Porous and granular materials

Count solid and pore volume

A container of grains occupies more space than the grains alone: there are gaps between them. A porous rock also contains solid material and pores. Two questions arise. How much fluid can fit in the gaps, and how readily can it move through them? A third question concerns the solids: what contact forces support the grains? We will answer small versions of these questions without treating every porous material as the same material.

A pore is a space within the chosen bulk sample. The bulk volume V includes both solid volume V_s and pore volume V_p, so V = V_s + V_p. Total porosity φ is the pore-volume fraction: φ = V_p/V. It has no unit. Porosity 0.40 means 40% of the bulk volume is pore space; it does not mean 40% of the material’s mass is missing.

Assign a cylindrical column with length L = 0.20 m and full cross-sectional area A = 0.010 m². Its bulk volume is AL = 0.0020 m³ = 2000 cm³ = 2 L. With φ = 0.40, the pore volume is 800 cm³ and the solid volume is 1200 cm³. The solid-volume fraction is 0.60. These are assigned teaching values, not measurements of a named rock or sand.

Porosity uses the bulk volume as its denominator. Another quantity, void ratio, uses solid volume: V_p/V_s. For this sample the void ratio is 800/1200 = 2/3, whereas porosity is 0.40. Changing the denominator changes the meaning. For a first reading, keep the volume fractions and the following two flow averages in view. Optional refreshers are “Units, standards, and physical constants” and “Newton’s laws and free-body diagrams”; the necessary terms are defined here.

Ask which pores carry flow

Having pore space does not guarantee a path across the sample. An isolated pore cannot carry fluid from one end to the other. A connected dead-end pore can exchange fluid with a passing flow yet contribute little to through-flow. Narrow connections and winding paths also influence resistance. Thus equal total porosity does not establish equal flow rate.

For the velocity calculation, define φ_e as the fraction of bulk volume occupied by the mobile pore fluid participating in the through-flow average. We call this the mobile effective porosity. Effective-porosity measurements and conventions can differ; merely being accessible to fluid does not prove that every part of a pore carries through-flow. In our assigned column all pores are mobile, connected and completely filled with the one fluid, so φ_e = φ = 0.40.

Saturated means that the pore space is fully filled with the stated fluid. Our column has one fluid phase and a fixed solid matrix: the arrangement of solid material is held unchanged. A partly filled sample requires further information about fluid distribution and interfaces. The capillary pressure of one ideal tube, studied in “Surface tension and capillarity”, is not by itself a flow law for a whole pore network.

Intrinsic permeability k measures the effective ability of the fixed pore structure to transmit a fluid in the slow-flow law used below. Its unit is m², but k is not the area of one visible opening. Here it is supplied independently as k = 1 × 10⁻¹² m². Pore connections, throat sizes and path geometry matter; total porosity alone does not determine k.

Divide flow by the right area

Volumetric flow Q is the fluid volume crossing a chosen face per unit time. Its unit is m³/s. Divide Q by the full bulk face area A, including the parts occupied by solids, to obtain Darcy flux q = Q/A. Its unit is m/s. Although it has velocity units, q is a volume-flow rate per whole area, not the local speed of fluid in a pore.

A different average divides by the mobile pore-volume fraction: v = q/φ_e = Q/(Aφ_e). Here v is the mean component of mobile pore velocity along the column, averaged over the participating pore fluid. The averaging is over a representative volume containing many pores. This is not a statement that every cross-section has an open area exactly Aφ_e, or that fluid has the same speed everywhere along a winding pore.

With Q = 0.10 mL/s = 1 × 10⁻⁷ m³/s and A = 0.010 m², q = 1 × 10⁻⁵ m/s = 0.010 mm/s. With φ_e = 0.40, v = 2.5 × 10⁻⁵ m/s = 0.025 mm/s. The same Q gives two different averages because they use different normalisations. The mean mobile value is 2.5 times the Darcy flux.

Do not use this mean alone to predict an individual particle’s travel time or a contaminant’s arrival time. Real paths and local speeds differ; mixing and dispersion need additional description. On a first reading, the important distinction is simply Q for volume per time, q for volume per full area per time, and v for the mobile-pore average.

Choose a positive flow direction

Choose x increasing from the inlet at x = 0 to the outlet at x = L. Pressure p is normal force per area, in Pa = N/m². At equal elevation, a pressure decreasing as x increases drives positive flow in our slow-flow model. The pressure gradient dp/dx means pressure change per distance. It is negative for that decreasing pressure.

