Fluidisation, minimum fluidisation velocity and bed expansion

The pressure drop–velocity curve of a fluidised bed, minimum fluidisation velocity from the Ergun and Wen-Yu equations, bed expansion, and particulate versus bubbling fluidisation.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Fluidised beds give excellent solids mixing, nearly uniform temperature and high heat- and mass-transfer rates, which is why they are used for fluid catalytic cracking, coal and biomass combustion, drying, granulation, coating and polymerisation. Designing one starts with three numbers: the pressure drop needed to support the bed, the minimum fluidisation velocity, and the terminal velocity above which particles are carried out.

Key ideas

From packed to fluidised. Pass fluid upward through a bed of particles and slowly raise the superficial velocity U:

  1. Below fluidisation the bed is fixed and ΔP rises with U according to the Ergun equation.
  2. At the minimum fluidisation velocity U_mf, the drag on the particles equals their effective (buoyancy-corrected) weight; the bed loosens to voidage ε_mf and becomes fluid-like.
  3. Above U_mf, ΔP stays essentially constant (equal to the bed's effective weight per unit area) while the bed expands and voidage increases.
  4. When U approaches the particle terminal velocity u_t, particles are carried out of the bed (elutriation, pneumatic transport). A plot of ΔP against U therefore rises, peaks slightly (interparticle forces must be overcome), then flattens. U_mf is read from the intersection of the fixed-bed line and the horizontal line, usually from data taken with decreasing velocity to avoid the hysteresis.

Pressure drop in a fluidised bed. Force balance on the bed: ΔP·A = (weight of solids − buoyancy) = A·L·(1 − ε)(ρ_s − ρ)g. Because L(1 − ε) is fixed by the mass of solids, ΔP is constant as the bed expands.

Minimum fluidisation velocity. Equate the Ergun pressure drop at ε_mf to the bed weight and solve the quadratic in U_mf. Limits:

  • Small particles (Re_mf < about 20): viscous term only, U_mf = (ρ_s − ρ)g·ε_mf³·(φ_s d_p)² / [150μ(1 − ε_mf)]. Note U_mf ∝ d_p² and ∝ 1/μ — a more viscous fluid fluidises particles at a lower velocity.
  • Large particles (Re_mf > about 1000): inertial term only, U_mf ∝ √d_p.
  • When ε_mf is unknown, the Wen–Yu correlation is used: Re_mf = √(33.7² + 0.0408·Ar) − 33.7, with Archimedes number Ar = d_p³ρ(ρ_s − ρ)g/μ².

Bed expansion. Since the solid volume L(1 − ε) is constant, L₂/L₁ = (1 − ε₁)/(1 − ε₂). For liquid-fluidised beds the Richardson–Zaki relation U = u_t·εⁿ gives voidage as a function of velocity (n ≈ 4.65 at low Re, about 2.4 at high Re).

Types of fluidisation.

  • Particulate (smooth): typical of liquid–solid systems; bed expands uniformly, no bubbles.
  • Aggregative (bubbling): typical of gas–solid systems; excess gas above U_mf passes as bubbles. Other regimes: slugging (bubbles span a narrow column), turbulent, fast fluidisation and pneumatic transport.
  • Geldart classification of powders: group A (aeratable, fine, e.g. FCC catalyst), B (sand-like, bubbles at U_mf), C (cohesive, hard to fluidise), D (coarse, spoutable).
  • Problems: channelling (fluid short-circuits through cracks), slugging, elutriation of fines, erosion of internals.

Operating range. Practical gas velocities lie between U_mf and u_t. The ratio u_t/U_mf is large (often 10–90) for small particles, giving a wide operating window.

Formulas

  • ΔP = L·(1 − ε)·(ρ_s − ρ)·g = M·g·(ρ_s − ρ)/(ρ_s·A) — fluidised-bed pressure drop (Pa); M mass of solids (kg), A bed area (m²).
  • (1 − ε_mf)(ρ_s − ρ)g = 150·μ·U_mf·(1 − ε_mf)²/[(φ_s d_p)²·ε_mf³] + 1.75·ρ·U_mf²·(1 − ε_mf)/(φ_s d_p·ε_mf³) — Ergun at incipient fluidisation.
  • U_mf = (ρ_s − ρ)·g·ε_mf³·(φ_s·d_p)²/[150·μ·(1 − ε_mf)] — small-particle limit (m/s).
  • Ar = d_p³·ρ·(ρ_s − ρ)·g/μ², Re_mf = √(33.7² + 0.0408·Ar) − 33.7 — Wen–Yu; Re_mf = ρU_mf d_p/μ.
  • L₁·(1 − ε₁) = L₂·(1 − ε₂) — bed expansion.
  • U = u_t·εⁿ — Richardson–Zaki.

