10.3 Mass Transfer Dimensionless Numbers (Sherwood, Schmidt, Peclet)

10.3 Mass Transfer Dimensionless Numbers (Sherwood, Schmidt, Peclet)

In chemical engineering transport phenomena, empirical mass transfer correlations are organized into dimensionless groups. This methodology allows laboratory data obtained on small-scale bench models using air-water systems to be scaled directly to multi-meter industrial reactors operating at high temperatures and pressures. On the NCEES PE Chemical Exam, mastery of mass transfer dimensionless numbers—and their profound analogies to heat and momentum transfer—enables engineers to solve complex rating problems rapidly without re-deriving microscopic transport balances.

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1. Core Mass Transfer Dimensionless Groups

+---------------------------------------------------------------------------------------+
|                         HEAT VS. MASS TRANSFER ANALOGIES                              |
+----------------------------+-----------------------------+----------------------------+
|      MOMENTUM TRANSFER     |        HEAT TRANSFER        |        MASS TRANSFER       |
+----------------------------+-----------------------------+----------------------------+
| Kinematic Viscosity (nu)   | Thermal Diffusivity (alpha) | Mass Diffusivity (D_AB)    |
| nu = mu / rho              | alpha = k / (rho * Cp)      | D_AB                       |
| Friction Factor (f / 2)    | Nusselt Number (Nu = h*L/k) | Sherwood Number (Sh=kc*L/D)|
| Reynolds Number (Re)       | Prandtl Number (Pr = nu/a)  | Schmidt Number (Sc = nu/D) |
| Colburn j_M = f / 2        | Colburn j_H = Nu/(Re*Pr^1/3)| Colburn j_D = Sh/(Re*Sc^1/3|
+----------------------------+-----------------------------+----------------------------+

1. Sherwood Number ($Sh$)

The Sherwood number is the dimensionless convective mass transfer coefficient, representing the ratio of total convective mass transfer to pure molecular diffusion across characteristic length $L$:

$Sh \equiv \frac{k_c L}{D_{AB}}$

Where:

• $k_c$ = convective mass transfer coefficient ($\text{m/s}$ or $\text{ft/hr}$).
• $L$ = characteristic geometric length (e.g., tube inner diameter $D$, particle diameter $d_p$, plate length $L$; in $\text{m}$ or $\text{ft}$).
• $D_{AB}$ = binary molecular diffusivity ($\text{m}^2/\text{s}$ or $\text{ft}^2/\text{hr}$).

Physical Meaning: If mass transfer occurred solely by pure steady-state molecular diffusion across a stagnant stagnant boundary film of thickness $\delta$, then $k_c = D_{AB} / \delta$. Substituting this yields $Sh = L / \delta$. Thus, the Sherwood number represents the ratio of the macroscopic system dimension to the concentration boundary layer thickness. It is the direct mass transfer analogue of the Nusselt number ($Nu = h L / k$).

2. Schmidt Number ($Sc$)

The Schmidt number is the ratio of momentum diffusivity (kinematic viscosity, $\nu$) to mass diffusivity ($D_{AB}$):

$Sc \equiv \frac{\mu}{\rho D_{AB}} = \frac{\nu}{D_{AB}}$

Where:

• $\mu$ = dynamic fluid viscosity ($\text{Pa}\cdot\text{s} = \text{kg/(m}\cdot\text{s)}$ or $\text{lb/(ft}\cdot\text{hr)}$).
• $\rho$ = fluid density ($\text{kg/m}^3$ or $\text{lb/ft}^3$).
• $\nu = \mu / \rho$ = kinematic viscosity / momentum diffusivity ($\text{m}^2/\text{s}$ or $\text{ft}^2/\text{hr}$).

Physical Meaning: The Schmidt number governs the relative rates of momentum and mass transport by molecular diffusion through fluid layers. It dictates the ratio of the hydrodynamic (velocity) boundary layer thickness ($\delta_u$) to the concentration boundary layer thickness ($\delta_c$):

$\frac{\delta_u}{\delta_c} \approx Sc^{1/3}$

Phase Contrast on the PE Exam:

• Gases: $\nu \approx 10^{-5}\text{ m}^2/\text{s}$, $D_{AB} \approx 10^{-5}\text{ m}^2/\text{s} \implies \mathbf{Sc \approx 0.5\text{ to }2.5}$.
  In gas flows, momentum and mass diffuse at comparable rates. The velocity boundary layer and concentration boundary layer are roughly the same thickness ($\delta_u \approx \delta_c$).
• Liquids: $\nu \approx 10^{-6}\text{ m}^2/\text{s}$, $D_{AB} \approx 10^{-9}\text{ m}^2/\text{s} \implies \mathbf{Sc \approx 500\text{ to }5,000}$.
  In liquid flows, momentum diffuses hundreds or thousands of times faster than mass! Consequently, $\delta_u / \delta_c \approx (1,000)^{1/3} = 10$. The concentration boundary layer is an ultra-thin film embedded deep within the viscous sublayer of the velocity boundary layer.

3. Mass Transfer Peclet Number ($Pe_m$)

The Peclet number for mass transfer characterizes the relative importance of convective bulk advection to molecular diffusion:

$Pe_m \equiv Re \cdot Sc = \left( \frac{\rho u L}{\mu} \right) \left( \frac{\mu}{\rho D_{AB}} \right) = \frac{u L}{D_{AB}}$

Where:

• $u$ = fluid superficial or bulk velocity ($\text{m/s}$ or $\text{ft/s}$).
• For $Pe_m \gg 1$: Bulk flow advection overwhelms molecular diffusion. Concentration changes are swept rapidly downstream.
• For $Pe_m \ll 1$: Molecular diffusion dominates advective flow. Solute diffuses upstream against the bulk current.

4. Lewis Number ($Le$)

The Lewis number compares thermal diffusivity ($\alpha$) to mass diffusivity ($D_{AB}$):

$Le \equiv \frac{Sc}{Pr} = \frac{\alpha}{D_{AB}} = \frac{k}{\rho C_p D_{AB}}$

Where $Pr = \nu / \alpha$ is the Prandtl number.
In air-water vapor mixtures at ambient atmospheric conditions, $Le \approx 1.0$. This thermodynamic coincidence means heat and mass transfer boundary layers grow at identical rates, allowing the wet-bulb psychrometric temperature to equal the thermodynamic adiabatic saturation temperature.

5. Stanton Number for Mass Transfer ($St_m$)

The mass transfer Stanton number represents the ratio of mass transfer flux to the convective flux carried by the main flow:

$St_m \equiv \frac{Sh}{Re \cdot Sc} = \frac{k_c}{u}$

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2. Transport Analogies: Reynolds and Chilton-Colburn

When mass transfer coefficients cannot be easily measured, chemical engineers rely on transport analogies to calculate $k_c$ directly from friction factors ($f$) or heat transfer coefficients ($h$).

1. Reynolds Analogy

Osborne Reynolds hypothesized that the mechanisms transferring momentum to a pipe wall are identical to those transferring heat or mass. For a smooth wall with zero pressure gradient:

$\frac{f}{2} = \frac{h}{\rho C_p u} = \frac{k_c}{u} \implies \frac{f}{2} = St_h = St_m$

Where $f$ is the Fanning friction factor.
Strict Limitation: The Reynolds analogy is valid only when $Pr \approx 1.0$ and $Sc \approx 1.0$, and where form drag (flow separation) is absent. It fails completely for liquids ($Sc \gg 1$).

2. Chilton-Colburn $j$-Factor Analogy

T. H. Chilton and A. P. Colburn modified the Reynolds analogy by introducing empirical exponent factors ($Sc^{2/3}$ and $Pr^{2/3}$) to account for boundary layer thickness differences across fluids:

$\mathbf{j_D \equiv St_m \cdot Sc^{2/3} = \left( \frac{k_c}{u} \right) Sc^{2/3} = \frac{Sh}{Re \cdot Sc^{1/3}}}$

$\mathbf{j_H \equiv St_h \cdot Pr^{2/3} = \left( \frac{h}{\rho C_p u} \right) Pr^{2/3} = \frac{Nu}{Re \cdot Pr^{1/3}}}$

$\mathbf{j_D = j_H = \frac{f}{2}}$

[!IMPORTANT] Fanning vs. Darcy Friction Factors in the Analogy:
The friction factor in the Chilton-Colburn analogy is strictly the Fanning friction factor ($f$). If the Moody / Darcy-Weisbach friction factor ($f_D = 4f$) is used, the relationship becomes: $j_D = \frac{f_D}{8}$ Confusing $f$ and $f_D$ is one of the most widespread numerical calculation errors on the NCEES PE Chemical exam, leading to a factor-of-four ($400%$) error!

Range of Validity: The Chilton-Colburn analogy provides remarkable accuracy (±10%) across turbulent regimes in pipes, flat plates, and conduits for:

• $0.6 \le Sc \le 3,000$
• $0.6 \le Pr \le 10,000$
• $Re > 10,000$

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3. Standard Empirical Correlations for Practical Engineering Geometries

1. Mass Transfer from a Sphere in Stagnant Fluid ($Re = 0$)

Consider a spherical pellet or droplet of diameter $d_p$ suspended in an infinitely large quiescent fluid ($u = 0$). In spherical coordinates, steady-state radial diffusion is governed by:

$\frac{d}{dr}\left( r^2 \frac{dC_A}{dr} \right) = 0$

Integrating from the sphere surface ($r = R = d_p / 2, C_A = C_{As}$) to infinity ($r \to \infty, C_A = C_{A\infty}$):

$C_A(r) = C_{A\infty} + (C_{As} - C_{A\infty}) \frac{R}{r}$

The molar flux leaving the surface is:

$N_{As} = -D_{AB} \left. \frac{dC_A}{dr} \right|_{r=R} = -D_{AB} \left[ -(C_{As} - C_{A\infty}) \frac{R}{R^2} \right] = \frac{D_{AB}}{R} (C_{As} - C_{A\infty}) = \frac{2 D_{AB}}{d_p} (C_{As} - C_{A\infty})$

Equating this to convective flux $N_{As} = k_c (C_{As} - C_{A\infty})$:

$k_c = \frac{2 D_{AB}}{d_p} \implies \mathbf{Sh = \frac{k_c d_p}{D_{AB}} = 2.0}$

[!NOTE] The Asymptotic Limit $Sh = 2.0$:
For a 3D sphere, the Sherwood number can never drop below $2.0$, no matter how slow the fluid moves! The value $Sh = 2.0$ represents the absolute lower thermodynamic bound for 3D steady-state radial molecular conduction.

2. Droplets, Bubbles, and Solid Spheres in Flow (Froessling / Ranz-Marshall)

When fluid flows past a sphere, forced convection augments molecular diffusion. The classic Froessling (or Ranz-Marshall) correlation superimposes convective boundary layer thinning onto the conductive limit:

$\mathbf{Sh = 2.0 + 0.60 \cdot Re_p^{1/2} \cdot Sc^{1/3}}$

Where:

• $Re_p = \frac{\rho u d_p}{\mu}$ = particle Reynolds number ($0 \le Re_p \le 200$).
• $Sc = \frac{\mu}{\rho D_{AB}}$ ($0.6 \le Sc \le 250$).
• This equation is identical in functional form to the Ranz-Marshall heat transfer equation: $Nu = 2.0 + 0.60 Re_p^{1/2} Pr^{1/3}$.

3. Flow Inside Circular Tubes

For mass transfer to or from the inner wall of a smooth cylindrical tube of inner diameter $D$:

• Laminar Flow ($Re < 2,100$):
  For fully developed velocity and concentration profiles with uniform wall concentration:
  $Sh = 3.66$ For uniform wall mass flux:
  $Sh = 4.36$ For the laminar entrance region, the Graetz-Leveque mass transfer equation applies:
  $Sh_{avg} = 1.62 \left( Re \cdot Sc \cdot \frac{D}{L} \right)^{1/3}$

• Turbulent Flow ($Re > 10,000, 0.7 \le Sc \le 160$):
  The Linton-Sherwood (or Gilliland) correlation relates pipe mass transfer to fluid properties:
  $\mathbf{Sh = 0.023 \cdot Re^{0.83} \cdot Sc^{0.44}}$ Alternatively, invoking the Chilton-Colburn analogy with the Blasius friction factor ($f = 0.079 Re^{-0.25}$): $Sh = 0.023 \cdot Re^{0.80} \cdot Sc^{1/3}$

4. Flow Over a Flat Plate

For boundary layer flow over a smooth flat plate of length $L$:

• Laminar Boundary Layer ($Re_L < 5 \times 10^5$):
  $Sh_{avg} = 0.664 \cdot Re_L^{1/2} \cdot Sc^{1/3}$
• Turbulent Boundary Layer ($Re_L > 5 \times 10^5$):
  $Sh_{avg} = 0.037 \cdot Re_L^{0.80} \cdot Sc^{1/3}$

5. Flow Through Packed Beds (Wakao and Kaguei)

For fluid flowing through a packed bed of spheres of diameter $d_p$:

$\mathbf{Sh = 2.0 + 1.1 \cdot Re_p^{0.60} \cdot Sc^{1/3}}$

Where $Re_p = \frac{\rho u_{superficial} d_p}{\mu}$ ($3 \le Re_p \le 10,000$).

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4. Summary Table of Dimensionless Groups and Analogies

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5. Comprehensive Worked Numerical Example: Sublimation from a Suspended Sphere

Problem Statement

To calibrate an aerodynamic test section, a smooth spherical pellet of pure naphthalene ($\text{C}_{10}\text{H}_8$, $M_A = 128.17\text{ g/mol}$) having a diameter of $d_p = 25.0\text{ mm} = 0.0250\text{ m}$ is suspended in a dry air stream inside a wind tunnel.

Operating Conditions:

• Air temperature: $T = 300.0\text{ K}$ ($26.85^\circ\text{C}$).
• Absolute tunnel pressure: $P = 1.01325\text{ bar} = 101,325\text{ Pa}$.
• Free-stream air velocity: $u = 4.50\text{ m/s}$.

Physical Properties at $300.0\text{ K}$:

• Air density: $\rho = 1.177\text{ kg/m}^3$.
• Air dynamic viscosity: $\mu = 1.85 \times 10^{-5}\text{ Pa}\cdot\text{s} = 1.85 \times 10^{-5}\text{ kg/(m}\cdot\text{s)}$.
• Air kinematic viscosity: $\nu = \mu / \rho = 1.5718 \times 10^{-5}\text{ m}^2/\text{s}$.
• Binary diffusivity of naphthalene in air: $D_{AB} = 6.11 \times 10^{-6}\text{ m}^2/\text{s}$.
• Vapor pressure of solid naphthalene: $P_A^* = 11.0\text{ Pa}$.
• Universal gas constant: $R = 8.31446\text{ J/(mol}\cdot\text{K)}$.

Calculate:

1. The particle Reynolds number ($Re_p$) and Schmidt number ($Sc$).
2. The mass transfer Peclet number ($Pe_m$).
3. The Sherwood number ($Sh$) using the Froessling correlation.
4. The convective mass transfer coefficient ($k_c$) in $\text{m/s}$.
5. The Chilton-Colburn $j_D$ factor.
6. The molar sublimation flux ($N_A$) in $\text{mol/(m}^2\cdot\text{s)}$ and total mass sublimation rate ($\dot{m}_A$) in $\text{mg/hr}$ from the sphere.

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Step 1: Dimensionless Group Evaluation ($Re_p$ and $Sc$)

Particle Reynolds number:

$Re_p = \frac{\rho u d_p}{\mu} = \frac{(1.177\text{ kg/m}^3) \times (4.50\text{ m/s}) \times (0.0250\text{ m})}{1.85 \times 10^{-5}\text{ Pa}\cdot\text{s}}$ $Re_p = \frac{0.1324125}{1.85 \times 10^{-5}} = \mathbf{7,157.4} \approx \mathbf{7,157}$

Schmidt number:

$Sc = \frac{\nu}{D_{AB}} = \frac{1.5718 \times 10^{-5}\text{ m}^2/\text{s}}{6.11 \times 10^{-6}\text{ m}^2/\text{s}} = \mathbf{2.5725} \approx \mathbf{2.573}$

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Step 2: Mass Transfer Peclet Number ($Pe_m$)

$Pe_m = Re_p \cdot Sc = 7,157.4 \times 2.5725 = \mathbf{18,412}$

Convective advection dominates molecular diffusion by more than four orders of magnitude.

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Step 3: Sherwood Number Determination

Applying the Froessling correlation for forced convection past a sphere:

$Sh = 2.0 + 0.60 \cdot Re_p^{1/2} \cdot Sc^{1/3}$

Evaluate components:

• $Re_p^{1/2} = \sqrt{7,157.4} = 84.601$
• $Sc^{1/3} = (2.5725)^{1/3} = 1.3702$

Substitute into correlation: $Sh = 2.0 + 0.60 \times 84.601 \times 1.3702 = 2.0 + 0.60 \times 115.920$ $Sh = 2.0 + 69.552 = \mathbf{71.552} \approx \mathbf{71.55}$

(Note: Stagnant conduction represents $2.0 / 71.55 = 2.8%$ of total transfer, while forced convection contributes $97.2%$).

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Step 4: Convective Mass Transfer Coefficient ($k_c$)

$Sh = \frac{k_c d_p}{D_{AB}} \implies k_c = \frac{Sh \cdot D_{AB}}{d_p}$

$k_c = \frac{71.552 \times (6.11 \times 10^{-6}\text{ m}^2/\text{s})}{0.0250\text{ m}} = \frac{4.3718 \times 10^{-4}}{0.0250} = \mathbf{0.017487\text{ m/s}} = \mathbf{1.749 \times 10^{-2}\text{ m/s}}$

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Step 5: Chilton-Colburn $j_D$ Factor

Mass transfer Stanton number: $St_m = \frac{k_c}{u} = \frac{0.017487\text{ m/s}}{4.50\text{ m/s}} = 3.8860 \times 10^{-3}$

Evaluate $Sc^{2/3}$: $Sc^{2/3} = (2.5725)^{2/3} = 1.8774$

Calculate $j_D$: $j_D = St_m \cdot Sc^{2/3} = (3.8860 \times 10^{-3}) \times 1.8774 = \mathbf{7.296 \times 10^{-3}} \approx \mathbf{7.30 \times 10^{-3}}$

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Step 6: Sublimation Flux and Mass Sublimation Rate

Surface vapor concentration of naphthalene: $C_{As} = \frac{P_A^*}{R T} = \frac{11.0\text{ Pa}}{(8.31446\text{ J/(mol}\cdot\text{K)}) \times (300.0\text{ K})} = \frac{11.0}{2,494.34} = \mathbf{4.40998 \times 10^{-3}\text{ mol/m}^3}$

Since bulk air contains zero naphthalene ($C_{A\infty} = 0$), the driving force is $\Delta C_A = C_{As}$:

$N_A = k_c \cdot C_{As} = (0.017487\text{ m/s}) \times (4.40998 \times 10^{-3}\text{ mol/m}^3) = \mathbf{7.7117 \times 10^{-5}\text{ mol/(m}^2\cdot\text{s)}}$

Total outer surface area of the sphere: $A_s = \pi d_p^2 = \pi \times (0.0250\text{ m})^2 = \mathbf{1.9635 \times 10^{-3}\text{ m}^2}$

Molar sublimation rate: $\dot{n}_A = N_A \cdot A_s = (7.7117 \times 10^{-5}\text{ mol/(m}^2\cdot\text{s)}) \times (1.9635 \times 10^{-3}\text{ m}^2) = 1.5142 \times 10^{-7}\text{ mol/s}$

Mass sublimation rate: $\dot{m}_A = \dot{n}_A \cdot M_A = (1.5142 \times 10^{-7}\text{ mol/s}) \times (128.17\text{ g/mol}) = 1.9407 \times 10^{-5}\text{ g/s} = 1.9407 \times 10^{-2}\text{ mg/s}$

Converting to hourly loss: $\dot{m}_A = 1.9407 \times 10^{-2}\text{ mg/s} \times 3,600\text{ s/hr} = \mathbf{69.87\text{ mg/hr}} \approx \mathbf{69.9\text{ mg/hr}}$

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6. Critical NCEES PE Exam Traps & Pitfalls

Trap 1: Confusing Fanning and Darcy Friction Factors in the Chilton-Colburn Analogy
The Chilton-Colburn relationship is defined strictly as $j_D = f / 2$, where $f$ is the Fanning friction factor (found on the PE Handbook mass transfer page and in chemical engineering fluid flow tables). In mechanical civil hydraulics tables, the Darcy friction factor is $f_D = 4f$. If you substitute $f_D$ directly into $j_D = f_D / 2$, you will calculate a mass transfer coefficient that is $400%$ too large! Remember: $j_D = f_{\text{Fanning}} / 2 = f_{\text{Darcy}} / 8$.

Trap 2: Forgetting the Asymptotic Lower Limit $Sh = 2.0$ for Spheres and Droplets
In spray scrubbers, fluid bed driers, or evaporative coolers, when relative droplet velocity approaches zero ($u \to 0, Re_p \to 0$), the convective contribution vanishes, but $Sh$ does not equal zero! It approaches $Sh = 2.0$ asymptotically due to 3D spherical molecular diffusion. Assuming $k_c \to 0$ when velocity stops will cause you to grossly miscalculate droplet drying or dissolution times.

Trap 3: Applying Gas-Phase Correlations Directly to Liquid-Phase Flows
Gas-phase mass transfer correlations often assume $Sc \approx 1$. If an engineer uses a gas-phase correlation for water flow ($Sc \approx 1,000$), the resulting coefficient will be completely erroneous. In liquids, because $Sc$ is so large, the mass transfer resistance is confined to an extremely thin sublayer ($\delta_c \approx 0.1 \delta_u$), requiring correlations with proper $Sc^{1/3}$ or $Sc^{0.44}$ dampening terms.