Heat Chap14134.docx

1、Heat Chap1413414-134 A circular pan filled with water is cooled naturally. The rate of evaporation of water, the rate of heat transfer by natural convection, and the rate of heat supply to the water needed to maintain its temperature constant are to be determined. Assumptions 1 The low mass flux mod

2、el and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 25C). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Radiation heat transfer is negligible. 4 Both air and water vapo

3、r are ideal gases.Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture at the average temperature of = (15+20)/2 = 17.5C = 290.5 K. The properties of dry air at 290.5 K and 1 atm are, from Table A-15, The mass diffusivity of water vap

4、or in air at the average temperature of 290.5 K is, from Eq. 14-15, The saturation pressure of water at 20C is Properties of water at 15C are (Table A-9). The specific heat of water at the average temperature of (15+20)/2 = 17.5C is Cp = 4.184 kJ/kg.C. The gas constants of dry air and water are Rair

5、 = 0.287 kPa.m3/kg.K and Rwater = 0.4615 kPa.m3/kg.K (Table A-1). Analysis (a) The air at the water surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (1.7051 kPa at 15C). The vapor pressure of air far from the water

6、 surface is determined from Treating the water vapor and the air as ideal gases and noting that the total atmospheric pressure is the sum of the vapor and dry air pressures, the densities of the water vapor, dry air, and their mixture at the water-air interface and far from the surface are determine

7、d to be At the surface: andAway from the surface: Note that, and thus this corresponds to hot surface facing down. The area of the top surface of the water and its perimeter is. Therefore, the characteristic length is Then using densities (instead of temperatures) since the mixture is not homogeneou

8、s, the Grashoff number is determined to be Recognizing that this is a natural convection problem with cold horizontal surface facing up, the Nusselt number and the convection heat transfer coefficients are determined to be (Eq. 14-13) and Then the rate of heat transfer from the air to the water by f

9、orced convection becomes (to water)(b) Utilizing the analogy between heat and mass convection, the mass transfer coefficient is determined the same way by replacing Pr by Sc. The Schmidt number is determined from its definition to be Therefore, the Sherwood number in this case is determined from Tab

10、le 14-13 to be Using the definition of Sherwood number, the mass transfer coefficient is determined to be Then the evaporation rate and the rate of heat transfer by evaporation become and (c) The net rate of heat transfer to the water needed to maintain its temperature constant at 15C is Discussion

11、Note that if no heat is supplied to the water (by a resistance heater, for example), the temperature of the water in the pan would drop until the heat gain by convection equals the heat loss by evaporation.14-135 Air is blown over a circular pan filled with water. The rate of evaporation of water, t

12、he rate of heat transfer by convection, and the rate of energy supply to the water to maintain its temperature constant are to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (ab

13、out 2 percent for saturated air at 25C). 2 The critical Reynolds number for flow over a flat plate is 500,000. 3 Radiation heat transfer is negligible. 4 Both air and water vapor are ideal gases.Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties fo

14、r the mixture at the average temperature of = (15+20)/2 = 17.5C = 290.5 K. The properties of dry air at 290.5 K and 1 atm are, from Table A-15, The mass diffusivity of water vapor in air at the average temperature of 290.5 K is, from Eq. 14-15,The saturation pressure of water at 20C is Properties of

15、 water at 15C are (Table A-9). Also, the gas constants of water is Rwater = 0.4615 kPa.m3/kg.K (Table A-1). Analysis (a) Taking the radius of the pan r0 = 0.15 m to be the characteristic length, the Reynolds number for flow over the pan is which is less than 500,000, and thus the flow is laminar ove

16、r the entire surface. The Nusselt number and the heat transfer coefficient are Then the rate of heat transfer from the air to the water by forced convection becomes (to water)(b) Utilizing the analogy between heat and mass convection, the mass transfer coefficient is determined the same way by repla

