Flow through packed beds, Ergun equation
Describing packed beds (porosity, sphericity, specific surface) and predicting their pressure drop with the Kozeny-Carman, Burke-Plummer and Ergun equations.
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Why it matters
Fixed-bed catalytic reactors, adsorbers, ion-exchange columns, sand filters, driers and packed absorption towers all force fluid through a bed of particles. The pressure drop sets blower or pump power and limits how small the catalyst can be (small particles give better effectiveness but higher pressure drop). The Ergun equation is the standard tool, and it also leads directly to the minimum fluidisation velocity in the next topic.
Key ideas
Describing the bed.
- Void fraction (porosity) ε: volume of voids / total bed volume. Randomly packed uniform spheres give ε ≈ 0.36–0.40; rings and saddles much higher. ε is higher near the wall (wall effect, significant if the column-to-particle diameter ratio is below about 10–20).
- Sphericity φ_s: surface area of a sphere of equal volume / actual surface area (1 for spheres, less for other shapes). The effective diameter is φ_s·d_p, with d_p the equal-volume sphere diameter.
- Specific surface of particles a_s = 6/(φ_s·d_p) (m²/m³ of solid); of the bed a = a_s(1 − ε).
- Superficial velocity V₀ = Q/A (based on the empty column); the average interstitial velocity in the voids is V₀/ε.
Physical model. The voids are pictured as a bundle of tortuous channels with hydraulic radius R_h = ε/[a_s(1 − ε)] = ε·φ_s·d_p/[6(1 − ε)]. Applying pipe-flow laws to these channels gives:
- Laminar (viscous) region — Kozeny–Carman equation: ΔP/L ∝ μV₀(1 − ε)²/(φ_s²d_p²ε³). Pressure drop is linear in velocity. The constant is 150 (Ergun's value) or 180 (Carman's value, often quoted as Kozeny constant 5); use the value in your textbook.
- Turbulent (inertial) region — Burke–Plummer equation: ΔP/L = 1.75ρV₀²(1 − ε)/(φ_s·d_p·ε³). Pressure drop grows as V₀².
- Ergun equation: the sum of the two, valid across the whole range.
Particle Reynolds number for beds. Re_p = ρV₀φ_s d_p/[μ(1 − ε)]. The viscous term dominates for Re_p < about 10; the inertial term for Re_p > about 1000. The bed friction factor f_p = (ΔP/L)·φ_s d_p ε³/[ρV₀²(1 − ε)] = 150/Re_p + 1.75.
Sensitivity. The (1 − ε)²/ε³ factor makes pressure drop extremely sensitive to porosity: a drop in ε from 0.40 to 0.35 raises the viscous term by about 75 %. Particle attrition, fines and poor loading therefore raise ΔP sharply in operating beds.
Assumptions and limits. Steady, incompressible flow (for gases with large ΔP, use average density or integrate along the bed), uniform random packing, rigid particles, no channelling or wall effect, Newtonian fluid. The constants 150 and 1.75 are empirical fits; real beds often deviate by ±20–30 %.
Formulas
ΔP/L = 150·μ·V₀·(1 − ε)²/(φ_s²·d_p²·ε³) + 1.75·ρ·V₀²·(1 − ε)/(φ_s·d_p·ε³)— Ergun; ΔP/L (Pa/m), μ (Pa·s), V₀ superficial velocity (m/s), d_p (m), ρ (kg/m³).ΔP/L = 150·μ·V₀·(1 − ε)²/(φ_s²·d_p²·ε³)— Kozeny–Carman (laminar), Re_p < ≈10.ΔP/L = 1.75·ρ·V₀²·(1 − ε)/(φ_s·d_p·ε³)— Burke–Plummer (turbulent), Re_p > ≈1000.Re_p = ρ·V₀·φ_s·d_p/[μ·(1 − ε)]— bed particle Reynolds number.f_p = 150/Re_p + 1.75— Ergun in friction-factor form.a_s = 6/(φ_s·d_p)— specific surface of particles (m⁻¹).V = V₀/ε— interstitial velocity (m/s).
Worked examples
Example 1 (standard). Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) flows at a superficial velocity of 0.1 m/s through a 1 m bed of 5 mm spheres with ε = 0.4. Find ΔP and the share of the inertial term.
- Viscous term: 150 × 0.001 × 0.1 × (0.6)² / [(0.005)² × (0.4)³] = 0.0054 / (1.6 × 10⁻⁶) = 3375 Pa/m.
