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Unformatted text preview: PROBLEM 16.3 KNOWN: Onedimensional system with prescribed surface temperatures and thickness. FIND: Heat ﬂux through system constructed of these materials: (a) pure aluminum, (b) plain carbon
steel, (c) AISI 316, stainless steel, (d) pyroceram, (e) teﬂon and (f) concrete. SCHEMATIC:
l‘*—’l— L =2.0mm
7;=325/< 7;=Z75/<
Mafer'ia/ of
known k l_"x ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction, (3) No heat
generation, (4) Constant thermal properties. PROPERTIES: The thermal conductivity is evaluated at the average temperature of the system, T =
(T1+T2)/2 = (325+275) K/2 = 300 K. Property values and table identiﬁcation are shown below. ANALYSIS: For this system, Fourier’s law can be written as dT T2 — T1
” = —k — = —k .
qx dx L
Substituting numerical values, the heat ﬂux is
275325 K
q; = —k(——T)— = +25005k
20 x 10 m m
where q; will have units W/m2 if k has units W/mK. The heat ﬂuxes for each system follow.
Thermal conductivity Heat ﬂux
Material Table k(W/mK) q; (kW/ m2)
—————_______—_
(a) Pure Aluminum HTl 237 593 <
(b) Plain carbon steel HTl 60.5 151
(c) AISI 316, SS. HT—1 13.4 33.5
(d) Pyroceram HTl 3.98 9.95
(e) Teflon HT2 0.35 0.88
(1) Concrete HT2 1.4 3.5 COMMENTS: Recognize that the thermal conductivity of these solid materials varies by more than
two orders of magnitude. PROBLEM 16.6 KNOWN: Steadystate conduction with uniform internal energy generation in a plane wall; temperature distribution has quadratic form. Surface at x = 0 is prescribed and boundary at x = L is
insulated. FIND: (a) Calculate the volumetric energy generation rate, q, by applying an overall energy balance to the wall, (b) Determine the coefﬁcients a, b, and c, by applying the boundary conditions to the
prescribed form of the temperature distribution; plot the temperature distribution. SCHEMATIC: T(x) = a + bx + cx2
k = 5 W/mK, q' T(0) = To = 120°C Tm: 20°C . If x , ’ = 2 ® '
h 500 W/m K is L_> x L=50mm Insulated
boundary ASSUMPTIONS: (l) Steadystate conditions, (2) Onedimensional conduction with constant
properties and uniform internal generation, and (3) Boundary at x = L is adiabatic. ANALYSIS: (a) The internal energy generation rate can be calculated from an overall energy balance
on the wall as shown in the schematic below. Ein — E[out + Egen = 0 Where Ein = qlconv
h(T°°—TO)+qL=O (1) q=—h(r°° —TO)/L=—500W/m2 K(20—120)°C/0.050 m=1.0><106 W/m3 < To 1’ V
q., a can.) Z
conv —> a. ""7?! d I i.‘ 4 1 %
Egan  —L> x = L L—> x L
(3) Overall energy balance (b) Surface energy balances (b) The coefﬁcients of the temperature distribution, T(x) = a + bx + cxz, can be evaluated by applying the boundary conditions at x = 0 and x = L. See Table 16.1 for representation of the boundary
conditions, and the schematic above for the relevant surface energy balances. Boundary condition at x = 0, convection surface condition Ein _ E[out = Ciiionv — (lit (0) = 0 Where (1x (0) = ‘k—
h(T°° —To)—[—k(0+b+20x)x=0] = 0 b=—h(T°°—TO)/k=—500W/m2.K(20—120)°C/5W/mK=1.0><104K/m < Continued . . . .. PROBLEM 16.6 (Cont.) Boundary condition at x = L, adiabatic or insulated surface . . , , dT
Ein —E0ut = —qx (L) = 0 where qx (L) =—k— dX =L
k[0+b+2cx]x=L =0 (3)
c=—b/2L=—1.0><104K/m/(2><0.050m)=—1.0><105K/m2 < Since the surface temperature at x = 0 is known, T(0) = T0 = 120°C, ﬁnd
T(0)=120°C=a+b0+c0 or a=120°C (4) < Using the foregoing coefﬁcients with the expression for T(x) in the Workspace of IHT, the
temperature distribution can be determined and is plotted in the graph below. 400
6
’_ 300
9
J
E
(T:
Q
g 200 ‘
F. 100 0 1O 20 30 4O 50 Wall position, x (mm) COMMENTS: The temperature distribution has a quadratic form as expected for steadystate
conduction with constant properties and uniform volumetric energy generation (see Eq. 16.43). The
maximum temperature is at the insulated (adiabatic) boundary, x = L, where the temperature gradient must be zero. The temperature gradient at x = 0 is, of course, ﬁnite since the conduction heat ﬂux
must be equal to the convective heat ﬂux to the ambient ﬂuid. PROBLEM 16.8 KNOWN: Dimensions of oven window materials as described in Example 16.1. The outside
convection coefﬁcient is increased to 35 W/m2 K , while all other conditions remain the same. FIND: (a) Temperature of the outer surface, Ts,o , and (b) Temperature at the interface between the
two plastics (A and B). SCHEMATIC:
LA = 41.5 mm L3 = 20.7 mm
Tsi = 335°C 5'" 5 Ts o “ Ts,i TAB Ts,o Too
' v ' q ———> WW—uvw.
