hw4_sol - PROBLEEI 3.101 KNO‘HN: Dimensions of a plate...

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Unformatted text preview: PROBLEEI 3.101 KNO‘HN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its ends. Net heat flux at top surface. FIND: {3) Differential equation which determines temperature distribution in plate, {b} Temperature distribution and heat loss to heat sinks. SCHEJ—IATIC : ASSUMPTIONS: [1) Steady-state, [2] One-dimensional conduction in x URLs-331}, [3) Constant properties, [4) Uniform surface heat this; [5] Adiabatic bottom, [6] Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume, qx + dq = q9; +dx, where £13....“ = (13 + (dqxi'dx) dx and {1‘31qu {VI - dx Hence, dqX dx}— a; “1:0. From Fourier‘s law. '51:.i = —lc{ t - W} dT.-"dx. Hence. the differential equation for the temperature distribution is —i [law £]—q3 ‘W=O dLT+q—°=0. <1 d); dx dxl kt (b) Integrating twice, the general solution is. t{x}=—q—°x3+c x+C- . 2h 1 4 and appropriate bormdary conditions are Tm) = To. and HI.) = To. Hence. TEJ = C2, and __ a3 2 . _ s31— Hence. the temperature distribution is QEL 2 T x} = — x — L); + T . < l ' 2m ° Applying Fourier’s law at x = U. and at x = L. . . - q” L q” as; U :—k Wt dT.-'d. :—kvv't ——0 . —— = — 0 ql l l- } WE” [ to] [K a] x=fl 2 q { L} = —k t w: }c1T.-’dx)x=]_ = aw: [— q—O] [x — E] = + ‘10 “I” ' ' ' ' kt 2 3:1 2 Hence the heat loss from the plates is q=2[q3WL—"2] = qSWL. ‘1 COMMENTS: (1) Note signs associated with 91(0) and qflL). [2) Note symmetry about it. = LIZ. Alternative boundary conditions are T63) = To and dT.-"dx)3=L.-'3=U. PROBLEIRI 3.125 KIN—DSTN: Dimensions and end temperatures of pin fins. FEST): (a) Heat transfer by convection from a single fin and (b) Total heat transfer from a l 2 . in surface u-‘ith fins mounted on 4mm centers. SCHEl-IATIC : ASSUMPTIONS: (1} Steady—state. (2) One—dimensional conduction along rod, {3) Constant properties. (4) No internal. heat generation, (5} Negligible radiation. PROPERTIES: “E is H- Copper. pure {323K}: 1-: a 4:30 w.-'m-I~:. ANALYSIS: (a) By applying conservation of energy to the fin. it follows that qconv = qcond,i _ qcondo where the conduction rates 111a},r be evaluated from knowledge of the temperature distribution. The general solution for the temperature distribution is 9{x}=CLem+C2 em SsT—Tm. The boundary conditions are BOD) E EIU = 100°C and ML} = ID. Hence 90 = C1 4-1:: :3: c1 eml +c: e'mL Therefore. C2 = C1 e:mL Jan .9sz 1_elmL 9 =—2 5 : l—e‘m:L and the temperature distribution has the form 9 So [emit _ elitism: _e1n1L The conduction heat rate can be evaluated by Fourier’s law, [19 ILA I9 an]; Clcond = ‘kAcE=—fim[em +e'"I m] or, 1with m = {lips-RAE _}1 1 , l.-"2 9 thA q qmnd : _ Oil Fl '2} [enm+eamL-nm]- _e—n]l' Continued PROBLEEI 3.] 25 (Cent) Hence at x = G, qconcl,i = _ atx=L _ 0,, {11me )l’": ( qconcLo = gmL niL l—e De Evaluating the fin parameters: 1P 1'52 4h “2 4 100 a“ 3 K “3 m=|: 1 :| =[—] =31.62111_1 l-cAE 1:13 400 W.-"m- F. X 0.001111 , 1.2 , 1.2 st II‘ 0‘ s w w _ w {hiatusc )1 a: —o3hk = —><{0.001m}' x100 x400— zassxio 3— 4 4 1113K m-K K niL =31.02 m‘1x0.025m=0.tsi. emL 2.204, .3sz =4.805 The conduction heat rates are —100K[s.s:«s><10'3 _ _ _ ClconcLi Z TXJBEO 21.30? W’ —100K(9.93x1fl'3 T and from the conservation relation. X4.403=1.1331V ClconcLD = qmm- 4.50? W—1.133 w =fl.3?4 w. <: {b} The total heat transfer rate is the heat transfer from N = 250x250 = 615%? rods and the heat transfer from the remaining {bare} surface (A = 11112 - NAE}. Hence. q =t: qcond’i +hA90 = 02,500 (1.50? w)+100w.