John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 41
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(6.32) instead of 0.332 (Problem 6.6).The local skin friction coefficient, or local skin drag coefficient, is defined asCf ≡0.664τw= 32Rexρu∞ /2(6.33)The overall skin friction coefficient, C f , is based on the average of theshear stress, τw , over the length, L, of the plate⌠L⌠L2ρu2∞ρu2∞ ⎮1⎮0.664ν3⌡dx = 1.328τ w = ⌡ τw dx =L 02L 0 u∞ x/ν2u∞ Lso1.328Cf = 3ReL(6.34)As a matter of interest, we note that Cf (x) approaches infinity at theleading edge of the flat surface. This means that to stop the fluid thatfirst touches the front of the plate—dead in its tracks—would requireinfinite shear stress right at that point. Nature, of course, will not allowsuch a thing to happen; and it turns out that the boundary layer analysisis not really valid right at the leading edge.In fact, the range x 5δ is too close to the edge to use this analysiswith accuracy because the b.l.
is relatively thick and v is no longer u.With eqn. (6.2), this converts tox > 600 ν/u∞for a boundary layer to existLaminar incompressible boundary layer on a flat surface§6.2or simply Rex 600. In Example 6.2, this condition is satisfied for allx’s greater than about 6 mm. This region is usually very small.Example 6.3Calculate the average shear stress and the overall friction coefficientfor the surface in Example 6.2 if its total length is L = 0.5 m.
Compare τ w with τw at the trailing edge. At what point on the surfacedoes τw = τ w ? Finally, estimate what fraction of the surface canlegitimately be analyzed using boundary layer theory.Solution.1.3281.328= 0.00607Cf = 3=347, 893Re0.5andτw =ρu2∞1.183(1.5)20.00607 = 0.00808 kg/m·s2Cf = 22N/m2(This is very little drag. It amounts only to about 1/50 ounce/m2 .)At x = L,3ρu2∞ /2 0.664 ReLτw (x) 13==2τw2ρu∞ /2 1.328 ReLx=Landτw (x) = τ wwhere0.6641.328√= √x0.5so the local shear stress equals the average value, wherex=18mor1x=L4Thus, the shear stress, which is initially infinite, plummets to τ w onefourth of the way from the leading edge and drops only to one-halfof τ w in the remaining 75% of the plate.The boundary layer assumptions fail whenx < 6001.566 × 10−5ν= 600= 0.0063 mu∞1.5Thus, the preceding analysis should be good over almost 99% of the0.5 m length of the surface.291292Laminar and turbulent boundary layers6.3§6.3The energy equationDerivationWe now know how fluid moves in the b.l.
Next, we must extend the heatconduction equation to allow for the motion of the fluid. This equationcan be solved for the temperature field in the b.l., and its solution can beused to calculate h, using Fourier’s law:h=Twqk∂T =−− T∞Tw − T∞ ∂y y=0(6.35)To predict T , we extend the analysis done in Section 2.1. Figure 2.4shows a volume containing a solid subjected to a temperature field. Wenow allow this volume to contain fluid with a velocity field u(x,y, z) in it,as shown in Fig. 6.12. We make the following restrictive approximations:• Pressure variations in the flow are not large enough to affect thermodynamic properties.
From thermodynamics, we know that thespecific internal energy, û, is related to the specific enthalpy asĥ = û + p/ρ, and that dĥ = cp dT + (∂ ĥ/∂p)T dp. We shall neglectthe effect of dp on enthalpy, internal energy, and density. This approximation is reasonable for most liquid flows and for gas flowsmoving at speeds less than about 1/3 the speed of sound.• Under these conditions, density changes result only from temperature changes and will also be small; and the flow will behave as if = 0 (Sect. 6.2).incompressible.
For such flows, ∇ · u• Temperature variations in the flow are not large enough to change ksignificantly. When we consider the flow field, we will also presumeµ to be unaffected by temperature change.• Potential and kinetic energy changes are negligible in comparisonto thermal energy changes. Since the kinetic energy of a fluid canchange owing to pressure gradients, this again means that pressurevariations may not be too large.• The viscous stresses do not dissipate enough energy to warm thefluid significantly.The energy equation§6.3293Figure 6.12 Control volume in aheat-flow and fluid-flow field.Just as we wrote eqn.
(2.7) in Section 2.1, we now write conservationof energy in the formddtρ û dR = −Rrate of internalenergy increasein RS·n dS(ρ ĥ) urate of internal energy andflow work out of R− dS +(−k∇T ) · nSnet heat conductionrate out of RRq̇ dR(6.36)rate of heatgeneration in R ·n dS represents the volume flow rate through anIn the third integral, uelement dS of the control surface.
