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Unformatted text preview: PROBLEM 3.71
KNOWN: Temperature distribution in a composite wall. FIND: (a) Relative magnitudes of interfacial heat ﬂuxes, (b) Relative magnitudes of thermal
conductivities, and (0) Heat flux as a function of distance x. SCHEMATIC: X ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties. ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant k),
the parabolic temperature distribution in C implies the existence of heat generation. Hence,
since dT/dx increases with decreasing x, the heat ﬂux in C increases with decreasing x.
Hence, 61% > £13
However, the linear temperature distributions in A and B indicate no generation, in which case
<13 = 015
(b) Since conservation of energy requires that qu = qic and dT/dx)B < dT/dx)c , it follows
from Fourier’s law that kB > kc.
Similarly, since £15, A = qi’B and dT/dx) A > dT/dx)B, it follows that
k A < kB. (0) It follows that the ﬂux distribution appears as shown below. If
X X; X 3 X4 0 x 92 COMMENTS: Note that, with dT/dx)4,(3 = 0, the interface at 4 is adiabatic. PROBLEM 3.46
KNOWN: Conditions associated with a composite wall and a thin electric heater. FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and' inner
heat ﬂows and conditions for which ratio is minimized. SCHEMATIC: "2r
' Mp '
(1,-4— / \ +qo / In (r21r1) In (r 3If2) (\hozﬂrar'l h-Z r '1
( ' 1' 1) 21th anA ASSUMPTIONS: (1) One-dimensional, steady-stare conduction, (2) Constant properties, (3) Isothermal
heater, (4) Negligible contact resistance(s). ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as
shown in the schematic. (b) Performing an energy balance for the heater, Bin = Bout , it follows that Th“ -00T Th_Tcoo
qh(2”12)=CIi+Cli-i= ——m——’—-—+ ’ <
—11nr2/I1 —1 Inn/r2
(hum 11) 22,1; (hams) +——*,E,kA)
(c) From the circuit,
_ —1 1n(r2/r1) ’ ' ,
q_, a, w (W) m .423— $1" in
__ ,, _
q; (Th— 4,0,) (huh 131+)—+111(13/12) ’ ”r g, 1 r3 ZﬂkA To reduce qb /q§ , one could increase k3, h,, and r3/r2, while reducing 19,, he and r2/r1. COMMENTS: Contact resistances between the heater and materials A and B could be important. ...
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- Spring '11