4.3 Multiphase Flow and Fluidization Principles

Key Takeaways

  • Two-phase gas-liquid flow regimes (bubble, slug, plug, churn, annular, mist) depend on phase superficial velocities and fluid physical properties; intermittent slug flow presents catastrophic water-hammer risks to piping and supports.
  • The Lockhart-Martinelli correlation evaluates two-phase frictional pressure drop via the parameter chi = sqrt[(dP/dz)_L / (dP/dz)_G] and the Chisholm two-phase multiplier phi_L^2 = 1 + C/chi + 1/chi^2, where C = 20 for turbulent-turbulent flow.
  • Frictional pressure drop through stationary packed beds is quantified across all flow regimes by the Ergun equation, which superimposes laminar viscous losses (Blake-Kozeny) and turbulent inertial losses (Burke-Plummer).
  • Incipient fluidization occurs when upward aerodynamic drag plus buoyancy balances the net weight of the bed particles, establishing the minimum fluidization velocity u_mf from the Ergun equation; once fluidized, the total bed pressure drop remains strictly constant regardless of further velocity increases.
  • The safe fluidization operating envelope is bounded between the minimum fluidization velocity u_mf and the single-particle terminal settling velocity u_t; operating at superficial velocities u_0 > u_t causes severe particle carryover (elutriation).
Last updated: September 2026

4.3 Multiphase Flow and Fluidization Principles

Multiphase flow phenomena—involving simultaneous movement of gas and liquid, liquid and solid, or gas and solid—are ubiquitous across chemical processing facilities. Examples include boiling reboilers, distillation column trays and packings, petroleum pipelines, catalytic fluidized bed reactors (FCCUs), slurry bubble columns, and pneumatic conveying lines. Because phases exhibit differing densities, viscosities, and velocities, multiphase hydrodynamics cannot be described by single-phase equations. This section covers two-phase gas-liquid pipe flow, the Ergun equation for packed beds, and fluidization principles essential for the NCEES PE Chemical exam.


1. Two-Phase Gas-Liquid Flow Regimes and Hydrodynamic Transitions

When a gas and a liquid flow concurrently through a conduit, the interfacial shear between phases generates distinct morphological configurations known as flow regimes (or flow patterns). Flow regimes govern phase holdup, heat transfer coefficients, and frictional pressure drops.

Horizontal Flow Regimes

In horizontal piping, gravity acts perpendicular to the flow direction, causing denser liquid to accumulate along the bottom of the pipe:

  1. Stratified Flow (Smooth): At low gas and liquid velocities, liquid flows along the pipe bottom while gas glides smoothly over the top across an undisturbed horizontal interface.
  2. Stratified-Wavy Flow: As gas velocity increases, interfacial shear creates capillary and gravity waves on the liquid surface.
  3. Plug Flow (Elongated Bubble): Liquid is continuous along the bottom and intermittent elongated gas plugs travel along the upper pipe wall.
  4. Slug Flow: Waves grow until their crests bridge the entire pipe diameter, forming high-velocity liquid slugs separated by large gas bubbles (Taylor bubbles). Slug flow is hazardous in industrial plants: the periodic impact of dense liquid slugs produces violent piping vibration, structural fatigue, and severe pressure shocks (water hammer) at pipe elbows and headers.
  5. Annular Flow: At high gas velocities, gas forms a high-speed core carrying atomized entrained droplets, while liquid forms a continuous annular film around the entire pipe circumference.
  6. Dispersed / Mist Flow: At ultra-high gas velocities, the liquid film is stripped from the pipe walls, dispersing into fine droplets suspended in a continuous gas matrix.

Vertical Upward Flow Regimes

In vertical pipes, gravity is aligned with the axis, producing axisymmetric flow patterns:

  • Bubbly Flow: Gas is dispersed as discrete bubbles in a continuous liquid continuum.
  • Slug / Plug Flow: Small bubbles coalesce into bullet-shaped Taylor bubbles that fill nearly the entire pipe cross-section, separated by liquid slugs.
  • Churn Flow: Highly chaotic, churning oscillatory flow occurring when gas momentum shatters liquid slugs before stable annular flow establishes.
  • Annular Flow: Liquid forms a thin film ascending the pipe wall, surrounding a high-speed gas core containing sheared liquid droplets.
  • Mist Flow: Complete atomization of the liquid phase into the gas core.

