Fixed-bed and fluidised-bed catalytic reactors
Design and comparison of fixed-bed and fluidised-bed catalytic reactors: catalyst-mass design equation, Ergun pressure drop, heat management, fluidisation pressure drop and bed expansion, with worked examples.
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Why it matters
Fixed beds and fluidised beds carry most of the world's solid-catalysed production: fixed beds for ammonia, methanol, hydrotreating and SO₂ oxidation; fluidised beds for catalytic cracking, acrylonitrile and phthalic anhydride. The choice between them is driven by heat removal, catalyst life and contacting efficiency, and the design rests on the catalytic design equation, the pressure drop and the flow model.
Key ideas
Fixed (packed) bed. Catalyst pellets (typically 2–10 mm) are stacked in a vessel or in many tubes; gas or liquid flows through. The flow is close to plug flow, so for a given catalyst mass it gives high conversion and good selectivity for intermediates.
- Design equation: W/FA0 = ∫dXA/(−r'A), the plug-flow equation written per mass of catalyst.
- Pressure drop: given by the Ergun equation; it rises steeply as particles get smaller, which is why pore diffusion (favouring small pellets) and pressure drop (favouring large ones) must be traded off.
- Heat management: the bed conducts heat poorly. Options: adiabatic beds in series with inter-stage exchangers or cold-shot quench (SO₂, ammonia); multitubular reactors with coolant on the shell side (ethylene oxide, phthalic anhydride); or diluting the feed. Exothermic multitubular reactors show a temperature peak (hot spot) near the inlet and can be parametrically sensitive: a small rise in coolant or feed temperature causes a large hot-spot rise (runaway).
- Deactivation: the whole bed must be regenerated in place or replaced; swing reactors allow continuous production.
Fluidised bed. Fine particles (typically 20–200 µm) are supported by upward gas flow. Above the minimum fluidisation velocity umf, the pressure drop equals the bed's buoyant weight per unit area and stays nearly constant. At higher velocities gas passes as bubbles (bubbling bed), then turbulent and fast-fluidised (riser) regimes; above the particle terminal velocity solids are carried out.
- Advantages: nearly isothermal bed (vigorous solids mixing), very high bed-to-wall heat-transfer coefficients, easy addition and removal of catalyst — essential when the catalyst deactivates in seconds (FCC riser plus regenerator).
- Disadvantages: gas bypasses in bubbles, so conversion is lower than in a fixed bed and can be worse than mixed flow; solids are back-mixed, which hurts selectivity in series reactions; attrition and entrainment of fines need cyclones; erosion of internals; harder scale-up.
- Models: the Kunii–Levenspiel bubbling-bed model divides gas into bubble, cloud and emulsion phases with interchange coefficients between them; take bubble-size and interchange correlations from your data book.
Bed expansion. The mass of solids is fixed, so L·(1 − ε) is constant: L2 = L1·(1 − ε1)/(1 − ε2).
Choosing. Fixed bed: slowly deactivating catalyst, moderate heat release, need for high conversion or plug-flow selectivity. Fluidised bed: strongly exothermic reactions needing tight temperature control, rapid deactivation with continuous regeneration, fine catalysts.
Formulas
W = FA0·∫(0→XA) dXA/(−r'A) (fixed bed, plug flow)
- W: catalyst mass (kg); −r'A: rate per catalyst mass (mol/kg·s).
W = (v0/k')·ln[1/(1 − XA)] (first order, −r'A = k'·CA, constant density)
- k': rate constant (m³/kg·s); v0: volumetric feed (m³/s).
ΔP/L = 150·μ·u·(1 − ε)²/(ε³·dp²) + 1.75·ρ·u²·(1 − ε)/(ε³·dp) (Ergun)
- ΔP: pressure drop (Pa); L: bed height (m); μ: fluid viscosity (Pa·s); u: superficial velocity (m/s); ε: bed voidage; dp: particle diameter (m); ρ: fluid density (kg/m³).
