John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook, страница 10

In particular, we must know how topredict h and how to evaluate the conductive resistance of bodies morecomplicated than plane passive walls. The evaluation of h is a matterthat must be deferred to Chapter 6 and 7. For the present, h values mustbe considered to be given information in any problem.The heat conduction component of most heat exchanger problems ismore complex than the simple planar analyses done in Chapter 1. Todo such analyses, we must next derive the heat conduction equation andlearn to solve it.Consider the general temperature distribution in a three-dimensionalbody as depicted in Fig. 2.1.

For some reason (heating from one side,in this case), there is a space- and time-dependent temperature field inthe body. This field T = T (x, y, z, t) or T (r , t), defines instantaneous4950Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.1Figure 2.1 A three-dimensional, transient temperature field.isothermal surfaces, T1 , T2 , and so on.We next consider a very important vector associated with the scalar,T .

The vector that has both the magnitude and direction of the maximumincrease of temperature at each point is called the temperature gradient,∇T :∇T ≡ i∂T∂T ∂T+ j+k∂x∂y∂z(2.1)Fourier’s law“Experience”—that is, physical observation—suggests two things aboutthe heat flow that results from temperature nonuniformities in a body.The heat diffusion equation§2.151These are:∇Tq=−|q||∇T | and ∇T are exactly opposite oneThis says that qanother in directionand|q| ∝ |∇T |This says that the magnitude of the heat flux is directly proportional to the temperature gradientNotice that the heat flux is now written as a quantity that has a specifieddirection as well as a specified magnitude.

Fourier’s law summarizes thisphysical experience succinctly as = −k∇Tq(2.2)which resolves itself into three components:qx = −k∂T∂xqy = −k∂T∂yqz = −k∂T∂zThe coefficient k—the thermal conductivity—also depends on positionand temperature in the most general case:k = k[r , T (r , t)](2.3)Fortunately, most materials (though not all of them) are very nearly homogeneous. Thus we can usually write k = k(T ).

The assumption thatwe really want to make is that k is constant. Whether or not that is legitimate must be determined in each case. As is apparent from Fig. 2.2 andFig. 2.3, k almost always varies with temperature. It always rises with Tin gases at low pressures, but it may rise or fall in metals or liquids. Theproblem is that of assessing whether or not k is approximately constantin the range of interest. We could safely take k to be a constant for ironbetween 0◦ and 40◦ C (see Fig. 2.2), but we would incur error between−100◦ and 800◦ C.It is easy to prove (Problem 2.1) that if k varies linearly with T , andif heat transfer is plane and steady, then q = k∆T /L, with k evaluatedat the average temperature in the plane.

If heat transfer is not planaror if k is not simply A + BT , it can be much more difficult to specify asingle accurate effective value of k. If ∆T is not large, one can still make areasonably accurate approximation using a constant average value of k.Figure 2.2 Variation of thermal conductivity of metallic solidswith temperature52Figure 2.3 The temperature dependence of the thermal conductivity of liquids and gases that are either saturated or at 1atm pressure.5354Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.1Figure 2.4 Control volume in aheat-flow field.Now that we have revisited Fourier’s law in three dimensions, we seethat heat conduction is more complex than it appeared to be in Chapter 1.We must now write the heat conduction equation in three dimensions.We begin, as we did in Chapter 1, with the First Law statement, eqn.

(1.3):Q=dUdt(1.3)This time we apply eqn. (1.3) to a three-dimensional control volume, asshown in Fig. 2.4.1 The control volume is a finite region of a conductingbody, which we set aside for analysis. The surface is denoted as S and thevolume and the region as R; both are at rest. An element of the surface,dS, is identified and two vectors are shown on dS: one is the unit normal (with |n| = 1), and the other is the heat flux vector, q = −k∇T ,vector, nat that point on the surface.We also allow the possibility that a volumetric heat release equal toq̇(r ) W/m3 is distributed through the region.

This might be the result ofchemical or nuclear reaction, of electrical resistance heating, of externalradiation into the region or of still other causes.With reference to Fig. 2.4, we can write the heat conducted out of dS,in watts, as(−k∇T ) · (ndS)(2.4)The heat generated (or consumed) within the region R must be added tothe total heat flow into S to get the overall rate of heat addition to R:+q̇ dR(2.5)Q = − (−k∇T ) · (ndS)S1RFigure 2.4 is the three-dimensional version of the control volume shown in Fig.