2. Two-Phase Frictional Pressure Drop: The Lockhart-Martinelli Method

The classical engineering framework for evaluating two-phase frictional pressure gradient $(-dP/dz)_{TP}$ is the Lockhart-Martinelli correlation.

The Lockhart-Martinelli Parameter ($\chi$)

Define the parameter $\chi$ as the square root of the ratio of the frictional pressure gradient if the liquid phase flowed alone in the conduit at its superficial rate, to the frictional pressure gradient if the gas phase flowed alone:

χ=(dP/dz)L(dP/dz)G\chi = \sqrt{\frac{(-dP/dz)_L}{(-dP/dz)_G}}

where $(-dP/dz)L$ and $(-dP/dz)G$ are computed using standard single-phase Darcy-Weisbach formulas based on the superficial velocities $u{SL} = Q_L / A_c$ and $u{SG} = Q_G / A_c$.

The Turbulent-Turbulent Regime ($\chi_{tt}$)

When both liquid and gas phases operate in the turbulent regime ($Re_{SL} > 4,000$ and $Re_{SG} > 4,000$), applying the Blasius friction factor ($f \propto Re^{-0.20}$) gives the explicit expression for $\chi_{tt}$ in terms of gas mass quality $x = \dot{m}_G / (\dot{m}_L + \dot{m}_G)$:

χtt=(1xx)0.9(ρGρL)0.5(μLμG)0.1\chi_{tt} = \left(\frac{1 - x}{x}\right)^{0.9} \left(\frac{\rho_G}{\rho_L}\right)^{0.5} \left(\frac{\mu_L}{\mu_G}\right)^{0.1}

Two-Phase Multipliers ($\phi_L^2$ and $\phi_G^2$)

The total two-phase frictional pressure drop is determined by multiplying the single-phase pressure drop by the two-phase multiplier (correlated by Chisholm):

(dPdz)TP=ϕL2(dPdz)L=ϕG2(dPdz)G\left(\frac{dP}{dz}\right)_{TP} = \phi_L^2 \left(\frac{dP}{dz}\right)_L = \phi_G^2 \left(\frac{dP}{dz}\right)_G

ϕL2=1+Cχ+1χ2\phi_L^2 = 1 + \frac{C}{\chi} + \frac{1}{\chi^2}

ϕG2=1+Cχ+χ2\phi_G^2 = 1 + C \chi + \chi^2

Chisholm Parameter ($C$) Selection Matrix

The dimensionless constant $C$ depends on the flow regime of each independent phase:

Liquid Phase RegimeGas Phase RegimeRegime DesignationChisholm Parameter ($C$)
Turbulent ($Re_L > 4,000$)Turbulent ($Re_G > 4,000$)$tt$$C = 20$
Viscous (Laminar) ($Re_L < 2,100$)Turbulent ($Re_G > 4,000$)$vt$$C = 12$
Turbulent ($Re_L > 4,000$)Viscous (Laminar) ($Re_G < 2,100$)$tv$$C = 10$
Viscous (Laminar) ($Re_L < 2,100$)Viscous (Laminar) ($Re_G < 2,100$)$vv$$C = 5$

3. Flow Through Packed Beds and the Ergun Equation

In chemical reactors (fixed-bed catalytic converters), absorbers, and filter media, fluid percolates through a stationary bed of granular solid particles. Sabri Ergun (1952) established that the total frictional pressure gradient through a packed bed is the sum of viscous laminar energy losses (viscous drag along particle surfaces) and turbulent kinetic energy losses (inertial form drag in tortuous interstitial expansions/contractions):

