Packed Bed vs. Fluidized Bed (Kunii–Levenspiel): 23-Chem-A4 Chemical Reactor Engineering, December 2018, Q4

Question 4: Packed Bed vs. Fluidized Bed (Kunii–Levenspiel) (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. First-order $A\rightarrow B$; lab packed bed $H_m = 10$ cm, $u_0 = 2$ cm/s, $X = 0.97$. Kunii–Levenspiel data for the fluidized bed: $u_{mf} = 3.2$ cm/s, $\varepsilon_{mf} = 0.5$, bubble diameter $d_b = 8$ cm, wake factor $\alpha = 0.34$, effective diffusivity $D_e = 3\times10^{-5}$ m²/s $= 0.30$ cm²/s, $g = 981$ cm/s².

Find. (a) the intrinsic rate constant $k_r$; (b) fluidized-bed conversion at $u_0=20$ cm/s, $H_m=100$ cm; (c) packed-bed conversion at the same $u_0$ and height.

Check: the source prints $D_e = 3\times10^{-5}$ cm/s, which is dimensionally inconsistent with a diffusivity. It is read here as $3\times10^{-5}\ \text{m}^2/\text{s} = 0.30\ \text{cm}^2/\text{s}$ (the physically sensible gas effective diffusivity), per exam instruction #1 on stating assumptions.

Figure 4. The lab packed bed (parts a, c) versus the scaled-up fluidized bed (part b). The packed bed is a plug-flow contactor; the fluidized bed suffers bubble bypassing modelled by the three-region Kunii–Levenspiel scheme.

Approach. Treat the packed bed as a first-order plug-flow reactor to back out $k_r$, then apply the Kunii–Levenspiel bubbling-bed model (bubble/cloud/emulsion resistances in series) for the fluidized bed, and finally re-use the plug-flow relation for the larger packed bed.

(a) Rate constant from the lab packed bed. First-order plug flow through a bed of voidage $\varepsilon_m$ gives $-\ln(1-X) = k_r\,(1-\varepsilon_m)\,H_m/u_0$. Solving for $k_r$: $$k_r = \frac{-\ln(1-X)\,u_0}{(1-\varepsilon_m)H_m} = \frac{\ln(1/0.03)\,(2)}{(0.5)(10)} = \frac{(3.507)(2)}{5}.$$ $$\boxed{k_r = 1.40\ \text{m}^3/(\text{m}^3_{\text{cat}}\cdot\text{s}).}$$
(b) Bubble rise velocity and phase fractions. Single-bubble rise $u_{br} = 0.711\sqrt{g\,d_b} = 0.711\sqrt{(981)(8)} = 63.0$ cm/s. Bubble velocity $u_b = u_0 - u_{mf} + u_{br} = 20 - 3.2 + 63.0 = 79.8$ cm/s, and the bubble fraction $\delta = (u_0-u_{mf})/u_b = 16.8/79.8 = 0.211$.
Interphase transfer coefficients. Bubble–cloud and cloud–emulsion exchange (Kunii–Levenspiel): $$K_{bc} = 4.5\frac{u_{mf}}{d_b} + 5.85\frac{D_e^{1/2}g^{1/4}}{d_b^{5/4}},\qquad K_{ce} = 6.77\sqrt{\frac{D_e\,\varepsilon_{mf}\,u_{br}}{d_b^{3}}}.$$ With the data (cm, s units), $K_{bc} = 1.80 + 1.33 = 3.13$ s$^{-1}$ and $K_{ce} = 0.920$ s$^{-1}$.
Solid hold-ups in each phase. Cloud-plus-wake solids per unit bubble volume $\gamma_c = (1-\varepsilon_{mf})\!\left[\frac{3\,u_{mf}/\varepsilon_{mf}}{u_{br}-u_{mf}/\varepsilon_{mf}} + \alpha\right]$, bubble solids $\gamma_b \approx 0.005$, and emulsion solids $\gamma_e = (1-\varepsilon_{mf})\frac{1-\delta}{\delta} - \gamma_b - \gamma_c$. Numerically $\gamma_c = 0.340$ and $\gamma_e = 1.530$.
Overall rate coefficient (resistances in series). $$K_f = \gamma_b k_r + \cfrac{1}{\cfrac{1}{K_{bc}} + \cfrac{1}{\gamma_c k_r + \cfrac{1}{\frac{1}{K_{ce}} + \frac{1}{\gamma_e k_r}}}} = 0.832\ \text{s}^{-1}.$$
(b) Expanded bed height. The bubbles rise through the fluidized bed, which is taller than the packed height $H_m$. With $1-\varepsilon_f = (1-\delta)(1-\varepsilon_{mf}) = (0.789)(0.5) = 0.395$, $$L_f = \frac{H_m(1-\varepsilon_m)}{1-\varepsilon_f} = \frac{(100)(0.5)}{0.395} = 126.7\ \text{cm}.$$ This is the height that makes the phase hold-ups account for all the catalyst: $\delta L_f(\gamma_b+\gamma_c+\gamma_e) = (1-\varepsilon_m)H_m$.
(b) Fluidized-bed conversion. Gas passes mainly through the bubble phase in plug flow, taking $L_f/u_b$ to cross the bed: $$X = 1 - \exp\!\left(-K_f\,\frac{L_f}{u_b}\right) = 1 - \exp\!\left(-0.832\times\frac{126.7}{79.8}\right) = 1 - e^{-1.321}.$$ $$\boxed{X_{\text{fluidized}} = 0.73\ (73\%).}$$
(c) Larger packed bed. The plug-flow relation depends only on the space time $ (1-\varepsilon_m)H_m/u_0$. Here $100/20 = 5$ s, identical to the lab bed’s $10/2 = 5$ s, so $$X = 1 - \exp\!\left(-k_r(1-\varepsilon_m)H_m/u_0\right) = 1 - e^{-3.507} = \boxed{0.97\ (97\%).}$$

The comparison is the lesson: at identical superficial velocity and ten times the height, the fluidized bed converts only 73% against the packed bed’s 97%. Gas short-circuits through the bubble phase, which holds little catalyst, so despite excellent heat control the fluidized bed is a markedly poorer contactor for a fast first-order reaction.

Quantity	Value
(a) Rate constant $k_r$	1.40 s$^{-1}$
Bubble rise $u_{br}$ / bubble velocity $u_b$	63.0 / 79.8 cm/s
Bubble fraction $\delta$	0.211
Overall coefficient $K_f$	0.832 s$^{-1}$
Expanded bed height $L_f$	126.7 cm
(b) Fluidized-bed conversion	0.73 (73%)
(c) Packed-bed conversion	0.97 (97%)

Concept & Theory

Topic. The Kunii–Levenspiel bubbling-bed model and packed-bed plug-flow conversion for a first-order catalytic reaction.

Key relations. Packed bed: $-\ln(1-X)=k_r(1-\varepsilon_m)H_m/u_0$. Fluidized bed: $u_{br}=0.711\sqrt{gd_b}$, $u_b=u_0-u_{mf}+u_{br}$, $\delta=(u_0-u_{mf})/u_b$; three-region resistance $K_f$ combining $K_{bc}$, $K_{ce}$ and phase solid hold-ups; $L_f = H_m(1-\varepsilon_m)/[(1-\delta)(1-\varepsilon_{mf})]$; $X = 1-e^{-K_f L_f/u_b}$.

Why this works. In a bubbling fluidized bed most gas rises as nearly solid-free bubbles; reactant must transfer from bubble → cloud → emulsion to reach catalyst, then products return. Modelling these as mass-transfer resistances in series with the intrinsic kinetics yields one lumped coefficient $K_f$ that plays the role of $k_r$ in a plug-flow expression based on bubble transit time.

Common pitfall. Using the superficial velocity instead of the bubble velocity $u_b$ in the exponent, or the packed height $H_m$ instead of the expanded height $L_f$ (here that would understate the conversion as 65%); forgetting the wake and cloud solid hold-ups (setting $\gamma_e$ negative if $\delta$ is mis-estimated); and unit slips in $D_e$ (cm²/s vs m²/s), which shift $K_{bc}$, $K_{ce}$.

Source. Kunii & Levenspiel, Fluidization Engineering, 2nd ed., Ch. 12; Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 20.

Question 4 of 5: Packed Bed vs. Fluidized Bed (Kunii–Levenspiel)

Question 4: Packed Bed vs. Fluidized Bed (Kunii–Levenspiel) (25 marks)