assignchap5-83-100

assignchap5-83-100 - PROBLEM 5.83 KNOWN Initial temperature...

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Unformatted text preview: PROBLEM 5.83 KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties of finger. F IN D: Whether copper or glass feels cooler to touch. SCHEMATIC: C apper' or glass" = 500K ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constant properties, (3) Negligible contact resistance. PROPERTIES: Skin (given): p = 1000 kg/m3, c : 4180 Jfkg-K, k = 0.625 me-K; Table A-] (T = 300K), Copper: 0 = 8933 kg/m3, c I 385 Jfkg-K, k = 401 me‘K; Table A-3 (T = 300K), Glass: p = 2500 kg/m3, e = 750 J/kg-K, k = 1.4 Wx’m-K. ANALYSIS: Which material feels cooler depends upon the contact temperature T5i given by Equation 5.63. For the three materials of interest, (kp of"? =(0_625 xioooxmao)”2 21,6161/m2 -1<-s“2 skin (kp cfif :(401x8933x385)”2 =37,137 11612 -K-sU2 (kp 6);; = (14x2500x750)”2 21,620 111162.195“? Since (kp >> (kp c )gfiss, the copper will feel much cooler to the touch. From Equation 5.63, 112 112 T _ (kp C)A TA’i +(kp C)B TBA S ‘ 1'12 1‘12 (kp c)A +(kp e)B 1,616 310 +37,137 300 < 1,616+ 37,137 1,616(310)+1,620(300) r, :——= 305.0 K. < SJlglaSS) 1,616+ 1,620 COMMENTS: The extent to which a material’s temperature is affected by a change in its thermal . . . . 1/2 . . . . enVIronment IS inversely proportional to (kpe) . large k applies an ability to Spread the effect by conduction, large 01.: implies a large capacity for thermal energy storage. PROBLEM 5.100 KNOWN: Plane wall. initially at a uniform temperature T. : 25°C. is suddenly exposed to convection with a fluid at Tm = 505C with a convection coefficient h = 75 Wimz-K at one surface. while the other is exposed to a constant heat flux (if) : 2000 W/ing. See also Problem 2.43. FIND: (a) Using spatial and time increments of Ax = 5 mm and At : 205. compute and plot the temperature distributions in the wall for the initial condition. the steady-state condition. and two intermediate times. (b) On (.1; —x coordinates, plot the heat flux distributions corresponding to the four temperature distributions represented in part (a). and (c) On q; -t coordinates. plot the heat flux at x = 0 and x = L. SCHEMATIC: Ttx,<0) = r: = 25°C , k : 1.5 Wim-K Heater as = 2000 me2\t = - 0‘ z “Km—E "‘23 W = i E 710 Tm-i ' Tm i Tm+1 a _"q;(L.t) _,.' l ° :._‘. = n: I I u g i I rm=10.000 K “W: l qxib I = 2' L:50mm in 7mm K Ax = 5mm ASSUMPTIONS: {1) One-dimensional. transient conduction and (2) Constant properties. ANALYSIS: (:1) Using the 1H)" Finitca—Dgfjtifrence Equations, ()ne-Diiiieii.ii0iiril. Transient Tool. the equations for determining the tern erature distribution were obtained and solved with a spatial increment ol Ax = 5 mm. Using the Looktip able functions. the temperature distributions were plotted as shown below. (b) The heat flux. (1’; (x.t). at each node can be evaluated considering the control volume shown with the schematic above « a : Tilh—l "Trix Trii ’TriiH p p qx (mlp): (Elma +Clx.b)/2 : k0 __ 'Tl' k(l)fix "— 2 : k(Tm—l _TIT}+1)/EAX Front knowledge of the temperature distribution. the heat flux at each node for the selected times is computed and plotted below. 