ﬂ 6311’} R’bd.A Rae "cw,
ifiggii/mzK kA = 0.15 W/mK k3 = 0.08 W/mK ASSUMPTIONS: (1) Steadystate conditions, (2) One dimensional conduction through windows, (3)
Negligible contact resistance between plastic materials, (4) Negligible radiation transfer through
windows, and (5) Constant properties. ANALYSIS: (a) From the thermal circuit above representing the composite window with its
boundary conditions, ﬁnd the outer surface temperature, Ts,o , from an energy balance q, _ Ts,i "Too _ Ts,o ‘ Too
R'cd,A + R'c':d,B + Rcv,o R'cvp The thermal resistances associated with conduction through the windows and convection from the
surface are Rad’A =L_A=_M=o_277 m2 .K/w kA 0.15 W/mK RgdB=£ﬁ=m=0259 m2.K/w
’ kB 0.08 W/mK
Rgv,0=1— 1 =0.0286 m2 K/W ho ’M Substituting numerical values in the energy balance expression, ﬁnd
(385—25)°C _ T“, —25°C (0.277+0.259+0.0286)m2 K/W _ 0.0286 m2 K/ w T“, =43°C <
(b) Referring to the thermal circuit, write the energy balance q" _ Ts,i "Too _ Ts,i ‘TAB
REd,A +R2d,B +R2v,o REd,A Continued ..... PROBLEM 16.8 (Cont) and substituting numerical values, ﬁnd TAB,
(38525)°C _ 385°C—TAB
0.5646 m2 K/W 0.277 m2 K/W
TAB = 208°C COMNIENTS: Note that Rgv’o is nearly 6 times smaller than the thermal resistances of the windows. As such, changes in the convection coefﬁcient will have a slight effect on the window
surface temperature, as well as on the heat rate through the window. PROBLEM 16.14 KNOWN: Drying oven wall having material with known thermal conductivity sandwiched between thin
metal sheets. Radiation and convection conditions prescribed on inner surface; convection conditions on
outer surface. FIND: (a) Thermal circuit representing wall and processes and (b) Insulation thickness required to
maintain outer wall surface at T0 = 40°C. SCHEMATIC: Insulation, k = 0.05 W/mK
To = 40°C Too i Ti To Too o Qi Raﬁ Rad vao = 25°C ..
h0 = 10 W/m2K qrad = 100 Wm2 M Ti Tm = 300°C
hi=30W/m2K \ L IE ,I ASSUNIPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction in wall, (3) Thermal
resistance of metal sheets negligible. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal
resistances and the relevant heat ﬂuxes. (b) Perform energy balances on the i and 0 nodes finding
Too,i Ti + T0 —Ti . ,, +(1" d = 0 (1)
Rcv,i cd ra
Ti —T0 +Too,oT0 =0 (2)
cd R'c’:v,o where the thermal resistances are 11ng =1/hi = 0.0333 m2 .K/w (3)
R301 =L/k=L/0.05 mZK/W (4)
Ram =1/h0 =0.0100 m2 K/W (5) Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, ﬁnd
L = 86 m < COMNIENTS: (l) The temperature at the inner surface can be found from an energy balance on the
inode using the value found for L. T ——T T —T
«391 l_+_____w’o 1 = 0 = II n n .
cv,o Rcd+ cv,1 It follows that Ti is close to Tm since the wall represents the dominant resistance of the system. (2) Verify that qi' = 50 W / m2 and q"O = 150 W / m2 . Is the overall energy balance on the system
satisﬁed? PROBLEM 16.19 KNOWN: Surface area and maximum temperature of a chip. Thickness of aluminum cover and
chip/cover contact resistance. Fluid convection conditions. FIND: Maximum chip power. SCHEMATIC:
QED = 1000 mm2
Tco = 25°C A“
h=1000Wlm2K ———p L=2mm
Al cover
Chip {c = 0.5x10'4 m2KNV pm TO M = 85cc ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional heat transfer, (3) Negligible heat
loss from sides and bottom, (4) Chip is isothermal. PROPERTIES: Table HT 1 , Aluminum (T z 325 K): k = 238 W/mK.
ANALYSIS: For a control surface about the chip, conservation of energy yields Pelec = (lo where the heat transfer rate from the chip, qc, can be expressed in terms of the conduction, contact
and thermal resistances, P _ (Tc — Too ) As
616° _ [(L/k) + Rgc +(1/h)] Substituting numerical values with T0 = Tmax, the maximum chip power is (85—25)°c(10‘4m2)
Pelec=——‘—“—T_——2_——
[(0.002/238)+0.5x10 +(1/1000)]m K/W
60x10‘4 OCm2
Pelee: "" _6 _" ———_3 2 "’—
(8.4x10 +0.5x10 +10 )m K/W Palec = 5.7 W. < COMMENTS: The dominant resistance is that due to convection (Rconv > Rt,c >> Roond ...
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 Spring '07
 Bruce

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