John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 43
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For, say, mercury at 100◦ C, Pr = 0.0162 and δt /δ = 3.952,which violates the condition by an intolerable margin. We therefore havea theory that is acceptable for gases and all liquids except the metallicones.The final step in predicting the heat flux is to write Fourier’s law:T − T∞ ∂Tw − T∞∂T Tw − T∞ = −k(6.56)q = −k∂y y=0δt∂(y/δt ) y/δt =0Using the dimensionless temperature distribution given by eqn.
(6.50),we getq = +kTw − T∞ 3δt2orh≡3k3k δq==∆T2δt2 δ δt(6.57)and substituting eqns. (6.54) and (6.31b) for δ/δt and δ, we obtain3hx3 Rex1/2=1.025 Pr1/3 = 0.3314 Rex Pr1/3Nux ≡k2 4.64Considering the various approximations, this is very close to the resultof the exact calculation, which turns out to be1/2Nux = 0.332 Rex Pr1/30.6 Pr 50(6.58)This expression gives very accurate results under the assumptions onwhich it is based: a laminar two-dimensional b.l. on a flat surface, withTw = constant and 0.6 Pr 50.Heat transfer coefficient for laminar, incompressible flow over a flat surface§6.5Figure 6.15 A laminar b.l.
in a low-Pr liquid. The velocity b.l.is so thin that u u∞ in the thermal b.l.Some other laminar boundary layer heat transfer equationsHigh Pr.isAt high Pr, eqn. (6.58) is still close to correct. The exact solution1/2Nux → 0.339 Rex Pr1/3 ,Pr → ∞(6.59)Low Pr. Figure 6.15 shows a low-Pr liquid flowing over a flat plate. Inthis case δt δ, and for all practical purposes u = u∞ everywhere withinthe thermal b.l. It is as though the no-slip condition [u(y = 0) = 0] andthe influence of viscosity were removed from the problem.
Thus, thedimensional functional equation for h becomes(6.60)h = fn x, k, ρcp , u∞There are five variables in J/K, m, and s, so there are only two pi-groups.They areNux =hxkand Π2 ≡ Rex Pr =u∞ xαThe new group, Π2 , is called a Péclét number, Pex , where the subscriptidentifies the length upon which it is based. It can be interpreted asfollows:Pex ≡ρcp u∞ ∆Theat capacity rate of fluid in the b.l.u∞ x=(6.61)=αaxial heat conductance of the b.l.k∆T305306Laminar and turbulent boundary layers§6.5So long as Pex is large, the b.l. assumption that ∂ 2 T /∂x 2 ∂ 2 T /∂y 2will be valid; but for small Pex (i.e., Pex 100), it will be violated and aboundary layer solution cannot be used.The exact solution of the b.l. equations gives, in this case:⎧⎪Pex ≥ 100and⎪⎪⎨1/21orPr 100(6.62)Nux = 0.565 Pex⎪⎪⎪⎩ Re ≥ 104xGeneral relationship.
Churchill and Ozoe [6.5] recommend the following empirical correlation for laminar flow on a constant-temperature flatsurface for the entire range of Pr:1/20.3387 Rex Pr1/3Nux = 1/41 + (0.0468/Pr)2/3Pex > 100(6.63)This relationship proves to be quite accurate, and it approximates eqns.(6.59) and (6.62), respectively, in the high- and low-Pr limits.
The calculations of an average Nusselt number for the general case is left as anexercise (Problem 6.10).Boundary layer with an unheated starting length Figure 6.16 showsa b.l. with a heated region that starts at a distance x0 from the leadingedge. The heat transfer in this instance is easily obtained using integralmethods (see Prob. 6.41).1/20.332 Rex Pr1/3Nux = 1/3 ,1 − (x0 /x)3/4x > x0(6.64)Average heat transfer coefficient, h. The heat transfer coefficient h, isthe ratio of two quantities, q and ∆T , either of which might vary with x.So far, we have only dealt with the uniform wall temperature problem.Equations (6.58), (6.59), (6.62), and (6.63), for example, can all be used tocalculate q(x) when (Tw − T∞ ) ≡ ∆T is a specified constant.
In the nextsubsection, we discuss the problem of predicting [T (x) − T∞ ] when q isa specified constant. That is called the uniform wall heat flux problem.Heat transfer coefficient for laminar, incompressible flow over a flat surface§6.5Figure 6.16 A b.l. with an unheated region at the leading edge.The term h is used to designate either q/∆T in the uniform wall temperature problem or q/∆T in the uniform wall heat flux problem. Thus, 1 L11 Lqq dx =h(x) dx=uniform wall temp.: h ≡∆T∆T L 0L 0(6.65)uniform heat flux:h≡qq= L∆T1∆T (x) dxL 0(6.66)The Nusselt number based on h and a characteristic length, L, is designated NuL . This is not to be construed as an average of Nux , which wouldbe meaningless in either of these cases.Thus, for a flat surface (with x0 = 0), we use eqn.
