AC Power Analysis
Download
Report
Transcript AC Power Analysis
AC Power Analysis
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
•
•
•
•
•
•
•
•
•
Introduction
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
Applications
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.
Instantaneous Power
T heinstantaneous power is defined as
p(t ) v(t )i (t ) watts(W)
Assuming sinusoidal excitation,
v(t ) Vm cos(t v )
i (t ) I m cos(t i )
p(t ) v(t )i (t ) Vm I m cos(t v ) cos(t i )
1
1
Vm I m cos( v i ) Vm I m cos(2t v i )
2
2
timeindependent
twice angular frequency
Average Power
T heaveragepower is theaverageof theinstantaneous power
over one period.
1
P
T
T
0
p (t )dt
v(t ) Vm cos(t v )
i (t ) I m cos(t i )
1 T1
1 T1
P Vm I m cos( v i )dt Vm I m cos(2t v i )dt
T 0 2
T 0 2
1
1 T
1
1 T
Vm I m cos( v i ) dt Vm I m cos(2t v i )dt
2
T 0
2
T 0
1
1
Vm I m cos( v i ) Re VI*
2
2
1 * 1
1
VI Vm I m v i Vm I m cos v i j sin v i
2
2
2
Average Power (Cont’d)
Im
V Vm
Z
( v i ) R jX cos( v i )
R
I Im
Vm
1
1
1 2
1 2
*
P Vm I m cos( v i ) Re VI I m R I R
2
2
2
2
Case 1 : v i for a purely resistivecircuit
1 2
P I R 0 , if R 0.
2
Case 2 : v i 90 for a purely reactivecircuit
1
P Vm I m cos90 0
2
A resistiveload absorbs power at all time,while a reactive
load ( L or C ) absorbs zero averagepower.
Example 1
Given that
v(t ) 120cos(377t 45)
i (t ) 10 cos(377t 10)
find theinstantaneous power and theaveragepower.
Sol :
p vi 1200cos(377t 45) cos(377t 10)
600cos55 cos(754t 35)
344.2 600cos(754t 35) W
1
P Vm I m cos( v i ) 344.2 W
2
Example 2
Find the theaveragepower absorbed by an impedance
Z (30 j 70) when a voltageV 1200 is applied
acrossit..
Sol :
V 1200
1200
I
1.57666.8
Z 30 j 70 76.16 66.8
1
1
*
P Re VI Re(120)(1.576) 66.8
2
2
1
(120)(1.576) cos(66.8) 37.24 W
2
Example 3
Find theaveragepower
supplied by thesource and
theaveragepower absorbed
by theresistor.
Sol :
V 530
I
Z 4 j2
530
4.472 26.57
1.11856.57
1
PV (5)(1.118) cos(30 56.57)
2
2.5 W
1
PR (1.118) 2 (4)
2
2.5 W
Maximum Power Transfer
Z Th RTh jX Th
Z L RL jX L
VTh
VTh
I
Z Th Z L ( RTh RL ) j ( X Th X L )
2
P
VTh RL
1 2
1
I RL
2
2 ( RTh RL ) 2 ( X Th X L ) 2
T o find theconditionwith maximumpower,
VTh RL ( X Th X L )
dP
dX L
( RTh RL ) 2 ( X Th X L ) 2
2
2
0
VTh ( RTh RL ) 2 ( X Th X L ) 2 2 RL ( RTh RL )
dP
0
2
2
2
dRL
2 ( RTh RL ) ( X Th X L )
2
V
X L X Th
Z L RTh jX Th Z*Th , Pmax Th
8 RTh
RL RTh
2
Maximum Power Transfer (Cont’d)
Z Th RTh jX Th
Z L RL
2
P
VTh RL
1 2
1
I RL
2
2 ( RTh RL ) 2 X Th 2
T o find thecondit ionwith maximumpower,
dP
dRL
VTh ( RTh RL ) 2 X Th 2 RL ( RTh RL )
2
2
2 ( RTh RL ) X Th
0
RL RTh X Th Z Th
2
2
2
2 2
Effective or RMS Value
T heeffective valueof a periodiccurrent
is thedc current that delivers thesame
power toa resistoras theperiodiccurrent.
T hepower absorbed by theresistorin the
ac circuit is
1 T 2
R T 2
1 T 2
P i Rdt i dt
v dt
0
0
0
T
T
RT
While thepower absorbed by theresistor
in thedc circuit is
2
V
2
P I eff
R eff
R
1 T 2
1
I eff
i dt I rms , Veff
0
T
T
T
0
v 2 dt Vrms
ac circuit
dc circuit
Effective or RMS Value (Cont’d)
For any periodicfunctionx(t ), the rms valueis given by
1 T 2
X rms
x dt
0
T
For thesinusoid i (t ) I m cost , therms valus is
1 T 2
2
I rms
I
cos
tdt
m
T 0
Similarly,for v(t ) Vm cost ,
I m2
T
T
0
1
1 cos 2t dt I m
2
2
Vm
Vrms
2
T heaveragepower can be writtenas
Vm I m
1
P Vm I m cos( v i )
cos( v i )
2
2 2
Vrms I rms cos( v i )
Example 1
Determinetherms value of i (t ).
