Potência instantânea, média, eficaz, complexa Factor de ... fileAnálise no domínio da...
Transcript of Potência instantânea, média, eficaz, complexa Factor de ... fileAnálise no domínio da...
Programa
1. Introdução aos circuitos eléctricos2. Grafos e circuitos resistivos lineares3. Circuitos dinâmicos lineares4. Regime forçado sinusoidal
Amplitudes complexas (Phasors)Diagramas vectoriaisPotência instantânea, média, eficaz, complexaFactor de potência
5. Análise no domínio da frequência complexa6. Circuitos resistivos não-lineares
STEADY-STATE POWER ANALYSISLEARNING GOALS
1. Instantaneous PowerFor the special case of steady state sinusoidal signals
2. Average PowerPower absorbed or supplied during one cycle
3. Maximum Average Power TransferWhen the circuit is in sinusoidal steady state
4. Effective or RMS ValuesFor the case of sinusoidal signals
5. Power FactorA measure of the angle between current and voltage phasors
Complex PowerMeasure of power using phasors
6. Power Factor CorrectionHow to improve power transfer to a load by “aligning” phasors
7. Single Phase Three-Wire CircuitsTypical distribution method for households and small loads
INSTANTANEOUS POWER
)cos()()cos()(
iM
vM
tItitVtvθωθω
+=+=
Statesteady In
)()()( titvtp =Impedance to
SuppliedPower ousInstantane
)cos()cos()( ivMM ttIVtp θωθω ++=
[ ])cos()cos(21coscos 212121 φφφφφφ ++−=
[ ])2cos()cos(2
)( ivivMM tIVtp θθωθθ +++−=
LEARNING EXAMPLE
)(),(302
),60cos(4)(
tptiZ
ttv
:Find
:AssumeΩ°∠=
°+= ω
)(302302604 A
ZVI °∠=
°∠°∠
==
))(30cos(2)( Atti °+= ω
°==°==
30,260,4
iM
vM
IV
θθ
)902cos(430cos4)( °++°= ttp ω
constant Twice thefrequency
AVERAGE POWER
For sinusoidal (and other periodic signals)we compute averages over one period
ωπ2
=T
[ ])2cos()cos(2
)( ivivMM tIVtp θθωθθ +++−=
)cos(2 iv
MM IVP θθ −=
If voltage and current are in phase
MMiv IVP21
=⇒=θθ
If voltage and current are in quadrature090 =⇒°±=− Piv θθ
Purelyresistive
Purely inductive orcapacitive
It does not matterwho leads
LEARNING EXAMPLE Find the averagepower absorbedby impedance
)(1553.34522
601022
6010 Aj
I °∠=°∠°∠
=+
°∠=
°=°=== 15,60,53.3,10 ivMM IV θθ
WP 5.12)45cos(3.35 =°=
∫+
=Tt
tdttp
TP
0
0
)(1−
+
RV
)(1506.7601022
2 Vj
VR °∠=°∠+
=
WP 53.306.721
×=
Since inductor does not absorb powerone can use voltages and currents acrossthe resistive part
LEARNING EXAMPLE Determine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source
)(45344512
1 AI °∠=°∠
=
WP 1831221
4 =×=Ω
If voltage and current are in phase
MMiv IVP21
=⇒=θθ 212
1MRI=
RVM
2
21
=
)(57.7136.537.265
451212
45122 A
jI °∠=
°−∠°∠
=−
°∠=
)(36.5221 2
2 WP ××=Ω W7.28=
