Coupled Heat and Mass Transfer
Real polymer processes rarely transfer heat or mass in isolation. Drying a coating, operating a vented extruder, or running a membrane distillation module all involve simultaneous heat and mass fluxes that interact. This chapter builds the framework for analyzing coupled problems: the Lewis number that connects the two fields, the Chilton–Colburn analogy that converts one coefficient into the other, and the engineering equations for three industrially important applications.
The Drying Film and Its Two-Way Street
Pull a freshly solvent-cast polymer film out of a bath and hang it in an oven. The surface dries as solvent molecules escape into the air. But evaporation requires energy, so the surface cools. A cooler surface has lower vapor pressure, which slows further evaporation. The air above supplies heat by convection, which warms the surface and accelerates evaporation again.
You cannot solve this problem by treating heat and mass transfer separately. Every change in the mass flux alters the temperature, and every change in temperature alters the mass flux. The two fields are coupled.
The same two-way coupling appears in a vented extruder removing solvent from a melt, in a membrane distillation unit separating saline water, and on the surface of every injection-moulded part as it cools in the mould. The tools in this chapter unlock all three.
Fig. 15.1 — Heat and mass fluxes at a drying polymer film surface. The two are coupled: evaporation cools the surface, changing the vapor pressure and hence the mass flux; the heat flux from the air controls how much evaporative cooling occurs.
15.1 The Heat–Mass Analogy
The governing equations for heat and mass transfer in a flowing fluid are structurally identical. Energy is convected and conducted; species are convected and diffuse. If the boundary conditions are analogous, the dimensionless solutions are identical — and the transfer coefficients are simply related.
Three dimensionless groups describe the relative importance of momentum, thermal, and species transport:
The Prandtl number compares how quickly momentum diffuses versus heat. The Schmidt number compares momentum diffusion versus mass diffusion. The Lewis number compares thermal diffusion directly to mass diffusion.
The Lewis Number
Le > 1 means thermal diffusivity exceeds mass diffusivity: heat spreads faster than species concentration in the fluid. Le < 1 means the opposite. For most vapour-in-gas systems, Le is close to 1 because both thermal conductivity and binary diffusivity scale similarly with molecular speed.
For water vapour in air at 25 °C: $\alpha_{\text{air}} \approx 2.21\times10^{-5}$ m2/s and $D_{\text{water/air}} \approx 2.56\times10^{-5}$ m2/s, giving Le = 0.86. For ethanol in air: Le ≈ 1.3. For heavy organic solvents (toluene, NMP) in air: Le ≈ 2–4.
In polymer melts, Le can be enormous. The thermal diffusivity of a melt is around 10−7 m2/s, while the diffusivity of a small solvent molecule through the melt is around 10−11 m2/s. Le ≈ 104: heat equilibrates ten thousand times faster than the solvent. Thermal gradients vanish quickly; concentration gradients persist.
During FDM printing, the melt at the nozzle tip has Le ≈ 103 for residual moisture in nylon or PLA. The printed bead equilibrates thermally within milliseconds, but moisture gradients across the bead persist for seconds or minutes. This is why wet filament produces bubbles long after the bead has cooled: moisture is still diffusing out.
Chilton–Colburn Relation
The Chilton–Colburn analogy extends Reynolds's earlier analogy between momentum and heat transfer to include mass transfer. In its most useful form, it states that the heat and mass transfer j-factors are equal:
Setting $j_H = j_M$ and solving for $h_m$ in terms of $h$:
Equation 15.3 is powerful because heat transfer correlations are abundant (flat plates, cylinders, packed beds, all flow regimes). You can look up $h$ from a Nusselt-number correlation and immediately compute $h_m$ without needing a separate mass transfer correlation.