ΔPL=150(1ε)2ε3μu0dp2Viscous Term (Blake-Kozeny)+1.751εε3ρu02dpInertial Term (Burke-Plummer)\frac{\Delta P}{L} = \underbrace{150 \frac{(1 - \varepsilon)^2}{\varepsilon^3} \frac{\mu u_0}{d_p^2}}_{\text{Viscous Term (Blake-Kozeny)}} + \underbrace{1.75 \frac{1 - \varepsilon}{\varepsilon^3} \frac{\rho u_0^2}{d_p}}_{\text{Inertial Term (Burke-Plummer)}}

where:

  • $\Delta P / L$ is the pressure drop per unit length of packed bed ($\text{Pa/m}$),
  • $\varepsilon = V_{void} / V_{bed}$ is the bed void fraction (porosity), typically $0.35 - 0.45$ for random spherical packings,
  • $u_0 = Q / A_c$ is the superficial fluid velocity (empty tube velocity, $\text{m/s}$),
  • $d_p$ is the effective particle diameter ($\text{m}$), defined for non-spherical particles as $d_p = \phi_s d_v$ (where $\phi_s$ is sphericity and $d_v$ is volume-equivalent sphere diameter),
  • $\mu$ is fluid dynamic viscosity ($\text{Pa}\cdot\text{s}$),
  • $\rho$ is fluid mass density ($\text{kg/m}^3$).

Asymptotic Regimes of the Ergun Equation

Define the particle Reynolds number based on superficial velocity:

Rep=ρu0dpμ(1ε)Re_p = \frac{\rho u_0 d_p}{\mu (1 - \varepsilon)}

  1. Laminar Flow ($Re_p < 10$): The viscous term dominates ($> 90%$ of total loss). The equation reduces to the Blake-Kozeny equation (analogous to Darcy's law for porous media): ΔPL=150(1ε)2ε3μu0dp2    ΔPu0\frac{\Delta P}{L} = 150 \frac{(1 - \varepsilon)^2}{\varepsilon^3} \frac{\mu u_0}{d_p^2} \implies \Delta P \propto u_0
  2. Highly Turbulent Flow ($Re_p > 1,000$): Inertial form drag dominates. The equation reduces to the Burke-Plummer equation: ΔPL=1.751εε3ρu02dp    ΔPu02\frac{\Delta P}{L} = 1.75 \frac{1 - \varepsilon}{\varepsilon^3} \frac{\rho u_0^2}{d_p} \implies \Delta P \propto u_0^2

4. Fluidization Fundamentals and Minimum Fluidization Velocity ($u_{mf}$)

Consider a vertical column packed with solid particles supported on a porous gas distributor plate. Fluid is pumped upward through the bed at steadily increasing superficial velocity $u_0$.

   Bed Pressure Drop (Delta P)
        ^
        |                     Fixed Bed                   Fluidized Bed (Bubbling/Slugging)
        |             (Delta P increases with u0)       (Delta P = CONSTANT = Buoyant Weight)
Delta P |                      /------------------------------------------------------------
   mf   |                    /*
        |                   /  *
        |                  /    * Peak (interlocking friction)
        |                 /
        |                /
        |               /
        +--------------+-------------------------------------------------------------> Superficial
                       u_mf                                                  u_t     Velocity (u0)
                 Incipient Fluidization                                  Elutriation Limit

The Incipient Fluidization Condition

As upward superficial velocity increases, the upward aerodynamic drag force and buoyant force exerted by the fluid on the particles increase. At a critical velocity known as the minimum fluidization velocity ($u_{mf}$), the net upward fluid forces exactly balance the downward gravitational weight of the solid particles:

Upward Pressure Force=Net Buoyant Weight of Bed\text{Upward Pressure Force} = \text{Net Buoyant Weight of Bed} ΔPbedAc=WsolidsFbuoyancy=(1εmf)AcLmf(ρpρf)g\Delta P_{bed} A_c = W_{solids} - F_{buoyancy} = (1 - \varepsilon_{mf}) A_c L_{mf} (\rho_p - \rho_f) g

Dividing by bed volume $A_c L_{mf}$ yields the fundamental bed pressure drop equation at fluidization:

ΔPbedLmf=(1εmf)(ρpρf)g\frac{\Delta P_{bed}}{L_{mf}} = (1 - \varepsilon_{mf}) (\rho_p - \rho_f) g

where:

  • $L_{mf}$ is bed height at minimum fluidization,
  • $\varepsilon_{mf}$ is void fraction at incipient fluidization (typically $0.40 - 0.45$),
  • $\rho_p$ is solid particle density,
  • $\rho_f$ is fluidizing fluid density.