Inductors and capacitors do not absorbpower in the average
WPtotal 7.2818+=
⇒= absorbedsupplied PP WP 7.46=supplied
Verification
°∠+°∠=+= 57.7136.545321 III)(10.6215.8 AI °∠=
)cos(2 iv
MM IVP θθ −=
)10.6245cos(15.81221
°−°××=suppliedP
LEARNING EXTENSION Find average power absorbed by each resistor
I
iZI °∠=
6012
iZ )4||4(2 j−+=
°−∠°−∠
=−−−
=−−
+=4524
6.713.2544
168844)4(42
jjj
jjZi
Ω°−∠= 6.2647.4iZ
)(6.8668.26.2647.4
6012 AI °∠=°−∠
°∠=
WRIP M 20.768.2221
21 22
2 =××==Ω
2I
°∠×°−∠
°−∠=
−−
= 6.8668.24524
90444
42 I
jjI
°∠= 6.4190.12I
)(90.1421 2
4 WP ××=Ω
LEARNING EXTENSION Find the AVERAGE power absorbed by each PASSIVE component and the total power supplied by the source
1I2I
°∠++
+= 3010
24324
1 jjI
)(62.4014.6301095.1528.757.2647.4
1 AI °∠=°∠°∠°∠
=
)(14.6321
21 22
3 WRIP M ××==Ω
°∠−°∠= 62.4014.630102I
)(05.1412.495.1528.7
30303010243
32
Aj
I
°∠=°∠
°∠=°∠
++=
)(12.4421 2
4 WP ××=Ω
)(02 WPj =Ω
Power supplied by source
Method 1. absorbedsupplied PP =
WPPP 50.9043 =+= ΩΩsupplied
Method 2: )cos(21
ivMM IVP θθ −=
−
+
sV
°∠== 62.4042.183 1IVs
)3062.40cos(1042.1821
°−°×××=P
LEARNING EXAMPLE Determine average power absorbed or supplied by each element
)(30623012
2 AI °∠=°∠
=
)(366221
21 22
2 WRIP M =××==Ω
01 =ΩjP
To determine power absorbed/suppliedby sources we need the currents I1, I2
)(19.3643.7
39.466639.101
0630123
A
jj
jj
I
°−∠=
−=−+
=°∠−°∠
=
°−∠=−=−++=+=
07.728.11)(39.12.1139.46320.5321 AjjjIII
)(54
)07.730cos(28.111221
3012
W
P
−=
°+°×××=°∠
1836 +=
)cos(21
ivMM IVP θθ −=
Power Average
WP 18)19.360cos(43.7621
06 =°+××=°∠
Passive sign convention
RVRIP M
M
22
21
21
==
resistorsFor
LEARNING EXTENSION Determine average power absorbed/supplied by eachelement
1I2I
)012(42304012
22
1
°∠++−=°∠°∠=
IIjI
Equations Loop
°−∠−+
=−
°∠−°∠=
57.2647.448246.3
24048304
2j
jI
)(20497.957.2647.443.17758.44
2 AI °∠=°−∠°∠
=
)(5.5492.19))(055.4108.912(4))(20497.912(4)(4 214
VVjVIIV
°−∠=−−=°∠+=+=Ω
WR
VP M 6.49492.19
21
21 22
4 =×==Ω
)(02 WP j =Ω−
)(4.69)05.54cos(1292.1921
012 WP −=°−°−×××−=°∠Ω4V
)(8.19)20430cos()97.9(421
304 WP =°−°××−=°∠
Check: Power supplied =power absorbed
jVI
jVV
2304
02
3044
012
42
44
−−°∠
=
=−
°∠−++°∠−
Ω
ΩΩ
Equations NodeAlternative Procedure
)cos(21
ivMM IVP θθ −=
PowerAverage
RVRIP M
M
22
21
21
==
resistorsFor
LEARNING EXTENSION Determine average power absorbed/supplied by eachelement
V
042
0122
024=+
°∠−+
°∠− Vj
VVEquationNode
4j×
0)12(2)24(2 =+−+− jVVVj
)(85.177.14)(125.788.14
1324192
3232
324824
4824)32(
VjV
jjj
jjV
jVj
+=°∠=
+=
−−
×++
=
+=+
1I
925.062.42
85.177.14242024
1 jjVI −=−−
=−°∠
=
)(32.1171.41 AI °−∠=
2Ijj
jj
jVI
−−
×+−
=−°∠
=2
85.177.14122
0122