Setting $j_H = j_M$:
$$\frac{h}{\rho c_p u}\,\text{Pr}^{2/3} = \frac{h_m}{u}\,\text{Sc}^{2/3}$$Cancel $u$ from both sides and rearrange:
$$h_m = \frac{h}{\rho c_p}\,\frac{\text{Pr}^{2/3}}{\text{Sc}^{2/3}} = \frac{h}{\rho c_p}\,\left(\frac{\text{Pr}}{\text{Sc}}\right)^{2/3}$$Since $\text{Le} = \text{Sc}/\text{Pr}$, we have $\text{Pr}/\text{Sc} = 1/\text{Le}$:
$$\boxed{h_m = \frac{h}{\rho c_p \,\text{Le}^{2/3}}}$$15.2 Drying of Polymer Films and Coatings
Coating lines for lithium-ion battery electrodes, optical films, and photoresist layers all require controlled drying of a wet polymer on a moving substrate. Drying too fast causes surface skin formation; too slowly wastes oven length and energy. The rate equation is straightforward when we treat the boundary layer as a mass transfer resistance.
External Drying Rate
The molar flux of solvent A evaporating from the film surface into the airstream is:
The surface concentration is set by vapour–liquid equilibrium. For a dilute solvent in air, $c_{A,s} = p_{\text{sat}}(T_s)/(RT_s)$ where $p_{\text{sat}}$ is evaluated at the actual surface temperature $T_s$, not the bulk air temperature.
The total drying rate from a film of surface area $A$ is $\dot{n}_A = N_A \cdot A$ (mol/s), or in mass units, $\dot{m}_A = N_A \cdot A \cdot M_A$ (kg/s).
Evaporative Cooling
Every mole of solvent that evaporates carries away a latent heat $\Delta H_{\text{vap}}$ (J/mol). The energy balance at the film surface is:
Substituting Eq. 15.4 and using Eq. 15.3 to replace $h_m$:
$$h\,(T_\infty - T_s) = \frac{h}{\rho c_p \,\text{Le}^{2/3}}\,(c_{A,s} - c_{A,\infty})\,\Delta H_{\text{vap}}$$Cancelling $h$:
This is precisely the equation that governs the wet-bulb depression on a psychrometer. When Le = 1 (approximately true for water/air), it reduces to the classical wet-bulb formula. The surface temperature $T_s$ is lower than the air temperature $T_\infty$ by an amount set by the evaporation rate and latent heat.
Worked Example 15.1 — Steady-State Drying Temperature
A coating of acetone (Le = 1.80, $\Delta H_{\text{vap}}$ = 31 000 J/mol, $M$ = 58 g/mol) dries in an airstream at $T_\infty$ = 50 °C. The acetone concentration at the film surface is $c_{A,s}$ = 0.60 mol/m3 and in the bulk air $c_{A,\infty}$ = 0. Air properties: $\rho c_p$ = 1130 J/(m3·K). Find the surface temperature.
Step 1. Compute the temperature depression using Eq. 15.6:
$$T_\infty - T_s = \frac{31\,000}{1130 \times (1.80)^{2/3}}\,(0.60 - 0)$$ $$= \frac{31\,000}{1130 \times 1.481}\times 0.60 = \frac{31\,000}{1673}\times 0.60 = 18.5 \times 0.60 = 11.1\;\text{K}$$Step 2. Surface temperature:
$$T_s = 50 - 11.1 = 38.9\;\text{°C}$$The film surface is about 11 °C cooler than the oven air. This matters for thick coatings: the polymer may never reach the nominal oven temperature until all solvent is gone.
Fig. 15.2 — The two drying stages. During the constant-rate period (CRP), the film surface is wet enough that external convection resistance limits the rate; Eq. 15.4 applies directly. During the falling-rate period (FRP), internal diffusion through the thickening dry polymer layer limits the rate.
15.3 Reactive Extrusion and Devolatilization
Devolatilization is the removal of residual monomer, solvent, or moisture from a polymer melt. It occurs in twin-screw extruders through open venting ports where a thin melt film is exposed to near-vacuum. The driving force is the difference between the solvent's partial pressure in the melt and the vapour pressure above the vent.
The key insight is that the process operates in the Lewis-number limit for melts. Because Le is enormous (103–104), heat transfer is essentially instantaneous: the melt temperature is uniform. The slow step is diffusion of solvent molecules through the viscous polymer to reach the melt surface.
