Energy- and exergy efficienciesof stationary LT and HT – fuel cell

systems

Summer school on electrochemical engineering, Palic, Republic of Serbia

Prof. a.D. Dr. Hartmut Wendt, TUD

VAILLANT

5 kW

PEMFC Hexis1,5 kWSOFC

UTC PC 25 C250 kWPAFC

FCE DF 300300 kWMCFC

MTU HOT MODULE250 kW, MCFC

Siemens Westinghouse100 kW SOFC

Do we need stationary fuel cells ?

Yes, if they are cheap enough. And why ?• Advantageous for distributed electricity

generation in absence of grids or in sparsely populated areas.

• If properly matched to demands (electricity and heat) they help to reduce carbon dioxide emission.

• They allow to tap biomass in the form of biogas

What is the fuel for stationary Cells ?

• Hydrogen is the natural fuel for fuel cells, because it is the most reactive of all fuels and can easily be activated by Pt and other electrocatalysts. – But hydrogen is expensive

• Natural gas is most convenient, in highly developed economies, where there is a NG – grid.

• In absence of NG –grids liquified gases (propane/butane), ethanol/methanol, and also diesel are viable options

Hydrocarbons ?

• They must be converted into hydrogen – containing gas mixtures

• Carbon dioxide and water vapour do not interfere (much) with hydrogen conversion at the fuel cell anode

• Steam reforming:

CH4 + 2 H2O —> CO2 + 4 H2

What is the fuel for stationary Cells ?

• Hydrogen is the natural fuel for fuel cells, because it is the most reactive of all fuels and can easily be activated by Pt and other electrocatalysts. – But hydrogen is expensive

• Natural gas is most convenient, in highly developed economies, where there is a NG – grid.

• In absence of NG –grids liquified gases (propane/butane), ethanol/methanol, and also diesel are viable options

Steam reforming of different fuels

• All these hydrocarbon fuels need chemical conversion into gas mixtures containing hydrogen

• STEAM REFORMING – established for hydrogen production from natural gas can be adopted for any of these fuels

• For stationary electricity production fuel cell systems are needed which are comprising

1. Heterogeneously catalysed steamreforming

2. Shift Conversion and for LT cells

3. CO -removal

FuelCell

NGanode exhaust20% of initial H2to reformerReformer Prox

Air,20% oxygen

Steam

Stack gas

PEMFC process scheme

Shift

Cathode exhaust(8% oxygen)to reformer

Reactors:

Again: Why fuel cells ?

• Fuel cells systems are able to achieve in small units (100 kW to 500 kW) electrical efficiencies approaching 50%.

• (Internal combustion:30 to 40%)• Fuel cells are environmentally more benign than

combustion engines with low toxic emissions, low noise level and clean waste-water.

• Fuel cells systems can operate on NG and biogas• But Fuel cells are (still awfully) expensive

Now : Thermodynamics

• The stress will be on gas processing of the whole process

• In order to decouple from cell thermodynamics and characteristics this assumption: The efficiency of the cells is independent on cell type: ήel = 50%

• ( respective to initially 1 mol of methane)

First Approach : Adiabatic

• Not Heat losses to environment considered, which are relatively low for large units (dispersed cogeneration 100 to 500 kW)

• And relatively higher for domestic cogeneration (1 to 5 kW)

Additional Assumptions

• Total process is adiabatic (1st approach) – no heat losses to the environment, lower heat of combustion amounts to: 802kJ/mol methane

• Only methane as a fuel• In the reforming and shift processes

CH4+H2O -> CO + 3H2 and

CO + H2O -> CO2 + H2

all heat is preserved, that means , that the residual fuel`s heat is stored in hydrogen

FuelCell

NGanodeexhaust20% of initial H2to reformerReformer Prox

Air,20% oxygen

Steam

Stack gas

PEMFC process scheme

Figure 1

Shift

Cathode exhaust(8% oxygen)to reformer

Reactors:

1200 K

650 K

470 K

360 K

CH4

HTshift

LTshift

SELOX

CH4

+ 3H2O

144kJ

Reformate gas to cell+ 145 kJ

Thermal Process Design of LT – fuel cell power plants

reformer

Figure 2

300 K 3 H2O - 145 kJ-145 kJ

process step enthalpy change kJ/ per mol CH4

01) 1mol methane 27 -> 127 C 302) 3 mol H2O 27 -> 127 C and steam generation 14403) 3 mol2 of H2O(steam) and 1 mol CH4 927C 16204) reforming 22705) cooling of reformate gas 927 -> 427 C -10206) HT steam shifting step -3707) Cooling of shift reacted gas to 327 C and endshifting 008) downcooling to 100 C -2509) electricity generation by cell reation -40010) heat generation by cell reaction -40011) heating of anode off-gas and air to 927 C 13612)burning of off-gas in reformer -22713) cooling of reformer stack gas to 100 C -14014) condensing 3 mols of water from the stack gas -144

Heat balance of LT fuel cell power plant – first law balance for a fuel cell power plant with postulated 50 % electric cell efficiency

First law heat balance: ok but 2nd law? Temperature niveau of off – heat is too low

• We need the evaporation heat for 3 mols of water to produce process steam - They must be supplied by burning additional methane (+17,7% of the lower heating value of methane) :postulated 50% eff. lowered to 42,3%

• Another 1% point is lost in the PROX – process by CO- and H2O-cooxidationPostulated 50% efficiency lowered to 41%

process step heat / kJ per mol CH4

01)*3mol H2O to 60 to 127 and steam production 12202) 1mol CH4 and 3mol steam to 650 C 12903) electricity generation 50% -40004) HT - heat gen.in-cell and by combustion -40005) Heating combust. air lamda=1,5 to 650C 28806) Cooling of hot off -gases to 150 C -35307) Cooling from 150 to 50 C -6508) condensing 3 mols of water -122*Only vaporisation, because hot condensate is vaporised

Heat balance of the HT Fuel cell power plant – first and second law balance for a cell with 50% efficiency producing 400 kJ electricity from 1 mol of methane.

First law heat balance: ok but 2nd law? Temperature niveau of off – heat high enough

•We need no combustion to produce process steam. Therefore:

postulated 50% eff. Not lowered• No Fuel is lost by the PROX – process because HT – cells are insensitive towards CO

Second Approach: Heat Losses

• HT – Fuel Cell systems are for dispersed power generation: 100 to 500 kWel Losses amount to only 5% relative of methane input:result: Total efficiency is 47%

• LT –Fuel systems are for domestic cogeneration ánd due to higher surface to volume ratio of the HT – parts have relatively higher losses: 10% with respect to input energy.Total energy efficiency is down to 35 to 37%

LT - process (3 to 5 kW el.) HT - process (100kW el.)lower heating value needed /kJ* 1029 922corrected for external losses/ kJ 1131 968exergy content of fuel /kJ 1090 932amount of methane /mol 1,4 1,2elctricity /kJ 400 400energy efficiency 35,40% 41,30%exergetic yield of electricity 36,50% 43,00%excess exit heat at 400 C /kJ 0 214exergy attributed to 400C - heat 0 123exergetic yield of HT - heat 0 14%exit heat at 60 C /kJ 560 71exergy of off heat at 60 C / kJ 61 7,8exergetic yield of LT heat 6% 1%

Exergy efficiencies in LT and HT fuel cell power plants

total exergetic yields: 42,5% 42,5%

Summary

• HT fuel cells are at an advantage compared to LT fuel cells as they can make use of internally produced high-temperature heat for their own gas process.

• LT-cell systems with methane redorming do not yield higher then 35% efficiencies – HT – systems may yield more than 50% efficiency