260 spring 2004.pdf

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FUNDAMENTALS OFADVANCED ENERGY CONVERSION

2.996 & 2.994, Spring 04A Ghoniem (IC), M Kazimi, Y Shao-Horn, J Teste

WHY ? CO2, Terawatts, Needs and Sources,

WHAT ? Few Examples:IGCC, fuel reforming and synthesisFuel Cells, fueling the fuel cell, FC3

Hydrogen Economy : generation and storagePhotoelectricity

HOW ?

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S Source: Internatinal Energy Agency

3035404550 2003The ENERGY REVOLUTION

(The Terawatt Challenge)14 Terawatts (4 TW in US)The Terawatt Challenge

R. Smalley* Rice University

25201510

50

1. ENERGY

2. WATER

3. FOOD n s s a l o G a s O i l

i s s i o a C m o F B i4. ENVIRONMENT

505. POVERTY45

6. TERRORISM & WAR 40357. DISEASE30

8. EDUCATION 25

9. DEMOCRACY20

15

10. POPULATION 1050

d H y

2050

30 -- 60 Terawatts

o n

a s a l s s O i l

C o G s i m a i s*Noble prize, Chemistry, 1996 o F B i o n /

s i H u F

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Energy Sources and Demand TW

50

40

30

20

10

0

Should be CO2 free

CO2 freeTo limit CO2 to 450

2000 2020 2040 2060 2080 YEAR Source: M.I. Hoffert et. al., Nature , 1998 , 395 , 881,

Wigley, Richels and Edmonds, ppmv of CO2, pre-industrial concentration is 3

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CO2 emissions andGlobal Temperature!

From Basic Research Need for aHydrogen Economy, Report of DOE BESWorkshop, May 13-15, 2003

CO2 emissionreduction:

Improve planefficiency

Use low C fu Use pure H2 Sequester CO RENEWAB

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FIGURE 1. GASIFICATION-BASED ENERGY PRODUCTION SYSTEM CONCEPTS

FIGURE 2. GASIFICATION-BASED ENERGY CONVERSION SYSTEM OPTIONS

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(Courtesy of Prof. Wilson and Prof. Korakianitis. Used with permission.)

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Optimal Hydrogen Utilization: the Fuel Cel

Basic Energy Needs for the Hydrogen Economy, May 2003, DO

The public may copy and use this information without charge, provided that this Notice and any statement of authorship areproduced on all copies. Neither the Government nor the University makes any warranty, express or implied, or assumesliability or responsibility for the use of this information. Courtesy of Los Alamos National Lab. Used with permission.

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How to Fuel the Fuel Cell Engine?

Especially for mobile applications, fuel cells may work with a reformer

(although direct methanol cells are also under development)

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HYDROGEN & THE HYDROGEN ECONO

Energy carrier : must be produced, stored, transported & c

Like electricity: expensive to produce, transport, not easy to store. Provides a good link with renewables or non-exhaustables.

Can be produced by:Oxygen or steam Reforming of hydrocarbon, or,Splitting water electrolytically or thermo-chemically.

Compression/liquefaction (20 K, 1 bar/293 K, 800 bar) energies are high. Has low volumetric energy density (even in liquid form)

Storage: metal fiber tanks, cryogenic container, or in metalhydrides (solids): through physical or chemical sorption.

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The Mobile

Storage Problem:

Fuel Cell may be theeasiest piece of the

puzzle! The Future of the Hydrogen Economy,

ABB) and Bossel (Fuel CellEliasson (

GMs

Basic Energy Needs for the Hydrogen Economy, May 2003, DOE

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Thermochemic Nuclear Hydrogen Production for Hydrogen Pr

Sulfur-iodine pQ@120C 2[ I 2 +SO2 +2 H 2O 2 HI + H 2SO2 Non-electrolytic wa

+Q@800950C 2 H 2SO4 2SO2 +2 H 2O+O2 + Q @ 850 C } 2 H 2O 2+Q@450C 2 2 HI I 2 + H 2][ 50%

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Solar Hydrogen Production Capture/conversion + Electrolysis

Basic Energy Needs for the HydrMay 2003, DOE.