Film Model for Devolatilization
The simplest model treats each vent pool as a thin melt film of thickness $\delta$ exposed to vacuum for an exposure time $t_e$. Fick's second law with a zero-concentration boundary condition at the film surface gives the average flux:
Because $D_{AB}$ in the melt is so small (compare 10−5 m2/s in air), achieving efficient devolatilization requires either a very long exposure time, or a very high surface renewal rate (many screw elements per vent), or both. This is the physical justification for multi-stage devolatilization in compounding extruders.
Each venting stage removes roughly 90% of the remaining volatile. Starting from 1000 ppm residual monomer, one stage gives ~100 ppm, two stages ~10 ppm, and three stages ~1 ppm. The logarithmic reduction follows directly from the exponential concentration profile in Fickian diffusion.
Worked Example 15.2 — Devolatilization Flux
A twin-screw extruder devolatilizes residual styrene from a polystyrene melt. The styrene diffusivity in the melt is $D_{AB}$ = 2.0×10−10 m2/s. Each screw revolution takes 0.50 s (exposure time $t_e$ = 0.25 s per half-revolution). The initial dissolved styrene concentration is $c_{A,0}$ = 300 mol/m3. Calculate the average molar flux from the melt surface.
Apply Eq. 15.7:
$$\bar{N}_A = 2 \times 300 \times \sqrt{\frac{2.0\times10^{-10}}{\pi \times 0.25}}$$ $$= 600 \times \sqrt{\frac{2.0\times10^{-10}}{0.785}} = 600 \times \sqrt{2.55\times10^{-10}}$$ $$= 600 \times 1.597\times10^{-5} = 9.6\times10^{-3}\;\text{mol/(m}^2\text{·s)}$$Over a vent zone of 0.010 m2 exposed area and at 2 revolutions/s, the mass rate of styrene removal is $9.6\times10^{-3} \times 0.010 \times 104 = 9.9\times10^{-3}$ g/s per vent.
15.4 Membrane Distillation
Membrane distillation (MD) uses a hydrophobic microporous membrane to separate water from a saline feed. The feed side is hot; the permeate side is cold. Liquid water cannot wet the hydrophobic pores, but water vapour can diffuse through them freely. The driving force is the vapour pressure difference created by the temperature difference.
MD interests polymer scientists for two reasons: the membrane is almost always a semicrystalline polymer (PP, PVDF, or PTFE), and the engineering performance depends sensitively on the membrane's thermal conductivity, pore geometry, and wettability.
Membrane Flux Equation
The water vapour flux through the membrane is proportional to the vapour pressure difference across it:
The saturation pressure depends strongly on temperature through the Antoine equation. At 60 °C, $p_{\text{sat}}$ = 19.9 kPa; at 20 °C, only 2.34 kPa. A 40 °C temperature differential provides about 17.6 kPa driving force.
Temperature Polarization
The actual temperature at the membrane surfaces is not the bulk temperature on each side. Thermal boundary layers develop on both sides: the hot boundary layer cools the membrane surface below the bulk feed temperature; the cold boundary layer warms the cold membrane surface above the bulk permeate temperature. This is temperature polarization.
Fig. 15.3 — Temperature polarization in membrane distillation. Boundary layers on both sides reduce the effective temperature difference at the membrane surfaces below the bulk temperature difference. The flux is driven by $\Delta T_{\text{effective}}$, not $\Delta T_{\text{bulk}}$.
The temperature polarization coefficient (TPC) quantifies how much of the bulk driving force reaches the membrane:
TPC = 1 means no polarization (perfectly mixed boundary layers). In practice, TPC = 0.4–0.8 in MD modules. Increasing feed velocity, adding turbulence promoters, or using spacers in the feed channel all improve TPC by thinning the boundary layers.
Worked Example 15.3 — MD Flux with Polarization
A PVDF membrane with $B_m$ = 4.0×10−7 kg/(m2·s·Pa) separates hot feed at 70 °C from cold permeate at 20 °C. Temperature polarization gives membrane surface temperatures of $T_{m,h}$ = 58 °C and $T_{m,c}$ = 32 °C. Vapour pressures: $p_{\text{sat}}$(58 °C) = 18.1 kPa, $p_{\text{sat}}$(32 °C) = 4.75 kPa. Find the flux and the TPC.