Deriving $u_{mf}$ from the Ergun Equation

Equating the hydrostatic buoyant bed weight gradient to the Ergun pressure drop at voidage $\varepsilon_{mf}$:

(1εmf)(ρpρf)g=150(1εmf)2εmf3μumfdp2+1.751εmfεmf3ρfumf2dp(1 - \varepsilon_{mf}) (\rho_p - \rho_f) g = 150 \frac{(1 - \varepsilon_{mf})^2}{\varepsilon_{mf}^3} \frac{\mu u_{mf}}{d_p^2} + 1.75 \frac{1 - \varepsilon_{mf}}{\varepsilon_{mf}^3} \frac{\rho_f u_{mf}^2}{d_p}

Dividing through by $(1 - \varepsilon_{mf})$ and expressing in dimensionless form yields a quadratic in terms of the particle Reynolds number at minimum fluidization ($Re_{p,mf} = \frac{\rho_f u_{mf} d_p}{\mu}$):

1.75εmf3Rep,mf2+150(1εmf)εmf3Rep,mf=Ar\frac{1.75}{\varepsilon_{mf}^3} Re_{p,mf}^2 + \frac{150 (1 - \varepsilon_{mf})}{\varepsilon_{mf}^3} Re_{p,mf} = Ar

where $Ar$ is the dimensionless Archimedes number, characterizing the ratio of gravitational/buoyancy forces to viscous forces:

Ar=dp3ρf(ρpρf)gμ2Ar = \frac{d_p^3 \rho_f (\rho_p - \rho_f) g}{\mu^2}

Analytical Solution for Small Particles (Laminar Flow, $Re_{p,mf} < 20$)

For fine catalyst particles and powders where $Ar < 1,000$, flow at incipient fluidization is strictly laminar ($Re_{p,mf} < 20$). The inertial quadratic term is negligible, yielding the direct formula for $u_{mf}$:

umf=dp2(ρpρf)g150μ(εmf31εmf)u_{mf} = \frac{d_p^2 (\rho_p - \rho_f) g}{150 \mu} \left(\frac{\varepsilon_{mf}^3}{1 - \varepsilon_{mf}}\right)


5. Bed Pressure Drop Dynamics and Expansion Above $u_{mf}$

A critical conceptual principle tested on the PE exam is the behavior of bed pressure drop $\Delta P_{bed}$ as superficial velocity increases beyond $u_{mf}$:

  1. Fixed Bed Regime ($u_0 < u_{mf}$): The particles remain stationary in physical contact. Bed voidage $\varepsilon$ and bed height $L$ remain constant. By the Ergun equation, $\Delta P_{bed}$ increases monotonically (linearly at low $u_0$, quadratically at higher $u_0$).
  2. Incipient Fluidization ($u_0 = u_{mf}$): Particles unlock and begin to float. A slight transient peak in $\Delta P$ often occurs just before $u_{mf}$ to overcome mechanical particle interlocking.
  3. Fluidized Bed Regime ($u_0 > u_{mf}$): As velocity increases beyond $u_{mf}$, the total bed pressure drop remains strictly constant! ΔPbed=Mbed,totalgAc(1ρfρp)=constant\Delta P_{bed} = \frac{M_{bed,total} g}{A_c} \left(1 - \frac{\rho_f}{\rho_p}\right) = \text{constant} The excess gas passes through the bed as bubbles (in bubbling beds). The bed expands vertically ($L > L_{mf}$), increasing the void fraction $\varepsilon$ such that the upward drag per unit particle remains exactly equal to particle weight.

6. Geldart Particle Classification System

D. Geldart (1973) categorized solids for gas fluidization into four distinct groups based on mean particle diameter $d_p$ and particle-fluid density difference $(\rho_p - \rho_g)$:

Geldart GroupDesignationParticle Size RangeFluidization CharacteristicsIndustrial Chemical Examples
Group AAeratable$30 - 100\ \mu\text{m}$ ($\rho_p < 1,400\text{ kg/m}^3$)Expands significantly and smoothly before bubbling begins ($u_{mb} > u_{mf}$); slow deaeration rate.Fluid Catalytic Cracking (FCC) catalysts; Fischer-Tropsch catalyst.
Group BSand-like / Bubbling$100 - 800\ \mu\text{m}$ ($\rho_p = 1,400 - 4,000$)Bubbles form immediately at incipient fluidization ($u_{mb} \approx u_{mf}$); moderate bed expansion.Coarse silica sand, coal gasifier granules, polymerization catalysts.
Group CCohesive / Fine$< 30\ \mu\text{m}$Very difficult to fluidize smoothly; interparticle Van der Waals forces cause channeling, lifting, and rat-holing.Flour, fine talc, face powders, submicron paint pigments, cement dust.
Group DSpoutable / Coarse$> 1,000\ \mu\text{m}$ ($> 1\text{ mm}$)Deep spouting beds; gas jets penetrate center; bubbles coalesce into large slugs.Coffee beans, roasted grains, polymer pellets, metal ores, roasted catalyst.

7. Terminal Settling Velocity ($u_t$) and Elutriation Limits

If gas velocity through a fluidized bed is increased continuously, particles eventually become entrained and are carried out of the vessel. The upper physical operating limit of a stationary fluidized bed is the terminal settling velocity ($u_t$) of a single particle.

Single-Particle Settling Balance

Setting drag force equal to net buoyant particle weight:

FD=CD(π4dp2)(12ρfut2)=(π6dp3)(ρpρf)gF_D = C_D \left(\frac{\pi}{4} d_p^2\right) \left(\frac{1}{2}\rho_f u_t^2\right) = \left(\frac{\pi}{6} d_p^3\right) (\rho_p - \rho_f) g

Solving for $u_t$:

ut=4dp(ρpρf)g3CDρfu_t = \sqrt{\frac{4 d_p (\rho_p - \rho_f) g}{3 C_D \rho_f}}

Regimes of Particle Settling

  1. Stokes' Law Regime ($Re_p = \frac{\rho_f u_t d_p}{\mu} < 1.0$): Viscous drag dominates; drag coefficient is $C_D = 24 / Re_p$: ut=gdp2(ρpρf)18μu_t = \frac{g d_p^2 (\rho_p - \rho_f)}{18 \mu}
  2. Intermediate Regime ($1.0 \le Re_p \le 1,000$): Transition drag: CD18.5Rep0.6    ut=0.153g0.71dp1.14(ρpρf)0.71ρf0.29μ0.43C_D \approx \frac{18.5}{Re_p^{0.6}} \implies u_t = 0.153 \frac{g^{0.71} d_p^{1.14} (\rho_p - \rho_f)^{0.71}}{\rho_f^{0.29} \mu^{0.43}}
  3. Newton's Law Regime ($1,000 < Re_p < 200,000$): Inertial turbulent form drag dominates; $C_D \approx 0.44$ (constant): ut=1.74dp(ρpρf)gρfu_t = 1.74 \sqrt{\frac{d_p (\rho_p - \rho_f) g}{\rho_f}}

The Fluidization Operating Window

A stable bubbling or turbulent fluidized bed reactor must operate within the strict velocity window:

umf<u0<utu_{mf} < u_0 < u_t

Operating at $u_0 \ge u_t$ triggers pneumatic transport (elutriation), carrying the entire catalyst inventory into overhead cyclones and downstream heat exchangers.


8. Summary Comparison of Gas-Solid Contacting Regimes

Contacting RegimeSuperficial Velocity ($u_0$)Bed Void Fraction ($\varepsilon$)Pressure Drop ($\Delta P$)Dominant Industrial Application
Fixed Bed$0 < u_0 < u_{mf}$Constant ($\approx 0.40$)Increases with $u_0$ (Ergun)Desulfurization hydrotreaters, packed absorption columns.
Particulate (Smooth)$u_{mf} \le u_0 < u_{mb}$Expands smoothly ($0.42 - 0.60$)Constant ($= W_{bed}/A_c$)Liquid-solid fluidization; Geldart Group A gas systems.
Bubbling Fluidized Bed$u_{mb} < u_0 < u_{ms}$Moderate expansion ($0.50 - 0.70$)Constant ($= W_{bed}/A_c$)Gas-phase polyethylene (Unipol), roasting, coal combustion.
Circulating Bed (CFB)$u_0 > u_t$High voidage ($0.85 - 0.95$)Low per meter; cyclone returnFluid catalytic cracking (FCC) risers, CFB boilers.
Pneumatic Conveying$u_0 \gg u_t$Dilute phase ($\varepsilon > 0.99$)Frictional pipe lossRailcar unloading of pellets, pneumatic ash transfer.

9. Step-by-Step Worked Numerical Example: Fluidized Bed Reactor Hydrodynamics

Problem Statement

A gas-phase catalytic fluidized bed reactor operates at $200^\circ\text{C}$ and $1.50\text{ MPa(a)}$ ($15.0\text{ bar(a)}$) to synthesize an organic intermediate. The fluidizing synthesis gas has a density of $\rho_f = 7.50\text{ kg/m}^3$ and dynamic viscosity of $\mu = 2.00 \times 10^{-5}\text{ Pa}\cdot\text{s} = 0.0200\text{ cP}$.

The solid catalyst particles are spherical with mean diameter $d_p = 100.0\ \mu\text{m} = 1.00 \times 10^{-4}\text{ m}$ and particle density $\rho_p = 1,600.0\text{ kg/m}^3$. The bed void fraction at minimum fluidization is measured as $\varepsilon_{mf} = 0.450$.

Calculate:

  1. The buoyant bed pressure drop gradient $\Delta P_{bed} / L_{mf}$ in $\text{kPa/m}$.
  2. The Archimedes number $Ar$.
  3. The minimum fluidization velocity $u_{mf}$ in millimeters per second ($\text{mm/s}$).
  4. The particle Reynolds number at minimum fluidization $Re_{p,mf}$ to verify the flow regime.
  5. The single-particle terminal settling velocity $u_t$ using Stokes' law.
  6. The allowable superficial operating velocity window for the reactor.

Solution

Step 1: Calculate bed pressure drop gradient at minimum fluidization.

ΔPbedLmf=(1εmf)(ρpρf)g\frac{\Delta P_{bed}}{L_{mf}} = (1 - \varepsilon_{mf}) (\rho_p - \rho_f) g ρpρf=1,600.0 kg/m37.50 kg/m3=1,592.5 kg/m3\rho_p - \rho_f = 1,600.0\text{ kg/m}^3 - 7.50\text{ kg/m}^3 = 1,592.5\text{ kg/m}^3 ΔPbedLmf=(10.450)×1,592.5 kg/m3×9.807 m/s2=0.550×15,617.65 N/m3=8,589.7 Pa/m=8.590 kPa/m\frac{\Delta P_{bed}}{L_{mf}} = (1 - 0.450) \times 1,592.5\text{ kg/m}^3 \times 9.807\text{ m/s}^2 = 0.550 \times 15,617.65\text{ N/m}^3 = 8,589.7\text{ Pa/m} = 8.590\text{ kPa/m}

Step 2: Calculate the Archimedes number ($Ar$).

Ar=dp3ρf(ρpρf)gμ2Ar = \frac{d_p^3 \rho_f (\rho_p - \rho_f) g}{\mu^2} dp3=(1.00×104 m)3=1.00×1012 m3d_p^3 = (1.00 \times 10^{-4}\text{ m})^3 = 1.00 \times 10^{-12}\text{ m}^3 μ2=(2.00×105 Pas)2=4.00×1010 Pa2s2\mu^2 = (2.00 \times 10^{-5}\text{ Pa}\cdot\text{s})^2 = 4.00 \times 10^{-10}\text{ Pa}^2\cdot\text{s}^2 Ar=(1.00×1012)×7.50×1,592.5×9.8074.00×1010=1.17132×1074.00×1010=292.83Ar = \frac{(1.00 \times 10^{-12}) \times 7.50 \times 1,592.5 \times 9.807}{4.00 \times 10^{-10}} = \frac{1.17132 \times 10^{-7}}{4.00 \times 10^{-10}} = 292.83

Because $Ar = 292.8 < 1,000$, flow through the bed at minimum fluidization is fully within the laminar regime ($Re_{p,mf} \ll 20$).

