a. (1) Assume T = 20 ºC = 293 K. Apply Equation 2.22 to find the resistivity of the brass in the disk with

Transcription

1 Aignmen #5 EE7 / Fall 0 / Aignmen Sluin.7 hermal cnducin Cnider bra ally wih an X amic fracin f Zn. Since Zn addiin increae he number f cnducin elecrn, we have cale he final ally reiiviy calculaed frm he imple Mahieen-Nrdheim rule in Equain. dwn by a facr (+X) (ee Quein.8) ha he reiiviy f he ally i [ CX ( X )]/( X ) in which C = 300 nω m and Cu 7 nω m. a. An 80 a.% Cu 0 a. % Zn bra dik f 0 mm diameer and 5 mm hickne i ued cnduc hea frm a hea urce a hea ink. () Calculae he hermal reiance f he bra dik. () If he dik i cnducing hea a a rae f 00 W, calculae he emperaure drp alng he dik. b. Wha huld be he cmpiin f bra if he emperaure drp acr he dik i be halved? Sluin a. () Aume = 0 ºC = 93 K. Apply Equain. find he reiiviy f he bra in he dik wih Cu = 7. n m and X Zn = 0.0: bra = Cu + C Zn in Cu X Zn ( X Zn ) i.e. bra = 7. n m + (300 n (0.0)( 0.0) bra = 65. n m We knw ha he hermal cnduciviy i given by / bra = C FWL where bra i he cnduciviy f he dik, C FWL i he Lrenz number and i he emperaure. hi equain can al be wrien a bra = C FWL ha = C FWL /. Applying hi equain, (0 C) = (. 0-8 W K - )(93 K) / ( (0 C) = 09.8 W K - m - he hermal reiance i = L/(A), where L i he hickne f he dik and A i he cr-ecinal area f he dik. = L/(A) = (5 0-3 /[(09.8 W K - m - )()( 0 - ] = K W - () Frm dq/d = A/x = / (x can be aken be he ame a L), and dq/d = P (pwer cnduced), we can ubiue bain: = P = (00 W)( K W - ) = 3.6 K r 3.6 C Ne: Change in emperaure i he ame in eiher Kelvin r degree Celiu, i.e. = = ( + 73) ( + 73). b. Since = P, ge half, we need half r duble r duble r half. We hu need / bra r / (65. n which can be aained if he bra cmpiin i X new ha new = Cu + C Zn in Cu X new ( X new )

2 EE7 / Fall 0 / Aignmen Sluin i.e. / (65. n = 7 n m + (300 n X new ( X new ) Slving hi quadraic equain we ge X new = 0.055, r 5.5% Zn. hu we need 9.5% Cu-5.5% Zn bra. Prblem 8, e-bkle Eenial Hea ranfer fr Elecrical Engineer hermal ime cnan Cnider he hermal circui f a hea inked ranir ha i hwn in Figure 9. Aume ha C i mall and C c and C are large ha while he emperaure f he uncin i riing, c remain cle he ambien emperaure. By cnidering he hea flw Q frm he uncin in C and al in θ c hw ha cq exp( / cc ) Pl hi funcin aking reanable value: = 5 C, Q = 0 W, and apprximaely, θ c 0 C/W, and C 0.0 J K -. Sluin Since C << C c and C, he hermal circui in Figure 9 can be implified a hwn in he figure belw. Q Q Q C θ c Δ = - We can herefre build a differenial equain u f hi circui a: C Q Q Q d Q C d Q d d d d he luin hi differenial equain i a fllw: c c Q c Q C

3 EE7 / Fall 0 / Aignmen Sluin Q c d d d ln cq A C cq C C cq A exp cc cq A exp cc c c d A c A exp( A) A a bundary cndiin, he emperaure difference beween he uncin and cae i aumed be zer a ime = 0. hu, 0 0 cq A exp cc A Q c cq cq exp cc cq exp cc cq exp cc he figure belw illurae he ranien repne f hi yem. 50 Q C emperaure, Celiu Q c C c 0.5 ime, ec

4 EE7 / Fall 0 / Aignmen Sluin Prblem, e-bkle Eenial Hea ranfer fr Elecrical Engineer Hea ranfer by radiain Cnider a 60 W, 0 V incandecen General Elecric ligh bulb. he ungen filamen i f lengh m and diameer 5.7 μm. ungen ha a deniy d = 9300 kgm -3, pecific hea capaciy c = 30 Jkg - K - ; reiiviy a rm emperaure ρ = 56.5 nωm, and ρ i prprinal. where i in Kelvin. If he meling emperaure f W i 3680 K, wha i he vlage ha guranee he blwing f he bulb? Hw lng will i ake blw he bulb if i i accidenally cnneced a 350 V upply? Sluin (a) Under eady ae cndiin, we have he fllwing relainhip: Rae f elecrical energy cnvered hea = Rae f hea l frm he filamen by radiain r P S Nw we are inereed in he vlage upply which lead he meling emperaure f he filamen. S, E R E S SR where R i he reiance f he filamen, R = ρl/a where L i he lengh and A i he cr-ecinal area f he filamen. Here, ρ i prprinal., ρ = ρ (/ ).. hu, he vlage blw he bulb i: L E dl ( d / ) L d (0.35)(5.670 Wm K )(0.533 ( ( K V. (b) Under ranien ae, we have he fllwing relainhip: Elecrical energy diipaed = Increae in hea cnen f filamen

5 r + Energy radiaed frm filamen urface V d mcd S R EE7 / Fall 0 / Aignmen Sluin where m i he ma f he filamen (deniy vlume). hu, lving he equain abve fr ime, we have V d mc S d R max V mc S d R. max L r dlc V rl d r. V r max cdlr L d L (30 Jkg K )(.930 kgm )(0.533 (.90 5 (350 V) ( (5.650 ( (0.533 (0.35)(5.670 Wm K ) m d. K (Ne: he inegrain huld be dne numerically.) d