ENV-28

Carman–Kozeny filter

ΔP/L = 180 μ (1−ε)² u /(dp² ε³). Laminar packed-bed headloss.

Reading speed

FiltrationCarman–Kozeny

Governing equation

\dfrac{\Delta P}{L}=180\dfrac{\mu(1-\varepsilon)^2 u}{d_p^2\varepsilon^3}

$u \in \left[0.1, 20\right]$

where

\mu: Viscosity (mPa·s)
\varepsilon: Porosity (—)
u: Approach velocity (mm/s)
d_p: Grain diameter (mm)
L: Bed depth (m)
\Delta P: Pressure drop (kPa)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

\dfrac{\Delta P}{L}=180\dfrac{\mu(1-\varepsilon)^2 u}{d_p^2\varepsilon^3}

$u \in \left[0.1, 20\right]$

$u = 0.1$ $\Delta P = 0.225$

Along the curve

②

Domain of the variable — blue

u_{\min} = 0.1,\quad u_{\max} = 20

$u \in \left[0.1, 20\right]$

$u = 0.1$ $u = 0.1$

Along the curve

③

Response curve — green

\dfrac{\Delta P}{L}=180\dfrac{\mu(1-\varepsilon)^2 u}{d_p^2\varepsilon^3}

$u \in \left[0.1, 20\right]$

$u = 0.1$ $\Delta P = 0.225$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}\Delta P}{\mathrm{d}u}

$u \in \left[0.1, 20\right]$

$u = 0.225$ $\Delta P^{\prime} = 2.25$

Along the curve

Lecture brief

Historical brief

Streeter–Phelps (1925) oxygen sag, settling theory and Guldberg–Waage kinetics made water and air quality a rate problem. The lab computes sag, overflow and a snapshot of reactor mass balance. This sheet (ENV-28 — Carman–Kozeny filter) is the form associated with Carman–Kozeny. Working symbols: $\mu$ , $\varepsilon$ , $u$ , $d_p$ , $L$ $\rightarrow$ $\Delta P$ . Kozeny treated the bed as a bundle of hydraulic capillaries; Carman set the constant at 180 for spheres.

Purpose

Purpose: compute

\Delta P

from

\mu

\varepsilon

u

d_p

L

in Environmental via

\dfrac{\Delta P}{L}=180\dfrac{\mu(1-\varepsilon)^2 u}{d_p^2\varepsilon^3}

ΔP/L = 180 μ (1−ε)² u /(dp² ε³). Laminar packed-bed headloss. Use it when a real environmental question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\mu = 1.000\,\mathrm{mPa·s}

\varepsilon = 0.400\,\mathrm{—}

u = 2.000\,\mathrm{mm/s}

d_p = 0.600\,\mathrm{mm}

L = 0.800\,\mathrm{m}

, the governing relation

\dfrac{\Delta P}{L}=180\dfrac{\mu(1-\varepsilon)^2 u}{d_p^2\varepsilon^3}

yields

\Delta P = 4.500\,\mathrm{kPa}

. A sand bed, a seepage arrow, a head drop. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Viscosity \mu1 mPa·sPorosity \varepsilon0.4 —Approach velocity u2 mm/sGrain diameter d_p0.6 mmBed depth L0.8 m

Outputs

Pressure drop \Delta P4.500 kPa

Reading speed

Watch on YouTube

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Narration of this film

A sand bed, a seepage arrow, a head drop.

Kozeny treated the bed as a bundle of hydraulic capillaries; Carman set the constant at 180 for spheres.

Reading speed

Watch on YouTube