ΔP = (1 − εmf)·(ρs − ρ)·g·Lmf (fluidised bed, at and above umf)
- ρs: particle density (kg/m³); g = 9.81 m/s²; Lmf: bed height at minimum fluidisation (m).
L·(1 − ε) = constant (bed expansion)
ρB = ρs·(1 − ε) (bulk density of a bed)
Worked examples
Example 1 (standard, Ergun pressure drop). Air (ρ = 1.2 kg/m³, μ = 1.8 × 10⁻⁵ Pa·s) flows at u = 0.1 m/s through a 1.5 m bed of 5 mm spheres with ε = 0.4. Find ΔP.
- Viscous term: 150 × 1.8 × 10⁻⁵ × 0.1 × 0.36/(0.064 × 2.5 × 10⁻⁵) = 9.72 × 10⁻⁵/1.6 × 10⁻⁶ = 60.75 Pa/m.
- Inertial term: 1.75 × 1.2 × 0.01 × 0.6/(0.064 × 0.005) = 0.0126/3.2 × 10⁻⁴ = 39.38 Pa/m.
- ΔP/L = 100.1 Pa/m → ΔP = 100.1 × 1.5 = 150 Pa.
ΔP ≈ 150 Pa.
Example 2 (GATE level, fluidised bed). A bed of catalyst (ρs = 2500 kg/m³) is 2.0 m tall at minimum fluidisation with εmf = 0.45; the gas has ρ = 1.2 kg/m³. Find the pressure drop when fluidised, and the bed height when the voidage rises to 0.60.
- ΔP = (1 − 0.45) × (2500 − 1.2) × 9.81 × 2.0 = 0.55 × 2498.8 × 19.62 = 26 965 Pa.
- Expansion: L = 2.0 × (1 − 0.45)/(1 − 0.60) = 2.0 × 0.55/0.40 = 2.75 m.
ΔP ≈ 27.0 kPa (unchanged as gas velocity rises); expanded height 2.75 m.
Example 3 (catalyst mass). A first-order gas reaction with k' = 0.002 m³/kg·s is to reach XA = 0.9 in a fixed bed at v0 = 0.5 m³/s (constant density): W = (0.5/0.002) × ln 10 = 250 × 2.303 = 576 kg.
Common mistakes
- Using a pipe friction-factor formula instead of Ergun for a packed bed.
- Using the interstitial velocity in Ergun — it uses the superficial velocity.
- Expecting fluidised-bed pressure drop to rise with gas velocity; it stays at the bed weight per unit area.
- Treating a bubbling fluidised bed as plug flow for the gas.
- Mixing catalyst mass and bed volume without the bulk density.
For GATE CH
Expect NAT questions on Ergun pressure drop, fluidised-bed pressure drop and bed expansion, catalyst mass from the design equation for first-order kinetics, and conceptual comparisons of fixed and fluidised beds (heat transfer, contacting, deactivation handling). Practise Ergun arithmetic carefully — the voidage cubed term is easy to get wrong.
Quick check
- Why is a fluidised bed nearly isothermal?
- Which term of Ergun dominates at low Reynolds number?
- A bed expands from ε = 0.5 to ε = 0.6; by what factor does its height change?
- Name one reason fluidised beds give lower conversion than fixed beds.
Answers: 1. vigorous solids circulation and high heat-transfer coefficients; 2. the viscous (150) term; 3. 0.5/0.4 = 1.25; 4. gas bypasses the catalyst in bubbles (and solids/gas back-mixing).
Interview questions
All Chemical Reaction Engineering interview questionsTry answering each one aloud before you open it.
1.What is a fixed-bed catalytic reactor?Concept
A fixed-bed catalytic reactor is a type of reactor where the catalyst is held in place and does not move with the fluid. The reactants flow over the catalyst bed, which is typically packed in a column. This setup is commonly used for gas-phase reactions and allows for efficient contact between the reactants and the catalyst.
2.Explain the working principle of a fluidised-bed catalytic reactor.Concept
In a fluidised-bed catalytic reactor, the catalyst particles are suspended in the fluid, typically by the upward flow of gas or liquid. This creates a fluid-like state for the solid particles, allowing for excellent mixing and heat transfer. The fluidised state enhances the contact between the reactants and the catalyst, improving reaction rates and efficiency.
