John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 21
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Determine the air temperature at the point where the leadhas just finished solidifying.d. Determine the height that the tower must have in order tofunction as desired. The heat transfer coefficient betweenthe air and the droplets is h = 318 W/m2 K.References[3.1] Tubular Exchanger Manufacturer’s Association. Standards ofTubular Exchanger Manufacturer’s Association. New York, 4th and6th edition, 1959 and 1978.[3.2] R. A. Bowman, A. C.
Mueller, and W. M. Nagle. Mean temperaturedifference in design. Trans. ASME, 62:283–294, 1940.[3.3] K. Gardner and J. Taborek. Mean temperature difference: A reappraisal. AIChE J., 23(6):770–786, 1977.[3.4] N. Shamsundar.A property of the log-mean temperaturedifference correction factor. Mechanical Engineering News, 19(3):14–15, 1982.[3.5] W. M. Kays and A. L.
London. Compact Heat Exchangers. McGrawHill Book Company, New York, 3rd edition, 1984.[3.6] J. Taborek. Evolution of heat exchanger design techniques. HeatTransfer Engineering, 1(1):15–29, 1979.[3.7] G. F. Hewitt, editor. Heat Exchanger Design Handbook 1998. BegellHouse, New York, 1998.[3.8] E. Fried and I.
E. Idelchik. Flow Resistance: A Design Guide forEngineers. Hemisphere Publishing Corp., New York, 1989.[3.9] R. H. Perry, D. W. Green, and J. Q. Maloney, editors. Perry’s Chemical Engineers’ Handbook. McGraw-Hill Book Company, New York,7th edition, 1997.[3.10] D. M. Considine. Energy Technology Handbook. McGraw-Hill BookCompany, New York, 1975.[3.11] A. P.
Fraas. Heat Exchanger Design. John Wiley & Sons, Inc., NewYork, 2nd edition, 1989.References[3.12] R. K. Shah and D. P. Sekulic. Heat exchangers. In W. M. Rohsenow,J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer,chapter 17. McGraw-Hill, New York, 3rd edition, 1998.[3.13] R. K. Shah and D. P. Sekulic. Fundamentals of Heat ExchangerDesign. John Wiley & Sons, Inc., Hoboken, NJ, 2003.137Part IIAnalysis of Heat Conduction1394.Analysis of heat conduction andsome steady one-dimensionalproblemsThe effects of heat are subject to constant laws which cannot be discoveredwithout the aid of mathematical analysis.
The object of the theory whichwe are about to explain is to demonstrate these laws; it reduces all physicalresearches on the propagation of heat to problems of the calculus whoseelements are given by experiment.The Analytical Theory of Heat, J. Fourier, 18224.1The well-posed problemThe heat diffusion equation was derived in Section 2.1 and some attention was given to its solution. Before we go further with heat conductionproblems, we must describe how to state such problems so they can really be solved. This is particularly important in approaching the morecomplicated problems of transient and multidimensional heat conduction that we have avoided up to now.A well-posed heat conduction problem is one in which all the relevantinformation needed to obtain a unique solution is stated.
A well-posedand hence solvable heat conduction problem will always read as follows:Find T (x, y, z, t) such that:1.∇ · (k∇T ) + q̇ = ρc∂T∂tfor 0 < t < T (where T can → ∞), and for (x, y, z) belonging to141142Analysis of heat conduction and some steady one-dimensional problems§4.1some region, R, which might extend to infinity.12. T = Ti (x, y, z)att=0This is called an initial condition, or i.c.(a) Condition 1 above is not imposed at t = 0.(b) Only one i.c. is required. However,(c) The i.c.
is not needed:i. In the steady-state case: ∇ · (k∇T ) + q̇ = 0.ii. For “periodic” heat transfer, where q̇ or the boundary conditions vary periodically with time, and where we ignorethe starting transient behavior.3. T must also satisfy two boundary conditions, or b.c.’s, for each coordinate. The b.c.’s are very often of three common types.(a) Dirichlet conditions, or b.c.’s of the first kind:T is specified on the boundary of R for t > 0. We saw suchb.c.’s in Examples 2.1, 2.2, and 2.5.(b) Neumann conditions, or b.c.’s of the second kind:The derivative of T normal to the boundary is specified on theboundary of R for t > 0. Such a condition arises when the heatflux, k(∂T /∂x), is specified on a boundary or when , with thehelp of insulation, we set ∂T /∂x equal to zero.2(c) b.c.’s of the third kind:A derivative of T in a direction normal to a boundary is proportional to the temperature on that boundary. Such a conditionmost commonly arises when convection occurs at a boundary,and it is typically expressed as∂T = h(T − T∞ )bndry−k∂x bndrywhen the body lies to the left of the boundary on the x-coordinate.
