Unsteady-State Diffusion in a Thick Slab. Repeat Example 7.1-2 but use a distribution coefficient K = 0.50 instead of 2.0. Plot the data.

Plot of Concentration Profile in Unsteady-State Diffusion. Using the same conditions as in Example 7.1-2, calculate the concentration at the points x = 0, 0.005, 0.01, 0.015, and 0.02 m from the surface. Also calculate c_Li in the liquid at the interface. Plot the concentrations in a manner similar to Fig. 7.1-3b, showing interface concentrations.

Unsteady-State Diffusion in Several Directions. Use the same conditions as in Example 7.1-1 except that the solid is a rectangular block 10.16 mm thick in the x direction, 7.62 mm thick in the y direction, and 10.16 mm thick in the z direction, and diffusion occurs at all six faces. Calculate the concentration at the midpoint after 10 h.

Drying of Moist Clay. A very thick slab of clay has an initial moisture content of c₀ = 14 wt %. Air is passed over the top surface to dry the clay. Assume a relative resistance of the gas at the surface of zero. The equilibrium moisture content at the surface is constant at c₁ = 3.0 wt %. The diffusion of the moisture in the clay can be approximated by a diffusivity of D_AB = 1.29 × 10⁻⁸ m²/s. After 1.0 h of drying, calculate the concentration of water at points 0.005, 0.01, and 0.02 m below the surface. Assume that the clay is a semi-infinite solid and that the Y value can be represented using concentrations of wt % rather than kg mol/m³. Plot the values versus x.

Unsteady-State Diffusion in a Cylinder of Agar Gel. A wet cylinder of agar gel at 278 K containing a uniform concentration of urea of 0.1 kg mol/m³ has a diameter of 30.48 mm and is 38.1 mm long with flat, parallel ends. The diffusivity is 4.72 × 10⁻¹⁰ m²/s. Calculate the concentration at the midpoint of the cylinder after 100 h for the following cases if the cylinder is suddenly immersed in turbulent pure water:

Drying of Wood. A flat slab of Douglas fir wood 50.8 mm thick containing 30 wt % moisture is being dried from both sides (neglecting ends and edges). The equilibrium moisture content at the surface of the wood due to the drying air blown over it is held at 5 wt % moisture. The drying can be assumed to be represented by a diffusivity of 3.72 × 10⁻⁶ m²/h. Calculate the time for the center to reach 10% moisture.

Flux and Conversion of Mass-Transfer Coefficient. A value of k_G was experimentally determined to be 1.08 lb mol/h · ft² · atm for A diffusing through stagnant B. For the same flow and concentrations it is desired to predict and the flux of A for equimolar counterdiffusion. The partial pressures are p_A1 = 0.20 atm, p_A2 = 0.05 atm, and P = 1.0 atm abs total. Use English and SI units.

Conversion of Mass-Transfer Coefficients. Prove or show the following relationships, starting with the flux equations:

Absorption of H₂S by Water. In a wetted-wall tower an air–H₂S mixture is flowing by a film of water that is flowing as a thin film down a vertical plate. The H₂S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and 30°C. A value for of 9.567 × 10⁻⁴ m/s has been predicted for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H₂S in the liquid at the liquid–gas interface is 2.0(10⁻⁵) and p_A of H₂S in the gas is 0.05 atm. The Henry's law equilibrium relation is p_A (atm) = 609x_A (mole fraction in liquid). Calculate the rate of absorption of H₂S. (Hint: Call point 1 the interface and point 2 the gas phase. Then calculate p_A1 from Henry's law and the given x_A. The value of p_A2 is 0.05 atm.)

Effect of High Flux on Mass-Transfer Coefficients. Using the data from Example 7.2-2, also calculate the flux ratios for x_A1 = 0.20 and x_A1 = 0.01. Tabulate these ratios for the three values of x_A1 and plot the ratio versus x_A1.

Mass Transfer from a Flat Plate to a Liquid. Using the data and physical properties of Example 7.3-2 calculate the flux for a water velocity of 0.152 m/s and a plate length of L = 0.137 m. Do not assume that x_BM = 1.0 but actually calculate its value.

Mass Transfer from a Pipe Wall. Pure water at 26.1°C is flowing at a velocity of 0.0305 m/s in a tube having an inside diameter of 6.35 mm. The tube is 1.829 m long, with the last 1.22 m having the walls coated with benzoic acid. Assuming that the velocity profile is fully developed, calculate the average concentration of benzoic acid at the outlet. Use the physical property data from Example 7.3-2. [Hint: First calculate the Reynolds number Dνρ/μ. Then calculate N_ReN_Sc(D/L)(π/4), which is the same as W/D_ABρL.]