)(73.12367.1)(385.1925.02
77.285.12 AAjjI °∠=+−=
+−=
)(18.2271.4221 2
2 WP =××=Ω
)(67.27488.14
21 2
4 WP =×=Ω
)(565.5)73.1230cos(67.11221
012 WP =°−°××−=°∠
)(42.55)32.110cos(71.42421
024 WP −=°+°×××−=°∠
)(42.55)(565.567.2718.22
WPWP
=++=
supplied
absorbed
:Check
)cos(21
ivMM IVP θθ −=
PowerAverage
RVRIP M
M
22
21
21
==
resistorsFor
MAXIMUM AVERAGE POWER TRANSFER
)cos(||||21
)cos(21
LL
LL
IVLL
IVLMLML
IV
IVP
θθ
θθ
−=
−=
THTHTH jXRZ +=
LLL jXRZ +=
L
LL
OCTHL
LL
ZVI
VZZ
ZV
=
+= |||| OC
THL
LL V
ZZZV+
=⇒
LIV
LLL
ZZVI
LL∠=−⇒∠−∠=∠⇒
θθ
LLL jXRZ +=
22)cos(
)tan(
LL
LIV
L
LL
XRR
RXZ
LL +=−∴
=∠⇒
θθθ
θ2tan1
1cos+
=
||||||
L
LL Z
VI =⇒
222
2
||||||
21
LL
L
THL
OCLL
XRR
ZZVZP
++=
22
2
)()(||
21
THLTHL
LOCL XXRR
RVP+++
=
⎩⎨⎧
=−=
⇒
⎪⎪⎭
⎪⎪⎬
⎫
=∂∂
=∂∂
THL
THL
L
L
L
L
RRXX
RPXP
0
0
*TH
optL ZZ =∴
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max
222 )()(||
)()(
THLTHLTHL
THLTHLTHL
XXRRZZ
XXjRRZZ
+++=+
+++=+
LEARNING EXAMPLEload the to suppliedpower average maximum the Compute
transfer.power averagemaximumfor Find LZ
Remove the load and determine the Thevenin equivalent of remaining circuit
1I °−∠=°∠
°∠=°∠
+×= 64.926.5
64.908.603204
1624
jVOC
Ω°∠=°∠°∠
=++
=
Ω+
=−+
=++
=+=
93.1647.164.908.657.2694.8
1648
371652
37)16)(48(
1648)12(||4
jj
jjjjjjZTH
Ω−=°−∠= 43.041.193.1647.1* jZL
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max*TH
optL ZZ =∴
)(45.241.14
26.521 2
max 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 averagemaximumfor Find LZ
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max*TH
optL 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)(20404)(24"
=−−+°∠=∠°−−++−
SCSC
SCx
IjIIIIIjV
CURRENT CIRCUITSHORTFOREQUATIONSLOOP
)(2" IIV SCx −=
VARIABLEGCONTROLLIN
4)22(244)44(−=−+−
=−+
SC
SC
IjIIIj
Substitute and rearrange
2)11( +−=⇒ SCIjI
[ ] 442)1()1(4 =−+−+ SCSC IIjj
°−∠=−−= 57.1165)(21 AjISC
°−∠=−−=−−=°∠−= 57.1611013411042 1 jjIVOC
Ω−=°−∠= 11452 jZTH Ω+=⇒ 11 jZ optL
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max*TH
optL ZZ =∴
)(25.14
)10(21 2
max WPL =×=
LEARNING EXTENSION
load the to suppliedpower average maximum the Computetransfer.power averagemaximumfor Find LZ
−
+
OCV
I
°−∠=−=−
=
+=°∠
4573.12)1(98
)22(36)22(036
jjI
Ij
186)1(9212
2012
jjjIjVOC
+=−×+−=
+°∠−=
)(57.71974.18 VVOC °∠=
Ω+
+−=+−=22
42)2||2(2jjjjjZTH
)(18
8822
4Ω−=
−=
+= jj
jZTH
)(1 Ω+= jZ optL
)(454
36021max WPL =×=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max*TH
optL ZZ =∴
360186|| 222 =+=OCV
LEARNING EXTENSION
load the to suppliedpower average maximum the Computetransfer.power averagemaximumfor Find LZ
⎟⎟⎠
⎞⎜⎜⎝
⎛=
TH
OCL R
VP4
||21 2
max*TH
optL ZZ =∴
