Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 71
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6.13).The Stanton number is given in nondimensional form byc1 (r0∗ )K (U ∗ )c21 1/2St = ∗1/2 Rex3L(U ∗ )c3 (r0∗ )2K dx3∗Lines: 1436 to 1486———(6.76)0where U ∗ = U (x)/U∞ , r0∗ = r0 /L, and x3∗ = x3 /L. Here the constants c1 , c2 , andc3 are given in Table 6.1.As an example, heat transfer from a heated sphere of radius a at a uniform temperature placed in an undisturbed flow characterized by U∞ can be calculated by thismethod. The velocity distribution in the inviscid flow isU=3x3U∞ sin φ = U∞ sin22a(6.77)The variation of Nux (U∞ a/ν)−1/2 downstream from the front stagnation point isseen in Fig.
6.12. The result of the similarity solution for the case of axisymmetricstagnation flow is also shown. As described by Cebeci and Bradshaw (1984), thissimilarity solution requires the transformation of the axisymmetric equations into anearly two-dimensional form through use of the Mangler transformation.6.4.11Heat Transfer in a Turbulent Boundary LayerThe two-dimensional boundary layer form of the time-averaged governing equations is∂ ū ∂ v̄+=0∂x∂yBOOKCOMP, Inc. — John Wiley & Sons / Page 469 / 2nd Proofs / Heat Transfer Handbook / Bejan[469], (31)(6.78)1.81322pt PgVar———Short PagePgEnds: TEX[469], (31)470∂ v̄dp̄∂ 2 ū∂ ū∂+ v̄=−+µ 2 −(ρu v )ρ ū∂x∂ydx∂y∂ydp̄∂τm∂+−(ρu v )dx∂y∂y(6.79)∂ 2 T̄∂∂∂qm(ρcp v T ) = −−(ρcp v T )−∂y 2∂y∂y∂y(6.80)=−ρcp∂ T̄∂ T̄ū+ v̄∂x∂y=kwhere τm = µ(∂ ū/∂y) and qm = −k(∂ T̄ /∂y) are the mean shear stress and heatflux, respectively.Axisymmetric Flows Equations (6.78) through (6.80) are derived for twodimensional flow, with the velocity component as well as all derivatives of timeaveraged quantities neglected in the z direction.
For axisymmetric flow, such as ina circular jet or within the boundary layer on a body of circular cross section, alsocalled a body of revolution (Fig. 6.13), Cebeci and Bradshaw (1984) show that thegoverning equations can be generalized:[470], (32)Lines: 1486 to 1512———2.48503pt PgVar2———Normal PagePgEnds: TEXPr = 101.6[470], (32)1.2( (⫺1/2SimilarityNux U⬁ av123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSPr = 10.8yritilamSi0.40020406080180 x (deg) r0100120Figure 6.12 Local heat transfer behavior for flow past a heated sphere at uniform surfacetemperature for various Pr values, using the integral method.
(From Cebeci and Bradshaw,1984.)BOOKCOMP, Inc. — John Wiley & Sons / Page 470 / 2nd Proofs / Heat Transfer Handbook / BejanHEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445471yMainstreamFluid␦u⬁rx=0roBoundaryLayer90Axis of SymmetryzFigure 6.13 Boundary layer development on a body of revolution.[471], (33)∂r N ū ∂r N v̄+=0∂x∂y ∂ ū∂ ū∂ v̄dp1 ∂Nµρ ū+ v̄=−+ Nr− ρu v∂x∂ydxr ∂y∂y∂ T̄∂ T̄1 ∂∂ T̄ρcp ū+ v̄= Nk− ρcp v T ∂x∂yr ∂y∂y(6.81)Lines: 1512 to 1539(6.82)(6.83)where N = 1 in axisymmetric flow and N = 0 in two-dimensional flow. Theturbulent components of the shear stress and the heat flux are defined asτT = −ρu v = ρ∂ ū∂yandqT = −ρcp v T = −ρcp H∂ T̄∂ywhere and H are, respectively, the turbulent or eddy diffusivities of momentum andheat. The governing two-dimensional equations with these incorporated become∂ ū∂ v̄∂1 d p̄∂ ūū+ v̄+= −(ν + )(6.84)∂xρ dx∂y∂y∂yand∂ T̄∂ T̄∂∂1 1 ∂ T̄∂ T̄ū+ v̄=(α + H )=ν+∂x∂y∂y∂y∂y Pr ν PrT ∂y(6.85)where PrT = /H is the turbulent Prandtl number.
