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Flow through packed and fluidised beds

Chemical Engineering (B.E./B.Tech) · Fluid and particle mechanics · free study notes for exams, placements and interviews

In short

A packed bed is a fixed column of solid particles through which a fluid is forced; a fluidised bed is the same column at a velocity high enough for the drag on the particles to support their weight, so the bed behaves like a boiling liquid. The design question in both cases is pressure drop versus velocity, which fixes blower/pump power, and the velocity window between minimum fluidisation and particle carry-over. The topic carries heavy weight in GATE-CH (Mechanical Operations and Fluid Mechanics) and in university Unit Operations papers, and is used daily in catalytic reactors, FCC units, AFBC/CFBC boilers, fluid-bed dryers and granulators, adsorbers, ion-exchange columns and absorption towers. Interviewers at Reliance, IOCL, BPCL, Tata Chemicals, UPL, Thermax, Cipla, Sun Pharma and GMM Pfaudler ask for the Ergun equation, u_mf and why bed pressure drop stays flat after fluidisation.

Explained

Geometry of a bed of particles

Pressure drop through a fixed packed bed

Onset of fluidisation

Regimes above u_mf, and Geldart classification

Design elements of a fluidised bed

Packed versus fluidised: choosing one

FeaturePacked (fixed) bedFluidised bed
ΔP with velocityRises (linear then quadratic)Constant after u_mf
Solids movementNoneVigorous mixing
Temperature profileHot spots, axial gradientNear isothermal
Solids handlingShutdown to changeContinuous feed/withdrawal
Gas-solid contactingNear plug flow, high conversionBubble bypassing lowers conversion
Particle sizePellets, 3-25 mm~0.03-3 mm
WearLowAttrition and erosion
Typical useFixed-bed catalysis, adsorption, absorption towersFCC, FBC boilers, drying, roasting, granulation

Key points

Formulas and examples

Ergun equation: ΔP/L = 150 μ u (1−ε)² / (φ_s² d_p² ε³) + 1.75 ρ u² (1−ε) / (φ_s d_p ε³)
ΔP = pressure drop (Pa), L = bed height (m), μ = fluid viscosity (Pa·s), u = superficial velocity (m/s, = Q/A on empty cross-section), ε = voidage (−), d_p = particle diameter (m), φ_s = sphericity (−), ρ = fluid density (kg/m³)
Kozeny-Carman (laminar, Re_p < 10): ΔP/L = 150 μ u (1−ε)² / (φ_s² d_p² ε³)
Burke-Plummer (turbulent, Re_p > 1000): ΔP/L = 1.75 ρ u² (1−ε) / (φ_s d_p ε³)
Dimensionless Ergun: f_p = 150/Re' + 1.75, with f_p = (ΔP/L)·d_p ε³ / (ρ u² (1−ε)) and Re' = d_p u ρ / (μ(1−ε))
Specific surface: a = 6(1−ε)/(φ_s d_p) (m²/m³ bed); Sauter mean d_sm = 1 / Σ(x_i/d_pi), x_i = mass fraction
Pressure drop of a fluidised bed: ΔP = L_mf (1−ε_mf)(ρ_s − ρ) g = W(1 − ρ/ρ_s) g/A
ρ_s = particle density (kg/m³), g = 9.81 m/s², W = mass of solids (kg), A = bed cross-section (m²)
u_mf, laminar form (Re_mf < 20): u_mf = [d_p² (ρ_s − ρ) g / (150 μ)] × [φ_s² ε_mf³ / (1 − ε_mf)]
u_mf, turbulent form (Re_mf > 1000): u_mf² = [d_p (ρ_s − ρ) g / (1.75 ρ)] × φ_s ε_mf³
Wen-Yu correlation: Re_mf = √(33.7² + 0.0408 Ar) − 33.7, where Ar = d_p³ ρ (ρ_s − ρ) g / μ², u_mf = Re_mf μ /(d_p ρ)
Terminal velocity: Stokes (Re < 0.3) u_t = d_p²(ρ_s − ρ)g/(18μ); intermediate/Allen (2 < Re < 500) u_t = 0.153 g0.71 d_p1.14 (ρ_s − ρ)0.71 / (ρ0.29 μ0.43); Newton (500 < Re < 2×10⁵) u_t = 1.74 √(g d_p(ρ_s − ρ)/ρ)
Bed expansion: L(1 − ε) = L_mf(1 − ε_mf); Richardson-Zaki: u/u_t = εn
Blower power: P = Q ΔP / η, Q = volumetric flow (m³/s), η = fan efficiency (−)

Example 1 - Ergun pressure drop in a catalyst bed. Spheres d_p = 2 mm, φ_s = 1, ε = 0.40, bed L = 1.2 m, column D = 0.6 m; air at 20 °C, ρ = 1.2 kg/m³, μ = 1.8×10⁻⁵ Pa·s, superficial u = 0.5 m/s.