If thecurrentis passed througha
2 - resistor,find theaverage
power absorbed by theresistor.
Sol :
T heperiodis T 4.
5t , 0 t 2
i (t )
10, 2 t 4
I rms
1
T
T
0
i 2 dt
4
1 2
2
2
(
5
t
)
dt
(
10
)
dt
2
4 0
I rms
4
1 t3 2
25 |0 100t |2 8.165 A
4 3
T heaveragepower absorbed is
2
P I rms
R (8.165) 2 (2) 133.3 W
Example 2
Find therms value and theaverage
power absorbed by a 10 - resistor.
Sol :
T heperiodis T 2 .
10sin t , 0 t
v(t )
0, t 2
1 T 2
2
Vrms v (t )dt
T 0
2
1
2
(
10
sin
t
)
dt
0
dt
2 0
1 100
(
1
cos
2
t
)
dt
2 0 2
50 sin 2t
t
|0 25
2
2
T heaveragepower absorbed
by a 10 - resistoris
Vrms 5 V
2
Vrms
52
P
2.5 W
R 10
Apparent Power and Power Factor
If thecurrentand voltageare
v(t ) Vm cos(t v )
i (t ) I m cos(t i )
V Vm v
or
I I m i
T heaveragepower is
1
P Vm I m cos( v i )
2
Vrms I rms cos( v i )
S cos( v i )
S pf
S Vrms I rms (unit VA) : Apparent power
where pf cos( v i )
: Power factor
: Power factor angle
v i
S and pf (Cont’d)
P ower FactorAngle Angle of theLoad Impedance
pf cos( v i )
V Vm v Vm
Z
( v i )
I I m i I m
V
Vrms 2 Vrms v
Since
I
I rms
I rms i
2
V Vrms Vrms
Z
( v i )
I I rms I rms
Leadingpf meanscurrentleads voltage a capacitiveload
Laggingpf meanscurrentlags voltage an inductiveload
Example 1
A series - connectedload draws a currentand voltageas given below
i(t ) 4 cos(100t 10)
v(t ) 120cos(100t 20)
Find theapparentpower and thepower factorof theload.
Determinetheelement values tahtform theseries - connectedload.
Sol :
T heapparentpower is
120 4
S Vrms I rms
240 VA
2 2
T hepower factoris
pf cos( v i )
cos(20 10) 0.886
V 120 20
Z
I
410
30 30
25.95 j15
1
X C 15
C
1
C
212.2 F
15
Example 2
Determinethepower factor
as seen by thesource.
Calculate theaveragepower
deliveredby thesource.
Sol :
T he totalimpedanceis
Z 6 4||(-j2) 6.8 j1.6
7 13.24
T hepower factoris
pf cos(13.24) 0.9734
Vrms
300
Z
7 13.24
4.28613.24
I rms
P Vrms I rms pf (30)(4.286)pf
125 W
Complex Power
Considering thecurrentand voltage
are given in phasorform as
V Vm v
I I m i
T hecomplexpower S absorbed by
theac load is given as
1 *
S VI VrmsI *rms
2
V
V
Vrms v
rms
2
where
I
I rms
I rms i
2
S Vrms I rms( v i )
Vrms I rms cos( v i ) j sin( v i )
Z
V Vrms Vrms
( v i )
I I rms I rms
2
V
2
S VrmsI *rms I rms
Z rms*
Z
Since Z R jX ,
2
S I rms
( R jX ) P jQ
2
P Re(S) I rms
R : Real power
2
Q
Im(
S
)
I
rms X : Reactivepower
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)
Summary
1 *
VI
2
Vrms I rms( v i )
ComplexP ower S P jQ
ApparentP ower S S Vrms I rms P 2 Q 2
Real P ower P Re(S) S cos( v i )
ReactiveP ower Q Im(S) S sin( v i )
P ower Factor
P
cos( v i )
S
Impedance Z
V Vrms Vrms
( v i )
I I rms I rms
Power Triangle
Power triangle
Impedance triangle
Power triangle
Example 1
T he volt age and currentacrossa load are given as
v(t ) 60 cos(t 10)
i (t ) 1.5 cos(t 50)
Find (a) thecomplexand apparentpowers,(b) thereal and reactive
powers,and (c) thepower factorand theload impedance.
