John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 47
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This equation may be used for either uniform Tw or uniform qw .The advantage of eqns. (6.120a) or (6.120c) is that, once φu or φum isknown, they will predict heat transfer from the laminar region, throughthe transition regime, and into the turbulent regime.327328Laminar and turbulent boundary layers§6.8Example 6.9After loading its passengers, a ship sails out of the mouth of a river,where the water temperature is 24◦ C, into 10◦ C ocean water. Theforward end of the ship’s hull is sharp and relatively flat.
If the shiptravels at 5 knots, find Cf and h at a distance of 1 m from the forwardedge of the hull.Solution. If we assume that the hull’s heat capacity holds it at theriver temperature for a time, we can take the properties of water atTf = (10 + 24)/2 = 17◦ C: ν = 1.085 × 10−6 m2 /s, k = 0.5927 W/m·K,ρ = 998.8 kg/m3 , cp = 4187 J/kg·K, and Pr = 7.66.One knot equals 0.5144 m/s, so u∞ = 5(0.5144) = 2.572 m/s.Then, Rex = (2.572)(1)/(1.085 × 10−6 ) = 2.371 × 106 , indicating thatthe flow is turbulent at this location.We have given several different equations for Cf in a turbulentboundary layer, but the most accurate of these is eqn. (6.102):0.455Cf (x) = !"2ln(0.06 Rex )0.455=<=2 = 0.003232ln[0.06(2.371 × 106 )]For the heat transfer coefficient, we can use either eqn.
(6.115)h(x) =k0.43· 0.032 Re0.8x Prx(0.5927)(0.032)(2.371 × 106 )0.8 (7.66)0.43(1.0)= 5, 729 W/m2 K=or its more complex counterpart, eqn. (6.111):Cf 24 h(x) = ρcp u∞ ·1 + 12.8 Pr0.68 − 1 Cf 2=998.8(4187)(2.572)(0.003232/2)!"31 + 12.8 (7.66)0.68 − 1 0.003232/2= 6, 843 W/m2 KThe two values of h differ by about 18%, which is within the uncertainty of eqn. (6.115).Heat transfer in turbulent boundary layers§6.8Example 6.10In a wind tunnel experiment, an aluminum plate 2.0 m in length iselectrically heated at a power density of 1 kW/m2 and is cooled onone surface by air flowing at 10 m/s. The air in the wind tunnel hasa temperature of 290 K and is at 1 atm pressure, and the Reynoldsnumber at the end of turbulent transition regime is observed to be400,000.
Estimate the average temperature of the plate.Solution. For this low heat flux, we expect the plate temperatureto be near the air temperature, so we evaluate properties at 300 K:ν = 1.578 × 10−5 m2 /s, k = 0.02623 W/m·K, and Pr = 0.713. At10 m/s, the plate Reynolds number is ReL = (10)(2)/(1.578×10−5 ) =1.267 × 106 . From eqn. (6.118), we getNuL = 0.037(0.713)0.43 (1.267 × 106 )0.8;− (400, 000)0.8 − 17.95(0.713)−0.097 (400, 000)1/2 = 1, 845soh=1845(0.02623)1845 k== 24.19 W/m2 KL2.0It follows that the average plate temperature isT w = 290 K +103 W/m2= 331 K.24.19 W/m2 KThe film temperature is (331+290)/2 = 311 K; if we recalculate usingproperties at 311 K, the h changes by less than 4%, and T w by 1.3 K.To take better account of the transition regime, we can use Churchill’sequation, (6.120c).
First, we evaluate φ:(1.267 × 106 )(0.713)2/35φ= !"1/2 = 9.38 × 102/31 + (0.0468/0.713)We then estimateφum = 1.875 · φ(ReL = 400, 000)(1.875)(400, 000)(0.713)2/35= !"1/2 = 5.55 × 101 + (0.0468/0.713)2/3329Chapter 6: Laminar and turbulent boundary layers330Finally,1/2NuL = 0.45 + (0.6774) 9.38 × 105⎧⎫1/2 ⎪⎪3/5⎨⎬59.38 × 10 /12, 500× 1+ !"2/5⎪⎪⎩⎭1 + (5.55 × 105 /9.38 × 105 )7/2= 2, 418which leads to2418(0.02623)2418 k== 31.71 W/m2 Kh=L2.0andT w = 290 K +103 W/m2= 322 K.31.71 W/m2 KThus, in this case, the average heat transfer coefficient is 33% higherwhen the transition regime is included.A word about the analysis of turbulent boundary layersThe preceding discussion has circumvented serious analysis of heat transfer in turbulent boundary layers.
In the past, boundary layer heat transfer has been analyzed in many flows (with and without pressure gradients, dp/dx) using sophisticated integral methods. In recent decades,however, computational techniques have largely replaced integral analyses. Various computational schemes, particularly those based on turbulent kinetic energy and viscous dissipation (so-called k-ε methods), arewidely-used and have been implemented in a variety of commercial fluiddynamics codes. These methods are described in the technical literatureand in monographs on turbulence [6.18, 6.19].We have found our way around analysis by presenting some correlations for the simple plane surface. In the next chapter, we deal withmore complicated configurations.
