For the gas-solid reaction A(g) + bB(s) → products [(s), (g)], calculate the solid holdup ( $W_B$ ) required in a reactor for $f_B = 0.80$ , if $f_{Bo} = 50 kg min^{-1}$ and $t_1 = 1.5 h$ . Assume the following apply: the solid particles are in PF, the gas is uniform in composition, the particles are spherical and all of one size, the reaction is a first-order surface reaction, and the SCM applies, with ash-layer-diffusion control .

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Accepted Answer

The holdup is calculated from the definition of mean residence time of solid particles;

[latex]\bar{t} = W_B /F_{Bo}[/latex] (22.2-8)

where [latex]F_{Bo} [/latex] is given and is [latex]\bar{t_B}[/latex] obtained from the performance equation, 22.2-5 or -7, together with the given value of [latex]f_B[/latex] and the kinetics model. Thus, from Table 9.1, for spherical particles and diffusion control, with [latex]t_1 = \rho _{Bm}R^2/6bc_{Ag}D_e[/latex],

[latex]1 - \bar{f_B} = 1 - \bar{f_B}(\bar{t}) ; o \lt \bar{t} \lt t_1 [/latex] (22.2-5)

[latex]= 0; \bar{t} \gt t_1 [/latex]

[latex]1 - \bar{f_B} = 1 - \bar{f_B}(\bar{t} , \bar{R})[/latex] (22.2-7)

[latex]t(f_B) \equiv \bar{t}_B = t_1[1 3(1 - \bar{f_B})^{2/3} + 2(1 - \bar{f_B})] = 0.561 h [/latex]

for [latex]t_1 = 1.5 h[/latex] and [latex]f_B = 0.80[/latex] . Then

[latex]W_B = \bar{t}_BF_{Bo} = 0.561(50)60 = 1680 kg[/latex]

The size of reactor is chosen to accommodate this holdup. The diameter is determined from the gas flow rate, 9, together with, for example, an allowable superficial linear gas velocity, u (in lieu of an allowable [latex](-AP) : D = (4q/ \pi u)^{1/2} [/latex] for a cylindrical vessel. The volume could be determined from an appropriate “bed” density, together with an overhead space for disengagement of solid and gas phases (we assume no carryover of solid in the gas exit stream).

If the particles in PF are not all of the same size, the performance equation, from equation 22.2-2 in the summation form of equation 22.2-4, is

[latex]1 - \bar{f_B}=\int_{R_{min}}^{R_{max}}{[1-f_B(t,R)]P(R)dR} [/latex] (22.2-2)

[latex]1 - \bar{f_B} = \sum\limits_{R_{min}}^{R_{max}}{\left\{\int_{0}^{t_1(R)}{[1 - \bar{f_B}(t,R)]E(t)dt} \right\} } \frac{F_{Bo}(R)}{F_{Bo}}[/latex] (22.2-4)

[latex]1 - \bar{f_B} = \sum\limits_{R(t_1 = \bar{t}_B)}^{R_{max}}{[1 - f_B(t,R)]\frac{F_{Bo}(R)}{F_{Bo}}} [/latex] (22.2-9)

where [latex]R(t_1 = \bar{t_B}) [/latex] is the smallest size of particle for which the time for complete conversion ([latex]t_1[/latex] ) is greater than the mean-residence time of solid particles ([latex]\bar{t_B}[/latex] ); particles smaller than this are completely converted in the time [latex]\bar{t_B}[/latex] , and do not contribute to [latex]1 - \bar{f_B}[/latex] .

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 22.1: For the gas-solid reaction A(g) + bB(s) → products [(s), (g)...

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