Heat Transfer · Chapter Fourteen

Mass Transfer

Where heat transfer is driven by temperature differences, mass transfer is driven by concentration differences. The two are so alike in form that nearly every heat transfer relation has a mass transfer twin.

This chapter builds the heat–mass analogy, develops Fick's law of diffusion, treats boundary conditions and steady and transient diffusion, diffusion in moving media, and mass convection — ending with combined heat and mass transfer as in evaporative cooling.

Learning objectives

Distinguish molecular diffusion, bulk advection, and convective mass transfer.
Use mass and molar concentration measures consistently.
Apply Fick's law to steady and transient one-dimensional diffusion.
Select dimensionless mass-transfer correlations only within their geometry and range.
Couple heat and mass balances and audit units, directions, and limiting behavior.

14–1

Introduction

Mass transfer is the movement of a chemical species from a region of high concentration to one of lower concentration. It requires a mixture of at least two species and a concentration gradient.

Mass Transfer vs. Bulk Flow

Mass transfer refers to the motion of one species relative to others due to a concentration difference — distinct from bulk fluid flow, which moves the whole mixture by a pressure difference. A drop of dye spreading in still water is mass transfer; the same water poured into a cup is bulk flow.

Physical meaning

Species may diffuse relative to the mixture even when total mass-average velocity is zero. The minus sign in Fick's law records that spontaneous diffusion points down the concentration gradient; it is not an arbitrary sign convention.

14–2

Analogy between Heat and Mass Transfer

Temperature corresponds to concentration, thermal diffusivity to mass diffusivity, and Nusselt number to Sherwood number. The analogy transfers a correlation only when geometry, boundary condition, velocity definition, and laminar/turbulent regime also match; replacing \(Pr\) by \(Sc\) is not permission to ignore a correlation's validity range.

The Heat–Mass Analogy

The governing equations for heat and mass transfer have identical mathematical form. Replacing temperature with concentration, the Prandtl number with the Schmidt number, and the Nusselt number with the Sherwood number converts a heat transfer solution into its mass transfer counterpart.

\[ Pr \leftrightarrow Sc = \frac{\nu}{D_{AB}}, \qquad Nu \leftrightarrow Sh = \frac{h_{\text{mass}}\,L_c}{D_{AB}} \]

Lewis number (links the two)

\[ Le = \frac{Sc}{Pr} = \frac{\alpha}{D_{AB}} \]

14–3

Mass Diffusion

Just as Fourier's law governs heat conduction, Fick's law governs mass diffusion: the diffusive flux of a species is proportional to its concentration gradient.

Fick's Law of Diffusion

The rate of mass diffusion of a species through a stationary medium is proportional to the concentration gradient and the diffusion coefficient, and flows from high to low concentration — the direct analogue of Fourier's law.

\[ \dot{m}_{\text{diff},A} = -\rho\,D_{AB}\,A\,\frac{dw_A}{dx} \]

Molar form and concentrations

\[ \dot{N}_{\text{diff},A} = -C\,D_{AB}\,A\,\frac{dy_A}{dx} \]

\(w_A\) is mass fraction, \(y_A\) mole fraction, \(D_{AB}\) the binary diffusion coefficient (m²/s).

For gases, \(D_{AB}\) depends strongly on absolute temperature and pressure; for liquids it is generally smaller and viscosity-sensitive. Evaluate supplied or tabulated diffusivity at the mean state and do not combine a mass-based flux with a molar concentration gradient.

14–4

Boundary Conditions

Mass transfer boundary conditions parallel thermal ones, but at a phase interface the concentration is set by equilibrium relations such as Henry's law (gas–liquid) and Raoult's law (vapor over a solution).

Henry's law (dilute gas dissolved in liquid)

\[ y_{A,\text{liquid side}} = \frac{P_{A,\text{gas side}}}{H} \]

\(H\) is Henry's constant; a larger \(H\) means a less soluble gas.

At an impermeable boundary the normal species flux is zero; at a prescribed surface concentration, concentration is fixed; at an interface, equilibrium supplies the interfacial value while flux continuity connects both phases. Henry constants appear in several conventions, so units and definition must accompany the number.

14–5

Steady Mass Diffusion through a Wall

Steady one-dimensional diffusion through a plane membrane mirrors steady conduction through a wall, with an analogous "mass resistance."