−
+
OCV
2jV
=°∠+−
= 024222
22 jj
jV j °∠9024
KVL
)(24129024012 VjVOC +=°∠+°∠=
7202412|| 222 =+=OCV
THZ
222)22(2)22(||2
jjjjjjZTH −+
−=−=
)(22 Ω+= jZTH
)(22 Ω−= jZ optL
)(4524
72021max WPL =
××=
EFFECTIVE OR RMS VALUES
)(ti
R
Rtitp )()( 2=
powerousInstantane
⎟⎟⎠
⎞⎜⎜⎝
⎛== ∫∫
++ Tt
t
Tt
tav dtti
TRdttp
TP
T0
0
0
0
)(1)(1 2
periodwith periodiciscurrent If
2
)(
dcdc
dc
RIP
Iti
=
= then)( DCiscurrent 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
signalsinusoidalaFor θω
If the current is sinusoidal the averagepower is known to be RIP Mav
2
21
=
22
21
Meff II =∴
)cos(21
ivMMav IVP θθ −= case sinusoidalFor
)cos( iveffeffav IVP θθ −=
square)mean(root rmseffective ≈
Nota: rede 230 V (AC) => Vm aprox 325 V
LEARNING EXAMPLE Compute the rms value of the voltage waveform
One period
⎪⎩
⎪⎨
⎧
≤<−−≤<≤<
=32)2(4
210104
)(tt
ttt
tv
3=T
∫∫∫ −+=3
2
21
0
2
0
2 ))2(4()4()( dttdttdttvT
The two integrals have the same value
332
3162)(
1
0
33
0
2 =⎥⎦⎤
⎢⎣⎡×=∫ tdttv
)(89.1332
31 VVrms =×=
∫+
=Tt
trms dttx
TX
0
0
)(1 2
)(ti
R
LEARNING EXAMPLE Compute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor
Ω= 2R
)(4 sT =
40;16)(2 <≤= tti ∫+
=Tt
trms dttx
TX
0
0
)(1 2
)(4 AIrms =
)(322 WRIP rmsav ==
LEARNING EXTENSION Compute rms value of the voltage waveform
tv 2=
4=T∫+
=Tt
trms dttx
TX
0
0
)(1 2
dttVrms ∫=2
0
2)2(41 )(
38
31 2
0
3 Vt =⎥⎦⎤
⎢⎣⎡=
)(ti
R
LEARNING EXTENSION
Ω= 4R
Compute the rms value for the current waveforms and usethem to determine average power supplied to the resistor
6=T
∫+
=Tt
trms dttx
TX
0
0
)(1 2
RIP rmsav2=
=⎥⎦
⎤⎢⎣
⎡++= ∫ ∫∫
4
2
6
4
2
0
2 416461 dtdtdtIrms 8
68328=
++ )(3248 WP =×=
8=T
8161681 6
4
2
0
2 =⎥⎦
⎤⎢⎣
⎡+= ∫∫ dtdtIrms )(32 WP =
THE POWER FACTOR
LZiMI θ∠
−∠
+
vMV θ
iθvθ
VI
izv
IZVZIVθθθ +=∠+∠=∠⇒=
zθ
)cos()cos(21
ivrmsrmsivMM IVIVP θθθθ −=−= rmsrms IVP =apparent
zivPPpf θθθ cos)cos( =−==
apparent
inductive pureinductiveor lagging
resistive
capacitiveor leadingcapacitive pure
°°<<°<<
°
°<<°−<<°−
90900
010
0109010
900
z
z
z
pf
pf
pf
θ
θ
θ
V
e)(capacitivleadscurrent
°<≤°− 090 zθ
)(inductivelagscurrent
°≤< 900 zθ
pfIVP rmsrms ××=
LEARNING EXAMPLE Find the power supplied by the power company.Determine how it changes if the power factor is changed to 0.9
Power company
pfIVP rmsrms ××=
rmsAWIrms )(3.259707.0480
)(1088 3=
××
=
rmsV )(480
rmsA)(453.259 °−∠
°−=⇒=⇒ 45707.0cos zz θθ
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)(000,88378.508.03.259 32
kWWPPkWRIP
lossesS
rmslosses
=+==×==
If pf=0.9
If pf=0.9
°−∠= 8.257.203rmsI
°−∠=+−= 82.049448009.747.14 jVSkWRIP
rmsAI
rmslosses
rms
32.3
)(7.2039.0480
000,88
2 ==
=×
=
Losses can be reduced by 2kW!
