Exercícios Resolvidos Incropera capítulos 10 ao 14

Exercícios Resolvidos Incropera capítulos 10 ao 14

(Parte 1 de 11)

PROBLEM 10.1

KNOWN: Water at 1 atm with Ts – Tsat = 10°C.

FIND: Show that the Jakob number is much less than unity; what is the physical significance of the result; does result apply to other fluids?

ASSUMPTIONS: (1) Boiling situation, Ts > Tsat. PROPERTIES: Table A-5 and Table A-6, (1 atm):

Water 22572029 373
Ethylene glycol812 2742* 470
Mercury301 135.5* 630
R-12165 1015* 243

hfg (kJ/kg)cp,v (J/kg×K)Tsat(K)

* Estimated based upon value at highest temperature cited in Table A-5.

ANALYSIS: The Jakob number is the ratio of the maximum sensible energy absorbed by the vapor to the latent energy absorbed by the vapor during boiling. That is,

( )p fg p,ve fgv JacT /hc T /h= D =D

For water with an excess temperature DTs = Te - T¥ = 10°C, find

Since Ja << 1, the implication is that the sensible energy absorbed by the vapor is much less than the latent energy absorbed during the boiling phase change. Using the appropriate thermophysical properties for three other fluids, the Jakob numbers are:

For ethylene glycol and R-12, the Jakob number is larger than the value for water, but still much less than unity. Based upon these example fluids, we conclude that generally we’d expect Ja to be much less than unity.

COMMENTS: We would expect the same low value of Ja for the condensation process since cp,g and cp,f are of the same order of magnitude.

PROBLEM 10.2

KNOWN: Horizontal 20 m diameter cylinder with DTe = Ts – Tsat = 5°C in saturated water, 1 atm.

FIND: Heat flux based upon free convection correlation; compare with boiling curve. Estimate maximum value of the heat transfer coefficient from the boiling curve.

ASSUMPTIONS: (1) Horizontal cylinder, (2) Free convection, no bubble information.

PROPERTIES: Table A-6, Water (Saturated liquid, Tf = (Tsat + Ts)/2 = 102.5 °C » 375K): rl = 956.9 kg/m3 , p,cl =

ANALYSIS: To estimate the free convection heat transfer coefficient, use the Churchill-Chu correlation, hD 0.387Ra Nu 0.60.

Substituting numerical values, with DT = DTe = 5°C, find

where a = k/r cp = (0.681 W/m×K/956.9 kg/m3 · 4220 J/kg×K) = 1.686 · 10-7 m2 /s. Note that RaD is within the prescribed limits of the correlation. Hence,

k 27.2 0.681W/ mK h Nu 928W/ m K.

Hence, 2 s fceqhT 4640W/m¢¢ =D=

From the typical boiling curve for water at 1 atm, Fig. 10.4, find at DTe = 5°C that

The free convection correlation underpredicts (by 1.8) the boiling curve. The maximum value of hbc can be estimated as

COMMENTS: (1) Note the large increase in h with a slight change in DTe. (2) The maximum value of h occurs at point P on the boiling curve.

PROBLEM 10.3

KNOWN: Spherical bubble of pure saturated vapor in mechanical and thermal equilibrium with its liquid.

FIND: (a) Expression for the bubble radius, (b) Bubble vapor and liquid states on a p-v diagram; how changes in these conditions cause bubble to collapse or grow, and (c) Bubble size for specified conditions.

ASSUMPTIONS: (1) Liquid-vapor medium, (2) Thermal and mechanical equilibrium.

PROPERTIES: Table A-6, Water (Tsat = 101°C = 374.15K): psat = 1.0502 bar; (Tsat = 100°C =

373.15K): psat = 1.0133 bar, s = 58.9 · 10-3 N/m.

ANALYSIS: (a) For mechanical equilibrium, the difference in pressure between the vapor inside the bubble and the liquid outside the bubble will be offset by the surface tension of the liquid-vapor interface. The force balance follows from the free-body diagram shown above (right),

Thermal equilibrium requires that the temperatures of the vapor and liquid be equal. Since the vapor inside the bubble is saturated, pi = psat,v (T). Since po < pi, it follows that the liquid outside the bubble must be superheated; hence, po = pl (T), the pressure of superheated liquid at T. Hence, we can write,

()bsat,vr2/pps=-l(2) <

(b) The vapor [1] and liquid [2] states are represented on the following p-v diagram. Thermal equilibrium requires both the vapor and liquid to be at the same temperature [3]. But mechanical equilibrium requires that the outside liquid pressure be less than the inside vapor pressure [4]. Hence the liquid must be in a superheated state. That is, its saturation temperature, Tsat(po) [5] is less than Tsat(pi); Tl = Tsat(po) and po = p.l

Continued …..

PROBLEM 10.3 (Cont.)

The equilibrium condition for the bubble is unstable. Consider situations for which the pressure of the surrounding liquid is greater or less than the equilibrium value. These situations are presented on portions of the p-v diagram

When oosat,vpp,T¢<<l and heat must be transferred out of the bubble and vapor condenses. Hence, the bubble collapses.

A similar argument for the condition oopp¢> leads to sat,vTT¢>l and heat is transferred into the bubble causing evaporation with the formation of vapor. Hence, the bubble begins to grow.

(c) Consider the specific conditions

m m

Note the small bubble size. This implies that nucleation sites of the same magnitude formed by pits and crevices are important in promoting the boiling process.

PROBLEM 10.4

KNOWN: Long wire, 1 m diameter, reaches a surface temperature of 126°C in water at 1 atm while dissipating 3150 W/m.

FIND: (a) Boiling heat transfer coefficient and (b) Correlation coefficient, Cs,f, if nucleate boiling occurs.

ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate boiling.

PROPERTIES: Table A-6, Water (saturated, 1 atm): Ts = 100°C, rl = 1/vf = 957.9 kg/m3, rf =

1/vg = 0.5955 kg/m3

, p,cl = 4217 J/kg×K, ml = 279 · 10-6 N×s/m2 , Prl = 1.76, hfg = 2257 kJ/kg, s =

ANALYSIS: (a) For the boiling process, the rate equation can be rewritten as h q/ T T / T

(Parte 1 de 11)

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