Recommended
More Related Content
What's hot
What's hot (15)
Viewers also liked
Viewers also liked (18)
Similar to Elizabeth Towle Coked Catalyst Regeneration Project
Similar to Elizabeth Towle Coked Catalyst Regeneration Project (20)
Elizabeth Towle Coked Catalyst Regeneration Project
- 1. Coked Catalyst Regeneration Project A Project Report for ChE 2013 submitted to the Faculty of the Department of Chemical Engineering Worcester Polytechnic Institute Worcester, MA 01609 March 4, 2016 The Not Coal Group “What’s that rock? It’s coal.” _____________________ Benjamin Drury _____________________ Weiran Gao _____________________ Natalie Thompson _____________________ Elizabeth Towle
- 2. To: Dr. M. Timko From: B. Drury, W. Gao, N. Thompson, E. Towle Subject: Reaction Equilibria Project Introduction The ThermoChem Company runs a process to catalytically crack hydrocarbons, but must periodically shut down their cracking process and pass H2 gas through a reactor to regenerate a catalyst that has been “coked” by carbon. This process begins when the catalyst has been coked with carbon weighing 10% of the catalyst. Hydrogen gas is passed over the carbon at 1000 K, 1.1 bar for nine hours with the expectation of recovering 95% of the carbon. It is the mission of this report to verify that after the given nine hours that no carbon remains coked upon the catalyst. This report also endeavors to calculate the viability of recycling H2 while recovering methane. Further, at the suggestion of our esteemed college Clem Chem, we have also analyzed a process using steam instead of pure hydrogen to recover CH4 from the coked carbon. These various processes will be evaluated for their viability. Methodology We considered a method involving the conversion of carbon to methane. Carbon recovery occurs when 105,000 lbs of catalyst have been coked with 10,500 lbs of carbon, which is equivalent to 867 lbmol. Hydrogen is passed through at 1000 lbmol/hr of pure hydrogen gas. Further, we considered the recovery of hydrogen and methane gas from the conversion. The total feed was still 1000 lbmol/hr, although methane was returned to the feed at no more than 4 mol%. This percentage was maintained by a purge on the recycle stream. We also considered the effect of using steam instead of hydrogen gas and what changes that would bring upon the overall conversion and recovery. This process produces hydrogen, carbon monoxide, and carbon dioxide, in addition to methane. It will be evaluated for its time and cost efficiency compared to the use of hydrogen, as well as its yield of methane and the health and safety issues it creates. The process was assumed to be 100% efficient and to reach equilibrium in the reactor. Further, it was assumed that neither hydrogen nor any other gas was dissolved into the liquid methane. Especially at this low pressure, this is not likely an oversimplification, and there is very low solubility of hydrogen in methane, and should not significantly affect the results. As the pressure was only 1.1 bar, it was assumed all gases behave ideally. MathCad was used for necessary calculations.
- 3. Results and Discussion Hydrogen Feed without Recycle It was calculated that it would take 10.452 hours to remove 95% of the carbon from the reactor. The details of these calculations can be found in Appendices 1 and 2. This data proves that the ThermoChem Company has not been shutting down the cracking process long enough to preserve the reduced state of the carbon. One possibility for this discrepancy between the expected time that ThermoChem was using and the calculated time we found, is that we used the Van’t Hoff equation and assumed ΔH is a function of T while ThermoChem might have assumed ΔH is constant. Because we assumed ΔH is a function of T and used mathcad, our calculated values can be adapted to account for a different flow rate, pressure, and temperature. One possible source of error could be failing to take into consideration the poynting factor. With such a low pressure, we assumed that it would be one but no system is ever ideal so this assumption could allow for a small margin of error.