- Inertial term: 1.75 × 1000 × (0.1)² × 0.6 / [0.005 × (0.4)³] = 10.5 / (3.2 × 10⁻⁴) = 32 812.5 Pa/m.
- Total ΔP/L = 36 187.5 Pa/m; over 1 m, ΔP ≈ 36.2 kPa.
- Re_p = 1000 × 0.1 × 0.005 / (0.001 × 0.6) = 833: transitional, inertia-dominated (91 % of ΔP). ΔP ≈ 36.2 kPa.
Example 2 (GATE level). Air (ρ = 1.2 kg/m³, μ = 1.8 × 10⁻⁵ Pa·s) passes at V₀ = 0.5 m/s through a 2 m catalyst bed of cylindrical pellets with equal-volume diameter 3 mm, sphericity 0.87 and ε = 0.38. Find ΔP, and by what factor it rises if V₀ is doubled.
- Effective diameter φ_s·d_p = 0.87 × 0.003 = 2.61 × 10⁻³ m.
- Viscous term = 150 × 1.8 × 10⁻⁵ × 0.5 × (0.62)² / [(2.61 × 10⁻³)² × (0.38)³] = 1388 Pa/m.
- Inertial term = 1.75 × 1.2 × 0.25 × 0.62 / [2.61 × 10⁻³ × (0.38)³] = 2273 Pa/m.
- ΔP/L = 3661 Pa/m; ΔP = 3661 × 2 = 7322 Pa ≈ 7.3 kPa (small compared with the inlet pressure, so treating air as incompressible is acceptable).
- Doubling V₀: viscous term doubles (2777), inertial term quadruples (9091); total 11 868 Pa/m, a factor of 3.24. ΔP ≈ 7.3 kPa; it rises about 3.2 times when V₀ doubles — between the purely laminar (2×) and purely turbulent (4×) limits.
Common mistakes
- Using interstitial velocity in the Ergun equation; it uses superficial velocity.
- Forgetting sphericity, or putting φ_s in the inertial term squared (it is φ_s² only in the viscous term).
- Mixing up ε³ and (1 − ε)³, or (1 − ε)² and (1 − ε).
- Using the pipe Reynolds number instead of the bed Re_p.
- Assuming pressure drop is linear in flow at all velocities.
For GATE CH
Expect Ergun numericals (ΔP, or V₀ given ΔP — solve the quadratic), Kozeny–Carman in the laminar limit, effect of changing particle size, porosity or flow on ΔP (ratio questions), specific surface and sphericity calculations, and the link to minimum fluidisation velocity. Practise solving the Ergun quadratic for velocity cleanly.
Quick check
- Which velocity appears in the Ergun equation?
- In the laminar limit, how does ΔP change if d_p is halved?
- What is the specific surface of 2 mm spheres?
- In the fully turbulent limit, how does ΔP vary with V₀?
Answers: 1. Superficial velocity V₀. 2. It increases four times. 3. 6/0.002 = 3000 m⁻¹. 4. ΔP ∝ V₀².
See it move
All Chemical animationsAdjust the particle diameter, void fraction, and fluid velocity to see how they affect the pressure drop in a packed bed. Observe the changes in the pressure drop per unit length.
Equations used
- ΔP/L = (150 * (1-ε)^2 * μ * v / (d_p^2 * ε^3)) + (1.75 * (1-ε) * ρ * v^2 / (d_p * ε^3)) — ΔP/L pressure drop per unit length, ε void fraction, μ dynamic viscosity, v superficial velocity, d_p particle diameter, ρ fluid density
Interview questions
All Fluid Mechanics interview questionsTry answering each one aloud before you open it.
1.What is a packed bed in the context of fluid mechanics?Concept
A packed bed is a type of reactor or column filled with solid particles, known as packing material. These particles can be of various shapes and sizes and are used to enhance contact between a fluid and the solid surface. Packed beds are commonly used in chemical engineering processes such as absorption, distillation, and catalytic reactions.
2.Explain the Ergun equation and its significance in fluid flow through packed beds.Concept
The Ergun equation is a fundamental equation used to calculate the pressure drop across a packed bed. It combines the effects of viscous and inertial forces and is expressed as ΔP/L = (150 * (1-ε)^2 * μ * v) / (d_p^2 * ε^3) + (1.75 * (1-ε) * ρ * v^2) / (d_p * ε^3), where ΔP is the pressure drop, L is the bed length, ε is the void fraction, μ is the fluid viscosity, v is the superficial velocity, d_p is the particle diameter, and ρ is the fluid density. This equation is significant because it helps in designing and optimizing packed bed reactors by predicting the pressure drop, which is crucial for energy efficiency and process control.
