steady-state power analysis learning goals instantaneous power for the special case of steady state...
TRANSCRIPT
STEADY-STATE POWER ANALYSISLEARNING GOALS
Instantaneous Power For the special case of steady state sinusoidal signals
Average Power Power absorbed or supplied during one cycle
Maximum Average Power Transfer When the circuit is in sinusoidal steady state
Effective or RMS Values For the case of sinusoidal signals
Power Factor A measure of the angle between current and voltage phasors
Complex Power Measure of power using phasors
Power Factor Correction How to improve power transfer to a load by “aligning” phasors
Single Phase Three-Wire Circuits Typical distribution method for households and small loads
INSTANTANEOUS POWER
)cos()(
)cos()(
iM
vM
tIti
tVtv
Statesteady In
)()()( titvtp Impedance to
SuppliedPower
ousInstantane
)cos()cos()( ivMM ttIVtp
)cos()cos(2
1coscos 212121
)2cos()cos(2
)( ivivMM t
IVtp
LEARNING EXAMPLE
)(),(
302
),60cos(4)(
tpti
Z
ttv
:Find
:Assume
)(302302
604A
Z
VI
))(30cos(2)( Atti
30,2
60,4
iM
vM
I
V
)902cos(430cos4)( ttp
constant Twice thefrequency
AVERAGE POWER
For sinusoidal (and other periodic signals)we compute averages over one period
2
T
)2cos()cos(2
)( ivivMM t
IVtp
)cos(2 iv
MM IVP
If voltage and current are in phase
MMiv IVP2
1
If voltage and current are in quadrature090 Piv
Purelyresistive
Purely inductive orcapacitive
It does not matterwho leads
LEARNING EXAMPLE Find the averagepower absorbedby impedance
)(1553.34522
6010
22
6010A
jI
15,60,53.3,10 ivMM IV
WP 5.12)45cos(3.35
Tt
t
dttpT
P0
0
)(1
RV
)(1506.7601022
2V
jVR
WP 53.306.72
1
Since inductor does not absorb powerone can use voltages and currents acrossthe resistive part
LEARNING EXAMPLEDetermine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source
)(4534
45121 AI
WP 183122
14
If voltage and current are in phase
MMiv IVP2
1 2
12
1MRI
R
VM2
2
1
)(57.7136.537.265
4512
12
45122 A
jI
)(36.522
1 22 WP W7.28
Inductors and capacitors do not absorbpower in the average
WPtotal 7.2818
absorbedsupplied PP WP 7.46supplied
Verification
57.7136.545321 III)(10.6215.8 AI
)cos(2 iv
MM IVP
)10.6245cos(15.8122
1 suppliedP
LEARNING EXTENSION Find average power absorbed by each resistor
I
iZI
6012
iZ )4||4(2 j
4524
6.713.25
44
1688
44
)4(42
j
jj
j
jZ i
6.2647.4iZ
)(6.8668.26.2647.4
6012AI
WRIP M 20.768.222
1
2
1 222
2I
6.8668.24524
904
44
42 I
j
jI
6.4190.12I
)(90.142
1 24 WP
LEARNING EXTENSIONFind the AVERAGE power absorbed by each PASSIVE component and the total power supplied by the source
1I2I
3010243
241 j
jI
)(62.4014.6301095.1528.7
57.2647.41 AI
)(14.632
1
2
1 223 WRIP M
62.4014.630102I
)(05.1412.4
95.1528.7
30303010
243
32
A
jI
)(12.442
1 24 WP
)(02 WPj
Power supplied by source
Method 1. absorbedsupplied PP
WPPP 50.9043 supplied
Method 2: )cos(2
1ivMM IVP
sV
