The Analogy Table: Fourier, Fick, and Newton Side by Side

In the last article, we built up why heat, mass, and momentum transport share a mathematical skeleton: a local flux law feeding into a global conservation statement. Now let's put that skeleton to work. This is the table that, once it actually makes sense instead of just being memorized, lets you transfer everything you know about heat transfer straight into a mass transfer problem (and vice versa) without re-deriving anything.

The three constitutive laws, lined up

$$q_x'' = -k\frac{dT}{dx} \qquad\qquad J_{A,x} = -D_{AB}\frac{dC_A}{dx} \qquad\qquad \tau_{xy} = -\mu \frac{du_x}{dy}$$

Same shape every time:

$$\text{flux} = -(\text{transport property})(\text{driving-force gradient})$$

Transport process	Flux	Transport property	Driving variable
Heat conduction	$q_x''$	$k$ (thermal conductivity)	$T$
Mass diffusion (A in B)	$J_{A,x}$	$D_{AB}$ (diffusivity)	$C_A$
Momentum diffusion	$\tau_{xy}$	$\mu$ (viscosity)	$u_x$

Physically, each property answers the same kind of question for its own quantity:

$k$ — how easily heat gets passed between neighboring molecules via collisions
$D_{AB}$ — how quickly molecules of A spread through B
$\mu$ — how strongly momentum gets shared between adjacent fluid layers

All three are, at bottom, measuring how efficiently random molecular motion transports something down a gradient. That's the real content of "the analogy" — not that the symbols rhyme, but that the same microscopic mechanism (molecules bumping into neighbors and handing off energy, species, or momentum) underlies all three.

Why the sign always means the same thing

This trips people up at first, especially with momentum. The minus sign in each law does not mean the flux is negative — it means the flux points opposite the gradient. Heat flows from hot to cold; species diffuse from high concentration to low; momentum "diffuses" from fast-moving fluid layers to slow-moving ones. The gradient points "uphill," and the flux points "downhill." That's universal across all three laws. (Newton's law of viscosity, written as $\tau_{yx} = \mu \frac{du_x}{dy}$ in most fluid mechanics textbooks, looks like the sign flipped — that's a stress-convention issue, not a physics contradiction. More on that in the next article.)

Diffusivities: same units, same job, very different sizes

Strip away the names and look at the units. $k/\rho c_p$ (thermal diffusivity, $\alpha$), $D_{AB}$ (mass diffusivity), and $\mu/\rho$ (kinematic viscosity, $\nu$) all carry units of m²/s. That's not a coincidence — all three measure how fast a disturbance spreads out spatially, and the diffusion time scale for any of them follows the same scaling:

$$t_d \sim \frac{L^2}{\mathscr{D}}$$

where $\mathscr{D}$ stands in for $\alpha$, $D_{AB}$, or $\nu$. Double the distance, and diffusion takes four times as long — not twice. This single relationship explains a lot of "why does this take so long" intuition in transport problems.

But here's the catch that the analogy table can hide if you're not careful: these diffusivities can differ from each other by orders of magnitude, even in the same fluid. For air at 300 K, $\nu \approx D_{AB} \approx \alpha \approx 10^{-5}\ \text{m}^2/\text{s}$ — they're all roughly the same size, which is part of why the analogy works so cleanly for gases. For water, though, $\nu \approx 10^{-6}\ \text{m}^2/\text{s}$ while $D_{AB} \approx 10^{-9}\ \text{m}^2/\text{s}$ — three orders of magnitude apart. Same equations, wildly different behavior. The math being analogous doesn't mean the physical scales are analogous.

The dimensionless numbers are just ratios of these diffusivities

Once you see $\alpha$, $D_{AB}$, and $\nu$ as the three "speeds of spreading," the classic dimensionless groups stop being arbitrary combinations of letters:

$$Pr = \frac{\nu}{\alpha} \qquad\qquad Sc = \frac{\nu}{D_{AB}} \qquad\qquad Le = \frac{\alpha}{D_{AB}} = \frac{Sc}{Pr}$$

Prandtl number compares how fast momentum diffuses to how fast heat diffuses.
Schmidt number compares how fast momentum diffuses to how fast species diffuse.
Lewis number compares heat diffusion directly to mass diffusion.