160 l— . ._ . ‘ .I . . . _. l s. 2000 . 6 140 ' _ 7 _. .i I 9' : . ' .. . ' . .. ' . l _ ;— 120 - 3 i500 2. 100 i? - i 1000 - E 80 Er a.) so >é % .3. 500 g 40 - E 20 I 0 -. 0 1t] 20 30 40 50 Wall coordinate, x (mm) wall coordinate. 7. (mm) + Initial condition. l<=OS -0- Initial condition. t<=o§ + Time = i505 + Time = 1508 —e- Time z 3005 -9— Time = 3005 —— Steady-state conditions, i>i200s —l—- Steady-stale conditions, t>i£005 (c) The heat fluxes for the locations x : 0 and x = L. are plotted as a function of time. At the x = 0 surface, the heat flux is constant. q: = 2000 Wimz. At the x = L surface. the heat flux is given by Newton’s law of cooling. q; (Lt) : h[T(L.t) — Tr,o ]: at t = 0, q; (L,0) = 4875 W/nil. For steady-state conditions. the heat flux q; (Leo) is everywhere constant at Continued... a, . I .1000 r l — - i ‘--——— i r -2000 _ - : ! 0 200 400 600 SUD 1000 1200 Elapsed time. t [5) —l— q"x[U,‘] - Healer ilux o"x[i_.1} - Convective llux Comments: The lH'l‘ workspace using the Finite—Difference Equations Toot to determine the temperature distributions and heat fluxes is shown below. Some lines of code were omitted to save space on the page. it Finite-Difference Equations. One-Dimensional, Transient Tool: ii Node 0 - Applied heater flux 1‘ Node 0: surface node (disorientation): transient conditions; e labeled t ‘f rho ‘ cp “ der(T0,t) : id_1d_sur_w(TO.T1 .k‘qdot,deltax,TiniO.h0.o"a0) q"aO : 2000 if Applied heat flux, “time; Time : 25 fr“ Fluid temperature. C; arbitrary value since he is zero: no convection process ht) = Ire-20 rt Convection coefficientr WimAEK: made zero since no convection process it Interior Nodes 1 — 9: 1* Node 1: interior node: e and w labeled 2 and D. “l rho‘cp‘der(T1 t) = fd_‘ld_int(T1,T2‘T0,k,qdot,deltax) 1* Node 2: interior node; e and w labeled 3 and 1. ‘l rho’cp‘derfl'E‘t) = td_‘l d_int(T2‘T3‘T‘I ,k,qdot,deltax) a" Node 9'. interior node; e and w labeled 10 and 8. ‘i rho’cp'cler(T9‘t) : td_t d_int(T9,TtO.T8.k,qdot.deltax) I! Node 10 - Convection process: a“ Node 10: surface node (e-orientation); transient conditions; w labeled 9. ‘i' r'no * cp ‘ dertTtUJ) = fd_1d_sur_e(T10,T9,k,qdot,deitax,Tintth,q"a) q"a = 0 iii Applied heat flux‘ Wimflz; zero flux shown if Heat Flux Distribution at Interior NodesI q“m: q"‘l : k r deltax * (TO - T2} r2 q"2 = kt deltax ' (T1 - T3) r2 q"9 = m deilax ' (TEl - T10) r2 iii Heat flux at boundary x: L, q”1l‘l q"xL x h ' (T10 A Tinf) it! Assigned Variables: deitax = 0.005 it Spatial increment rn k : 15 ii“ thermal conductivity, WimiK alpha a 159-63 if Thermal diffusivity. meals cp : 1000 ii” Specific heat, Jr'kgK: arbitrary value alpha 2 k t (rho ' cp) if Detintion from which rho is calculated qdot z t] if Volumetric heat generation rate. Wimr’fl Ti = 25 if Initial temperature. C: used also tor plotting initial distribution Tint 2 50 if Fluid temperature, K h = 75 if Convection COEfiiClen‘L wrmAETK it! Solver Conditions: integrated tirom [i to 1200 with ‘l 3 step, log every 2nd value ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at University of Wisconsin.

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assignchap5-83-100 - PROBLEM 5.83 KNOWN Initial temperature...

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