62.4042.183 1IVs
)3062.40cos(1042.182
1 P
LEARNING EXAMPLEDetermine average power absorbed or supplied by each element
)(3062
30122 AI
)(36622
1
2
1 222 WRIP M
01 jP
To determine power absorbed/suppliedby sources we need the currents I1, I2
)(19.3643.7
39.466639.10
1
0630123
A
jj
j
jI
07.728.11
)(39.12.1139.46320.5321 AjjjIII
)(54
)07.730cos(28.11122
13012
W
P
)1836(
)cos(2
1ivMM IVP
Power Average
WP 18)19.360cos(43.762
106
Passive sign convention
R
VRIP M
M
22
2
1
2
1
resistorsFor
LEARNING EXTENSIONDetermine average power absorbed/supplied by eachelement
1I
2I
)012(42304
012
22
1
IIj
I
Equations Loop
57.2647.4
48246.3
24
0483042
j
jI
)(20497.957.2647.4
43.17758.442 AI
)(5.5492.19))(055.4108.912(4
))(20497.912(4)(4 214
VVj
VIIV
WR
VP M 6.49
4
92.19
2
1
2
1 22
4
)(02 WP j
)(4.69)05.54cos(1292.192
1012 WP
4V
)(8.19)20430cos()97.9(42
1304 WP
Check: Power supplied =power absorbed
j
VI
j
VV
2
304
02
304
4012
42
44
Equations Node
Alternative Procedure
)cos(2
1ivMM IVP
Power Average
R
VRIP M
M
22
2
1
2
1
resistorsFor
LEARNING EXTENSIONDetermine average power absorbed/supplied by eachelement
V
042
012
2
024
V
j
VV
Equation Node
4j
0)12(2)24(2 jVVVj
)(85.177.14
)(125.788.14
13
24192
32
32
32
4824
4824)32(
Vj
V
j
j
j
j
jV
jVj
1I
925.062.42
85.177.1424
2
0241 j
jVI
)(32.1171.41 AI
2Ij
j
j
j
j
VI
2
85.177.1412
2
0122
)(73.12367.1)(385.1925.02
77.285.12 AAj
jI
)(18.2271.422
1 22 WP
)(67.274
88.14
2
1 2
4 WP
)(565.5)73.1230cos(67.1122
1012 WP
)(42.55)32.110cos(71.4242
1024 WP
)(42.55
)(565.567.2718.22
WP
WP
supplied
absorbed
:Check
)cos(2
1ivMM IVP
Power Average
R
VRIP M
M
22
2
1
2
1
resistorsFor
MAXIMUM AVERAGE POWER TRANSFER
)cos(||||2
1
)cos(2
1
LL
LL
IVLL
IVLMLML
IV
IVP
THTHTH jXRZ
LLL jXRZ
L
LL
OCTHL
LL
Z
VI
VZZ
ZV
|||| OC
THL
LL V
ZZ
ZV
LIV
LLL
Z
ZVI
LL
LLL jXRZ
22)cos(
)tan(
LL
LIV
L
LL
XR
R
R
XZ
LL
2tan1
1cos
||
||||
L
LL Z
VI
222
2
||
||||
2
1
LL
L
THL
OCLL
XR
R
ZZ
VZP
22
2
)()(
||
2
1
THLTHL
LOCL
XXRR
RVP
THL
THL
L
L
L
L
RR
XX
R
PX
P
0
0
*TH
optL ZZ
TH
OCL R
VP
4
||
2
1 2max
222 )()(||
)()(
THLTHLTHL
THLTHLTHL
XXRRZZ
XXjRRZZ
LEARNING EXAMPLEload the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
Remove the load and determine the Thevenin equivalent of remaining circuit
1I
64.926.5
64.908.6
03204
142
24
jVOC
93.1647.164.908.6
57.2694.8
16
48
37
1652
37
)16)(48(
16
48)12(||4
j
j
jjj
j
jjZTH
43.041.193.1647.1* jZL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
)(45.241.14
26.5
2
1 2max WPL
We are asked for the value of thepower. We need the Thevenin voltage
LEARNING EXAMPLEload the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
Circuit with dependent sources!
SC
OCTH I
VZ
1'
1'
2
)42(04
IV
IjV
X
x
KVL
)(45707.04524
04
)4524()44(04
1
11
AI
IIj
KVL
5.1611013411042 1 jjIVOC
Next: the short circuit current ...
LEARNING EXAMPLE (continued)...
Original circuit
02)(204
04)(24"
SCSC
SCx
IjII
IIIjV
CURRENT CIRCUIT
SHORT FOR EQUATIONS LOOP
)(2" IIV SCx
VARIABLEGCONTROLLIN
4)22(2
44)44(
SC
SC
IjI
IIj
Substitute and rearrange
2)11( SCIjI
442)1()1(4 SCSC IIjj
57.1165)(21 AjISC
57.1611013411042 1 jjIVOC
11452 jZTH 11 jZ optL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
)(25.14
)10(
2
1 2max WPL
LEARNING EXTENSION
load the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
OCV
I
4573.12)1(98
)22(36
)22(036
jj
I
Ij
186
)1(9212
2012
j
jj
IjVOC
)(57.71974.18 VVOC
22
42)2||2(2
j
jjjjZTH
)(18
88
22
4
jj
jZTH
)(1 jZ optL
)(454
360
2
1max WPL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
360186|| 222 OCV
LEARNING EXTENSION
load the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
OCV
2jV
024222
22 jj
jV j 9024
KVL
)(24129024012 VjVOC
7202412|| 222 OCV
THZ
222
)22(2)22(||2
jj
jjjjZTH
)(22 jZTH
)(22 jZ optL
)(4524
720
2
1max WPL
EFFECTIVE OR RMS VALUES
)(ti
R
Rtitp )()( 2
power ousInstantane
Tt
t
Tt
tav dtti
TRdttp
TP
T
0
0
0
0
)(1
)(1 2
period with periodic iscurrent If
2
)(
dcdc
dc
RIP
Iti
then )( DC iscurrent If dcaveff PPI :
Tt
teff dtti
TI
0
0
)(1 22
Tt
teff dtti
TI
0
0
)(1 2
The effective value is the equivalent DCvalue that supplies the same average power
Definition is valid for ANY periodicsignal with period T
2
)cos()(
Meff
M
XX
tXtx
is valueeffective the
signal sinusoidal aFor
If the current is sinusoidal the averagepower is known to be RIP Mav
2
2
1
22
2
1Meff II
)cos(2
1ivMMav IVP case sinusoidalFor
)cos( iveffeffav IVP
square) mean(root rmseffective
LEARNING EXAMPLECompute the rms value of the voltage waveform
One period
32)2(4
210
104
)(
tt
t
tt
tv
3T
3
2
21
0
2
0
2 ))2(4()4()( dttdttdttvT
The two integrals have the same value
3
32
3
162)(
1
0
33
0
2
tdttv
)(89.13
32
3
1VVrms
Tt
trms dttx
TX
0
0
)(1 2
)(ti
R
LEARNING EXAMPLECompute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor
2R
)(4 sT
40;16)(2 tti
Tt
trms dttx
TX
0
0
)(1 2
)(4 AIrms
)(322 WRIP rmsav
LEARNING EXTENSIONCompute rms value of the voltage waveform
tv 2
4T
Tt
trms dttx
TX
0
0
)(1 2
dttVrms 2
0
2)2(4
1VVt 633.1)(
3
8
3
12
0
3
)(ti
R
LEARNING EXTENSION
4R
Compute the rms value for the current waveforms and usethem to determine average power supplied to the resistor
6T
Tt
trms dttx
TX
0
0
)(1 2
RIP rmsav2
4
2
6
4
2
0
2 41646
1dtdtdtIrms 8
6
8328
)(3248 WP
8T
816168
1 6
4
2
0
2
dtdtIrms )(32 WP
THE POWER FACTOR
LZiMI
vMV
iv
VI
izv
IZVZIV
z
)cos()cos(2
1ivrmsrmsivMM IVIVP rmsrms IVP apparent
zivP
Ppf cos)cos(
apparent
inductive pure
inductiveor lagging
resistive
capacitiveor leading
capacitive pure
90
900
0
10
01
09010
900
z
z
z
pf
pf
pf
V
e)(capacitiv
leadscurrent 090 z
)(inductive
lagscurrent 900 z
pfIVP rmsrms
LEARNING EXAMPLEFind the power supplied by the power company.Determine how it changes if the power factor is changed to 0.9
Power company
pfIVP rmsrms
rmsAW
Irms )(3.259707.0480
)(1088 3
rmsV )(480
rmsA)(453.259
45707.0cos zz
Current lags the voltage
480453.25908.0
08.0
LrmsS VIVrms
)(7.14957.147.494
480)4.1834.183(08.0
Vj
jVrmsS
)(378.93)(000,88
378.508.03.259 32
kWWPP
kWRIP
lossesS
rmslosses
If pf=0.9
If pf=0.9
8.257.203rmsI
82.049448009.747.14 jVS
kWRIP
rmsAI
rmslosses
rms
32.3
)(7.2039.0480
000,88
2
Losses can be reduced by 2kW!
Examine also the generated voltage
LEARNING EXTENSION
1.0
707.0,)(480,100
line
LL
R
pfrmsVVkWP
Determine the power savings if the power factor can be increased to 0.94
pfIVP rmsrms
22
22 1
pfV
RPRIP
pfV
PI
rms
linelinermslosses
rmsrms
kW
WpfPlosses
34.42
)(707.0
1
480
1.010)707.0( 22
10
kW
WpfPlosses
34.413.1
)(94.0
1
480
1.010)94.0( 22
10
kWkWPsaved 77.334.487.0
COMPLEX POWER*rmsrms IVS
PowerComplex of Definition
ivrmsrms
irmsvrms
IVS
IVS
*
)sin()cos( ivrmsrmsivrmsrms IVjIVS
The units of apparent and reactive power are Volt-Ampere
2* ||)( rmsrmsrmsrmsrms IZIZISZIV Another useful form
2
2
||
||
rms
rms
IXQ
IRPjXRZ
PActive Power
Q
Reactive Power
inductive
capacitive
LEARNING EXAMPLE
HzfjZrmsVlaggingpfkWP LLL 60,3.009.0,0220,8.0,20 :Given
Determine the voltage and power factor at the input to the line
}Re{SP pfSS iv ||)cos(||
inductive
capacitive
inductive
kVAQPSQ L 15222 87.3625)(1520 kVAjSL
*LLL IVS
)(86.3664.1130220
87.36000,25**
AV
SI
L
LL
)(220)18.6891.90)(3.009.0(
0220)3.009.0(
VjjV
IjV
S
LS
)(18.6891.90220
000,15000,20*
Ajj
IL
86.453.24914.2163.248 jVS
86.4
86.36
SV
LI
746.0)72.41cos( sourcepf
lagging
kVApf
PSL 25||
LEARNING EXAMPLECompute the average power flow between networksDetermine which is the source
1
012030120
jZ
VVI BA
rmsVVA )(30120 rmsVVB )(0120
1j
rmsAjj
jI )(08.1660
120)6092.103(
rmsAI )(1512.62
rmsBB VAIVS 15454,71512.620120)( *
)(200,7)165cos(454,7 WPA
)(200,7)15cos(454,7 WPB
A supplies 7.2kW average power to B
rmsAA VAIVS 165454,719512.6230120)( * Passive sign convention.Power received by A
LEARNING EXTENSION
laggingpf
kW
84.0
40
1.0 25.0j
Determine real and reactive power losses andreal and reactive power supplied
inductive
capacitive}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* ||)( LlineLLline IZIIZS losses
VAj 713,11685,4
Balance of power
kVAjjj
SSS
552.37685.44839.2540713.11685.4
loadlossessupplied
2)45.216)(25.01.0( jS losses
laggingpf
kW
85.0
60
12.0 18.0j
LEARNING EXTENSIONDetermine line voltage and power factor at the supply end
pfIVS LLiv ||||)cos(||
rmsL
L ApfV
PI )(86.320
||||
}Re{SP *LLL IVS
)(cos 1 pfiv 79.31iv rmsL AI )(79.3186.320 rmsAj )(03.16972.272
LLS VIZV line 220)03.16972.272)(18.012.0( jj
rmsS VjV )(81.2815.283 rmsV )(81.561.284
792.0)6.37cos( sourcepf
lagging
81.5
79.31
SV
LILVThe phasor diagram helps in visualizing
the relationship between voltage and current
Low power factors increase losses and are penalized by energy companies
Typical industrial loadsare inductive
Simple approach to power factor correction
)cos(
||
oldold
oldoldoldoldold
:capacitorWithout
pf
SjQPS
)cos(
||
newnew
newnew
capacitoroldold
capacitoroldnew
capacitor With
pf
S
jQjQP
SS S CV
IVQ
L
L
2||
||||
capacitorcapacitor
capacitorICj
VL 1
old
capacitoroldnew P
QQ tan
2tan1
1cos
POWER FACTOR CORRECTION
LEARNING EXAMPLE
laggingpf
VkW rmsL
8.0
0220,50
Roto-molding
process
lagging 0.95 to
factorpower the increase to requiredcapacitor the Determine
.60Hzf
}Re{SP pfSS iv ||)cos(||
kVApf
PSold 5.62
80.
50|| ).(5.37|||| 22 kVAPSQ oldold
329.01
tan95.0cos2
new
newnewnew pf
pf kVAPQ
P
Qnew
new 43.16329.0
kVAQQQ newoldcapacitor 07.2143.165.37
CV
IVQ
L
L
2||
||||
capacitorcapacitorFF
V
QC
L
capacitor
1156)(001156.0)602()220(
1007.21
|| 2
3
2
LEARNING EXTENSION
HzfR
pfrmsVVkWP
line
LL
60,1.0
707.0,)(480,100
Determine the capacitor necessary to increase the power factor to 0.94
}Re{SP pfSS iv ||)cos(||
kVApf
PSold 44.141
707.
100|| ).(02.100|||| 22 kVAPSQ oldold
363.01
tan94.0cos2
new
newnewnew pf
pf kVAPQ
P
Qnew
new 3.36363.0
kVAQQQ newoldcapacitor 72.633.3602.100
CV
IVQ
L
L
2||
||||
capacitorcapacitorFF
V
QC
L
capacitor
733)(000733.0)602()480(
1072.63
|| 2
3
2
Power circuit normallyused for residencial supply
Line-to-line usedto supply majorappliances (AC, dryer). Line-to-neutral for lightsand small appliances
Basic circuit.
