CHE 318 Lecture 16

Dimensionless Numbers In Mass Transfer

Author

Dr. Tian Tian

Published

February 11, 2026

Note

Learning Outcomes

After today’s lecture, you will be able to:

  • Recall the nature behind dimensionless numbers
  • Describe how to correlate mass transfer behaviour with dimensionless numbers
  • Understand the usage of dimensionless numbers in different geometries

Recap: Boundary Layer Theory

The global mass transfer coefficient in a tube:

\[\begin{align} k_c' = \frac{0.664 D_{AB}}{L} N_{Re}^{0.5} N_{Sc}^{1/3} \end{align}\]
  • \(N_{Re}\): Reynolds number
  • \(N_{Sc}\): Schmidt number

Why Do We Need These Dimensionless Numbers?

  • Expressing fluxes using \(k\) coefficients are easy
  • But do we need to measure \(k\) for each system specifically?
    • Of course NO!
  • We can correlate the values of \(k\) measured in different geometries, velocities using dimensionless numbers
  • Similar treatment exists in heat and momentum (fluid) transfer

Dimensionless Numbers In Mass Transfer

  • General form: \(N_{\text{name}} = \dfrac{\text{Scale of effect 1}}{\text{Scale of Effect 2}}\)

  • Schmidt number (ratio between momentum diffusivity and molecular diffusivity)

\[\begin{align} N_{\text{Sc}} = \frac{\mu}{\rho D_{AB}} \end{align}\]
  • Sherwood number (ratio between convective mass transfer and molecular mass transfer)
\[\begin{align} N_{\text{Sh}} = \frac{k_c' L}{D_{AB}} \end{align}\]
  • Reynolds number (ratio between kinetic vs viscous forces of fluid flow)
\[\begin{align} N_{\text{Re}} = \frac{L v \rho}{\mu} \end{align}\]
  • \(L\): characteristic length of system
  • Location specific \(N_{\text{Re}, x}\) also used

How Are \(k_c'\) Correlated By Dimensionless Numbers?

  • Idea: mass transfer in flowing fluid described by \(v\), \(\rho\), \(\mu\), \(c\), \(D_{AB}\) and geometry (characteristic length \(L\))
  • The combinations of these properties –> dimensionless number groups (\(N_{\text{Sc}}\), \(N_{\text{Sh}}\), \(N_{\text{Re}}\))
  • The Chilton-Colburn \(j\)-factor analog has most successful use in mass transfer
\[\begin{align} j_D &= f/2 \\ &= \frac{k_c'}{v_{av}} (N_{\text{Sc}})^{2/3} \\ &= \boxed{\frac{k_c' L}{D_{AB}}} \boxed{\frac{\rho D_{AB}}{\mu}} \boxed{\frac{\mu}{L v_{av} \rho}} (N_\text{sc})^{2/3} \\ &= N_{\text{Sh}} N_{\text{Sc}}^{-1} N_{\text{Re}}^{-1} (N_\text{sc})^{2/3} \\ &= \frac{N_{\text{Sh}}}{N_{\text{Re}} N_{\text{Sc}}^{1/3}} \end{align}\]
  • \(f\) is the Fanning Friction Factor (can be found in table)

General Procedure To Calculate \(k_c'\)

  1. Calculate Reynolds number \(N_{Re}\) from fluid properties + geometry

  2. Determine flow regime (liquid)

  • \(N_{Re} < 2100\) β†’ laminar flow
  • \(N_{Re} \ge 2100\) β†’ turbulent flow
  1. Evaluate friction factor \(f\)
  • Laminar flow:
    \[ f = \frac{16}{N_{Re}} \]
  • Turbulent flow:
    \[ f = \frac{\tau_s}{\tfrac12 \rho v^2},\qquad \tau_s = \frac{\Delta P_f\,\pi R^2}{2\pi R\,\Delta L} \]
  1. Compute mass-transfer \(j\)-factor
    \[ j_D = \frac{f}{2} \]

  2. Obtain mass-transfer coefficient
    \[ k_c' = j_D\,v_{av}\,N_{Sc}^{-2/3} \]

Use of Empirical Mass Transfer Laws

  • In many systems, flux and / or concentration profiles become hard to have simple form
  • Luckily we can simplify typical mass transfer problems as different geometries
    • Cyliner / Pipe
    • Parallel plates
    • Flow around sphere
    • Packed bed
  • We will show a few case studies for different geometries
  • Dimensionless numbers (\(N_{Re}\), \(N_{Sc}\), \(N_{Sh}\)) help determine governing equations

Case 1: Mass Transfer for Flow Inside Pipes

  • Usually use the Linton & Sherwood chart
  • Valid for gas / liquid in both laminar & turbulent regimes

Flow Inside Pipes: Solution Procedure

  • Governing dimensionless quantity:
\[\begin{align} \frac{W}{D_{AB}\rho L} &= N_{Re}N_{Sc} \frac{D}{L} \frac{\pi}{4} \\ &= \frac{\text{[Total Forced Flow]} \text{(kg/s)}}{\text{[Total Diffusive Flow]} \text{(kg/s)}} \end{align}\]
  • If gas πŸ‘‰ use the β€œrodlike flow” line
  • If liquid, distinguish 2 cases
    • parabolic flow (\(N_{Re} < 2100;\ \frac{W}{D_{AB}\rho L} > 400\))
    • turbulent flow (\(N_{Re} > 2100;\ 0.6 < N_{Sc} < 3000\))

Flow Inside Pipes: Solution For Liquid

Parabolic flow

\[\begin{align} \frac{c_A - c_{A,s}}{c_{A,i} - c_{A,s}} &= 5.5 \left[ \frac{W}{D_{AB}\,\rho\,L} \right]^{-\tfrac{2}{3}} \end{align}\]
  • \(c_A\): exit concentration

  • \(c_{A,i}, c_{A,s}\): inlet & surface concentration

  • \(W\): flow rate in (kg/s)

  • \(k_c'\) can be calculated by \(j_D\)

Turbulent flow

\[\begin{align} N_{Sh} &= k_c'\left(\frac{D}{D_{AB}}\right) \\ &= \frac{k_c\,p_{BM}}{P} \left(\frac{D}{D_{AB}}\right) \\ &= 0.023 \left(\frac{\rho D v}{\mu}\right)^{0.83} \left(\frac{\mu}{\rho D_{AB}}\right)^{0.33} \\ &= 0.023\, N_{Re}^{0.83}\, N_{Sc}^{0.33} \end{align}\]
  • Similar to the \(j_D\) analog
  • Just need \(N_{Re}\) and \(N_{Sc}\) to determine \(k_c'\)
  • Characteristic length \(D\) is pipe diameter!

Case 2: Flow Past Parallel Plates

  • Can be used for gases or evaporation of liquid
  • Distinguished between laminar & turbulent flow
  • \(N_{Re}\) regime cutoff different in gas & liquid!
    • Characteristic length \(L\): length of plate in flow direction

Flow Past Parallel Plates: Results

Laminar flow (\(N_{Re} < 15,000\))

\[\begin{align} j_D &= 0.664 N_{Re, L}^{-0.5} \\ \frac{k_c' L}{D_{AB}} &= 0.664 N_{Re, L}^{0.5} N_{Sc}^{1/3} \end{align}\]
  • This follows our derivation of boundary layer theory

Turbulent flow

  • Gas: $15,000 < N_{Re}< 300,000 $
\[\begin{align} j_D = 0.036 N_{Re, L}^{-0.2} \end{align}\]
  • Liquid: \(600 < N_{Re}< 50,000\)
\[\begin{align} j_D = 0.99 N_{Re, L}^{-0.5} \end{align}\]

Case 3: Flow Past Single Sphere

  • Frequent geometry in particle solutions
  • Low Reynolds regime πŸ‘‰ solution for stagnant diffusion on spherical surface
  • High Reynolds regime πŸ‘‰ correct \(N_{Sh}\) and back calculate \(k_c'\)

Flow Past Single Sphere: Results

Low Reynolds (\(N_{Re} < 2\))

\[\begin{align} N_A &= \boxed{\frac{2 D_{AB}}{D_p}} (c_{A1} - c_{A2}) \\ &= k_c (c_{A1} - c_{A2}) \\ &= \frac{k_c'}{x_{Bm}}(c_{A1} - c_{A2}) \\ \end{align}\]
  • For \(x_{Bm} \approx 1\), we have:

\[ k_c' = \frac{2D_{AB}}{D_p} \]

  • Sherwood number: \(N_{Sh} = 2\)

High Reynolds (\(N_{Re} > 2\))

