Bed Pressure Drop Calculation in Fluidized Beds

Reference ID: MET-CE8B | Process Engineering Reference Sheets Calculation Guide | Printed from myengineeringtools.com

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Introduction & Context

The pressure drop across a fluidized bed is a critical design parameter in chemical, food, and pharmaceutical processing. It dictates the fan or blower size, the energy demand of the gas-handling system, and the mechanical stresses on support grids and filters. A reliable estimate of the total pressure loss—combining the resistance of the particle bed itself plus ancillary items such as distributors, filters, and ducting—ensures that the specified gas velocity can be maintained without excessive power consumption or operational instability.

The fastest conservative estimate is to treat the bed as a packed bed at incipient fluidization:

Measure or assume the minimum fluidization voidage ε_mf (typically 0.4–0.6 for Geldart B powders).
Use the Ergun equation with ε = ε_mf and particle diameter d_p to obtain ΔP_bed ≈ (ρ_p – ρ_g)(1 – ε_mf)gH_mf, where H_mf is the static bed height.
Add 10–15 % to cover distributor and cyclone return losses; this gives a first-pass value for blower or compressor sizing.

A drop below ΔP_mf usually signals particle attrition or elutriation:

Fines are carried out, increasing average d_p and decreasing bed weight.
Static bed height H_mf shrinks; the pressure gradient falls proportionally.
Check downstream filters and cyclone catch pots for solids accumulation; replenish fines or adjust cyclone diplegs to restore inventory.

Temperature and pressure change gas density ρ_g and viscosity μ, but the dominant term is solids weight:

ΔP_bed = (ρ_p – ρ_g)(1 – ε)gH; ρ_g is small compared with ρ_p, so ΔP_bed is largely insensitive to gas conditions.
High-pressure units see a slight increase because ρ_g rises, reducing buoyancy.
High-temperature units see a slight decrease because μ rises, increasing ε at the same velocity, but the effect is < 5 % for typical 400–600 °C ranges.

Use differential pressure taps that span the dense phase only:

Bottom tap 50–100 mm above the distributor to avoid jet and plenum effects.
Top tap 100–200 mm below the splash-zone transition; too high includes lean-phase solids and gives falsely low ΔP.
Install purged, inverted-L tubes to prevent clogging; compare with load-cell data during commissioning to confirm calibration.

Worked Example – Estimating Fan Pressure Rise for a 60 °C Air-Fluidized Sand Bed

A small pilot unit is being designed to study the drying kinetics of pharmaceutical granules. The bed is a 0.6 m ID cylinder charged with 250 mm of 850 µm Geldart B particles (ρ_p = 1580 kg m^–3). Air at 60 °C and 1 atm will fluidize the solids at a superficial velocity of 0.35 m s^–1. Determine the pressure rise the fan must deliver, including a 15 % safety factor.

Knowns

Bed diameter, D_bed = 0.6 m
Static bed height, L_bed = 0.25 m
Particle diameter, d_p = 850 µm = 0.00085 m
Particle density, ρ_p = 1580 kg m^–3
Air temperature, T = 60 °C (333 K)
Air density, ρ_air = 1.06 kg m^–3
Air viscosity, μ = 2 × 10^–5 Pa s
Superficial velocity, v₀ = 0.35 m s^–1
Bed voidage at incipient fluidization, ε = 0.43
Allowances for distributor, filter, and duct, ΔP_anc = 600 Pa
Safety factor, SF = 1.15
Fan efficiency, η_fan = 65 %
Belt drive efficiency, η_belt = 95 %

Step-by-step calculation

Calculate the bed cross-sectional area: \[ A_{\text{bed}} = \frac{\pi}{4} D_{\text{bed}}^{2} = \frac{\pi}{4}(0.6)^{2} = 0.283\ \text{m}^{2} \]
Compute the two Ergun terms for pressure gradient: \[ \text{term1} = 150\ \frac{\mu v_{0}}{d_{\text{p}}^{2}}\ \frac{(1 - \varepsilon)^{2}}{\varepsilon^{3}} \] \[ = 150\ \frac{(2 \times 10^{-5})(0.35)}{(0.00085)^{2}}\ \frac{(1 - 0.43)^{2}}{0.43^{3}} = 5939\ \text{Pa m}^{-1} \] \[ \text{term2} = 1.75\ \frac{\rho_{\text{air}} v_{0}^{2}}{d_{\text{p}}}\ \frac{(1 - \varepsilon)}{\varepsilon^{3}} \] \[ = 1.75\ \frac{(1.06)(0.35)^{2}}{0.00085}\ \frac{(1 - 0.43)}{0.43^{3}} = 1917\ \text{Pa m}^{-1} \]
Total pressure gradient and bed pressure drop: \[ \frac{\Delta P_{\text{bed}}}{L} = \text{term1} + \text{term2} = 5939 + 1917 = 7855\ \text{Pa m}^{-1} \] \[ \Delta P_{\text{bed}} = \left(\frac{\Delta P}{L}\right) L_{\text{bed}} = 7855 \times 0.25 = 1964\ \text{Pa} \approx 200\ \text{mm WG} \]
Add ancillary losses (distributor, filter, duct): \[ \Delta P_{\text{total}} = \Delta P_{\text{bed}} + \Delta P_{\text{anc}} = 1964 + 600 = 2564\ \text{Pa} \]
Apply safety factor to obtain fan pressure rise: \[ \Delta P_{\text{fan}} = \Delta P_{\text{total}} \times \text{SF} = 2564 \times 1.15 = 2948\ \text{Pa} \approx 301\ \text{mm WG} \]
Estimate fan power: Volumetric flow rate: \[ Q = v_{0} A_{\text{bed}} = 0.35 \times 0.283 = 0.099\ \text{m}^{3}\text{s}^{-1}\ (356\ \text{m}^{3}\text{h}^{-1}) \] Ideal power: \[ P_{\text{ideal}} = Q \Delta P_{\text{fan}} = 0.099 \times 2948 = 292\ \text{W} \] Shaft power: \[ P_{\text{shaft}} = \frac{P_{\text{ideal}}}{\eta_{\text{fan}} \eta_{\text{belt}}} = \frac{292}{0.65 \times 0.95} = 473\ \text{W} \approx 0.47\ \text{kW} \]

Final answer

The fan must supply a pressure rise of 2948 Pa (≈ 301 mm WG) and requires 0.47 kW of shaft power under the stated conditions.

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