v   i )
2
4
Average Power
Resistive load (R) absorbs power all the time.
For a purely resistive circuit, the voltage and
the current are in phase (v = i).
1
1 2
1 2
P  Vm I m  I m R  | I | R
2
2
2
5
Average Power
Reactive load (L or C) absorbs zero average
power.
For a purely reactive circuit, the voltage and the
current are out of phase by 90o (v - i = ±90).
1
o
P  Vm I m cos 90  0
2
6
Exercise 11.3
Find the average power supplied by the source and the average
power absorb by the resistor
7
Solution
The current I is given by
530 o
I
 1.118 56.57 o A
4  j2
The average power supplied by the voltage source is
1
P  (5)(1.118) cos(30o  56.57 o )  2.5W
2
8
Solution
The current through the resistor is
I R  I  1.11856.57 o A
The voltage across resistor is
VR  4I R  4.47256.57 o V
The average power absorbed by the resistor is
1
P  (4.472)(1. 118)  2.5W
2
Notice that the average power supplied by the voltage source is same
as the power absorbed by the resistor.
This result shows the capacitor absorbed zero average power.
9
Practice Problem 11.3
Calculate the average power absorbed by the resistor and the
inductor. Then find the average power supplied by the voltage source
10
MAXIMUM AVERAGE POWER
TRANSFER
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Maximum Power Transfer
For maximum power transfer, the load
impedance ZL must equal to the complex
conjugate of the Thevenin impedance Zth
Z L  Z th *
R L  jX L  R th  jX th
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Maximum Average Power
The current through the load is
Vth
Vth
I

Z th  ZL (R th  jX th )  (R L  jX L )
The Maximum Average Power delivered to
the load is
1 2
| I | RL
2
| VTh |2 R L
1

2 (R Th  R L ) 2  (X Th  X L ) 2
P
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Maximum Average Power
By setting RL = Rth and XL = -Xth , the maximum average power is
Pavg,max
1 | VTh |2 | VTh |2


2 4R Th
8R Th
In a situation in which the load is purely real, the load resistance
must equal to the magnitude of the Thevenin impedance.
R L | Z th |
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Exercise 11.5
Determine the load impedance ZL that maximize the power drawn
and the maximum average power.
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Solution
First we obtain the Thevenin equivalent
To find Zth, consider circuit (a)
ZTh  j5  4 || (8  j6)
 (2.933  j4.467) Ω
To find Vth, consider circuit (b)
VTh 
(8 - j6)
(100o )
4  (8 - j6)
 7.454   10.3o V
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Solution
From the result obtained, the load impedance draws the maximum
power from the circuit when
Z L  ZTh *  (2.933  j4.467) Ω
The maximum average power is
Pmax
| VTh |2 (7.454) 2


 2.368W
8R Th
8(2.933)
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Practice Problem 11.5
Determine the load impedance ZL that absorbs the maximum
average power. Calculate the maximum average power.
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Example 11.6
Find the value of RL that will absorbs maximum average power.
Then calculate that power.
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Solution
First we obtain the Thevenin equivalent
Find Zth
Z Th  j20 || (40  j30)
 (9.412  j22.35) Ω
Find Vth
By using voltage divider
j20
VTh 
(15030 o )
j20  40  j30
 72.76134 o V
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Solution
The value of RL that will absorb the maximum average power is
R L | ZTh | (9.412) 2  (22.35) 2
 24.25
The current through the load is
VTh
72.76 134 o
I

Z Th  R L (9.412  j22.35)  24.25
 1.8100.42 o A
The maximum average power is
Pmax
1 2
1
 | I | R L  (1.8) 2 24.25  39.29W
2
2
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Practice Problem 11.6
Find the value of RL that will absorbs maximize average power,
Then calculate the power.
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COMPLEX POWER
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Complex Power
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Apparent Power, S (VA)
Real Power, P (Watts)
Reactive Power, Q (VAR)
Power Factor, cos 
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Complex Power
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Complex power is the product of the rms
voltage phasor and the complex conjugate
of the rms current phasor.
Measured in volt-amperes or VA
As a complex quantity
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Its real part is real power, P
Its imaginary part is reactive power, Q
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Complex Power (Derivation)
1
S  VI *
2
S  Vrms I rms *
Vrms
V