Mass-Transfer Coefficient for Various Geometries. It is desired to estimate the mass-transfer coefficient k_G in kg mol/s · m²· Pa for water vapor in air at 338.6 K and 101.32 kPa flowing in a large duct past solids of different geometries. The velocity in the duct is 3.66 m/s. The water vapor concentration in the air is small, so the physical properties of air can be used. Water vapor is being transferred to the solids. Do this for the following geometries:

Mass Transfer to Definite Shapes. Estimate the value of the mass-transfer coefficient in a stream of air at 325.6 K flowing in a duct past the following shapes made of solid naphthalene. The velocity of the air is 1.524 m/s at 325.6 K and 202.6 kPa. The D_AB of naphthalene in air is 5.16 × 10⁻⁶ m²/s at 273 K and 101.3 kPa.

Mass Transfer to Packed Bed and Driving Force. Pure water at 26.1°C is flowing at a rate of 0.0701 ft³/h through a packed bed of 0.251-in. benzoic-acid spheres having a total surface area of 0.129 ft². The solubility of benzoic acid in water is 0.00184 lb mol benzoic acid/ft³ solution. The outlet concentration c_A2 is 1.80 × 10⁻⁴ lb mol/ft³. Calculate the mass-transfer coefficient k_c.

Mass Transfer in Liquid Metals. Mercury at 26.5°C is flowing through a packed bed of lead spheres having a diameter of 2.096 mm with a void fraction of 0.499. The superficial velocity is 0.02198 m/s. The solubility of lead in mercury is 1.721 wt %, the Schmidt number is 124.1, the viscosity of the solution is 1.577 × 10⁻³ Pa · s, and the density is 13 530 kg/m³.

Mass Transfer from a Pipe and Log Mean Driving Force. Use the same physical conditions as in Problem 7.3-2, but the velocity in the pipe is now 3.05 m/s. Do as follows:

Derivation of Relation Between J_D and N_Sh. Equation (7.3-3) defines the Sherwood number and Eq. (7.3-5) defines the J_D factor. Derive the relation between N_Sh and J_D in terms of N_Re and N_Sc.

Driving Force to Use in Mass Transfer. Derive Eq. (7.3-42) for the log mean driving force to use for a fluid flowing in a packed bed or in a tube. (Hint: Start by making a mass balance and a diffusion rate balance over a differential area dA as follows:

Maximum Oxygen Uptake of a Microorganism. Calculate the maximum possible rate of oxygen uptake at 37°C of microorganisms having a diameter of μm suspended in an agitated aqueous solution. It is assumed that the surrounding liquid is saturated with O₂ from air at 1 atm abs pressure. It will be assumed that the microorganism can utilize the oxygen much faster than it can diffuse to it. The microorganism has a density very close to that of water. Use physical-property data from Example 7.4-1. (Hint: Since the oxygen is consumed faster than it is supplied, the concentration c_A2 at the surface is zero. The concentration c_A1 in the solution is at saturation.)

Mass Transfer of O₂ in Fermentation Process. A total of 5.0 g of wet microorganisms having a density of 1100 kg/m³ and a diameter of 0.667 μm are added to 0.100 L of aqueous solution at 37°C in a shaker flask for a fermentation. Air can enter through a porous stopper. Use physical property data from Example 7.4-1.

Sum of Molar Fluxes. Prove the following equation using the definitions in Table 7.5-1:

Proof of Derived Relation. Using the definitions from Table 7.5-1, prove the following:

Different Forms of Fick's Law. Using Eq. (1), prove Eq. (2):

Equation 1

Equation 2

Other Form of Fick's Law. Show that the following form of Fick's law is valid:

Different Form of Equation of Continuity. Starting with Eq. (7.5-12),

Equation 7.5-12

Equation 1

[Hint: From Table 7.5-1, substitute n_A = j_A + ρ_Aν into Eq. (7.5-12). Note that (∇ · ν) = 0 for constant ρ. Then substitute Fick's law in terms of j_A.]

Diffusion and Reaction at a Surface. Gas A is diffusing from a gas stream at point 1 to a catalyst surface at point 2 and reacts instantaneously and irreversibly as follows:

Unsteady-State Diffusion and Reaction. Solute A is diffusing at unsteady state into a semi-infinite medium of pure B and undergoes a first-order reaction with B. Solute A is dilute. Calculate the concentration c_A at points z = 0, 4, and 10 mm from the surface for t = 1 × 10⁵ s. Physical property data are D_AB = 1 × 10⁻⁹ m²/s, k' = 1 × 10⁻⁴ s⁻¹, c_A0 = 1.0 kg mol/m³. Also calculate the kg mol absorbed/m².

Multicomponent Diffusion. At a total pressure of 202.6 kPa and 358 K, ammonia gas (A) is diffusing at steady state through an inert, nondiffusing mixture of nitrogen (B) and hydrogen (C). The mole fractions at z₁ = 0 are x_A1 = 0.8, x_B1 = 0.15, and x_C1 = 0.05; and at z₂ = 4.0 mm, x_A2 = 0.2, x_B2 = 0.6, and x_C2 = 0.2. The diffusivities at 358 K and 101.3 kPa are D_AB = 3.28 × 10⁻⁵ m²/s and D_AC = 1.093 × 10⁻⁴ m²/s. Calculate the flux of ammonia.

