Fuel Cell Design

Chemical Engineering Senior DesignSpring 2005

UTC

Technical and EconomicAspects of a 25 kW Fuel Cell

Chris BoudreauxWayne JohnsonNick Reinhardt

Technical and EconomicAspects 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

Outline

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

Introduction

Overall Reaction

Methane + Air --> Electricity + Hydrogen

+ Heat + CO2

Introduction

Pressure SwingAdsorption

Fuel Cell

ReformerGas

Hydrogen

ElectricityAirHeat

SynGas

POR

Water

Exhaust

Fuel Cell-ChemistrySynGas

Air

O- O-

H2 H2O CO CO2

POR

O2 N2“Air”

Solid Oxide Electrolyte

is porous to O-

H2 + CO

Fuel Cell-ElectricitySynGas

Air

O- O-

H2 H2O CO CO2

POR

O2 N2“Air”

Electrons

Load

Fuel Cell-ChallengesSynGas

Air

O- O-

H2 H2O CO CO2

POR

O2 N2“Air”

H2 + COHot SynGas

Hot Air

Recover H2

Recover Heat

Process Description

Turn it over to Nick Reinhardt

Process Description

Fuel Preparation

Air Preparation

Post Processing

Fuel Cell

Fuel Preparation

Air Preparation

Post Processing

Fuel Cell

Fuel Preparation - 100

Fuel Preparation

Air Preparation

Post Processing

Fuel Cell

Air Preparation - 200

Fuel Preparation

Air Preparation

Post Processing

Fuel Cell

Fuel Cell - 300

Fuel Preparation

Air Preparation

Post Processing

Fuel Cell

Post Processing - 400

Process and Equipment Design

Turn it over to Chris Boudreaux

Pure Natural Gas25°C0.33 kmol/hrCH4 = 100%

Sulfur Purge25°C0.0002 kmol/hrH2S = 100%

Natural Gas Inlet25°C0.33 kmol/hrCH4 = 99.9%H2S = 0.001%

Desulfurizer

Heat Exchangers

• A=q/UFΔTlm

• F = 0.9• U = 30 W/m2°C• ΔTlm = (ΔT2 – ΔT1) / [ ln(ΔT2 / ΔT1) ]

Recycled Water5°C0.37 kmol/hrH2O = 100% Cooled POC

283°C3.51 kmol/hrN2 = 86%O2 = 9%H2O = 4%CO2 =1%

Humidified NG273°C0.67 kmol/hrH2O = 56% CH4 = 44%

Pure NG25°C0.3 kmol/hrCH4 = 100%

POC Vent26°C

Fuel Humidifier

Area = 2.6 m2

q = 1.8 kW

Heated HNG840°C

Cooled POR479°C

POR850°C1.3 kmol/hrH2O = 47% H2 = 29%CO2 = 23%CO = 1%

Humidified NG273°C

Fuel Preheater

Area = 6.3 m2

q = 5.3 kW

Heated HNG840°C0.67 kmol/hrH2O = 56%CH4 = 44%

SynGas734°C1.26 kmol/hrH2 = 73%CO = 21%H2O = 3%CO2 = 2%

ReformerR-104

q = 17 kW

R-104COMB-105

Heated HNG SynGas

POCDepleted AirPure NG

CH4 + H2O → CO + 3H2

CH4 + 2H2O → CO2 + 4H2

CombustorCOMB-105

Depleted Air850°C3.48 kmol/hrN2 = 87%O2 = 11%H2O = 2%

POC784°C3.51 kmol/hrN2 = 86%O2 = 9%H2O = 4%CO2 =1%

Pure NG25°C0.03 kmol/hrCH4 = 100%

q = -17 kW

R-104COMB-105

CH4 + 2O2 → CO2 + 2H2O

SynGas

POC

Heated HNG

Depleted AirPure NG

Cooled POR480°C1.3 kmol/hrH2O = 47%H2 = 29%CO2 = 23%CO = 1%

WGS Exhaust480°C1.26 kmol/hrH2O = 46.5%H2 = 30%CO2 = 23.2%CO = 0.3%

Water Gas Shift Reactor

CO + H2O → CO2 + H2

POR850°C1.3 kmol/hrH2O = 47% H2 = 29%CO2 = 23%CO = 1%

Depleted Air850°C3.48 kmol/hrO2 = 11.5%

Heated Air650°C3.88 kmol/hrO2 = 21%

SynGas750°C1.26 kmol/hrH2 = 73%CO = 21%H2O = 3%CO2 = 2%

Fuel Cell

CO + ½ O2 → CO2

H2 + ½ O2 → H2O

H Exhaust25°C0.38 kmol/hrH2 = 100%

Purge25°C0.43 kmol/hrCO2 = 68%

Uncondensed Gases5°C0.68 kmol/hrH2 = 56%CO2 = 43%

Air Inlet25°C0.13 kmol/hr

Pressure Swing Adsorber

Economic Analysis

Turn it over to Wayne Johnson

Economic Components

• Capital Costs• Operating Costs• Income Generated• Payback Period• Return on Investment

Capital Cost Assumptions

• Cap Cost Program– Analysis, Synthesis, and Design of Chemical

Processes– Compares to Peters and Timmerhaus

• Stainless Steel

Equipment CostsEquipment Name Capital Cost ($)

Solid Oxide Fuel Cell $10,000PSA $10,000

Reformer/Combustor $6,300Fuel Preheater $3,400

Chiller $3,000Fuel Humidifier $2,800

Desulfurizer $1,200Water Gas Shift Reactor $1,200

Air Compressor $950Air Side Heat Recovery $670

Air Preheater $290Water Condenser $150Water Purification $100

$40,000

Lang Factor

• Fluid Processing = 4.74• Includes:

– Construction material and overhead– Labor– Contract engineering– Contingency– Site development

• $40,000 X 4.74 = $190,000

Operating Costs

Fuel $17,000Utilties $1,000

Maintenance $1,000Labor $1,000Total $20,000

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

Income• Electricity = 25kW• Price = $0.10/kWhr

• Hydrogen = 0.38 kmol/hr= .76 kg/hr

• Tennessee Valley industrial rate

= $11.64/kg

Electricity $22,000Hydrogen $76,000

Total $98,000

Income

Total Income vs. Expense

Costs $20,000Income $98,000Total $78,000

Income vs Expense

Investment Results

Non-discounted Payback = 2.4 Years

Return on Investment = 41%

Conclusions

• Rate of return and payback period are interesting

• Emerging technology means cost may decrease

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?