Solution

Consider the reaction stoichiometry and the equilibrium constant for the oxidation of SO₂:

$\begin{array}{l}{\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2\iff {\mathrm{SO}}_3\hfill \\ K_{\mathrm{p}}=\frac{P_{{\mathrm{SO}}_3}}{P_{{\mathrm{SO}}_2}·P_{{\mathrm{O}}_2}^{0.5}}\hfill \\ {\log }_{10}(K_{\mathrm{p}})=\frac{4956}{T}-4.678\hfill \end{array}$

Assume 1 kmol of dry air. The mass balances for the pyrite roasting and the sulfur dioxide conversion are as follows (Table 7E6.1):

The fraction of SO₂ in the gas that enters the converter is given by:

$\%{\mathrm{SO}}_2=\frac{2a}{1-0.75a}\to a=\frac{\%{\mathrm{SO}}_2}{2+0.75\cdot \%{\mathrm{SO}}_2}$

In the equilibrium, the total number of moles is given by the equation below:

$n_{\mathrm{T}}=2\cdot a\cdot (1-X)+0.79+0.21-2.75a-a\cdot X+2\cdot a\cdot X=1-0.75\cdot a-a\cdot X$

Therefore, the partial pressures for the species involved are the following:

$\begin{array}{l}{P}_{{\mathrm{SO}}_2}=\frac{2\cdot a\cdot (1-X)}{1-0.75a-a\cdot X}{P}_{\mathrm{T}}\hfill \\ P_{{\mathrm{SO}}_3}=\frac{2\cdot a\cdot (X)}{1-0.75a-a\cdot X}{P}_{\mathrm{T}}\hfill \\ P_{{\mathrm{O}}_2}=\frac{0.21-2.75\cdot a-a\cdot X}{1-0.75a-a\cdot X}{P}_{\mathrm{T}}\hfill \end{array}$

The equilibrium constant is given by this equation:

$\begin{array}{cc}{K}_{\mathrm{p}}& =\frac{P_{{\mathrm{SO}}_3}}{P_{{\mathrm{SO}}_2}\cdot P_{{\mathrm{O}}_2}^{0.5}}=\frac{\left[\frac{2·a·(X)}{1-0.75a-a\cdot X}{P}_{\mathrm{T}}\right]}{\left[\frac{2·a·(1-X)}{1-0.75a-a·X}{P}_{\mathrm{T}}\right]{\left[{\displaystyle \frac{0.21-2.75·a-a·X}{1-0.75a-a·X}}{P}_{\mathrm{T}}\right]}^{0.5}}\\ & ={\displaystyle \frac{X·{[1-0.75a-a·X]}^{0.5}}{(1-X){[0.21-2.75·a-a·X]}^{0.5}{P_{\mathrm{T}}}^{0.5}}}\end{array}$

Fig. 7E6.1 shows the effect of feed composition, operating pressure, and temperature on the conversion.

Solution

Using the results presented in Example 7.5, with a=0.1, we compute the equilibrium line. As before, from sulfur, and considering 1 kmol of dry air, we have the mass balances as seen in Table 7E7.1.

The moles in the equilibrium are as follows:

$n_{\mathrm{T}}=a-a·X+0.79+0.21-a-0.5·a·X+a·X=1-0.5·a·X$

The equilibrium constant can be computed as below (and plotted in Fig. 7E7.1):

$\begin{array}{cc}{K}_{\mathrm{p}}& =\frac{P_{{\mathrm{SO}}_3}}{P_{{\mathrm{SO}}_2}·P_{{\mathrm{O}}_2}^{0.5}}=\frac{\left[{\displaystyle \frac{a·(X)}{1-{\displaystyle \frac{a\cdot X}{2}}}}{P}_{\mathrm{T}}\right]}{\left[{\displaystyle \frac{a·(1-X)}{1-{\displaystyle \frac{a·X}{2}}}}{P}_{\mathrm{T}}\right]{\left[{\displaystyle \frac{0.21-a-{\displaystyle \frac{a·X}{2}}}{1-{\displaystyle \frac{a·X}{2}}}}{P}_{\mathrm{T}}\right]}^{0.5}}\\ & =\frac{X·{\left[1-{\displaystyle \frac{a·X}{2}}\right]}^{0.5}}{(1-X){\left[0.21-a-{\displaystyle \frac{a·X}{2}}\right]}^{0.5}{P}_{\mathrm{T}}^{0.5}}\end{array}$

