食品工程原理：Chapter 10 Mass Transfer

1、10.1 A droplet of water is falling through 20 0C air at the terminal velocity. The relative humidity of the air is 10% and the droplet is at the wet bulb temperature. The diffusivity of water vapor in air is 0.210-4 m2/s. Estimate the convective mass transfer coefficient for a 100 m diameter droplet

2、.Given: the dry bulb temperature Ta= 20 , =10%; the droplet is at the wet bulb temperature;The diffusivity of water vapor in air D= 0.2 10-4 m 2 /s. See Example 10.4Approach: Estimate the convective mass transfer coefficient for a 100 m diameter droplet.Solution:For mass transfer from a freely falli

3、ng liquid droplet, the following expression is recommended.1. The Reynolds number,T=20 , p=998.2 kg/m3, f=1.164 kg/m3; for air, =18.24010-6 Pa s;ut=0.298 m/sthe dry bulb temperature Ta= 20 , =10%, from Figure A.5.2, the droplet is at the wet bulb temperature Tw=7.6, Viscosity of water: 150 10-5 Pa.s

4、estimated;At temperature 7.6, for water, =999.8kg/m3;NRe=19.862. The Schmidt number isNSc=7.5010-23. The Sherwood number can be obtained from Equation (10.38)NSh=3.1284. The mass transfer coefficientkm=0.626 m/s10.2 The Sherwood Number for vapor transport from the surface of a high moisture food pro

5、duct to the surrounding air is 2.78.Compute the convective mass transfer coefficient, when the dimension of the product in the direction of air movement is 15 cm, and the mass diffusivity for water vapor in air is 1.810 -5 m 2/s.Given: The Sherwood Number for vapor transport from the surface of a hi

6、gh moisture food product to the surrounding air Nsh= 2.78.Approach: Compute the convective mass transfer coefficient, when the dimension of the product in the direction of air movement dc=15 cm=0.15m;the mass diffusivity for water vapor in air D= 1.8 10-5 m 2 /s.Solution:Using the formula 10.16 for

7、calculationkm=3.33610-4 m/s10.3 A flavor compound is held within a 5 mm diameter time release sphere. The sphere is placed in a liquid food and will release the flavor compound after one month of storage at 20 0C. The concentration of flavor within the sphere is 100% and the mass diffusivity for the

8、 flavor compound within the liquid food is 7.810-9 m2/s. Estimate the steady-state mass flux of flavor into the liquid food from the surface of the time-release sphere. The convective mass transfer coefficient is 50 m/s.Given: 5 mm diameter time release sphere, the storage temperature T=20;The conce

9、ntration of flavor within the sphere is 100% and the mass diffusivity for the flavor compound within the liquid food D=7.8 10-9 m2 /s.Approach: Estimate the steady-state mass flux of flavor into the liquid food from the surface of the time-release sphere. The convective mass transfer coefficient km

10、=50 m/s.Solution:1. For steady-state diffusion of gases (and Liquids) through solids;Assuming the mass diffusivity does not depend on concentration, from Equation (10.1) we obtainwhere DAB is mass diffusivity for gas A (or liquid A) in a solid B.Subscript A for m and c represents a gas or liquid dif

11、fusing through a solid. In reality, DAB represents an effective diffusivity through solids., At T=20, the saturated water vapor density =0.01834 kg/m3, 0, CA20;mA=4r1DABCA1=43.142.510-37.810-90.01834=4.4910-12 kg/s2.The convective mass transfer coefficient k m is defined as the rate of mass transfer

12、 per unit area per unit concentration difference. Thus,w here m B is the mass flux (kg/s); c is concentration of component B, mass per unit volume (kg/m 3); A is area (m 2). The units of k m are m 3/m2 s or m/s. The coefficient represents the volume (m3) of component B transported across a boundary

13、of one square meter per second; the surface area of the sphere A=d2=(510-3)2=7.8510-5 m2;mB=kmA(CB1-CB2)=507.8510-50.01834=7.2210-5 kg/s10.4 A desiccant is being used to remove water vapor from a stream of air with 90% RH at 50 0C. The desiccant is a flat plate with 25 cm length and 10 cm width, and

14、 maintains a water activity of 0.04 at the surface exposed to the air stream. The stream of air has velocity of 5 m/s moving over the surface in the direction of the 25 cm length. The mass diffusivity for water vapor in air is 0.1810-3 m 2/s. Compute the mass flux of water vapor from the air to the

15、desiccant surface.Given: the dry bulb temperature Ta= 50 , =90%; The desiccant flat plate with 25 cm length and 10 cm width, water activity aw=0.04;The air moving velocity u=5 m/s, the mass diffusivity for the flavor compound within the liquid food D=0.18 10-3 m2 /s.Approach: Compute the mass flux o

16、f water vapor from the air to the desiccant surface.We will first determine the Reynolds number and then use an appropriate dimensionless correlation to obtain the mass transfer coefficient and the water evaporation rate.see example 10.1Solution:1. u=5 m/s, from Table A.4.4, T=50 ,viscosity of air:

17、19.515 10-6 Pa s, =1.057kg/m3;NRe=6.77104NRe 100From Figure 4.40,=0.42.Thenf=1.067105s=29.63h3. Using Figure 4.42 at NBi 100,jm =0.74.Using the diffusion rate equation:log(0.2-0.15)=-t=17.45h10.6 An apple slice with 1 cm thickness is exposed to 80% RH for one week. After one week, the water activity

18、 of the apple has increased to 0.6, from an initial value of 0.1. The convective mass transfer coefficient at the product surface is 810-3 m/s. Estimate the mass diffusivity for water vapor in the apple.Given: An apple slice with 1 cm thickness; exposed to =80% for one week;After one week, water activity from aw=0.1 to aw=0.6;The convective mass transfer coefficient at the product surface km =810-3 m /s;see Example 10.7.Approach: Estimate the mass diffusiv