17、cing Pr by Sc. The Schmidt number is determined from its definition to be Therefore, the Sherwood number in this case is determined from Table 14-13 to be Using the definition of Sherwood number, the mass transfer coefficient is determined to be The air at the water surface is saturated, and thus th

18、e vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (1.7051 kPa at 15C). The vapor pressure of air far from the water surface is determined from Treating the water vapor and the air as ideal gases, the vapor densities at the water-air interface and f

19、ar from the surface are determined to beAt the surface: Away from the surface: Then the evaporation rate and the rate of heat transfer by evaporation become and (c) The net rate of heat transfer to the water needed to maintain its temperature constant at 15C is Discussion Note that if no heat is sup

20、plied to the water (by a resistance heater, for example), the temperature of the water in the pan would drop until the heat gain by convection equals the heat loss by evaporation. Also, the rate of evaporation could be determined almost as accurately using mass fractions of vapor instead of vapor fr

21、actions and the average air density from the relation.14-136 A spherical naphthalene ball is hanged in a closet. The time it takes for the naphthalene to sublimate completely is to be determined. Assumptions 1 The concentration of naphthalene in the air is very small, and the low mass flux condition

22、s exist so that the Chilton-Colburn analogy between heat and mass transfer is applicable (will be verified). 2 Both air and naphthalene vapor are ideal gases. 3 The naphthalene and the surrounding air are at the same temperature. 4 The radiation effects are negligible.Properties The molar mass of na

23、phthalene is 128.2 kg/kmol. Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 298 K and 1 atm, at which, , and (Table A-15).Analysis The incoming air is free of naphthalene, and thus the mass fraction of naphthalene at free stream cond

24、itions is zero, wA, = 0. Noting that the vapor pressure of naphthalene at the surface is 11 Pa, the mass fraction of naphthalene on the air side of the surface is Normally we would expect natural convection currents to develop around the naphthalene ball because the amount of naphthalene near the su

25、rface is much larger, and determine the Nusselt number (and its counterpart in mass transfer, the Sherwood number) from Eq. 14-16, But the mass fraction value determined above indicates that the amount of naphthalene in the air is so low that it will not cause any significant difference in the densi

26、ty of air. With no density gradient, there will be no natural convection and thus the Rayleigh number can be taken to be zero. Then the Nusselt number relation above will reduce to Nu = 2 or its equivalent Sh = 2. Then using the definition of Sherwood number, the mass transfer coefficient can be exp

27、ressed as The mass of naphthalene ball can be expressed as. The rate of change of the mass of naphthalene is equal to the rate of mass transfer from naphthalene to the air, and is expressed as Simplifying and rearranging, Integrating from at time t = 0 to D = 0 (complete sublimation) at time t = t g

28、ives Substituting, the time it takes for the naphthalene to sublimate completely is determined to be 14-137E A swimmer extends his wet arms into the windy air outside. The rate at which water evaporates from both arms and the corresponding rate of heat transfer by evaporation are to be determined. A

29、ssumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 60F). 2 The arm can be modeled as a long cylinder.Properties Because of low mass flux conditions, we can use

30、 dry air properties for the mixture at the average temperature of (40 + 80)/2 = 60F and 1 atm, for which = 0.15910-3 ft2/s , and = 0.077lbm / ft3 (Table A-15E). The saturation pressure of water at 40F is 0.1217 psia. Also, at 80F, the saturation pressure is 0.5073 psia and the heat of vaporization i

31、s 1048 Btu/lbm (Table A-9E). The molar mass of water is R = 0.5956 psia.ft3/lbm.R (Table A-1E). The mass diffusivity of water vapor in air at 60F = 520 R = 288.9 K is determined from Eq. 14-15 to beAnalysis The Reynolds number for flow over a cylinder is The Schmidt number in this case is Then utili

32、zing the analogy between heat and mass convection, the Sherwood number is determined from Eq. 10-32 by replacing Pr number by the Schmidt number to be Using the definition of Sherwood number, the mass transfer coefficient is determined to be The air at the water surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (0.5073 p