A few of these configurations will beamenable to elementary analyses, but for others we shall only be able topresent the best data correlations available.Problems6.1Verify that eqn. (6.13) follows from eqns. (6.11) and (6.12).Problems3316.2The student with some analytical ability (or some assistancefrom the instructor) should complete the algebra between eqns.(6.16) and (6.20).6.3Use a computer to solve eqn. (6.18) subject to b.c.’s (6.20). Todo this you need all three b.c.’s at η = 0, but one is presentlyat η = ∞. There are three ways to get around this:• Start out by guessing a value of ∂f /∂η at η = 0—say,∂f /∂η = 1. When η is large—say, 6 or 10—∂f /∂η willasymptotically approach a constant.
If the constant > 1,go back and guess a lower value of ∂f /∂η, or vice versa,until the constant converges on unity. (There are manyways to automate the successive guesses.)• The correct value of df /dη is approximately 0.33206 atη = 0.
You might cheat and begin with it.• There exists a clever way to map df /dη = 1 at η = ∞back into the origin. (Consult your instructor.)6.4Verify that the Blasius solution (Table 6.1) satisfies eqn. (6.25).To do this, carry out the required integration.6.5Verify eqn.
(6.30).6.6Obtain the counterpart of eqn. (6.32) based on the velocity profile given by the integral method.6.7Assume a laminar b.l. velocity profile of the simple form u/u∞ =y/δ and calculate δ and Cf on the basis of this very rough estimate, using the momentum integral method. How accurateis each? [Cf is about 13% low.]6.8√In a certain flow of water at 40◦ C over a flat plate δ = 0.005 x,for δ and x measured in meters.
Plot to scale on a commongraph (with an appropriately expanded y-scale):• δ and δt for the water.• δ and δt for air at the same temperature and velocity.6.9A thin film of liquid with a constant thickness, δ0 , falls downa vertical plate. It has reached its terminal velocity so thatviscous shear and weight are in balance and the flow is steady.Chapter 6: Laminar and turbulent boundary layers332The b.l. equation for such a flow is the same as eqn.
(6.13),except that it has a gravity force in it. Thus,u∂u1 dp∂2u∂u+v=−+g+ν∂x∂yρ dx∂y 2where x increases in the downward direction and y is normalto the wall. Assume that the surrounding air density 0, sothere is no hydrostatic pressure gradient in the surroundingair. Then:• Simplify the equation to describe this situation.• Write the b.c.’s for the equation, neglecting any air dragon the film.• Solve for the velocity distribution in the film, assumingthat you know δ0 (cf. Chap. 8).(This solution is the starting point in the study of many processheat and mass transfer problems.)6.10Develop an equation for NuL that is valid over the entire rangeof Pr for a laminar b.l.
over a flat, isothermal surface.6.11Use an integral method to develop a prediction of Nux for alaminar b.l. over a uniform heat flux surface. Compare yourresult with eqn. (6.71). What is the temperature difference atthe leading edge of the surface?6.12Verify eqn. (6.118).6.13It is known from flow measurements that the transition to turbulence occurs when the Reynolds number based on mean velocity and diameter exceeds 4000 in a certain pipe.
Use the factthat the laminar boundary layer on a flat plate grows accordingto the relation2νδ= 4.92xumax xto find an equivalent value for the Reynolds number of transition based on distance from the leading edge of the plate andumax . (Note that umax = 2uav during laminar flow in a pipe.)Problems3336.14Execute the differentiation in eqn. (6.24) with the help of Leibnitz’s rule for the differentiation of an integral and show thatthe equation preceding it results.6.15Liquid at 23◦ C flows at 2 m/s over a smooth, sharp-edged,flat surface 12 cm in length which is kept at 57◦ C. Calculateh at the trailing edge (a) if the fluid is water; (b) if the fluid isglycerin (h = 346 W/m2 K).
(c) Compare the drag forces in thetwo cases. [There is 23.4 times as much drag in the glycerin.]6.16Air at −10◦ C flows over a smooth, sharp-edged, almost-flat,aerodynamic surface at 240 km/hr. The surface is at 10◦ C.Find (a) the approximate location of the laminar turbulent transition; (b) the overall h for a 2 m chord; (c) h at the trailing edgefor a 2 m chord; (d) δ and h at the beginning of the transitionregion. [δxt = 0.54 mm.]6.17Find h in Example 6.10 using eqn.
(6.120c) with Reu = 105 and2 × 105 . Discuss the results.6.18For system described in Example 6.10, plot the local value ofh over the whole length of the plate using eqn. (6.120c). Onthe same graph, plot h from eqn. (6.71) for Rex < 400, 000 andfrom eqn. (6.115) for Rex > 200, 000. Discuss the results.6.19Mercury at 25◦ C flows at 0.7 m/s over a 4 cm-long flat heaterat 60◦ C. Find h, τ w , h(x = 0.04 m), and δ(x = 0.04 m).6.20A large plate is at rest in water at 15◦ C. The plate is suddenlytranslated parallel to itself, at 1.5 m/s. The resulting fluidmovement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from anedge.
The governing transient momentum equation, Du/Dt =ν(∂ 2 u/∂y 2 ), takes the form∂2u1 ∂u=ν ∂t∂y 2Determine u at 0.015 m from the plate for t = 1, 10, and1000 s. Do this by first posing the problem fully and thencomparing it with the solution in Section 5.6. [u 0.003 m/safter 10 s.]Chapter 6: Laminar and turbulent boundary layers3346.21Notice that, when Pr is large, the velocity b.l.
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