EXAMINATION IN 2C1134 Electrotechnical design December

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Brian Price
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1 SOLUTIONS: EXAMINATION IN C4 Electotechnical esign Decembe --6 Solution to oblem : The temeatue ise o staight conuctos buie in goun is given b Li D Ti Pi Pk πλ πλ g i k i g ik ik whee L i is thei eth une goun, P i the heat issiation e unit length o the conuctos, an λ g is the heat conuctivit o the goun. The isces ik an D ik ae illustate in the igue below. L D D Goun L The temeatue ise is maximal o the mile conucto, because it has two neighbou conuctos that contibutes the temeatue ise. We then can concentate on to teat the mile conucto. B using the given values one obtains.5 D π..5 π. Fom the igue it is evient that b g D L We then get.5.46 an o ANSWER: The minimal allowe hoizontal seaation is.7 cm

2 Solution to oblem : ~ j j j The maximum is oun when, which gives khz khz 6 max max max max ANSWER: At khz the loss gent is at its maximum. The value is.75 Solutions to oblem Fist estimate the equivalent aius o each hase conuctos. Fo mm iamete conuctos that can be cicumscibe b a cicle o aius cm we have eq The coniguation o owe line can be awn accoing to the igue below Goun 9 4 D

3 The isce to the mio image cuents, which coesons to the eal hase cuents can be calculate accoing to Cason s omula ρ D m 5 The isces, an between the conuctos ae calculate accoing to m The comonents o the imeance matix can now be calculate. The sel an mutual inucces o the hases is ist calculate as μ 46 L L L. μh/m π μ 46 6 L L.95 [H/m] π 6.4 μ 46 6 L L.5 [H/m] π 8 μ 46 6 L L.95 [H/m] π 6.4 The imeance matix will the be Z i Ω I a shot cicuit occus between ajacent hases we have U J Z -J Z -J a U Z Z b g b g U a U Z J Z J Z J Z J c a U a U J Z Z Z Z h 4 4 a J.649 A..95 i..8 π 5. ANSWER: The shot-cicuit cuent is 649 A

4 Solution to oblem 4 Fist name the ats o the smission such that these ae,, an om let to ight. The smission coeicients then ae α.5 an α.5 The voltage at both sies o the election oints A an B must be the same i. e. u i u u t. Deinition o smission an election coeicients gives Z Z Z at A u βu an u αu..; α, β i u u t -u i at B u u t -u i t i Z Z β u α u u α u β.5 i i i i β u α u u α u β.5 i i i i Z Z Futhe we b einition have β - β Now we can use the election iagam metho with α.5, α.5, β.5 β.5 With Hs being the Heavsie s ste uction the signal at B then will be α α H t β β α α τ H t τ β β α α H t 5τ β β α α H t 7τ n β β α α H t n τ U Ate an ininite numbe o elections we get β β.5.5 u α α β β U α α U U U n.5.5. n ββ.5.5.5u.65u τ.5u τ 5τ t ANSWER: The voltage ate an ininite numbe o elections will be.u

5 Solution to oblem 5 Fist estimate the equivalent aius o each hase conuctos. Fo mm iamete conuctos that can be cicumscibe b a cicle o aius cm we have eq The coniguation o owe line can be awn accoing to the igue below 9m Goun D D D The isces, an between the conuctos ae calculate accoing to m The couling coeicients ae then calculate accoing to H 9 k k m F.98 [ / ] π.844 π H 4 k D k k k k.4 [ m / F ] π 8.8 k k.84 [ m/ F] π 8.4 [ m/ F] π.844 π To get caacices one nees to invet the couling matix to obtain the coeicients o caacice an the coeicients o inuction. k k k K k k k k k k such that c c c c c c K c c c e. i. K K Now the caacices between the hases can be calculate as

6 ANSWER: C C C c c c caacice e mete between hase an caacice e mete between hase an caacice e mete between hase an C c c c C c c c C c c c Caacice between hase an goun Caacice between hase an goun Caacice between hase an goun Solutions to oblem 6 The single Thomson coil tun has a mean aius o R. The wie aius is. The tuns can be staightene out. The eective length o the Thomson coil wie then is l NπR Intouce a xz Catesian co-oinate sstem accoing to the igue below with.,,- x In a oint, between the centes o the wies the magnetic lux ensit that oiginates om a cuent J in oint, is B J μ π J In a oint, between the centes o the wies the magnetic lux ensit that oiginates om a cuent -J in oint,- is B J μ π J It is now ossible to calculate the geneate lux e length geneate b a loo comising the two thin conuctos o aiuses <<. En eects ae neglecte. J μ x π J μ π J μ π Jμ J J π Accoing to the thumb ule the lux is iecte in the negative iection. The oute inucce now can be calculate. L μ H oute π m x

7 The oce e length unit o the Thomson coil wie can be calculate b use o the incile o vitual motion. The inne inucce is ineenent o the osition o the wies. This means that a eivation o L with esect to onl will give non zeo contibutions om the oute inucce. LJ W L L F With One gets L π L μ μ π Jμ π μ F J π μ π μ μ F J J π π << gives Initiall. With lπr we inall get the initial oce accoing to R F J μ Fo J5, R.5, N, an.5 we get F68N ANSWER: 68N

PHY 213. General Physics II Test 2.

PHY 213. General Physics II Test 2. Univesity of Kentucky Depatment of Physics an Astonomy PHY 3. Geneal Physics Test. Date: July, 6 Time: 9:-: Answe all questions. Name: Signatue: Section: Do not flip this page until you ae tol to o so.