电机拖动课后题答案汇总.doc

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1、Chapter1Problems：1-1solution:(a)(b)(c)1-2solution:Chapter2Question2-14:transformer:480=k60,sok=8Whenf=50Hz,Load:U1=k*50=8x50=400v<415v,okProblems：2-1solution:2-2solution:(a)30x240=7200va(b)7200va(c)Ip=7200/2400=3A2-3solution:(a)P=kvaxpowerfactor=5x1.0=5kw(b)P=5x0.8=4kw

2、(c)P=5x0.3=1.5kw(d)I=va/E=5000/120=41.7A2-4solution:(a)0.8powerfactor,lagging1)Withasareference,thefull-loadsecondarycurrentis2)Theprimaryvoltageseenfromthesecondaryside:3)(b)0.8powerfactor,leading.1)2)3)2-5solution:Averysmallamountofironlossinthetransformerisnegligibl

3、e.Togettheprimaryandsecondaryvaluesseparately,theprimaryandequivalentsecondaryvaluesareassumedtobeequal:thus:02-6solution:(1)theexcitingparametersare:ThemagneticresistanceisThecurrentoflossesisThemagnetizationcurrentisThemagnetizationinductanceis(2)(a)theequivalentprim

4、aryimpedance(b)theequivalentprimaryresistance(c)theequivalentprimaryreactance2-7Solution:(a)Powerfactorforshort-circuitcondition=Therefore,thephaseangleforshort-circuitconditions=Therefore,(b)usingper-unitvalues2-8solution：voltageofeachturn，andU1=E1U2=E2changecm2tom2hi

5、ghvoltagewindinglowvoltagewinding2-9solution:theeffectivesectionareaofthecorereferredtotheprimarysidetheeffectivesectionareaofthecorereferredtothesecondaryside2-10solution:thetwotransformerhavethesamevoltageandthesameN1andneglectZ1distributethevoltageinseriestheprimary

6、sidethesecondarysideChapter3Problems：3-1solution:Whenfrequencyis50HzNumberofpoles2468101214Fieldspeed300015001000750600500428.57Whenfrequencyis60HzNumberofpoles2468101214Fieldspeed360018001200900720600514.29Whenfrequencyis400HzNumberofpoles2468101214Fieldspeed240001200

7、080006000480040003428.63-2solution:3-3solution:（1）τ=Z1/2p=24/4=6（2）α=p×360°/Z1=2×360°/24=30°q=Z1/(2pm)=24/(2×2×3)=2（3）（4）（5）TheEMFstaris:131719234568910111224222120181716151423aU1U2V1W2W1V2（6）Thewindingconnectiondiagramis:3-4solution:(1)PulsateMMF(2)EllipseMMF3-5soluti

8、on:TheinputpowerWAInordertooperateatunitypowerfactor3-6solution:Formtheno-loadtestwefind:Fromthelocked-rotortestwefin