Worked examples

Example 1 (standard). A column 0.5 m in diameter holds 100 kg of sand (ρ_s = 2600 kg/m³) fluidised by water (ρ = 1000 kg/m³). At minimum fluidisation ε_mf = 0.42. Find the bed height at U_mf, the pressure drop across the bed, and the voidage when the bed has expanded to 1.5 times this height.

  1. A = (π/4)(0.5)² = 0.1963 m².
  2. Solid volume = 100/2600 = 0.03846 m³; L_mf = 0.03846/[(1 − 0.42) × 0.1963] = 0.338 m.
  3. ΔP = L(1 − ε)(ρ_s − ρ)g = 0.338 × 0.58 × 1600 × 9.81 = 3075 Pa.
  4. Expanded bed: (1 − ε₂) = 0.58/1.5 = 0.387, so ε₂ = 0.613; ΔP is unchanged. L_mf ≈ 0.338 m; ΔP ≈ 3.07 kPa; ε ≈ 0.61 after expansion.

Example 2 (GATE level). Find U_mf for the same sand, as spheres of 0.3 mm diameter, in water (μ = 0.001 Pa·s), with ε_mf = 0.42.

  1. Effective weight per unit volume: (1 − ε_mf)(ρ_s − ρ)g = 0.58 × 1600 × 9.81 = 9104 Pa/m.
  2. Viscous coefficient: 150μ(1 − ε)²/(d_p²ε³) = 150 × 0.001 × 0.3364 / (9 × 10⁻⁸ × 0.07409) = 7.568 × 10⁶ Pa·s/m².
  3. Inertial coefficient: 1.75ρ(1 − ε)/(d_p ε³) = 1.75 × 1000 × 0.58 / (3 × 10⁻⁴ × 0.07409) = 4.567 × 10⁷ Pa·s²/m³.
  4. Solve 4.567 × 10⁷·U² + 7.568 × 10⁶·U − 9104 = 0: U_mf = 1.19 × 10⁻³ m/s.
  5. Check: Re_mf = 1000 × 0.00119 × 0.0003 / 0.001 = 0.36 — viscous regime, so the small-particle formula gives almost the same result: 1.20 × 10⁻³ m/s. U_mf ≈ 1.2 mm/s.

Common mistakes

  • Using the Stokes terminal-velocity formula as U_mf; they are different velocities (u_t is many times U_mf).
  • Forgetting buoyancy in the bed weight: use (ρ_s − ρ), which matters a lot for liquid fluidisation.
  • Expecting ΔP to keep rising above U_mf.
  • Assuming a more viscous fluid needs a higher U_mf; in the viscous regime U_mf ∝ 1/μ.
  • Mixing up particulate and aggregative fluidisation.

For GATE CH

Expect numericals on fluidised-bed pressure drop from the mass of solids, U_mf from the Ergun equation (often with the viscous term only), bed height and voidage changes on expansion, and conceptual questions on the ΔP–U curve, Geldart groups, and particulate versus bubbling behaviour. The operating window between U_mf and u_t also appears.

Quick check

  1. What happens to ΔP when U is increased above U_mf?
  2. 50 kg of particles (ρ_s = 2500 kg/m³) are fluidised by air in a 0.2 m² bed. Estimate ΔP (neglect air density).
  3. A bed expands from ε = 0.45 to ε = 0.6. By what factor does its height increase?
  4. In the viscous regime, how does U_mf change if particle diameter doubles?

Answers: 1. It stays nearly constant. 2. 50 × 9.81/0.2 = 2453 Pa. 3. 0.55/0.4 = 1.375. 4. It increases four times.

Interview questions

All Fluid Mechanics interview questions

Try answering each one aloud before you open it.

