Darcy porous-medium energy equation — dimensions and non-dimensionalization: 20-Bio-A7 Bioinstrumentation, December 2019, Q1

Question 1: Darcy porous-medium energy equation — dimensions and non-dimensionalization (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The PDE above governing 2-D convective–diffusive heat transport in a Darcy-flow porous medium, with the symbol dimensions (SI, mass–length–time–temperature basis {M, L, T, Θ}):

Symbol	Meaning	Dimensions
ρ	density	M L⁻³
c_p	specific heat	L² T⁻² Θ⁻¹
k	thermal conductivity	M L T⁻³ Θ⁻¹
μ	dynamic viscosity	M L⁻¹ T⁻¹
p	pressure	M L⁻¹ T⁻²
T, x, y	temperature, coordinates	Θ; L; L

Find. (a) the dimensions of the permeability σ; (b) the non-dimensional form of the PDE using scaling constants L (length), U (velocity), ρ (density), T₀ (reference temperature), and the dimensionless parameter(s) that appear.

Approach. Every additive term of a physically valid PDE must carry identical dimensions (dimensional homogeneity), so (a) follows by forcing the convective term's net dimensions to match the diffusive term $k\,\partial^2T/\partial y^2$; for (b), introduce dimensionless coordinates/fields built from the four given scaling constants, substitute, and collect the surviving coefficient into a single dimensionless group.

Fix the reference dimensions from the diffusion term. Since $k\,\partial^2T/\partial y^2$ must have the same dimensions as the whole equation, $$[k][T]/[y]^2 = (\text{M L T}^{-3}\Theta^{-1})(\Theta)(\text{L}^{-2}) = \text{M L}^{-1}\text{T}^{-3}$$ This is the target dimension every other term (including the convective term) must match.
Collect the convective term's dimensions apart from σ/μ. $$[\rho][c_p][p/x][T/x] = (\text{M L}^{-3})(\text{L}^2\text{T}^{-2}\Theta^{-1})(\text{M L}^{-2}\text{T}^{-2})(\Theta\,\text{L}^{-1}) = \text{M}^2\text{L}^{-4}\text{T}^{-4}$$ (the Θ exponents cancel between $c_p$ and $T$, as they must for the term to end up temperature-homogeneous with the diffusion term).
Solve for [σ/μ], then [σ]. Matching Step 1 to Step 2 times [σ/μ]: $$[\sigma/\mu] = \frac{\text{M L}^{-1}\text{T}^{-3}}{\text{M}^2\text{L}^{-4}\text{T}^{-4}} = \text{M}^{-1}\text{L}^{3}\text{T}$$ $$\boxed{[\sigma] = [\sigma/\mu]\,[\mu] = (\text{M}^{-1}\text{L}^{3}\text{T})(\text{M L}^{-1}\text{T}^{-1}) = \text{L}^{2}\ \ (\text{i.e. SI units m}^2)}$$ σ carries dimensions of area — exactly the textbook units of Darcy permeability, recovered here purely from the stated PDE's own internal dimensional consistency (no separate appeal to Darcy's law $u=-(\sigma/\mu)\nabla p$ is needed).
Introduce dimensionless variables from the four scaling constants (part b). With only $L,U,\rho,T_0$ supplied, the natural choices are the length scale $L$ for both coordinates, the reference temperature $T_0$ for $T$, and the dynamic-pressure scale $\rho U^2$ for $p$ (the standard choice when a velocity and a density, but not a pressure, are given — White §5.3): $$x=Lx^*,\quad y=Ly^*,\quad T=T_0T^*,\quad p=\rho U^2p^*$$ so that $\partial p/\partial x=(\rho U^2/L)\,\partial p^*/\partial x^*$, $\partial T/\partial x=(T_0/L)\,\partial T^*/\partial x^*$ (and likewise in $y$), and $\partial^2T/\partial y^2=(T_0/L^2)\,\partial^2T^*/\partial y^{*2}$.
Substitute and collect the dimensionless equation. Substituting into the original PDE, $$\rho c_p\frac{\sigma}{\mu}\left(\frac{\rho U^2 T_0}{L^2}\right)\!\left[\frac{\partial p^*}{\partial x^*}\frac{\partial T^*}{\partial x^*}+\frac{\partial p^*}{\partial y^*}\frac{\partial T^*}{\partial y^*}\right] + k\frac{T_0}{L^2}\frac{\partial^2T^*}{\partial y^{*2}} = 0$$ Dividing through by $kT_0/L^2$ leaves $$\boxed{\ \Pi\left[\frac{\partial p^*}{\partial x^*}\frac{\partial T^*}{\partial x^*}+\frac{\partial p^*}{\partial y^*}\frac{\partial T^*}{\partial y^*}\right] + \frac{\partial^2T^*}{\partial y^{*2}} = 0, \qquad \Pi=\frac{\rho^2 c_p\,\sigma\,U^2}{\mu k}\ }$$ a single dimensionless group $\Pi$ multiplies the convective terms, and the dimensional-consistency check of Step 3 confirms $\Pi$ is indeed dimensionless (its net {M,L,T,Θ} exponents all vanish).
Interpret Π as a product of two named dimensionless numbers. Writing $\ell=\sqrt{\sigma}$ as the permeability-based (Darcy) length scale, $$Re_\sigma=\frac{\rho U\ell}{\mu}, \qquad Pe_\sigma=\frac{\rho c_p U\ell}{k}=Re_\sigma\,Pr, \qquad \Pi=Re_\sigma\cdot Pe_\sigma=Re_\sigma^2\,Pr$$ so $\Pi$ is the product of a permeability-based Reynolds number and a permeability-based Péclet number (equivalently $Re_\sigma^2$ times the Prandtl number $Pr=\mu c_p/k$): when $\Pi\ll1$ conduction dominates and the temperature field is nearly one-dimensional in $y$; when $\Pi\gg1$ pressure-driven (Darcy) advection of heat dominates over conduction.