For the first bed, we perform an energy balance:

$\begin{array}{l}\sum _{i\in \mathrm{out}}{n}_i·\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT-\sum _{i\in \mathrm{in}}{n}_i·\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT=\mathrm{\Delta }{H}_{\mathrm{r}}\hfill \\ {n_{{\mathrm{SO}}_3}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT+n_{{\mathrm{SO}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT+n_{{\mathrm{N}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT+n_{{\mathrm{O}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT|}_{\mathrm{out}}-\hfill \\ {n_{{\mathrm{SO}}_3}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT+n_{{\mathrm{SO}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT+n_{{\mathrm{N}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT+n_{{\mathrm{O}}_2}\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT|}_{\mathrm{in}}=\mathrm{\Delta }{H}_{\mathrm{r}}\hfill \\ n_{{\mathrm{SO}}_2}X\underset{T_{\mathrm{in}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT+n_{{\mathrm{SO}}_2}(1-X)\underset{T_{\mathrm{ref}}}{\overset{T_{\mathrm{in}}}{\int }}{c}_{p}dT+n_{{\mathrm{N}}_2}\underset{T_{\mathrm{in}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT+\left(n_{{\mathrm{O}}_2}-n_{{\mathrm{SO}}_2}X\right)\underset{T_{\mathrm{in}}}{\overset{T_{\mathrm{out}}}{\int }}{c}_{p}dT=\mathrm{\Delta }{H}_{\mathrm{r}}\hfill \end{array}$

Assuming constant heat capacity for the mixture and the fact that the moles of gas are almost constant in the bed, the balance becomes linear:

$\begin{array}{l}{n}_{\mathrm{T},1}·c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=(1-0.5·a·X)·c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·X·23,200\hfill \\ c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·X·23,200\hfill \\ T=\frac{1}{c_{p,\mathrm{mix}}}a\cdot X\cdot 23,200+T_{\mathrm{o}}=(400+273)+293.7X\hfill \end{array}$

This straight line reaches the equilibrium line. For the sake of argument, we fit the equilibrium curve to a polynomial as a function of temperature in order to solve the conversion at each bed. X is the conversion, and T is the temperature in K:

$\begin{array}{ll}X\hfill & =-4.7106·{10}^{-16}{T}^6+2.3838·{10}^{-12}{T}^5-4.9151·{10}^{-09}{T}^4+5.2914·{10}^{-06}{T}^3\\ & -3.1449·{10}^{-03}{T}^2+9.8094·{10}^{-01}T–1.2473·{10}^2\hfill \end{array}$

By solving the energy balance and the equilibrium line simultaneously, we have X=0.7 and T=881K; see Fig. 7E7.2. However, the conversion is only 97% of that: 0.686.

For the second bed, the gas is cooled with no progress in the conversion; see the horizontal line in Fig. 7E7.3. The energy balance uses the flow rate exiting the first bed as a starting point and assumes that there is not a large change in the flow across the bed. Note that the equilibrium is computed with respect to total conversion and therefore X is the global conversion, not the one reached at each bed. Thus, the energy balance becomes (Fig. 7E7.3):

$\begin{array}{l}{n}_{\mathrm{T},2\mathrm{ini}}·c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-X_1)·23,200\hfill \\ 0.965c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-0.686)·23,200\hfill \\ T=\frac{1}{0.965\cdot c_{p,\mathrm{mix}}}a·(X-0.686)·23,200+T_{\mathrm{o}}=(450+273)+304.1(X-0.686)\hfill \end{array}$

By solving the equilibrium and the energy balance we have X=0.91 at T=787K. Again, only 97% of the equilibrium is reached; thus, X=0.89. For the third bed, the procedure is repeated. The energy balance is as follows:

$\begin{array}{l}{n}_{T,\text{3ini}}·c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-X_2)·23,200\hfill \\ 0.956c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-0.89)·23,200\hfill \\ T=\frac{1}{0.956\cdot c_{p,\mathrm{mix}}}a·(X-0.89)\cdot 23200+T_{\mathrm{o}}=(450+273)+307(X-0.89)\hfill \end{array}$

The conversion becomes 0.97 at 739K. Thus 0.94 is the actual conversion (Fig. 7E7.4).

