408 CHAPTER 12  Introduction to convective mass transfer

It may be noted that Eq. (12.47) is used for both constant wall temperature and constant wall heat flux conditions. Eq. (12.48) is used for both constant wall mass concentration and constant wall mass flux conditions.

The Chilton-Colburn analogy for tube flow is expressed as

where f is the Darcy friction factor.

12.7  A note on the convective heat and mass analogy

The analogy between the convective heat transfer and the convective mass transfer is applicable for the cases in which the flow rate of species undergoing mass transfer is low compared to the total flow rate of the gas or liquid mixture so that the flow velocity is not affected significantly by the species mass transfer between the surface and the fluid. For example, in evaporative cooling or the evaporation at the free surface of a water pool, the mole fraction of water vapor in the air-water vapor mixture at the water surface is very small, and so the use of the convective heat and mass transfer analogy will not result in significant errors. However, for the cases such as boilers, condensers, and fuel droplet evaporation in combustors, wherein the vapor fraction approaches unity, the convective heat and mass transfer analogy cannot be used.

12.8  Simultaneous heat and mass transfer

The simultaneous transfer of heat and mass occurs in many engineering applications, such as wet bulb thermometers, cooling towers, humidifiers, and dehumidifiers. Water evaporates when air blows over the water surface. The energy required for evaporation comes from the latent heat of vaporization of the water. Under steady state conditions, the latent heat transferred by the water is equal to the heat transferred to the water from the air blowing over the surface, and in the process, the air gets cooled. Using the energy balance at the water surface, as shown in Fig. 12.4, we have

12.8 Simultaneous heat and mass transfer 409

FIGURE 12.4  Energy transfer at the air-water interface.

In the above equation, the air properties, ρ, cp, and Le, correspond to the film

If the mass concentration of species B (air) is very large compared to the mass concentration of species A (water vapor),

where P = PA + PB

Eq. (12.53) can now be written as

Let T1 and T2 be the dry bulb and wet bulb temperatures, respectively, and the corresponding partial pressures of water vapor be Pv1 and Pv2, respectively.

Specific humidity, w, is the mass of water vapor per kg of dry air.

Pa = P − Pv ≈ P as Pv Pa

410 CHAPTER 12  Introduction to convective mass transfer

Eq. (12.55) can now be written as

Using Eq. (12.56) with Ra = RB and Rv = RA , Eq. (12.54) becomes

For elaborate illustrations and more examples on convective mass transfer, readers can refer to Lienhard IV and Lienhard V (2020), Bergman et al. (2011), Cengel (2003), and Bejan (1993).

Example 12.1: Pressurized hydrogen gas is stored at 298 K in a 1.2 m long cylindrical container made of iron. The inner and outer radii of the cylinder are 0.18 m and 0.19 m, respectively. The molar concentration of hydrogen in iron at the inner surface is 0.1 kmol/m3 and that at the outer surface is negligible. Calculate the mass flow rate of hydrogen by diffusion through the cylindrical wall of the container, assuming the mass diffusion to be steady and one-dimensional.

Solution:

The binary diffusion coefficient for hydrogen in iron at the specified temperature is

2.6× 10−13 m2/s (Table 12.1).

The molar flow rate of the hydrogen through the cylindrical wall by diffusion is

Example 12.2: An outdoor swimming pool contains water at 30 °C. The length and breadth of the pool are 24 m and 4 m, respectively. The ambient temperature and the relative humidity are 35 °C and 30%, respectively. Wind with a speed of 0.7 m/s blows in the direction of the length of the pool. The atmospheric pressure is 101 kPa. Determine the following.

(a)The mole fraction and mass fraction of water vapor at the water surface.

(b)The molar concentration and mass concentration of the water vapor at the water surface.

(c)The average mass transfer coefficient.

(d)The rate of evaporation of water from the pool.

12.8 Simultaneous heat and mass transfer 411

Solution:

(a) The air will be saturated at the water surface. Hence the partial pressure of water vapor in the air at the water surface will be the saturated pressure of water at 30 °C. The mole fraction of water vapor,

y = Psat (30 °C)

vapor ,s Ptotal

= 4.24 kPa

101kPa = 0.042

The molecular weight of dry air is 28.97 kg/kmol.

The mass fraction of water vapor,

=0.0265

(b)The molar concentration of water vapor at the water surface is

Cvapor ,s = PRTvapor

= 4.24

8.314 × 303

= 1.683 ×10−3 kmol/m3

The mass concentration of water vapor at the water surface is

ρvapor,s = MH2OCvapor,s

=18 ×1.683 ×10−3 kg/m3

=0.03kg/m3

(c)Film temperature,

Tf = Ts + T∞

2

= 30 + 35 2

= 32.5 °C

The properties of air at 1 atm and 32.5 °C are

ρ = 1.14 kg/m3, cp = 1.007 kJ/kgK, ν = 1.63 × 10−5 m2 /s, = 1.86 × 10−5 Ns/m2 , Pr = 0.706

From Table 12.1, DAB = 2.63 × 10−5 m2 /s

Since ReL > 5 × 105, the flow is turbulent.

