John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 39
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is pictured in Fig. 6.5. Now, with reference to this picture, we equate the heat conducted away from the wallby the fluid to the same heat transfer expressed in terms of a convectiveheat transfer coefficient:∂T = h(Tw − T∞ )(6.5)−kf∂y y=0conductioninto the fluidwhere kf is the conductivity of the fluid.
Notice two things about thisresult. In the first place, it is correct to express heat removal at the wallusing Fourier’s law of conduction, because there is no fluid motion in thedirection of q. The other point is that while eqn. (6.5) looks like a b.c. ofthe third kind, it is not. This condition defines h within the fluid insteadof specifying it as known information on the boundary. Equation (6.5)can be arranged in the formTw − T ∂hLTw − T∞ == NuL , the Nusselt number(6.5a)∂(y/L)kfy/L=0§6.1Some introductory ideas275Figure 6.5 The thermal boundary layerduring the flow of cool fluid over a warmplate.where L is a characteristic dimension of the body under consideration—the length of a plate, the diameter of a cylinder, or [if we write eqn. (6.5)at a point of interest along a flat surface] Nux ≡ hx/kf .
From Fig. 6.5 wesee immediately that the physical significance of Nu is given byNuL =Lδt(6.6)In other words, the Nusselt number is inversely proportional to the thickness of the thermal b.l.The Nusselt number is named after Wilhelm Nusselt,3 whose work onconvective heat transfer was as fundamental as Prandtl’s was in analyzingthe related fluid dynamics (see Fig. 6.6).We now turn to the detailed evaluation of h. And, as the precedingremarks make very clear, this evaluation will have to start with a development of the flow field in the boundary layer.3Nusselt finished his doctorate in mechanical engineering at the Technical University in Munich in 1907. During an indefinite teaching appointment at Dresden (1913 to1917) he made two of his most important contributions: He did the dimensional analysis of heat convection before he had access to Buckingham and Rayleigh’s work. In sodoing, he showed how to generalize limited data, and he set the pattern of subsequentanalysis.
He also showed how to predict convective heat transfer during film condensation. After moving about Germany and Switzerland from 1907 until 1925, he wasnamed to the important Chair of Theoretical Mechanics at Munich. During his earlyyears in this post, he made seminal contributions to heat exchanger design methodology. He held this position until 1952, during which time his, and Germany’s, greatinfluence in heat transfer and fluid mechanics waned. He was succeeded in the chairby another of Germany’s heat transfer luminaries, Ernst Schmidt.276Laminar and turbulent boundary layers§6.2Figure 6.6 Ernst Kraft Wilhelm Nusselt(1882–1957).
This photograph, providedby his student, G. Lück, shows Nusselt atthe Kesselberg waterfall in 1912. He wasan avid mountain climber.6.2Laminar incompressible boundary layer on a flatsurfaceWe predict the boundary layer flow field by solving the equations thatexpress conservation of mass and momentum in the b.l. Thus, the firstorder of business is to develop these equations.Conservation of mass—The continuity equationA two- or three-dimensional velocity field can be expressed in vectorialform: + jv + kw = iuuwhere u, v, and w are the x, y, and z components of velocity. Figure 6.7shows a two-dimensional velocity flow field. If the flow is steady, thepaths of individual particles appear as steady streamlines.
The streamlines can be expressed in terms of a stream function, ψ(x, y) = constant, where each value of the constant identifies a separate streamline,as shown in the figure. is directed along the streamlines so that no flow canThe velocity, u,cross them. Any pair of adjacent streamlines thus resembles a heat flow§6.2Laminar incompressible boundary layer on a flat surfaceFigure 6.7 A steady, incompressible, two-dimensional flowfield represented by streamlines, or lines of constant ψ.channel in a flux plot (Section 5.7); such channels are adiabatic—no heatflow can cross them.
Therefore, we write the equation for the conservation of mass by summing the inflow and outflow of mass on two faces ofa triangular element of unit depth, as shown in Fig. 6.7:ρv dx − ρu dy = 0(6.7)If the fluid is incompressible, so that ρ = constant along each streamline,then−v dx + u dy = 0(6.8)But we can also differentiate the stream function along any streamline,ψ(x, y) = constant, in Fig.
6.7:∂ψ ∂ψ dy = 0dx +(6.9)dψ =∂x y∂y xIf we compare eqns. (6.8) and (6.9), we immediately see that the coefficients of dx and dy must be the same, so∂ψ∂ψ and u =(6.10)v=−∂x y∂y x277278Laminar and turbulent boundary layers§6.2Furthermore,∂2ψ∂2ψ=∂y∂x∂x∂yso it follows that∂v∂u+=0∂x∂y(6.11)This is called the two-dimensional continuity equation for incompressible flow, because it expresses mathematically the fact that the flow iscontinuous; it has no breaks in it. In three dimensions, the continuityequation for an incompressible fluid is=∇·u∂v∂w∂u++=0∂x∂y∂zExample 6.1Fluid moves with a uniform velocity, u∞ , in the x-direction.