Examine also the generated voltage
LEARNING EXTENSION
Ω====
1.0707.0,)(480,100
line
LL
RpfrmsVVkWP
Determine the power savings if the power factor can be increased to 0.94
pfIVP rmsrms ××=
22
22 1
pfVRPRIP
pfVPI
rms
linelinermslosses
rmsrms
×==
×=
kW
WpfPlosses
34.42
)(707.01
4801.010)707.0( 22
10
×=
××
==
kW
WpfPlosses
34.413.1
)(94.01
4801.010)94.0( 22
10
×=
××
==
kWkWPsaved 77.334.487.0 =×=
COMPLEX POWER*rmsrms IVS =
PowerComplex ofDefinition
[ ]ivrmsrms
irmsvrms
IVSIVSθθ
θθ−∠=
°∠×°∠= *
)sin()cos( ivrmsrmsivrmsrms IVjIVS θθθθ −+−=
The units of apparent and reactive power are
Volt-Ampere
2* ||)( rmsrmsrmsrmsrms IZIZISZIV ==⇒=
Another useful form
⎩⎨⎧
==
⇒+= 2
2
||||
rms
rms
IXQIRPjXRZ
PActive Power
QReactive Power
inductive
capacitive
| || |
rms msS V IS P pf=
= ×
ANALYSIS OF BASIC COMPONENTS
RESISTORS
∴ 0Q⇒ =
INDUCTORS
CAPACITORS
Supplies reactive power!!
WARNING ( )0IF X ≠
LEARNING EXAMPLE
HzfjZrmsVlaggingpfkWP LLL 60,3.009.0,0220,8.0,20 =Ω+=°∠===:Given
Determine the voltage and power factor at the input to the line
}Re{SP = pfSS iv ×=−= ||)cos(|| θθ
inductive
capacitive
inductive
kVAQPSQ L 15222 =⇒−= °∠=+= 87.3625)(1520 kVAjSL
*LLL IVS =
)(86.3664.1130220
87.36000,25 **
AVSI
L
LL °−∠=⎥⎦
⎤⎢⎣⎡
°∠°∠
=⎥⎦
⎤⎢⎣
⎡=⇒
)(220)18.6891.90)(3.009.0(0220)3.009.0(
VjjVIjV
S
LS
+−+=°∠++=
)(18.6891.90220
000,15000,20 *
AjjIL −=⎥⎦⎤
⎢⎣⎡ +
=
°∠=+= 86.453.24914.2163.248 jVS
°86.4
°− 86.36
SV
LI
746.0)72.41cos( =°=sourcepflagging
LEARNING EXAMPLE Compute the average power flow between networksDetermine which is the source
1012030120
jZVVI BA °∠−°∠
=−
=
rmsVVA )(30120 °∠= rmsVVB )(0120 °∠=
Ω1j
rmsAjjjI )(08.1660120)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
laggingpfkW
84.040
=
Ω1.0 Ω25.0j
Determine real and reactive power losses andreal and reactive power supplied
inductive
capacitive}Re{SP = pfSS iv ×=−= ||)cos(|| θθ
kVApfPSL 62.47
84.40|| ===
rmsL
LL A
VSIVIS )(45.216
||||||* ==⇒=
°=−⇒−= 86.32)cos( ivivpf θθθθ
)(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
laggingpfkW
85.060
=
Ω12.0 Ω18.0j
LEARNING EXTENSION Determine line voltage and power factor at the supply end
pfIVS LLiv ××=−= ||||)cos(|| θθ
rmsL
L ApfV
PI )(86.320||
|| =×
=
}Re{SP =*LLL IVS =
⇒=− − )(cos 1 pfiv θθ °=− 79.31iv θθrmsL 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( =°=sourcepflagging
°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
θθ
=∠=+=
pfSjQPS
)cos(||
newnew
newnew
capacitoroldold
capacitoroldnew
capacitorWith
θθ
=∠=
−+=
+=
pfS
jQjQP
SS S CV
IVQ
L
L
ω2||
||||
=
= capacitorcapacitor
capacitorICj
VL ω1
=
old
capacitoroldnew P
QQ −=θtan
θθ
2tan11cos
+=
POWER FACTOR CORRECTION
LEARNING EXAMPLE Economic Impact of Power Factor Correction
Operating Conditions
Current Monthly Utility Bill
Correcting to pf=0.9
New demand charge
Additional energy charge due to capacitor bank is negligible
Monthly savings are approx $1853 per month! A reasonable capacitor bank should Itself in about a year
LEARNING EXAMPLE
laggingpfVkW rmsL
8.00220,50
=°∠=Roto-molding
process
lagging 0.95 tofactorpower the increase to requiredcapacitor the Determine
