1. Fuel Cell Design Chemical Engineering Senior Design Spring 2005 UTC

2. Technical and Economic Aspects of a 25 kW Fuel Cell Chris Boudreaux Wayne Johnson Nick Reinhardt

3. Technical and Economic Aspects of a 25 kW Fuel Cell Investigate the design of --a 25 kW Fuel Cell --Coproduce Hydrogen --Grid parallel --Solid Oxide Electrolyte Chemical and Thermodynamic Aspects Our Competence Not Our Competence

4. Outline Introduction to the project Process Description Process & Equip. Design Economic Analysis

5. Introduction Overall Reaction Methane + Air --> Electricity + Hydrogen + Heat + CO 2

6. Introduction Pressure Swing Adsorption Fuel Cell Reformer Gas Hydrogen Electricity Air Heat SynGas POR Water Exhaust

7. Fuel Cell-Chemistry SynGas Air O-O- O-O- H2H2 H2OH2O CO CO 2 POR O 2 N 2 “Air” Solid Oxide Electrolyte is porous to O - H2H2 + CO

8. Fuel Cell-Electricity SynGas Air O-O- O-O- H2H2 H2OH2O CO CO 2 POR O 2 N 2 “Air” Electrons Load

9. Fuel Cell-Challenges SynGas Air O-O- O-O- H2H2 H2OH2O CO CO 2 POR O 2 N 2 “Air” H2H2 + CO Hot SynGas Hot Air Recover H 2 Recover Heat

10. Process Description Turn it over to Nick Reinhardt

11. Process Description Fuel Preparation Air Preparation Post Processing Fuel Cell

12. Fuel Preparation Air Preparation Post Processing Fuel Cell Fuel Preparation - 100

13. Fuel Preparation Air Preparation Post Processing Fuel Cell Air Preparation - 200

14. Fuel Preparation Air Preparation Post Processing Fuel Cell Fuel Cell - 300

15. Fuel Preparation Air Preparation Post Processing Fuel Cell Post Processing - 400

16. Process and Equipment Design Turn it over to Chris Boudreaux

17. Pure Natural Gas 25°C 0.33 kmol/hr CH 4 = 100% Sulfur Purge 25°C 0.0002 kmol/hr H 2 S = 100% Natural Gas Inlet 25°C 0.33 kmol/hr CH 4 = 99.9% H 2 S = 0.001% Desulfurizer

18. Heat Exchangers A=q/UFΔT lm F = 0.9 U = 30 W/m 2 °C ΔT lm = (ΔT 2 – ΔT 1 ) / [ ln(ΔT 2 / ΔT 1 ) ]

19. Recycled Water 5°C 0.37 kmol/hr H 2 O = 100% Cooled POC 283°C 3.51 kmol/hr N 2 = 86% O 2 = 9% H 2 O = 4% CO 2 =1% Humidified NG 273°C 0.67 kmol/hr H 2 O = 56% CH 4 = 44% Pure NG 25°C 0.3 kmol/hr CH 4 = 100% POC Vent 26°C Fuel Humidifier Area = 2.6 m 2 q = 1.8 kW

20. Heated HNG 840°C Cooled POR 479°C POR 850°C 1.3 kmol/hr H 2 O = 47% H 2 = 29% CO 2 = 23% CO = 1% Humidified NG 273°C Fuel Preheater Area = 6.3 m 2 q = 5.3 kW

21. Heated HNG 840°C 0.67 kmol/hr H 2 O = 56% CH 4 = 44% SynGas 734°C 1.26 kmol/hr H 2 = 73% CO = 21% H 2 O = 3% CO 2 = 2% Reformer R-104 q = 17 kW R-104 COMB-105 Heated HNG SynGas POC Depleted Air Pure NG CH 4 + H 2 O → CO + 3H 2 CH 4 + 2H 2 O → CO 2 + 4H 2

22. Combustor COMB-105 Depleted Air 850°C 3.48 kmol/hr N 2 = 87% O 2 = 11% H 2 O = 2% POC 784°C 3.51 kmol/hr N 2 = 86% O 2 = 9% H 2 O = 4% CO 2 =1% Pure NG 25°C 0.03 kmol/hr CH 4 = 100% q = -17 kW R-104 COMB-105 CH 4 + 2O 2 → CO 2 + 2H 2 O SynGas POC Heated HNG Depleted Air Pure NG

23. Cooled POR 480°C 1.3 kmol/hr H 2 O = 47% H 2 = 29% CO 2 = 23% CO = 1% WGS Exhaust 480°C 1.26 kmol/hr H 2 O = 46.5% H 2 = 30% CO 2 = 23.2% CO = 0.3% Water Gas Shift Reactor CO + H 2 O → CO 2 + H 2

24. POR 850°C 1.3 kmol/hr H 2 O = 47% H 2 = 29% CO 2 = 23% CO = 1% Depleted Air 850°C 3.48 kmol/hr O 2 = 11.5% Heated Air 650°C 3.88 kmol/hr O 2 = 21% SynGas 750°C 1.26 kmol/hr H 2 = 73% CO = 21% H 2 O = 3% CO 2 = 2% Fuel Cell CO + ½ O 2 → CO 2 H 2 + ½ O 2 → H 2 O

25. H Exhaust 25°C 0.38 kmol/hr H 2 = 100% Purge 25°C 0.43 kmol/hr CO 2 = 68% Uncondensed Gases 5°C 0.68 kmol/hr H 2 = 56% CO 2 = 43% Air Inlet 25°C 0.13 kmol/hr Pressure Swing Adsorber

26. Economic Analysis Turn it over to Wayne Johnson

27. Economic Components Capital Costs Operating Costs Income Generated Payback Period Return on Investment

28. Capital Cost Assumptions Cap Cost Program –Analysis, Synthesis, and Design of Chemical Processes –Compares to Peters and Timmerhaus Stainless Steel

29. Equipment Costs

30. Lang Factor Fluid Processing = 4.74 Includes: –Construction material and overhead –Labor –Contract engineering –Contingency –Site development $40,000 X 4.74 = $190,000

31. Operating Costs Fuel: 0.33 kmol/hr = 260,000 BTU/hr = 0.26 therms/hr Tennessee Valley industrial rate = $7.70/therm Labor included at site

32. Income Electricity = 25kW Price = $0.10/kWhr Hydrogen = 0.38 kmol/hr =.76 kg/hr Tennessee Valley industrial rate = $11.64/kg

33. Total Income vs. Expense

34. Investment Results Non-discounted Payback = 2.4 Years Return on Investment = 41%

35. Conclusions Rate of return and payback period are interesting Emerging technology means cost may decrease

36. Questions for the Board What areas require more detail? What locations should be investigated? Should we enlist an electro-chemistry team? Should we enlist an electrical engineering team?