Step 3: Calculate minimum fluidization velocity ($u_{mf}$). Using the laminar Blake-Kozeny form of the Ergun equation:

umf=dp2(ρpρf)g150μ(εmf31εmf)u_{mf} = \frac{d_p^2 (\rho_p - \rho_f) g}{150 \mu} \left(\frac{\varepsilon_{mf}^3}{1 - \varepsilon_{mf}}\right)

Evaluate the physical grouping:

dp2(ρpρf)g150μ=(1.00×104 m)2×1,592.5 kg/m3×9.807 m/s2150×(2.00×105 Pas)=1.561765×1043.00×103=0.052059 m/s\frac{d_p^2 (\rho_p - \rho_f) g}{150 \mu} = \frac{(1.00 \times 10^{-4}\text{ m})^2 \times 1,592.5\text{ kg/m}^3 \times 9.807\text{ m/s}^2}{150 \times (2.00 \times 10^{-5}\text{ Pa}\cdot\text{s})} = \frac{1.561765 \times 10^{-4}}{3.00 \times 10^{-3}} = 0.052059\text{ m/s}

Evaluate the voidage function:

εmf31εmf=(0.450)310.450=0.0911250.550=0.165682\frac{\varepsilon_{mf}^3}{1 - \varepsilon_{mf}} = \frac{(0.450)^3}{1 - 0.450} = \frac{0.091125}{0.550} = 0.165682

Multiply the two factors:

umf=0.052059 m/s×0.165682=0.008625 m/s=8.625 mm/s8.63 mm/su_{mf} = 0.052059\text{ m/s} \times 0.165682 = 0.008625\text{ m/s} = 8.625\text{ mm/s} \approx 8.63\text{ mm/s}

Step 4: Verify the particle Reynolds number at $u_{mf}$.

Rep,mf=ρfumfdpμ=7.50 kg/m3×0.008625 m/s×1.00×104 m2.00×105 Pas=6.4688×1062.00×105=0.323Re_{p,mf} = \frac{\rho_f u_{mf} d_p}{\mu} = \frac{7.50\text{ kg/m}^3 \times 0.008625\text{ m/s} \times 1.00 \times 10^{-4}\text{ m}}{2.00 \times 10^{-5}\text{ Pa}\cdot\text{s}} = \frac{6.4688 \times 10^{-6}}{2.00 \times 10^{-5}} = 0.323

Because $Re_{p,mf} = 0.323 \ll 20$, the laminar simplification is verified with extreme precision.

Step 5: Calculate single-particle terminal settling velocity ($u_t$). Assuming the Stokes' law regime ($Re_p < 1.0$):

ut,Stokes=gdp2(ρpρf)18μ=9.807×(1.00×104)2×1,592.518×(2.00×105)=1.561765×1043.60×104=0.4338 m/s=433.8 mm/su_{t,Stokes} = \frac{g d_p^2 (\rho_p - \rho_f)}{18 \mu} = \frac{9.807 \times (1.00 \times 10^{-4})^2 \times 1,592.5}{18 \times (2.00 \times 10^{-5})} = \frac{1.561765 \times 10^{-4}}{3.60 \times 10^{-4}} = 0.4338\text{ m/s} = 433.8\text{ mm/s}

Check Reynolds number at terminal velocity:

Rep,t=ρfutdpμ=7.50×0.4338×1.00×1042.00×105=1.63Re_{p,t} = \frac{\rho_f u_t d_p}{\mu} = \frac{7.50 \times 0.4338 \times 1.00 \times 10^{-4}}{2.00 \times 10^{-5}} = 1.63