If $Pr > 1$, momentum diffuses faster than heat, so the velocity boundary layer grows faster than the thermal one — meaning the thermal boundary layer ends up thinner. Same logic for $Sc > 1$ and the concentration boundary layer. This is the entire reason boundary layer thickness comparisons show up constantly in convection problems: it's just Pr and Sc doing their job.

When Pr > 1 (or Sc > 1 for mass transfer), momentum diffuses outward faster than heat or species, so the velocity boundary layer always wraps around a thinner thermal or concentration boundary layer.

The transfer-coefficient analogy: h and k_c

The same logic extends past the diffusivities into the convective transfer coefficients:

Quantity	Heat transfer	Mass transfer
Convective coefficient	$h$ (W/m²·K)	$k_c$ (m/s)
Driving force	$T_s - T_\infty$	$C_{A,s} - C_{A,\infty}$
Dimensionless group	$Nu = \dfrac{hL}{k}$	$Sh = \dfrac{k_c L}{D_{AB}}$

Nusselt number and Sherwood number play identical roles: they're both "how much better is convection doing than pure diffusion alone would?" And because both $Nu$ and $Sh$ depend on the flow field in the same way, you get correlations like:

$$Nu_L = C\, Re_L^m Pr^n \qquad\qquad Sh_L = C\, Re_L^m Sc^n$$

Same $C$, $m$, $n$. If you know the heat transfer correlation for flow over a cylinder, you already know the mass transfer correlation — just swap $Pr \to Sc$ and $Nu \to Sh$. This is the single most useful practical payoff of the whole analogy: half of what looks like "new material" in a mass transfer unit is actually a heat transfer result you already memorized, wearing a different hat.

Boundary conditions follow the same pattern too

Type	Heat transfer	Mass transfer	Meaning
Specified value	$T = T_s$	$C_A = C_{A,s}$	Surface value fixed
Specified flux	$-k\dfrac{dT}{dx} = q_s''$	$-D_{AB}\dfrac{dC_A}{dx} = J_{A,s}$	Flux fixed at surface
Convective (3rd kind)	$-k\dfrac{dT}{dx} = h(T_s - T_\infty)$	$-D_{AB}\dfrac{dC_A}{dx} = k_c(C_{A,s} - C_{A,\infty})$	Surface exchanges with bulk fluid

Same three flavors of boundary condition, every time. Once you've solved one conduction problem with a convective boundary condition, you've effectively solved the mass-transfer-through-a-stagnant-film problem too — the math doesn't know the difference.

Heat transfer also gets its own electrical-circuit analogy

One more layer worth knowing, because it shows up everywhere in heat transfer problems specifically: thermal resistance, modeled exactly like Ohm's law.

$$\dot{Q} = \frac{\Delta T}{R_{total}} \qquad\qquad \text{(compare to)} \qquad\qquad I = \frac{\Delta V}{R_e}$$

Geometry	Conduction resistance	Convection resistance
Plane wall (thickness $L$, area $A$)	$\dfrac{L}{kA}$	$\dfrac{1}{hA}$
Cylindrical wall (radii $r_1, r_2$, length $L$)	$\dfrac{\ln(r_2/r_1)}{2\pi k L}$	$\dfrac{1}{h(2\pi r L)}$
Spherical wall (radii $r_1, r_2$)	$\dfrac{1}{4\pi k}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$	$\dfrac{1}{h(4\pi r^2)}$

Resistances in series add directly; resistances in parallel add as reciprocals — exactly like circuits. This is why multi-layer wall problems (igloo domes, windows with storm panes, insulated pipes) all reduce to "draw the circuit, add up the resistances, solve for $\dot{Q}$."

Why this all matters

The payoff of internalizing this table isn't passing a quiz on definitions. It's that you stop solving three subjects and start solving one subject three times. Every correlation, every boundary condition, every diffusivity ratio you learn in heat transfer has a near-identical twin waiting in mass transfer — you just need to know where to swap the letters.