Download Extra Chapter 3 - AC Power Analysis

AC POWER ANALYSIS
Tunku Muhammad Nizar Bin Tunku Mansur
Pegawai Latihan Vokasional
Pusat Pengajian Kejuruteraan Sistem Elektrik
Content




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
Solution
The current I is given by
845o
I
 2.5326.57 o A
3 j
For the resistor
I R  I  2.5326.57 o A
VR  3I  7.5926.57 o V
1
1
PR  Vm I m  (7.59)(2.53)  9.6W
2
2
11
Solution
For the inductor
I L  I  2.5326.57 o A
VL  jI L  2.53116.57 o V
1
PL  (2.53)(2.53)cos(90o )  0W
2
The average power supplied by the voltage source is
1
P  (8)(2.53)cos(45o  26.57 o )  9.6W
2
Notice that the average absorbed by the resistor is same as the power
supplied by the voltage source.
This result shows the inductor also absorbed zero average power.
12
MAXIMUM AVERAGE POWER
TRANSFER
13
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
14
Maximum Average Power
The current through the load is
Vth
Vth
I

Zth  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
15
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 |
16
Exercise 11.5
Determine the load impedance ZL that maximize the power drawn
and the maximum average power.
17
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
18
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)
19
Practice Problem 11.5
Determine the load impedance ZL that absorbs the maximum
average power. Calculate the maximum average power.
20
Solution
First we obtain the Thevenin equivalent
To find Zth, consider circuit (a)
Z Th  5 || (8  j4  j10)
 (3.415  j0.7317) Ω
To find Vth, consider circuit (b)
By using current divider
8  j4
I
(2)
8  j4  5  j10
 1.249   51.34o A
 VTh  5I  6.25  51.34 o V
21
Solution
From the result obtained, the load impedance draws the maximum
power from the circuit when
Z L  ZTh *  (3.415  j0.7317) Ω
The maximum average power is
Pmax
| VTh |2
(6.25) 2


 1.429 W
8R Th
8(3.415)
22
Example 11.6
Find the value of RL that will absorbs maximum average power.
Then calculate that power.
23
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
24
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
25
Practice Problem 11.6
Find the value of RL that will absorbs maximize average power,
Then calculate the power.
26
Solution
First we obtain the Thevenin equivalent
To find Zth let
Z1  80  j60
and Z 2  90 || ( j30)  (9  j27)
Then
ZTh  Z1 || Z 2  (17.181  j24.57) Ω
To find Vth
By using voltage divider
VTh 
Z2
(12060 o )
Z1  Z2
 35.98  31.91o V
27
Solution
The value of RL that will absorb the maximum average power is
R L | ZTh | 30
The current through the load is
VTh
35.98 - 31.91o
I

Z Th  R L (17.181  j24.57)  30
 0.6764   4.4o A
The maximum average power is
Pmax
1 2
1
 | I | R L  (0.6764) 2 (30)  6.863W
2
2
28
COMPLEX POWER
29
Complex Power




Apparent Power, S (VA)
Real Power, P (Watts)
Reactive Power, Q (VAR)
Power Factor, cos 
30
Complex Power



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


Its real part is real power, P
Its imaginary part is reactive power, Q
31
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 )
32
Complex Power (Derivation)
SI
2
rms
Z
S  I 2 rms (R  jX)
 I 2 rms R  jI 2 rms X
33
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
34
Real or Average Power




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
35
Reactive Power



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



Q = 0 for resistive load (unity pf)
Q < 0 for capacitive load (leading pf)
Q > 0 for inductive load (lagging pf)
36
Apparent Power



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
37
Power Factor


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
38
Power Factor



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.
39
Power Triangular
Comparison between the power triangular (a) and the impedance triangular (b).
40
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.

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.
41
Solution
a)
S = VI* = (22030o)( 0.5-60o)
= 110-30o VA = 95.26 – j55 VA
c)
Apparent power = 110 VA
Real Power = 95.26 W
Reactive Power = -55 VAR
pf is leading because current leads voltage
b)
S = VI* = (250-10o)(6.225o)
= 155015o VA = 1497.2 + j401.2 VA
Apparent power = 1550 VA
Real Power = 1497.2 W
Reactive Power = 401.2 VAR
pf is lagging because current lags voltage
S = VI* = (1200o)( 2.415o)
= 28815o VA = 278.2 + j74.54 VA
Apparent power = 288 VA
Real Power = 278.2 W
Reactive Power = 74.54 VAR
pf is lagging because current lags voltage
d)
S = VI* = (16045o)(8.5-90o)
= 1360-45o VA = 961.7 – j961.7 VA
Apparent power = 1360 VA
Real Power = 961.7 W
Reactive Power = -961.7 VAR
pf is leading because current leads voltage
42
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)
43
Solution
a) Given P = 269W, Q = 150VAR (capacitive)
Complex power,
S  P  jQ  (269  j150)VA
 308   29.14 o VA
b) Given Q = 2000VAR, pf = 0.9 (leading)
pf  cos  0.9    25.84 o
Q
2000
Q  S sin   S 