 Vrms θ v
2
I rms 
I
 I rms θ i
2
S  Vrms I rms θ v  θi
 Vrms I rms cos(θ v  θi )  jVrms I rms sin( θ v  θi )
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Complex Power (Derivation)
SI
2
rms
Z
S  I 2 rms (R  jX)
 I 2 rms R  jI 2 rms X
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Complex Power (Derivation)
From derivation, we notice that the real power is
P  Vrms I rms cos(θ v  θi )
or
P  I 2 rms R
and also the reactive power
Q  Vrms I rms sin( θ v  θi )
or
Q  I 2 rms X
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Real or Average Power
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The real power is the average power
delivered to a load.
Measured in watts (W)
The only useful power
The actual power dissipated by the load
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Reactive Power
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The reactive power, Q is the imaginary parts
of complex power.
The unit of Q is volt-ampere reactive (VAR).
It represents a lossless interchange between
the load and the source
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Q = 0 for resistive load (unity pf)
Q < 0 for capacitive load (leading pf)
Q > 0 for inductive load (lagging pf)
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Apparent Power
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The apparent power is the product of rms
values of voltage and current
Measured in volt-amperes or VA
Magnitude of the complex power
| S | Vrms I rms  P  Q
2
2
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Power Factor
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Power factor is the cosine of the phase
difference between voltage and current.
It is also cosine of the angle of the load
impedance.
P
pf   cos( v   i )
S
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Power Factor
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The range of pf is between zero and unity.
For a purely resistive load, the voltage
and current are in phase so that v- i = 0
and pf = 1, the apparent power is equal
to average power.
For a purely reactive load, v- i = 90 and
pf = 0, the average power is zero.
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Power Triangular
Comparison between the power triangular (a) and the impedance triangular (b).
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Problem 11.46
For the following voltage and current phasors, calculate the
complex power, apparent power, real power and reactive
power. Specify whether the pf is leading or lagging.
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a)
V = 22030o Vrms, I = 0.560o Arms.
b)
V = 250-10o Vrms, I = 6.2-25o Arms.
c)
V = 1200o Vrms, I = 2.4-15o Arms.
d)
V = 16045o Vrms, I = 8.590o Arms.
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Problem 11.48
Determine the complex power for the following cases:
a)
P = 269 W, Q = 150 VAR (capacitive)
b)
Q = 2000 VAR, pf = 0.9 (leading)
c)
S = 600 VA, Q = 450 VAR (inductive)
d)
Vrms = 220 V, P = 1 kW, |Z| = 40  (inductive)
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Problem 11.42
A 110Vrms, 60Hz source is applied to a load impedance Z. The
apparent power entering the load is 120VA at a power factor
of 0.707 lagging. Calculate
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a)
The complex power
b)
The rms current supplied to the load.
c)
Determine Z
d)
Assuming that Z = R + j L, find the value of R and L.
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Problem 11.83
Oscilloscope measurement indicate that the voltage across a
load and the current through is are 21060o V and 825o A
respectively. Determine
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a)
The real power
b)
The apparent power
c)
The reactive power
d)
The power factor
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POWER FACTOR CORRECTION
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Power Factor Correction
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The process of increasing the power
factor without altering the voltage or
current to the original load.
It may be viewed as the addition of a
reactive element (usually capacitor) in
parallel with the load in order to make
the power factor closer to unity.
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Power Factor Correction
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Normally, most loads are inductive.
Thus power factor is improved or
corrected by installing a capacitor in
parallel with the load.
In circuit analysis, an inductive load is
modeled as a series combination of an
inductor and a resistor.
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Implementation of Power Factor Correction
42
Calculation
If the original inductive load has apparent power S1, then
P = S1 cos 1
and
Q1 = S1 sin 1 = P tan 1
If we desired to increased the power factor from
cos1 to cos2 without altering the real power,
then the new reactive power is
Q2 = P tan 2
The reduction in the reactive power is caused by the shunt capacitor is given by
QC = Q1 – Q2 = P (tan 1 - tan 2)
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Calculation
The value of the required shunt capacitance is determined by the
formula
QC
P(tanθ1  tanθ 2 )
C

2
2
ωV rms
ωV rms
Notice that the real power, P dissipated by the load is not affected
by the power factor correction because the average power due to
the capacitor is zero
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Example 11.15

When connected to a 120V (rms), 60Hz power line, a
load absorbs 4 kW at a lagging power factor of 0.8.
Find the value of capacitance necessary to raise the
pf to 0.95.
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Solution
If the pf = 0.8 then,
cos1 = 0.8

1 = 36.87o
where 1 is the phase difference between the voltage and current.
We obtained the apparent power from the real power and the pf as shown below.
P
4000
S1 

 5000 VA
cos1
0.8
The reactive power is
Q1  S1 sin 1  5000 sin 36.87  3000 VAR
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Solution
When the pf raised to 0.95,
cos2 = 0.95

2 = 18.19o
The real power P has not changed. But the apparent power has changed. The
new value is
P
4000
S2 

 4210 .5VA
cos 2 0.95
The new reactive power is
Q2  S 2 sin  2  1314 .4VAR
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Solution
The difference between the new and the old reactive power is due to the parallel
addition of the capacitor to the load.
The reactive power due to the capacitor is
QC  Q1  Q2  3000  1314 .4  1685 .6VAR
The value of capacitance added is
QC
1685 .6
C

 310 .5μF
2
2
V rms 2 (60)(120)
48
Practice Problem 11.15

Find the value of parallel capacitance needed to
correct a load of 140 kVAR at 0.85 lagging pf to
unity pf. Assume the load is supplied by a 110V
(rms) 60Hz power line.
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Problem 11.82
A 240Vrms, 60Hz source supplies a parallel
combination of a 5 kW heater and a 30 kVA
induction motor whose power factor is 0.82.
Determine

a)
b)
c)
d)
The system apparent power
The system reactive power
The kVA rating of a capacitor required to adjust the
system power factor to 0.9 lagging
The value of capacitance required
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