- 4. Hydrogen Feed with Recycle By adding a recycle, the time to remove 95% of the carbon increases to 19.631 hours, over twice the original allotted time of 9 hours. However, it reduces the amount of hydrogen necessary to 273 lbmol/hr from 1000 lbmol/hour. This will reduce the costs of H2, from about 105 lbmol over 10 hours to 5400 lbmol over 19 hours, but the additional 9 hours needed every time carbon is to be recovered will detract from the productivity of the plant. One reason this takes so long might be that the recycle stream contains a bit of the product, CH4 which is not advantageous for the equilibrium reaction to shift to the right. This could be addressed by purging more methane from the stream, although this would also reduce the amount of H2 recycled. It can be generalized that any amount of recycle will decrease the H2 costs but increase the required time. The potential sources of error include potentially flawed assumptions. The process was assumed to be 100% efficient and to reach equilibrium in the reactor. Further, it was assumed that no hydrogen was dissolved into the liquid methane. Especially at this low pressure, this is not likely an oversimplification, and there is very low solubility of hydrogen in methane, and should not significantly affect the results. As the pressure was only 1.1 bar, it was assumed all gases behave ideally.
- 5. Steam Feed, No Recycle If 1000 lbmol/hr of steam is used instead of hydrogen, the process of removing 95% carbon from the catalyst takes only 1.062 hours. This would reduce the downtime of the plant, as well as significantly reducing the cost of materials. steam is much cheaper than hydrogen, and 1088 lbmol of steam is much more by weight than the amount of hydrogen needed to run the recycle, and significantly less than the current practice uses. These results can be extended to generalize that using steam instead of hydrogen will be more efficient to remove carbon from a catalyst. This efficiency stems from multiple reactions happening at the same time, pushing the equilibrium in a favorable direction. This practice is extremely time and cost efficient, but does however create exhausts like CO2 and CO. As will be addressed more later, these gases have toxic effects on both humans and the environment, and all necessary precautions should be taken. The potential sources of error include potentially flawed assumptions. The process was assumed to be 100% efficient and to reach equilibrium in the reactor. As the pressure was only 1.1 bar, it was assumed all gases behave ideally. We are confident in these results, and they are very reliable.
- 6. Health and safety The chemicals used in our processes are thankfully not incredibly harmful. Carbon should not be ingested and hydrogen and methane shouldn’t be allowed to reach more than 10% concentration in the volume they are contained in, as if oxygen contaminates the feed there is a possibility of fire and explosions. Carbon monoxide is toxic and has the ability to form potentially explosive chemicals when introduced into the atmosphere. Carbon dioxide can cause asphyxiation if concentrations greater than or equal to 10% are inhaled. In regards to the amount of carbon dioxide being created, it is recommended that this company capture, recycle, or convert the carbon dioxide into a more environmentally friendly compound before being released into the atmosphere.
- 7. Conclusion and Recommendations Upon completion of our studious research and calculations we determined that the ThermoChem Company has not been passing hydrogen gas through the reactor long enough for all of the carbon to be converted to methane. Further, we would like to recommend Clem Chem’s proposed idea of using steam instead of hydrogen, as it will reduce downtime and material costs. With the necessary health and safety precautions, this can be a significantly more economically viable option, increasing production by the ThermoChem Company. Without equations for the costs of time, supplies, and waste removal, it is impossible to say with absolute certainty that this will be the most cost effective method, but we are nevertheless confident in our analysis that using steam instead of hydrogen for just 1.088 hours will be the best way to remove carbon from the catalyst.