We have already used such a b.c. in Step 4 of Example2.6, and we have discussed it in Section 1.3 as well.(x, y, z) might be any coordinates describing a position r: T (x, y, z, t) = T (r , t).Although we write ∂T /∂x here, we understand that this might be ∂T /∂z, ∂T /∂r ,or any other derivative in a direction locally normal to the surface on which the b.c. isspecified.12The general solution§4.2Figure 4.1 The transient cooling of a body as it might occur,subject to boundary conditions of the first, second, and thirdkinds.This list of b.c.’s is not complete, by any means, but it includes a greatnumber of important cases.Figure 4.1 shows the transient cooling of body from a constant initialtemperature, subject to each of the three b.c.’s described above.
Noticethat the initial temperature distribution is not subject to the boundarycondition, as pointed out previously under 2(a).The eight-point procedure that was outlined in Section 2.2 for solvingthe heat diffusion equation was contrived in part to assure that a problemwill meet the preceding requirements and will be well posed.4.2The general solutionOnce the heat conduction problem has been posed properly, the first stepin solving it is to find the general solution of the heat diffusion equation.We have remarked that this is usually the easiest part of the problem.We next consider some examples of general solutions.143144Analysis of heat conduction and some steady one-dimensional problems§4.2One-dimensional steady heat conductionProblem 4.1 emphasizes the simplicity of finding the general solutions oflinear ordinary differential equations, by asking for a table of all generalsolutions of one-dimensional heat conduction problems.
We shall workout some of those results to show what is involved. We begin the heatdiffusion equation with constant k and q̇:∇2 T +1 ∂Tq̇=kα ∂t(2.11)Cartesian coordinates: Steady conduction in the y-direction. Equation(2.11) reduces as follows:∂ 2 T q̇∂2T ∂2T+++ =222∂x ∂y∂z k=0=01 ∂Tα ∂t= 0, since steadyTherefore,q̇d2 T=−2dykwhich we integrate twice to getT =−q̇ 2y + C1 y + C22kor, if q̇ = 0,T = C1 y + C2Cylindrical coordinates with a heat source: Tangential conduction.This time, we look at the heat flow that results in a ring when two pointsare held at different temperatures.
We now express eqn. (2.11) in cylindrical coordinates with the help of eqn. (2.13):∂T1 ∂T1 ∂2T∂ 2 T q̇1 ∂r+ 2++ =22r∂r∂rr∂φ∂zkα∂t =0r =constant=0= 0, since steadyTwo integrations giver 2 q̇ 2φ + C1 φ + C2(4.1)2kThis would describe, for example, the temperature distribution in thethin ring shown in Fig. 4.2. Here the b.c.’s might consist of temperaturesspecified at two angular locations, as shown.T =−The general solution§4.2145Figure 4.2 One-dimensional heat conduction in a ring.T = T(t only)If T is spatially uniform, it can still vary with time.
In such cases1 ∂Tq̇∇2 T + =kα∂t =0and ∂T /∂t becomes an ordinary derivative. Then, since α = k/ρc,q̇dT=dtρc(4.2)This result is consistent with the lumped-capacity solution described inSection 1.3. If the Biot number is low and internal resistance is unimportant, the convective removal of heat from the boundary of a body can beprorated over the volume of the body and interpreted asq̇effective = −h(Tbody − T∞ )AW/m3volume(4.3)and the heat diffusion equation for this case, eqn. (4.2), becomeshAdT=−(T − T∞ )dtρcV(4.4)The general solution in this situation was given in eqn. (1.21).
[A particular solution was also written in eqn. (1.22).]146Analysis of heat conduction and some steady one-dimensional problems§4.2Separation of variables: A general solution of multidimensionalproblemsSuppose that the physical situation permits us to throw out all but one ofthe spatial derivatives in a heat diffusion equation. Suppose, for example,that we wish to predict the transient cooling in a slab as a function ofthe location within it.
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