Diffusion through a plane wall

\[ \dot{N}_{\text{diff},A} = D_{AB}\,A\,\frac{C_{A,1} - C_{A,2}}{L} = \frac{C_{A,1} - C_{A,2}}{R_{\text{mass}}}, \quad R_{\text{mass}} = \frac{L}{D_{AB}A} \]

EXAMPLE 1Oxygen Transport through a Sensor Film

Problem. Oxygen diffuses through a 0.80-mm polymer sensor film of area 24 cm². Its dissolved molar concentrations are 0.036 and 0.011 mol/m³ at the two faces, and \(D_{AB}=1.6\times10^{-10}\) m²/s. Find molar rate and direction.

Strategy

A stationary plane film with fixed face concentrations is steady one-dimensional equimolar diffusion, so integrated Fick's law applies.

Givens

\(L=8.0\times10^{-4}\) m, \(A=24\times10^{-4}\) m².
\(C_{A,1}=0.036,\ C_{A,2}=0.011\) mol/m³; \(D=1.6\times10^{-10}\) m²/s.

Physical/model interpretation

Constant \(D\), no reaction, \(+x\) from face 1 to face 2; the linear concentration decline makes flux positive.

Numbered derivation steps

1. Integrate Fick's law. \(\dot N_A=D A(C_{A,1}-C_{A,2})/L\).

2. Find gradient magnitude. \((0.036-0.011)/(8.0\times10^{-4})=31.25\ {\rm mol/m^4}\).

3. Find flux. \(N_A''=(1.6\times10^{-10})(31.25)=5.00\times10^{-9}\ {\rm mol/m^2s}\).

4. Find rate. \(\dot N_A=(5.00\times10^{-9})(2.4\times10^{-3})=1.20\times10^{-11}\ {\rm mol/s}\).

Unit/sign/limit checks

Units: In \(DA\Delta C/L\), \(({\rm m^2/s})({\rm m^2})({\rm mol/m^3})/{\rm m}={\rm mol/s}\); thus the result is dimensionally a molar rate rather than a molar flux.

Signs: Flow is from face 1, the higher concentration, toward face 2.

Limits: Equal face concentrations would give zero rate, and doubling thickness would halve it.

Final answer: \(\boxed{\dot N_A=1.20\times10^{-11}\ {\rm mol/s},\ 1\to2}\).

Engineering insight

The very small rate is consistent with a dense polymer; real sensor response may also be limited by interfacial partitioning and reaction kinetics.

14–6

Transient Mass Diffusion

Short-time penetration of a species into a thick medium uses the same error-function solution as transient heat conduction in a semi-infinite solid — the analogy again.

Semi-infinite medium (error-function solution)

\[ \frac{C_A(x,t) - C_{A,i}}{C_{A,s} - C_{A,i}} = \operatorname{erfc}\!\left(\frac{x}{2\sqrt{D_{AB}\,t}}\right) \]

Used for case hardening (carbon into steel), doping, and moisture uptake.

EXAMPLE 2Salt Penetration into a Hydrogel

Problem. A thick hydrogel initially contains no salt. At \(t=0\), its surface concentration is held at 1.00 kg/m³. With \(D=9.0\times10^{-10}\) m²/s, estimate concentration 5.0 mm below the surface after 20 h; use \(\operatorname{erfc}(0.3106)=0.6605\).

Strategy

The penetration depth is small relative to the slab thickness (stated thick), so use the constant-surface-concentration semi-infinite solution.

Givens

\(C_i=0,\ C_s=1.00\) kg/m³, \(x=5.0\times10^{-3}\) m.
\(D=9.0\times10^{-10}\) m²/s, \(t=72{,}000\) s.

Physical/model interpretation

One-dimensional diffusion, constant \(D\), no swelling or reaction; \(x\) is measured inward from the maintained surface.

Numbered derivation steps

1. Convert time. \(20\ {\rm h}=72{,}000\ {\rm s}\).

2. Form similarity coordinate. \(\eta=x/[2\sqrt{Dt}]=0.0050/[2\sqrt{(9.0\times10^{-10})(72{,}000)}]=0.3106\).

3. Apply the solution. \((C-C_i)/(C_s-C_i)=\operatorname{erfc}(\eta)=0.6605\).

4. Resolve concentration. \(C=0+0.6605(1.00-0)=0.661\ {\rm kg/m^3}\).

Unit/sign/limit checks

Units: \(Dt\) has m², making \(\eta\) dimensionless.

Signs: Concentration rises inward from its zero initial value.

Limits: \(0<C<C_s\); as \(t\to0\) at fixed \(x\), \(C\to C_i\).

Final answer: \(\boxed{C(5.0{\rm\ mm},20{\rm\ h})=0.661\ {\rm kg/m^3}}\).

Engineering insight

Confirm that the opposite face remains several diffusion lengths away; otherwise a finite-slab solution is required.

14–7

Diffusion in a Moving Medium

When the medium itself moves (or when diffusion induces a bulk flow), the total flux includes both diffusion and advection. A classic case is evaporation into a stagnant gas column — Stefan flow.