For the last bed we have the following:

$\begin{array}{l}{n}_{T,4\mathrm{ini}}·c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-X_3)·23,200\hfill \\ 0.955c_{p,\mathrm{mix}}(T-T_{\mathrm{o}})=a·(X-0.94)·23,200\hfill \\ T=\frac{1}{0.955·c_{p,\mathrm{mix}}}a·(X-0.94)\cdot 23,200+T_{\mathrm{o}}=(430+273)+307.4(X-0.94)\hfill \end{array}$

The computed conversion is 0.98 at 722K. The actual one is 0.95; see Fig. 7E7.5.

The summary of the mass balances for each of the beds is presented below. Note that the total molar flow is the one used as n_T,ini for each bed (Table 7E7.2).

Solution

We assume a plug flow reactor where the sulfur dioxide comes from sulfur burning. With the information provided, and a=0.11, the mass balance for the sulfur burner and the first bed is as shown in Table 7E8.1.

Kinetics of the reactor:

$F_{A_{\mathrm{o}}}\frac{dX}{dW}=-{r\prime }_{\mathrm{A}}$

If the conversion is below 5%, the reaction rate does not depend on the conversion, and is therefore given as:

$\begin{array}{l}x\le 0.05\hfill \\ -r_{{\mathrm{SO}}_2}=-r_{{\mathrm{SO}}_2}(X=0.05)\hfill \end{array}$

The reaction rate is given by:

$-{r\prime }_{{\mathrm{SO}}_2}=k\sqrt{\frac{P_{{\mathrm{SO}}_2}}{P_{{\mathrm{SO}}_3}}}\left[P_{{\mathrm{O}}_2}-{\left(\frac{P_{{\mathrm{SO}}_3}}{K_{\mathrm{p}}{P}_{{\mathrm{SO}}_2}}\right)}^2\right]$

The equilibrium constant is given as:

$K_{\mathrm{p}}=0.0031415\exp \left(\frac{42.311}{1.987·(1.8·T-273.15)+491.67}-11.24\right)$

K_p [=] Pa^−0.5, T [=] K

For the rate constant (based on Eklund’s (1956) results) we have:

$\begin{array}{cc}k& =9.86\cdot {10}^{-6}\\ & \exp \left[\frac{-176008}{((1.8\cdot T-273.15)+491.67)}-110.1\ln (((1.8\cdot T-273.15)+491.67))+912.8\right]\end{array}$

k is in mol of SO₂/kg cat S·Pa, while T is in K.

The stoichiometrics of the reaction are as follows:

$\begin{array}{l}{\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2+{\mathrm{N}}_2\leftrightarrow {\mathrm{SO}}_3+{\mathrm{N}}_2\hfill \\ F_{A_{\mathrm{o}}}{F}_{B_{\mathrm{o}}}{F}_{C_{\mathrm{o}}}\hfill \\ F_{A_{\mathrm{o}}}(1-X)F_{B_{\mathrm{o}}}-\frac{1}{2}{F}_{A_{\mathrm{o}}}X{F}_{C_{\mathrm{o}}}{F}_{A_{\mathrm{o}}}X\hfill \\ {\mathit{\Theta }}_i=\frac{F_{i_{\mathrm{o}}}}{F_{A_{\mathrm{o}}}}\hfill \\ F_{\mathrm{T}}=F_{A_{\mathrm{o}}}\left[(1-x)+{\mathit{\Theta }}_{{\mathrm{O}}_2}-\frac{1}{2}X+{\mathit{\Theta }}_{{\mathrm{N}}_2}+{\mathit{\Theta }}_{{\mathrm{SO}}_3}+X\right]=F_{A_{\mathrm{o}}}(1+{\mathit{\Theta }}_{{\mathrm{O}}_2}+{\mathit{\Theta }}_{{\mathrm{N}}_2}+{\mathit{\Theta }}_{{\mathrm{SO}}_3})+F_{A_{\mathrm{o}}}\delta X\hfill \\ \delta =-\frac{1}{2}\hfill \\ F_{\mathrm{T}}=F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X\hfill \end{array}$

Assuming ideal gases:

$\begin{array}{l}{C}_{\mathrm{T}}=\frac{F_{\mathrm{T}}}{v}=\frac{P}{RT}\hfill \\ C_{T_{\mathrm{o}}}=\frac{F_{T_{\mathrm{o}}}}{v_{\mathrm{o}}}=\frac{P_{\mathrm{o}}}{R{T}_{\mathrm{o}}}\hfill \end{array}$

