Flow Through a Packed Bed

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

11 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Flow Through a Packed Bed within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what flow through a packed bed describes physically and when it applies.
State every one of the 11 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Flow Through a Packed Bed is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: flow through a packed bed.
Capstone Studio instructional photograph

Fluid Mechanics — Flow Through a Packed Bed: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 11 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

L	Quantity produced by "L = length of particle bed (m)" — read its definition and unit from the handbook line directly above the equation.
Dp	Quantity produced by "Dp = average particle diameter (m)" — read its definition and unit from the handbook line directly above the equation.
Φs	Quantity produced by "Φs = sphericity of particles, dimensionless (0–1)" — read its definition and unit from the handbook line directly above the equation.
ε	Quantity produced by "ε = porosity or void fraction of the particle bed, dimensionless (0–1)" — read its definition and unit from the handbook line directly above the equation.
DP	Quantity produced by "DP = 150vo n _1 − f i + 1.75t v o _1 − f i" — read its definition and unit from the handbook line directly above the equation.
∆P	Quantity produced by "∆P = pressure loss across packed bed (Pa)" — read its definition and unit from the handbook line directly above the equation.
vo	Quantity produced by "vo = superficial (flow through empty vessel) fluid velocity (m/s)" — read its definition and unit from the handbook line directly above the equation.
ρ	Quantity produced by "ρ = fluid density (kg/m3)" — read its definition and unit from the handbook line directly above the equation.
µ	Quantity produced by "µ = fluid viscosity [kg/(m•s)]" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = A2v2 = CcCv A 2g _h1 − h2 i = CA 2g _h1 − h2 i" — read its definition and unit from the handbook line directly above the equation.
v2	Quantity produced by "v2 = velocity of fluid exiting orifice" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A porous, fixed bed of solid particles can be characterized by
The Ergun equation can be used to estimate pressure loss through a packed bed under laminar and turbulent flow conditions.
2 2
L 2 2 3 UsDp f3
U D f
s p
where
Submerged Orifice Operating under Steady-Flow Conditions:
Vennard, J.K., Elementary Fluid Mechanics, 6th ed., J.K. Vennard, 1954.
in which the product of Cc and Cv is defined as the coefficient of discharge of the orifice.
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.130 m/s through a 2.2 m deep bed of 9.0 mm media with a porosity of 0.39. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 9.0 mm
ε = 0.39
V_s = 0.130 m/s
L = 2.2 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.39)^{2} (0.130) / [0.3 9^{3} (0.0090)^{2}] = 1, 510 P a / m$
Inertial term
$1.75 (1000) (1 - 0.39) (0.130)^{2} / [0.3 9^{3} (0.0090)] = 33, 792 P a / m$
Gradient — Δp/L = 1,510 + 33,792 = 35,302 Pa/m
Total — Δp = 35,302(2.2) = 77.67 kPa

Answer: Δp/L = 35,302 Pa/m, Δp = 77.67 kPa over the bed

Why the other options are there

3.32 kPa (inertial term dropped)
199.1 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 2

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (2)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.040 m/s through a 0.5 m deep bed of 4.0 mm media with a porosity of 0.36. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 4.0 mm
ε = 0.36
V_s = 0.040 m/s
L = 0.5 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.36)^{2} (0.040) / [0.3 6^{3} (0.0040)^{2}] = 3, 292 P a / m$
Inertial term
$1.75 (1000) (1 - 0.36) (0.040)^{2} / [0.3 6^{3} (0.0040)] = 9, 602 P a / m$
Gradient — Δp/L = 3,292 + 9,602 = 12,894 Pa/m
Total — Δp = 12,894(0.5) = 6.45 kPa

Answer: Δp/L = 12,894 Pa/m, Δp = 6.45 kPa over the bed

Why the other options are there

1.65 kPa (inertial term dropped)
17.91 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 3

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (3)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.110 m/s through a 0.8 m deep bed of 20.0 mm media with a porosity of 0.39. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 20.0 mm
ε = 0.39
V_s = 0.110 m/s
L = 0.8 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.39)^{2} (0.110) / [0.3 9^{3} (0.0200)^{2}] = 258.8 P a / m$
Inertial term
$1.75 (1000) (1 - 0.39) (0.110)^{2} / [0.3 9^{3} (0.0200)] = 10, 888 P a / m$
Gradient — Δp/L = 258.8 + 10,888 = 11,146 Pa/m
Total — Δp = 11,146(0.8) = 8.92 kPa

Answer: Δp/L = 11,146 Pa/m, Δp = 8.92 kPa over the bed

Why the other options are there

0.21 kPa (inertial term dropped)
22.86 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 4

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (4)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.080 m/s through a 1.8 m deep bed of 6.0 mm media with a porosity of 0.41. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 6.0 mm
ε = 0.41
V_s = 0.080 m/s
L = 1.8 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.41)^{2} (0.080) / [0.4 1^{3} (0.0060)^{2}] = 1, 684 P a / m$
Inertial term
$1.75 (1000) (1 - 0.41) (0.080)^{2} / [0.4 1^{3} (0.0060)] = 15, 980 P a / m$
Gradient — Δp/L = 1,684 + 15,980 = 17,663 Pa/m
Total — Δp = 17,663(1.8) = 31.79 kPa

Answer: Δp/L = 17,663 Pa/m, Δp = 31.79 kPa over the bed

Why the other options are there

3.03 kPa (inertial term dropped)
77.55 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 5