- 8. References MSDS C "Material Safety Data Sheet." Actp-12-18-2013. Cleartech, 19 Dec. 2013. Web. 2 Mar. 2016. <http://www.dynamicaqua.com/msds/activatedcarbon.pdf>. MSDS H2 "Material Safety Data Sheet." Hydrogen. Air Products, June 1994. Web. 2 Mar. 2016. <http://avogadro.chem.iastate.edu/MSDS/hydrogen.pdf>. MSDS CH4 "Material Safety Data Sheet." Methane. Air Products, July 1999. Web. 3 Mar. 2016. <http://avogadro.chem.iastate.edu/msds/methane.pdf>. MSDS H2O "Material Safety Data Sheet Water MSDS." Msds.php. Science Lab.com, 21 May 2013. Web. 3 Mar. 2016. <http://www.sciencelab.com/msds.php?msdsId=9927321>. MSDS CO "Safety Data Sheet Carbon Monoxide." 001014.pdf. Airgas, 12 May 2015. Web. 3 Mar. 2016. <https://www.airgas.com/msds/001014.pdf>. MSDS CO2 "Material Safety Data Sheet." Carbon Dioxide. Air Products, Sept. 1998. Web. 3 Mar. 2016. <http://avogadro.chem.iastate.edu/MSDS/carbon_dioxide.pdf>.
- 9. Appendices Appendix 1: Calculation of a set of independent reactions H2+½ O2 H2O H2+½ O2 H2O = 0 [1]→ − C+½ O2 CO C+½ O2 CO = 0 [2]→ − C+O2 CO2 C+O2 CO2 = 0 [3]→ − C+2H2 CH4 C+2H2 CH4 = 0 [4] [3’]→ − [1] - [2] H2+½ O2 H2O - (C+½ O2 CO) = 0− − H2 H2O - C+CO = 0 [1’]− [3]- 2*[2] C+O2 CO2 - 2*(C+½ O2 CO) = 0− − -C CO2 + 2CO = 0 [2’]− [1’], [2’], [3’] give (1) H2O + C H2+CO⇋ (2) 2CO C+CO2⇋ (3) C+2H2 CH4⇋
- 10. Appendix 2: Mathcad calculations for Parts 1 and 2 Part 1 C (s) + 2H2 (g) CH4 (g)⇋ I 10500/12.01 1000 0 C - ξ -2 ξ ξ F 10500/12.01 - ξ 1000 -2 ξ ξ Total 1000-ξ yi (1000 -2 ξ)/(1000-ξ) ξ/(1000-ξ) KT = (yCH4*P)/(yH2 2 * P2 ) P*KT =[ ξ*(1000-ξ)]/(1000 -2 ξ)2 Part 2 C (s) + 2H2 (g) CH4 (g)⇋ I 10500/12.01 960 40 C - ξ -2 ξ ξ F 10500/12.01 - ξ 960 -2 ξ 40+ξ Total 1000-ξ yi (960 -2 ξ)/(1000-ξ) (40+ξ)/(1000-ξ) KT = (yCH4*P)/(yH2 2 * P2 ) P*KT =[(40+ξ)*(1000-ξ)]/(960 -2 ξ)2
- 12. Appendix 3: Mass balance for Part 1 H2 (1), CH4 (2), C (3) Finding M2 M2 = (1000 - 2ξ1) + (ξ1) M2 = (1000 - 2(78.805)) + (78.805) M2 = 921.2 lbmol/hr Finding x1 x1 = mH2/M2 = (1000 - 2ξ1)/(921.2) x1 = 0.91 Finding x2 x2 = 1 - n1 x2 = 0.09
