John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 69
Текст из файла (страница 69)
Approximately what is the maximumspacing?Solution. The mercury is most susceptible to Taylor instabilitywhen the spacing reaches the wavelength given by eqn. (9.6):2λd1√= 2π 32√σ= 2π 3g(ρf − ρg )0.487= 0.021 m = 2.1 cm9.8(13600)(Actually, this spacing would give the maximum√ rate of collapse. Itcan be shown that collapse would begin at 1 3 times this value, orat 1.2 cm.)Prediction of qmaxGeneral expression for qmax The heat flux must be balanced by thelatent heat carried away in the jets when the liquid is saturated. Thus,we can write immediatelyqmax = ρg hfg ugAjAh(9.9)where Aj is the cross-sectional area of a jet and Ah is the heater area thatsupplies each jet.For any heater configuration, two things must be determined.
Oneis the length of the particular disturbance in the jet wall, λH , which willtrigger Helmholtz instabilityand fix ug in eqn. (9.8) for use in eqn. (9.9).The other is the ratio Aj Ah . The prediction of qmax in any pool boilingconfiguration always comes down to these two problems.qmax on an infinite horizontal plate. The original analysis of this typewas done by Zuber in his doctoral dissertation at UCLA in 1958 (see [9.17]).He first guessed that the jet radius was λd1 4. This guess has receivedcorroboration by subsequent investigators, and (with reference to Fig. 9.9)Peak pool boiling heat flux§9.3479it givesAjcross-sectional area of circular jetAharea of the square portion of the heater that feeds the jetπ (λd1 /4)2π(9.10)==(λd1 )216=Lienhard and Dhir ([9.18, 9.19, 9.20]) guessed that the Helmholtz-unstablewavelength might be equal to λd1 , so eqn.
(9.9) became>2?? 2π σg(ρf − ρg )π1@√×qmax = ρg hfgρg 2π 3σ16or31/2qmax = 0.149 ρg hfg44g(ρf − ρg )σ(9.11)Equation (9.11) is compared with available data for large flat heaters,with vertical sidewalls to prevent any liquid sideflow, in Fig. 9.12. Solong as the diameter or width of the heater is more than about 3λd1 , theprediction is quite accurate. When the width or diameter is less thanthis, there is a small integral number of jets on a plate which may belarger or smaller in area than 16/π per jet.
When this is the case, theactual qmax may be larger or smaller than that predicted by eqn. (9.11)(see Problem 9.13).The form of the preceding prediction is usually credited to Kutateladze [9.21] and Zuber [9.17]. Kutateladze (then working in Leningradand later director of the Heat Transfer Laboratory near Novosibirsk, Siberia) recognized that burnout resembled the flooding of a distillationcolumn. At any level in a distillation column, alcohol-rich vapor (for example) rises while water-rich liquid flows downward in counterflow. Ifthe process is driven too far, the flows become Helmholtz-unstable andthe process collapses.
The liquid then cannot move downward and thecolumn is said to “flood.”Kutateladze did the dimensional analysis of qmax based on the flooding mechanism and obtained the following relationship, which, lacking acharacteristic length and being of the same form as eqn. (9.11), is reallyvalid only for an infinite horizontal plate:4 1/2qmax = C ρg hfg 4 g ρf − ρg σ3Readers are reminded that√nx ≡ x 1/n .480Heat transfer in boiling and other phase-change configurations§9.3Figure 9.12 Comparison of the qmax prediction for infinitehorizontal heaters with data reported in [9.18].He then suggested that C was equal to 0.131 on the basis of data fromconfigurations other than infinite flat plates (horizontal cylinders, for example).
Zuber’s analysis yielded C = π /24 = 0.1309, which was quiteclose to Kutateladze’s value but lower by 14% than eqn. (9.11). We therefore designate the Zuber-Kutateladze prediction as qmaxz . However, weshall not use it directly, since it does not predict any actual physical configuration.4 1/2(9.12)qmaxz ≡ 0.131 ρg hfg 4 g ρf − ρg σIt is very interesting that C.
F. Bonilla, whose qmax experiments in theearly 1940s are included in Fig. 9.12, also suggested that qmax shouldbe compared with the column-flooding mechanism. He presented theseideas in a paper, but A. P. Colburn wrote to him: “A correlation [of theflooding velocity plots with] boiling data would not serve any great purpose and would perhaps be very misleading.” And T.
H. Chilton—anothereminent chemical engineer of that period—wrote to him: “I venture tosuggest that you delete from the manuscript…the relationship betweenboiling rates and loading velocities in packed towers.” Thus, the technicalconservativism of the period prevented the idea from gaining acceptancefor another decade.Peak pool boiling heat flux§9.3Example 9.5Predict the peak heat flux for Fig. 9.2.Solution.
We use eqn. (9.11) to evaluate qmax for water at 100◦ C onan infinite flat plate:44g(ρf − ρg )σ44= 0.149(0.597)1/2 (2, 257, 000) 9.8(958.2 − 0.6)(0.0589)1/2qmax = 0.149 ρg hfg= 1.260 × 106 W/m2= 1.260 MW/m2Figure 9.2 shows qmax 1.160 MW/m2 , which is less by only about8%.Example 9.6What is qmax in mercury on a large flat plate at 1 atm?Solution. The normal boiling point of mercury is 355◦ C.
