John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 63
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g(x) is the component of gravity or other body force in the x-direction.437438Natural convection in single-phase fluids and during film condensation§8.5• R(x) is a radius of curvature about the vertical axis. In Fig. 8.13a, itis a constant that factors out of eqn. (8.61). In Fig. 8.13c, R is infinite.Since it appears to the same power in both the numerator and thedenominator, it again can be factored out of eqn. (8.61). Only inaxisymmetric bodies, where R varies with x, need it be included.When it can be factored out,xg 4/3geff reduces to xg 1/3 dx(8.62)0• ge is earth-normal gravity. We introduce ge at this point to distinguish it from g(x).Example 8.7Find Nux for laminar film condensation on the top of a flat surfacesloping at θ ◦ from the vertical plane.Solution.
In this case g = ge cos θ and R = ∞. Therefore, eqn. (8.61)or (8.62) reduces to4/3geff(cos θ)4/3x= ge cos θ=1/31/3ge (cos θ)dxxge0as we might expect. Then, for a slanting plate,⎤1/4⎡ρf (ρf − ρg )(ge cos θ)hfg x 3⎦Nux = 0.707 ⎣µk(Tsat − Tw )(8.63)Example 8.8Find the overall Nusselt number for a horizontal cylinder.Solution. There is an important conceptual hurdle here. The radiusR(x) is infinity, as shown in Fig. 8.13c—it is not the radius of the cylinder.
It is also very easy to show that g(x) is equal to ge sin(2x/D),where D is the diameter of the cylinder. Then4/3geff =xge (sin 2x/D)4/3x1/3ge(sin 2x/D)1/3 dx0Film condensation§8.5439and, with h(x) from eqn. (8.57),⌠ π D/22 ⎮1 k⎮√⎮h=πD ⌡2x0⎡⎤1/4 ⎢ ρf ρf − ρg h x 3 xg (sin 2x/D)4/3 ⎥fg⎥⎢x e⎥⎢⎦⎣ µk (Tsat − Tw )1/3dx(sin 2x/D)dx0This integral can be evaulated in terms of gamma functions. Theresult, when it is put back in the form of a Nusselt number, is⎤1/4 ρf ρf − ρg ge hfg D 3⎦NuD = 0.728 ⎣µk (Tsat − Tw )⎡(8.64)for a horizontal cylinder. (Nusselt got 0.725 for the lead constant, buthe had to approximate the integral with a hand calculation.)Some other results of this calculation include the following cases.Sphere of diameter D:⎤1/4 ρf ρf − ρg ge hfg D 3⎦NuD = 0.828 ⎣µk (Tsat − Tw )⎡(8.65)This result9 has already been compared with the experimental data inFig.
8.10.Vertical cone with the apex on top, the bottom insulated, and a coneangle of α◦ :⎡⎤ 3 1/4ρ−ρhxgρgefffg⎦Nux = 0.874 [cos(α/2)]1/4 ⎣µk (Tsat − Tw )(8.66)Rotating horizontal disk 10 : In this case, g = ω2 x, where x is thedistance from the center and ω is the speed of rotation. The Nusseltnumber, based on L = (µ/ρf ω)1/2 , is⎤1/4⎡ µ ρf − ρg hfg⎦Nu = 0.9034 ⎣= constantρf k (Tsat − Tw )9(8.67)There is an error in [8.33]: the constant given there is 0.785. The value of 0.828given here is correct.10This problem was originally solved by Sparrow and Gregg [8.38].440Natural convection in single-phase fluids and during film condensation§8.5This result might seem strange at first glance.
It says that Nu ≠ fn(x or ω).The reason is that δ just happens to be independent of x in this configuration.The Nusselt solution can thus be bent to fit many complicated geometric figures. One of the most complicated ones that have been dealtwith is the reflux condenser shown in Fig. 8.14. In such a configuration,cooling water flows through a helically wound tube and vapor condenseson the outside, running downward along the tube.