1.8.The heat diffusion equation§2.155The rate of energy increase of the region R isdU=dt ρcR∂T∂tdR(2.6)where the derivative of T is in partial form because T is a function ofboth r and t.Finally, we combine Q, as given by eqn. (2.5), and dU /dt, as given byeqn. (2.6), into eqn. (1.3). After rearranging the terms, we obtainSk∇T · ndS= ∂T− q̇ dRρc∂tR(2.7)To get the left-hand side into a convenient form, we introduce Gauss’stheorem, which converts a surface integral into a volume integral. Gauss’s is any continuous function of position, thentheorem says that if AS · ndSA=R dR∇·A(2.8) with (k∇T ), eqn. (2.7) reduces toTherefore, if we identify A R∂T∇ · k∇T − ρc+ q̇∂tdR = 0(2.9)Next, since the region R is arbitrary, the integrand must vanish identically.2 We therefore get the heat diffusion equation in three dimensions:∇ · k∇T + q̇ = ρc∂T∂t(2.10)The limitations on this equation are:• Incompressible medium.

(This was implied when no expansionwork term was included.)• No convection. (The medium cannot undergo any relative motion.However, it can be a liquid or gas as long as it sits still.)Consider f (x) dx = 0. If f (x) were, say, sin x, then this could only be trueover intervals of x = 2π or multiples of it. For eqn. (2.9) to be true for any range ofintegration one might choose, the terms in parentheses must be zero everywhere.256Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.1If the variation of k with T is small, k can be factored out of eqn. (2.10)to get∇2 T +1 ∂Tq̇=kα ∂t(2.11)This is a more complete version of the heat conduction equation [recalleqn. (1.14)] and α is the thermal diffusivity which was discussed aftereqn.

(1.14). The term ∇2 T ≡ ∇ · ∇T is called the Laplacian. It arises thusin a Cartesian coordinate system:∂∂ ∂+ j+k∇ · k∇T k∇ · ∇T = k i∂x∂y∂x ∂T∂T ∂T· i+ j+k∂x∂y∂zor∇2 T =∂2T∂2T∂2T++∂x 2∂y 2∂z2(2.12)The Laplacian can also be expressed in cylindrical or spherical coordinates. The results are:• Cylindrical:∇2 T ≡1 ∂r ∂rr∂T∂r+1 ∂2T∂2T+22r ∂θ∂z2(2.13)• Spherical:∂T1∂1 ∂ 2 (r T )1∂2Tsinθ++(2.14a)∇ T ≡r ∂r 2r 2 sin θ ∂θ∂θr 2 sin2 θ ∂φ22or1 ∂≡ 2r ∂rr2 ∂T∂r1∂T1∂2T∂+ 2sin θ+r sin θ ∂θ∂θr 2 sin2 θ ∂φ2(2.14b)where the coordinates are as described in Fig. 2.5.Figure 2.5 Cylindrical and spherical coordinate schemes.5758Heat conduction, thermal resistance, and the overall heat transfer coefficient2.2§2.2Solutions of the heat diffusion equationWe are now in position to calculate the temperature distribution and/orheat flux in bodies with the help of the heat diffusion equation.

In everycase, we first calculate T (r , t). Then, if we want the heat flux as well, wedifferentiate T to get q from Fourier’s law.The heat diffusion equation is a partial differential equation (p.d.e.)and the task of solving it may seem difficult, but we can actually do alot with fairly elementary mathematical tools. For one thing, in onedimensional steady-state situations the heat diffusion equation becomesan ordinary differential equation (o.d.e.); for another, the equation is linear and therefore not too formidable, in any case. Our procedure can belaid out, step by step, with the help of the following example.Example 2.1Basic MethodA large, thin concrete slab of thickness L is “setting.” Setting is anexothermic process that releases q̇ W/m3 .

The outside surfaces arekept at the ambient temperature, so Tw = T∞ . What is the maximuminternal temperature?Solution.Step 1. Pick the coordinate scheme that best fits the problem and identify the independent variables that determine T. In the example,T will probably vary only along the thin dimension, which we willcall the x-direction. (We should want to know that the edges areinsulated and that L was much smaller than the width or height.If they are, this assumption should be quite good.) Since the interior temperature will reach its maximum value when the process becomes steady, we write T = T (x only).Step 2. Write the appropriate d.e., starting with one of the forms ofeqn.

(2.11).∂2T1 ∂T∂ 2 T q̇∂2T+++ =222∂x∂y∂zk α ∂t =0, sinceT ≠ T (y or z)= 0, sincesteadyTherefore, since T = T (x only), the equation reduces to theSolutions of the heat diffusion equation§2.2ordinary d.e.d2 Tq̇=−dx 2kStep 3. Obtain the general solution of the d.e. (This is usually theeasiest step.) We simply integrate the d.e. twice and getT =−q̇ 2x + C1 x + C22kStep 4. Write the “side conditions” on the d.e.—the initial and boundary conditions. This is always the hardest part for the beginningstudents; it is the part that most seriously tests their physicalor “practical” understanding of problems.Normally, we have to make two specifications of temperatureon each position coordinate and one on the time coordinate toget rid of the constants of integration in the general solution.(These matters are discussed at greater length in Chapter 4.)In this case there are two boundary conditions:T (x = 0) = Twand T (x = L) = TwVery Important Warning: Never, never introduce inaccessibleinformation in a boundary or initial condition.