Coupled Heat & Mass Transfer
Coupled Heat & Mass Transfer
When a fluid carries both thermal energy and dissolved species, we need two coupled transport equations: the heat equation and a mass-species equation. If the flow velocity depends on temperature (buoyancy) or on concentration (solutal convection), the velocity field becomes a third unknown — a full multiphysics problem.
For a 1-D simplification with constant velocity $u$, temperature $T(x,t)$ and species concentration $c(x,t)$ obey convection–diffusion equations. These are paradigm PDEs for chemical reactors, drying, combustion, and heat-exchanger design.
This lesson presents the governing equations, the dimensionless numbers that classify regimes (Péclet, Lewis, Sherwood), and worked examples for a simple slab.
$$\frac{\partial T}{\partial t} + u\frac{\partial T}{\partial x} = \alpha\frac{\partial^2 T}{\partial x^2} + \frac{Q}{\rho c_p}$$
$$\frac{\partial c}{\partial t} + u\frac{\partial c}{\partial x} = D\frac{\partial^2 c}{\partial x^2} - k_r c$$
where $\alpha = k/(\rho c_p)$ is thermal diffusivity and $D$ is species diffusivity.
Péclet: $\text{Pe}_T = uL/\alpha$ (thermal), $\text{Pe}_M = uL/D$ (mass). Lewis: $\text{Le} = \alpha/D$.
$\text{Pe} \gg 1$: advection dominates. $\text{Pe} \ll 1$: diffusion dominates.
Water at $u = 0.1\,\text{m/s}$ flows in a $L = 0.05\,\text{m}$ slab. $\alpha = 1.4\times 10^{-7}\,\text{m}^2/\text{s}$. Find $\text{Pe}_T$.
$\text{Pe}_T = (0.1)(0.05)/(1.4\times 10^{-7}) = 0.005/1.4\times 10^{-7} \approx 3.6\times 10^{4}$. Highly advection-dominated; heat moves with the bulk flow.
$\alpha_{\text{air}} \approx 2.2\times 10^{-5}\,\text{m}^2/\text{s}$, $D_{\text{CO}_2\text{-air}} \approx 1.6\times 10^{-5}\,\text{m}^2/\text{s}$. Find Le.
$\text{Le} = 2.2\times 10^{-5}/1.6\times 10^{-5} \approx 1.38$. Heat diffuses slightly faster than CO₂, so the thermal boundary layer is thicker than the mass one.
Neglect convection in a thin stagnant film of thickness $L$. Species obeys $D\,c'' - k_r c = 0$ with $c(0) = c_0$, $c(L) = 0$. Find $c(x)$.
General solution $c = A\cosh(\mu x) + B\sinh(\mu x)$ with $\mu = \sqrt{k_r/D}$.
Apply BCs: $A = c_0$, and $c_0 \cosh(\mu L) + B\sinh(\mu L) = 0 \Rightarrow B = -c_0\coth(\mu L)$.
Closed form: $c(x) = c_0 \dfrac{\sinh[\mu (L - x)]}{\sinh(\mu L)}$.
Practice Problems
Show Answer Key
1. $\alpha = k/(\rho c_p)$.
2. Advective and diffusive time scales are equal.
3. Heat diffuses faster than species.
4. $\partial_{t^*}T^* + u^* \partial_{x^*} T^* = (1/\text{Pe})\,\partial_{x^*x^*} T^*$.
5. Even a single scalar couples to the momentum field (via buoyancy) or to a reaction rate that depends on $T$.
6. Drying paper, catalytic converters, fuel cells, food processing, HVAC dehumidification.