 4588 .31
o
sin  sin( 25.84 )
P  S cos  4129 .48
Complex power,
S  P  jQ  (4129  j2000)VA
 4588  - 25.84 o VA
44
Solution
c) Given S = 600VA, Q = 450VAR (inductive)
Q 450
Q  S sin   sin   
 0.75
S 600
  48 .59 o
pf  cos  0.6614
P  S cos  396 .86
Complex power,
S  P  jQ  (396.9  j450)VA
 600 48.59 o VA
45
Solution
d) Given Vrms = 220V, P = 1kW, |Z| = 40 (inductive)
| V |2 220 2
S

 1210
| Z|
40
P 1000
P  S cos  cos  
 0.8264
S 1210
  34.26 o
Q  S sin   681 .25
Complex power,
S  P  jQ  (1000  j681.2)VA
 1210 34.26 o VA
46
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

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.
47
Solution
Given S = 120VA,
pf = 0.707 = cos

 = 45o
a) the complex power
S  S cos  jS sin   84.84  j84.84VA
b) the rms current supplied to the load
S  Vrms I rms
S
120
 I rms 

 1.091A
Vrms 110
48
Solution
c) the impedance Z
S  I rms Z
S
Z 
 (71.278  j 71.278)
2
I rms
2
d) value of R and L
If Z = R + jL
then
Z = 71.278 + j 71.278
 R  71.278Ω
ωL  71.278
71.278
L 
 0.1891H
2f
49
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

a)
The real power
b)
The apparent power
c)
The reactive power
d)
The power factor
50
Solution
a) the real power
1
1
S  VI*  (21060 o )(8  25o )
2
2
 (84035o )VA  (688.1  j481.8)VA
P  S cos(35o )  840 cos(35o )  688.1W
b) the apparent power
S  840 VA
c) the reactive power
Q  S sin (35o )  840 sin (35o )  481.8VAR
d) the power factor
pf 
P
 cos(35o )  0.8191(lag ging)
S
51
POWER FACTOR CORRECTION
52
Power Factor Correction


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.
53
Power Factor Correction


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.
54
Implementation of Power Factor Correction
55
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)
56
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
57
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.
58
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
59
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
60
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)
61
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.
62
Solution
If the pf = 0.85 then,
cos1 = 0.85

1 = 31.79o
where 1 is the phase difference between the voltage and current.
We obtained the apparent power from the reactive power and the pf as shown
below.
S1 
Q1
140kVAR

 265.8kVA
sin 1 sin( 31.79)
The real power is
P  S1 cos1  265 .8 cos(31.79)  225 .93kW
63
Solution
When the pf raised to 1 (unity),
cos2 = 1

2 = 0o
The real power P has not changed. But the apparent power has changed. The
new value is
S2 
P
225.93kW

 225.93kVA
cos 2
1
The new reactive power is
Q2  S 2 sin  2  0
64
Solution
The difference between the new and the old reactive power due to the parallel
addition of the capacitor to the load.
The reactive power due to the capacitor is
QC  Q1  Q2  140000  0  140 kVAR
The value of capacitance is
QC
140kVAR
C

 30.69mF
2
2
V rms 2 (60)(110)
65
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
66
Solution
For the heater
P1 = 5000
Q1 = 0
For the 30kVA induction motor, the pf = 0.82 then,
cos1 = 0.82

1 = 34.92o
The real and the reactive power for the induction motor
P2  S 2 cos(34.91)  30000 x 0.82  24600 W
Q2  S 2 sin( 34.91)  17171kVAR
67
Solution
The total system complex power
Stotal = S1 + S2 = (P1 + P2) + j (Q1 + Q2) = 29600 + j17171
The system apparent power
S = |Stotal| = 34.33kVA
The system reactive power
Q = 17171 kVAR
The system power factor
pf 
P 29600

 0.865
S 34220
68
Solution
The system pf = 0.865 then,
cos1 = 0.865
The new system pf = 0.9 then,
cos2 = 0.9

1 = 30.12o

2 = 25.84o
The rating for the capacitance required to adjust the power factor to 0.9
QC = P (tan 1 + tan 2) = 29600 (tan 30.12 + tan 25.84) = 2833kVAR
69
Solution
The value of capacitance is
QC
2833
C

 130.46μF
2
2
V rms 2 (60)(240)
70
Problem 11.91
The nameplate of an electric motor has the
following information.







Line voltage: 220 Vrms
Line current: 15 Arms
Line frequency: 60 Hz
Power: 2700 W
Determine the power factor (lagging) of the
motor.
Find the value of the capacitance, C that must
be connected across the motor to raise the pf
to unity.
71
Additional Problem
The nameplate of an electric motor has the
following information.








Line voltage: 230 Vrms
Line current: 13 Arms
Line frequency: 60 Hz
Reactive Power: 1500 VAR
Determine the power factor (lagging) of the
motor.
Find the reactive power required to increase
the pf to 0.95.
Find the value of the capacitance, C.
72