The horizontal Darcy law is q = −(k/μ)(dp/dx). Dynamic viscosity μ measures the fluid’s viscous resistance and has unit Pa·s. This μ is a fluid property; the later symbol μ_s will instead denote a dimensionless static contact-friction coefficient. The minus sign makes a negative pressure gradient give positive q. For a uniform column with a linear pressure profile, q = k(p_up − p_down)/(μL).

Elevation matters when the column is not horizontal. Define hydraulic head h = z + p/(ρg), where z is elevation relative to one datum, ρ is the fluid density and g is gravitational acceleration. Both terms have unit m: pressure divided by weight per volume becomes a height. In this slow model, head combines pressure and elevation contributions; a kinetic-energy head is not included.

The corresponding one-dimensional law is q = −K dh/dx, with hydraulic conductivity K = kρg/μ. Fluid flows toward decreasing head under these assumptions. For example, equal pressure at two different elevations does not mean equal head. Conversely, the hydrostatic pressure gradient can exactly offset the elevation change and give constant head and zero Darcy flux.

These laws describe steady saturated flow relative to a fixed matrix, averaged over many pores, with constant fluid properties and a homogeneous isotropic scalar k. Incompressible means that fluid density is held constant in this model. For a Newtonian fluid in simple shear, viscous shear stress (tangential force per area) is proportional to shear rate (relative layer speed divided by layer separation), with constant μ here. We also assume a regime dominated by viscous resistance, so fluid inertia is neglected. Calling a flow laminar alone does not establish every Darcy assumption; no universal speed or Reynolds-number threshold is assigned here.

Calculate a saturated column

Use the horizontal column already specified: L = 0.20 m, A = 0.010 m², k = 1 × 10⁻¹² m² and φ_e = 0.40. Assign a fluid with μ = 0.001 Pa·s = 1 mPa·s and ρ = 1000 kg/m³. Let p_up = 3000 Pa and p_down = 1000 Pa, measured from the same pressure reference. These numerical properties are supplied for the model; no particular fluid is being identified.

The pressure decreases by 2000 Pa over 0.20 m, so dp/dx = (1000 − 3000)/0.20 = −10000 Pa/m. Then q = −(10⁻¹²/0.001)(−10000) = 1 × 10⁻⁵ m/s. Multiplying by A gives Q = 1 × 10⁻⁷ m³/s = 0.10 mL/s = 6 mL/min. Dividing q by φ_e gives v = 0.025 mm/s. This calculation supplies the flow used in the preceding comparison.

The unit check is useful: (m²/(Pa·s)) × (Pa/m) = m/s. Multiplying by m² then gives m³/s. Keep millimetres, millilitres and SI units separate until the calculation is complete; 1 m³ contains 10⁶ mL.

Take z = 0 along this horizontal column and g = 9.81 m/s². The downstream-minus-upstream head difference is (1000 − 3000)/(1000 × 9.81) = −0.203874 m after rounding. K = 9.81 × 10⁻⁶ m/s and dh/dx ≈ −1.019368 give the same positive q. A hydraulic gradient is dimensionless because it is a head difference divided by a distance.

The pressure profile below is linear because the supplied area and permeability are constant, the flow is steady and there are no internal sources or sinks. It is a bulk pressure profile, not a claim that microscopic pore speed is uniform. The pressure axis should include its stated reference and the position axis its unit.

Separate permeability from conductivity

Intrinsic permeability k characterises the fixed pore structure within this model. Hydraulic conductivity K = kρg/μ combines that structure with the fluid and the assigned gravitational acceleration. Their units differ: k has m², whereas K has m/s. Neither quantity is the same as porosity, which is a dimensionless volume fraction.

Compare two supplied fluids in the same unchanged column at the same 2000 Pa pressure drop. Fluid 1 has μ = 1 mPa·s; fluid 2 has μ = 2 mPa·s. Both have ρ = 1000 kg/m³. Assume neither fluid alters or blocks the matrix. The larger viscosity halves K, q, Q and v. It leaves the assigned intrinsic permeability k unchanged.

The pressure-gradient form makes a further distinction visible. For horizontal flow at fixed pressure gradient, q = −(k/μ) dp/dx contains no separate density factor. The head-gradient form contains density through K and through the conversion of pressure to head. Do not change one formulation’s inputs while silently holding the other formulation’s gradient fixed.