Dividing both:

$v=v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{\mathrm{T}}}{F_{T_{\mathrm{o}}}}=v_o\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}{F_{T_{\mathrm{o}}}}=v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\left(1+\frac{F_{A_{\mathrm{o}}}}{F_{T_{\mathrm{o}}}}\delta X\right)$

The concentration of the species i is as follows:

$\begin{array}{l}{C}_i=\frac{F_i}{v}=\frac{F_i}{v_{\mathrm{o}}\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_{\mathrm{T}}}{F_{T_{\mathrm{o}}}}}=C_{T_{\mathrm{o}}}\frac{F_i}{F_{\mathrm{T}}}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \\ C_i=C_{T_{\mathrm{o}}}\frac{F_i}{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}=C_{T_{\mathrm{o}}}\frac{F_{A_{\mathrm{o}}}({\mathit{\Theta }}_i+v_{i}X)}{F_{T_{\mathrm{o}}}+F_{A_{\mathrm{o}}}\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \end{array}$

Dividing both:

$\begin{array}{l}{C}_i=C_{T_{\mathrm{o}}}\frac{F_{A_{\mathrm{o}}}}{F_{T_{\mathrm{o}}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\left(F_{A_{\mathrm{o}}}/F_{T_{\mathrm{o}}}\right)\delta X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}=C_{A_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}\frac{P}{P_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{T}\hfill \\ \epsilon =y_{A_{\mathrm{o}}}\delta =\left(F_{A_{\mathrm{o}}}/F_{T_{\mathrm{o}}}\right)\delta \hfill \end{array}$

Thus, the partial pressure of component i is the following:

$\begin{array}{l}{P}_i=C_{i}RT=C_{A_{\mathrm{o}}}\frac{T_{\mathrm{o}}}{P_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}R\cdot P=P_{A_{\mathrm{o}}}\frac{({\mathit{\Theta }}_i+v_{i}X)}{1+\epsilon X}\frac{P}{P_{\mathrm{o}}}\hfill \\ P_{A_{\mathrm{o}}}=C_{A_{\mathrm{o}}}R{T}_{\mathrm{o}}\hfill \end{array}$

Therefore, the kinetics of the reaction become:

$-{r\prime }_{{\mathrm{SO}}_2}=k\sqrt{\frac{P_{{\mathrm{SO}}_2}}{P_{{\mathrm{SO}}_3}}}\left[P_{{\mathrm{O}}_2}-{\left(\frac{P_{{\mathrm{SO}}_3}}{K_{\mathrm{p}}\cdot P_{{\mathrm{SO}}_2}}\right)}^2\right]$

$\frac{dx}{dW}=\frac{-{r\prime }_{{\mathrm{SO}}_2}}{F_{A_{\mathrm{o}}}}=\frac{k}{F_{A_{\mathrm{o}}}}\sqrt{\frac{P_{{\mathrm{SO}}_2,\mathrm{o}}\left({\displaystyle \frac{1-X}{1+\epsilon X}}\right)\frac{P}{P_{\mathrm{o}}}}{P_{{\mathrm{SO}}_2,\mathrm{o}}\left({\displaystyle \frac{0+X}{1+\epsilon X}}\right)\frac{P}{P_{\mathrm{o}}}}}\left[P_{{\mathrm{SO}}_2,\mathrm{o}}\left(\frac{{\mathit{\Theta }}_{{\mathrm{O}}_2}-\frac{1}{2}X}{1+\epsilon X}\right)\frac{P}{P_{\mathrm{o}}}-{\left({\displaystyle \frac{P_{{\mathrm{SO}}_2,\mathrm{o}}\left({\displaystyle \frac{0+X}{1+\epsilon X}}\right)\frac{P}{P_{\mathrm{o}}}}{K_{\mathrm{p}}{P}_{{\mathrm{SO}}_2,\mathrm{o}}\left({\displaystyle \frac{1-X}{1+\epsilon X}}\right)\frac{P}{P_{\mathrm{o}}}}}\right)}^2\right]=\frac{k}{F_{A_{\mathrm{o}}}}\sqrt{\frac{1-X}{(X)}}\left[P_{{\mathrm{SO}}_2,\mathrm{o}}(\frac{{\mathit{\Theta }}_{{\mathrm{O}}_2}-\frac{1}{2}X}{1+\epsilon X})\frac{P}{P_{\mathrm{o}}}-{\left(\frac{X}{K_{\mathrm{p}}(1-X)}\right)}^2\right]$

