(a) Estimate the amount of catalyst (Wcat/kg) required for a fluidized-bed reactor, according to the Kunii-Levenspiel bubbling-bed model, for the production of 60,000 Mg year-¹ of acrylonitrile by the ammoxidation of propylene with air. Data and assumptions: The heat transfer configuration within

Question

(a) Estimate the amount of catalyst ([latex]W_{cat}/kg[/latex]) required for a fluidized-bed reactor, according to the Kunii-Levenspiel bubbling-bed model, for the production of 60,000 Mg year-¹ of acrylonitrile by the ammoxidation of propylene with air.

Data and assumptions:

The heat transfer configuration within the bed (for the exotheimic reaction) and other internal features are ignored.
Only [latex]C_3H_3N[/latex] is formed (with water).
The feed contains [latex]C_3H_6[/latex] and NH, in the stoichiometric ratio and 20% excess air (79 mole % [latex]N_2[/latex] , 21% 0,); there is no water in the feed.
Conversion based on [latex]C_3H_6(A)[/latex] is 70%.
T = 400°C ; P = 2 bar.
The annual stream service factor (fraction of time in operation) is 0.94.
[latex]d_b = 0.1 m ; d_p = 0.05 mm ; \rho_p = 2500 kg m^{-3} ; \mu_f = 1.44 kg h^{-1} m^{-1} [/latex]
[latex]\blacklozenge _mf = 0.002 m s^{-1} ; \alpha = 0.6 ; D_m = 0.14 m^2 h^{-1} at 400^\circ C ; \epsilon _{pa} = 0.5[/latex]
[latex]\epsilon _{mf} = 0.6 ; k_A = 1.0 s^{-1} ; u_{fl} = 720 m h^{-1} ; \gamma _b = 0.004[/latex]

(b) Calculate the vessel diameter and the bed depth (fluidized) in m.
(c) For comparison, calculate [latex]W_{cat}[/latex], for the two cases (assume constant density for both):

(i) The reactor is a PFR.
(ii) The reactor is a CSTR.

Accepted Answer

(a) To calculate [latex]W_{cat}[/latex], we use equation 23.4-8 in conjunction with 23.4-5 and -6 (file ex23-3.msp). For this purpose, we also need to calculate other quantities, as indicated below. The reaction is

[latex]k_{overall} = \gamma _bk_A + \frac{1}{\frac{1}{K_{bc}} + \frac{1}{\gamma _{cw}k_A + \frac{1}{\frac{1}{K_{ce}} + \frac{1}{\gamma _ek_A}}}}[/latex] (23.4-5)

[latex]1 - f_A = c_A/c_{Ao} = \exp(-k_{overall}L_{fl}/u_b)[/latex] (23.4-6)

[latex]W_{cat} = \frac{ ho_pq(1 - \epsilon _{mf})u_{br}}{k_{overall}u_{fl}}\ln\left(\frac{c_{Ao}}{c_A} ight) [/latex] (23.4-8)

[latex]C_3H_6(A) + NH_3 + \frac{3}{2}O_2 ightarrow C_3H_3N(B) + 3H_2O [/latex]

Because of the small change in moles on reaction, and the dilution with [latex]N_2(inert)[/latex] , together with an anticipated (but not calculated) relatively small (-AZ’), the assumption of constant density is a reasonable one. It can be calculated that for every mole of A entering, the total moles entering is 10.57 and the total leaving is 10.92, an increase of only 3.3%. The actual molar feed rates of A ([latex]F_{Ao}[/latex]) and total gas ([latex]F_{to} [/latex]) are:

[latex]F_{Ao} = \left(60,000(1000)\frac{kg B}{yr} ight) \left(\frac{1}{53}\frac{kmol B}{kg B} ight)\left(1\frac{kmol A (reacted)}{kmol B (formed)} ight) \left(\frac{1}{0.70}\frac{kmol A (in feed)}{kmol A (reacted)} ight) imes \frac{1}{365(24)3600(0.94)}\frac{yr}{s} = 0.0546 kmol s^{-1}[/latex]

[latex]F_{to} = 10.57(0.0546) = 0.577 kmol s^{-1} [/latex]

The total volumetric feed rate is :

[latex]q_{o} = \frac{F_{to}RT}{P} = \frac{0.577(1000)8.314(673)}{2(105)} = 16^1 m^3 s^{-1}[/latex]

To calculate koverall in equation 23.4-5, we require [latex]K_{bc}[/latex] , [latex]K_{ce}[/latex] , [latex]\gamma _{cw}[/latex] , and [latex]\gamma_e[/latex] ; these, in turn, require calculations of [latex]u_{br}[/latex] , [latex]u_b[/latex] , [latex]f_b[/latex] , and ([latex]f_c + f_w [/latex]), as follows (with each equation indicated):