On a return reading, consider what happens if the pore structure varies with direction or changes during flow. An anisotropic permeability needs a tensor rather than our scalar k; deformation or pore blockage can change its value. Unsaturated or multi-phase flow needs phase and interface information. The straight lines in the proposed comparison apply only while the stated constant-property Darcy model remains appropriate.

Replace fluid shear with dry contacts

Now remove the flowing fluid from the mechanical example. A dry granular material is a collection of separate solid grains. Each grain can touch other grains or a boundary. At a contact, a normal force acts perpendicular to the contact and a tangential force acts along it. The many contacts transmit loads through an arrangement that can change when grains move.

A cohesionless contact has no assigned adhesive pull: its compressive normal force N is non-negative. For an ideal static Coulomb contact, the tangential friction magnitude F must satisfy |F| ≤ μ_sN, where μ_s is a supplied dimensionless coefficient. This is a capacity test. Static friction adjusts to the force needed for equilibrium; it need not equal μ_sN.

For one grain at rest, both the total force and the total moment must vanish. Moment measures the turning tendency of a force about a chosen point. A force balance alone is insufficient if the forces would still turn the grain. “Torque, angular momentum, and gyroscopes” is an optional refresher; the short geometric moment check is worked out in the next section.

Real contact friction can depend on load, surface condition and motion history. A constant μ_s is a stipulated simplification, not a universal value. A bulk granular shear response also depends on the contact arrangement and grain geometry. A coefficient at one grain–support contact does not directly give a bulk friction angle, a heap’s angle of repose or an avalanche prediction.

Contact conditions distinguish further regimes. Persistent contacts are central to a solid-like dense packing; collisions matter when grains repeatedly separate and impact. A surrounding liquid introduces hydrodynamic forces, and liquid bridges or adhesive contacts can introduce cohesion. Our dry, static, non-adhesive calculation includes none of those added interactions. It is a transparent local test of support forces, rather than a complete granular-flow law.

Balance one angular grain

Use an ideal rigid angular grain with a rectangular flat base on a rough plane inclined by θ = 30°. Assign mass m = 20 g = 0.020 kg, base length b = 20 mm along the plane and height H = 10 mm normal to it. The grain is homogeneous, so its centre of mass is at its geometric centre, H/2 above the base. Gravity and the support contact are the only forces; there is no adhesion, fluid force or interlocking.

Choose downhill and outward-normal directions. Weight mg points vertically downward. Its downhill component is mg sin θ and its inward-normal component is mg cos θ. For rest, the required uphill friction is F = mg sin θ and the normal support is N = mg cos θ. These trigonometric components are explained in “Geometry, trigonometry, and coordinate systems” if you need a refresher.

With g = 9.81 m/s², mg = 0.1962 N, F = 0.098100 N and N ≈ 0.169914 N. The required ratio is F/N = tan 30° ≈ 0.577350. For μ_s = 0.60, the capacity μ_sN ≈ 0.101949 N exceeds the demand, so the friction condition allows rest. For μ_s = 0.50, the capacity ≈ 0.084957 N is too small; rest is inadmissible under these assumptions.

Check turning as well. A distributed flat-base contact can place its resultant at a position s downhill of the base centre. About the grain’s centre of mass, the normal force contributes a moment sN and uphill friction contributes the opposite moment (H/2)F. Zero total moment therefore requires s = (H/2)(F/N) = (H/2)tan θ. At 30°, s ≈ 2.886751 mm, inside the base’s ±10 mm footprint. The combined support force then has a vertical line of action through the centre of mass.

Thus the μ_s = 0.60 case satisfies both the stated friction condition and a feasible resultant location for this ideal flat-base contact. Do not replace the shape with a freely rolling sphere while retaining the same moment argument. When μ_s = 0.50 fails the static test, we have not specified kinetic friction, contact dynamics or a later trajectory, so no acceleration or rolling conclusion follows.

The angle comparison below keeps the same mass, shape and μ_s = 0.60. Its demand ratio grows as tan θ. Even at 40° the required resultant remains inside the base, but the friction capacity is exceeded. The friction and moment checks answer different questions; passing one does not guarantee the other.

