A bed of coal (assumed to be pure carbon) in a coal gasifier is fed with steam and air and produces a gas stream containing H2, CO, O2, H2O, CO2, and N2. If the feed to the gasifier consists of 1 mol of steam and 2.38 mol of air, calculate the equilibrium composition of the gas stream at P = 20

Question

A bed of coal (assumed to be pure carbon) in a coal gasifier is fed with steam and air and produces a gas stream containing [latex] \mathrm{H}_2, \mathrm{CO}, \mathrm{O}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CO}_2 ext {, and } \mathrm{N}_2 [/latex]. If the feed to the gasifier consists of 1 mol of steam and 2.38 mol of air, calculate the equilibrium composition of the gas stream at P = 20 bar for temperatures of 1000, 1100, 1200, 1300, 1400, and 1500 K. Available data are listed in the following table.

T/ K	[latex] \Delta G_f^{\circ} / \mathrm{J} \cdot \mathrm{mol}^{-1} [/latex]
T/ K	[latex]H_2O[/latex]	CO	[/latex]CO_2[/latex]
1000	−192,420	−200,240	−395,790
1100	−187,000	−209,110	−395,960
1200	−181,380	−217,830	−396,020
1300	−175,720	−226,530	−396,080
1400	−170,020	−235,130	−396,130
1500	−164,310	−243,740	−396,160

Accepted Answer

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

[latex] \mathrm{O}_2:(0.21)(2.38)=0.5 \mathrm{~mol} \quad \mathrm{~N}_2:(0.79)(2.38)=1.88 \mathrm{~mol} [/latex]

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

[latex] \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 ightarrow \mathrm{H}_2 \mathrm{O} [/latex] (I)

[latex] \mathrm{C}+\frac{1}{2} \mathrm{O}_2 ightarrow \mathrm{CO} [/latex] (II)

[latex] \mathrm{C}+\mathrm{O}_2 ightarrow \mathrm{CO}_2 [/latex] (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),

[latex] \prod_i\left(\frac{\hat{f_i}}{f_i^0} ight)^{ u_{i, j}}=K_j [/latex] (14.38)

the fugacity ratio of the pure carbon, is  [latex] \hat{f}_C / f_C^{\circ}=f_C / f_C^{\circ} [/latex] , 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 [latex] \hat{f}_C / f_C^{\circ}=1 [/latex] , 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,

[latex] \prod_i\left(y_i ight)^{ u_{i, j}}=\left(\frac{P}{P^0} ight)^{- u_j} K_j [/latex] (14.40)

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

[latex] 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}} ight)^{-1 / 2} \quad K_{\mathrm{II}}=\frac{y_{\mathrm{CO}}}{y_{\mathrm{O}_2}^{1 / 2}}\left(\frac{P}{P^{\circ}} ight)^{-1 / 2} \quad K_{\mathrm{III}}=\frac{y_{\mathrm{CO}_2}}{y_{\mathrm{O}_2}} [/latex]

The reaction coordinates for the three reactions are designated [latex] \varepsilon_{\mathrm{I}}, \varepsilon_{\mathrm{II}} ext {, and } \varepsilon_{\mathrm{III}} ext {. } [/latex]

For the initial state,

[latex] 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 [/latex]

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

[latex] u_{\mathrm{I}}=-\frac{1}{2} \quad u_{\mathrm{II}}=\frac{1}{2} \quad u_{\mathrm{III}}=0 [/latex]

Applying Eq. (14.7) to each species gives:

[latex] y_i=\frac{n_{i_0}+\sum_j u_{i, j} \varepsilon_j}{n_0+\sum_j u_j \varepsilon_j} \quad(i=1,2, \ldots, N) [/latex] (14.7)

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

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

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

Substitution of these expressions for [latex]y_i[/latex] into the equilibrium equations yields:

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

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

[latex] K_{\mathrm{III}}=\frac{2 \varepsilon_{\mathrm{III}}}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}} ight)} [/latex]

where [latex] n \equiv 3.38+\frac{\varepsilon_{ ext {II }}-\varepsilon_{\mathrm{I}}}{2} [/latex]

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

For example, at 1500 K,

[latex] \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 [/latex]

[latex] \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 [/latex]

[latex] \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} [/latex]

With each [latex]K_j [/latex] so large, the quantity [latex] 1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}-2 \varepsilon_{\mathrm{III}} [/latex] 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 [latex]O_2 [/latex] from the formation reactions. For this, Eq. (I) is combined, first with Eq. (II), and then with Eq. (III).
This provides two equations:

[latex] \mathrm{C}+\mathrm{CO}_2 ightarrow 2 \mathrm{CO} [/latex] (a)

[latex] \mathrm{H}_2 \mathrm{O}+\mathrm{C} ightarrow \mathrm{H}_2+\mathrm{CO} [/latex] (b)

The corresponding equilibrium equations are:

[latex] K_a=\frac{y_{\mathrm{CO}}^2}{y_{\mathrm{CO}_2}}\left(\frac{P}{P^{\circ}} ight) \quad K_b=\frac{y_{\mathrm{H}_2} y_{\mathrm{CO}}}{y_{\mathrm{H}_2 \mathrm{O}}}\left(\frac{P}{P^{\circ}} ight) [/latex]

The input stream is specified to contain 1 mol [latex]H_2[/latex], 0.5 mol [latex]O_2[/latex], and 1.88 mol [latex]N_2[/latex]. Because [latex]O_2[/latex] has been eliminated from the set of reaction equations, we replace the 0.5 mol of [latex]O_2[/latex] in the feed by 0.5 mol of [latex]CO_2[/latex]. The presumption is that this amount of [latex]CO_2[/latex] has been formed by prior reaction of the 0.5 mol [latex]O_2[/latex] 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 [latex]CO_2[/latex], and 1.88 mol [latex]N_2[/latex], and the application of Eq. (14.7) to Eqs. (a) and (b) gives:

[latex] 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} [/latex]

[latex] 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} [/latex]

[latex] y_{\mathrm{N}_2}=\frac{1.88}{3.38+\varepsilon_a+\varepsilon_b} [/latex]

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

[latex] 0 \leq \varepsilon_b \leq 1 \quad-0.5 \leq \varepsilon_a \leq 0.5 [/latex]

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

[latex] K_a=\frac{\left(2 \varepsilon_a+\varepsilon_b ight)^2}{(0.5-\varepsilon)\left(3.38+\varepsilon_a+\varepsilon_b ight)}\left(\frac{P}{P^0} ight) [/latex] (A)

[latex] K_b=\frac{\varepsilon_b\left(2 \varepsilon_a+\varepsilon_b ight)}{\left(1-\varepsilon_b ight)\left(3.38+\varepsilon_a+\varepsilon_b ight)}\left(\frac{P}{P^0} ight) [/latex] (B)

For reaction (a) at 1000 K,

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

and by Eq. (14.11),

[latex] \ln K_a=\frac{4690}{(8.314)(1000)}=0.5641 \quad K_a=1.758 [/latex]

Similarly, for reaction (b),

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

and

[latex] \ln K_b=\frac{7820}{(8.314)(1000)}=0.9406 \quad K_b=2.561 [/latex]

Equations (A) and (B) with these values for [latex] K_a ext { and } K_b [/latex] and with (P∕P°) = 20 constitute two nonlinear equations in unknowns [latex] \varepsilon_a ext { and } \varepsilon_b [/latex]. 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.

T/K	[latex]K_a[/latex]	[latex]K_b[/latex]	[latex]ε_a[/latex]	[latex]ε_b[/latex]
1000	1.758	2.561	−0.0506	0.5336
1100	11.405	11.219	0.1210	0.7124
1200	53.155	38.609	0.3168	0.8551
1300	194.430	110.064	0.4301	0.9357
1400	584.85	268.76	0.4739	0.9713
`1500	1514.12	583.58	0.4896	0.9863

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.

T/K	[latex]yH_2[/latex]	yCO	[latex]yH_2O[/latex]	[latex]yCO_2[/latex]	[latex]yN_2[/latex]
1000	0.138	0.112	0.121	0.143	0.486
1100	0.169	0.226	0.068	0.090	0.447
1200	0.188	0.327	0.032	0.040	0.413
1300	0.197	0.378	0.014	0.015	0.396
1400	0.201	0.398	0.006	0.005	0.390
1500	0.203	0.405	0.003	0.002	0.387

At the higher temperatures the values of [latex] \varepsilon_a ext { and } \varepsilon_b [/latex] 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 [latex] CO_2 and H_2O[/latex] approach zero, and for the product species,

[latex] y_{\mathrm{H}_2}=\frac{1}{3.38+0.5+1.0}=0.205 [/latex]

[latex] y_{\mathrm{CO}}=\frac{1+1}{3.38+0.5+1.0}=0.410 [/latex]

[latex] y_{\mathrm{N}_2}=\frac{1.88}{3.38+0.5+1.0}=0.385 [/latex]

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.

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

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