.60Hzf =
}Re{SP = pfSS iv ×=−= ||)cos(|| θθ
kVApfPSold 5.62
80.50|| === ).(5.37|||| 22 kVAPSQ oldold =−=
329.01
tan95.0cos2
=−
=⇒=new
newnewnew pf
pfθθ kVAPQ
PQ
newnew 43.16329.0 =×=⇒=
kVAQQQ newoldcapacitor 07.2143.165.37 =−=−=∴
CV
IVQ
L
L
ω2||
||||
=
= capacitorcapacitorFF
VQ
CL
capacitor μπω
1156)(001156.0)602()220(
1007.21|| 2
3
2 ==××
×==∴
LEARNING EXTENSION
HzfRpfrmsVVkWP
line
LL
60,1.0707.0,)(480,100
=Ω====
Determine the capacitor necessary to increase the power factor to 0.94
}Re{SP = pfSS iv ×=−= ||)cos(|| θθ
kVApfPSold 44.141
707.100|| === ).(02.100|||| 22 kVAPSQ oldold =−=
363.01
tan94.0cos2
=−
=⇒=new
newnewnew pf
pfθθ kVAPQ
PQ
newnew 3.36363.0 =×=⇒=
kVAQQQ newoldcapacitor 72.633.3602.100 =−=−=∴
CV
IVQ
L
L
ω2||
||||
=
= capacitorcapacitorFF
VQ
CL
capacitor μπω
733)(000733.0)602()480(
1072.63|| 2
3
2 ==××
×==∴
Power circuit normallyused for residencialsupply
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 EXAMPLE Determine energy use over a 24-hour period and the costif the rate is $0.08/kWh
rmsA1= Lights on
rmsA2.0=
Stereo on
rmsA30=
LSnN
RSbB
RLaA
IIIIII
III
−=−−=
+=KCL
Assume allresistive
rmsrms IVP =
∫ ×== TimePdttpEnergy average)(
kWhHrkWHrkWElights 8.1712.0812.0 =×+×=
kWhHrkWErange 8.28)112(2.7 =++×=
kWhHrkWEstereo 192.0)35(024.0 =+×=
kWhEdaily 792.30= dayCost /46.2$= ∫∫ == 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
Ω150Ω150
Wetskin
Ω400
Limbstrunk
Ω≈1
mAIbody 171701120
==
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 thenfault)atodue(e.g.,different becomeandIf 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 EXAMPLE A 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++
=27200
0=bodyI
dangerous! Very mAI 460≈
LEARNING EXAMPLE Find the maximum cord length
CASE 1: 16-gauge wire ft
mΩ4 CASE 2: 14-gauge wireft
mΩ5.2
Ω==×
382.051.13
cord
cord
RVAR
ftRLft
mcord 5.95
4==
Ω ftRLft
mcord 8.152
5.2==
Ω
Minimum voltage for properoperation
Working with RMS values the problem is formally the same as a DC problem
LEARNING EXAMPLE Light 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 5.119120241240
=×=
AC in normal operation
rmsAN VV 115≈
LEARNING EXAMPLE DRYER HEATING AND TEMPERATURE CONTROL
Temperature is controlled by disconnecting the heatingelement and letting it cool off.
Vary power to thermostatheater
If temperature from dryer is highthan the one from thermostat thopen switch and allow heater tocool off
Safety switch in case controlfails.
LEARNING BY DESIGN Analysis of single phase 3-wire circuit installations
Option 1
Option 2
||1205000|| HL II ==
67.412×=nI
°−∠= 9.3667.41mI°−∠+°∠+°∠= 9.3667.41067.41067.41aI )(1.124.119 A°−∠=
mb II =
[ ]kVAjkVA
SA
3141.1238.141.124.1190120 *
+=°∠=°−∠×°∠=
[ ]kVAjkVA
SB
349.3659.3667.410120 *
+=°∠=°−∠×°∠=
°−∠+°∠= 9.3667.41067.41aI )(4.1807.79 A°−∠=
ab II =
HL II =0=nI
[ ]kVAjkVA
SS BA
394.185.94.1807.790120 *
+=°∠=°∠×°∠==
( ) kWIIIP
R
nbalosses
lines
147.1||||||05.0
05.0222 =++×=
Ω=
( ) kWIIP
R
balosses
lines
625.0||||05.0
05.022 =+×=
Ω=
)/$08.0(@366/$522.0
kWhyearkWPsaved
==
Steady-statePower Analysis