As the condensateflows, centripetal forces sling the liquid outward at a downward angle.This complicated flow was analyzed by Karimi [8.39], who found that⎡ ⎤3 1/4dhd cos α ⎣ ρf − ρg ρf hfg g(d cos α) ⎦=,BNu ≡fnkµk∆TD(8.68)where B is a centripetal parameter:B≡ρf − ρg cp ∆T tan2 αρfhfgPrand α is the helix angle (see Fig. 8.14).
The function on the righthand sideof eqn. (8.68) was a complicated one that must be evaluated numerically.Karimi’s result is plotted in Fig. 8.14.Laminar–turbulent transitionThe mass flow rate of condensate per unit width of film, ṁ, is more commonly designated as Γc (kg/m · s). Its calculation in eqn.
(8.50) involvedsubstituting eqn. (8.48) inδṁ or Γc = ρfu dy0Equation (8.48) gives u(y) independently of any geometric features. [Thegeometry is characterized by δ(x).] Thus, the resulting equation for themass flow rate is still ρf ρf − ρg gδ3(8.50a)Γc =3µThis expression is valid for any location along any film, regardless of thegeometry of the body. The configuration will lead to variations of g(x)and δ(x), but eqn. (8.50a) still applies.Film condensation§8.5441Figure 8.14 Fully developed film condensation heat transferon a helical reflux condenser [8.39].It is useful to define a Reynolds number in terms of Γc . This is easyto do, because Γc is equal to ρuav δ.Rec =ρf (ρf − ρg )gδ3Γc=µ3µ 2(8.69)It turns out that the Reynolds number dictates the onset of film instability, just as it dictates the instability of a b.l.
or of a pipe flow.11 WhenRec 7, scallop-shaped ripples become visible on the condensate film.When Rec reaches about 400, a full-scale laminar-to-turbulent transitionoccurs.Gregorig, Kern, and Turek [8.40] reviewed many data for the filmcondensation of water and added their own measurements. Figure 8.15shows these data in comparison with Nusselt’s theory, eqn.
(8.60). Thecomparison is almost perfect up to Rec 7. Then the data start yieldingsomewhat higher heat transfer rates than the prediction. This is becauseTwo Reynolds numbers are defined for film condensation: Γc /µ and 4Γc /µ. Thelatter one, which is simply four times as large as the one we use, is more common inthe American literature.11442Natural convection in single-phase fluids and during film condensation§8.5Figure 8.15 Film condensation on vertical plates.
Data are forwater [8.40].the ripples improve heat transfer—just a little at first and by about 20%when the full laminar-to-turbulent transition occurs at Rec = 400.Above Rec = 400, NuL begins to rise with Rec . The Nusselt numberbegins to exhibit an increasingly strong dependence on the Prandtl number in this turbulent regime. Therefore, one can use Fig.
8.15, directly asa data correlation, to predict the heat transfer coefficient for steam condensating at 1 atm. But for other fluids with different Prandtl numbers,one should consult [8.41] or [8.42].Two final issues in natural convection film condensation• Condensation in tube bundles. Nusselt showed that if n horizontaltubes are arrayed over one another, and if the condensate leaveseach one and flows directly onto the one below it without splashing,thenNuDforn tubes=NuD1 tuben1/4(8.70)This is a fairly optimistic extension of the theory, of course. Inaddition, the effects of vapor shear stress on the condensate and ofpressure losses on the saturation temperature are often importantin tube bundles. These effects are discussed by Rose et al.
[8.42]and Marto [8.41].Problems443• Condensation in the presence of noncondensable gases. When thecondensing vapor is mixed with noncondensable air, uncondensedair must constantly diffuse away from the condensing film and vapor must diffuse inward toward the film. This coupled diffusionprocess can considerably slow condensation.