$\begin{array}{l}\epsilon =\delta y_{A_{\mathrm{o}}}=-\frac{1}{2}\frac{0.10}{0.1+0.11+0.79}=-0.05\hfill \\ P_{{\mathrm{SO}}_2,\mathrm{o}}=P_{\mathrm{T}}{y}_{{\mathrm{SO}}_2,\mathrm{o}}=202650\cdot 0.1=20265\mathrm{Pa}\hfill \\ {\mathit{\Theta }}_{{\mathrm{O}}_2}=\frac{0.11}{0.1}=1.1\hfill \\ {\mathit{\Theta }}_{{\mathrm{N}}_2}=\frac{0.79}{0.1}=7.9\hfill \end{array}$

F_Ao=51.42 mol/s

$\frac{dX}{dW}=\frac{-{r\prime }_{{\mathrm{SO}}_2}}{F_{A_{\mathrm{o}}}}=\frac{k}{0.00237}\sqrt{\frac{1-X}{(X)}}\left[20,265\left(\frac{1.1-\frac{1}{2}X}{1-0.05X}\right)\frac{P}{P_{\mathrm{o}}}-{\left(\frac{X}{K_{\mathrm{p}}(1-X)}\right)}^2\right]$

The energy balance is as follows:

$\begin{array}{l}\stackrel{·}{Q}-W_{\mathrm{s}}-F_{A_{\mathrm{o}}}\sum _{i=1}^n\underset{T_{i\mathrm{o}}}{\overset{T}{\int }}{\mathit{\Theta }}_i{c}_{p,i}dT-\left[\mathrm{\Delta }{H}_{\mathrm{R}}(T_{\mathrm{R}})+\underset{T_{\mathrm{R}}}{\overset{T}{\int }}\mathrm{\Delta }{c}_{p}dT\right]F_{A_{\mathrm{o}}}X=0\hfill \\ \frac{d\stackrel{·}{Q}}{dV}-F_{A_{\mathrm{o}}}\left(\sum _{i=1}^n{\mathit{\Theta }}_i{c}_{p,i}+X\mathrm{\Delta }{c}_p\right)\frac{dT}{dV}-\left[\mathrm{\Delta }{H}_{\mathrm{R}}(T_{\mathrm{R}})+\underset{T_{\mathrm{R}}}{\overset{T}{\int }}\mathrm{\Delta }{c}_{p}dT\right]F_{A_{\mathrm{o}}}\frac{dX}{dV}=0\hfill \\ -r_{\mathrm{A}}=F_{A_{\mathrm{o}}}\frac{dX}{dV}\hfill \end{array}$

Assuming adiabatic operation of the bed:

$\frac{dT}{dW}=\frac{(-r_{\mathrm{A}})\left[\mathrm{\Delta }{H}_{\mathrm{R}}(T_{\mathrm{R}})+\underset{T_{\mathrm{R}}}{\overset{T}{\int }}\mathrm{\Delta }{c}_{p}dT\right]}{F_{A_{\mathrm{o}}}\left(\sum _{i=1}^n{\mathit{\Theta }}_i{c}_{p,i}+X\mathrm{\Delta }{c}_p\right)}$

Thus the heat of reaction is computed as:

$\begin{array}{l}\mathrm{\Delta }{H}_{\mathrm{R}}(298\mathrm{K})=-98480J/\text{mol}{\mathrm{SO}}_2\hfill \\ c_{p,{\mathrm{SO}}_2}=23.852+66.989\cdot {10}^{-3}T-4.961\cdot {10}^{-5}{T}^2+13.281\cdot {10}^{-9}{T}^3\hfill \\ c_{p,{\mathrm{O}}_2}=28.106-3.680\cdot {10}^{-6}T+17.459\cdot {10}^{-6}{T}^2-1.065\cdot {10}^{-8}{T}^3\hfill \\ c_{p,{\mathrm{SO}}_3}=16.370+14.591\cdot {10}^{-2}T-1.120\cdot {10}^{-4}{T}^2+32.324\cdot {10}^{-9}{T}^3\hfill \\ c_{p,{\mathrm{N}}_2}=31.150-1.357\cdot {10}^{-2}T+26.796\cdot {10}^{-6}{T}^2-1.168\cdot {10}^{-8}{T}^3\hfill \end{array}$

where c_p is in J/mol K, and temperature (T) is in K (Sinnot, 1999):