The position of R is not changing intime, so we can bring the time derivative inside the first integral. If wethen we call in Gauss’s theorem [eqn. (2.8)] to make volume integrals ofthe surface integrals, eqn. (6.36) becomes R ∂(ρ û) ĥ − ∇ · k∇T − q̇ dR = 0+ ∇ · ρu∂tBecause the integrand must vanish identically (recall the footnote onpg. 55 in Chap. 2) and because k depends weakly on T , ∂(ρ û) ĥ − k∇2 T − q̇ = 0+ ∇ · ρu∂t · ∇ĥ + ĥ∇ · (ρ u)= ρu294Laminar and turbulent boundary layers§6.3Since we are neglecting pressure effects, we may introduce the followingapproximation:d(ρ û) = d(ρ ĥ) − dp ≈ d(ρ ĥ) = ρdĥ + ĥ dρThus, collecting and rearranging terms ∂ρ∂ ĥ · ∇ĥ + ĥ = k∇2 T + q̇+u+ ∇ · ρuρ∂t ∂tneglectThe term involving density derivatives may be neglected on the basis thatdensity changes are small and the flow is nearly incompressible (but seeProblem 6.36 for a more general result).Upon substituting dĥ ≈ cp dT , we obtain our final result:ρcp∂T · ∇T+ u ∂tenergystorageenthalpyconvection=k∇2 T + heatconductionq̇(6.37)heatgenerationThis is the energy equation for a constant pressure flow field.
It is thesame as the corresponding equation (2.11) for a solid body, except for · ∇T .the enthalpy transport, or convection, term, ρcp uConsider the term in parentheses in eqn. (6.37):∂T∂T∂T∂T∂TDT · ∇T =+u+u+v+w≡∂t∂t∂x∂y∂zDt(6.38)DT /Dt is exactly the so-called material derivative, which is treated insome detail in every fluid mechanics course. DT /Dt is the rate of changeof the temperature of a fluid particle as it moves in a flow field.In a steady two-dimensional flow field without heat sources, eqn. (6.37)takes the form∂T∂2T∂2T∂T+v=α+u(6.39)∂x∂y∂x 2∂y 2Furthermore, in a b.l., ∂ 2 T /∂x 2 ∂ 2 T /∂y 2 , so the b.l.
form isu∂T∂2T∂T+v=α∂x∂y∂y 2(6.40)The energy equation§6.3295Heat and momentum transfer analogyConsider a b.l. in a fluid of bulk temperature T∞ , flowing over a flat surface at temperature Tw . The momentum equation and its b.c.’s can bewritten as⎧ u ⎪⎪=0⎪⎪u∞ y=0⎪⎪⎪ ⎪⎨ u ∂∂2uuu∂=1+v=νu⎪u∞ y=∞∂x u∞∂y u∞∂y 2 u∞⎪⎪⎪⎪⎪u∂⎪⎪=0⎩∂y u∞ y=∞(6.41)And the energy equation (6.40) can be written in terms of a dimensionlesstemperature, Θ = (T − Tw )/(T∞ − Tw ), as⎧Θ(y = 0) = 0⎪⎪⎪⎪⎪⎨∂Θ∂2Θ∂ΘΘ(y = ∞) = 1(6.42)+v=αu2⎪∂x∂y∂y⎪⎪ ∂Θ ⎪⎪=0⎩ ∂y y=∞Notice that the problems of predicting u/u∞ and Θ are identical, withone exception: eqn.
(6.41) has ν in it whereas eqn. (6.42) has α. If ν andα should happen to be equal, the temperature distribution in the b.l. isfor ν = α :T − Tw= f (η)T∞ − Twderivative of the Blasius functionsince the two problems must have the same solution.In this case, we can immediately calculate the heat transfer coefficientusing eqn. (6.5):k∂f ∂η∂(T − Tw ) h==kT∞ − Tw∂y∂η ∂y η=0y=03but (∂ 2 f /∂η2 )η=0 = 0.33206 (see Fig.
6.10) and ∂η/∂y = u∞ /νx, so3hx= Nux = 0.33206 Rexkfor ν = α(6.43)Normally, in using eqn. (6.43) or any other forced convection equation,properties should be evaluated at the film temperature, Tf = (Tw +T∞ )/2.296Laminar and turbulent boundary layers§6.4Example 6.4Water flows over a flat heater, 0.06 m in length, at 15 atm pressureand 440 K. The free stream velocity is 2 m/s and the heater is held at460 K. What is the average heat flux?Solution. At Tf = (460 + 440)/2 = 450 K:ν = 1.725 × 10−7 m2 /sα = 1.724 × 10−7 m2 /sTherefore, ν α, and we can use eqn.
(6.43). First, we must calculatethe average heat flux, q. To do this, we set ∆T ≡ Tw − T∞ and write 91 L∆T L kk∆T L u∞Nux dx = 0.332dxq=(h∆T ) dx =L 0L 0 xL 0 νx√=2u∞ L/νsok3q = 2 0.332ReL ∆T = 2qx=LLNote that the average heat flux is twice that at the trailing edge, x = L.Using k = 0.674 W/m·K for water at the film temperature,20.6742(0.06)q = 2(0.332)(460 − 440)0.061.72 × 10−7= 124, 604 W/m2 = 125 kW/m2Equation (6.43) is clearly a very restrictive heat transfer solution.
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