3.What are the main differences between fixed-bed and fluidised-bed catalytic reactors?Concept
The main differences are in the catalyst arrangement and flow dynamics. In fixed-bed reactors, the catalyst is stationary, leading to potential issues with heat and mass transfer. In fluidised-bed reactors, the catalyst is suspended, allowing for better mixing and heat transfer. Fluidised beds can handle larger volumes and are more suitable for reactions with significant heat effects.
4.Why is a fluidised-bed reactor preferred for exothermic reactions?Application
Fluidised-bed reactors are preferred for exothermic reactions because they offer excellent heat transfer properties. The fluid-like movement of the catalyst particles ensures uniform temperature distribution, preventing hot spots that can lead to catalyst deactivation or unwanted side reactions. This makes them ideal for managing the heat generated in exothermic processes.
5.What happens if the gas velocity is too low in a fluidised-bed reactor?Application
If the gas velocity is too low in a fluidised-bed reactor, the catalyst particles will not be adequately fluidised and will settle at the bottom of the reactor. This can lead to poor mixing, reduced contact between the reactants and the catalyst, and inefficient reaction rates. It may also cause channeling, where the gas flows through paths of least resistance, bypassing much of the catalyst.
6.How does catalyst deactivation affect the performance of a fixed-bed reactor?Application
Catalyst deactivation in a fixed-bed reactor can significantly reduce the reactor's performance. As the catalyst loses activity, the reaction rate decreases, leading to lower conversion rates and potentially incomplete reactions. This may require more frequent catalyst replacement or regeneration, increasing operational costs and downtime.
7.Estimate the pressure drop across a fixed bed 1.5 m high of 5 mm spheres with voidage 0.4, for air (ρ = 1.2 kg/m³, μ = 1.8 × 10⁻⁵ Pa·s) at a superficial velocity of 0.1 m/s.Numerical
Use the Ergun equation: ΔP/L = 150·μ·u·(1 − ε)²/(ε³·dp²) + 1.75·ρ·u²·(1 − ε)/(ε³·dp). The viscous term is 150 × 1.8 × 10⁻⁵ × 0.1 × 0.36/(0.064 × 2.5 × 10⁻⁵) = 60.8 Pa/m and the inertial term is 1.75 × 1.2 × 0.01 × 0.6/(0.064 × 0.005) = 39.4 Pa/m. The total is about 100 Pa/m, so ΔP ≈ 150 Pa for 1.5 m. An interviewer will check that you used the superficial velocity and the cube of voidage.
8.What are the advantages of using a fixed-bed reactor for liquid-phase catalytic reactions?Application
A fixed bed is simple, has no moving catalyst and so no attrition, and gives nearly plug flow, which means high conversion per unit catalyst and good selectivity for intermediates. Liquid-phase heat capacities are high, so temperature rise per unit conversion is modest and adiabatic or lightly cooled beds are often adequate. Catalyst separation from product is automatic, unlike slurry reactors. Limitations are pore diffusion in large pellets, pressure drop with small pellets, and the need to shut down or swing beds for catalyst replacement.
9.Explain how catalyst attrition can be a problem in fluidised-bed reactors.Application
Catalyst attrition in fluidised-bed reactors occurs due to the constant movement and collision of catalyst particles. This can lead to the breakdown of particles into fines, which may be carried out of the reactor with the fluid. Attrition reduces the effective catalyst volume and can lead to operational issues such as increased pressure drop and loss of catalyst activity.
10.A fluidised bed is 2 m tall at minimum fluidisation with voidage 0.45. At a higher gas velocity the voidage becomes 0.60. What is the expanded bed height?Numerical
The mass of solids is fixed, so L·(1 − ε) stays constant. L = 2 m × (1 − 0.45)/(1 − 0.60) = 2 × 0.55/0.40 = 2.75 m. Bed height is not proportional to gas velocity; the velocity sets the voidage through bubbling or expansion correlations, and the solids balance then gives the height.
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