Heatup path equilibrium

This value becomes apparent when the equilibrium curves of this chapter are combined with the approach-to-equilibrium heatup paths of Chapters 11 and 12. 

Fig. 7.5 also shows that S02 oxidation and gas heat up end when the heatup path reaches Fig. 7.4 s equilibrium % S02 oxidized curve. 

Fig. 7.5. Heatup path for gas descending the Fig. 7.1 catalyst bed. It begins at the feed gas s input temperature and 0% SO2 oxidized. Its temperature rises as S02 oxidizes. Maximum attainable S02 oxidation is predicted by the heatup path-equilibrium curve intercept, 69% oxidized at 893 K in this case. This low % SO> oxidized confirms that efficient SO-> oxidation cannot be obtained in a single catalyst bed. Multiple catalyst beds with gas cooling between must be used.

Fig. 11.1. Heatup path for S02, 02, N2 gas descending a catalyst bed. The S02 and 02 in feed gas react to form S03, Eqn. (1.1). The gas is heated by the exothermic heat of reaction. The result is a path with increasing % S02 oxidized and increasing gas temperature. Notice how the feed gas s heatup path approaches its Chapter 10 equilibrium curve.

Fig. 11.7. Heatup paths and equilibrium curve for 10 volume% SO2, 11 volume% 02, 79 volume% N2 feed gas. Notably, the 660 K heatup path will reach the equilibrium curve at a higher % S02 oxidized value than the 690 K heatup path. 660 K is about the lowest feed gas temperature that will keep V, alkali metal, S, 0, Si02 catalyst active and S02 oxidation rapid, Table 8.1.

Chapters 12 onwards combine these heatup paths with Chapter 10 s % SO2 oxidized equilibrium curves to show how ... 

For an intercept calculation to be valid, its heatup path and equilibrium curve must be for the same feed gas. Each intercept calculation must, therefore, specify a feed gas... 

Table 12.1 shows heatup path and equilibrium curve % S02 oxidized-temperature points near a heatup path-equilibrium curve intercept. They are for ... 

Temperature, K Heatup path % S02 oxidized, Equilibrium % S02 oxidized, E... 

This indicates that at 893 K and below, S02 + /202 —> S03 oxidation can proceed further up the heatup path towards equilibrium, Fig. 12.1. 

At 894 K and above, however, heatup path % S02 oxidized is greater than equilibrium % S02 oxidized. This is, of course, impossible because equilibrium % S02 oxidized cannot be exceeded up a heatup path. 

Fig. 12.1. Plot of Table 12.1 heatup path points and equilibrium curve, expanded from Fig. 11.7. Below the equilibrium curve, S02 is being oxidized, gas temperature is increasing and equilibrium is being approached up the heatup path. Maximum (equilibrium) oxidation is attained where the heatup path meets the equilibrium curve.

Fig. 12.3. Heatup paths, equilibrium curves and intercepts for 7, 10, and 13 volume% S02 feed gas. Volume% 02/volume% S02 ratio =1.1. Intercept temperature increases with increasing S02 strength. Intercept % S02 oxidized decreases with increasing S02 strength. The intercepts have been calculated as described in Appendix J.

These pressure differences have no effect on heatup paths, Fig. 12.4 - and little effect on equilibrium curves and intercepts. Intercept temperature and % S02 oxidized both increase slightly with increasing pressure. 

C02 has no effect on S02 + /202 — S03 equilibrium curves, Appendix F. It does, however, have a small effect on heatup paths and intercepts, Fig. 12.6. 

Fig. 12.7. Equilibrium curves, heatup paths and intercepts for 12 volume% S02 feed gas. 690 K feed gas gives a 915 K intercept temperature, in the catalyst degradation range. 660 K feed gas gives a 900 K intercept temperature, avoiding degradation.

This chapter assumes that equilibrium is attained in an acid plant s 1st catalyst bed, i.e. that a feed gas s heatup path always intercepts its equilibrium curve. 

Fig. 12.8. Heatup path, equilibrium curve and intercept for 660 K, 13 volume% S02 feed gas. With 13 volume% S02 and higher, catalyst degradation is likely even with 660 K feed gas. 660 K is about the lowest temperature at which V, alkali metal, S, 0, Si02 catalyst is fully active.

Catalyst bed S02 + /202 — S03 oxidation is represented by heatup paths and equilibrium curves. Maximum S02 oxidation occurs where a feed gas s heatup path intercepts its equilibrium curve. 

Now determine the %S02 oxidized - temperature point at which the Problem 11.4 heatup path intercepts the Problem 10.4 equilibrium curve. 

This suggests that you should calculate your equilibrium curve and heatup path points at -904,905.911 K. 

Fig. 13.2. 1st catalyst bed heatup path, equilibrium curve and intercept point, from Fig. 12.1. The 1st catalyst bed s exit gas is its intercept gas, Section 12.12. It is cooled and fed to a 2nd catalyst bed for more S02 oxidation.

Table 15.1. 2nd catalyst bed % S02 ox/d/zed/temperature points near heatup path-equilibrium curve intercept. They have been calculated as described in Appendices K and D. 

Interpolation suggests that Table 15.1 s heatup path intercepts its equilibrium curve at ... 

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