电气专业毕业设计外文翻译.docx
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电气专业毕业设计外文翻译.docx
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电气专业毕业设计外文翻译
附录A
AfewexampleswillrefreshyourmemoryaboutthecontentofChapter8andthegeneralapproachtoanodal-analysissolution.
EXAMPLE17.12DeterminethevoltageacrosstheinductorforthenetworkofFig.
Solution:
Steps1and2areasndicatedinFig.17.22.
Step3:
NoteFig.17.23fortheapplicationofKirchhoff’scurrentlawtonodeV1:
Fig.17.22Fig.17.23
∑Ii=∑I00=I1+I2+I3V1-E/Z1+(V1/Z2)+(V1-V2)/Z3=0
Rearrangingterms:
V1[1/Z1+1/Z2+1/Z3]-V2[1/Z3]=E1/Z1(17.1)
NoteFig.17.24fortheapplicationofKirchhoff’scurrentlawtonodeV2:
0=I3+I4+I
V2-V1/Z3+V2/Z4+I=0
Rearrangingterms:
V2[1/Z3+1/Z4]-V1[1/Z3]=-I(17.2)
Fig.17.24
Groupingequations:
V1[1/Z1+1/Z2+1/Z3]-V2[1/Z3]=E1/Z1
V1[1/Z3]-V2[1/Z3+1/Z4]=I
1/Z1+1/Z2+1/Z3=1/0.5kΩ+1/10kΩ+1/2kΩ=2.5mS∠-2.29°
1/Z3+1/Z4=1/2kΩ+1/-5kΩ=0.539mS∠21.80°
andV1[2.5ms∠-2.29°]-V2[0.5mS∠0°]=24mΑ∠0°
V1[0.5mS∠0°]-V2[0.539mS∠21.80°]=4mΑ∠0°
with24mΑ∠0°-0.5mS∠0°
4mΑ∠0°-0.539mS∠21.80°
V1=2.5ms∠-2.29°-0.5mS∠0°
0.5mS∠0°-0.539mS∠21.80°
=(24mΑ∠0°)(-0.539mS∠21.80°)+(0.5mS∠0°)(4mΑ∠0°)/[(2.5ms∠
-2.29°)(-0.539mS∠21.80°)+(0.5mS∠0°)(0.5mS∠0°)]
=-10.01ν-j4.81ν/-1.021-j0.45=11.106ν∠-154.33°/1.116∠-156.21°
V1=9.95∠1.88°
MathCadThelengthandcomplexityoftheabovemathematicaldevelopmentstronglysuggesttheuseofanalternativeapproachsuchasMathCad.NoteinMathCad17.2thattheequationsareenteredinthesameformatasEqs.(17.1)and(17.2).BothV1andV2weregenerated,butbecauseonlyV1wasaskedfor,itwastheonlysolutionconvertedtothepolarform.Inthelowersolutionthecomplexitywassignificantlyreducedbysimplyrecognizingthatthecurrentisinmilliamperesandtheimpedancesinkilohms.Theresultwillthenbeinvolts.
K:
=10³m:
=0.01rad:
=1
V1:
=1+jV2:
=1+jdeg:
=π/180
Given
V1·[1/5·k+1/10j·k+1/2·k]-V2·1/2·k≈24·m
V1·[1/2·k]-V2[1/2·k+1/-5j·k]≈4·m
Find(V1,V2)=9.944+0.319jVolts
1.786-0.396jVolts
V1:
=9.944+0.319jV1=9.949arg(V1)=1.837·deg
RecognizingthatcurrentinmAresultsehenZisinkilohmns,analternativeformatfollows:
Given
V1·[1/5+1/10j+1/2]-V2·1/2≈24
V1·1/2-V2[1/2+1/-5j]≈4
Find(V1,V2)=9.944+0.319jVolts
1.786-0.396jVolts
V1:
=9.944+0.319jV1=9.949arg(V1)=1.837·deg
MATHCAD17.2
DependentCurrentSourcesFordependentcurrentsources,theprocedureismodifiedasfollows:
Steps1and2arethesameasthoseappliedforindependentsources.
Step3ismodifiedasfollows:
TreateachdependentcurrentsourcelikeanindependentsourcewhenKirchhoff’scurrentlawappliedtoeachdefinednode.However,oncetheequationsareestablished,substitutetheequationforthecontrollingquantitytoensurethattheunknownsarelimitedsolelytothechosennodalvoltages.
1.Step4isasbefore.
EXAMPLE17.13WritethenodalequationsforthenetworkofFig.17.25havingadependentcurrentsource.
Solution:
Steps1and2areasdefinedinFig.17.25.
Fig.17.25.
