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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 semiinﬁnite solids, (2) Constant
properties, (3) Negligible contact resistance. PROPERTIES: Skin (given): p = 1000 kg/m3, c : 4180 JfkgK, k = 0.625 meK; Table A]
(T = 300K), Copper: 0 = 8933 kg/m3, c I 385 JfkgK, k = 401 me‘K; Table A3 (T = 300K),
Glass: p = 2500 kg/m3, e = 750 J/kgK, k = 1.4 Wx’mK. 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 KsU2
(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 WimzK 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 steadystate 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 WimK Heater as = 2000 me2\t =  0‘ z “Km—E "‘23 W
= i
E 710 Tmi ' 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) Onedimensional. transient conduction and (2) Constant properties.
ANALYSIS: (:1) Using the 1H)" Finitca—Dgfjtifrence Equations, ()neDiiiieii.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
—— Steadystate conditions, i>i200s —l— Steadystale 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 steadystate 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 FiniteDifference Equations. OneDimensional, 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) = Ire20 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 (eorientation); 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 15963 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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 Fall '08
 Rothamer
 Heat Transfer

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