John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 59
Текст из файла (страница 59)
A correctexpression for uc will eventually depend upon these variables, but wewill not attempt to make uc fit this particular condition. Doing so wouldyield two equations, (8.13) and (6.47), in a single unknown, δ(x). It wouldbe impossible to satisfy both of them. Instead, we shall allow the velocityprofile to violate this condition slightly and writeuc (x) = C1βg |Tw − T∞ | 2δ (x)ν(8.18)Then we shall solve the two integrated conservation equations for thetwo unknowns, C1 (which should ¼) and δ(x).The dimensionless temperature and velocity profiles are plotted inFig. 8.4. With them are included Schmidt and Beckmann’s exact calculation for air (Pr = 0.7), as presented in [8.4].
Notice that the integral approximation to the temperature profile is better than the approximationto the velocity profile. That is fortunate, since the temperature profileexerts the major influence in the heat transfer solution.When we substitute eqns. (8.15) and (8.17) in the momentum equa-407408Natural convection in single-phase fluids and during film condensation§8.3tion (8.13), using eqn. (8.18) for uc (x), we getC12gβ |Tw − T∞ |ν2 1ydy 2y 45δd1−dxδ 0 δ δ1= gβ |Tw − T∞ | δ01= 105 yy 2d1−δδ 1=3− C1 gβ |Tw∂− T∞ | δ(x) ∂ y δyδy1−δ2 =1(8.19)yδ =0where we change the sign of the terms on the left by replacing (Tw − T∞ )with its absolute value. Equation (8.19) then becomes1dδ1 2 gβ |Tw − T∞ |C= − C1δ321 1dxν23or1−C1dδ43=gβ |Tw − T∞ |dxC12ν284Integrating this with the b.c., δ(x = 0) = 0, gives184− C13δ4 =gβ |Tw − T∞ |C12xν2(8.20)Substituting eqns.
(8.15), (8.17), and (8.18) in eqn. (6.47) likewise gives(Tw 1y 4ygβ |Tw − T∞ | dy3δ1−− T∞ ) C1dνdxδ δ0 δ1= 30dTw − T∞= −αδd(y/δ)y1−δ=−22 y/δ=0Laminar natural convection on a vertical isothermal surface§8.3or3C1 dδ4C1 3 dδδ==30dx40 dx2gβ |Tw − T∞ |Prν2Integrating this with the b.c., δ(x = 0) = 0, we getδ4 =80xgβ|Tw − T∞ |C1 Prν2(8.21)Equating eqns. (8.20) and (8.21) for δ4 , we then obtain21201− C113x=xgβ |Tw − T∞ |gβ |Tw − T∞ |C1Prν2ν2orC1 =Pr20+ Pr321(8.22)Then, from eqn.
(8.21):20240+ Pr21δ4 =xgβ |Tw − T∞ |Pr2ν2or0.952 + Pr 1/4 1δ= 3.9361/4xPr2Grx(8.23)Equation (8.23) can be combined with the known temperature profile,eqn. (8.15), and substituted in Fourier’s law to find q:T − T∞ d∂T k(Tw − T∞ )k∆TTw − T∞ q = −k(8.24)=−=2y∂y y=0δδdδy/δ=0=−2409410Natural convection in single-phase fluids and during film condensation§8.3so, writing h = q |Tw − T∞ | ≡ q/∆T , we obtain41/4x2qxPr1/4=2 =Nux ≡(PrGrx )∆T kδ3.9360.952 + Pror1/4Nux = 0.508 RaxPr0.952 + Pr1/4(8.25)This is the Squire-Eckert result for the local heat transfer from a verticalisothermal wall during laminar natural convection. It applies for eitherTw > T∞ or Tw < T∞ .The average heat transfer coefficient can be obtained fromLLq(x) dxh(x) dx= 0h= 0L∆TLThus,1hL=NuL =kkL04kNux dx =Nux x3x=Lor1/4NuL = 0.678 RaLPr0.952 + Pr1/4(8.26)All properties in eqn.
