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CHAPTER II
AC POWER CALCULATIONS
1
Contents
• Introduction
• Instantaneous and Average Power
• Effective or RMS Value
• Apparent Power
• Complex Power
• Conservation of AC Power
• Power Factor and Power Factor Correction
• Maximum Average Power Transfer
• Applications
2
Introduction
• Every electrical device has a power rating that indicates
how much power the equipment requires; exceeding the
power rating can cause permanent damage.
• The choice of power delivery in 50- or 60-Hz ac form is
due to the allowed high-voltage power transformation.
3
Instantaneous Power
frequencyangular cet twiindependen time
)2cos(2
1)cos(
2
1
)cos()cos()()()(
)cos()(
)cos()(
,excitation sinusoidal Assuming
(W) watts)()()(
as defined ispower ousinstantane The
ivmmivmm
ivmm
im
vm
tIVIV
ttIVtitvtp
tIti
tVtv
titvtp
4
Average Power
)cos(2
1
)2cos(1
2
11)cos(
2
1
)2cos(2
11)cos(
2
11
)cos()(
)cos()( )(
1
period. oneover
power ousinstantane theof average theispower average The
00
00
0
ivmm
T
ivmm
T
ivmm
T
ivmm
T
ivmm
im
vmT
IV
dttT
IVdtT
IV
dttIVT
dtIVT
P
tIti
tVtvdttp
TP
5
Average Power (Cont’d)
power. average zero absorbs )or ( load
reactive a while time,allat power absorbs load resistiveA
090cos2
1
circuit reactivepurely afor 90 :2 Case
.0 if , 02
1
circuit resistivepurely afor :1 Case
2
1
2
1Re
2
1)cos(
2
1
)cos()(
2
22*
CL
IVP
RRP
RRIIVP
RV
IjXR
I
V
mm
iv
iv
mivmm
m
miviv
m
m
I
IVI
I
VZ
6
Example 1
W2.344)cos(2
1
W)35754cos(6002.344
)35754cos(55cos600
)10377cos()45377cos(1200
:Sol
power. average theandpower ousinstantane thefind
)10377cos(10)(
)45377cos(120)(
Given that
ivmmIVP
t
t
ttvip
tti
ttv
7
Example 2
W24.37)8.66cos()576.1)(120(2
1
8.66)576.1)(120(Re2
1Re
2
1
8.66576.18.6616.76
0120
7030
0120
:Sol
it.. across
applied is 0120 voltagea when )7030(
impedancean by absorbedpower average the theFind
*
VI
Z
VI
VZ
P
j
j
8
Example 3
W5.2
)4()118.1(2
1
W5.2
)57.5630cos()118.1)(5(2
1
2
R
V
P
P
57.56118.1
57.26472.4
305
24
305
:Sol
resistor. by the
absorbedpower average the
and source by the supplied
power average theFind
jZ
VI
9
Effective or RMS Value
rms0
2
effrms0
2
eff
2
eff2
eff
0
2
0
2
0
2
1 ,
1
iscircuit dc in the
resistor by the absorbedpower theWhile
11
iscircuit ac
in theresistor by the absorbedpower The
current. periodic theasresistor a power to
same thedeliversat current th dc theis
current periodic a of The
VdtvT
VIdtiT
I
R
VRIP
dtvRT
dtiT
RRdti
TP
valueeffective
TT
TTT
ac circuit
dc circuit
10
Effective or RMS Value (Cont’d)
)cos(
)cos(22
)cos(2
1
as written becan power average The
2
,cos)(for Similarly,
22cos1
2
1cos
1
is valusrms the,cos)( sinusoid For the
1
bygiven is valuerms the,)(function periodicany For
rmsrms
rms
0
2
0
22
rms
0
2
rms
iv
ivmm
ivmm
m
m
mT
mT
m
m
T
IV
IVIVP
VV
tVtv
Idtt
T
ItdtI
TI
tIti
dtxT
X
tx
11
Complex Power
)sin()cos(
)(
2
2 where
2
1
asgiven is load ac the
by absorbed power complex The
as formphasor in given are
voltageandcurrent thegConsiderin
rmsrms
rmsrms
rmsrms
rmsrms
*
rmsrms
*
iviv
iv
i
v
im
vm
jIV
IV
I
V
I
V
S
II
VV
IVVIS
S
I
V
power Reactive: )Im(
power Real: )Re(
)(
, Since
)(
2
rms
2
rms
2
rms
*
2
rms2
rms
*
rmsrms
rms
rms
rms
rms
XIQ
RIP
jQPjXRI
jXR
VI
I
Viv
S
S
S
Z
ZZIVS
I
V
I
VZ
12
Complex Power (Cont’d)
• P is the average or real power.
– The power delivered to the load
– The actual power dissipated by the load
• Q is the reactive or quardrature power.