Step3:
AtnodeV1,I=I1+I2
V1/Z1+V1-V2/Z2-I=0
andV1[1/Z1+1/Z2]-V2[1/Z2]=I
AtnodeV2,I2+I3+ΚI=0
V2-V1/Z2+V2/Z3+Κ[V1-V2/Z2]=0
andV1[1-Κ/Z2]-V2[1-Κ/Z2+1/Z3]=0
resultingintwoequationsandtwounknowns.
IndependentVoltageSourcesbetweenAssignedNodesForindependentvoltagesourcesbetweenassignednodes,theprocedureismodifiedasfollows:
1.Steps1and2arethesameasthoseappliedforindependentsources.
2.Step3ismodefiedasfollows:
Treateachsourcebetwwendefinednodesasashortcircuit(recallthesupernodeclassificationofChapter8),andwritethenodalequationsforeachremainingindependentnode.Thenrelatethechosennodalvoltagestotheindependentvoltagesourcetoensurethattheunknownsoftheginalequationsarelimitedsolelytothenodalvoltages.
3.Step4isasbefore.
EXAMPLE17.14WritethenodalequationsforthenetworkofFig.17.26havinganindependentsourcebetweentwoassignednodes.
Solution:
Steps1and2definedinFig.17.26.
Step3:
ReplacingtheindependentsourceEwithashort-circuitequivalentresultsinasupernodethatwillgeneratethefollowingequationwhenKirchhoff’scurrentlawisappliedtonodeV1:
I1=V1/Z1+V2/Z2+I2
withV2-V1=E
Fig.17.26.
andwehavetwoequationsandtwounknowns.
DependentVoltageSourcebetweenDefinedNodesFordependentvoltagesourcesbetweendefinednodes,theprocedureismodifiedasfollows:
1.Steps1and2arethesameasthoseappliedforindependentvoltagesources.
2.Step3ismodifiedasfollows:
Theprocedureisessentiallythesameasthatappliedforindependentvoltagesources,exceptnowthedependentsourcesshavingtobedefinedintermsofthechosennodalvoltagetoensurethatthefinalequationshaveonlynodalvoltageastheirunknownquantities.
3.Step4isasbefore.
EXAMPLE17.15
WritethenodalequationsforthenetworkofFig.17.27havingadependentvoltagesourcebetweentwodefinednodes.
Solution:
Steps1and2aredefinedinFig.17.27.
Fig.17.27.
Step3:
ReplacingthedependentsourceμVxwithashort-circuitequivalentwillresultinthefollowingequationwhenKirchhoff'scurrentlawisappliedatnodeV1:
I=I1+I2
V1/Z1+(V1-V2)/Z2-I=0
andV2=μVx=μ[V1-V2]
orV2=μV1/1+μ
resultingintwoequationsandtwounknowns.NotethatbecausetheimpedanceZ3isinparallelwithavoltagesource,itdoesnotappearintheanalysis.Itwill,however,affectthecurrentthroughthedependentvoltagesource.
FormatApproach
AcloseexaminationofEqs.(17.1)and(17.2)inExample17.12willrevealthattheyarethesameequationsthatwouldhavebeenobtainedusingtheformatapproachintroduceinChapter8.Recallthattheapproachrequiredthatthevoltagesourcefirstbeconvertedtoacurrentsource,butthewritingoftheequationswasquitedirectandminimizedanychancesofanerrorduetolostsignormissingterm.
Thesequenceofstepsrequiredtoapplytheformatapproachisthefollowing:
1.Chooseareferencenodeandassignasubscriptedvoltagelabletothe(N-1)remainingindependentnodesofthenetwork.
2.Thenumberofequationsrequiredforacompletesolutionisequaltothenumberofsubcriptedvoltages(N-1).Column1ofeachequationisformedbysummingtheadmittancestiedtothenodeofinterestandmultiplyingtheresultbythatsubscriptednodalvoltage.
3.Themutualtermsarealwayssubtractedfromthetermsofthefirstcolumn.Itispossibletohavemorethanonemutualtermifthenodalvoltageofinteresthasanelementincommonwithmorethanoneothernodalvoltage.Eachmutualtermisproductofthemutualadmittanceandtheothernodalvoltagetiedtothatadmittance.
4.Thecolumntotherightoftheequalitysignisthealgebraicsumofthecurrentsourcestiedtothenodeofinterest.Acurrentsourceisassignedapositivesignifitsuppliescurrenttoanode,andanegativesignifitdrawscurrentfromthenode.
Solveresultingsimultaneousequationsforthedesirednodalvoltages.Thecommentsofferedformeshanalysisregardingindependentanddependentsourcesapplyherealso.
EXAMPLE17.16
Usingtheformatapproachtonodalanalysis,findthevoltageacrossthe4-ΩresistorinFig.17.28.
Fig.17.28.
Solution:
Choosingnodes(Fig.17.29)andwritingthenodalequations,wehave
Z1=R=4ΩZ2=jXl=j5ΩZ3=-jXc=-j2Ω
Fig.17.29
V1(Y1+Y2)-V2(Y2)=-I1
V2(Y3+Y2)-V1(Y2)=+I2
orV1(Y1+Y2)-V2(Y2)=-I1
-V1(Y2)+V2(Y3+Y2)=+I2
Y1=1/Z1Y2=1/Z2Y3=1/Z3
Usingdeterminantsyields
-I1-Y2
+I2Y2+Y3
V1==-(Y3+Y2)I1+I2Y2/(Y1+Y2)(Y3+Y2)-Y2Y2
Y1+Y2-Y2
-Y2Y3+Y2
=-(Y3+Y2)I1+I2Y2/Y1Y3+Y2Y3+Y1Y2
Substitutingnumericalvalues,wehave
V1=-[(1/-j2Ω)+(1/j5Ω)]6Α∠0°+4Α∠0°(1/j5Ω)/(1/4Ω)(1/-j2Ω)+(1/j5Ω)(1/-j2Ω)+(1/4Ω)(1/j5Ω)
=-(+j0.5-j0.2)6∠0°+4∠0°(-j0.2)/(1/j8)+(1/10)+(1/j20)
=(-0.3∠90°)(6∠0°)+4∠0°(-j0.2)/j0.125+0.1-j0.05
=-1.8∠90°+0.8∠-90°/0.1+j0.075=2.6ν∠-90°/0.125∠36.87°
V1=20.80ν∠-126.87°
MathCadUsingMathCadandthematrixformatwiththeadmittanceparameterswillquicklyprovideasolutionforV1inExample17.16,asshowninMathCad17.3.
Z1:
=4Z2:
=5jZ3:
=-2jrad:
=1deg:
=π/180
Y:
=[1/Z1+1/Z2]-1/Z2I:
=-6-1/Z2[1/Z2+1/Z3]4
I/Y=-12.48-16.64jVolts
8.32-2.24jVolts
V1:
=-12.48-16.64jV1=20.8arg(V1)=-126.87·deg
V2:
=8.32-2.24jV2=8.616arg(V2)=-15.068·deg
MATHCAD17.3
EXAMPLE17.17Usingtheformatapproach,writethenodalequationsforthenetworkofFig.17.30.
Fig.17.30.
Solution:
ThecircuitisredrawninFig.17.31,where
Z1=R1+jXl1=7Ω+j8ΩE1=20ν∠0°Z3=-jXc=-j10Ω
Z2=R2+jXl2=4Ω+j5ΩI1=10Α∠20°Z4=R3=8Ω
Convertingthevoltagesourcetoacurrentsourceandchoosingnodes,weobtainFig.17.32.Notethe“neat”appearanceofthenetworkusingthesubscriptedimpedances.WorkingdirectlywithFig.17.30wouldbemoredifficultandcouldproduceerrors.
Writethenodalequations:
V1(Y1+Y2+Y3)-V2(Y3)=+I2
V2(Y3+Y4)-V1(Y3)=+I1
Y1=1/Z1Y2=1/Z2Y3=1/Z3Y4=1/Z4
whicharerewrittenasV1(Y1+Y2+Y3)-V2(Y3)=+I2
-V1(Y3)+V2(Y3+Y4)=+I1
EXAMPLE17.18WritethenodalequationsforthenetworkofFig.17.33.Donotsolve.
Solution:
Choosenodes(Fig.17.34):
Z1=R1Z2=jXl1Z3=R2-jXc2
Z4=-jXc1Z5=R3Z6=jXl2
andwritethenodalequations:
V1(Y1+Y2)-V2(Y2)=+I1
V2(Y2+Y3+Y4)-V1(Y2)-V3(Y4)=-I2
V3(Y4+Y5+Y6)-V2(Y4)=+I2
whicharerewrittenasV1(Y1+Y2)-V2(Y2)+0=+I1
-V1(Y2)+V2(Y2+Y3+Y4)-V3(Y4)=-I2
0-V2(Y4)+V3(Y4+Y5+Y6)=+I2
Y1=1/R1Y2=1/jXl1Y3=1/R2-jXc2Y4=-1/jXc1Y5=1/R3Y6=1/jXl2
Fig.17.31
Notethesymmetryaboutthediagonalforthisexampleandthoseprecedingitinthissection.
EXAMPLE17.19ApplynodalanalysistothenetworkofFig.17.35.DeterminethevoltageVl.
Solution:
Inthiscasethereisnoneedforasourceconversion.ThenetworkisredrawninFig.17.36withthechosennodalvoltageandsubscriptedimpedances.
Applytheformatapproach:
Y1=1/Z1=1/4kΩ=0.25mS∠0°=G1∠0°
Y2=1/Z2=1/1kΩ=1mS∠0°=G1∠0°
Y3=1/Z3=1/2kΩ∠90°=0.5mS∠-90°=-j0.5mS=-jBl
V1:
(Y1+Y2+Y3)V
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