(8.26)and the preceding equations should be evaluated at T = (Tw + T∞ ) 2 except in gases, where β should be evaluatedat T∞ .Example 8.1A thin-walled metal tank containing fluid at 40◦ C cools in air at 14◦ C;h is very large inside the tank. If the sides are 0.4 m high, computeh, q, and δ at the top. Are the b.l. assumptions reasonable?Solution.βair = 1 T∞ = 1 (273 + 14) = 0.00348 K−1 .RaL =gβ∆T L3ναThen9.8(0.00348)(40 − 14)(0.4)3 = = 1.645 × 1081.566 × 10−5 2.203 × 10−5Recall that, in footnote 2, we anticipated that Nu would vary as Gr1/4 .
We now seethat this is the case.4Laminar natural convection on a vertical isothermal surface§8.3and Pr = 0.711, where the properties are evaluated at 300 K = 27◦ C.Then, from eqn. (8.26),8NuL = 0.678 1.645 × 101/4 0.7110.952 + 0.7111/4= 62.1soh=62.1(0.02614)62.1k== 4.06 W/m2 KL0.4andq = h ∆T = 4.06(40 − 14) = 105.5 W/m2The b.l. thickness at the top of the tank is given by eqn. (8.23) atx = L:0.952 + 0.711 1/41δ= 3.936 1/4 = 0.0430L0.7112RaL PrThus, the b.l. thickness at the end of the plate is only 4% of the height,or 1.72 cm thick. This is thicker than typical forced convection b.l.’s,but it is still reasonably thin.Example 8.2Large thin metal sheets of length L are dipped in an electroplatingbath in the vertical position. Their average temperature is initiallycooler than the liquid in the bath.
How rapidly will they come up tobath temperature, Tb ?Solution. We can probably take Bi 1 and use the lumped-capacityresponse equation (1.20). We obtain h for use in eqn. (1.20) fromeqn. (8.26):kh = 0.678LPr0.952 + Pr1/4 gβL3αν1/4∆T 1/4call this BSince h ∝ ∆T 1/4 , with ∆T = Tb − T , eqn. (1.20) becomesd(Tb − T )BA=−(Tb − T )5/4dtρcV411412Natural convection in single-phase fluids and during film condensation§8.3where V /A = the half-thickness of the plate, w.
Integrating this between the initial temperature of the plate, Ti , and the temperature attime t, we gettTd(Tb − T )Bdt5/4 = −Ti (Tb − T )0 ρcwsoTb − T =1(Tb − Ti )1/4Bt+4ρcw−4(Before we use this result, we should check Bi = Bw∆T 1/4 k to becertain that it is, in fact, less than unity.) The temperature can be putin dimensionless form as−4B (Tb − Ti )1/4Tb − Tt= 1+Tb − Ti4ρcwwhere the coefficient of t is a kind of inverse time constant of theresponse. Thus, the temperature dependence of h in natural convection leads to a solution quite different from the exponential responsethat resulted from a constant h [eqn.
(1.22)].Comparison of analysis and correlations with experimental dataChurchill and Chu [8.3] have proposed two equations for the data correlated in Fig. 8.3. The simpler of the two is shown in the figure. It isNuL = 0.68 + 0.671/4RaL0.4921+Pr9/16 −4/9(8.27)which applies for all Pr and for the range of Ra shown in the figure. TheSquire–Eckert prediction is within 1.2% of this correlation for high Pr andhigh RaL , and it differs by only 5.5% if the fluid is a gas and RaL > 105 .Typical Rayleigh numbers usually exceed 105 , so we conclude that theSquire–Eckert prediction is remarkably accurate in the range of practicalinterest, despite the approximations upon which it is built.
The additiveconstant of 0.68 in eqn. (8.27) is a correction for low RaL , where the b.l.1/4assumptions are inaccurate and NuL is no longer proportional to RaL .At low Prandtl numbers, the Squire-Eckert prediction fails and eqn.(8.27) has to be used. In the turbulent regime, Gr 109 [8.6], eqn. (8.27)Laminar natural convection on a vertical isothermal surface§8.3predicts a lower bound on the data (see Fig. 8.3), although it is reallyintended only for laminar boundary layers. In this correlation, as ineqn.