Solution of (6.84) and (6.85)requires modeling of the turbulent shear stress and heat flux.Analogy Solutions The total shear stress and heat flux are written as a combination of the mean and turbulent components:BOOKCOMP, Inc. — John Wiley & Sons / Page 471 / 2nd Proofs / Heat Transfer Handbook / Bejan———1.56715pt PgVar———Normal Page* PgEnds: Eject[471], (33)472123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSτ = τm + τT = ρ(ν + )∂ ū∂yq = qm + qT = −ρcp (α + H )(6.86)∂ T̄∂y(6.87)These can be integrated to yieldyū =0yT0 − T̄ =0τdyρ(ν + )(6.88)q dyρcp (α + H )(6.89)[472], (34)6.4.12Algebraic Turbulence ModelsThe simplest class of turbulence closure models is the zero-equation model or firstorder closure model.
Prandtl’s mixing length model is an example of this class wherethe eddy diffusivities and H are modeled in terms of the gradients of the mean flow: 2 ∂ ū 2 ∂ ū = and H = C (6.90)∂y∂yLines: 1539 to 1601———7.28726pt PgVar———Normal PagePgEnds: TEXwhere is the mixing length. Here C is found from PrT :C=H1=PrT[472], (34)where PrT is the turbulent Prandtl number.6.4.13Near-Wall Region in Turbulent FlowThe mean momentum equation can be written near the wall asρū∂ ū∂ ū ∂τdp̄+ ρv̄−+=0∂x∂y∂ydx(6.91)In a region close to the wall, the approximationsρū∂ ū≈0∂xū = ū(y)(6.92b)v̄ = v0(6.92c)(6.92a)and the normal velocity at the wall,BOOKCOMP, Inc.
— John Wiley & Sons / Page 472 / 2nd Proofs / Heat Transfer Handbook / BejanHEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445473can be made. With these, the momentum equation becomesρv0dp̄∂ ū ∂τ−+=0∂y∂ydxwhich can be integrated using τ = τ0 and ū = 0 at y = 0:τρv0 ūdp̄ y=1++τ0τ0dx τ0(6.93)Near the wall, a friction velocity v ∗ can be defined asτ0U2= Cf= (v ∗ )2ρ2[473], (35)which yieldsv∗ =τ0ρ1/2(6.94)Moreover, normalized variables can be defined near the wall:ūū/U=v∗Cf /2u+ =yv ∗νv0v0+ = ∗vµ(dp̄/dx)p+ =3/2ρ1/2 τ0v+ =———Normal Page(6.95a) * PgEnds: Eject(6.95c)(6.95d)and substitution of eqs. (6.95) into eq. (6.93) yields(6.96)In the absence of pressure gradient and transpiration (p+ = 0, v0+ = 0),τ=1τ0τ0 = ρ(ν + )ord ūdy(6.97)In a region very close to the wall (called the viscous sublayer), ν and eq. (6.97)can be integrated from the wall to a nearby location (say, 0 ≤ y + ≤ 5) in the flow0ūd ū =τ0µy0BOOKCOMP, Inc.
— John Wiley & Sons / Page 473 / 2nd Proofs / Heat Transfer Handbook / Bejandyoru+ = y +———0.94249pt PgVar(6.95b)τ= 1 + v + u+ + p + y +τ0Lines: 1601 to 1659(6.98)[473], (35)474123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSIn a region farther out from the wall, turbulence effects become important. However,the near-wall region still has effectively constant total shear stress and total heat flux:τ = τ0 + ρ(ν + )∂ ū ∂ ū∂ ū ∂ ū∂ ū= ρ ν + l2= ρ ν + κ2 y 2∂y∂y ∂y∂y ∂y(6.99)where the mixing length near the wall is taken as = κy, with κ = 0.40 being thevon Kármán constant. In the fully turbulent outer region, ν , resulting inτ0 = ρκ y2 2∂ ū∂y2or∂ ū=κy∂yτ0ρ1/2= v∗(6.100)This can be expressed as∂u+κy + +∂y[474], (36)=1Lines: 1659 to 1704which can be integrated to yield the law of the wall:1u+ = ln y + + Cκ———(6.101)With κ = 0.4 and C = 5.5, this describes the data reasonably well for y + > 30,as seen in Fig.
