John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 42
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Wenow want to find how to evaluate q when ν does not equal α.6.4The Prandtl number and the boundary layerthicknessesDimensional analysisWe must now look more closely at the implications of the similarity between the velocity and thermal boundary layers. We first ask what dimensional analysis reveals about heat transfer in the laminar b.l. We knowby now that the dimensional functional equation for the heat transfercoefficient, h, should beh = fn(k, x, ρ, cp , µ, u∞ )The Prandtl number and the boundary layer thicknesses§6.4We have excluded Tw − T∞ on the basis of Newton’s original hypothesis,borne out in eqn. (6.43), that h ≠ fn(∆T ) during forced convection. Thisgives seven variables in J/K, m, kg, and s, or 7 − 4 = 3 pi-groups.
Notethat, as we indicated at the end of Section 4.3, there is no conversionbetween heat and work so it we should not regard J as N·m, but ratheras a separate unit. The dimensionless groups are then:Π1 =hx≡ NuxkΠ2 =ρu∞ x≡ Rexµand a new group:Π3 =µcpν≡≡ Pr, Prandtl numberkαThus,Nux = fn(Rex , Pr)(6.44)in forced convection flow situations. Equation (6.43) was developed forthe case in which ν = α or Pr = 1; therefore, it is of the same form aseqn. (6.44), although it does not display the Pr dependence of Nux .To better understand the physical meaning of the Prandtl number, letus briefly consider how to predict its value in a gas.Kinetic theory of µ and kFigure 6.13 shows a small neighborhood of a point of interest in a gasin which there exists a velocity or temperature gradient.
We identify themean free path of molecules between collisions as and indicate planesat y ± /2 which bracket the average travel of those molecules found atplane y. (Actually, these planes should be located closer to y ± for avariety of subtle reasons. This and other fine points of these argumentsare explained in detail in [6.4].)The shear stress, τyx , can be expressed as the change of momentumof all molecules that pass through the y-plane of interest, per unit area: change in fluidmass flux of molecules·τyx =velocityfrom y − /2 to y + /2The mass flux from top to bottom is proportional to ρC, where C, themean molecular speed of the stationary fluid, is u or v in incompressible flow. Thus, Ndudu(6.45)and this also equals µτyx = C1 ρC2dy mdy297Laminar and turbulent boundary layers298§6.4Figure 6.13 Momentum and energy transfer in a gas with avelocity or temperature gradient.By the same token,qy = C2 ρcv CdTdyand this also equals − kdTdywhere cv is the specific heat at constant volume.
The constants, C1 andC2 , are on the order of unity. It follows immediately that soν = C1 Cµ = C1 ρCandk = C2 ρcv Csoα = C2Cγwhere γ ≡ cp /cv is approximately a constant on the order of unity for agiven gas. Thus, for a gas,Pr ≡ν= a constant on the order of unityαMore detailed use of the kinetic theory of gases reveals more specificinformation as to the value of the Prandtl number, and these points areborne out reasonably well experimentally, as you can determine fromAppendix A:2• For simple monatomic gases, Pr = 3 .§6.4The Prandtl number and the boundary layer thicknesses• For diatomic gases in which vibration is unexcited (such as N2 and5O2 at room temperature), Pr = 7 .• As the complexity of gas molecules increases, Pr approaches anupper value of unity.• Pr is most insensitive to temperature in gases made up of the simplest molecules because their structure is least responsive to temperature changes.In a liquid, the physical mechanisms of molecular momentum andenergy transport are much more complicated and Pr can be far fromunity.
For example (cf. Table A.3):• For liquids composed of fairly simple molecules, excluding metals,Pr is of the order of magnitude of 1 to 10.• For liquid metals, Pr is of the order of magnitude of 10−2 or less.• If the molecular structure of a liquid is very complex, Pr might reachvalues on the order of 105 . This is true of oils made of long-chainhydrocarbons, for example.Thus, while Pr can vary over almost eight orders of magnitude incommon fluids, it is still the result of analogous mechanisms of heat andmomentum transfer. The numerical values of Pr, as well as the analogyitself, have their origins in the same basic process of molecular transport.Boundary layer thicknesses, δ and δt , and the Prandtl numberWe have seen that the exact solution of the b.l.
equations gives δ = δtfor Pr = 1, and it gives dimensionless velocity and temperature profilesthat are identical on a flat surface. Two other things should be easy tosee:• When Pr > 1, δ > δt , and when Pr < 1, δ < δt . This is true becausehigh viscosity leads to a thick velocity b.l., and a high thermal diffusivity should give a thick thermal b.l.• Since the exact governing equations (6.41) and (6.42) are identicalfor either b.l., except for the appearance of α in one and ν in theother, we expect thatνδt= fnonlyδα299Laminar and turbulent boundary layers300§6.5Therefore, we can combine these two observations, defining δt /δ ≡ φ,and getφ = monotonically decreasing function of Pr only(6.46)The exact solution of the thermal b.l.
