The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 4
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Raithby and K.G. Terry HollandsIntroductionNatural convection heat transfer occurs when the convective fluid motion is induced by density differencesthat are themselves caused by the heating. An example is shown in Figure 4.2.1(A), where a body atsurface temperature Ts transfers heat at a rate q to ambient fluid at temperature T¥ < Ts.FIGURE 4.2.1 (A) Nomenclature for external heat transfer. (A) General sketch; (B) is for a tilted flat plate, and(C) defines the lengths cal for horizontal surfaces.In this section, correlations for the average Nusselt number are provided from which the heat transferrate q from surface area As can be estimated. The Nusselt number is defined asNu =hc LqL=kAs DT k(4.2.1)where DT = Ts – T¥ is the temperature difference driving the heat transfer.
A dimensional analysis leadsto the following functional relation:Nu = f (Ra, Pr, geometric shape, boundary conditions)(4.2.2)For given thermal boundary conditions (e.g., isothermal wall and uniform T¥), and for a given geometry(e.g., a cube), Equation (4.2.2) states that Nu depends only on the Rayleigh number, Ra, and Prandtlnumber, Pr. The length scales that appear in Nu and Ra are defined, for each geometry considered, ina separate figure. The fluid properties are generally evaluated at Tf , the average of the wall and ambienttemperatures. The exception is that b, the temperature coefficient of volume expansion, is evaluated atT¥ for external natural convection(Figures 4.2.1 to 4.2.3) in a gaseous medium.The functional dependence on Pr is approximately independent of the geometry, and the followingPr-dependent function will be useful for laminar heat transfer (Churchill and Usagi, 1972):(Cl = 0.671 1 + (0.492 Pr )9 16 4 9)CtV and CtH are functions that will be useful for turbulent heat transfer:© 1999 by CRC Press LLC(4.2.3)4-15Heat and Mass TransferFIGURE 4.2.2 Nomenclature for heat transfer from planar surfaces of different shapes.FIGURE 4.2.3 Definitions for computing heat transfer from a long circular cylinder (A), from the lateral surfaceof a vertical circular cylinder (B), from a sphere (C), and from a compound body (D).(CtV = 0.13Pr 0.22 1 + 0.61Pr 0.81)1 + 0.0107Pr öCtH = 0.14æè 1 + 0.01Pr ø0.42(4.2.4)(4.2.5)The superscripts V and H refer to the vertical and horizontal surface orientation.The Nusselt numbers for fully laminar and fully turbulent heat transfer are denoted by Nu, and Nut,respectively.
Once obtained, these are blended (Churchill and Usagi, 1972) as follows to obtain theequation for Nu:(mNu = (Nu l ) + (Nu t )m 1m)(4.2.6)The blending parameter m depends on the body shape and orientation.The equation for Nu, in this section is usually expressed in terms of NuT, the Nusselt number thatwould be valid if the thermal boundary layer were thin. The difference between Nul and NuT accountsfor the effect of the large boundary layer thicknesses encountered in natural convection.It is assumed that the wall temperature of a body exceeds the ambient fluid temperature (Ts > T¥).For Ts < T¥ the same correlations apply with (T¥ – Ts) replacing (Ts – T¥) for a geometry that is rotated© 1999 by CRC Press LLC4-16Section 4180° relative to the gravitational vector; for example, the correlations for a horizontal heated upwardfacing flat plate applies to a cooled downward-facing flat plate of the same planform.Correlations for External Natural ConvectionThis section deals with problems where the body shapes in Figures 4.2.1 to 4.2.3 are heated whileimmersed in a quiescent fluid.
Different cases are enumerated below.1. Isothermal Vertical (f = 0) Flat Plate, Figure 4.2.1B. For heat transfer from a vertical plate(Figure 4.2.1B), for 1 < Ra < 1012,Nu T = Cl Ra 1 42.0ln 1 + 2.0 Nu TNu l =((Nu t = CtV Ra 1 3 1 + 1.4 ´ 10 9 Pr Ra(4.2.7)))Cl and CtV are given by Equations (4.2.3) and (4.2.4). Nu is obtained by substituting Equation(4.2.7) expressions for Nul and Nut into Equation (4.2.6) with m = 6.2. Vertical Flat Plate with Uniform Heat Flux, Figure 4.2.1B.
If the plate surface has a constant(known) heat flux, rather than being isothermal, the objective is to calculate the average temperature difference, DT, between the plate and fluid. For this situation, and for 15 < Ra* < 105,15( )Nu T = Gl Ra *Nu l =Gl =1.83ln 1 + 1.83 Nu T()6æPrö÷ç5 è 4 + 9 Pr + 10 Pr ø34* 14( ) (Ra )Nu t = CtV(4.2.8a)15(4.2.8b)Ra* is defined in Figure 4.2.1B and CtV is given by Equation (4.2.4).
Find Nu by inserting theseexpressions for Nul and Nut into Equation (4.2.6) with m = 6. The Gl expression is due to Fujiiand Fujii (1976).3. Horizontal Upward-Facing (f = 90°) Plates, Figure 4.2.1C. For horizontal isothermal surfacesof various platforms, correlations are given in terms of a lengthscale L* (Goldstein et al., 1973),defined in Figure 4.2.1C. For Ra ³ 1,Nu T = 0.835Cl Ra 1 4Nu l =2.0ln 1 + 1.4 Nu T()Nu t = CtH Ra 1 3(4.2.9)Nu is obtained by substituting Nul and Nut from Equation 4.2.9 into Equation 4.2.6 with m = 10.For non-isothermal surfaces, replace DT by DT.4.