6.14. Minor variations in the κ and C values are found in the literature,as indicated in Fig. 6.14. In the range 5 < y + < 30, the buffer region, both νFigure 6.14 Velocity profiles in turbulent flow past a flat surface in the wall coordinates.(Data from the literature, as reported by Kays and Crawford, 1993.)BOOKCOMP, Inc. — John Wiley & Sons / Page 474 / 2nd Proofs / Heat Transfer Handbook / Bejan0.68515pt PgVar———Long PagePgEnds: TEX[474], (36)HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445475and are important and the experimental data from various investigators can beapproximated byu+ = 5 ln y + − 3.056.4.14Analogy Solutions for Boundary Layer FlowConsider the apparent shear stress and heat flux in turbulent flat plate boundary layerflow:τ = ρ(ν + )∂ ū∂yand∂ ūτ=∂yρ(ν + )∂ T̄∂yandq = −ρcp (α + H )(6.102)∂ T̄q =−∂yρcp (α + H )(6.103)These expressions can be integrated from the wall to a location in the free stream,u =0T0 − T =0τdyρ(ν + )(6.104)[475], (37)Lines: 1704 to 1765———2.31464pt PgVar———Long Page(6.105)* PgEnds: Ejectq dyρcp (α + H )Under the assumptions thatPr = PrT = 1[475], (37)q τ=q0τ0andthenT 0 − T =0q q dy = 0τ0ρcp (ν + )0τdyρcp (ν + )(6.106)Upon division by the velocity expressionq T0 − T= 0ucp τ0thenq0 xτ0= 2T0 − T kρuρu1 xµµcpkorFor moderate RexCf =BOOKCOMP, Inc.
— John Wiley & Sons / Page 475 / 2nd Proofs / Heat Transfer Handbook / Bejan0.058Re0.2xNux =CfRex · Pr2(6.107)476123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSand for Pr = PrT = 1 and no pressure gradient,Nux = 0.0296Re0.80x(6.108)Mixed Boundary Conditions Often, the boundary layer may start as laminar,undergo transition, and become turbulent along the plate length.
The average heattransfer coefficient in such a case can be determined by assuming an abrupt transitionat a location xT : xT L1h̄ =hL dx +hT dx(6.109)L 0xTwhere hL and hT are the local heat transfer coefficient variations for the laminarand turbulent boundary layers, respectively. A critical Reynolds number ReT can beselected and eq. (6.109) can be normalized to obtain the average Nusselt number0.8NuL = 0.664Re0.50+ 0.036 Re0.8TL − ReT(6.110)[476], (38)Lines: 1765 to 1821———7.43932pt PgVar———Three-Layer Model for a “Physical Situation” Considering the total shearNormal Pagestress and heat flux in eqs.
(6.102) and (6.103), division and introduction of theturbulent scales leads to* PgEnds: Ejectτ ∂u+= 1+τ0ν ∂y +q /ν ∂T +1+=q0Pr PrT ∂y +(6.111a)(6.111b)The turbulent boundary layer may be considered to be divided into an inner region,a buffer region, and an outer region. For the inner region (0 ≤ y + < 5), = 0 andq = q0 , which results in1=1 ∂T +Pr ∂y +orT + = Pr · y +(6.112)At the outer edge of the inner layer, Ts+ = 5Pr.For the buffer region (5 ≤ y + < 30),τ = τ0q = q0u+ = 5 + 5 lny+5(6.113a)or∂u+5= +∂y +yBOOKCOMP, Inc. — John Wiley & Sons / Page 476 / 2nd Proofs / Heat Transfer Handbook / Bejan(6.113b)[476], (38)HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445477Substituting eq.
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