Because $Re_{p,t} = 1.63$ is slightly above $1.0$, applying the Schiller-Naumann intermediate drag correction ($C_D = \frac{24}{Re}(1 + 0.15 Re^{0.687})$ with factor $1 + 0.15(1.63)^{0.687} = 1.21$):

ut,actualut,Stokes1.21=0.43381.10=0.394 m/s=394 mm/su_{t,actual} \approx \frac{u_{t,Stokes}}{\sqrt{1.21}} = \frac{0.4338}{1.10} = 0.394\text{ m/s} = 394\text{ mm/s}

Step 6: Determine the operational window. The fluidization operating window is:

umf=8.63 mm/s<u0<ut394 mm/su_{mf} = 8.63\text{ mm/s} < u_0 < u_t \approx 394\text{ mm/s}

The ratio of terminal velocity to minimum fluidization velocity is $u_t / u_{mf} = 394 / 8.63 = 45.7$. Typically, commercial bubbling beds operate at $u_0 \approx 3 - 6 \times u_{mf}$ ($25 - 50\text{ mm/s}$), ensuring vigorous fluidization and high catalyst contacting without risking particle elutriation.


10. Common PE Exam Traps in Multiphase Flow and Fluidization

  1. Assuming Bed Pressure Drop Keeps Rising After Fluidization: In fixed beds ($u_0 < u_{mf}$), pressure drop increases continuously. But once the bed fluidizes ($u_0 \ge u_{mf}$), the total pressure drop remains strictly constant because the buoyant weight of solids does not change. Excess gas merely forms bubbles.
  2. Confusing Superficial Velocity with Interstitial Pore Velocity: Superficial velocity $u_0 = Q / A_c$ is based on the empty column cross-sectional area. The true interstitial pore velocity through the voids is $u_{pore} = u_0 / \varepsilon$. The Ergun equation uses superficial velocity $u_0$.
  3. Omitting the Buoyancy Term in Bed Calculations: Using particle density $\rho_p$ instead of buoyant density difference $(\rho_p - \rho_f)$. In high-pressure gas fluidization (where gas density can exceed $20\text{ kg/m}^3$) or liquid fluidization, neglecting $\rho_f$ creates major errors.
  4. Applying Stokes' Law Outside Its Validity Range: Using $u_t = g d_p^2 (\rho_p - \rho_f) / (18 \mu)$ when $Re_p > 1.0$. For large particles ($d_p > 1\text{ mm}$), the settling regime shifts to Newton's law ($C_D \approx 0.44$), where $u_t \propto \sqrt{d_p}$ rather than $d_p^2$.
Test Your Knowledge

A chemical engineer monitors an experimental fluidization column containing a 1.5 m settled bed of cracking catalyst. Air is introduced at the bottom distributor plate, and superficial gas velocity u_0 is steadily increased from zero to 2.5 times the minimum fluidization velocity (u_0 = 2.5 * u_mf). How does the total bed pressure drop (Delta P_bed) respond across this velocity progression?

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Test Your Knowledge

A catalytic reactor contains spherical catalyst beads (diameter d_p = 150 µm = 1.50 × 10⁻⁴ m, density rho_p = 2,000 kg/m³) fluidized by nitrogen gas (density rho_g = 1.20 kg/m³, viscosity µ = 1.80 × 10⁻⁵ Pa·s) at standard atmospheric conditions (g = 9.81 m/s²). The bed void fraction at incipient fluidization is epsilon_mf = 0.40. Assuming laminar flow through the bed, what is the minimum fluidization velocity u_mf?

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Test Your Knowledge

A hydrocarbon transport pipeline operates with co-current gas-liquid flow in the turbulent-turbulent (tt) regime. Process measurements yield a Lockhart-Martinelli parameter of chi = 2.50. Using the Chisholm correlation (phi_L² = 1 + C / chi + 1 / chi²) with C = 20, what is the two-phase liquid frictional multiplier phi_L², and if the single-phase liquid pressure drop at superficial velocity is Delta P_L = 12.0 kPa, what is the total two-phase frictional pressure drop Delta P_TP?

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