(Kunii and Levenspiel, 1991, p. 298) Phthalic anhydride is produced in the following process: naphthalene (A) →^k1 phthalic anhydride (B) →^k2 CO2 + H2O The reaction occurs in a fluidized-bed reactor, with sufficient heat exchange to ensure isothermal operation. The bed (before fluidization)

Question

(Kunii and Levenspiel, 1991, p. 298) Phthalic anhydride is produced in the following process:

[latex] naphthalene (A) \overset{k_1}{\longrightarrow } phthalic anhydride (B) \overset{k_2}{\longrightarrow } CO_2 + H_2O [/latex]

The reaction occurs in a fluidized-bed reactor, with sufficient heat exchange to ensure isothermal operation. The bed (before fluidization) is 5 m deep ([latex]L_{pa}[/latex]), with a voidage ([latex]\epsilon_{pa}[/latex]) of 0.52. The reaction rate constants for the two steps are [latex]k_1 = 1.5 m^3 gas (m^3 cat)^{-1} s^{-1} [/latex] and [latex]k_2 = 0.010 m^3 gas (m^3 cat)^{-1} s^{-1} [/latex]. Additional data are:

[latex]u_{mf} = 0.005 m s^{-1} d_b = 0.05 m D_A = 8.1 imes 10^{-6} m^2 s^{-1} [/latex]

[latex]\epsilon_{mf} = 0.57 u_b = 1.5 m s^{-1} D_B = 8.4 imes 10^{-6} m^2 s^{-1} [/latex]

[latex]u_{fl} = 0.45 m s^{-1} \gamma_b = 0.005 f_w/f_b = 0.6 [/latex]

Determine the overall fractional conversion of naphthalene, and the selectivity to phthalic anhydride.

Accepted Answer

To solve the problem, continuity equations are written for A and B in all three regions of the fluidized bed. First, however, the exchange parameters ([latex]K_{bc}[/latex] and [latex]K_{ce} [/latex]) for A and B are determined, along with [latex]u_{br}[/latex] , [latex]f_b[/latex] , [latex]f_c[/latex] , [latex]\gamma_{cw} [/latex] , and [latex]\gamma _e[/latex] .

[latex]u_{br} = 0.711(gd_b)^{0.5} = 0.711[(9.8)(0.05)]^{0.5} = 0.498 m s^{-1}[/latex] (23.3-2)

Since [latex]u_{fl} \gg u_{mf} [/latex], equation 23.3-6 is used to determine [latex]f_b[/latex] :

[latex]f_b = u_{fl}/u_b = 0.45/1.5 = 0.30[/latex] (23.3-6)

From equation 23.3-7

[latex]f_c = \frac{3u_{mf}}{(\epsilon_{mf} u_{br} - u_{mf})}f_b[/latex] (23.3-7)

[latex]f_c /f_b = 3u_{mf}/(\epsilon_{mf} u_{br} - u_{mf}) = 3(0.005)/[(0.57)(0.498) - 0.005] = 0.0538 [/latex]

The catalyst fractions in the cloud + wake and emulsion regions are determined using
equations 23.3-12 and 23.3-10, respectively:

[latex]\gamma _b + \gamma_{cw} + \gamma_e \equiv \frac{m^3 total solid }{m^3 bubbles} \equiv \left(\frac{m^3 total solid }{m^3 bed at mf}\right) \left(\frac{m^3 (c + w + e) }{m^3 fl bed}\right) \left(\frac{m^3 fl bed}{m^3 bubbles}\right) = (1 - \epsilon _{mf})(f_c + f_w + f_e)/f_b = (1 - \epsilon_{mf})(1 - f_b)/f_b[/latex] (23.3-10)