Diffusion in Liquid Metals and Variable Diffusivity. The diffusion of tin (A) in liquid lead (B) at 510°C was carried out by using a 10.0-mm-long capillary tube and maintaining the mole fraction of tin at x_A1 at the left end and x_A2 at the right end of the tube. In the range of concentrations of 0.2 ≤ x_A ≤ 0.4 the diffusivity of tin in lead has been found to be a linear function of x_A (S7):

Diffusion and Chemical Reaction of Molten Iron in Process Metallurgy. In a steelmaking process using molten pig iron containing carbon, a spray of molten iron particles containing 4.0 wt % carbon falls through a pure-oxygen atmosphere. The carbon diffuses through the molten iron to the surface of the drop, where it is assumed that it reacts instantly at the surface because of the high temperature, as follows, according to a first-order reaction:

Effect of Slow Reaction Rate on Diffusion. Gas A diffuses from point 1 to a catalyst surface at point 2, where it reacts as follows: 2A → B. Gas B diffuses back a distance δ to point 1.

Diffusion and Heterogeneous Reaction on Surface. In a tube of radius R m filled with a liquid, dilute component A is diffusing in the nonflowing liquid phase represented by

Derive the differential equation for unsteady state for diffusion and reaction for this system. [Hint: First make a mass balance for A for a Δz length of tube as follows: rate of input (diffusion) + rate of generation (heterogeneous) = rate of output (diffusion) + rate of accumulation.]

Knudsen Diffusivities. A mixture of He (A) and Ar (B) is diffusing at 1.013 × 10⁵ Pa total pressure and 298 K through a capillary having a radius of 100 Å.

Transition-Region Diffusion. A mixture of He (A) and Ar (B) at 298 K is diffusing through an open capillary 15 mm long with a radius of 1000 Å. The total pressure is 1.013 × 10⁵ Pa. The molecular diffusivity D_AB at 1.013 × 10⁵ Pa is 7.29 × 10⁻⁵ m²/s.

Diffusion in a Pore in the Transition Region. Pure H₂ gas (A) at one end of a noncatalytic pore of radius 50 Å and length 1.0 mm (x_A1 = 1.0) is diffusing through this pore with pure C₂H₆ gas (B) at the other end at x_A2 = 0. The total pressure is constant at 1013.2 kPa. The predicted molecular diffusivity of H₂–C₂H₆ is 8.60 × 10⁻⁵ m²/s at 101.32 kPa and 373 K. Calculate the Knudsen diffusivity of H₂ and flux N_A of H₂ in the mixture at 373 K and steady state.

Transition-Region Diffusion in Capillary. A mixture of nitrogen gas (A) and helium (B) at 298 K is diffusing through a capillary 0.10 m long in an open system with a diameter of 10 μm. The mole fractions are constant at x_A1 = 1.0 and x_A2 = 0. See Example 7.6-2 for physical properties.

Numerical Method for Unsteady-State Diffusion. A solid slab 0.01 m thick has an initial uniform concentration of solute A of 1.00 kg mol/m³. The diffusivity of A in the solid is D_AB = 1.0 × 10⁻¹⁰ m²/s. All surfaces of the slab are insulated except the top surface. The surface concentration is suddenly dropped to zero concentration and held there. Unsteady-state diffusion occurs in the one x direction with the rear surface insulated. Using a numerical method, determine the concentrations after 12 × 10⁴ s. Use Δx = 0.002 m and M = 2.0. The value of K is 1.0.

c₂ = 0.3125 kg mol/m³ (x = 0.002 m)

c₃ = 0.5859 (x = 0.004 m)

c₄ = 0.7813 (x = 0.006 m)

c₅ = 0.8984 (x = 0.008 m)

c₆ = 0.9375 (insulated surface, x = 0.01 m)

Digital Computer and Unsteady-State Diffusion. Using the conditions of Problem 7.7-1, solve that problem by digital computer. Use Δx = 0.0005 m. Write the spreadsheet program and plot the final concentrations. Use the explicit method, M = 2.

Numerical Method and Different Boundary Condition. Use the same conditions as for Example 7.7-1, but in this case the rear surface is not insulated. At time t = 0 the concentration at the rear surface is also suddenly changed to c₅ = 0 and held there. Calculate the concentration profile after 2500 s. Plot the initial and final concentration profiles and compare with the final profile of Example 7.7-1.

Dimensional Analysis in Mass Transfer. A fluid is flowing in a vertical pipe and mass transfer is occurring from the pipe wall to the fluid. Relate the convective mass-transfer coefficient to the variables D, ρ, μ, ν, D_AB, g, and Δρ, where D is pipe diameter, L is pipe length, and Δρ is the density difference.

Mass Transfer and Turbulence Models. Pure water at a velocity of 0.11 m/s is flowing at 26.1°C past a flat plate of solid benzoic acid where L = 0.40 m. Do as follows: