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Unformatted text preview: ST EADYSTATE
CONDUCTION—
ONE DIMENSION I 21 INTRODUCTION We now wish to examine the applications of Fourier‘s law of heat conduction
to calculation of heat ﬂow in some simple onedimensional systems. Several
different physical shapes may fall in the category of onedimensional systems:
cylindrical and spherical systems are onedimensional when the temperature
in the body is a function only of radial distance and is independent of azimuth
angle or axial distance. In some twodimensional problems the effect of a
secondspace coordinate may be so small as to justify its neglect, and the
multidimensional heatﬂow problem may be approximated with a onedimen
sional analysis. in these cases the differential equations are simpliﬁed, and we
are led to a much easier solution as a result of this simpliﬁcation. I 22 THE PLANE WALL First consider the plane wall where a direct application of Fourier's law [Eq.
(171)] may be made. Integration yields 0 I — —(T~ * Ti) (21) when the thermal conductivity is considered constant. The wall thickness is
Ar, and T. and T1 are the wall—face temperatures. If the thermal conductivity
varies with temperature according to some linear relatioa k : k0(] + ,BT), the
resultant equation for the heat ﬂow is
El” “ kﬁé [(Tz _TI) +
x E
2 ll (T22 — TH] (22) If more than one material is present, as in the multilayer wall shown in Fig.
21, the analysis would proceed as follows: The temperature gradients in the
three materials are shown, and the heat ﬂow may be written ‘— T2_T] T3_T2 T4_T3
= —kA——= —kA———: —kA——
q A AxA B AXE C AIC
Note that the heat ﬂow must be the same through all sections.
Solving these three equations simultaneously, the heat ﬂow is written
,_ T — T
q = ——*'_4____ (23)
A.YA/kAA ‘i' AXE/kiwi ‘i‘ Axc/kcA
At this point we retrace our development slightly to introduce a different
conceptual vieWpoint for Fourier‘s law. The heattransfer rate may be consid
ered as a flow, and the combination of thermal conductivity, thickness of
material, and area as a resistance to this ﬂow. The temperature is the potential,
or driving, function for the heat ﬂow, and the Fourier equation may be written
thermal ote tial d'fference
Heat flow = —p—L_‘—_ (2—4)
thermal resistance
a relation quite like Ohm’s law in electriccircuit theory. In Eq. (21) the thermal
resistance is Ax/kA, and in Eq. (23) it is the sum of the three terms in the
denominator. We should expect this situation in Eq. (23) because the three
walls side by side act as three thermal resistances in series. The equivalent
electric circuit is shown in Fig. 2—1.1). The electrical analogy may be used to solve more complex problems in — A, overall
‘1 — t (25)
ER.h
Temperature ————__._ '2?"
F]
q q pro Ie RA RE RC
I ;
Ti MA I 2 ma 73 mg. 74
AAA REA RCA
(a) (b) Fig. 21 Onedimensional heat transfer through a composite wall and electrical analog. (bl Fig. 2~2 Series and paraltel onedimensional heat transfer through a com
posite wall and electrical analog. 'where the Rim are the thermal resistances of the various materials. The units
for the thermal resistance are “(I/W or “F  hlBtu. It is well to mention that in some systems like that in Fig. 2«2 twodimensional
heat ﬂow may result if the thermal conductivities of materials B, C. and D
differ by an appreciable amount. In these cases other techniques must be
employed to effect a solution. 23 INSULATION AND H VALUES In Chap. 1 we noted that the thermal conductivities for a number of insulating
materials are given in Appendix A. In classifying the performance of insulation,
it is a common practice in the building industry to use a term called the R
value, which is deﬁned as g
'EJ/A The units for R are °C  mZ/W or “F  ft2  h/Btu. Note that this differs from
the thermalresistance concept discussed above in that a heat flow per unit
area is used. At this point it is worthwhile to classify insulation materials in terms of their
application and allowable temperature ranges. Table 21 furnishes such infor
mation and may be used as a guide for the selection of insulating materials. R = (2—6) Table 21 Insulation Types and Applications
Thermal
Temperature conductivity, Density,
Type range, c’C mW/m  °C kg/m3 Application
1 Lindc evacuated ~240—1100 (LOUIS—0.72 Variable Many
superinsulation
2 Urethane foam 7 1807150 16—20 25—48 Hot and cold pipes
3 Urethane foam — 1707110 1640 32 Tanks
4 Cellular glass blocks — 200—200 29108 I 10—150 Tanks and pipes
5 Fiber—glass blanket —80—290 22—78 10—50 Pipe and pipe ﬁttings
for wrapping
6 Fiberglass blankets —170—230 25—86 10~50 Tanks and
equipment
7 Fiberglass preformed —50—230 3245 1050 Piping
shapes
8 Elastomeric sheets —40—100 36—39 70—100 Tanks
9 Fiberglass mats 60—370 30—55 l0~50 Pipe and pipe ﬁttings
10 Elastorneric 740—100 36—39 70—l()0 Pipe and ﬁttings