Generalbalancedcase
General case by sourcesuperposition
Neutral current is zero
Neutral currentis zero
An exercise insymmetry
SINGLE-PHASE THREE-WIRE CIRCUITS
LEARNING EXAMPLEDetermine energy use over a 24-hour period and the costif the rate is $0.08/kWh
rmsA1 Lights on
rmsA2.0
Stereo on
rmsA30
LSnN
RSbB
RLaA
III
III
III
KCL
Assume allresistive
rmsrms IVP
TimePdttpEnergy average)(
kWhHrkWHrkWElights 8.1712.0812.0
kWhHrkWErange 8.28)112(2.7
kWhHrkWEstereo 192.0)35(024.0
kWhEdaily 792.30 dayCost /46.2$ dtIVpE rmsrmssuppliedsupplied
Outline of verification
aAIA31
A1
SAFETY CONSIDERATIONS
Average effect of 60Hz current from hand tohand and passing the heart
Required voltage depends on contact, personand other factors
Typical residential circuit with ground andneutral
Ground conductor is not needed fornormal operation
Increased safety due to grounding
When switched on the tool case is energized
without the ground connector theuser can be exposed to the fullsupply voltage!
Conducting due to wet floor
If case is grounded then the supply is shorted and the fuse acts to openthe circuit
LEARNING EXAMPLE
More detailed numbers in a related case study
LEARNING EXAMPLE
Ground prong removed
R(dry skin) 15kOhmR(wet skin) 150OhmR(limb) 100OhmR(trunk) 200Ohm
Suggested resistancesfor human body
150150
Wet skin
400
Limbstrunk
1
mAIbody 171701
120
Can cause ventricular fibrillation
LEARNING EXAMPLE Ground Fault Interrupter (GFI)
In normal operating mode the two currents induce canceling magnetic fluxes
No voltage is induced in the sensing coil
coil sensing the in induced voltagea is there then
fault) a to due (e.g.,different become and If 21 ii
LEARNING EXAMPLE
xGroundfault
Vinyl lining (insulator)
Circuit formed when boy inwater touches boy holding grounded rail
A ground fault scenario
While boy is alone in the poolthere is no ground connection
LEARNING EXAMPLE Accidental groundingOnly return path in normaloperation
New path created by the grounding
Using suggested values of resistance thesecondary path causes a dangerous currentto flow through the body
mA150
LEARNING EXAMPLE A grounding accident
After the boom touches the live line the operator jumps down and starts walkingtowards the pole
7200 V
Ground is not a perfectconductor
10m
720V/m
One step applies 720 Volts to theoperator
LEARNING EXAMPLEA 7200V power line falls on the car and makes contact with it
Wet Road
Option 1.Driver opens door and steps down
Option 2: Driver stays inside the car
7200V
Car body is good conductor
Tires are insulators
R(dry skin) 15kOhmR(wet skin) 150OhmR(limb) 100OhmR(trunk) 200Ohm
Suggested resistancesfor human body
trunklimbskindry body RRR
I
2
72000bodyI
dangerous! Very mAI 460
LEARNING EXAMPLE Find the maximum cord length
CASE 1: 16-gauge wire ft
m4 CASE 2: 14-gauge wire
ft
m5.2
382.0
51.13
cord
cord
R
VAR
ftR
Lft
mcord 5.95
4
ftR
Lft
mcord 8.152
5.2
Minimum voltage for properoperation
Working with RMS values the problem is formally the same as a DC problem
LEARNING EXAMPLELight dimming when AC starts
Typical single-phase 3-wire installation
rmslight AI 5.0
Circuit at start of AC unit. Current demand is very high
A40
rmsAN VV 100
AC off
rmsAN VV 5.119120241
240
AC in normal operation
rmsAN VV 115
LEARNING BY DESIGNAnalysis of single phase 3-wire circuit installations
Option 1
Option 2
||120
5000|| HL II
67.412nI
9.3667.41mI 9.3667.41067.41067.41aI )(1.124.119 A
mb II
kVAjkVA
SA
3141.1238.14
1.124.1190120 *
kVAjkVA
SB
349.365
9.3667.410120 *
9.3667.41067.41aI )(4.1807.79 A
ab II
HL II 0nI
kVAjkVA
SS BA
394.185.9
4.1807.790120 *
kWIIIP
R
nbalosses
lines
147.1||||||05.0
05.0222
kWIIP
R
balosses
lines
625.0||||05.0
05.022
)/$08.0(@366/$
522.0
kWhyear
kWPsaved
Steady-statePower Analysis