  1. 1.What is fluidisation in the context of fluid mechanics?Concept

    Fluidisation is the process by which a granular material is transformed into a fluid-like state through the introduction of a fluid (liquid or gas) flowing through the material. This occurs when the upward drag force exerted by the fluid equals the gravitational force on the particles, causing them to become suspended and behave like a fluid.

  2. 2.Explain the concept of minimum fluidisation velocity.Concept

    The minimum fluidisation velocity is the lowest velocity at which a fluid can be passed through a bed of particles to initiate fluidisation. At this velocity, the drag force from the fluid balances the weight of the particles, causing them to lift and become suspended, transitioning the bed from a packed state to a fluidised state.

  3. 3.What factors affect the minimum fluidisation velocity?Concept

    U_mf comes from equating the Ergun pressure drop to the buoyant weight of the bed, so it depends on particle size, shape (sphericity) and density, the fluid density and viscosity, and the voidage at minimum fluidisation. Larger and denser particles need a higher U_mf (U_mf ∝ d_p² for fine particles, ∝ √d_p for coarse ones). A more viscous fluid exerts more drag, so it fluidises fine particles at a lower velocity (U_mf ∝ 1/μ in the viscous regime). A higher ε_mf also lowers the velocity needed.

  4. 4.How does bed expansion occur in a fluidised bed?Concept

    Bed expansion occurs when the fluid velocity exceeds the minimum fluidisation velocity, causing the particles to move further apart and the bed to expand. As the velocity increases, the bed height increases due to the increased spacing between particles, while the pressure drop across the bed remains relatively constant.

  5. 5.Why is fluidisation used in industrial processes?Application

    Fluidisation is used in industrial processes because it enhances heat and mass transfer, provides uniform temperature distribution, and allows for easy handling of solid particles. It is commonly used in processes like catalytic cracking, drying, and coating due to these advantages.

  6. 6.What happens if the fluid velocity is much higher than the minimum fluidisation velocity?Application

    If the fluid velocity is much higher than the minimum fluidisation velocity, the bed may enter a turbulent or slugging regime, where large bubbles form and disrupt the uniformity of the fluidised state. This can lead to poor mixing, uneven temperature distribution, and potential damage to equipment.

  7. 7.How would you determine the minimum fluidisation velocity experimentally?Application

    To determine the minimum fluidisation velocity experimentally, gradually increase the fluid velocity through a bed of particles and measure the pressure drop across the bed. The minimum fluidisation velocity is identified at the point where the pressure drop becomes constant, indicating the onset of fluidisation.

  8. 8.Estimate the minimum fluidisation velocity for 0.5 mm spheres of density 2500 kg/m³ in water (ρ = 998 kg/m³, μ = 0.001 Pa·s), taking ε_mf = 0.4.Numerical

    Set the Ergun pressure drop at ε_mf equal to the bed weight (1 − ε)(ρ_s − ρ)g ≈ 8840 Pa/m. That gives a quadratic in U_mf whose positive root is about 2.6 × 10⁻³ m/s (2.6 mm/s). Then Re_mf = 998 × 0.0026 × 0.0005 / 0.001 ≈ 1.3, so the viscous term dominates and the simple formula (ρ_s − ρ)gε³d²/[150μ(1 − ε)] gives almost the same answer. Without an ε_mf value, the Wen–Yu correlation gives about 2.2 mm/s.

  9. 9.A fluidised bed reactor operates with a gas flow rate of 0.1 m/s. If the minimum fluidisation velocity is 0.05 m/s, describe the state of the bed.Application

    Since the gas flow rate of 0.1 m/s is higher than the minimum fluidisation velocity of 0.05 m/s, the bed is in a fluidised state. The particles are suspended and exhibit fluid-like behavior. However, care must be taken to ensure the velocity does not lead to turbulent or slugging conditions.

  10. 10.What are the consequences of operating a fluidised bed below the minimum fluidisation velocity?Application

    Operating below the minimum fluidisation velocity means the bed remains in a packed state, with particles not fully suspended. This results in poor mixing, reduced heat and mass transfer efficiency, and potential channeling of the fluid, which can lead to uneven processing and reduced performance.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousFlow through packed beds, Ergun equationNext Flow measurement: orifice, venturi, pitot tube, rotameter