Diffusion Mass Transfer. Chapter 14 Sections 14.1 through 14.7. Lecture 19. Physical Origins and Rate Equations Mass Transfer in Nonstationary Media Conservation Equation and Diffusion through Stationary Media Diffusion and Concentrations at Interfaces
Physical Origins and Rate Equations Mass Transfer in Nonstationary Media Conservation Equation and Diffusion through Stationary Media Diffusion and Concentrations at Interfaces Diffusion with Homogenous Reactions Transient Diffusion
3. Conservation Equation and Diffusion through Stationary Media Similar to 1st Law of thermodynamics Law of conservation of specie Control volume & control surface
Conservation of Species
Mass Diffusion Equations
Mass Diffusion Equations For mass concentration in Cartesian coordinates: For molar concentration in Cartesian coordinates:
Mass Diffusion Equations For mass concentration in Cartesian coordinates (DAB and are constant): For molar concentration in Cartesian coordinates (DAB and C are constant):
Mass Diffusion Equations For molar concentration in Cylindrical coordinates: For molar concentration in Spherical coordinates:
Boundary and Initial Conditions B.C.’s: B.C.-1: at x=0, xA(0,t) = xA,s B.C.-2: Special case for B.C.-2 Impermeable surface
Stationary Media with Specific Surface Concentrations 1-D diffusion, No homogeneous reaction, Steady-state, Planar medium: For stationary medium:
Stationary Planar Medium Concentration distribution: Molar flux: Diffusion resistance:
Stationary Media with Specific Surface Concentrations 1-D diffusion, No homogeneous reaction, Steady-state, Planar medium: 1-D diffusion, No homogeneous reaction, Steady-state, Cylindrical medium: 1-D diffusion, No homogeneous reaction, Steady-state, Spherical medium:
Diffusion in Stationary Medium
Example 1 You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of 0.1-mm-thick soft rubber and has a diameter of 15 cm when inflated. The pressure and temperature inside the balloon are initially 110 kPa and 25°C. The permeability of rubber to helium, oxygen, and nitrogen at 25°C are 9.4*10-13, 7.05*10-13, and 2.6*10-13 kmol/m·s·bar, respectively.
Example 1 Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first 5 h assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be 100 kPa and 25°C.
Balloon Air He 25C 110 kPa He diffusion Example 1 Known:Permeability of helium, nitrogen and oxygen, dimension of balloon and gas conditions Find: Initial rate of diffusion of gases, mass fraction of helium escaped in 5 hours. Schematic:
Example 1 Assumptions: 1 Constant helium pressure. 2 Steady-state. 3 Mass diffusion is one-dimensional . 4 No chemical reactions. 5 Ideal gases. 6. The balloon wall can be treated as a plane layer. Properties: The permeability of rubber to helium, oxygen, and nitrogen at 25C are given to be 9.410-13, 7.0510-13, and 2.610-13 kmol/m.s.bars, respectively. The molar mass of helium is M = 4 kg/kmol and its gas constant is R = 2.0709 kPa.m3/kg.K.
Example 1 Analysis: (1). We can consider the total molar concentration to be constant (C = CA + CB CB = constant), and the balloon to be a stationary medium since there is no diffusion of rubber molecules (NB=0 ) and the concentration of the helium in the balloon is extremely low (CA << 1). The partial pressures of oxygen and nitrogen in the air are
Example 1 The partial pressure of helium in the air is negligible. Since the balloon is filled with pure helium gas at 110 kPa, the initial partial pressure of helium in the balloon is 110 kPa, and the initial partial pressures of oxygen and nitrogen are zero. When permeability data is available, the molar flow rate of a gas through a solid wall of thickness L under steady one-dimensional conditions can be determined from the following equation: Where is the permeability and PA,1 and PA,2 are the partial pressures of gas A on the two sides of the wall (Note that the balloon can be treated as a plain layer since its thickness is very small compared to its diameter). Noting that the surface area of the balloon is A= D2= (0.015)2=0.07069 m2, the initial rates of diffusion of helium, oxygen, and nitrogen at 25ºC are determined to be:
Example 1 The initial mass flow rate and amount of helium that escape during the first 5 hours are:
Example 1 The initial mass of helium in the balloon is: Therefore, the fraction of helium that escapes the balloon during the first 5 h is:
Example 2 Oxygen gas is maintained at pressure of 2 bars and 1 bar on opposite sides of a rubber membrane that is 0.5 mm thick, and the entire system is at 25 C. What is the molar diffusive flux of O2 through the membrane? What are the molar concentration of O2 on both sides of the membrane (outside the rubber)?
Example 3 (Example 14.3) The efficacy of pharmaceutical products is reduced by prolonged exposure to high temperature, light, and humidity. For water vapor-sensitive consumer products that are in tablet or capsule form, and might be stored in humid environments such as bathroom medicine cabinets, blister packaging is used to limit the direct exposure of the medicine to humid conditions until immediately before their use. Consider tablets that are contained in a blister package composed of a flat lidding sheet and a second, formed sheet that includes troughs to hold each tablet. The formed sheet is L = 50 mm thick and is fabricated of a polymer material. Each trough is of diameter D = 5 mm and depth h= 3 mm. The lidding sheet is fabricated of aluminum foil. The binary diffusion coefficient for water vapor in the polymer is D=6*10-14 m2/s while the aluminum may be assumed to be impermeable to water vapor. For molar concentrations of water vapor in the polymer at the outer and inner surfaces of CA,si= 4.5*10-3kmol/m3 and CA,si=0.5*10-3kmol/m3 , respectively, determine the rate at which water vapor is transferred through the trough wall to the tablet.
Example 3 (Example 14.3) Known:Molar concentration of water vapor at inner and outer surface of a polymer sheet and trough geometry. Find: Rate of water vapor molar diffusivity transfer through the wall Schematic:
Example 3 (Example 14.3) Assumptions: Steady-state, 1-D conditions Stationary medium No chemical reaction Polymer sheet is thin relative to the dimensions of trough, and diffusion may be analyzed as though it occurs through a plan wall Analysis: The total water vapor transfer is the summation of the transfer rate through the cylindrical walls of the trough and the bottom, circular surface of the trough.
Example 3 (Example 14.3) From equation 14.54 we may write Hence Lecture 19