$\begin{array}{l}\mathrm{\Delta }{H}_{\mathrm{R}}=\mathrm{\Delta }{H}_{\mathrm{R}}(T_{\mathrm{R}})+\underset{T_{\mathrm{R}}}{\overset{T}{\int }}\mathrm{\Delta }{c}_{p}dT\hfill \\ =\mathrm{\Delta }{H}_{\mathrm{R}}(T_{\mathrm{R}})+\mathrm{\Delta }\alpha (T-T_{\mathrm{R}})+\frac{\mathrm{\Delta }\beta }{2}(T^2-T_{\mathrm{R}}^2)+\frac{\mathrm{\Delta }\gamma }{3}(T^3-T_{\mathrm{R}}^3)+\frac{\mathrm{\Delta }\xi }{4}(T^4-T_{\mathrm{R}}^4)\hfill \\ \mathrm{\Delta }\alpha ={\alpha }_{{\mathrm{SO}}_3}-\frac{1}{2}{\alpha }_{{\mathrm{O}}_2}-{\alpha }_{{\mathrm{SO}}_2}=16.370-0.5(28.106)-23.852=-21.535\hfill \\ \mathrm{\Delta }\beta =0.07892\hfill \\ \mathrm{\Delta }\gamma =-7112\cdot {10}^{-5}\hfill \\ \mathrm{\Delta }\xi =24,467\cdot {10}^{-8}\hfill \\ \mathrm{\Delta }{H}_{\mathrm{R}}=-98,480-21.535(T-298)+0.0395(T^2-{298}^2)-2.371\cdot {10}^{-5}(T^3-{298}^3)\hfill \\ +6.11675\cdot {10}^{-9}(T^4-{298}^4)\hfill \\ \mathrm{\Delta }{c}_p=\mathrm{\Delta }\alpha +\mathrm{\Delta }\beta T+\mathrm{\Delta }\gamma T^2=-21.535+0.0789T-7.112\cdot {10}^{-5}{T}^2+2.447\cdot {10}^{-8}{T}^3\hfill \\ \sum {\mathit{\Theta }}_i{c}_{p,i}=300.85-0.0402T+0.00018T^2-9.071\cdot {10}^{-8}{T}^3\hfill \end{array}$

The pressure drop along the tube is given by the Ergun equation:

$\frac{dP}{dz}=-\frac{G(1-\varphi )}{\rho D_{\mathrm{p}}{\varphi }^3}\left[\frac{150(1-\varphi )\mu }{D_{\mathrm{p}}}+1.752G\right]$

G is the superficial mass velocity (kg/m² s), ϕ is the porosity of the bed, D_p is the particle diameter (m), μ is the viscosity of the gas (Pa·s), and ρ is the gas density (kg/m³) that can be found in Fogler (1997):

$\begin{array}{l}\frac{dP}{dz}=-\frac{G(1-\varphi )}{{\rho }_{\mathrm{o}}{g}_{\mathrm{c}}{D}_{\mathrm{p}}{\varphi }^3}\left[\frac{150(1-\varphi )\mu }{D_{\mathrm{p}}}+1.752G\right]\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\frac{F_T}{F_{T_{\mathrm{o}}}}\hfill \\ G=\frac{\sum _i{F}_{i\mathrm{o}}{M}_i}{A_{\mathrm{c}}}=0.433\frac{\mathrm{kg}}{{\mathrm{m}}^2\mathrm{s}}\hfill \\ {\rho }_{\mathrm{o}}=0.866{\text{kg}/m}^3\hfill \\ \mu =3.72\cdot {10}^{-5}\text{Pa}·s\hfill \\ A_{\mathrm{c}}=\pi D^{2}4\hfill \\ W={\rho }_{\mathrm{b}}{A}_{\mathrm{c}}z\frac{dP}{dW}=-\frac{G(1-\varphi )(1+\epsilon X)}{{\rho }_{\mathrm{b}}{A}_{\mathrm{c}}{\rho }_{\mathrm{o}}{g}_{\mathrm{c}}{D}_{\mathrm{p}}{\varphi }^3}\left[\frac{150(1-\varphi )\mu }{D_{\mathrm{p}}}+1.752G\right]\frac{P_{\mathrm{o}}}{P}\frac{T}{T_{\mathrm{o}}}\hfill \\ \mu \approx cte\hfill \end{array}$

To solve the problem we use MATLAB. For further notes on the use of MATLAB for solving chemical reactors we refer the reader to Martín (2014).