Stefan flow (evaporation through a stagnant gas)

\[ \dot{N}_A = \frac{C\,D_{AB}\,A}{L}\,\ln\!\left(\frac{1 - y_{A,L}}{1 - y_{A,0}}\right) \]

14–8

Mass Convection

Mass convection between a surface and a moving fluid parallels heat convection, with a mass transfer coefficient \(h_{\text{mass}}\) in place of \(h\) and a concentration difference as the driving force.

Convective mass transfer and Sherwood correlations

\[ \dot{m}_{\text{conv}} = h_{\text{mass}}\,A_s\,(\rho_{A,s} - \rho_{A,\infty}), \qquad Sh = C\,Re^{m}\,Sc^{1/3} \]

The Sherwood correlations are the same as the Nusselt ones with \(Pr \to Sc\).

Problem-solving strategy

Identify stationary diffusion, Stefan flow, external/internal convection, or coupled heat–mass transfer.
Choose mass or molar basis and define positive species-flux direction.
Evaluate \(D_{AB}\), \(\nu\), and other properties at the stated mean temperature and pressure.
Check \(Re\), \(Sc\), geometry, and boundary-condition range before using a Sherwood correlation.
Audit dimensions, high-to-low diffusion direction, interface equilibrium, and conservation.

14–9

Simultaneous Heat and Mass Transfer

Evaporation cools a surface because the escaping vapor carries away latent heat. Many processes — drying, cooling towers, sweating — couple heat and mass transfer through this mechanism.

Energy balance with evaporative cooling

\[ h(T_\infty - T_s) = \dot{m}_{\text{evap}}\,h_{fg} = h_{\text{mass}}(\rho_{A,s} - \rho_{A,\infty})\,h_{fg} \]

Course Note

This single balance explains why you feel cold stepping out of a pool, how cooling towers work, and why deserts get cold at night. The Lewis number \(Le \approx 1\) for air–water vapor is what makes the wet-bulb thermometer and the whole psychrometric analysis so clean — heat and mass transfer track each other almost exactly.

EXAMPLE 3Evaporative Load on a Coated Tray

Problem. A ventilated coating tray of area 1.20 m² evaporates water into air. At the film temperature, take \(h_m=0.012\) m/s, surface vapor density 0.030 kg/m³, bulk vapor density 0.010 kg/m³, and \(h_{fg}=2400\) kJ/kg. Find evaporation rate and latent cooling load.

Strategy

A supplied convective mass-transfer coefficient and surface-to-bulk density difference give rate directly; latent enthalpy then couples mass to heat.

Givens

\(A=1.20\) m², \(h_m=0.012\) m/s.
\(\rho_{v,s}=0.030,\rho_{v,\infty}=0.010\) kg/m³; \(h_{fg}=2.400\times10^6\) J/kg.

Physical/model interpretation

Steady uniform coefficient; positive flux is outward. Values are supplied at the film state, avoiding an unsupported property lookup.

Numbered derivation steps

1. Driving difference. \(\Delta\rho_v=0.030-0.010=0.020\ {\rm kg/m^3}\).

2. Mass flux. \(\dot m''=h_m\Delta\rho_v=(0.012)(0.020)=2.40\times10^{-4}\ {\rm kg/m^2s}\).

3. Total evaporation. \(\dot m=(2.40\times10^{-4})(1.20)=2.88\times10^{-4}\ {\rm kg/s}=1.04\ {\rm kg/h}\).

4. Latent load. \(\dot Q_{lat}=\dot m h_{fg}=(2.88\times10^{-4})(2.400\times10^6)=691\ {\rm W}\).

Unit/sign/limit checks

Units: (m/s)(kg/m³)(m²)=kg/s and (kg/s)(J/kg)=W.

Signs: Surface concentration exceeds bulk, so vapor and latent energy leave the tray.

Limits: At saturated bulk air, the density difference and predicted evaporation vanish.

Final answer: \(\boxed{\dot m=1.04\ {\rm kg/h},\ \dot Q_{lat}=0.691\ {\rm kW}}\).

Engineering insight

A heater must supply at least this latent load plus sensible and environmental losses to hold surface temperature.

Chapter recap

Fick's law describes molecular transport relative to a chosen mixture velocity.
Boundary equilibrium and flux continuity are separate interface requirements.
Dimensionless analogies are powerful only inside the source correlation's validity range.
Before moving on: choose a concentration basis, state direction and model, evaluate properties, solve, and audit conservation and limits.

← Chapter 13 · Heat Exchangers Chapter 15 · Cooling of Electronics →