Quantity	Value
(a) Dimensions of σ	L² (SI: m²)
(b) Dimensionless PDE	$\Pi\left[\partial_{x^}p^\,\partial_{x^}T^+\partial_{y^}p^\,\partial_{y^}T^\right]+\partial^2_{y^}T^=0$
(b) Dimensionless group	$\Pi = \rho^2 c_p\sigma U^2/(\mu k) = Re_\sigma\cdot Pe_\sigma$

Topic. Dimensional homogeneity and non-dimensionalization of a governing transport equation.

Key relations. Dimensional homogeneity (every additive PDE term shares one dimension); dynamic-pressure non-dimensionalization $p^*=p/(\rho U^2)$; permeability-based Reynolds/Péclet numbers $Re_\sigma=\rho U\sqrt{\sigma}/\mu$, $Pe_\sigma=Re_\sigma Pr$.

Why this works. A dimensionally consistent equation lets an unknown coefficient's dimensions be read off directly by forcing every term to balance against a fully known term (here the conduction term) — this is exactly how permeability's area dimensions emerge without separately invoking Darcy's law. Non-dimensionalizing with a fixed, problem-independent set of scaling constants ($L,U,\rho,T_0$) converts every remaining physical coefficient into a single dimensionless multiplier in front of the terms it governs; because that multiplier is dimensionless by construction (Buckingham Pi), its value alone — not the individual dimensional constants — controls which physical mechanism (advection vs. conduction) dominates the solution.

Common pitfall. Choosing an inconsistent pressure scale (e.g. a viscous scale $\mu U/L$) when the problem supplies $\rho$ rather than $\mu$ as a base scaling constant — the scale must be built only from the constants actually given ($L,U,\rho,T_0$), which forces the inertial/dynamic choice $p^*=p/(\rho U^2)$ here. A second common error is dropping the Θ (temperature) dimension bookkeeping when $c_p$ and $T$ dimensions are multiplied, leading to a spurious leftover $\Theta$ exponent in $[\sigma]$.

Source. White, Fluid Mechanics, 8th ed., §5.3 (non-dimensionalizing the basic equations); Bear, Dynamics of Fluids in Porous Media (Darcy permeability and its dimensions).

Question 1 of 4: Darcy porous-medium energy equation — dimensions and non-dimensionalization

Question 1: Darcy porous-medium energy equation — dimensions and non-dimensionalization (25 marks)