SteadyCond_1 - ST EADY-STATE CONDUCTION— ONE DIMENSION I...

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Unformatted text preview: ST EADY-STATE CONDUCTION— ONE DIMENSION I 2-1 INTRODUCTION We now wish to examine the applications of Fourier‘s law of heat conduction to calculation of heat flow in some simple one-dimensional systems. Several different physical shapes may fall in the category of one-dimensional systems: cylindrical and spherical systems are one-dimensional when the temperature in the body is a function only of radial distance and is independent of azimuth angle or axial distance. In some two-dimensional problems the effect of a second-space coordinate may be so small as to justify its neglect, and the multidimensional heat-flow problem may be approximated with a one-dimen- 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 2-2 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) (2-1) 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] (2-2) If more than one material is present, as in the multilayer wall shown in Fig. 2-1, 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 heat-transfer 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 electric-circuit theory. In Eq. (2-1) the thermal resistance is Ax/kA, and in Eq. (2-3) it is the sum of the three terms in the denominator. We should expect this situation in Eq. (2-3) 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 (2-5) 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. 2-1 One-dimensional heat transfer through a composite wall and electrical analog. (bl Fig. 2~2 Series and paraltel one-dimensional 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 two-dimensional 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. 2-3 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 thermal-resistance 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 2-1 furnishes such infor- mation and may be used as a guide for the selection of insulating materials. R = (2—6) Table 2-1 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 29-108 I 10—150 Tanks and pipes 5 Fiber—glass blanket —80—290 22—78 10—50 Pipe and pipe fittings for wrapping 6 Fiber-glass blankets —170—230 25—86 10~50 Tanks and equipment 7 Fiber-glass preformed —50—230 3245 10-50 Piping shapes 8 Elastomeric sheets —40—100 36—39 70—100 Tanks 9 Fiber-glass 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 Fiber-glass 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 2-9 CONDUCTION-CONVECTION SYSTEMS convection. 1n heat-exchanger applications a finned-tube 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 conduction-convection systems is very important from a practical standpoint. surrounding fluid at a temperature 1'"m as shown in Fig. 2-9. 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. 2-9 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. (2-300) 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. (2-30b) 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 2-32 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 (2-33b) 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 ' (2-35) 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. 2-10 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.

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SteadyCond_1 - ST EADY-STATE CONDUCTION— ONE DIMENSION I...

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