John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 40
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It is exact, and we shall sketch itonly briefly. First we introduce the stream function, ψ, into eqn. (6.15).This reduces the number of dependent variables from two (u and v) tojust one—namely, ψ. We do this by substituting eqns. (6.10) in eqn. (6.15):∂ψ ∂ 2 ψ∂3ψ∂ψ ∂ 2 ψ−=ν∂y ∂y∂x∂x ∂y 2∂y 3(6.16)It turns out that eqn. (6.16) can be converted into an ordinary d.e.with the following change of variables:9√u∞y(6.17)ψ(x, y) ≡ u∞ νx f (η) where η ≡νx5Blasius achieved great fame for many accomplishments in fluid mechanics and thengave it up. He is quoted as saying: “I decided that I had no gift for it; all of my ideascame from Prandtl.”§6.2Laminar incompressible boundary layer on a flat surfacewhere f (η) is an as-yet-undertermined function. [This transformation israther similar to the one that we used to make an ordinary d.e.
of theheat conduction equation, between eqns. (5.44) and (5.45).] After somemanipulation of partial derivatives, this substitution gives (Problem 6.2)fd3 fd2 f+2=0dη2dη3and1v3=2u∞ ν/xdfu=u∞dη(6.18)dfη−fdη(6.19)The boundary conditions for this flow areu(y = 0) = 0u(y = ∞) = u∞v(y = 0) = 0⎫⎪df ⎪=0 ⎪or⎪⎪⎪dη η=0⎪⎪⎬df or=1 ⎪⎪⎪dη η=∞⎪⎪⎪⎪⎪or f (η = 0) = 0 ⎭(6.20)The solution of eqn. (6.18) subject to these b.c.’s must be done numerically. (See Problem 6.3.)The solution of the Blasius problem is listed in Table 6.1, and thedimensionless velocity components are plotted in Fig.
6.10. The u component increases from zero at the wall (η = 0) to 99% of u∞ at η = 4.92.Thus, the b.l. thickness is given byδ4.92 = 3νx/u∞or, as we anticipated earlier [eqn. (6.2)],4.924.92δ=3=3xRexu∞ x/νConcept of similarity. The exact solution for u(x, y) reveals a mostuseful fact—namely, that u can be expressed as a function of a singlevariable, η: 9u∞u= f (η) = f yu∞νx283284Laminar and turbulent boundary layers§6.2Table 6.1 Exact velocity profile in the boundary layer on a flatsurface with no pressure gradient3y u∞ /νxη0.000.200.400.600.801.002.003.004.004.9186.008.00f (η)3v x/νu∞(ηf − f ) 2f (η)0.000000.066410.132770.198940.264710.329790.629770.846050.955520.990000.998981.00000−0.000000.003320.013220.029810.052830.082110.304760.570670.758160.833440.857120.860390.332060.331990.331470.330080.327390.323010.266750.161360.064240.018370.002400.00001u u∞f (η)0.000000.006640.026560.059740.106110.165570.650031.396822.305763.201694.279646.27923This is called a similarity solution. To see why, we solve eqn.
(6.2) for94.92u∞=νxδ(x)3and substitute this in f (y u∞ /νx). The result isuyf =(6.21)= fnu∞δ(x)The velocity profile thus has the same shape with respect to the b.l.thickness at each x-station. We say, in other words, that the profile issimilar at each station. This is what we found to be true for conduction√into a semi-infinite region. In that case [recall eqn. (5.51)], x/ t alwayshad the same value at the outer limit of the thermally disturbed region.Boundary layer similarity makes it especially easy to use a simpleapproximate method for solving other b.l. problems.
This method, calledthe momentum integral method, is the subject of the next subsection.Example 6.2Air at 27◦ C blows over a flat surface with a sharp leading edge at11.5 m/s. Find the b.l. thickness 2 m from the leading edge. Check theb.l. assumption that u v at the trailing edge.§6.2Laminar incompressible boundary layer on a flat surfaceFigure 6.10 The dimensionless velocity components in a laminar boundary layer.Solution. The dynamic and kinematic viscosities are µ = 1.853 ×10−5 kg/m·s and ν = 1.566 × 10−5 m2 /s. ThenRex =1.5(0.5)u∞ x== 47, 893ν1.566 × 10−5The Reynolds number is low enough to permit the use of a laminarflow analysis.
Then4.92x4.92(0.5)= 0.01124 = 1.124 cmδ= 3= 3Rex47, 893(Remember that the b.l. analysis is only valid if δ/x 1. In this case,δ/x = 1.124/50 = 0.0225.) From Fig. 6.10 or Table 6.1, we observethat v/u is greatest beyond the outside edge of the b.l, at large η.Using data from Table 6.1 at η = 8, v at x = 0.5 m is0.8604= 0.8604v=3x/νu∞2(1.566)(10−5 )(1.5)(0.5)= 0.00590 m/s285Laminar and turbulent boundary layers286§6.2or, since u/u∞ → 1 at large ηv0.00590v== 0.00393=uu∞1.5Since v grows larger as x grows smaller, the condition v u is not satisfied very near the leading edge.
There, the b.l. approximations themselves break down. We say more about this breakdown after eqn. (6.34).Momentum integral method.6 A second method for solving the b.l. momentum equation is approximate and much easier to apply to a widerange of problems than is any exact method of solution. The idea is this:We are not really interested in the details of the velocity or temperatureprofiles in the b.l., beyond learning their slopes at the wall.