The resulting h caneasily be cut by a factor of five if there is as little as 5% by massof air mixed into the steam. This effect was first analyzed in detailby Sparrow and Lin [8.43]. More recent studies of this problem arereviewed in [8.41, 8.42].Problems8.1Show that Π4 in the film condensation problem can properlybe interpreted as Pr Re2 Ja.8.2A 20 cm high vertical plate is kept at 34◦ C in a 20◦ C room.Plot (to scale) δ and h vs. height and the actual temperatureand velocity vs.
y at the top.8.3Redo the Squire-Eckert analysis, neglecting inertia, to get ahigh-Pr approximation to Nux . Compare your result with theSquire-Eckert formula.8.4Assume a linear temperature profile and a simple triangularvelocity profile, as shown in Fig. 8.16, for natural convectionon a vertical isothermal plate. Derive Nux = fn(Pr, Grx ), compare your result with the Squire-Eckert result, and discuss thecomparison.8.5A horizontal cylindrical duct of diamond-shaped cross section(Fig. 8.17) carries air at 35◦ C.
Since almost all thermal resistance is in the natural convection b.l. on the outside, take Twto be approximately 35◦ C. T∞ = 25◦ C. Estimate the heat lossper meter of duct if the duct is uninsulated. [Q = 24.0 W/m.]8.6The heat flux from a 3 m high electrically heated panel in awall is 75 W/m2 in an 18◦ C room.
What is the average temperature of the panel? What is the temperature at the top? at thebottom?444Chapter 8: Natural convection in single-phase fluids and during film condensationFigure 8.16 Configuration for Problem 8.4.Figure 8.17 Configuration forProblem 8.5.8.7Find pipe diameters and wall temperatures for which the filmcondensation heat transfer coefficients given in Table 1.1 arevalid.8.8Consider Example 8.6.
What value of wall temperature (if any),or what height of the plate, would result in a laminar-to-turbulenttransition at the bottom in this example?8.9A plate spins, as shown in Fig. 8.18, in a vapor that rotates synchronously with it. Neglect earth-normal gravity and calculateNuL as a result of film condensation.8.10A laminar liquid film of temperature Tsat flows down a verticalwall that is also at Tsat . Flow is fully developed and the filmthickness is δo . Along a particular horizontal line, the walltemperature has a lower value, Tw , and it is kept at that temperature everywhere below that position.
Call the line wherethe wall temperature changes x = 0. If the whole system isProblems445Figure 8.18 Configuration forProblem 8.9.immersed in saturated vapor of the flowing liquid, calculateδ(x), Nux , and NuL , where x = L is the bottom edge of thewall. (Neglect any transition behavior in the neighborhood ofx = 0.)8.11Prepare a table of formulas of the formh (W/m2 K) = C [∆T ◦ C/L m]1/4for natural convection at normal gravity in air and in waterat T∞ = 27◦ C. Assume that Tw is close to 27◦ C. Your tableshould include results for vertical plates, horizontal cylinders,spheres, and possibly additional geometries.
Do not includeyour calculations.8.12For what value of Pr is the conditiongβ(Tw − T∞ )∂2u =2∂y y=0νsatisfied exactly in the Squire-Eckert b.l. solution? [Pr = 2.86.]8.13The overall heat transfer coefficient on the side of a particularhouse 10 m in height is 2.5 W/m2 K, excluding exterior convection. It is a cold, still winter night with Toutside = −30◦ C andTinside air = 25◦ C. What is h on the outside of the house? Isexternal convection laminar or turbulent?8.14Consider Example 8.2.
The sheets are mild steel, 2 m long and6 mm thick. The bath is basically water at 60◦ C, and the sheets446Chapter 8: Natural convection in single-phase fluids and during film condensationare put in it at 18◦ C. (a) Plot the sheet temperature as a functionof time. (b) Approximate h at ∆T = [(60 + 18)/2 − 18]◦ C andplot the conventional exponential response on the same graph.8.15A vertical heater 0.15 m in height is immersed in water at 7◦ C.Plot h against (Tw − T∞ )1/4 , where Tw is the heater temperature, in the range 0 < (Tw − T∞ ) < 100◦ C.
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