[latex]u_{br} = 0.711(gd_b)^{1/2} = 0.711[9.81(0.1)]^{1/2} = 0.704 m s^{-1} [/latex] (from 23.3-2)

[latex]u_b = u_{fl} - u_{mf} + u_{br} = (720/3600) - 0.002 + 0.704 = 0.902 m s^{-1}[/latex] (from 23.3-4)

[latex]K_{bc} = 4.5(u_{mf}/d_b) + 5.85(D_m^{1/2}g^{1/4}/d_b^{5/4})[/latex]

[latex] = 4.5(0.002/0.1) + 5.85[(0.14/3600)^{1/2}(9.81)^{1/4}(0.1)^{5/4}] = 1.24 s^{-1}[/latex] (from 23.3-13)

[latex]K_{ce} = 6.77(\epsilon _{mf}D_mu_{br}/d_b^3)^{1/2} = 6.77[0.6(0.14/3600)0.704/(0.1)^3]^{1/2} = 0.868 s^{-1}[/latex] (from 23.3-14)

[latex]f_b = u_{fl}/u_b = 0.2/0.902 = 0.222[/latex] (from 23.3-6)

[latex]f_c = \frac{3u_{mf}f_b}{\epsilon _{mf}u_{br} - u_{mf}} = \frac{3(0.002)0.222}{(0.60)(0.704) - 0.002} = 0.0032[/latex] (from 23.3-7)

[latex]f_w = \alpha f_b = (0.60)(0.222) = 0.133[/latex] (from 23.3-8)

[latex]f_e = 1 - 0.222 - 0.0032 - 0.133 = 0.642 [/latex] (from 23.3-9)

[latex]\gamma _{cw} = (1 - \epsilon _{mf})(f_c + f_w)/f_b = (1 - 0.60)(0.0032 + 0.133)/0.222 = 0.245[/latex] (from 23.3-12)

[latex]Ye = \frac{(1 - \epsilon _{mf})(1 - f_b)}{f_b} - \gamma _b - \gamma _{cw} [/latex] (from 23.3-10)

[latex]= \frac{(1 -0.60)(1 - 0.222)}{0.222} - 0.004 - 0.245 = 1.153 [/latex]

[latex]k_{overall} = \gamma _bk_A + \frac{1}{\frac{1}{K_{bc}} + \frac{1}{\gamma _{cw}k_A + \frac{1}{\frac{1}{K_{ce}} + \frac{1}{\gamma _ek_A}}}} [/latex] (from 23.4-5)

[latex] = (0.004)(1.0) + \frac{1}{\frac{1}{1.24} + \frac{1}{(0.245)(1.0) + \frac{1}{\frac{1}{0.868} + \frac{1}{(1.153)(1.0)}}}} = 0.467 s^{-1}[/latex]

The bed depth, [latex]L_{fl}[/latex] can now be determined using 23.4-6:

[latex]L_{fl} = \frac{-[\ln(1 - f_A)]u_b}{k_{overall}} = \frac{-[\ln(1 - 0.70)](0.902 m s^{-1})}{0.467 s^{-1}} = 233 m[/latex]

The catalyst requirement can be determined from 23.4-8:

[latex]W_{cat} = \frac{2500(16.1)(1 - 0.60)0.704[- \ln(1 - 0.070)]}{0.467(720/3600)} = 1.46 imes 10^5 kg[/latex]

(b) From [latex]q = u_{fl}A_c = u_{fl} \pi D^2/4[/latex] ,

[latex]D = (4q/\pi u_{fl})^{1/2} = [4(16.1)/\pi (720/3600)] ^{1/2} = 10.1 m[/latex]

From (a), [latex]L_{fl} = 2.33 m[/latex]

(c) (i) For a PFR,

[latex]W_{cat} = ho V_{cat} = ( ho_pq/k_A)\int_{0}^{f_A}{df_A/(1 - f_A)}[/latex]

[latex] = [2500(16.1)/1.0][ - \ln(1 - 0.70)] = 0.48 imes 10^5 kg[/latex]

(ii) For a CSTR,

[latex]W_{cat} = ho _{p}V_{cat} = ( ho_pq/k_A)[f_A/(1 - f_A)][/latex]

[latex]= [2500(16.1)/1.0][0.70/(1 - 0.70)] = 0.94 imes 10^5 kg[/latex]

The amount of catalyst required in (a) is even greater than that required for a CSTR, which may be accounted for by the “bypassing effect” (Section 23.1.3).

Question 23.3: (a) Estimate the amount of catalyst (Wcat/kg) required for a...

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