Horizontal Downward-Facing (f = –90°) Plates, Figure 4.2.1C. For horizontal downward-facingplates of various planforms, the main buoyancy force is into the plate so that only a very weakforce drives the fluid along the plate; for this reason, only laminar flows have been measured. Forthis case, the following equation applies for Ra < 1010, Pr ³ 0.7:Nu T = Hl Ra 1 5Hl =0.527[1 + (1.9 Pr )Hl fits the analysis of Fujii et al. (1973).© 1999 by CRC Press LLC9 10 2 9]Nu =2.45ln 1 + 2.45 Nu T()(4.2.10)4-17Heat and Mass Transfer5.
Inclined Plates, Downward Facing (–90° £ f £ 0), Figure 4.2.1B. First calculate q from Case 1with g replaced by g cos f; then calculate q from Case 4 (horizontal plate) with g replaced by gsin (–f), and use the maximum of these two values of q.6. Inclined Plates, Upward Facing (0 £ f £ 90), Figure 4.2.1B.
First calculate q from Case 1 withg replaced by g cos f; then calculate q from Case 3 with g replaced by g sin f, and use themaximum of these two values of q.7. Vertical and Tilted Isothermal Plates of Various Planform, Figure 4.2.2. The line of constant cin Figure 4.2.2 is the line of steepest ascent on the plate. Provided all such lines intersect theplate edges just twice, as shown in the figure, the thin-layer (NuT) heat transfer can be found bysubdividing the body into strips of width Dc, calculating the heat transfer from each strip, andadding. For laminar flow from an isothermal vertical plate, this results inNu T = C1Cl Ra 1 4æ L1 4C1 º çè AWòS340ödc÷ø(4.2.11)Symbols are defined in Figure 4.2.2, along with L and calculated C1 values for some plate shapes.If the plate is vertical, follow the procedure in Case 1 above (isothermal vertical flat plate) exceptreplace the expression for NuT in Equation (4.2.7) by Equation (4.2.11).
If the plate is tilted,follow the procedure described in Case 5 or 6 (as appropriate) but again use Equation (4.2.11)for NuT in Equation (4.2.7)8. Horizontal Cylinders, Figure 4.2.3A. For a long, horizontal circular cylinder use the followingexpressions for Nul and Nut:Nu T = 0.772Cl Ra 1 4Nu l =2f1 + 2 f Nu T(Nu t = Ct Ra 1 3)(4.2.12)Ct is given in the table below. For Ra > 10–2, f = 0.8 can be used, but for 10–10 < Ra < 10–2 usef = 1 – 0.13/(NuT)0.16. To find Nu, the values of Nul and Nut from Equation (4.2.12) are substitutedinto Equation (4.2.6) with m = 15 (Clemes et al., 1994).Ct for Various Shapes and Prandtl NumbersPr®0.010.0220.100.712.06.0501002000Horizontal cylinderSpheres0.0770.0740.810.0780.900.0880.1030.1040.1080.1100.1090.1110.1000.1010.0970.970.0880.0869.
Vertical Cylinders (f = 90°), Figure 4.2.3B. For high Ra values and large diameter, the heattransfer from a vertical cylinder approaches that for a vertical flat plate. Let the NuT and Nulequations for a vertical flat plate of height L, Equation (4.2.7), be rewritten here as Nu Tp andNup, respectively. At smaller Ra and diameter, transverse curvature plays a role which is accountedfor in the following equations:Nu l =0.9xNu pln(1 + 0.9x)x=2L DNu Tp(4.2.13)These equations are valid for purely laminar flow.
To obtain Nu, blend Equation (4.2.13) for Nulwith Equation (4.2.7) for Nut using Equation (4.2.6) with m = 10.© 1999 by CRC Press LLC4-18Section 410. Spheres, Figure 4.2.3C. For spheres use Equation (4.2.6), with m = 6, and withNu l = 2 + 0.878Cl Ra 1 4 and Nu t = Ct Ra 1 3(4.2.14)The table above contains Ct values.11. Combined Shapes, Figure 4.2.3D.
For combined shapes, such as the cylinder in Figure 4.2.3Dwith spherical end caps, calculate the heat transfer from the cylinder of length L (Case 8), theheat transfer from a sphere of diameter D (Case 10) and add to obtain the total transfer. Othershapes can be treated in a similar manner.Correlations for Open CavitiesExamples of this class of problem are shown in Figure 4.2.4. Walls partially enclose a fluid region(cavity) where boundary openings permit fluid to enter and leave.
Upstream from its point of entry, thefluid is at the ambient temperature, T¥. Since access of the ambient fluid to the heated surfaces isrestricted, some of the heated surface is starved of cool ambient to which heat can be transferred. Asthe sizes of the boundary openings are increased, the previous class of problems is approached; forexample, when the plate spacing in Figure 4.2.4A (Case 12) becomes very large, the heat transfer fromeach vertical surface is given by Case 1.FIGURE 4.2.4 Nomenclature for various open-cavity problems.12.