[These slopesgive us the shear stress at the wall, τw = µ(∂u/∂y)y=0 , and the heatflux at the wall, qw = −k(∂T /∂y)y=0 .] Therefore, we integrate the b.l.equations from the wall, y = 0, to the b.l. thickness, y = δ, to make ordinary d.e.’s of them. It turns out that while these much simpler equationsdo not reveal anything new about the temperature and velocity profiles,they do give quite accurate explicit equations for τw and qw .Let us see how this procedure works with the b.l. momentum equation. We integrate eqn.
(6.15), as follows, for the case in which there isno pressure gradient (dp/dx = 0):δδ 2δ∂u2∂(uv)∂ udy +dy = νdy2∂y0 ∂x00 ∂yAt y = δ, u can be approximated as the free stream value, u∞ , and otherquantities can also be evaluated at y = δ just as though y were infinite:⎡⎤δ∂u∂u2∂u⎢⎥dy + (uv)y=δ − (uv)y=0 = ν ⎣−⎦ ∂y y=δ∂y y=00 ∂x=u∞ v∞=00(6.22)The continuity equation (6.11) can be integrated thus:δ∂udyv∞ − vy=0 = − 0 ∂x(6.23)=06This method was developed by Pohlhausen, von Kármán, and others. See the discussion in [6.3, Chap. XII].§6.2Laminar incompressible boundary layer on a flat surfaceMultiplying this by u∞ givesδu∞ v∞ = −0∂uu∞dy∂xUsing this result in eqn.
(6.22), we obtainδ∂∂u [u(u − u∞ )] dy = −ν∂y y=00 ∂xFinally, we note that µ(∂u/∂y)y=0 is the shear stress on the wall, τw =τw (x only), so this becomes7ddx δ(x)0u(u − u∞ ) dy = −τwρ(6.24)Equation (6.24) expresses the conservation of linear momentum inintegrated form. It shows that the rate of momentum loss caused by theb.l. is balanced by the shear force on the wall. When we use it in place ofeqn. (6.15), we are said to be using an integral method. To make use ofeqn. (6.24), we first nondimensionalize it as follows: 1ν ∂(u/u∞ ) uyud=−−1 dδdxδu∞ δ ∂(y/δ) y=00 u∞ u∞=−τw (x)12 ≡ − 2 Cf (x)ρu∞(6.25)where τw /(ρu2∞ /2) is defined as the skin friction coefficient, Cf .Equation (6.25) will be satisfied precisely by the exact solution (Problem 6.4) for u/u∞ . However, the point is to use eqn. (6.25) to determineu/u∞ when we do not already have an exact solution.
To do this, werecall that the exact solution exhibits similarity. First, we guess the solution in the form of eqn. (6.21): u/u∞ = fn(y/δ). This guess is madein such a way that it will fit the following four things that are true of thevelocity profile:⎫⎪• u/u∞ = 0 at y/δ = 0⎪⎪⎪⎬• u/u∞ 1 at y/δ = 1(6.26) ⎪⎪uy⎪• d 0 at y/δ = 1 ⎪d⎭u∞δ7The interchange of integration and differentiation is consistent with Leibnitz’s rulefor differentiation of an integral (Problem 6.14).287Laminar and turbulent boundary layers288•§6.2and from eqn.
(6.15), we know that at y/δ = 0:∂u∂u∂2u u=ν+ v2∂x∂y∂yy=0 =0so=0∂ 2 (u/u∞ ) =02∂(y/δ) y/δ=0(6.27)If fn(y/δ) is written as a polynomial with four constants—a, b, c,and d—in it, 2 3yyyu=a+b +c+d(6.28)u∞δδδthe four things that are known about the profile give• 0 = a, which eliminates a immediately• 1=0+b+c+d• 0 = b + 2c + 3d• 0 = 2c, which eliminates c as wellSolving the middle two equations (above) for b and d, we obtain d = − 123and b = + 2 , so3y1u=−u∞2 δ2yδ3(6.29)This approximate velocity profile is compared with the exact Blasiusprofile in Fig. 6.11, and they prove to be equal within a maximum errorof 8%.
The only remaining problem is then that of calculating δ(x). Todo this, we substitute eqn. (6.29) in eqn. (6.25) and get, after integration(see Problem 6.5): dν393−δ=−(6.30)dx280u∞ δ 2or−39280 21 dδ2ν=−32 dxu∞§6.2Laminar incompressible boundary layer on a flat surfaceFigure 6.11 Comparison of the third-degree polynomial fitwith the exact b.l. velocity profile. (Notice that the approximateresult has been forced to u/u∞ = 1 instead of 0.99 at y = δ.)We integrate this using the b.c. δ2 = 0 at x = 0:δ2 =280 νx13 u∞(6.31a)orδ4.64=3xRex(6.31b)This b.l.
thickness is of the correct functional form, and the constant islow by only 5.6%.The skin friction coefficientThe fact that the function f (η) gives all information about flow in the b.l.must be stressed. For example, the shear stress can be obtained from it289Laminar and turbulent boundary layers290§6.2by using Newton’s law of viscous shear: ∂η∂u ∂ dfτw =µu∞ f =µ= µu∞∂y y=0∂ydη ∂y y=0y=0√2u∞ d f =µu∞ √νx dη2 η=0But from Fig. 6.10 and Table 6.1, we see that (d2 f /dη2 )η=0 = 0.33206,soµu∞ 3Rex(6.32)τw = 0.332xThe integral method that we just outlined would have given 0.323 for theconstant in eqn.
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