[latex]\gamma_{cw} \equiv \frac{m^3 solid in (c + w)}{m^3 bubbles} \equiv \frac{m^3 solid in c}{m^3 bubbles} + \frac{m^3 solid in w}{m^3 bubbles} \equiv \left(\frac{m^3 total solid}{m^3 bed at mf}\right) \left(\frac{m^3 bed in c}{m^3 fl bed }\right) \left(\frac{m^3 fl bed}{m^3 bubbles}\right) + \left(\frac{m^3 total solid}{m^3 bed at mf}\right) \left(\frac{m^3 bed in w}{m^3 fl bed }\right) \left(\frac{m^3 fl bed}{m^3 bubbles}\right) = (1 - \epsilon _{mf})(f_c + f_w)/f_b[/latex] (23.3-12)

[latex]Y_{cw} = (1 - \epsilon _{mf})(f_c + f_w)/f_b = (1 - 0.57)(0.0538 + 0.60) = 0.281 [/latex]

[latex]\gamma _e = (1 - \epsilon _{mf})\frac{1 - f_b}{f_b} - \gamma _{cw} - \gamma _b = ( 1 - 0.57) \frac{1 - 0.30}{0.30} - 0.281 - 0.005 = 0.717 [/latex]

[latex]K_{bc} [/latex] and [latex]K_{ce} [/latex] are determined for naphthalene (A) and phthalic anhydride (B) using equations 23.3-13 and -14. The resulting values are: [latex]K_{bcA} = 1.70 s^{-1} [/latex] ; [latex]K_{bcB} = 1.72 s^{-1} [/latex] ; [latex]K_{ceA} = 0.92 s^{-1} [/latex]; and [latex]K_{ceB} = 0.93 s^{-1} [/latex] .
Continuity equations for A and B are written similar to equations 23.4-11 to -16 (in thiscase we have a series network rather than a parallel network):

Continuity equation for A in the bubble region:

[latex]u_b\frac{dc_{Ab}}{dx} = - \gamma_bk_1c_{Ab} - K_{bcA}(c_{Ab} - c_{Ac})[/latex] (A)

Continuity equation for A in the cloud + wake region:

[latex]K_{bcA}(c_{Ab} - c_{Ac}) = \gamma _{cw}k_1c_{Ac} + K_{ceA}(c_{Ac} - c_{Ae})[/latex] (B)

Continuity equation for A in the emulsion region:

[latex] K_{ceA}(c_{Ac} - c_{Ae}) = \gamma_ek_1c_{Ae}[/latex] (C)

Continuity equation for B in the bubble region:

[latex]u_b\frac{dc_{Bb}}{dx} = \gamma_bk_1c_{Ab} - \gamma_bk_2c_{Bb} - K_{bcB}(c_{Bb} - c_{Bc})[/latex] (D)

Continuity equation for B in the cloud + wake region:

[latex]K_{bcB}(c_{Bb} - c_{Bc}) = - \gamma_{cw}k_1c_{Ac} + \gamma_{cw}k_2c_{Bc} + K_{ceB}(c_{Bc} - c_{Be})[/latex] (E)

Continuity equation for B in the emulsion region:

[latex]K_{ceB}(c_{Bc} - c_{Be}) = -\gamma_ek_1c_{Ae} + \gamma_ek_2c_{Be}[/latex] (F)

Equations (A) and (D) are numerically integrated from x = 0 to [latex]x = L_{fl}[/latex] , with simultaneous solution of equations (B), (C), (E), and (F) to obtain [latex]c_{Ab}[/latex] and [latex]c_{Bb} [/latex] (see E-Z Solve file ex23-4.msp). From equation 23.4-7b,

[latex]L_{fl} = \frac{1 - \epsilon _{pa}}{(1 - \epsilon _{mf})(1 - f_b)}L_{pa} = \frac{1 - 0.52}{(1 - 0.57)(1 - 0.30)}(5 m) = 7.97 m[/latex]

The resulting conversion is 0.96, with a selectivity to phthalic anhydride of 0.95.

Question 23.4: (Kunii and Levenspiel, 1991, p. 298) Phthalic anhydride is p...

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