- 13. Appendix 4: Mass Balance for Part 2 H2 (1), CH4 (2), C (3) x1,2 = 0.96, x2,2 = 0.04, M2 = 1000 lbmol/hr Finding M3 M3x1,3 = M2x1,2 - 2ξ1 = (960) - (2(41.957)) = 876 lbmol/hr M3x2,3 = M2x2,2 + ξ1 = 40 + 41.957= 82 lbmol/hr M3 = M3x2,3 + M3x1,3 = 958 lbmol/hr Finding x1,3 x1,3 = M3x1,3/M3 = 876/958 = 0.914 Finding x2,3 x2,3 = M3x2,3/M3 = 82/958 = 0.086 Finding PCH4 sat ln(PCH4 sat (mmHg)) = 15.2243- = 15.2243- = 4.11697.84/(T(K) .16)8 − 7 97.84/(88K .16)8 − 7 PCH4 sat = exp(4.116) = 61.43 mmHg /(750mmHg/bar) = 0.0819 bar Finding x2,5 Assume x2 →1, therefore ɣ2→1 P * y2 = x2*ɣ2*PCH4 sat 1.5 bar *y2 = 1*1*0.0819 bar y2 = x2,5 = 0.055 Finding x1,5 x1,5 = 1 - x2,5 = 1 - 0.055 = 0.945 Finding M5 M5*x1,5 = M3*x1,3 M5*0.945 = 958*0.914 M5 = 927 lbmol/hr M4 = M3 - M5 = 31 lbmol/hr x1,5 = x1,6 = x1,7 x2,5 = x2,6 = x2,7 x2,7 * M7 = 40 lbmol/hr (design specification such that methanol is 4% of M2) 0.055*M7 = 40; M7 = 727 lbmol/hr M5 - M7 = M6 = 200 lbmol/hr M2 - M7 = M1 = 273 lbmol/hr
- 14. Appendix 5: Mathcad calculations for Part 3 H2O + C H2+CO ξ1⇋ K1T= [(yH2*P)*(yCO*P)]/(yH2*P) 2CO C+CO2 ξ2⇋ K2T= (yCO2*P)/(yCO*P)2 C+2H2 CH4 ξ3⇋ K3T = (yCH4*P)/(yH2 2 * P2 ) H2O C H2 CO CO2 CH4 I 1000 10500/12.01 0 0 0 0 C - ξ1 - ξ1+ξ2- ξ3 ξ1-2ξ3 ξ1-2 ξ2 ξ2 ξ3 F 1000 - ξ1 ξ1-2ξ3 ξ1-2 ξ2 ξ2 ξ3 Total 1000-ξ1+ ξ1-2ξ3+ ξ1-2 ξ2+ ξ2 + ξ3=1000+ξ1-ξ2-ξ3 yi (1000 - ξ1)/(Total) (ξ1-2ξ3)/(Total) (ξ1-2ξ2)/(Total) ξ2/(Total) ξ3/(Total) (K1T/P)= [(ξ1-2ξ3)*(ξ1-2ξ2)]/[(1000 - ξ1)*(1000+ξ1-ξ2-ξ3)] (K2T*P)= [(ξ2)*(1000+ξ1-ξ2-ξ3)]/( ξ1-2 ξ2 )2 (K3T*P)= [(ξ3)*(1000+ξ1-ξ2-ξ3)]/( ξ1-2 ξ3 )2
- 17. Appendix 6: Mass Balance for Part 3 H2 (1), CH4 (2), CO (3), CO2 (4), H2O (5) M1 = 1000 lbmol/hr, x1,1 = 1 Finding M2 M2x1,2 = ξ1 - 2ξ3 = (873.415) - 2(38.238) M2x1,2 = 796.939 lbmol/hr M2x2,2 = ξ3 = 38.238 lbmol/hr M2x3,2 = ξ1 - 2ξ2 = (873.415) - 2(136.024) M2x3,2 = 601.367 lbmol/hr M2x4,2 = ξ2 = 136.024 lbmol/hr M2x5,2 = 1000 - ξ1 = 1000 - 873.415 M2x5,2 = 126.585 lbmol/hr M2 = M2x1,2 + M2x2,2 + M2x3,2 + M2x4,2 + M2x5,2 M2 = 1699.153 lbmol/hr Finding x1,2 x1,2 = M2x1,2/M2 x1,2 = 0.469 Finding x2,2 x2,2 = M2x2,2/M2 x2,2 = 0.023 Finding x3,2 x3,2 = M2x3,2/M2 x3,2 = 0.354 Finding x4,2 x4,2 = M2x4,2/M2 x4,2 = 0.080 Finding x5,2 x5,2 = M2x5,2/M2 x5,2 = 0.074