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (5)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.110 m/s through a 1.4 m deep bed of 16.0 mm media with a porosity of 0.47. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 16.0 mm
ε = 0.47
V_s = 0.110 m/s
L = 1.4 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.47)^{2} (0.110) / [0.4 7^{3} (0.0160)^{2}] = 174.4 P a / m$
Inertial term
$1.75 (1000) (1 - 0.47) (0.110)^{2} / [0.4 7^{3} (0.0160)] = 6, 756 P a / m$
Gradient — Δp/L = 174.4 + 6,756 = 6,930 Pa/m
Total — Δp = 6,930(1.4) = 9.70 kPa

Answer: Δp/L = 6,930 Pa/m, Δp = 9.70 kPa over the bed

Why the other options are there

0.24 kPa (inertial term dropped)
20.64 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 6

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (6)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.080 m/s through a 2.3 m deep bed of 16.0 mm media with a porosity of 0.39. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 16.0 mm
ε = 0.39
V_s = 0.080 m/s
L = 2.3 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.39)^{2} (0.080) / [0.3 9^{3} (0.0160)^{2}] = 294.0 P a / m$
Inertial term
$1.75 (1000) (1 - 0.39) (0.080)^{2} / [0.3 9^{3} (0.0160)] = 7, 198 P a / m$
Gradient — Δp/L = 294.0 + 7,198 = 7,492 Pa/m
Total — Δp = 7,492(2.3) = 17.23 kPa

Answer: Δp/L = 7,492 Pa/m, Δp = 17.23 kPa over the bed

Why the other options are there

0.68 kPa (inertial term dropped)
44.19 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 7

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (7)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.180 m/s through a 2.0 m deep bed of 10.0 mm media with a porosity of 0.46. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 10.0 mm
ε = 0.46
V_s = 0.180 m/s
L = 2.0 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.46)^{2} (0.180) / [0.4 6^{3} (0.0100)^{2}] = 808.9 P a / m$
Inertial term
$1.75 (1000) (1 - 0.46) (0.180)^{2} / [0.4 6^{3} (0.0100)] = 31, 456 P a / m$
Gradient — Δp/L = 808.9 + 31,456 = 32,265 Pa/m
Total — Δp = 32,265(2.0) = 64.53 kPa

Answer: Δp/L = 32,265 Pa/m, Δp = 64.53 kPa over the bed

Why the other options are there

1.62 kPa (inertial term dropped)
140.3 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 8

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (8)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.120 m/s through a 2.4 m deep bed of 6.0 mm media with a porosity of 0.37. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 6.0 mm
ε = 0.37
V_s = 0.120 m/s
L = 2.4 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.37)^{2} (0.120) / [0.3 7^{3} (0.0060)^{2}] = 3, 918 P a / m$
Inertial term
$1.75 (1000) (1 - 0.37) (0.120)^{2} / [0.3 7^{3} (0.0060)] = 52, 238 P a / m$
Gradient — Δp/L = 3,918 + 52,238 = 56,156 Pa/m
Total — Δp = 56,156(2.4) = 134.8 kPa

Answer: Δp/L = 56,156 Pa/m, Δp = 134.8 kPa over the bed

Why the other options are there

9.40 kPa (inertial term dropped)
364.3 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 9

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (9)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.190 m/s through a 1.6 m deep bed of 4.0 mm media with a porosity of 0.36. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 4.0 mm
ε = 0.36
V_s = 0.190 m/s
L = 1.6 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.36)^{2} (0.190) / [0.3 6^{3} (0.0040)^{2}] = 15, 638 P a / m$
Inertial term
$1.75 (1000) (1 - 0.36) (0.190)^{2} / [0.3 6^{3} (0.0040)] = 216, 650 P a / m$
Gradient — Δp/L = 15,638 + 216,650 = 232,287 Pa/m
Total — Δp = 232,287(1.6) = 371.7 kPa

Answer: Δp/L = 232,287 Pa/m, Δp = 371.7 kPa over the bed

Why the other options are there

25.02 kPa (inertial term dropped)
1,032 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Example 10

Ergun pressure drop through a packed filter bed — Flow Through a Packed Bed (10)

Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) passes at a superficial velocity of 0.020 m/s through a 0.6 m deep bed of 13.0 mm media with a porosity of 0.35. Use the Ergun equation to find the pressure gradient and the total pressure drop.

Given

D_p = 13.0 mm
ε = 0.35
V_s = 0.020 m/s
L = 0.6 m

Find

Pressure gradient and total Δp

Start with the thinking

The Ergun equation adds a viscous (laminar) term and an inertial (turbulent) term — evaluate them separately.
Superficial velocity uses the empty-column area, not the pore area.

Step-by-step solution

Formula — \dfrac{\Delta p}{L} = \dfrac{150\mu(1-\varepsilon)^2 V_s}{\varepsilon^3 D_p^2} + \dfrac{1.75\rho(1-\varepsilon)V_s^2}{\varepsilon^3 D_p}
Viscous term
$150 (0.001) (1 - 0.35)^{2} (0.020) / [0.3 5^{3} (0.0130)^{2}] = 174.9 P a / m$
Inertial term
$1.75 (1000) (1 - 0.35) (0.020)^{2} / [0.3 5^{3} (0.0130)] = 816.3 P a / m$
Gradient — Δp/L = 174.9 + 816.3 = 991.3 Pa/m
Total — Δp = 991.3(0.6) = 0.59 kPa

Answer: Δp/L = 991.3 Pa/m, Δp = 0.59 kPa over the bed

Why the other options are there

0.10 kPa (inertial term dropped)
1.70 kPa (porosity applied twice)

Reference: FE Reference Handbook — Fluid Mechanics → Flow Through a Packed Bed

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Flow Through a Packed Bed contains 11 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.