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Spring 04, subject is offered as 2.996 (G) and 2.994 (U)

FUNDAMENTALS OF ADVANCED ENERGY CONVERSION (2.60j, 2.62j,10.392j, 22.40j) Perquisite: 2.006 or permission of instructor G (Spring) 4-0-8 H-LEVEL Grad Credit,

A.F. Ghoniem (IC), M. Kazimi, Y. Shao-Horn, J. Tester.

Fundamentals of thermodynamics, chemistry, transport processes in energy sy stemAnalysis of energy conv ersion in thermo-mechanical, thermo-chemical, electrochand photoelectric processes in existing and future power and transportation systememphasis on efficiency, environmental impact and performance. Systems utilizingfuels, hydrogen, nuclear and renewable resources, over a range of sizes and scales

discussed. App lications include fuel reforming, hydrogen and synthetic fuel produfuel cells and batteries, combustion, catalysis, supercritical and combined cycles, photovoltaics, etc. Energy storage and transmission. Optimal source utilization anlife cycle analysis.

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From matter and Energy (1912) by Frederick Soddy -Noble Prize in Chemistry, 1921.

The laws expressing the relations between energy andmatter are not solely of importance in pure science.They necessarily come first.. In the whole record ofhuman experience, and they control, in the last resort,

the rise or fall of political systems, the freedom orbondage of nations, the movements of commerce andindustry, the origin of wealth and poverty and the

physical welfare of the race. If this has beenimperfectly recognized in the past, there is no excuse,now that these physical laws have become incorporatedinto everyday habits of thought, for neglecting toconsider them first in questions relating to the future

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2.996 Fundamentals of Advanced Energy Conversion Lecture Memo

Lecture number: 1Date: February 4 th , 2004

Needs: Terawatt challenge

Energy source and demand

CO 2 emissions and global temperature

Thermal efficiencies of different turbines and combined cycles

Fuel cell: storage problem, storage in the solid state

Nuclear hydrogen production

Solar hydrogen production

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The first law of thermodynamics: Work and heat interactionsThe stored energy

First law for an open system

The second law of thermodynamics: Entropy generation, efficiency

The combined statement of the first and second laws: Availability

Efficiency and second law efficiency

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Entropy in an open system

Isentropic process

Brayton cycles : Compressor and turbine efficienciesCycle efficiency, w netRecuperative cycle

Steam cycles : Conventional, superheating, reheat and regeneration cycle

Combined cycles

Gas mixtures: Molar and mass fractions, molecular weight, partial pressureInternal energy, enthalpy and ehthropy.

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Mole, mole fraction, mass fraction, average molecular weight

Gas mixture: Partial pressure and volumeInternal energy, enthalpy and entropycp change with temperature

Entropy generation due to mixing

Separation of gases: CO 2 sequestrationH 2 production

Minimum work for separation (Ex. C 6H 6)

Reacting mixtures: IrreversibilityEndothermic and exothermicStoichiometry

Energy balance: Enthalpy of formationAdiabatic flame temperature

IC engine: Introduction to Otto cycle

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Chemical thermodynamics: Adiabatic flame temperatureEnthalpy of reactionLow and high heating values

Availability of a chemical reaction: Gibbs free energy, efficiency

Direct conversion: Fuel cellEfficiency change with respect to operating temperatureReactions in fuel cellsElectrolyzer

Indirect conversion: Carnot cycle efficiencyInternal combustion engine (SI)Otto cycle efficiencies1) Air standard cycle2) Fuel/air cycle3) Complete/incomplete combustions

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History of Thermodynamics: Thompson, Joule, Clausius, Carnot

Chemical potential in fuel cells

Review of Chemical EquilibriumConstant U,V constraints: Maximization of entropyConstant T,P constraints: Minimization of gibbs free energyLaw of mass actionEquilibrium constantEndothermic and exothermic reactions

Fuel reformingSteam reformingWater gas shift

Introduction to a chemical equilibrium program(Equil).

