CHE 318 Lecture 08

Steady State Mass Transfer: More Examples

Dr. Tian Tian

2026-01-21

Recap

  • Mass transfer equations in other geometries
    • Use M.B. for \(\overline{N}_A\), not \(N_A\)
    • Be careful with the coordinate system (sphere? cylinder?)
  • Solving examples with pseudo steady state assumption

Learning Outcomes

(Continue from Lecture 07) After today’s lecture, you will be able to:

  • Solve P.S.S problems in spherical coordinates
  • Recall difference between steady and unsteady state solutions
  • Analyze time-dependent concentration profiles for U.S.S situations.

P.S.S Examples (continued)

Example 1: Diffusion Through Stagnant B with Changing Path Length

Adapted from Geankoplis 6.2-3

Water vapor diffuses through a stagnant gas in a narrow vertical tube, dry air is constantly blown at the top of tube.
At time \(t\), the liquid level is a distance \(z\) from the tube top (i.e., the diffusion path length is \(z\)).
As diffusion proceeds, the liquid level drops slowly, so \(z\) increases with time. The liquid has density \(\rho_A\), and molecular weight \(M_A\)

  1. Derive an expression for the time \(t_F\) required for the level to drop such that the diffusion path length changes from \(z=z_0\) at \(t=0\) to \(z=z_F\) at \(t=t_F\).

Example 1: solutions

Tip

  1. Use pseudo steady-state assumption
  2. \(N_A\) change over time!

Answer \[ t_F = \frac{\rho_A (z_F^2 - z_0^2) RT p_{Bm}}{2 D_{AB} M_A p_{T}} \frac{1}{(p_{A1} - p_{A2})} \]

Example 2: Determine \(D_{AB}\) Through Evaporation

Adapted from Griskey 10-2

Sample setup as example 4, a vertical tube of diameter \(D=0.01128\) m contains a liquid volatile species \(A\) (chloropicrin, \(CCl_3NO_2\)) evaporating into stagnant air (\(B\)) at 1 atm. The gas-phase diffusion of \(A\) occurs through the air column above the liquid surface.

At \(t=0\), the distance from the tube top to the liquid surface is \(z_0 = 0.0388\) m, after \(t=1\) day, the distance is \(z_1 = 0.0412\) m.

  • Vapor pressure at the interface: \(p_{A1} = 3178.3\) Pa
  • Liquid density: \(\rho_A = 1650\) kg/m\(^3\)
  • Molecular weight: \(M_A = 164.39\) kg/kmol
  1. Use your expression from example 4, determine the binary diffusivity \(D_{AB}\) of \(A\) in air.

Example 2: solutions

Tip

Pseudo steady state solution and assuming \(N_A=\text{const}\) solution differ very little. Why?

Answer:

  1. Pseudo-steady state: \(D_{AB} = 8.56\times{}10^{-6}\ \text{m}^2/\text{s}\)
  2. \(N_A\) constant: \(D_{AB} = 8.75\times{}10^{-6}\ \text{m}^2/\text{s}\) (\(+2.2\%\) error)

Example 3: P.S.S For Diffusion Through Sphere

Adapted from Geankoplis Ex 6.2-4

A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and \(1.01325 \times{}10^5\) Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mm Hg. The \(D_AB\) of naphthalene in air at 318 K is \(6.92\times{}10^{−6}\ \text{m}^2/\text{s}\).

  1. Calculate the rate of evaporation of naphthalene from the surface.
  2. Write the expression for the time \(t_F\) to evaporate a sphere from radius \(r_0\) to \(r_F\). The solid density for naphthalene is \(\rho\) and molecular weight is \(M_A\).
  3. What is the \(t_F\) value when the sphere is completely evaporated?

Example 3: solutions

Tip

Similar setup as example 5. \(N_A\) is time-dependent

Answer:

  1. \(N_A = 9.68\times{}10^{-8}\ \text{kg mol}/\text{m}^2/\text{s}\)
  2. Expression for \(t_F{r=r_F}\): \[ t_F(r=r_F) = \frac{\rho RT p_{Bm}}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} (r_0^2 - r_F^2) \]
  3. Expression for \(t_F(r=0)\): \[ t_F(r=0) = \frac{\rho RT p_{Bm} r_0^2}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} \]

Compare the solutions with Example 4. We can also measure the diffusivity of volatile organic molecules using the sphere evaporation method!

Summary

  • Pseudo-steady state solutions to diffusion-evaporation problems
  • We will discuss about unsteady state mass transfer next lecture!