Flux (Eq. 15.8):
$$J = 4.0\times10^{-7}\times(18\,100 - 4\,750) = 4.0\times10^{-7}\times13\,350 = 5.34\times10^{-3}\;\text{kg/(m}^2\text{·s)}$$ $$= 5.34\times10^{-3}\times3600 = 19.2\;\text{kg/(m}^2\text{·h)}$$TPC (Eq. 15.9):
$$\text{TPC} = \frac{58 - 32}{70 - 20} = \frac{26}{50} = 0.52$$Only 52% of the bulk driving force reaches the membrane. If the module were perfectly mixed (TPC = 1), the membrane surfaces would be at 70 and 20 °C, giving a much higher flux. Improving the hydrodynamics is the primary route to better MD performance for this membrane.
Interactive: Drying Rate Calculator
Use the calculator below to explore how the Lewis number and boundary-layer conditions control the drying rate and surface temperature of a solvent-coated polymer film.
Concept Check
Answer each question in your own words, then click Get AI Feedback to receive Socratic guidance.
State in physical terms what Le > 1 means. Give a concrete example involving a polymer process where Le > 1 matters, and explain what practical consequence it has for the process design.
Explain in one paragraph why the surface of a drying polymer film is cooler than the surrounding air. Identify the two competing fluxes and explain which one sets the surface temperature at steady state.
In a vented twin-screw extruder, the solvent diffusivity in the melt is $D_{AB}$ = 5×10−11 m2/s and the thermal diffusivity of the melt is $\alpha$ = 1×10−7 m2/s. Calculate Le. Based on this value, which property — temperature or solvent concentration — reaches equilibrium faster in the melt? What does this imply for the design of vented extruder sections?
A membrane distillation module has TPC = 0.40. A second module uses turbulence promoters and achieves TPC = 0.80. If all other parameters are equal, by what factor does the flux increase? Explain physically why turbulence promoters improve TPC, and identify a trade-off the engineer must accept when using them.
A polymer scientist proposes to dry a PMMA film (glass transition temperature Tg = 105 °C) in an oven at 120 °C to speed up solvent removal. Using the concepts in this chapter (evaporative cooling, CRP vs. FRP, Lewis number), argue whether this is a good or bad idea. Consider what happens to the surface temperature during the constant-rate period and what happens once drying enters the falling-rate period.
Problem Set
Air at 60 °C has thermal diffusivity $\alpha$ = 2.50×10−5 m2/s. Ethanol vapour diffuses through air at this temperature with binary diffusivity $D_{AB}$ = 1.25×10−5 m2/s. Calculate the Lewis number Le for the ethanol/air system.
Le = $\alpha / D_{AB}$. Divide the thermal diffusivity by the binary diffusivity.
Le = $2.50\times10^{-5} / 1.25\times10^{-5}$ = 2.00. A Lewis number of 2 means heat diffuses twice as fast as species concentration in this mixture.
Air flows over a drying polymer film. A Nusselt-number correlation gives a heat transfer coefficient $h$ = 25 W/(m2·K). Air properties: density $\rho$ = 1.20 kg/m3, specific heat $c_p$ = 1005 J/(kg·K). The Lewis number for the solvent/air system is Le = 2.00. Use the Chilton–Colburn analogy to calculate the mass transfer coefficient $h_m$. Express your answer in mm/s.
From Eq. 15.3: $h_m = h / (\rho c_p \,\text{Le}^{2/3})$.
$\text{Le}^{2/3} = 2.00^{2/3} = e^{(2/3)\ln 2} = 1.587$.
$h_m = 25 / (1.20 \times 1005 \times 1.587) = 25 / 1583 = 0.01580$ m/s = 15.8 mm/s.
A polymer coating (molar mass of solvent $M$ = 150 g/mol) is dried on a pilot coater. The mass transfer coefficient is $h_m$ = 8.0×10−3 m/s. The solvent vapour concentration at the film surface is $c_{A,s}$ = 1.20 mol/m3 and the bulk airflow is essentially solvent-free ($c_{A,\infty}$ = 0). The film area exposed to air is 0.40 m2. Calculate the solvent evaporation rate in g/s.
