Question 14.13: A bed of coal (assumed to be pure carbon) in a coal gasifier...

The feed stream to the coal bed consists of 1 mol of steam and 2.38 mol of air, containing:

The species present at equilibrium are \mathrm{C}, \mathrm{H}_2, \mathrm{CO}, \mathrm{O}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CO}_2 \text {, and } \mathrm{N}_2 , and N2. The formation reactions for the compounds present are:

\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}     (I)

\mathrm{C}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}     (II)

\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2       (III)

Because the elements hydrogen, oxygen, and carbon are themselves presumed to be present in the system, this set of three independent reactions is a complete set.

All species are present as gases except carbon, which is a pure solid phase. In the equilibrium expression, Eq. (14.38),

\prod_i\left(\frac{\hat{f_i}}{f_i^0}\right)^{\nu_{i, j}}=K_j       (14.38)

the fugacity ratio of the pure carbon, is  ​​ \hat{f}_C / f_C^{\circ}=f_C / f_C^{\circ} , the fugacity of carbon at 20 bar divided by the fugacity of carbon at 1 bar. Because the effect of pressure on the fugacity of a solid is very small, negligible error is introduced by the assumption that this ratio is unity. The fugacity ratio for carbon is then \hat{f}_C / f_C^{\circ}=1 , and it is omitted from the equilibrium expression. With the assumption that the remaining species are ideal gases, Eq. (14.40) is written for the gas phase only,

\prod_i\left(y_i\right)^{\nu_{i, j}}=\left(\frac{P}{P^0}\right)^{-\nu_j} K_j     (14.40)

and it provides the following equilibrium expressions for reactions (I) through (III):

K_{\mathrm{I}}=\frac{y_{\mathrm{H}_2 \mathrm{O}}}{y_{\mathrm{O}_2}^{1 / 2} y_{\mathrm{H}_2}}\left(\frac{P}{P^{\circ}}\right)^{-1 / 2} \quad       K_{\mathrm{II}}=\frac{y_{\mathrm{CO}}}{y_{\mathrm{O}_2}^{1 / 2}}\left(\frac{P}{P^{\circ}}\right)^{-1 / 2} \quad         K_{\mathrm{III}}=\frac{y_{\mathrm{CO}_2}}{y_{\mathrm{O}_2}}

The reaction coordinates for the three reactions are designated \varepsilon_{\mathrm{I}}, \varepsilon_{\mathrm{II}} \text {, and } \varepsilon_{\mathrm{III}} \text {. }

For the initial state,

n_{\mathrm{H}_2}=n_{\mathrm{CO}}=n_{\mathrm{CO}_2}=0 \quad n_{\mathrm{H}_2 \mathrm{O}}=1 \quad n_{\mathrm{O}_2}=0.5 \quad n_{\mathrm{N}_2}=1.88

Moreover, because only the gas-phase species are considered,

\nu_{\mathrm{I}}=-\frac{1}{2} \quad \nu_{\mathrm{II}}=\frac{1}{2} \quad \nu_{\mathrm{III}}=0

Applying Eq. (14.7) to each species gives:

y_i=\frac{n_{i_0}+\sum_j \nu_{i, j} \varepsilon_j}{n_0+\sum_j \nu_j \varepsilon_j} \quad(i=1,2, \ldots, N)       (14.7)

y_{\mathrm{H}_2}=\frac{-\varepsilon_{\mathrm{I}}}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2} \quad y_{\mathrm{CO}}=\frac{\varepsilon_{\mathrm{II}}}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2}

y_{\mathrm{O}_2}=\frac{\frac{1}{2}\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}\right)-\varepsilon_{\mathrm{III}}}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2} \quad y_{\mathrm{H}_2 \mathrm{O}}=\frac{1+\varepsilon_{\mathrm{I}}}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2}

y_{\mathrm{CO}_2}=\frac{\varepsilon_{\mathrm{III}}}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2} \quad \quad y_{\mathrm{N}_2}=\frac{1.88}{3.38+\left(\varepsilon_{\mathrm{II}}-\varepsilon_{\mathrm{I}}\right) / 2}

Substitution of these expressions for y_i into the equilibrium equations yields:

K_{\mathrm{I}}=\frac{\left(1+\varepsilon_{\mathrm{I}}\right)(2 n)^{1 / 2}\left(P / P^°\right)^{-1 / 2}}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}}\right)^{1 / 2}\left(-\varepsilon_{\mathrm{I}}\right)}

K_{\mathrm{II}}=\frac{\sqrt{2} \varepsilon_{\mathrm{II}}\left(P / P^°\right)^{1 / 2}}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}}\right)^{1 / 2} n^{1 / 2}}

K_{\mathrm{III}}=\frac{2 \varepsilon_{\mathrm{III}}}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}}\right)}

where                              n \equiv 3.38+\frac{\varepsilon_{\text {II }}-\varepsilon_{\mathrm{I}}}{2}

Numerical values for the K_j calculated by Eq. (14.11) are found to be very large.

For example, at 1500 K,

\ln K_{\mathrm{I}}=\frac{-\Delta G_{\mathrm{I}}^{\circ}}{R T}=\frac{164,310}{(8.314)(1500)}=13.2 \quad K_{\mathrm{I}} \sim 10^6

\ln K_{\mathrm{II}}=\frac{-\Delta G_{\mathrm{II}}^{\circ}}{R T}=\frac{243,740}{(8.314)(1500)}=19.6 \quad K_{\mathrm{II}} \sim 10^8

\ln K_{\mathrm{III}}=\frac{-\Delta G_{\mathrm{III}}^{\circ}}{R T}=\frac{396,160}{(8.314)(1500)}=31.8 \quad K_{\mathrm{III}} \sim 10^{14}

With each K_j so large, the quantity 1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}} in the denominator of each equilibrium equation must be nearly zero. This means that the mole fraction of oxygen in the equilibrium mixture is very small. For practical purposes, no oxygen is present.

