hw1Soln - nw an 1 Samuel/x ME we F03 PROBLEM 1 KNOWN...

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Unformatted text preview: nw an: 1 Samuel/x ME we F03 PROBLEM 1 KNOWN: Inner'surface temperature and thermal conductivity of a concrete wall, FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C. SCHEMA'I‘IC: A = 20 m2 k = 1, 0.75. or 1.25 Wlm-K T1 = 25 “0 T2 = 4510 38 DC ' To!) —>L H. = 0.30 rn X ASSUMPTIONS: (1) One—dimensional conduction in the x-direction, (2) Steady-state conditions. (3) Constant properties, (4) Outside wall temperature is that of the ambient air. ANALYSIS: From Fourier's law, it is evident that the gradient, dT/dx = ~—q;'/lt, is a constant, and hence the temperature distribution is linear, if q’x' and k are each constant. The heat flux must be constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T: = -15°C are err T —T 25'C—(—15“c) , “:4: =1: 1 2=1w vI{——-—-—-—=133.3W ~. ' 1 q“ dx L m 0.30m /m ( ) q‘=q§xA=133.3W/m3x20m2=2667W. (g)< Combining Eqs. (1) and (2), the heat rate q,‘ can be determined for the range of ambient temperature, -15 S T: S 38°C, with different wall thermal conductivities, it. :15ch 2500 '15 a: 1500 n .2 a 500 r I son 4590 -2D -1D D 19 20 30 40 Amblenl air tamporaiurn, T! (C) - I: - Wall thermal cunfluctlvlly. k a: 1.25 WimK ——-~ I:- 1 WimK. concraia wall 4— k u 0.75 Wink For the concrete wall. I: = i W/m-K, the heat loss varies linearin from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity. COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear. impingiihi 5.,;..‘.:g::;" I PROBLEM «at 2 KNO‘W 1“ : Chip Width and maximum alIGWable temperature. Coolant conditions. FIND: Maximum allowable chip power. for air and liquid coolents. 5 CHEMIATEC: #Wfi'nrm 'é’ a E . . .. 739:5 C —-i‘> Air, i7=200i/‘£/m3-K _ GNP: latex: ‘35 C Dieisci'rlc fluid, a”: ' I1: 3000 W/mz-K I ------- — 4' ASSUZ‘JIFTEONS: [1) Steady-state conditions, (2) Negligible heat transfer from sides and button, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air. AJVALYSIS: All of the electrical power dissipated in the chip is transferred by convection to. the coolant. Hence, ‘ P = q and from Newton's law of cooling, P = new — T9,) = hW2(T — Tm). In oz'r, - PM, = 900 “vii/n12 £01005 mflss - 15) ° C = 0.35 W. <1 In the dielectric liquid Pm, = 3000 W/m’J-Kmnns m)2(85 -— 15) ° C = 5.25 W. <1 DOWNTS: Relative to liquids, air is a. poor heat transfer fluid. Hence, in air the chip can dissipate far less energ}r than in the dielectric liquid. 3 . For a medium in which the heat conduction equation is given in its simplest by i; = ii : ax‘ Ct at (:7) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conduativity is constant, For a medium in which the heat conduction equation is given by l a Er?" a BT . ~“(h'T-]+T[k*:-J+ age“ = 0! r 51' or o: .. (:1) Heat transfer is steady, (b) it is two-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable. Psoerara 4&3 KNOWN: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature. FWD: (:1) Rate of surface radiation emission, ([3) Net rate of radiation exchange between surface and chamber walls. SCHEMATIC: ASSUMPTIONS: (1) Area of the enclosed surface is much less than that of chamber walls. ANALYSIS: (3.) From Eq. 1.5, the rate at which radiation is emitted by the surface is qenfit=q;nit -A=£AUT§ clam = 0-8(0-5 1112) 5.67 x 10'8 W/mz-K4 [(150 + 273)K]4 gem, = 726 w'. ' <1 (13) From Eq. 1.7, the net rate at which radiation is transferred from the surface to the chamber walls is q = e A o (Ti—T3... q= 0.3(0.5 mg) 5.57 x 10'“8 W/mZ-K4 [(423104 -— (298194] q=547w. . <1 COMMENTS: The foregoing result gives the net heat loss from the surface which occurs at the instant the surface is placed in the chamber. The surface would, of course, cool due to this heat loss and its temperature, as well as the heat loss, would decrease with increasing time. Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings. ' PROBLEM r it 4 KNOWN: Onedimensional system with prescribed thermal conductivity and thiokness. - ' ‘ FIND: Unknowns for various temperature conditions and sketch distribution. - SCHEMATIC: ASSUMIPTEDNS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are u (11' £1; __ T2 _T1 qx # ——k — and dx dx L ' (1’2) Using Eqs. (1) and (2), the unknown quantities for each case can be determined. dT _ (—20 —— 50) K (a) dx 0.25m —280 / n W . K - =—5 — —280——— =. . W 1112. qx Gm'Kx[ m] 140k / (b) £=W=30Kfin dx 0.25m - H W K = —50 —— so— = —-i 2 qx mDK x[ m] OkW/rn n W K ‘ = —50 —— 1'60 — = -— . 2 (c) qx WK >< [ m] s o kw/m _ T2 =L-—dl + T1 = 0.25111><[160E +70 " 0. dx 111 T2 = 110 " c. H W K d =—5o —— ksoh— = . 2 Us); WK x m 40kW/m T1 =T2 . dx 111 T1 a C. Ir W ‘K I = —50 —. 200—— = — . 2 (e) qx WK x [ m] 10 o kW/nti T1 = T2 —— Lac—13 = 30 ° c — 0.2542005] = —20 "to. dx m a PROBLEM e5” KNOWN: Temperature distribution in a one-dimensional wall with prescribed thickness and thermal conductivity. FIND: (a) The heat generation rate, q, in the wall, (in) Heat fluxes at the wall faces and relation ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) Constant properties. ‘ ' ANALYSES: (a) The appropriate form of the heat equation for steady-State, one-dimensional conditions with constant properties is Eq. 2.15 re-written as - . d dT = _k .— ‘1 d3; dx ] Using the form for the temperature distribution, evaluate the gradient giving, - _‘ _ d _d_ 2 - .. _d_ .— -9 q .. 1: dx dx (a+bx )] — k dx [2bx] — "bit q = —2(—2000°C/m2) x 50 W/nn-K = 2.0x105 W/m3 . . 4 ‘ (b) The heat fluxes at the wall faces can be evaluated from-Fourier’s'law, n dT QXCX)=“k a x Using the temperature distribution T(x) to evaluate the gradient, find _ q;(x) = —k % [a—t-bx2] = —2kbx . The flux at the face x=0, is then . q;(0)=o <1 and at x = L, q;(L) = —2kbL = ---2 x SOW/m-K (—2000°C/m2) x 0.050m gm): 10,000 W/mz. - < - ; COMMENTS: _ From an overall energy balance on the wall, it follows that '. _ E0111 + 1‘31:} = 0 "' + 611‘ = 0 E q- nO 2— ‘ 9 Q=M=W=g_0xlofiw/m3. _ ...
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hw1Soln - nw an 1 Samuel/x ME we F03 PROBLEM 1 KNOWN...

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