Keep the model scales distinct

The same bed of grains can be viewed as a porous medium for fluid flow and as a contact network for solid support. These views ask different questions. Porosity counts space; permeability enters a flow relation; contact forces and moments support the solids. A sample can have a large pore volume but poor through-connections, or a contact arrangement that changes while its total solid volume stays the same.

Before a calculation, state which phase is moving, what is averaged and what remains fixed. A fixed-matrix Darcy calculation cannot predict grain rearrangement or deformation. A dry one-grain friction calculation cannot predict saturated pressure, seepage forces or cohesion from liquid bridges. Coupling fluid and solid motion requires additional balances and material laws; it is not obtained by substituting one of these coefficients for another.

These distinctions organise questions about filters, groundwater-bearing materials and collections of grains: is the limiting information pore connectivity, the fluid’s viscosity, or the forces and geometry of the contacts? The numbers here illustrate reasoning, not a measured material or a design limit. A practical prediction requires material measurements, boundary conditions and a model valid for the relevant scale and regime.

For a first pass, explain the three different flow quantities and why static friction is a limit rather than an automatic force. On a return pass, derive the head conversion and the grain’s moment balance. Later lessons on fluid mechanics, coupled continua and jamming develop the additional equations and collective behaviour. Preserve the distinctions learned here when those descriptions become more detailed.

Worked examples

Try it yourself

1. A bulk sample has volume 2000 cm³ and solid volume 1200 cm³. Find pore volume and total porosity. Does this determine intrinsic permeability?

   Show the answer and reasoning

   Pore volume 800 cm³; total porosity 0.40 = 40%. Intrinsic permeability is not determined.

   Subtract solid from bulk volume, then divide pore volume by bulk volume. Connectivity and pore-path resistance require further information.

2. A different assigned sample has total porosity 0.40 and mobile effective porosity 0.20. Its measured Q is held at 0.10 mL/s through A = 0.010 m². Find q and v. Do not infer a new permeability from the changed mobile fraction.

   Show the answer and reasoning

   q = 0.010 mm/s and v = 0.050 mm/s.

   The supplied Q/A is unchanged. The mobile mean is q/0.20, so it doubles relative to the φ_e = 0.40 case. These supplied data do not determine how k changed.

3. Keep the original horizontal column, viscosity 1 mPa·s and positive x direction. Reverse the end pressures to 1000 Pa at x = 0 and 3000 Pa at x = 0.20 m. Find signed q and Q.

   Show the answer and reasoning

   q = −0.010 mm/s and Q = −0.10 mL/s, flowing opposite to positive x.

   dp/dx = +10000 Pa/m. The Darcy minus sign therefore gives a negative flux. Direction comes from the chosen axis, not from a negative physical volume.

4. At the original 2000 Pa pressure drop, double viscosity from 1 to 2 mPa·s without changing matrix, density or g. What happens to k, K and q?

   Show the answer and reasoning

   k stays 10⁻¹² m²; K becomes 4.905 × 10⁻⁶ m/s; q becomes 0.005 mm/s.

   K = kρg/μ and horizontal q = kΔp/(μL) are inversely proportional to μ under these conditions. The unchanged pore structure retains its assigned intrinsic permeability.

5. For the original 20 g flat-base grain at 30°, μ_s = 0.50. Required uphill friction is 0.098100 N and N ≈ 0.169914 N. Is rest admissible? Would setting friction to μ_sN solve the equilibrium equation?

   Show the answer and reasoning

   No. The capacity is approximately 0.084957 N, below the required 0.098100 N; setting friction to its capacity does not balance the downhill weight component.

   Static friction is a capacity-limited response, not always a prescribed maximum. The available resultant location does not repair a failed force balance. Subsequent acceleration needs a dynamic contact law.

What to carry forward

• Porosity counts pore volume; flow-active mobile porosity selects an averaging fraction; neither alone fixes intrinsic permeability.
• Q is volume per time, q = Q/A uses the whole face, and v = q/φ_e is the mean mobile-pore component.
• Darcy flow goes toward decreasing hydraulic head in the stated slow saturated fixed-matrix model. Keep the positive direction and minus sign explicit.
• Intrinsic permeability k has unit m²; hydraulic conductivity K = kρg/μ has unit m/s and includes fluid properties.
• A dry grain at rest needs both force and moment balance. Required static friction must lie within its capacity, and one contact model does not determine a whole granular material’s behaviour.