We therefore reformulate the problem by eliminating O_2 from the formation reactions. For this, Eq. (I) is combined, first with Eq. (II), and then with Eq. (III).
This provides two equations:

\mathrm{C}+\mathrm{CO}_2 \rightarrow 2 \mathrm{CO}     (a)

\mathrm{H}_2 \mathrm{O}+\mathrm{C} \rightarrow \mathrm{H}_2+\mathrm{CO}     (b)

The corresponding equilibrium equations are:

K_a=\frac{y_{\mathrm{CO}}^2}{y_{\mathrm{CO}_2}}\left(\frac{P}{P^{\circ}}\right) \quad       K_b=\frac{y_{\mathrm{H}_2} y_{\mathrm{CO}}}{y_{\mathrm{H}_2 \mathrm{O}}}\left(\frac{P}{P^{\circ}}\right)

The input stream is specified to contain 1 mol H_2, 0.5 mol O_2, and 1.88 mol N_2. Because O_2 has been eliminated from the set of reaction equations, we replace the 0.5 mol of O_2 in the feed by 0.5 mol of CO_2. The presumption is that this amount of CO_2 has been formed by prior reaction of the 0.5 mol O_2 with carbon. In this problem, the quantity of carbon or its ratio to other elements is not constrained, so this introduces no new complications. Thus the equivalent feed stream contains 1 mol H2, 0.5 mol CO_2, and 1.88 mol N_2, and the application of Eq. (14.7) to Eqs. (a) and (b) gives:

y_{\mathrm{H}_2}=\frac{\varepsilon_b}{3.38+\varepsilon_a+\varepsilon_b} \quad            y_{\mathrm{CO}}=\frac{2 \varepsilon_a+\varepsilon_b}{3.38+\varepsilon_a+\varepsilon_b}

y_{\mathrm{H}_2 \mathrm{O}}=\frac{1-\varepsilon_b}{3.38+\varepsilon_a+\varepsilon_b} \quad         y_{\mathrm{CO}_2}=\frac{0.5-\varepsilon_a}{3.38+\varepsilon_a+\varepsilon_b}

y_{\mathrm{N}_2}=\frac{1.88}{3.38+\varepsilon_a+\varepsilon_b}

Because values of y_i must lie between zero and unity, the two expressions on the left and the two on the right show that:

0 \leq \varepsilon_b \leq 1 \quad-0.5 \leq \varepsilon_a \leq 0.5

Combining the expressions for the yi with the equilibrium equations gives:

K_a=\frac{\left(2 \varepsilon_a+\varepsilon_b\right)^2}{(0.5-\varepsilon)\left(3.38+\varepsilon_a+\varepsilon_b\right)}\left(\frac{P}{P^0}\right)     (A)

K_b=\frac{\varepsilon_b\left(2 \varepsilon_a+\varepsilon_b\right)}{\left(1-\varepsilon_b\right)\left(3.38+\varepsilon_a+\varepsilon_b\right)}\left(\frac{P}{P^0}\right)       (B)

For reaction (a) at 1000 K,

\Delta G_{i 000}^{\circ}=2(-200,240)-(-395,790)=-4690 \mathrm{~J} \cdot \mathrm{mol}^{-1}

and by Eq. (14.11),

\ln K_a=\frac{4690}{(8.314)(1000)}=0.5641 \quad K_a=1.758

Similarly, for reaction (b),

\Delta G_{1000}^{\circ}=(-200,240)-(-192,420)=-7820 \mathrm{~J} \cdot \mathrm{mol}^{-1}

and

\ln K_b=\frac{7820}{(8.314)(1000)}=0.9406 \quad K_b=2.561

Equations (A) and (B) with these values for K_a \text { and } K_b and with (P∕P°) = 20 constitute two nonlinear equations in unknowns \varepsilon_a \text { and } \varepsilon_b . An ad hoc iteration scheme can be devised for their solution, but Newton’s method for solving an array of nonlinear algebraic equations is attractive. It is described and applied to this example in App. H. The results of calculations for all temperatures are shown in the following table.

Values for the mole fractions yi of the species in the equilibrium mixture are calculated by the equations already given. The results of all such calculations appear in the following table and are shown graphically in Fig. 14.6.

At the higher temperatures the values of \varepsilon_a \text { and } \varepsilon_b are approaching their upper limiting values of 0.5 and 1.0, indicating that reactions (a) and (b) are proceeding nearly to completion. In this limit, which is approached even more closely at still higher temperatures, the mole fractions of CO_2  and  H_2O approach zero, and for the product species,

y_{\mathrm{H}_2}=\frac{1}{3.38+0.5+1.0}=0.205

y_{\mathrm{CO}}=\frac{1+1}{3.38+0.5+1.0}=0.410

y_{\mathrm{N}_2}=\frac{1.88}{3.38+0.5+1.0}=0.385

We have here assumed a sufficient depth of coal so that the gases reach equilibrium while in contact with incandescent carbon. This need not be the case; if oxygen and steam are supplied at too high a rate, the reactions may not attain equilibrium or may reach equilibrium after they have left the coal bed. In this event, carbon is not present at equilibrium, and the problem must again be reformulated.