Electrochemical energy conversion and storageReadings, objective and scope

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Lecture number: 8Date: March 1 st , 2004

Efficiency revisited: First law, second law, electrochemical, fuel efficiencies

Faradays law, Faradic efficiency

Basic operations of electrochemical cells: anode and cathode

Conservation of charge

Cell types: Galvanic, Equilibrium and Electrolytic

Pb-acid rechargeable battery

Complete cell diagram

Standard reference potential

The combined 1 st and 2 nd laws with generalized work

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Lecture number: 10Date: March 8 th , 2004

Examples for cell potential calculations: H 2 /O 2 and Zn/Cu cells

Review of Gibbs-Faraday equilibrium equation

Review of fugacity and activity

Nernst equation for ideal gases, liquid solutions and solids

Overall redox reaction in a galvanic cell

Chemical kinetics

Rate processes: chemical reaction rate, transport rate and flowrateconvection

Elementary, global and chain reactions.

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Power ratings of different applications

Chemical kinetics of H 2 and O 2 reaction

Reaction rate equation

Forward and backward reactions

Activation energy

Ozone reaction

Single step kinetics

Well stirred reactor: Governing equations, T-Q diagram, efficiency andblowout

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Power densities in IC engines, lithium batteries and fuel cells

Over potential

1) Ohmic overpotential

2) Activation overpotential

3) Mass transfer overpotential

Forward and backward reactions in a simple redox process

Standard activation energy for the anodic and cathodic reactions

Transfer coefficient

1) Changes in the free energies

2) Changes in the forward and backward reaction rates

Electrochemical reaction rate

Butler-Volmer Equation

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Cell voltage profile for PEM: Cathode activation, cell resistance, anodeactivation and mass transport

Butler Volmer model :Tafel relationship

Example: Copper plating

Oxygen reduction

Multistep mechanisms

Introduction to mass transport

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Lecture number: 16Date: April 5 th , 2004

Kinetic theory of gasesRate of change of momentumMean free pathMolecular transport

Transport equationFouriers lawFicks law

Species conservation equation

Energy conservation equation

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Flames in SI engines

Laminar and turbulent burning velocities

Microscopic flame structure

Two zone model

Dependency of Su on thermal diffusivity and characteristic time scale

Dependency of Su on equivalence ratio

Stratification and turbulence generation in engines

Knock

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Brayton cycle

Gas turbine combined cycle

Supercritical steam cycle

Coal gasification

Fuel cell/gas turbine cycle

CO2 capture

Nuclear power conversion: Boiling water reactor and pressurized waterreactor

Fission reaction

Neutron number vs atomic number and binding energy per neutron

Energy spectrum of fission neutrons

Supercritical water reactor

Gas cooled reactors

Multiple parameter optimization of a CO2 supercritical cycle to advancednuclear reactors.

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Hydrogen characteristics

Current and projected hydrogen powered cells

Projected effects of switching to hydrogen powered cars

Alternative ways of H2 production today and tomorrow

Nuclear technology in the hydrogen economy

Availability

Availability diagram for water

Component irreversibility

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Changes in the utilization factor with respect to turbine inlet pressure

Irreversibility analysis for a feed pump and the primary heat exchanger

Optimization of Supercritical CO 2 Power Cycle

- Efficiencies of various cycles

- Various cycle losses

- Thermodynamic optimization process in

A simple Brayton cycle with recuperation

Brayton Cycles with Reheat Recompression

Brayton Cycle with Intercooling

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Electrolysis

- Diffusion and migration

- Limiting current

- Applied voltage vs. current

Photovoltaics

- Opportunity

- Market segments

- Cost

- Solar panel and cell

- Market share by technology

- Crystalline silicon technology

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Lecture number: 25Date: May 10th, 2004

Battery types: Primary, secondary and reserve batteries World battery production Major considerations in battery selection Fuel cell and battery voltages Voltage profiles of common battery systems Specific capacity of electrode materials Alkaline Zn/MnO2 and lithium primary batteries Battery design Energy and power density of batteries Secondary batteries (Lead acid, Ni-Cd, NiMH and Lithium batteries):

Advantages and disadvantagesApplications

Material degradation

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Fundamentals of Advanced Energy ConversionSpring 2004

Homework Set #1

Problem 1. The purpose of this problem is to familiarize you with the units of energyand to compare some of the characteristics of various energy sources.