  • Gas:
\[\begin{align} &N_{Sh} = 2 + 0.552 N_{Re}^{0.53} N_{Sc}^{1/3} \\ &0.6 < N_{Sc} < 2.7\;\ N_{Re} < 48000 \end{align}\]
  • Liquid:
\[\begin{align} N_{Sh} &= 2 + 0.95 N_{Re}^{0.5} N_{Sc}^{1/3};\ N_{Re} < 2000\\ N_{Sh} &= 0.347 N_{Re}^{0.62} N_{Sc}^{1/3};\ 2000 < N_{Re} < 17000 \end{align}\]
  • Back calculate \(k_c' = N_{Sh} \frac{D_{AB}}{D_p}\)

Case 4: Mass Transfer for Packed Beds

  • Very common geometry for chemical engineering
    • Adsorption and desorption through solid particles (gases and liquids)
    • Catalytic processes with very large surface area
  • Geometry characteristics: void fraction \(\varepsilon\): \[ \varepsilon = \frac{\text{void space}}{\text{total space}} = \frac{\text{void space}}{\text{void space} + \text{solid space}} \]
    • Typically \(0.3 < \varepsilon < 0.5\)
    • Void fraction is difficult to measure experimentally

Correlation Equations In Packed Bed

Correlation 1, applicable to:

  • gase with \(10 < N_{Re} <10,000\)
  • liquid with \(10 < N_{Re} < 1500\)
\[\begin{align} j_D &= j_H = \frac{0.4548}{\varepsilon} N_{Re}^(-0.4069) \\ N_{Re} &= \frac{D_p v' \rho}{\mu} \end{align}\]
  • \(D_p\): (average) particle diameter
  • \(v’\): superficial velocity in the tube without packing

Correlation Equations In Packed Bed (II)

Correlation 2, applicable to:

  • liquid with \(0.0016 < N_{Re} < 55\), \(165 < N_{Sc} < 70000\)
\[\begin{align} j_D = \frac{1.09}{\varepsilon} N_{Re}^(-2/3) \end{align}\]
  • liquid with \(55 < N_{Re} < 1500\), \(165 < N_{Sc} < 10690\)
\[\begin{align} j_D = \frac{0.250}{\varepsilon} N_{Re}^(-0.31) \end{align}\]

Correlation Equations In Packed Bed (III)

Correlation 3, applicable to fluidized beds

  • \(10 < N_{Re} < 4000\) (gas & liquid)
\[\begin{align} j_D = \frac{0.4548}{\varepsilon} N_{Re}^(-0.4069) \end{align}\]
  • \(1 < N_{Re} < 10\) (liquid only)
\[\begin{align} j_D = \frac{1.1068}{\varepsilon} N_{Re}^(-0.72) \end{align}\]

Packed Bed Calculation Steps

  1. Known value from operational column: \(\varepsilon\), \(V_b\) (total volume), \(D_p\), \(D_{AB}\), \(\mu\), \(\rho\), etc.
  2. Depend on the operational range, calculate \(N_{Re}\), \(N_Sc\) πŸ‘‰ choose the equation for \(j_D\)
  3. Obtain \(k_c\) from \(j_D\) value
  4. Calculate flux \(N_A\)
  5. Estimate effective area \(A_{eff}\) inside the columne πŸ‘‰ \(\overline{N}_A = A_{eff} N_A\)

Caveats In Packed Bed Problems (1)

  • Estimate the effective area?
    • First calculate the effective surface area per volume \(a\) then \(A_{eff}\)
    \[\begin{align} a &= \frac{6 (1 - \varepsilon)}{D_p} \\ A_{eff} &= a V_b \end{align}\]

    Caveats In Packed Bed Problems (2)

  • Use log-mean driving force correction
\[\begin{align} \overline{N}_A &= A_{eff} N_A \\ &= A_{eff} k_c \frac{ (c_{A, i} - c_{A1}) - (c_{A, i} - c_{A2})} { \ln\!\frac{(c_{A, i} - c_{A1})}{(c_{A, i} - c_{A2})} } \end{align}\]

where

  • \(c_{A, i}\): surface concentration
  • \(c_{A1}, c_{A2}\): in- and outlet concentrations

Caveats In Packed Bed Problems (3)

  • Mass-flow balance
\[\begin{align} \overline{N}_A &= A_{eff} N_A \\ &= V (c_{A2} - c_{A1}) \end{align}\]

where \(V\) is the volumetric flow rate.

These equations will give rise to solving the flow in packed bed problem.

Summary

  • Dimensionless numbers can be used to correlate mass transfer problems in different flow rate, dimension etc
  • Typically, start with a known geometry (pipe? parallel plate? sphere? packed bed?)
  • Find the correlation with dimensionless numbers \(N_{Re}\), \(N_{Sc}\)
  • Calculate the final mass transfer rate
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