电气专业毕业设计外文翻译.docx

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