preformed shapes
11 Fiber glass with vapor —5—70 29—45 10—32 Refrigeration lines
barrier blanket
12 Fiber glass without to 250 29—45 24—48 Hot piping
vapor barrier jacket
13 Fiberglass boards 20—450 33—372 25—100 Boilers, tanks, heat
exchangers
14 Cellular glass blocks 20—500 29—108 1107150 Hot piping
and boards
15 Urethane foam blocks 100—150 16—20 24—65 Piping
and boards
16 MineraI ﬁber to 650 35—91 125460 Hot piping
preformed shapes
17 Mineral ﬁber blankets to 750 37781 125 Hot piping
18 Mineral W001 blocks 450—1000 52—130 175—290 Hot piping
19 Calcium silicate 230—1000 32—85 100—160 Hot piping, boilers,
blocks, boards chimney linings
20 to 1100 52430 210 Mineral ﬁber blocks Boilers and tanks 29 CONDUCTIONCONVECTION SYSTEMS convection. 1n heatexchanger applications a ﬁnnedtube arrangement might
be used to remove heat from a hot liquid. The heat transfer from the liquid to
the ﬁnned tube is by convection. The heat is conducted through the material
and ﬁnally dissipated to the surroundings by convection. Obviously, an analysis of combined conductionconvection systems is very important from a practical
standpoint. surrounding ﬂuid at a temperature 1'"m as shown in Fig. 29. The temperature
of the base of the ﬁn is Tu. We approach the problem by making an energy
balance on an element of the ﬁn of thickness dx as shown in the ﬁgure. Thus Energy in left face 2 energy out right face + energy lost by convection The deﬁning equation for the convection heat—transfer coefﬁcient is recalled as '5 = MW... i =..) (2—29) where the area in this equation is the surface area for convection. Let the cross sectional area of the ﬁn be A and the perimeter be P. Then the energy quantities
are . _ dT
Energy in left face = qx = —kA d—
x Fig. 29 Sketch illustrating one
dimensional conduction and con
vection through a rectangular fin. . dT
Energy out right face 2 Erick = —kA .\'+d.r
:11“ (FT
— —kA + dxz (it) Energy lost by convection = In” dx(T k L) Here it is noted that the differential surface area for convection is the product
of the perimeter of the ﬁn and the differential length dx. When we combine the
quantities, the energy balance yields (PT hP
_ _ a w : 0 _
dxz M (T T) (2 30a)
Let 9 2 T — L. Then Eq. (2300) becomes
0'26 [1P
7 _ 9 = 2—
(1x2 104 0 ( 30b) One boundary condition is
6:190:1“0—1"m atx:0 The other boundary condition depends on the physical situation. Several cases
may be considered: CASE1 The ﬁn is very long, and the temperature at the end of the fin is es
sentially that of the surrounding ﬂuid.
CASE 2 The ﬁn is of ﬁnite length and loses heat by convection from its end. CASES The end of the ﬁn is insulated so that dT/dx : 0 at x = L. If we let m2 = hP/kA, the general solution for Eq. (230b) may be written 9 = Cle‘m" + Cze'“ (2—31)
For case 1 the boundary conditions are 9 = 60 at x = 0 (9 = 0 at x = 00
and the solution becomes 6 T—Tm — = —— = ‘mx 232
60 To _* TD: 8 ( )
For case 3 the boundary conditions are
3 = 60 at x = 0
d6
— 0 atx : L E;_ Thus 90 = C! + C2
0 : m(—C1€—’"L + CgemL) Solving for the constants Cl and C2, We obtain 6 {1‘ m 'm'" emx
_ : —_ + u
60 1 + e—EmL l + esz cosh [m(L — x)]
cosh mi. N (233b) The hyperbolic functions are deﬁned as . e“ — e“ e“ + a?"r
smh x = ——,)— cosh x : je— sinh x 6‘" — e' " cosh x e“ + 6"" tanh x 2 The solution for case 2 is more involved algebraically, and the result is T Tm i cosh m (L — x) + (h/mk) sinh m (L — x) (2 34)
To — T,c cosh mL + (him/t) sinh mL All of the heat lost by the ﬁn must be conducted into the base at x = 0.
Using the equations for the temperature distribution, we can compute the heat
loss from dT .— : _ A _
q k dx]x=0 An alternative method of integrating the convection heat loss could be used:
L L
z, =L hP(T a 12,) dx =L [11°de In most cases, however, the ﬁrst equation is easier to apply. For case I, 21‘: —kA(—mﬂoe‘mm)J : VthA I90 ' (235)
For case 3,
‘“‘ e —kA6 1 — 1
q — 9m 1 + eilmL l + e+2mL 2 VthA 60 tanh mL The heat ﬂow for case 2 is _ , sinh mL + (ii/mk) cosh ml.
2 V — w g—‘E _37
q 11PM a" T ) cosh m1. + (him/c) sinh mL (2 ) lie) or) lg) (I! i n) Fig. 210 Dilierent types of tinned surfaces. according to Kern and Kraus [8]: (a) Longitudinal
fin of rectangular profile; (b) cylindrical lube equipped with iins oi rectangular profile; (c) longi
tudinal tin of trapezoidal profile; (d) longitudinal fin of parabolic profile; (8} cylindrical tube equipped
with radial tin of rectangular profile; (i) cylindrical tube equipped with radial tin of truncated conical
profile; (g) cylindrical spine; (h) truncated conical spine; (i) parabolic spine. ...
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This note was uploaded on 01/23/2012 for the course AOE 3044 taught by Professor Schetz,j during the Fall '08 term at Virginia Tech.
 Fall '08
 Schetz,J

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