A 1000 MWe power plant is being planned for the rapidly expanding city of Las Vegas.Consider three possible options for the plant: nuclear, coal and solar fueled. The totalsystem should have a capacity factor of 90% . Answer the following questions.

1.1 If an advanced nuclear power reactor is to be used, what amount of natural uraniumwill be needed to fuel the plant annually? The thermal to electrical energy conversionefficiency using a water-steam Rankine cycle with sub-critical pressure is 35%. Thenuclear to thermal energy conversion efficiency in the reactor is 95%, due to theemissions of neutrinos which are not captured in the core. For simplicity assume U235 is the only fissionable material. By weight, U-235 constitutes 0.71% of naturaluranium, and the rest is U-238. The reactor fresh fuel requires a higher fraction ofU-235 (typically 4.4% in todays light water cooled reactors) which is obtained byisotopic enrichment of the natural uranium. However, due to the energy intensity ofthe operation of the enrichment plants when the U-235 fraction gets to be small, thediscarded uranium in the enrichment process (called uranium tails) has 0.25% U-235fraction. The fission of one gram of U-235 produces 1 MW-day worth of nuclearenergy.

1.2 If the plant is to be of an advanced pulverized coal fired plant with a supercriticalwater-steam Rankine cycle, what is the average daily amount of coal (in kg/day)consumed to power the plant? Consult the notes from the first lecture (or othersources) for the thermal efficiency of such plants. Assume the combustion of 1kg ofcoal provides 27,800 Btu of energy (this is an average value, the real heat ofcombustion varies with the type of coal). 1J = 9.48x10 -4 Btu = 0.239cal.

1.3 If an advanced solar photovoltaic plant is to be used, with the best available solarflux-to-electricity conversion efficiency of 12%, what is the total land area required to

provide the needed power? You may assume that the land needed is 2 times the flat

panel area. The daily total (direct and diffuse radiation) solar energy flux near thecity of Las Vegas has an annual average of 500 cal /cm 2 per day. Assume that thesolar plant will have to also store sufficient energy during the day to meet anequivalent demand at night.

The 90% capacity factor means the plant on average over a long time (like a year) will operate 90% of thetime.

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Problem 2 This problem illustrates the concept of specific work, the effect of fluiddensity on the work required for its compression and the effect of intercooling as ameans to reduce the work needed for compressing gases.

2.1 Find the pressure at which intercooling should be performed to achieve the greatestreduction in the work needed by a two-stage compressor of air from atmospheric pressureand temperature to 100 atmospheres. The air can be assumed to behave as an ideal gas,with temperature independent specific heat. Assume that each stage of the compression isisentropic and that the intercooling occurs at constant pressure returning the air to itsinitial temperature prior to entering the second stage.2.2 What is the work supplied per unit mass of air being compressed?2.3 What would be the work needed per unit mass if no intercooling was applied?2.4 Compare the work above to the work needed to compress a unit mass of liquid waterfrom one to 100 atmospheres. You may ignore the density and temperature changes ofwater during the compression.2.5 What values would you expect for the actual compressor work in parts 2.2 and 2.3 if a

practical device of reasonable size would be used instead of the ideal ones?

Problem 3 This problem introduces the concept of total energy efficiency of atransportation system.

According to Faradys Law, the heat formation of a molecule, H, can also be expressedas an electrochemical potential:U = - H/ n e F,where n e is the number of electrons participating in the chemical bonding and F is theFarady constant = 96,485 Coulomb/mol.Only a fraction of the energy of formation is available for reversible energy conversion,which is determined by the Gibbs free energy, G.To produce hydrogen from water by electrolysis, H has to be invested ( in this case 286kJ/mol at 25 C) .However, when hydrogen combines in a fuel cell with oxygen to produce water, only thereversible energy G is obtained (237kJ/mol). Since two electrons participate in the

bonding of a hydrogen molecule, the theoretical voltage that can be produced in a fuelcell is 1.23 volts. In reality, there are losses due to resistance to ion motions and

polarization at the electrode interfaces, so that the actual voltage obtained is much less in practice, about 0.7 volts.3.1 What is the power-plant to wheels efficiency of a fuel cell powered transportationsystem, if the hydrogen is produced by electricity with a thermal- to electrical powerconversion efficiency in the plant is 0.5?3.2 What should the internal combustion engine (ICE) efficiency of the cars to allow theconventional transportation power system to have the same energy utilization efficiencyas the electrolysis/fuel cell system? You may assume that the energy efficiency ofgasoline and diesel manufacturing plants (i.e. refinery) is 96%.3.2 If the hydrogen is obtained by methane steam reformation, which currently has 0.7energy efficiency, what would the answers to questions 3.1 and 3.2 be?

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Homework I Solution

Problem 11.1

Natural Uranium : X (g/yr)Enriched Uranium :Y (g/yr)Discarded Uranium: Z (g/yr)

Mass Conservation: X=Y+Z Eq.(1)Mass Conservation of U-235:0.71X=4.4Y+0.25Z Eq.(2)

Using Eqs. (1) and (2), we getY=0.1108X Eq.(3)

Hence, we lose almost 90% of the natural uranium in the enrichment process.

Since U-235 in the enriched Uranium is 4.4/100*Y, we can represent the used Uraniumfor fission from the natural Uranium as

4.4/100*0.111X=0.004877X. Eq.(4)

Hence, only 0.4877% of the natural Uranium is actually used for fission.

Then, power can be represented as

0.004877X* rankine * nuclear *1MW-day/g=1000MW*365 day* capacity

where rankine =0.35, nuclear =0.95 and capacity =0.9. Then, we get X=2.026*108

g/yr=202.6 ton/yr=0.56ton/day.

1.2

Average daily amount of coal used: X (kg/day)

X* steam *27800BTU/kg*1J/9.48*10-4BTU*1day/(24*3600sec) =1000*10 6 W* capacity

where steam =0.471. Then, we get X=6.04*10 6kg/day=5640 ton/day.

1 M.M. El-Wakil, Power Plant Technology, McGraw Hill, 2984, page 72

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1.3

Area for the flat panel : X (cm 2)

500cal/(cm 2 day)/(0.239cal/J)* conv *X*1day/(24*3600sec)=1000*106W

where conv =0.12. Then, X=3.441*1011cm 2. Also, total area required is 2X=68.8km 2.

Problem 2

2.1P1=1atm, T1=300K Isentropic compression I P2,T2P2,T2 Intercooling P2=P2,T2=T1=300P2,T2 Isentropic compression II P3=100atm,T3

In the compressions I and II, T and P have the following relationships

)1/(

1

2

1

2

=

k k

T T

P P

and)1/(

1

3

2

3

=

k k

T T

P P

Eq. (5)

Also, work can be represented as

)()( 1213 T T cT T cw p p += Eq.(6)

Using Eqs. (5) and (6), we get

+

=

2/)1(

2

3

/)1(

1

21

k k k k

p P P

P P

T cw Eq.(7)

By differentiating w by 2 P , we get

312 P P P = =10atm.

2.2

From Eq. (5), we get T 2=T3=579K. Using Eq. (6) and cp~1kJ/kg K at T~450K, we get558kJ/kg.

2.3

P1=1atm, T1=300K Isentropic compression P2=100atm,T2

Using Eq. (5) again, we get T2=1118K. Using )( 13 T T cw p = and cp~1.1kJ/kg K atT~700K, we get =w 900kJ/kg.

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It is clear that intercooling reduces the work required for compression significantly.

2.4w=vdP=0.001m 3/kg*(10130kPa-101.3kPa)=10kJ/kg.

The compression work of liquids is typically 1-2% of that of gases.