Find thestream function and see if it gives plausible behavior (see Fig. 6.8).Solution. u = u∞ and v = 0. Therefore, from eqns. (6.10)∂ψ ∂ψ u∞ =and 0 =∂y x∂x yIntegrating these equations, we getψ = u∞ y + fn(x) and ψ = 0 + fn(y)Comparing these equations, we get fn(x) = constant and fn(y) =u∞ y+ constant, soψ = u∞ y + constantThis gives a series of equally spaced, horizontal streamlines, as we wouldexpect (see Fig. 6.8). We set the arbitrary constant equal to zero in thefigure.§6.2Laminar incompressible boundary layer on a flat surfaceFigure 6.8 Streamlines in a uniformhorizontal flow field, ψ = u∞ y.Conservation of momentumThe momentum equation in a viscous flow is a complicated vectorial expression called the Navier-Stokes equation.
Its derivation is carried outin any advanced fluid mechanics text (see, e.g., [6.3, Chap. III]). We shalloffer a very restrictive derivation of the equation—one that applies onlyto a two-dimensional incompressible b.l. flow, as shown in Fig. 6.9.Here we see that shear stresses act upon any element such as to continuously distort and rotate it. In the lower part of the figure, one suchelement is enlarged, so we can see the horizontal shear stresses4 andthe pressure forces that act upon it.
They are shown as heavy arrows.We also display, as lighter arrows, the momentum fluxes entering andleaving the element.Notice that both x- and y-directed momentum enters and leaves theelement. To understand this, one can envision a boxcar moving downthe railroad track with a man standing, facing its open door. A childstanding at a crossing throws him a baseball as the car passes.
Whenhe catches the ball, its momentum will push him back, but a componentof momentum will also jar him toward the rear of the train, becauseof the relative motion. Particles of fluid entering element A will likewiseinfluence its motion, with their x components of momentum carried intothe element by both components of flow.The velocities must adjust themselves to satisfy the principle of conservation of linear momentum. Thus, we require that the sum of theexternal forces in the x-direction, which act on the control volume, A,must be balanced by the rate at which the control volume, A, forces x4The stress, τ, is often given two subscripts. The first one identifies the directionnormal to the plane on which it acts, and the second one identifies the line along whichit acts.
Thus, if both subscripts are the same, the stress must act normal to a surface—itmust be a pressure or tension instead of a shear stress.279280Laminar and turbulent boundary layers§6.2Figure 6.9 Forces acting in a two-dimensional incompressibleboundary layer.directed momentum out. The external forces, shown in Fig. 6.9, are∂τyx∂pdy dx − τyx dx + p dy − p +dx dyτyx +∂y∂x∂τyx∂p=−dx dy∂y∂xThe rate at which A loses x-directed momentum to its surroundings is∂ρu2∂ρuv22ρu +dx dy − ρu dy + u(ρv) +dy dx∂x∂y∂ρu2∂ρuv− ρuv dx =+dx dy∂x∂y§6.2Laminar incompressible boundary layer on a flat surfaceWe equate these results and obtain the basic statement of conservation of x-directed momentum for the b.l.:∂τyxdp∂ρu2∂ρuvdy dx −dx dy =+dx dy∂ydx∂x∂yThe shear stress in this result can be eliminated with the help of Newton’slaw of viscous shear:τyx = µ∂u∂yso the momentum equation becomes∂udp∂ρu2∂ρuv∂µ−=+∂y∂ydx∂x∂yFinally, we remember that the analysis is limited to ρ constant, andwe limit use of the equation to temperature ranges in which µ constant.Then∂uv1 dp∂2u∂u2+=−+ν∂x∂yρ dx∂y 2(6.12)This is one form of the steady, two-dimensional, incompressible boundary layer momentum equation.
Although we have taken ρ constant, amore complete derivation reveals that the result is valid for compressible flow as well. If we multiply eqn. (6.11) by u and subtract the resultfrom the left-hand side of eqn. (6.12), we obtain a second form of themomentum equation:u∂u∂u1 dp∂2u+v=−+ν∂x∂yρ dx∂y 2(6.13)Equation (6.13) has a number of so-called boundary layer approximations built into it:• ∂u/∂x is generally ∂u/∂y .• v is generally u.• p ≠ fn(y)281Laminar and turbulent boundary layers282§6.2The Bernoulli equation for the free stream flow just above the boundary layer where there is no viscous shear,u2p+ ∞ = constantρ2can be differentiated and used to eliminate the pressure gradient,1 dpdu∞= −u∞ρ dxdxso from eqn.
(6.12):∂(uv)∂2udu∞∂u2+= u∞+ν∂x∂ydx∂y 2(6.14)And if there is no pressure gradient in the flow—if p and u∞ are constantas they would be for flow past a flat plate—then eqns. (6.12), (6.13), and(6.14) become∂(uv)∂u∂u∂2u∂u2+=u+v=ν∂x∂y∂x∂y∂y 2(6.15)Predicting the velocity profile in the laminar boundary layerwithout a pressure gradientExact solution. Two strategies for solving eqn. (6.15) for the velocityprofile have long been widely used. The first was developed by Prandtl’sstudent, H. Blasius,5 before World War I.
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