Consider evaporation of water contained in a tube or tank and its subsequent diffusion through the stagnant layer of air over it.
Assumptions:
(i) The system is under steady state and isothermal conditions
(ii) The total pressure within the tube is constant and equal to the outside pressure
ADVERTISEMENTS:
(iii) Both air and water vapours behave as perfect gases
(iv) Air which flows across the open end of the tube or tank has negligible solubility in water, and
(v) The slight movement of air over the top of the tube does not bring about any change in the concentration profile of air. However, the movement is just sufficient to carry the water vapours which diffuse to that point.
The water evaporates and diffuses upward through the air. Under stipulations of steady state, the upward movement must be balanced by a downward diffusion of air so that concentration at any location from the water surface remains constant,
ADVERTISEMENTS:
Downward mass diffusion of air,
where A is cross-sectional area of the tube and dpa/dx is the partial pressure gradient of air Since there can be no net mass movement of air downward at the surface of water, there will be a bulk mass movement upward with a velocity just large enough to compensate for the mass diffusion of air downward.
Bulk mass transfer of air is equal to
where v is the bulk mass velocity upward
Combination of equations 14.16 and 14.17 gives:
The mass transfer of water is due to (i) upward mass diffusion of water and (ii) water vapours carried upward along with upward bulk movement of moving air. Thus the total mass transfer of water vapour is
Equation 14.22 is referred to as Stefan’s law for diffusion of an ideal gaseous component through a practically stagnant and ideal constituent of the binary system.
The expression 14.22 for Stefan’s law may be integrated between the planes x1 and x2;
Introducing the concept of log mean partial pressure of air (LMPA);
When the partial pressure of water vapour does not change appreciably compared with the total pressure of water vapour air mixture, then the mass of water vapours can be calculated (without appreciable error) by using arithmetic mean partial pressure of air (pa1 + pa2)/2 instead of log mean partial pressure.
It follows from equation 14.23 that at any point x between x1 and x –
The variation of pressure pw and pa in the direction of x has been plotted in Fig. 14.5. These curves also represent the variation of concentration in the direction of x.
Example 1:
Derive an expression for the steady state diffusion of a gas A through another stagnant gas B.
Oxygen is diffusing through stagnant carbon monoxide at 0°C and 760 mm of Hg pressure under steady state conditions. The partial pressure of oxygen at two planes 0.3 cm apart is 100 mm of Hg and 25 mm of Hg respectively. Calculate the rate of diffusion of oxygen in gm-moles through cm2 area. It may be presumed that-
Diffusivity of oxygen in carbon monoxide = 0.185 cm2/s
Gas constant R = 82.06 cm2 atm/gm mole K.
Solution:
Let subscripts o and c refer to oxygen and carbon monoxide.
Partial pressures of oxygen on the given planes are:
Example 2:
Estimate the diffusion coefficient of carbon tetrachloride into air from the following data recorded in a Stefan-tube experiment with carbon tetrachloride and oxygen-
Diameter of tube and its length above liquid surface: 1 cm and 15 cm respectively
Temperature and pressure maintained: 0°C and 760 mm of mercury
Evaporation of carbon tetrachloride: 0.03 gm in 10 hour
Vapour pressure of carbon tetrachloride: 33 mm of mercury
Solution:
The molecular weights of carbon tetrachloride Mc and oxygen Mo are-
Mc = 12 + 4 (35.5) = 154 and Mo = 32
Partial pressures of these elements at bottom and top are:
