Lawrence LabLouisiana State UniversityLab website →
Home › Chapter 13: Convective Mass Transfer
Chapter 13

Convective Mass Transfer

Diffusion alone is slow. A stagnant boundary layer next to a membrane surface can limit mass transfer far more than the membrane itself. This chapter introduces the film model, the Sherwood number, and the heat-mass analogy, giving you the tools to predict convective mass transfer coefficients in tubes, over flat plates, and in membrane modules under concentration polarization.

📖 ~45 min read 📝 10 equations 🧠 3 worked examples 🔎 5 concept checks
Why This Chapter Matters

The Dialysis Dilemma

A kidney dialysis machine must remove urea from blood at a rate of about 180 mg per hour. The dialysis membrane itself is permeable enough. Yet clinical data show that simply increasing the membrane area beyond a certain point barely improves clearance. Why? Because the limiting resistance is not the membrane: it is the thin, nearly-stagnant layer of blood flowing slowly past the membrane surface. Urea diffuses through this film from the bulk blood to the membrane, and that film is the bottleneck.

The same phenomenon controls oxygen transfer in bioreactors, solvent stripping in polymer devolatilizers, salt rejection in reverse osmosis, and drug release from coated tablets. In every case, convective mass transfer in the fluid phase determines the rate, and you need the Sherwood number to predict it.

Before you read: If you doubled the flow rate of blood past a dialysis membrane, would the urea clearance rate double, increase by less than a factor of two, or stay the same? Why?

13.1 Film Theory: The Mass Transfer Coefficient

Chapter 11 established that diffusion in a stagnant fluid gives a flux proportional to the concentration gradient. Near any solid surface in a flowing fluid, there exists a thin concentration boundary layer (the film) of thickness $\delta$ in which convection is suppressed and diffusion dominates. Film theory replaces the actual, complicated concentration profile in this layer with a linear approximation, giving a simple rate equation:

(13.1) $$N = k_m \bigl(C_\infty - C_s\bigr)$$
$N$ [mol/(m²·s)] is the molar flux from bulk to surface; $k_m$ [m/s] is the mass transfer coefficient; $C_\infty$ [mol/m³] is the bulk (far-field) concentration; $C_s$ [mol/m³] is the surface concentration. This equation has the same form as Newton's Law of Cooling, $q = h(T_\infty - T_s)$.

Within the film, molecular diffusion carries the solute across thickness $\delta$. Applying Fick's first law to a linear profile:

(13.2) $$k_m = \frac{D}{\delta}$$
$D$ [m²/s] is the molecular diffusion coefficient of the solute in the solvent; $\delta$ [m] is the effective film thickness. A thinner film (higher flow velocity, turbulence) or a higher $D$ (smaller solute, lower viscosity) gives a larger $k_m$ and faster mass transfer. Film theory does not predict $\delta$ directly: that requires a hydrodynamic analysis captured by Sherwood number correlations in Section 13.3.
membrane film, δ $C_s$ $C_\infty$ $N = k_m(C_\infty - C_s)$ film region well-mixed bulk concentration C distance from membrane →

Figure 13.1: Film theory concentration profile. In the thin stagnant film of thickness $\delta$, concentration decreases linearly from $C_\infty$ (bulk) to $C_s$ (membrane surface). The mass transfer coefficient $k_m = D/\delta$ is set by film thickness and diffusivity. A higher flow velocity thins $\delta$ and increases $k_m$.

Resistances in Series

When a solute must cross a fluid film, a solid membrane, and a second fluid film (as in dialysis or hollow-fiber modules), the resistances add in series, just as thermal resistances do. The overall mass transfer coefficient $K_{\rm ov}$ [m/s] is:

(13.3) $$\frac{1}{K_{\rm ov}} = \frac{1}{k_{m1}} + \frac{L}{P} + \frac{1}{k_{m2}}$$
$k_{m1}$ and $k_{m2}$ [m/s] are the film-side mass transfer coefficients on each side of the membrane; $L$ [m] is the membrane thickness; $P$ is the membrane permeability in consistent units. The resistance $1/k_{m1}$ dominates when the film is thick or diffusivity is low. In dialysis, the blood-side film often controls the overall rate.
Analogy with heat transfer. Equations 13.1 and 13.3 are structurally identical to Newton's law of cooling ($q = h\Delta T$) and the series thermal resistance model from Chapter 2. Every concept carries over: if one resistance is much larger than the others, it controls the overall rate. Improving a non-controlling resistance has little effect on performance.
Example 13.1

Film mass transfer coefficient in a kidney dialyzer

Blood flows past a dialysis membrane. The diffusion coefficient of urea in plasma is $D = 1.5\times10^{-9}$ m²/s and the estimated film thickness is $\delta = 150\;\mu\mathrm{m}$. Find $k_m$ and the urea flux if $C_\infty = 10\;\mathrm{mmol/L}$ and dialysate maintains $C_s \approx 0$.

Solution

Convert: $\delta = 1.5\times10^{-4}$ m. Apply Eq. 13.2:

$$k_m = \frac{D}{\delta} = \frac{1.5\times10^{-9}}{1.5\times10^{-4}} = 1.0\times10^{-5}\;\mathrm{m/s}$$

Convert: $C_\infty = 10\;\mathrm{mmol/L} = 10\;\mathrm{mol/m^3}$. Apply Eq. 13.1:

$$N = k_m(C_\infty - C_s) = (1.0\times10^{-5})(10 - 0) = 1.0\times10^{-4}\;\mathrm{mol/(m^2{\cdot}s)}$$

The flux is $0.1\;\mathrm{mmol/(m^2{\cdot}s)}$. With a typical dialyzer surface area of $1.5\;\mathrm{m^2}$, total urea removal is $1.5\times10^{-4}$ mol/s, consistent with clinical requirements.

13.2 Dimensionless Groups

Just as heat transfer uses the Nusselt number to correlate convective heat transfer data, mass transfer uses three analogous dimensionless groups. The correlations in Section 13.3 express $k_m$ in terms of these groups.

Sherwood Number: the Mass Transfer Nusselt Number

The Sherwood number is the ratio of convective mass transfer to diffusive mass transfer across a characteristic length $d$:

(13.4) $$\mathrm{Sh} = \frac{k_m\, d}{D}$$
$k_m$ [m/s] is the mass transfer coefficient; $d$ [m] is the characteristic length (tube diameter, plate length, etc.); $D$ [m²/s] is the diffusion coefficient. $\mathrm{Sh} = 1$ means mass transfer equals pure diffusion across the characteristic length. Higher Sh means convection enhances the rate. Analogous to: $\mathrm{Nu} = hd/k_{\rm thermal}$.

Schmidt Number: the Mass Transfer Prandtl Number

The Schmidt number compares momentum diffusivity to mass diffusivity:

(13.5) $$\mathrm{Sc} = \frac{\nu}{D}$$
$\nu$ [m²/s] is the kinematic viscosity of the solvent; $D$ [m²/s] is the diffusion coefficient of the solute. For gases, $\mathrm{Sc} \approx 1$. For liquids, $\mathrm{Sc}$ ranges from about 100 to $10^5$ because $D$ is orders of magnitude smaller than $\nu$. Analogous to: $\mathrm{Pr} = \nu/\alpha$.

The Reynolds number is the same as in fluid mechanics and heat transfer:

(13.6) $$\mathrm{Re} = \frac{u\, d}{\nu}$$
$u$ [m/s] is the characteristic velocity; $d$ [m] is the characteristic length. $\mathrm{Re} < 2300$ is laminar in a tube; $\mathrm{Re} > 10{,}000$ is turbulent. The Sherwood number correlations have exactly the same form as Nusselt correlations, with Sc replacing Pr.
Why Sc is so large in liquids. Water at 25°C has $\nu = 10^{-6}$ m²/s. A typical small solute like urea has $D \approx 10^{-9}$ m²/s, giving $\mathrm{Sc} = 1000$. A dissolved polymer chain might have $D \approx 10^{-11}$ m²/s, giving $\mathrm{Sc} = 10^5$. This enormous Sc means the concentration boundary layer is much thinner than the momentum boundary layer, and Sh correlations become more sensitive to flow conditions.

13.3 Sherwood Number Correlations

The Sherwood number depends on the flow geometry, the flow regime, and entry-length effects. The following correlations are the mass-transfer analogs of the Nusselt number correlations for heat transfer.

Tube Flow

For laminar, fully developed flow in a tube ($\mathrm{Re}\,\mathrm{Sc}\,(d/L) \lesssim 10$, constant wall concentration):

(13.7a) $$\mathrm{Sh} = 3.66 \quad \bigl(\text{fully developed laminar, constant wall concentration}\bigr)$$
The mass-transfer analog of $\mathrm{Nu} = 3.66$ for heat transfer. Applies when the Graetz number $\mathrm{Gz} = \mathrm{Re}\,\mathrm{Sc}\,(d/L) \ll 1$. Note: $\mathrm{Sh} = 3.66$ means $k_m = 3.66\,D/d$. Smaller tube diameter increases $k_m$: this is why dialyzers use thousands of fine fibers.

In the entry region (developing concentration profile, high Graetz number):

(13.7b) $$\mathrm{Sh} = 1.86\,\mathrm{Re}^{1/3}\,\mathrm{Sc}^{1/3}\!\left(\frac{d}{L}\right)^{1/3}$$
The Graetz (Leveque) solution for the entry region, where $\mathrm{Gz} \gg 1$. The $d/L$ factor means shorter tubes give higher average Sh because fresh concentration gradient is always present. Used for hollow-fiber modules where $L/d$ is moderate and Sc is large.

For turbulent flow ($\mathrm{Re} > 10{,}000$), the Dittus-Boelter mass-transfer analog:

(13.8) $$\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$$
Valid for $\mathrm{Re} > 10{,}000$, $0.6 < \mathrm{Sc} < 160$, and $L/d > 10$. The 0.8 exponent on Re means doubling the flow velocity increases Sh (and $k_m$) by $2^{0.8} = 1.74$, not twofold. Analogous to: $\mathrm{Nu} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Pr}^{1/3}$.

Flat Plate

For laminar boundary layer flow over a flat plate of length $L$ ($\mathrm{Re}_L < 5\times10^5$):

(13.9) $$\overline{\mathrm{Sh}}_L = 0.664\,\mathrm{Re}_L^{1/2}\,\mathrm{Sc}^{1/3}$$
This gives the average Sherwood number over the plate length $L$. The average $k_m = \overline{\mathrm{Sh}}\,D/L$. Analogous to: $\overline{\mathrm{Nu}}_L = 0.664\,\mathrm{Re}_L^{1/2}\,\mathrm{Pr}^{1/3}$.
Range checks matter. Each correlation has explicit Re and Sc limits. Using Eq. 13.8 (turbulent tube) at Re = 500 gives a wildly wrong answer. Always verify that flow regime and Sc fall within the stated limits before applying a correlation.
Example 13.2

Laminar tube: km in a hollow-fiber dialyzer

Plasma flows through a hollow fiber of inner diameter $d = 200\;\mu\mathrm{m}$ and length $L = 0.20$ m in laminar, fully developed flow. The diffusion coefficient of urea in plasma is $D = 2.0\times10^{-9}$ m²/s. Find $k_m$.

Solution

Convert: $d = 2.0\times10^{-4}$ m. For fully developed laminar flow, use Eq. 13.7a:

$$\mathrm{Sh} = 3.66$$ $$k_m = \frac{\mathrm{Sh}\cdot D}{d} = \frac{3.66 \times 2.0\times10^{-9}}{2.0\times10^{-4}} = \frac{7.32\times10^{-9}}{2.0\times10^{-4}} = 3.66\times10^{-5}\;\mathrm{m/s}$$

This $k_m = 36.6\;\mu\mathrm{m/s}$ is about 3.7 times larger than the value from Example 13.1. Smaller tube diameter increases $k_m$ because $k_m = \mathrm{Sh}\cdot D/d$: halving $d$ doubles $k_m$. This explains why commercial dialyzers use thousands of very fine hollow fibers.

13.4 The Heat-Mass Transfer Analogy

The governing equations for heat transfer and mass transfer in a laminar boundary layer are mathematically identical when Pr is replaced by Sc and Nu by Sh. This analogy is more than a mnemonic: any Nusselt number correlation converts to a Sherwood number correlation by substituting $\mathrm{Pr} \to \mathrm{Sc}$.

Quantity Heat Transfer Mass Transfer
Driving potential $\Delta T$ [K] $\Delta C$ [mol/m³]
Flux $q = h\,\Delta T$ [W/m²] $N = k_m\,\Delta C$ [mol/(m²·s)]
Transport coefficient $h$ [W/(m²·K)] $k_m$ [m/s]
Fluid diffusivity ratio $\mathrm{Pr} = \nu/\alpha$ $\mathrm{Sc} = \nu/D$
Dimensionless rate $\mathrm{Nu} = hd/k_{\rm th}$ $\mathrm{Sh} = k_m d/D$
Laminar flat plate $\overline{\mathrm{Nu}} = 0.664\,\mathrm{Re}^{1/2}\,\mathrm{Pr}^{1/3}$ $\overline{\mathrm{Sh}} = 0.664\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}$
Turbulent tube $\mathrm{Nu} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Pr}^{1/3}$ $\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$

The Chilton-Colburn j-factor extends the analogy to include the friction factor, connecting momentum, heat, and mass transport:

(13.10) $$j_D = \mathrm{St}_m\,\mathrm{Sc}^{2/3} = \frac{f}{2}$$
$j_D$ is the Colburn j-factor for mass transfer; $\mathrm{St}_m = k_m/u$ is the Stanton number for mass transfer; $f$ is the Fanning friction factor. This relation means that if you know the pressure drop in a tube, you can estimate $k_m$ without a separate mass-transfer experiment. Holds for fully turbulent pipe flow and flat plates.

13.5 Concentration Polarization in Membrane Filtration

In pressure-driven membrane processes (reverse osmosis, ultrafiltration, nanofiltration), solvent is pushed through the membrane by hydraulic pressure while solute is partially or fully rejected. The rejected solute accumulates at the membrane surface, creating a concentration polarization layer. The wall concentration $C_w$ exceeds the bulk $C_b$, reducing flux and potentially causing fouling or gel formation.

membrane permeate bulk, $C_b$ polarization layer $C_b$ $C_w$ $J$ (solvent flux) $k_m$ back-diffusion concentration C distance → (toward membrane)

Figure 13.2: Concentration polarization at a membrane surface. Solvent flux $J$ carries solute toward the membrane; rejected solute diffuses back toward the bulk at rate $k_m$. At steady state, $C_w/C_b = \exp(J/k_m)$. High $J/k_m$ ratios cause severe polarization and eventual gel formation.

At steady state, convective transport of solute toward the membrane equals the diffusive back-flux toward the bulk. Solving the 1-D mass balance in the polarization layer gives:

(13.11) $$\frac{C_w}{C_b} = \exp\!\left(\frac{J}{k_m}\right)$$
$C_w$ [mol/m³] is the wall (membrane surface) concentration; $C_b$ [mol/m³] is the bulk concentration; $J$ [m/s] is the volumetric permeate flux (m³/(m²·s)); $k_m$ [m/s] is the mass transfer coefficient in the feed channel. When $J \ll k_m$, polarization is negligible. When $J/k_m \approx 1$, $C_w \approx 2.7\,C_b$: severe polarization.
Concentration polarization limits performance. As $C_w$ rises, the osmotic pressure at the membrane surface increases, reducing the effective driving force. In extreme cases, $C_w$ reaches the gel concentration $C_{\rm gel}$ and the flux is limited at $J_{\rm lim} = k_m \ln(C_{\rm gel}/C_b)$, regardless of applied pressure. Operating at high $k_m$ (turbulent flow, spacers, pulsation) is the engineering response.
Example 13.3

Concentration polarization in an ultrafiltration module

An ultrafiltration module operates with permeate flux $J = 2.0\times10^{-5}$ m/s and mass transfer coefficient $k_m = 2.0\times10^{-4}$ m/s. The bulk protein concentration is $C_b = 5.0$ g/L. Find $C_w$ and assess whether gel formation is a concern (gel concentration $\approx 100$ g/L for this protein).

Solution

$$\frac{J}{k_m} = \frac{2.0\times10^{-5}}{2.0\times10^{-4}} = 0.10$$ $$C_w = C_b \exp\!\left(\frac{J}{k_m}\right) = 5.0 \times e^{0.10} = 5.0 \times 1.105 = 5.5\;\mathrm{g/L}$$

The wall concentration is $5.5$ g/L, only 10% above bulk and far below the gel concentration of 100 g/L: polarization is mild. If flux were increased to $J = 2.0\times10^{-4}$ m/s ($J/k_m = 1.0$), then $C_w = 5.0 \times e = 13.6$ g/L, still below gel but rising rapidly. The ratio $J/k_m$ is the key diagnostic for polarization severity.

Interactive: Sherwood Number Explorer

Adjust the Reynolds number, Schmidt number, and geometry to see how Sherwood number and mass transfer coefficient respond. Build intuition for when the turbulent correlation takes over and how large Sc amplifies mass transfer dependence on Re.

Concept Check

Five question pools, three variants each. The pool advances after a correct answer.

Question 1: Film theory and km

Doubling the flow velocity past a membrane surface will:

In a dialysis module, the blood-side film resistance is $1/k_{m1} = 5\times10^4$ s/m and the membrane resistance is $L/P = 1\times10^4$ s/m. The overall resistance is controlled by:

Film theory assumes the concentration profile within the boundary layer is:

Concept Summary

Film theory simplifies the concentration boundary layer to a stagnant film with a linear concentration profile. The mass transfer coefficient $k_m = D/\delta$ increases when the film is thinned by higher velocity or turbulence. In series resistance models ($1/K_{\rm ov} = 1/k_{m1} + L/P + 1/k_{m2}$), the largest resistance controls the rate. Improving a non-controlling resistance yields negligible benefit.

Question 2: Dimensionless groups

Water ($\nu = 1.0\times10^{-6}$ m²/s) contains a dissolved polymer with $D = 1.0\times10^{-11}$ m²/s. The Schmidt number is:

$\mathrm{Sh} = k_m d / D = 50$ means:

In mass transfer correlations, the Schmidt number plays the same role as which heat transfer group?

Concept Summary

The three key dimensionless groups: $\mathrm{Sh} = k_m d/D$ (like Nu), $\mathrm{Sc} = \nu/D$ (like Pr), $\mathrm{Re} = ud/\nu$ (unchanged). For gases, $\mathrm{Sc} \approx 1$. For small solutes in liquids, $\mathrm{Sc} \approx 10^2$ to $10^3$. For dissolved polymers, $\mathrm{Sc}$ can reach $10^5$. The heat-mass analogy: substitute $\mathrm{Pr} \to \mathrm{Sc}$ and $\mathrm{Nu} \to \mathrm{Sh}$ in any Nusselt correlation.

Question 3: Sherwood number correlations

For turbulent flow ($\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$), doubling the velocity increases $k_m$ by a factor of:

For fully developed laminar flow (Sh = 3.66), increasing tube diameter by 4× while keeping velocity fixed:

A dialyzer designer wants to increase $k_m$ for laminar flow in hollow fibers. Which change gives the greatest improvement?

Concept Summary

For laminar fully developed tube flow, $\mathrm{Sh} = 3.66$ (constant), so $k_m = 3.66\,D/d$ decreases with diameter. For turbulent tube flow, $\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$: doubling velocity increases $k_m$ by $2^{0.8} = 1.74$. For laminar flat plates, $\overline{\mathrm{Sh}} = 0.664\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}$. In laminar hollow-fiber devices, reducing fiber diameter is the most impactful lever.

Question 4: Heat-mass analogy

Converting a Nusselt number correlation to a Sherwood correlation requires substituting:

The Chilton-Colburn analogy $j_D = f/2$ is most useful for:

For air ($\mathrm{Sc} \approx 1$) flowing over a flat plate, the concentration and velocity boundary layers are:

Concept Summary

The heat-mass analogy converts any Nusselt correlation to a Sherwood correlation via $\mathrm{Nu} \to \mathrm{Sh}$ and $\mathrm{Pr} \to \mathrm{Sc}$ (Re unchanged). The Chilton-Colburn analogy $j_D = \mathrm{St}_m\cdot\mathrm{Sc}^{2/3} = f/2$ connects mass transfer to fluid friction: measured pressure drop lets you estimate $k_m$. Concentration and velocity boundary layers have the same thickness when $\mathrm{Sc} = 1$ (gases); for $\mathrm{Sc} \gg 1$ (liquids), the concentration layer is much thinner.

Question 5: Concentration polarization

In a reverse osmosis system, $J/k_m = 1.0$. The concentration polarization ratio $C_w/C_b$ is approximately:

To reduce concentration polarization in an ultrafiltration module, the most effective strategy is to:

The gel polarization model predicts that when $C_w = C_{\rm gel}$, the limiting flux is:

Concept Summary

Concentration polarization: $C_w/C_b = \exp(J/k_m)$. Small $J/k_m$ means mild polarization. At $J/k_m = 1$, $C_w \approx 2.7\,C_b$. At the gel limit, $J_{\rm lim} = k_m\ln(C_{\rm gel}/C_b)$ regardless of applied pressure. To reduce polarization: increase $k_m$ (crossflow, turbulence promoters, spacers) or reduce $J$ (lower pressure). Increasing pressure beyond the gel limit does not increase flux.

Problem Set : Chapter 13

Part A — Short Answer

C1
Easy

A dialysis engineer proposes to improve urea clearance by increasing the blood flow rate by 50%. A colleague argues that halving the hollow-fiber diameter would be more effective. Using film theory and Sh = 3.66, evaluate both proposals qualitatively.

C2
Easy

In a reverse osmosis system operating near the gel polarization limit, an operator doubles the transmembrane pressure. Predict what happens to the permeate flux and explain using the concentration polarization model.

C3
Medium

A gas ($\mathrm{Sc} \approx 1$) and a dissolved polymer ($\mathrm{Sc} \approx 10^5$) flow at the same Re in a turbulent tube. By what factor does $k_m$ differ between the two cases? Which has the higher $k_m$?

C4
Medium

The Chilton-Colburn $j$-factor analogy states $j_D = \mathrm{St}_m\,\mathrm{Sc}^{2/3} = f/2$, linking mass transfer to the Fanning friction factor. Explain in physical terms why a higher friction factor should accompany a higher mass transfer coefficient. Under what flow conditions does this analogy break down?

C5
Hard

In a crossflow ultrafiltration module, an engineer observes that increasing transmembrane pressure above a certain threshold does not increase permeate flux. Using the concentration polarization framework, explain this observation. Identify two operating changes the engineer could make to recover flux, and explain the mechanism by which each change acts.

Part B — Calculations

P1
Easy

A stagnant plasma film ($D_{\rm urea} = 1.5\times10^{-9}$ m²/s) of thickness $\delta = 150\;\mu\mathrm{m}$ coats a dialysis membrane. Calculate the mass transfer coefficient $k_m$ in μm/s.

$k_m$ = μm/s
Setup

Use $k_m = D/\delta$. Convert: $\delta = 150\;\mu\mathrm{m} = 1.5\times10^{-4}$ m.

Calculate

$k_m = 1.5\times10^{-9}/1.5\times10^{-4} = 1.0\times10^{-5}$ m/s $= 10.0\;\mu\mathrm{m/s}$.

P2
Easy

Using $k_m$ from P1, compute the urea flux $N$ in mmol/(m²·s) when the bulk urea concentration is $C_\infty = 100$ mol/m³ and the dialysate maintains $C_s = 0$.

$N$ = mmol/(m²·s)
Setup

$N = k_m(C_\infty - C_s) = (1.0\times10^{-5})(100)$ mol/(m²·s).

Convert

$N = 1.0\times10^{-3}$ mol/(m²·s) $= 1.0$ mmol/(m²·s).

P3
Medium

A hollow-fiber dialyzer operates under conditions where the Sherwood number has been determined to be Sh = 20. The fiber inner diameter is $d = 2.0$ mm and the diffusivity of urea is $D = 2.0\times10^{-9}$ m²/s. Find $k_m$ in μm/s.

$k_m$ = μm/s
Setup

Rearrange the Sherwood number definition: $k_m = \mathrm{Sh}\cdot D/d$.

Calculate

$k_m = 20 \times 2.0\times10^{-9} / (2.0\times10^{-3}) = 4.0\times10^{-8}/2.0\times10^{-3} = 2.0\times10^{-5}$ m/s $= 20.0\;\mu\mathrm{m/s}$.

P4
Easy

A polymer solution has $D = 1.0\times10^{-11}$ m²/s and water kinematic viscosity $\nu = 1.0\times10^{-6}$ m²/s. Calculate the Schmidt number. (Enter as a plain integer.)

Sc = (dimensionless)
Setup

$\mathrm{Sc} = \nu/D = 1.0\times10^{-6}/1.0\times10^{-11}$.

P5
Medium

A dialysis module has three resistances in series: a blood-side film ($k_{m1} = 2.0\times10^{-5}$ m/s), a membrane ($P/L = 4.0\times10^{-5}$ m/s), and a dialysate-side film ($k_{m2} = 4.0\times10^{-5}$ m/s). Calculate the overall mass transfer coefficient $K_{\rm ov}$ in μm/s.

$K_{\rm ov}$ = μm/s
Setup

Add resistances in series: $1/K_{\rm ov} = 1/k_{m1} + L/P + 1/k_{m2}$.

Substitute

$1/K_{\rm ov} = 1/(2.0\times10^{-5}) + 1/(4.0\times10^{-5}) + 1/(4.0\times10^{-5})$ $= 5.0\times10^4 + 2.5\times10^4 + 2.5\times10^4 = 1.0\times10^5$ s/m.

Invert

$K_{\rm ov} = 1.0\times10^{-5}$ m/s $= 10.0\;\mu\mathrm{m/s}$.

Chapter Summary

Film Theory

$N = k_m(C_\infty - C_s)$ and $k_m = D/\delta$. Mass transfer coefficient equals diffusivity divided by film thickness. Higher velocity thins $\delta$ and increases $k_m$. Series resistances: $1/K_{\rm ov} = 1/k_{m1} + L/P + 1/k_{m2}$. The largest resistance controls.

Dimensionless Groups

$\mathrm{Sh} = k_m d/D$ (like Nu), $\mathrm{Sc} = \nu/D$ (like Pr), $\mathrm{Re} = ud/\nu$ (unchanged). For gases, $\mathrm{Sc} \approx 1$. For dissolved polymers, $\mathrm{Sc}$ can reach $10^5$. Heat-mass analogy: substitute $\mathrm{Pr} \to \mathrm{Sc}$ and $\mathrm{Nu} \to \mathrm{Sh}$.

Tube Correlations

Laminar fully developed: $\mathrm{Sh} = 3.66$ (so $k_m = 3.66\,D/d$). Graetz entry: $1.86\,\mathrm{Re}^{1/3}\mathrm{Sc}^{1/3}(d/L)^{1/3}$. Turbulent: $0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$. Flat plate: $0.664\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}$. Always check flow regime and Sc range.

Chilton-Colburn Analogy

$j_D = \mathrm{St}_m\,\mathrm{Sc}^{2/3} = f/2$. The j-factor connects mass transfer to the Fanning friction factor. Measured pressure drop lets you estimate $k_m$ without separate mass transfer experiments. Applies to turbulent flow and flat plates.

Concentration Polarization

$C_w/C_b = \exp(J/k_m)$. The ratio $J/k_m$ governs severity. Gel limit: $J_{\rm lim} = k_m\ln(C_{\rm gel}/C_b)$. Increasing pressure beyond the gel limit does not increase flux. To reduce polarization: raise $k_m$ via crossflow or turbulence promoters.

Design Rules

Hollow fibers: halving $d$ doubles $k_m$ (for fixed $D$ in laminar Sh = 3.66). Turbulent flow: doubling velocity increases $k_m$ by $2^{0.8} = 1.74$. Identify the controlling resistance first before investing effort in the wrong part of the system.

Key Equations
  • Film flux: $N = k_m(C_\infty - C_s)$
  • Film coefficient: $k_m = D/\delta$
  • Series resistance: $1/K_{\rm ov} = 1/k_{m1} + L/P + 1/k_{m2}$
  • Sherwood: $\mathrm{Sh} = k_m d/D$
  • Schmidt: $\mathrm{Sc} = \nu/D$
  • Turbulent tube: $\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}$
  • Laminar flat plate: $\overline{\mathrm{Sh}} = 0.664\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}$
  • Polarization: $C_w/C_b = \exp(J/k_m)$
  • Chilton-Colburn: $j_D = \mathrm{St}_m\,\mathrm{Sc}^{2/3} = f/2$

AI Tutor

Connecting…

Ask me anything about Chapter 13: film theory, Sherwood number, Schmidt number, Sh correlations, the heat-mass analogy, or concentration polarization.