– Unit: volt-ampere reactive (VAR)
– A measure of the energy exchange between the source and
the reactive part of the load
– Q = 0 for resistive loads (unity pf)
– Q < 0 for capacitive loads (leading pf)
– Q > 0 for inductive loads (lagging pf)
13
Summary
)(Impedance
)sin()Im(Power Reactive
)cos()Re(Power Real
PowerApparent
)(
2
1PowerComplex
rms
rms
rms
rms
22
rmsrms
rmsrms
*
iv
iv
iv
iv
I
V
SQ
SP
QPIVS
IV
jQP
I
V
I
VZ
S
S
S
VIS
14
Power Triangle
Power triangle Impedance triangle
Power triangle
15
Apparent Power and Power Factor
angleorPower fact
orPower fact
owerApparent pIVS
S
S
IV
IVP
I
V
tIti
tVtv
iv
iv
iv
iv
ivmm
im
vm
im
vm
:
: )cos(pf
:VA)(unit
re whe
pf
)cos(
)cos(
)cos(2
1
ispower average The
or )cos()(
)cos()(
are voltageandcurrent theIf
rmsrms
rmsrms
I
V
16
S and pf (Cont’d)
load an voltagelagscurrent means pf
load a voltageleadscurrent means pf
)(
2
2 Since
)(
)cos(pf
Impedance Load theof Angle AngleFactor Power
rms
rms
rms
rms
rmsrms
rmsrms
m
m
m
m
inductiveLagging
capacitiveLeading
I
V
I
V
I
V
I
V
iv
i
v
iv
i
v
iv
I
V
I
VZ
II
VV
I
VZ
17
CONSERVATION OF AC POWER
Whether the loads are connected in series or in parallel (or in general),
the total power supplied by the source equals the total power delivered to
the load. Thus, in general, for a source connected to N loads,
18
lagging0.84pf
40kW
1.0 25.0jDetermine real and reactive power losses
And real and reactive power supplied
}Re{SP pfSS iv ||)cos(||
kVApf
PSL 62.47
84.
40||
rmsL
LL A
V
SIVIS )(45.216
||
||||
*
86.32)cos( ivivpf
)(839,25||||22
VAPSQ LL
rmsL AI )(86.3245.216
2*losses ||)( LlineLLline IZIIZS
VAj 713,11685,4
Balance of power
kVAjjj
SSS
552.37685.44839.2540713.11685.4
loadlossessupplied
2losses )45.216)(25.01.0( jS
LEARNING EXAMPLE
19
kVAj
S
839.2540
load
Power Factor Correction
2
1
IIII
VI
IV
I
CL
C
LL
Cj
LjR
pf
correction
Most loads
are inductive.
• It is the process of increasing the power factor without
altering the voltage or current to the original load.
20
pf Correction (Cont’d)
zero. is because correction pf the
by affectednot is power real that theNote
)tan(tan
But
)tan(tan
giveson conservatipower ac theApplying
tansin
tansin
coscos
power, real thealtering without cos
tocos from pf increase todesire weIf
2
rms
21
2
rms
2
rms
2
rms
*
2
rms
2121
2222
11111
222111
2
1
C
C
C
C
C
P
P
V
P
V
QC
CVX
VQ
V
PQQQ
PSQ
PSQ
PSPSP
ZS
21
Example
0.95. topf theraise tonecessary ecapacitanc theFind
0.8. offactor power lagging 1at kW -4 absorbs load a
line,power Hz-60 V(rms),-120 a toconnectedWhen
VAR
.sinsinSQ
VA cos
PScosSP
..cospf
.pf
W P
rad/s )(
V V
have We
:Sol
rms
3000
87365000
5000
873680
80
4000
120602
120
111
1111
11
F 5.310 120120
6.1685
VAR 6.1685
4.13143000
VAR 4.1314
19.18sin5.4210sin
VA 5.4210cos
19.1895.0cos
0.95, toraised is pfWhen
22
rms
21
212
2
2
22
V
QC
QQQ
SQ
PS
C
C
22
23
Example
Example
P.F = cos (7.89+40.43) =0.665
S=VI* = 301.37*403<7.89+40.43 VA
S=121452 <48.32
S=80762 + j 90709
New p.f =0.95 , s2 = 18.1948
Qc = P(Tans1 – Tans2)=64156 VAR
FVS
QC
capacitor
1878)602()37.301(
64156
|| 22
24
Calculate the source complex power and power factor,
Show how the power factor can be improved to 0.95
Example
973402413
24137
6186246
.).cos(pf
is factor power The
.
.j.)-j(||Z
is impedance total The
:Sol
source.theby delivered
power average the Calculate
source.theby seenas
factor power the Determine
W125
pf)286.4)(30(pf
24.13286.4
24.137
030
rmsrms
rmsrms
IVP
Z
VI
25
Maximum Power Transfer
Th
2
Th
maxTh*
ThTh
Th
Th
22
Th
2
Th
Th
2
Th
2
Th
2
Th
22
Th
2
Th
Th
2
Th
2
Th
2
Th
2
Th2
ThTh
Th
Th
Th
ThThTh
8 ,
0)()(2
)(2)()(
0)()(
)(
power, maximumwith condition thefind To
)()(2
1
2
1
)()(
RPjXR
RR
XX
XXRR
RRRXXRR
dR
dP
XXRR
XXR
dX
dP
XXRR
RRP
XXjRR
jXR
jXR
L
L
L
LL
LLLL
L
LL
LL
L
LL
L
L
LLL
LLL
VZZ
V
V
VI
V
ZZ
VI
Z
Z
26
Maximum Average Power Transfer
LLL X j R Z
THTHTH X j R Z
The maximum average power can be transferred to the load if
XL = –XTH and RL = RTH
TH
2
TH
maxR 8
V P
If the load is purely resistive, thenTH
2
TH
2
THL Z X R R
27
28
Find the value of ZL in the circuit shown for maximum power transfer.
29
Solution