412 CHAPTER 12  Introduction to convective mass transfer

(d) RH = 30%

= 0.0121kg/m3

The rate of evaporation is

mvapor = hmL A(ρvapor ,s − ρvapor ,∞ )

=1.4 ×10−3 × 24 × 4 × (0.03 − 0.0121)

=2.406 ×10−3 kg/s

=2.406 ×10−3 × 24 × 3600 kg/day

=207.86 kg/day

Example 12.3: Consider Example 12.2. If the wind blows in the direction of the breadth of the pool, determine the average mass transfer coefficient and the rate of evaporation of water from the pool.

Solution:

Since ReL < 5 × 105, the flow is laminar.

ShL = 0.664 Re1/2L Sc1/3

=0.664 × (1.71 × 105 )1/2 × 0.6191/3

=234

12.8 Simultaneous heat and mass transfer 413

hmL = ShL DLAB

= × 2.63 ×10−5 234

4

= 1.538 ×10−3 m/s

The rate of evaporation is

mvapor = hmL A(ρvapor ,s − ρvapor ,∞ )

=1.538 ×10−3 × 24 × 4 × (0.03 − 0.0121)

=2.64 ×10−3 kg/s

=228.3 kg/day

Example 12.4: Ethanol flows through a long tube of diameter 2 mm. Later, in order to remove the ethanol from the inner surface of the tube, water at 1 atm and 298 K is forced to flow through the tube with a mean velocity of 0.4 m/s. Assuming fully developed flow, evaluate the mass transfer coefficient inside the tube.

Solution:

From Table 12.1, the value of DAB for ethanol diffusing through water at 1 atm and 298 K is 1.2 × 10-9 m2/s.

Since ReD < 2300, the flow is laminar. The Sherwood number, in this case, is 3.66. Therefore, the convective mass transfer coefficient can be calculated as

hm = ShDAB = 3.66 ×1.2 ×10−9 = 2.19 ×10−6 m/s D 0.002

Example 12.5: The dry bulb temperature and wet bulb temperature of an air stream are 30 °C and 20 °C, respectively. Estimate the specific humidity of the air.

Solution:

The properties of air at 1 atm and 25 °C are

ρ = 1.16 kg/m3, cp = 1.007 kJ/kg K, ν = 1.58 × 10−5 m2 /s, and α = 22.5 × 10−6 m2 /s. From Table 12.1, DAB = 2.5 ×10−5 m2 /s,

414 CHAPTER 12  Introduction to convective mass transfer

Specific humidity corresponding to the wet-bulb temperature is

hfg corresponding to 20 °C is 2451 kJ/kg.

ω1 = ω2 − (T1 − T2 ) cp (Le)2/3

hfg

=0.0146 − (30 − 20) 12451.007 (0.9)2/3

=0.0146 − 0.00382

=0.0107 kg/kg dry air

Problems

12.1Air at 1 atm and 298 K flows with a velocity of 1 m/s inside a tube of 20 mm diameter and 2 m length. There is a deposit of iodine on the inner surface of the tube. Determine the mass transfer coefficient of iodine from the surface into the air.

12.2A 1.5 m long and 1.5 m wide tub contains water at 20 °C. Air at 1 atm and 30 °C with a relative humidity of 40% blows past the water surface with a velocity of 1.6 m/s. Calculate the rate of evaporation of water.

12.3The dry bulb and wet bulb temperatures recorded by a thermometer are 40 °C and 30 °C, respectively.

a.Determine the relative humidity of the air.

b.If the air is completely dry and has the same dry bulb temperature of 40 °C, determine the wet-bulb temperature.

12.4A bottled drink needs to be chilled by wrapping it in a wet cloth and blowing air over it. If the ambient air is at 1 atm and 35 °C with a relative humidity of 45%, what will be the steady state temperature attained by the drink?

12.5A cooling tower in a power plant cools 5000 kg/h of water from 40 °C to 32 °C. If the air enters the tower at 28 °C with a relative humidity of 50% and leaves at 36 °C with a relative humidity of 97%, determine the water evaporation rate or the mass flow rate of makeup water. Assume the ambient pressure to be 1 atm.

References 415

References

Barrer, R.M., 1941. Diffusion in and Through Solids. Macmillan, New York. Bejan, A., 1993. Heat Transfer, John Wiley & Sons, New York.

Bergman, T.L., Lavine, A.S., Incropera F.P., Dewitt, D.P., 2011. Fundamentals of Heat and Mass Transfer, seventh ed. John Wiley & Sons, NJ.

Black, Van L., 1980. Elements of Material Science and Engineering. Addison-Wesley, Reading, MA.

Cengel, Y. A., 2003. Heat Transfer: A Practical Approach, second ed. McGraw-Hill, New York. Geankoplis, C.J., 1972. Mass Transport Phenomena. Holt, Rinehart, and Winston, New York. Lienhard IV, J.H., Lienhard V, J.H., 2020. A Heat Transfer Textbook. Courier Dover

Publications,­ Mineola, NY.

Mills, A.F., 1995. Basic Heat and Mass Transfer. Richard D. Irwin, Burr Ridge, IL.

Perry, J.H. (Ed.), 1963. Chemical Engineer’s Handbook. fourth ed. McGraw-Hill, New York. Reid, R.D., Prausnitz, J.M., Sherwood, T.K., 1977. The Properties of Gases and Liquids, third

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Thomas, L.C., 1991. Mass Transfer Supplement-Heat Transfer. Prentice Hall, Englewood Cliffs, NJ.