(6.58) in eqn. (6.65)to get1h=LL00.332 k Pr1/3h(x) dx = Lkx9u∞νL √0x dxxNux1/2= 0.664 ReLPr1/3 kL(6.67)Thus, h = 2h(x = L) in a laminar flow, andNuL =hL1/2= 0.664 ReL Pr1/3k(6.68)Likewise for liquid metal flows:1/2NuL = 1.13 PeL(6.69)307308Laminar and turbulent boundary layers§6.5Some final observations. The preceding results are restricted to thetwo-dimensional, incompressible, laminar b.l. on a flat isothermal wall atvelocities that are not too high. These conditions are usually met if:• Rex or ReL is not above the turbulent transition value, which istypically a few hundred thousand.• The Mach number of the flow, Ma ≡ u∞ /(sound speed), is less thanabout 0.3.
(Even gaseous flows behave incompressibly at velocitieswell below sonic.) A related condition is:• The Eckert number, Ec ≡ u2∞ /cp (Tw − T∞ ), is substantially less thanunity. (This means that heating by viscous dissipation—which wehave neglected—does not play any role in the problem. This assumption was included implicitly when we treated J as an independent unit in the dimensional analysis of this problem.)It is worthwhile to notice how h and Nu depend on their independentvariables:11h or h ∝ √ or √ ,xL3Nux or NuL ∝ x or L,√u∞ , ν −1/6 , (ρcp )1/3 , k2/3√u∞ , ν −1/6 , (ρcp )1/3 , k−1/3(6.70)Thus, h → ∞ and Nux vanishes at the leading edge, x = 0.
Of course,an infinite value of h, like infinite shear stress, will not really occur atthe leading edge because the b.l. description will actually break down ina small neighborhood of x = 0.In all of the preceding considerations, the fluid properties have beenassumed constant. Actually, k, ρcp , and especially µ might all vary noticeably with T within the b.l. It turns out that if properties are all evaluated at the average temperature of the b.l. or film temperature Tf =(Tw + T∞ )/2, the results will normally be quite accurate. It is also worthnoting that, although properties are given only at one pressure in Appendix A, µ, k, and cp change very little with pressure, especially in liquids.Example 6.5Air at 20◦ C and moving at 15 m/s is warmed by an isothermal steamheated plate at 110◦ C, ½ m in length and ½ m in width.
Find theaverage heat transfer coefficient and the total heat transferred. Whatare h, δt , and δ at the trailing edge?Heat transfer coefficient for laminar, incompressible flow over a flat surface§6.5Solution. We evaluate properties at Tf = (110+20)/2 = 65◦ C. ThenPr = 0.707andReL =15(0.5)u∞ L== 386, 600ν0.0000194so the flow ought to be laminar up to the trailing edge. The Nusseltnumber is then1/2NuL = 0.664 ReLPr1/3 = 367.8andh = 367.8367.8(0.02885)k== 21.2 W/m2 KL0.5The value is quite low because of the low conductivity of air.
The totalheat flux is thenQ = hA ∆T = 21.2(0.5)2 (110 − 20) = 477 WBy comparing eqns. (6.58) and (6.68), we see that h(x = L) = ½ h, so1h(trailing edge) = 2 (21.2) = 10.6 W/m2 KAnd finally,δ(x = L) = 4.92L34.92(0.5)= 0.00396 mReL = 3386, 600= 3.96 mmand3.96δ= √= 4.44 mmδt = √33Pr0.707The problem of uniform wall heat fluxWhen the heat flux at the heater wall, qw , is specified instead of thetemperature, it is Tw that we need to know.
We leave the problem offinding Nux for qw = constant as an exercise (Problem 6.11). The exactresult is1/2Nux = 0.453 Rex Pr1/3forPr 0.6(6.71)309Laminar and turbulent boundary layers310§6.5where Nux = hx/k = qw x/k(Tw − T∞ ). The integral method gives thesame result with a slightly lower constant (0.417).We must be very careful in discussing average results in the constantheat flux case. The problem now might be that of finding an averagetemperature difference (cf. (6.66)):dx1 Lqw x1 L3√Tw − T∞ =(Tw − T∞ ) dx =1/3L 0L 0 k(0.453 u∞ /ν Pr ) xorTw − T∞ =qw L/k1/20.6795 ReLPr(6.72)1/31/21/3(although thewhich can be put into the form NuL = 0.6795 ReL PrNusselt number yields an awkward nondimensionalization for Tw − T∞ ).Churchill and Ozoe [6.5] have pointed out that their eqn.
(6.63) will describe (Tw − T∞ ) with high accuracy over the full range of Pr if the constants are changed as follows:1/20.4637 Rex Pr1/3Nux = 1/41 + (0.02052/Pr)2/3Pex > 100(6.73)Example 6.6Air at 15◦ C flows at 1.8 m/s over a 0.6 m-long heating panel. Thepanel is intended to supply 420 W/m2 to the air, but the surface cansustain only about 105◦ C without being damaged.
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