Sol :
(b) S 45 60 22.5 j 38.97
60
1.5
(a) Vrms
10 , I rms
50
P 22.5 W
2
2
Q 38.97 VAR
T hecomplexpower is
(c) pf cos(60) 0.5 (leading)
*
S VrmsI rms 45 60 VA
V 60 10
Z
40 60
T heapparentpower S S 45 VA
I
1.550
Example 2
A load Z draws 12 - kVA at pf 0.856lagging froma 120- V rms
sinusoidal source.
Calculate: (a) theaverageand reactivepowersdelivered to theload,
(b) thepeak current,and (c) theload impedance.
S
10272 j 6204
Vrms
1200
Sol :
(a) pf cos 0.856
31.13
S 12000VA
I
P S cos 10272W
Q S sin 6204 VAR
T hepeak currentis
(b) T hecomplexpower is
S P jQ 10272 j 6204 VA
(c) Z
S VrmsI *rms
*
rms
10031.13
I rms 100 31.13
I m 2 I rms 141.4 A
Vrms
1200
I rms 100 31.13
1.231.13
Conservation of AC Power
KCL gives I I1 I 2
KVL gives V V1 V2
T hecomplexpower supplied by
T hecomplexpower supplied by
thesource is
thesource is
1 * 1
S VI V (I1* I *2 )
2
2
1
1
VI1* VI*2 S1 S 2
2
2
S1, 2 is thepower delivered to Z1,2.
1 * 1
S VI (V1 V2 )I *
2
2
1
1
V1I * V2 I * S1 S 2
2
2
S1, 2 is thepower delivered to Z1,2.
Summary
• The total power supplied by the source equals
the total power delivered to the loads.
S S1 S2 S N
• The complex, real, and reactive powers of the
sources equal the respective sums of the
complex, real, and reactive powers of the
individual loads.
Example
Find thereal power and reactive
power absorbed by : (a) thesource,
(b) theline, and (c) theload.
Sol :
T he totalimpedanceis
Z (4 j 2) (15 j10)
19 j8 20.62 22.83
Vs
2200
I
Z 20.62 22.83
10.6722.83
S line VlineΙ *
(47.7249.4)(10.67 22.83)
455.4 j 227.7 VA (lagging)
(c) VL (15 j10)I 192.38 10.87
(a) S s Vs Ι *
S L VL Ι *
(2200)(10.67 22.83)
2163.5 j 910.8 VA (leading)
(b) Vline (4 j 2)I 47.7249.4
(192.38 10.87)(10.67 22.83)
1708 j1139 VA (leading)
Note that S s S line S L
Power Factor Correction
• It is the process of increasing the power factor without
altering the voltage or current to the original load.
pf
correction
Most loads
are inductive.
V
I L 1
R jL
I C j C V
IL
I I L IC I 2
pf Correction (Cont’d)
If we desire to increasepf fromcos1 to
cos 2 without alteringthereal power,
P1 S1 cos1 P2 S 2 cos 2 P
Q1 S1 sin 1 P1 tan1
Q2 S 2 sin 2 P tan 2
Applyingtheac power conservaton
i gives
QC Q1 Q2 P (tan1 tan 2 )
2
2
Vrms
Vrms
2
But S * QC
CVrms
Z
XC
QC
P (tan1 tan 2 )
C
2
2
Vrms
Vrms
Note that thereal power P is not affect edby
the pf correctionbecause PC is zero.
Example
Whenconnectedto a 120- V(rms),60 - Hz power line,
a load absorbs 4 - kW at 1 lagging power factorof 0.8.
Find thecapacitance necessaryto raise thepf to 0.95.
Sol :
When pf is raised to 0.95,
Vrms 120 V
2 (60) 120 rad/s
We have
P 4000 W
pf 0.8
pf cos1 0.8 1 36.87
cos 2 0.95 2 18.19
P S1 cos1 S1
P
5000VA
cos1
Q1 S1 sin 1 5000sin 36.87
3000VAR
S2
P
4210.5 VA
cos 2
Q2 S1 sin 2 4210.5 sin 18.19
1314.4 VAR
QC Q1 Q2 3000 1314.4
1685.6 VAR
QC
1685.6
C
310.5 F
2
2
Vrms 120 120
Applications: Power Measurement
P
Low impedance
High impedance
Vm
V
v
v(t ) Vm cos(t v ) rms
2
I
i(t ) I m cos(t i )
I rms m i
2
1
P Vrms I rms cos( v i ) Vm I m cos( v i )
2
Example
Find the wattmeterreading.
Sol :
Vrms
I rms
Vm
v
2
Im
i
2
1500
150
I rms
(12 j10) (8 j 6) 20 j 4
150(8 j 6)
Vrms I rms (8 j 6)
20 j 4
150(8 j 6) 150
*
S VrmsI rms
20 j 4 20 j 4
423.7 j 324.6 VA
P Re(S) 423.7 W