  1. Re_p = d_p u ρ/μ = (0.002 × 0.5 × 1.2)/1.8×10⁻⁵ = 66.7 → transition, both terms needed.
  2. Viscous term = 150 × 1.8×10⁻⁵ × 0.5 × (0.6)² / (0.002² × 0.40³) = 4.86×10⁻⁴ / 2.56×10⁻⁷ = 1,898 Pa/m.
  3. Inertial term = 1.75 × 1.2 × 0.5² × 0.6 / (0.002 × 0.064) = 0.315 / 1.28×10⁻⁴ = 2,461 Pa/m.
  4. ΔP/L = 4,359 Pa/m → ΔP = 4,359 × 1.2 = 5.23 kPa (≈ 533 mm water gauge).
  5. A = (π/4)(0.6)² = 0.2827 m², Q = 0.1414 m³/s. Blower power at η = 0.60: P = 0.1414 × 5,231/0.60 = 1.23 kW. At 8,000 h/yr and ₹8/kWh the fan alone costs about ₹0.79 lakh a year — which is why the inertial term, and hence velocity, is cut first.

Example 2 - u_mf of sand in air, two methods. Sand d_p = 300 µm, ρ_s = 2,600 kg/m³, ε_mf = 0.40, φ_s = 1; air as above.

  1. Laminar Ergun form: d_p²(ρ_s − ρ)g/(150μ) = (9×10⁻⁸ × 2,598.8 × 9.81)/(2.7×10⁻³) = 2.294×10⁻³/2.7×10⁻³ = 0.850 m/s.
  2. × ε_mf³/(1 − ε_mf) = 0.064/0.60 = 0.1067 → u_mf = 0.091 m/s. Check Re_mf = 0.0003 × 0.091 × 1.2/1.8×10⁻⁵ = 1.8 (< 20, so the laminar form is legitimate).
  3. Wen-Yu: Ar = (2.7×10⁻¹¹ × 1.2 × 2,598.8 × 9.81)/(3.24×10⁻¹⁰) = 8.26×10⁻⁷/3.24×10⁻¹⁰ = 2,550.
  4. Re_mf = √(1,135.7 + 0.0408×2,550) − 33.7 = √1,239.7 − 33.7 = 35.21 − 33.70 = 1.51 → u_mf = 1.51 × 1.8×10⁻⁵/(3×10⁻⁴ × 1.2) = 0.076 m/s.
  5. The two differ by ~20% purely through the assumed ε_mf — report the range 7.6-9.1 cm/s and design the blower on the higher figure.

Example 3 - fluidised bed ΔP and expansion. Same sand, L_mf = 0.5 m, A = 0.2827 m².

  1. ΔP = L_mf(1 − ε_mf)(ρ_s − ρ)g = 0.5 × 0.60 × 2,598.8 × 9.81 = 7.65 kPa — constant for all u > u_mf.
  2. Solids mass = L_mf(1 − ε_mf)ρ_s A = 0.5 × 0.6 × 2,600 × 0.2827 = 220.5 kg; check ΔP = W(1 − ρ/ρ_s)g/A = 220.5 × 0.99954 × 9.81/0.2827 = 7.65 kPa ✓.
  3. At ε = 0.55: L = L_mf(1 − ε_mf)/(1 − ε) = 0.5 × 0.60/0.45 = 0.667 m, i.e. 33% expansion — set freeboard above this.

Example 4 - upper limit of the operating window. Terminal velocity of the same sand in air.

  1. Try Stokes: u_t = 2.294×10⁻³/(18 × 1.8×10⁻⁵) = 7.08 m/s → Re = 142, far outside Stokes' range, so reject.
  2. Allen (intermediate) law: u_t = 0.153 × 9.810.71 × (3×10⁻⁴)1.14 × 2,598.80.71 / (1.20.29 × (1.8×10⁻⁵)0.43) = 0.01982/9.61×10⁻³ = 2.06 m/s. Check Re = 41 — inside 2-500 ✓.
  3. u_t/u_mf = 2.06/0.076 ≈ 27, a wide window; an operating velocity of 0.3-0.6 m/s (4-8 × u_mf) gives vigorous bubbling with little carry-over of the main cut, though fines below ~80 µm will elutriate and need a cyclone.

Example 5 - mixed sizes. Mass fractions 0.30 at 1.0 mm, 0.50 at 0.50 mm, 0.20 at 0.20 mm. Σ(x_i/d_pi) = 0.30/1.0 + 0.50/0.50 + 0.20/0.20 = 2.30 mm⁻¹ → d_sm = 0.435 mm. Using the mass-weighted arithmetic mean (0.59 mm) instead would under-predict Ergun's viscous ΔP by about 45%.

Velocity rangeBed stateΔP behaviour
u < u_mfFixed bedRises with u (Ergun)
u = u_mfIncipient fluidisationEquals buoyant bed weight
u_mf < u < u_tBubbling / turbulent bedConstant
u > u_tPneumatic transportFalls as solids leave, then gas-only ΔP

Where it is used

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Related topics

Filtration, sedimentation and fluidisation · Size reduction and screening · Conveying and storage of solids · Fluid flow and pumps in process plants · Dimensional analysis and similitude · Heterogeneous catalytic reactor design

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