A kinetics study of the reaction A(g) + 2B(s) → products (s, g) (problem 9-21) yielded the following values of parameters for 2-cm (diameter) spherical particles at 700 K and pA = 200 kPa : tl = 4.92 X lo^5 s; kAg = 9.2 X lo^5 m s-¹ ,kAs = 8.85 X lo^5 m s-¹; De = 8.37 X lo^-7 m² s-¹ For the value

Question

A kinetics study of the reaction A(g) + 2B(s) → products (s, g) (problem 9-21) yielded
the following values of parameters for 2-cm (diameter) spherical particles at 700 K and
[latex] P_A = 200 kPa[/latex] :

[latex] t_1 = 4.92 imes 10^{5} s ; k_{Ag} = 9.2 imes 10^{-5} m s^{-1}[/latex]

[latex] k_{As} = 8.85 imes 10^{-5} m s^{-1} ; D_e = 8.37 imes 10^{-7} m^2 s^{-1} [/latex]

For the value of [latex]t_1[/latex] , obtained, determine, using the SCM model, the mean fractional conversions ([latex]\bar{f_B}[/latex]) for mean residence times ([latex]\bar{t}[/latex]) of(i) 50 h, and (ii) 100 h, assuming

(a) reaction rate-control;
(b) ash-layer diffusion control;
(c) gas-film mass transfer control;
(d) resistance contributed by all three processes

Accepted Answer

Although it is possible to obtain analytical expressions for the reactor model for spherical particles for parts (a), (b), and (c) using the approach in Example 22-3(a) and (b) for cylindrical particles (expressions for [latex]\bar{f_B}[/latex] for spherical particles are given by Levenspiel,1972, pp. 384-5), we use the numerical procedure described in Example 22-3 (c) for all four parts of this example, (a) to (d), since this procedure must be used for (d).

For numerical solution with the E-Z Solve software (file ex22-5.msp), a procedure equivalent to that in Example 22-3(c) is used. First, equation 22.2-13 is expressed as a differential equation:

[latex]1 - \bar{f_B} = \frac{1}{\bar{t}} \overset{\underset{}{ t_1} }{\underset{I_{0}}{}}[1 - f_B(t)]e^{- t/\bar{t}}dt[/latex] (22.2-13)

[latex]\frac{d\bar{f_B}}{dt} = -\frac{1}{\bar{t}}e^{-t/\bar{t}}[1 - f_B(t)][/latex] (22.2-22)

Equation 22.2-22 is then numerically integrated simultaneously with solution of the algebraic equation [latex]t = t({f_B})[/latex] given below for each of the cases (a) to (d). The integration is from t = 0 to [latex]t = t_1[/latex], subject to the initial condition that [latex]\bar{f_B} = 0[/latex] at t = 0. The expressions for t([latex]t({f_B})[/latex]) are taken from Table 9.1, as follows :

(a) [latex]t = t_1[1 - (1 - f_B)^{1/3}][/latex]

where [latex]t_1 = \rho_{Bm} R/bc_{Ag}k_{As} [/latex]

(b) [latex]t = t_1[1 3(1 - f_B)^{2/3} + 2(1 - f_B)][/latex]

where [latex]t_1 = \rho_{Bm} R^2/6bc_{Ag}D_e[/latex]

(c) [latex]t = t_1f_B[/latex]

where [latex]t_1 = \rho_{Bm} R/3bc_{Ag}k_{Ag}[/latex]

(d) [latex]t = \frac{\rho_{Bm}R}{bc_{Ag}}\left\{\frac{f_B}{3k_{Ag}} + \frac{R}{6D_e}[1 - 3(1 - f_B)^{2/3} + 2(1 - f_B)] + \frac{1}{k_{As}}[1 - ( 1 - f_B)^{1/3}]\right\}[/latex] (22.2-23)

Table 22.1 Results for Example 22-5

[latex]\bar{t} / h[/latex]	[latex]\bar{f_B}[/latex]
[latex]\bar{t} / h[/latex]	(a)	(b)	(c)	(d)
50	0.569	0.669	0.342	0.515
100	0.734	0.793	0.545	0.692

File ex22-5.msp illustrates the implementation of the solution in each case. Note that, to avoid evaluation of zero to a fractional power, [latex]{f_B}[/latex] is bounded between zero and 0.9999. Note also that tt is not used for case (d). The results for [latex]\bar{f_B}[/latex] for all cases, (a) to (d), are given in Table 22.1. Because of the way the problem is stated, with [latex]t_1[/latex] the same for each case, the result for case (d), all three rate processes involved, is an average of some sort of the results for cases (a) to (c), each of which involves only one rate process. The assumption of a single rate-controlling process introduces significant error in each of cases (a) to (c),relative to case (d).

Table 9.1 SCM: Summary of [latex]t(f_B) [/latex] ffor various shapes of particle '

Particle shape	Size Parameters	[latex]f_B[/latex](size parameters	[latex]t(f_B)[/latex]
flat plate (one face permeable)	length l (of Zone unreacted ) L(of particale)	[latex]1 - \frac{1}{L}[/latex]	[latex]\frac{\rho _{Bm}Lf_B}{bc_{Ag}}\left(\frac{1}{k_{Ag}} + \frac{Lf_B}{2D_e} + \frac{1}{k_{As}}\right) [/latex]
cylinder (ends sealed)	radius [latex]r_c[/latex] (of core unreacted R(of particle)	[latex]1 - \frac{r_c}{_0R}^2[/latex]	[latex]\frac{\rho _{Bm}R}{bc_{Ag}}\left\{\frac{f_B}{2k_{Ag}} + \frac{R}{4D_e}\left[f_B + (1 - f_B)\ln(1 f_B)\right] + \frac{1}{k_{As}}\left[1 - (1 - f_B)^{1/2}\right] \right\} [/latex]
sphere	same as for cylinder	[latex]1-\left(\frac{r_c}{R}\right) ^3[/latex]	[latex]\frac{\rho _{Bm}R}{bc_{Ag}}\left\{\frac{f_B}{3k_{Ag}} + \frac{R}{6D_e}[1 - 3(1 f_B)^{2/3} + 2(1 -f_B) + 1\frac{1}{k_{As}}1\left[ - (1 - f_B)^{1/3}\right] \right\}[/latex] (9.1-28)
Reaction : A(g k bB(s) -) product (s),(g)]; first order with respect to A at core surface. Particle(B) : constant-size (L, R constant); isothermal. For [latex]t_1[/latex] (time for complete reaction of particle), set [latex]f_B = 1 ; [(1 f_B)\ln(1 f_B) \rightarrow 0 ][/latex] . Symbols : see text and Nomenclature.

Question 22.5: A kinetics study of the reaction A(g) + 2B(s) → products (s,...

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