-'m3-K[0.9511:13] 100K q = 9.4:2x104 *0,-'+0.9:<.x104 W=1.O3?x105 w. CUMMENTS: (1) The fins, which cover only 5% of the surface area, provide for more than 90% of the heat transfer from the surface. {2) The fin effectiveness, E“ E qmndi hAcflc, is E = 192. and the fin efficiency, It; (qcom. .-"hJ'E DLHD }, is n =O_43_ {3) The temperature distribution, filial-"BO, and the conduction term, elem-1m. could have been obtained directly fiom Eqs. 3.??and 3.?3, respectively. {4) Heat transfer by convection from a single fin could also have been obtained fiom Eq. 3 5'3. PROBL ElI 3.130 IC'M'OW'N: Arrangement of fins between parallel plates. Temperature and convection coefficient of air flow in finned passages. I’daxarnuni allowable plate temperatures. FIND: (a) Expressions relating fin heat transfer rates to end temperatures, (b) E-Iaxiimun power dissipation for each plate. SCHEl-IATIC : PE=4QOK Fish? dime :15 ions: D=I05mm xW=200 mm {512 mm aflfi— i inf—p E=5%DV$}:;LK x f=fmm; n¢=50 9”: 4‘ ' m" tT: 350K I. ASSUMPTIONS: [1) Steady-state conditions, (2} One-dunensional conduction in fins, [3] Constant properties: [4-) Negligible radiation, {5) All of the heat is dissipated to the air, {6) Uniform h, {T} Negligible variation in Too, (8) Negligible contact resistance. PRoPERIrEs: m“; A‘- - Aluminum (pm—e}. 3T5 K: k = 240 W-"ni-K. ANALYSIS: [a] The general solution for the temperature distribution in fin is I9 ET{x}—Tm = Clem +C'2e_mX Boundaryr conditions: 91:9}:90 =T0 —Tc.,. I9 {L]=9L =TL —Tm. Hence so = Cl +c3 9L = Cleml' +c2c'mL 9L 2 ClemL +{so —c1}e'mL C1 2 9L —90e_mL Cr: 2 90 _ 9L —I90e_mL : age-“1L —sLI ELrnl. _ E-rnL ‘ einL _ E-rnL ernL _€-rnL in x—L in La. _ Hence elxl=9Lem_90e { l+eoe l )_9Le m - EInL _ -1nL 90[ 111{l.—2i] _ —rn|:L—x]]+9L (Erna _ E-tmr] MK}: nL m]. eI —e_ 9‘: } Bosinh tn[L—x}+9Lsinh 111K x 2—. ' sinh mL The fin heat transfer rate is then elf =—kAC£=—th —_9icosh1n{L—X}+_SL—mcosh rm: . dx sinl‘i inL ' sinh inL Hence qfa =th Hui— tar—m ‘: ' tanth smth 5' In H In =th U ——L <1 qfl [ sinh mL tanl‘i n1]. } Continued PROBLEEI 3.130 (Cent) (13} = = 35.5. m‘1 use r, 2 1113 I: 50 Elfin“ -K(2XC|.1 m+2><|DfiDl In) in: 340 “r.-"'111-K‘><U.l 111x 9.091 :11 mL =355 m‘1x0012 m: 0.43 sinh mL=0439 mman 0.401 90 =100 K BL :50 K leo = 240 Vii-"in -K‘><U.l nDCDDDl 111 ’ 0.401 0.439 L00 Kx35.5 m‘l _ 50 11:65.5. m‘1 ] qf’g : 1154 W (PC-J the top plate) 4; a -1 q —1 em = 240 a-‘.-'m-Kx0.1mx0001m 100 Km“ 1“ — "U “355 m * 0.439 0.401 qf’L = 3?.3 W'. (“I 29 the bottom plate) Maximum power dissipations are therefore Clomax = Nf‘lio +(1W ‘thlflheo Clomax =5UX115.4 ‘W‘F'lUEUU —5DXU.GUI}111XU.111D<150 Ell-"1112 -KX100 K 51.111133 = 5WD W+225 W = 995 W .5: LJ1 ClLJ‘flflK = ‘Nf‘lfiL +1“: _ I"lift theo cm.me = —SUXB?.B‘W +[fl209—50X0flflllmXUJ le 50 Ell-"m3 -K><50 K ELL,an = —4390 w+112w = 4133 w. r: C'DMILIEXIS: [1) It is of interest to deteimine the air velocity needed to prevent excessive heating of the air as it passes between the plates. If the air temperature change is restricted to flTm = 5 K, its flcwrate must be qmt _ 1T1? W ma1r = — — = O. 34 kgr’s. cpnTm 100? 15kg -K Xi H. Its mean velocity is then Vair _ mm: = 0.34 kg‘ls = 1153 III-"s. PairAc 1.16 lag-"ml3 x0012 mffll —50x0.flfll]m Such a velocityu-‘ould be impossible to maintain. To reduce it to a reasonable value, e. g. 1!