M file: ReactorSO₂

[a,b]=ode15s(‘ReacSI’,[0 5000],[0,750,202650]);

plot(a,b(:,1))

xlabel(‘W (kg)’)

ylabel(‘X’)

figure

plot(a,b(:,2))

xlabel(‘W (kg)’)

ylabel(‘T (K)’)

figure

plot(a,b(:,3))

xlabel(‘W (kg)’)

ylabel(‘P (Pa)’)

M file: Reac

function Reactor=ReacSI(w,x)

X=x(1);

T=x(2);

Presion=x(3);

k=9.8692e−3*exp(−176008/(1.8*(T−273.15)+491.67) −110.1*log((1.8*(T-273.15)+491.67))+912.8);

Kp=0.0031415*exp(42311/(1.987*(1.8*(T−273.15)+491.67) )−11.24);

Pto=202650;

ySO2o=0.11;

yO2o=0.1;

yN2o=0.79;

PhiO2=yO2o/ySO2o;

PSO2o=Pto*ySO2o;

Fto=467.1292;

Fao=Fto*ySO2o;

epsilon=-0.055;

G=0.433;

ph=0.45;

rhoo=0.866;

Dp=0.00457;

visc=3.72e-5;

Ac=3.14*(6.650/2)^2;

rhob=542;

To=750;

sumCp=272.77−0.0303*T+0.000158*T^2−8.016e−8*T^3;

dCp=−21.535+0.0789*T−7.112*10^(−5)*T^2+2.447e−8*T^3;

deltaHr=−98480−21.535*(T−298)+0.0395*(T^2−298^2)−2.371*10^(−5)*(T^3−298^3)+6.117*10^(−9)*(T^4−298^4);

if X<0.05;

  ra=−k*((1−0.05)/0.05)^(0.5)*(PSO2o*((PhiO2−0.5*X)/(1+epsilon*0.05))*(Presion/Pto)−(0.05/(Kp*(1−0.05)))^2);

  else

  ra=−k*((1-X)/X)^(0.5)*(PSO2o*((PhiO2−0.5*X)/(1+epsilon*X))*(Presion/Pto)−(X/(Kp*(1−X)))^2);

end

Reactor(1,1)=−ra/Fao;

Reactor(2,1)=((−ra)*(−deltaHr))/(Fao*(sumCp+X*dCp));

Reactor(3,1)=−G*(1−ph)*(1+epsilon*X)*Pto*T*(150*(1−ph)*visc/Dp+1.75*G)/(rhob*Ac*rhoo*Dp*ph^3*To*Presion);

The solution is summarized in the two figures below for the conversion and the temperature profiles. We need 1600 kg of catalyst. We see that the temperature increases 120K since it operates adiabatically (Fig. 7E8.1).

Based on the geometry of the converter and the density, we need:

$\begin{array}{l}{A}_{\mathrm{c}}=\pi D^{2}4\hfill \\ W={\rho }_{\mathrm{b}}{A}_{\mathrm{c}}z\hfill \\ z=0.09\mathrm{m}\hfill \end{array}$

Solution

The kinetics are as follows:

$-r_{\mathrm{A}}=k_2\left[{\mathrm{SO}}_3\right]\left[{\mathrm{H}}_2\mathrm{O}\right]=k_2[C_{{\mathrm{SO}}_3}-C_{{\mathrm{SO}}_3,\mathrm{o}}X][C_{{\mathrm{H}}_2\mathrm{O}}-C_{{\mathrm{SO}}_3,\mathrm{o}}X]$

However, the reaction rate is limited by diffusion:

$-r_{\mathrm{A}}=\frac{r{D}_{\mathrm{L}}}{Z_{\mathrm{L}}}(C_{{\mathrm{SO}}_3,i}-C_{{\mathrm{SO}}_3,\mathrm{L}})$

where r is the ratio between the film thickness for absorption and that for chemical reaction. According to Krevelen and Hoftyzer (Goodhead and Abowei, 2014), r is defined as:

$r=\frac{{(k_2{D}_{\mathrm{L}}{C}_{{\mathrm{H}}_2\mathrm{O},\mathrm{L}})}^{1/2}}{K_{\mathrm{L}}}$

where K_L is the mass transfer coefficient, and D_L is the SO₃ diffusivity. Therefore, the reaction rate becomes:

$-r_{\mathrm{A}}=k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3}·C_{{\mathrm{H}}_2\mathrm{O},\mathrm{L}}^{1/2}$

Based on basic reaction kinetics of a bimolecular reaction, the converted mass is computed as $C_{{\mathrm{SO}}_3}X$. Thus, the equation becomes:

$-r_{\mathrm{A}}=k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_{3,}\mathrm{o}}^{3/2}{\left(\frac{C_{{\mathrm{H}}_2\mathrm{O},\mathrm{o}}}{C_{{\mathrm{SO}}_3,\mathrm{o}}}-X\right)}^{1/2}(1-X)$

Assuming that the tower behaves as a plug flow reactor, the design equation becomes:

$-r_{\mathrm{A}}=\frac{-d{F}_{\mathrm{A}}}{dV}=F_{A_{\mathrm{o}}}\frac{dX}{dV}=k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3,o}^{3/2}{\left(\frac{C_{{\mathrm{H}}_2\mathrm{O},\mathrm{o}}}{C_{{\mathrm{SO}}_3,\mathrm{o}}}-X\right)}^{1/2}(1-X)$

Rearranging the terms:

$F_{A_{\mathrm{o}}}\frac{dX}{k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3,\mathrm{o}}^{3/2}{\left(\frac{C_{{\mathrm{H}}_2\mathrm{O},\mathrm{o}}}{C_{{\mathrm{SO}}_3,\mathrm{o}}}-X\right)}^{1/2}(1-X)}=dV$

$\frac{F_{A_{\mathrm{o}}}}{k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3,\mathrm{o}}^{3/2}}\int \frac{dX}{{\left(\frac{C_{{\mathrm{H}}_2\mathrm{O},\mathrm{o}}}{C_{{\mathrm{SO}}_3,\mathrm{o}}}-X\right)}^{1/2}(1-X)}=V$

$\begin{array}{l}\frac{F_{A_{\mathrm{o}}}}{k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3,\mathrm{o}}^{3/2}}\left[\frac{2{(m-X)}^{1/2}}{1-X}\right]=V=\frac{\pi D_{\mathrm{R}}^2{L}_{\mathrm{R}}}{4}\hfill \\ \frac{4F_{Ao}}{\pi D_{\mathrm{R}}^2·k_2^{1/2}·D_{\mathrm{L}}^{1/2}{C}_{{\mathrm{SO}}_3,\mathrm{o}}^{3/2}}\left[\frac{2{(m-X)}^{1/2}}{1-X}\right]=L_{\mathrm{R}}\hfill \end{array}$

Table 7E9.1 gives information on the parameters needed for the tower design.

Table 7E9.1

Data for the Column Operation

$C_{{\mathrm{SO}}_3,\mathrm{o}}$	15	mol/m3
k2	0.3	1/s
$F_{{\mathrm{SO}}_3,\mathrm{o}}$	4	mol/s
DR	0.1	m
m	1
DL	17	m2/s
Vo	2.3×10−4	m3/s

It is assumed that the tower operates isothermally. Thus, the energy to be removed is computed as:

$Q=(-\mathrm{\Delta }{H}_{\mathrm{r}})F_{{\mathrm{SO}}_3,\mathrm{o}}X$

Solving the problem, we have L=34.7 m.

Solution

We can determine the final composition using the integral heat of dilution, but it is easier to use an enthalpy diagram (see below). We locate the points for both solutions in the diagram. The lever rule can be used as a rough estimate to identify whether there will be water evaporation or not. We see that for a 1:1 mixture the product will be above the boiling point curve. Therefore, the energy generated is enough to boil the water.

Performing a mass balance we compute the point if no water is evaporated:

$F_1·x_1+F_2·x_2=F_m·x_m\Rightarrow x_m=\frac{1}{F_m}(F_1·x_1+F_2·x_2)=\frac{1}{2}(1·1+1·0.15)=0.575$