$N_A = h_m (c_{A,s} - c_{A,\infty}) = 8.0\times10^{-3} \times 1.20 = 9.6\times10^{-3}$ mol/(m2·s).
$\dot{n}_A = N_A \times A = 9.6\times10^{-3} \times 0.40 = 3.84\times10^{-3}$ mol/s.
$\dot{m}_A = \dot{n}_A \times M = 3.84\times10^{-3} \times 150 = 0.576$ g/s.
A direct-contact membrane distillation (DCMD) module uses a polypropylene membrane with $B_m$ = 3.5×10−7 kg/(m2·s·Pa). The hot feed side is at 60 °C (saturation vapour pressure $p_{\text{sat}}$ = 19.9 kPa) and the cold permeate side is at 20 °C ($p_{\text{sat}}$ = 2.34 kPa). Assume no temperature polarization. Calculate the water flux in kg/(m2·h).
$\Delta p = p_{\text{sat},h} - p_{\text{sat},c} = 19\,900 - 2\,340 = 17\,560$ Pa.
$J = B_m \Delta p = 3.5\times10^{-7} \times 17\,560 = 6.15\times10^{-3}$ kg/(m2·s).
$J = 6.15\times10^{-3} \times 3600 = 22.1$ kg/(m2·h).
The DCMD module in P4 operates with bulk temperatures of 60 °C (hot) and 20 °C (cold). Due to temperature polarization, the actual temperature at the hot membrane surface is 52 °C and the cold membrane surface is 28 °C. Calculate the temperature polarization coefficient (TPC). Then calculate the ratio of the actual flux to the ideal flux (from P4) and comment on whether improving fluid mixing would be worthwhile.
$\text{TPC} = (T_{m,h} - T_{m,c})/(T_{h,\text{bulk}} - T_{c,\text{bulk}}) = (52 - 28)/(60 - 20) = 24/40 = 0.60$.
Use $p_{\text{sat}}(52\,°C) \approx 13.5$ kPa and $p_{\text{sat}}(28\,°C) \approx 3.78$ kPa. $J_{\text{actual}} = 3.5\times10^{-7} \times (13\,500 - 3\,780) = 3.5\times10^{-7} \times 9\,720 = 3.40\times10^{-3}$ kg/(m2·s) = 12.2 kg/(m2·h).
$J_{\text{actual}}/J_{\text{ideal}} = 12.2/22.1 = 0.55$. Polarization reduces flux by 45%. Improving mixing to raise TPC from 0.60 toward 1.0 would nearly double the flux, making it very worthwhile.
Chapter Summary
Coupled heat and mass transfer problems are unified by the Lewis number Le = α/DAB, which measures how much faster thermal diffusivity spreads compared to mass diffusivity. When Le ≈ 1 (vapours in air), the Chilton–Colburn analogy connects the heat and mass transfer coefficients through the simple relation $h_m = h / (\rho c_p \,\text{Le}^{2/3})$, allowing mass transfer coefficients to be inferred directly from heat transfer correlations.
During the constant-rate drying period, evaporation is limited by the external boundary-layer resistance. The molar flux is $N_A = h_m(c_{A,s} - c_{A,\infty})$ and the surface temperature is depressed below the air temperature by evaporative cooling. Once the film enters the falling-rate period, internal Fickian diffusion through the polymer matrix limits the rate.
In polymer melts Le is enormous (103–104), so heat equilibrates far faster than concentration. In vented extruders, devolatilization is therefore always mass-transfer limited, and the penetration-theory flux $\bar{N}_A = 2 c_{A,0}\sqrt{D_{AB}/(\pi t_e)}$ governs performance.
Membrane distillation exploits a vapour pressure difference created by a temperature gradient across a hydrophobic membrane. The flux $J = B_m(p_{\text{sat},h} - p_{\text{sat},c})$ can be much lower than the ideal value because boundary layers reduce the effective temperature difference at the membrane surfaces. The temperature polarization coefficient TPC = ΔTmembrane/ΔTbulk quantifies the loss.