3.Why is the void fraction (ε) important in the Ergun equation?Application
The void fraction, ε, represents the fraction of the bed volume that is occupied by the fluid. It is important in the Ergun equation because it affects both the viscous and inertial components of the pressure drop. A higher void fraction means more space for the fluid to flow, resulting in a lower pressure drop. Conversely, a lower void fraction indicates less space for fluid flow, leading to a higher pressure drop. Accurate knowledge of the void fraction is essential for precise calculations of pressure drop in packed beds.
4.What happens to the pressure drop if the particle diameter in a packed bed is decreased?Application
If the particle diameter in a packed bed is decreased, the pressure drop across the bed generally increases. This is because smaller particles create more surface area and resistance to fluid flow, leading to higher frictional losses. According to the Ergun equation, the pressure drop is inversely proportional to the square of the particle diameter for the viscous term and inversely proportional to the particle diameter for the inertial term. Therefore, reducing the particle diameter results in a significant increase in pressure drop.
5.How does fluid viscosity affect the pressure drop in a packed bed?Application
Fluid viscosity directly affects the pressure drop in a packed bed. According to the Ergun equation, the viscous term of the pressure drop is directly proportional to the fluid viscosity. This means that as the viscosity of the fluid increases, the pressure drop also increases. High-viscosity fluids experience greater resistance to flow, leading to higher energy consumption to maintain the same flow rate.
6.In what scenarios would you prefer using a packed bed reactor over other types of reactors?Application
Packed beds suit solid-catalysed gas or liquid reactions where you want near plug flow, a high catalyst loading per unit volume and no need to separate catalyst from product. They are simple, cheap and scale well, and the catalyst suffers little attrition. They are less suitable for strongly exothermic reactions, because radial heat removal through a packed bed is poor and hot spots form (multitubular designs or fluidised beds are used instead), and for catalysts that deactivate quickly and need frequent regeneration.
7.Calculate the pressure drop across a 1 m packed bed of 5 mm spheres with void fraction 0.4, for water (μ = 0.001 Pa·s, ρ = 1000 kg/m³) at a superficial velocity of 0.1 m/s.Numerical
Ergun: the viscous term is 150 × 0.001 × 0.1 × 0.6² / (0.005² × 0.4³) = 3375 Pa/m, and the inertial term is 1.75 × 1000 × 0.1² × 0.6 / (0.005 × 0.4³) = 32 812 Pa/m. The total is about 36 200 Pa/m, so ΔP ≈ 36.2 kPa over 1 m. The bed Reynolds number ρV₀d/[μ(1 − ε)] ≈ 833, so the inertial term carries about 90 % of the loss.
8.What are the limitations of the Ergun equation?Concept
The Ergun equation assumes that the flow is steady and incompressible, which may not be valid for all fluid types or flow conditions. It also assumes that the packing is uniform and that the particles are spherical, which may not be the case in real-world applications. Additionally, the equation may not accurately predict pressure drops for very high or very low Reynolds numbers, where other flow regimes might dominate. These limitations mean that while the Ergun equation is a useful tool, it should be applied with an understanding of its assumptions and potential inaccuracies.
9.How would you modify the design of a packed bed to minimize pressure drop?Application
Use larger particles (ΔP falls as 1/d² in the viscous term and 1/d in the inertial term), or shapes that pack with higher voidage such as rings or trilobes, since ΔP is very sensitive to (1 − ε)²/ε³. Lower the superficial velocity by using a larger diameter, shorter bed or radial-flow design. Keep fines out and load the bed carefully to avoid local low voidage. Each option trades against something, such as lower catalyst effectiveness for large particles or poorer flow distribution in wide, shallow beds.
10.A packed bed of 10 mm spheres with void fraction 0.35 shows a pressure drop of 5000 Pa per metre of bed for a liquid of density 800 kg/m³. If the viscous term is negligible, what is the superficial velocity?Numerical
With only the Burke–Plummer (inertial) term, ΔP/L = 1.75ρV₀²(1 − ε)/(d_p ε³). Rearranging, V₀ = √[(ΔP/L)·d_p·ε³ / (1.75ρ(1 − ε))] = √[5000 × 0.01 × 0.042875 / (1.75 × 800 × 0.65)] ≈ 0.049 m/s. You should then check the bed Reynolds number to confirm that neglecting the viscous term was reasonable.
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