At this temperature, hfg = 292, 500 J/kg, ρf = 13, 400 kg/m3 , ρg = 4.0 kg/m3 ,and σ 0.418 kg/s2 , so44qmax = 0.149(4.0)1/2 (292, 500) 9.8(13, 400 − 4)(0.418)= 1.334 MW/m2The result is very close to that for water. The increases in density andsurface tension have been compensated by a much lower latent heat.Peak heat flux in other pool boiling configurationsThe prediction of qmax in configurations other than an infinite flat heaterwill involve a characteristic length, L.
Thus, the dimensional functionalequation for qmax becomes qmax = fn ρg , hfg , σ , g ρf − ρg , Lwhich involves six variables and four dimensions: J, m, s, and kg, where,once more in accordance with Section 4.3, we note that no significantconversion from work to heat is occurring so that J must be retainedas a separate unit. There are thus two pi-groups. The first group can481482Heat transfer in boiling and other phase-change configurations§9.3arbitrarily be multiplied by 24/π to giveqmaxqmax4Π1 ==1/2qmaxz(π /24) ρg hfg 4 σ g(ρf − ρg )(9.13)Notice that the factor of 24/π has served to make the denominator equalto qmaxz (Zuber’s expression for qmax ).
Thus, for qmax on a flat plate, Π1equals 0.149/0.131, or 1.14. The second pi-group is√ LLΠ2 = 4 = 2π 3≡ L(9.14)λd1σ g(ρf − ρg )The latter group, Π2 , is the square root of the Bond number, Bo, which isused to compare buoyant force with capillary forces.Predictions and correlations of qmax have been made for several finitegeometries in the form qmax= fn L(9.15)qmaxzThe dimensionless characteristic length in eqn. (9.15) might be a dimensionless radius (R ), a dimensionless diameter (D ), or a dimensionlessheight (H ). The graphs in Fig. 9.13 are comparisons of several of theexisting predictions and correlations with experimental data.
These predictions and others are listed in Table 9.3. Notice that the last three itemsin Table 9.3 (10, 11, and 12) are general expressions from which severalof the preceding expressions in the table can be obtained.The equations in Table 9.3 are all valid within ±15% or 20%, which isvery little more than the inherent scatter of qmax data. However, they aresubject to the following conditions:• The bulk liquid is saturated.• There are no pathological surface imperfections.• There is no forced convection.Another limitation on all the equations in Table 9.3 is that neither thesize of the heater nor the relative force of gravity can be too small.
WhenL < 0.15 in most configurations, the Bond number isBo ≡ L =2g(ρf − ρg )L3σL=buoyant forcecapillary force<144In this case, the process becomes completely dominated by surface tension and the Taylor-Helmholtz wave mechanisms no longer operate. AsL is reduced, the peak and minimum heat fluxes cease to occur and theFigure 9.13 The peak pool boiling heat flux on several heaters.483484Small flat heaterHorizontal cylinderLarge horizontal cylinderSmall horizontal cylinderLarge sphereSmall sphere2.3.4.5.6.7.√−3.44 R 1/4 1/2Constant (L )1/2Characteristiclength, LTransverseperimeter, P1.4/(P )1/411.
Small slender cylinderof any cross section12. Small bluff bodyCharacteristiclength, LHeight of side, HHeight of side, HSphere radius, RSphere radius, RCylinder radius, RCylinder radius, RCylinder radius, RHeater width or diameterHeater width or diameterBasis for L∼ 0.901.4/(H ) 1/41.18/(H )1/41.734/(R )0.840.94/(R )0.900.89 + 2.27e1.14(λd1 /Aheater )1.14qmax /qmaxz10. Any large finite body9. 1 side insulated8. plainSmall horizontal ribbonoriented verticallyInfinite flat heater1.Situationcannot specifygenerally; L 40.15 ≤ P ≤ 5.86cannot specifygenerally; L 40.15 ≤ H ≤ 5.860.15 ≤ H ≤ 2.960.15 ≤ R ≤ 4.26R ≥ 4.260.15 ≤ R ≤ 1.2R ≥ 1.2R ≥ 0.159 < L < 20[9.20][9.20][9.20][9.20][9.20][9.23][9.23][9.20][9.20][9.22][9.19][9.19]L ≥ 27SourceRange of LTable 9.3 Predictions of the peak pool boiling heat flux(9.27)(9.26)(9.25)(9.24)(9.23)(9.22)(9.21)(9.20)(9.19)(9.18)(9.17)(9.16)Eqn.
No.Peak pool boiling heat flux§9.3boiling curve becomes monotonic. When nucleation occurs on a verysmall wire, the wire is immediately enveloped in vapor and the mechanism of heat removal passes directly from natural convection to filmboiling.Example 9.7A spheroidal metallic body of surface area 400 cm2 and volume 600cm3 is quenched in saturated water at 1 atm. What is the most rapidrate of heat removal during the quench?Solution. As the cooling process progresses, it goes through theboiling curve from film boiling, through qmin , up the transitional boiling regime, through qmax , and down the nucleate boiling curve. Cooling is finally completed by natural convection.
One who has watchedthe quenching of a red-hot horseshoe will recall the great gush ofbubbling that occurs as qmax is reached. We therefore calculate therequired heat flow as Q = qmax Aspheroid , where qmax is given by eqn.(9.25) in Table 9.3:1/2qmax = 0.9 qmaxz = 0.9(0.131)ρg hfg44gσ (ρf − ρg )so1/2Q = 0.9(0.131)(0.597)× 400 × 10−4 m2442(2, 257, 000) 9.8(0.0589)(958) W/morQ = 39, 900 W or 39.9 kWThis is a startingly large rate of energy removal for such a small object.To complete the calculation, it is necessary to check whether ornot R is large enough to justify the use of eqn.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