} 1115's, AC 1a-‘cmld have to be increased substantially by increasing W (and hence the space between fins} and by increasing L. The present configuration is impractical fiom the standpoint that I“? W could not be transferred to air in such a small volume. (2) A negative value of qux implies that heat must be transferred from the bottom plate to the air to maintain the plate at 350 K. PROBLEM 3.132 ICN'U‘WS: Width and maximum allowable temperature of an electronic chip. Thermal contact resistance between chip and heat sink. Dimensions and thermal contluctit'it}= of heat sink. Temperature and convection coeff1c1ent associated Wll'h air flow through the heat sink. FIND: (a) Marrimum allowable chip power for heat sink with prescribed number of fins. fin thickness, and tin pitch. and (1:) Effect of fin thickne ss-"nmnber and convection coefficient on performance. SCHEI‘LIATIC: / ———- Tc = 85°C w 2 2” mm \ at; 2x1o-h‘ mz-KI’W / F, — k = 180 wrm-K L b: 3 mm : l —;~ <— t Tc Rtb Tm Lr=15mm r-WN‘W'W' L / fl qc Rt Rt .C .D (gal-x Tngooc S—l‘i—H K—5=1.Etmm If M h=100me2-K ASSUMPTIONS: [l] Steady—state. [2] One—dimensional heat transfer. [3] Isothermal chip. [4] Negligible heat transfer from top Isurface of chip, Negligible temperature rise for air flow. [6] Uniform convection coefficient as soc1ated With air flow through channels and over outer surfaces of heat sink, (T) Negligible radiation. ANALYSIS: [a] From the thermal circuit. _ Tc —Tc., _ Tc —Tc., Ell: — Rtot Rt; + Rtb + Rm 3 '3' 6 where atC = an w- = 2 r10_ 111'; - K.-"W :‘(uozmjr' = 0.005 w and all, = Lb .-'it[a-") =0.003m:‘180 w 'm-Kfflfllmll = 0.042 KIW. From Eqs. [3.103), (3.102], and {3.99) 1 11'021—NA1" a = . 1—t , AzNA +A t__o nohAt Art if} t r b where Af= EWLf = 2 x 0.03m >< 0.015111 = 5 x 10-”1 n1L and Ab = W" — NEtW] = [0.03m}; — 11(0132 1 . a :1 a x 10" m x 0.02111] 2 3.6 X 104111“. With 111ng {Eh-*lajl" L5: {zoo w.-'nr-K-'1ao W.-"m-K >< 0.132 x 10431131" {0.015m) = 1.1't', tanh mILf= 0.824 and Eq. (3.8?) yields tanh mLf _ 0.82-4- 1an 1.1? = 0.?04 Hf = It follows that at = 6.96 x 10'3 in]. no = oars. R15 = zoo K—‘W, and {35 — 2o}°c _— = 31.8 w <1 (o.oo:+ 0.042 +2.00}K.-' w '51:: {b} The following results are obtained from parametric calculations performed to explore the effect of decreasing the number of fins and increasing the fin thickness. Continued . PROBLEM 3.132 (Cont) N rim) 0:" Rm tic-m qt (w) A. (ml) 6 1.1333 0.95? 2.?6 23.2 000328 ?‘ 1.314 0.941 2.40 26.6 0.00442 3 0.925 0.919 2.15 29.}r 0.00505 9 0.622 0.885 1.9? 32.2 0.00565I 10 0.330 0.826 1.139 33.5 0.00632 11 0.182 0.204 2.00 31.23 0.00696 Although m [and no} increases With decreasing N [increasing 1]. there is a reduction in A; which yields a ininiinmn in RE. and hence a maximum value of qt, for N = 10. For N = 11. the effect of h on the performance of the heat sink is shown below. Heatra:e~ as a function of convection coefficienttN=1 1'; m III-III 3:: 1 on E" {T cu" E *5 g 50 D III-III- 100 200 300 400 500 500 7’00 300 9001000 Convection coefficient. htwmzfi} With increasing h from 100 to 1000 WIm‘K, Rm decreases fiom 2.00 to 0.4}' K.-'W._ despite a decrease in nf (and no] from 0.7104 (0.?19) to 0.269 {0.309). The corresponding increase in qc is significant. COMMENTS: The heat sink significantly increases the allowable heat dissipation. If 11' were not used and heat was simply transferred convection from the surface ofthewchip with h = 100 WEmLI-K, Rm,T = 2.05 K-‘VJ from Part (a) would be replaced by Rm. = liltw; = 25 K-‘W', yielding q: = 2.60 W. ...
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hw4_sol - PROBLEEI 3.101 KNO‘HN: Dimensions of a plate...

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