(8.26), the thermal properties should all be evaluated at a film temperature, Tf = (T∞ + Tw )/2, except for β, which is to be evaluated at T∞if the fluid is a gas.Example 8.3Verify the first heat transfer coefficient in Table 1.1. It is for air at20◦ C next to a 0.3 m high wall at 50◦ C.Solution. At T = 35◦ C = 308 K, we findPr = 0.71, ν = 16.45 ×10−6 m2 /s, α = 2.318×10−5 m2 /s, and β = 1 (273+20) = 0.00341 K−1 .ThenRaL =9.8(0.00341)(30)(0.3)3gβ∆T L3== 7.10 × 107αν(16.45)(0.2318)10−10The Squire-Eckert prediction gives1/4 NuL = 0.678 7.10 × 1070.710.952 + 0.711/4= 50.3soh = 50.30.0267k= 50.3= 4.48 W/m2 K.L0.3And the Churchill-Chu correlation gives 7.10 × 107NuL = 0.68 + 0.67 1/41 + (0.492/0.71)9/164/9 = 47.88so0.0267h = 47.880.3= 4.26 W/m2 KThe prediction is therefore within 5% of the correlation.
We shoulduse the latter result in preference to the theoretical one, although thedifference is slight.413414Natural convection in single-phase fluids and during film condensation§8.3Variable-properties problemSparrow and Gregg [8.7] provide an extended discussion of the influenceof physical property variations on predicted values of Nu. They foundthat while β for gases should be evaluated at T∞ , all other propertiesshould be evaluated at Tr , whereTr = Tw − C (Tw − T∞ )(8.28)and where C = 0.38 for gases. Most books recommend that a simplemean between Tw and T∞ (or C = 0.50) be used.
A simple mean seldomdiffers much from the more precise result above, of course.It has also been shown by Barrow and Sitharamarao [8.8] that whenβ∆T is no longer 1, the Squire-Eckert formula should be corrected asfollows:1/43Nu = Nusq−Ek 1 + 5 β∆T + O(β∆T )2(8.29)This same correction can be applied to the Churchill-Chu correlation orto other expressions for Nu. Since β = 1 T∞ for an ideal gas, eqn. (8.29)gives only about a 1.5% correction for a 330 K plate heating 300 K air.Note on the validity of the boundary layer approximationsThe boundary layer approximations are sometimes put to a rather severe test in natural convection problems.
Thermal b.l. thicknesses areoften fairly large, and the usual analyses that take the b.l. to be thin canbe significantly in error. This is particularly true as Gr becomes small.Figure 8.5 includes three pictures that illustrate this. These pictures areinterferograms (or in the case of Fig. 8.5c, data deduced from interferograms). An interferogram is a photograph made in a kind of lightingthat causes regions of uniform density to appear as alternating light anddark bands.Figure 8.5a was made at the University of Kentucky by G.S. Wang andR. Eichhorn.
The Grashof number based on the radius of the leadingedge is 2250 in this case. This is low enough to result in a b.l. that islarger than the radius near the leading edge. Figure 8.5b and c are fromKraus’s classic study of natural convection visualization methods [8.9].Figure 8.5c shows that, at Gr = 585, the b.l. assumptions are quite unreasonable since the cylinder is small in comparison with the large regionof thermal disturbance.a.
A 1.34 cm wide flat plate with arounded leading edge in air. Tw =46.5◦ C, ∆T = 17.0◦ C, Grradius 2250b. A square cylinder with a fairly lowvalue of Gr. (Rendering of an interferogram shown in [8.9].)c. Measured isotherms around a cylinderin airwhen GrD ≈ 585 (from [8.9]).Figure 8.5 The thickening of the b.l. during natural convection at low Gr, as illustrated by interferograms made ontwo-dimensional bodies.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