equations proves this to be preciselytrue.The fact that φ is independent of x will greatly simplify the use ofthe integral method. We shall establish the correct form of eqn. (6.46) inthe following section.6.5Heat transfer coefficient for laminar,incompressible flow over a flat surfaceThe integral method for solving the energy equationIntegrating the b.l. energy equation in the same way as the momentumequation gives δt δt 2 δt∂T∂T∂ Tuvdydy +dy = α2∂x∂y∂y000And the chain rule of differentiation in the form xdy ≡ dxy − ydx,reduces this to δt δt δt δt δt∂T ∂u∂v∂uT∂vTdy −dy +dy −dy = αTT∂x∂x∂y∂y∂y00000or δt0∂uTdy +∂x δtvT 0 =T∞ v|y=δt −0− δtT0∂v∂u+∂x∂y= 0, eqn. (6.11)dy⎡ ⎤∂T ∂T ⎦ −= α⎣∂y δt ∂y 0 =0We evaluate v at y = δt , using the continuity equation in the form ofeqn. (6.23), in the preceeding expression: δt1∂∂T u(T − T∞ ) dy =−k= fn(x only)ρcp∂y 00 ∂x§6.5Heat transfer coefficient for laminar, incompressible flow over a flat surfaceorddx δt0u(T − T∞ ) dy =qwρcp(6.47)Equation (6.47) expresses the conservation of thermal energy in integrated form.
It shows that the rate thermal energy is carried away bythe b.l. flow is matched by the rate heat is transferred in at the wall.Predicting the temperature distribution in the laminar thermalboundary layerWe can continue to paraphrase the development of the velocity profile inthe laminar b.l., from the preceding section. We previously guessed thevelocity profile in such a way as to make it match what we know to betrue.
We also know certain things to be true of the temperature profile.The temperatures at the wall and at the outer edge of the b.l. are known.Furthermore, the temperature distribution should be smooth as it blendsinto T∞ for y > δt . This condition is imposed by setting dT /dy equalto zero at y = δt . A fourth condition is obtained by writing eqn. (6.40)at the wall, where u = v = 0. This gives (∂ 2 T /∂y 2 )y=0 = 0. These fourconditions take the following dimensionless form:⎫T − T∞⎪⎪= 1 at y/δt = 0⎪⎪⎪Tw − T∞⎪⎪⎪⎪⎪⎪⎪T − T∞⎪= 0 at y/δt = 1⎪⎪⎪⎬Tw − T∞(6.48)⎪d[(T − T∞ )/(Tw − T∞ )]⎪⎪= 0 at y/δt = 1⎪⎪⎪d(y/δt )⎪⎪⎪⎪⎪⎪2⎪⎪∂ [(T − T∞ )/(Tw − T∞ )]⎪⎪=0aty/δ=0⎭t2∂(y/δt )Equations (6.48) provide enough information to approximate the temperature profile with a cubic function.
2 3T − T∞yyy=a+b+c+d(6.49)Tw − T∞δtδtδtSubstituting eqn. (6.49) into eqns. (6.48), we geta=1−1=b+c+d0 = b + 2c + 3d0 = 2c301302Laminar and turbulent boundary layers§6.5which gives3a=1c=0b = −2d=12so the temperature profile is3yT − T∞1=1−+Tw − T∞2 δt2yδt3(6.50)Predicting the heat flux in the laminar boundary layerEquation (6.47) contains an as-yet-unknown quantity—the thermal b.l.thickness, δt . To calculate δt , we substitute the temperature profile,eqn.
(6.50), and the velocity profile, eqn. (6.29), in the integral form ofthe energy equation, (6.47), which we first express asu∞ (Twd− T∞ )dx1δt0uu∞ T − T∞ydTw − T∞δtT − T∞dα(Tw − T∞ )Tw − T∞=−d(y/δt )δt(6.51)y/δt =0There is no problem in completing this integration if δt < δ. However,if δt > δ, there will be a problem because the equation u/u∞ = 1, insteadof eqn. (6.29), defines the velocity beyond y = δ.
Let us proceed for themoment in the hope that the requirement that δt δ will be satisfied.Introducing φ ≡ δt /δ in eqn. (6.51) and calling y/δt ≡ η, we get⎡⎤1331 3 31 3d ⎢3α⎥1 − η + η dη ⎦ =ηφ − η φδt⎣δtdx2222u∞0 2(6.52)33φ− 280 φ3= 20Since φ is a constant for any Pr [recall eqn. (6.46)], we separate variables:2δtdδ2t3α/u∞dδt==33dxdx3φ−φ20280§6.5Heat transfer coefficient for laminar, incompressible flow over a flat surfaceFigure 6.14 The exact and approximate Prandtl number influence on the ratio of b.l.
thicknesses.Integrating this result with respect to x and taking δt = 0 at x = 0, weget2δt =3αxu∞:233φ−φ320280(6.53)3But δ = 4.64x/ Rex in the integral formulation [eqn. (6.31b)]. We divideby this value of δ to be consistent and obtainδt≡ φ = 0.9638δ4 Pr φ 1 − φ2 /14Rearranging this gives11δt=1/3 δ1.025 Pr1/31.025 Pr1/3 1 − (δ2t /14δ2 )(6.54)The unapproximated result above is shown in Fig.
6.14, along with theresults of Pohlhausen’s precise calculation (see Schlichting [6.3, Chap. 14]).It turns out that the exact ratio, δ/δt , is represented with great accuracy303Laminar and turbulent boundary layers304§6.5byδt= Pr−1/3δ0.6 Pr 50(6.55)So the integral method is accurate within 2.5% in the Prandtl numberrange indicated.Notice that Fig. 6.14 is terminated for Pr less than 0.6. The reason fordoing this is that the lowest Pr for pure gases is 0.67, and the next lowervalues of Pr are on the order of 10−2 for liquid metals. For Pr = 0.67,δt /δ = 1.143, which violates the assumption that δt δ, but only by asmall margin.
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