2.5 Assuming c ~0.7-0.9, work in 2.2 and 2.3 should increase by 10-40%.

Problem 3

3.1thermal =0.5 : thermal efficiency

voltage =0.7/1.23=0.569 : Voltage drop

2 H =237/286=0.829 : H2 to electric power conversion efficiency

tot = thermal * voltage * 2 H =0.236

3.2

tot =0.96* ICE

ICE =0.246

3.3

0.7* 2 H * voltage =0.330

0.330=0.96 ICE

ICE =0.344

Without considering the refinery efficiency of fuels, it becomes

ICE =0.330

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MASSACHESUTS ISNTITUTE OF TECHNOLOGYFUNDAMNETALS OF ADVANCED ENEFRGY CONVERSION

SPRING 04HOMEWORK IIDUE DATE, FEBRARY 23, 2004

The rising atmospheric concentration of carbon dioxide, a product of hydrocarboncombustion and a known green house gas, is a major environmental concern.Means to reduce the rate of CO2 emission into the atmosphere include improvingenergy conversion efficiency and using carbon free energy sources. Another way issequestration, that is to separate CO2 from the combustion products and store it insome form away from the atmosphere. This technology is still in its infancy, buteffort to evaluate the energy penalty of sequestration, and potential long-termsuccess of storage is underway (see sequestration.mit.edu for more information andsome useful links). In this exercise, we will explore the energy penalty of CO2sequestration from advanced power plants, using different separation strategies.

Consider a case in which benzene(C 6H6) is used as a fuel, and calculate the totalseparation and liquefaction work required in each of the following cases, as a fraction ofthe available energy in the fuel, and of the work produced in the cycle withoutsequestration.

Make the following assumptions:

(a) The maximum allowable cycle temperature is 1300 C. This is the temperature inthe combustor, achieved by burning the fuel stoichiometrically, and then mixing the

products with extra gases to reduce the temperature.(b) The power cycle is an advanced combined cycle with efficiency 50%, based on the

total work produced without sequestration, and the fuel enthalpy of reaction.(c) The products exit the cycle at 1 atm and 300 K.(d) The environment is at 300 K and 1 atm.(e) The final state of sequestered CO2 is liquid at 300 K and 100 atm.

I. In the first design, it is suggested that we burn benzene in air, then remove CO2from the products. The pressure ratio across the gas turbine compressor is 25, and the

compressor efficiency is 85%. Separate CO2 from the products in gaseous form first. Next liquefy CO2.

(1) What is the Second Law efficiency of the power cycle, without sequestration?(2) Calculate the molar fuel/air ratio in this case.(3) Calculate the work transfer in the separation and liquefaction steps of CO2.

The total sequestration work is the sum of both. First assume that bothgaseous separation and liquefaction are ideal, i.e. both require the minimumwork. (Use attached T-s diagram of CO 2 for calculation)

(4) If the Second Law efficiencies of both steps are 80%, what is the fraction ofsequestration work of the power cycle work? Note that the Second law

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efficiency for separation and liquefaction is the ratio between the minimumwork required and the actual work required.

(5) What thermodynamic processes should be used in liquefaction to achieveideal minimum work? Why are these ideal minimum work processes difficultto achieve in practice?

(6) What processes would you use instead in liquefaction, and why do theyrequire more work? How would you modify these more practical liquefaction

process to bring them closer to the ideal processes, and what additional cost isneeded to achieve those modifications?

II. It has been proposed (see attachment) that separating nitrogen from air before burning is perhaps more economical, since in this case the products of combustion areonly CO2 and O2. Repeat the calculations for the separation and liquefaction work, andcompare the results with the first case. Note that in this case, some of O2 is circulated inthe cycle to keep the maximum temperature at 1300 C. Separating N2 before combustionmay have some emissions benefits.

III. Yet another proposal is to burn in air, at atmospheric pressure, but use a catalyticcombustion unit so that the products of combustion in this case are limited toCO2+N2+H2. Next CO2 is separated isothermally at the products temperature, and theH2+N2 mixture is used in a fuel cell to produce electricity directly. Calculate the totalsequestration work in this case, and compare with the previous two cases..

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(Courtesy of Prof. Steven Penoncello, Professor of Mechanical Engineering. Used with permission.)

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Homework II Solution

(1) The efficiency in the ideal case (Carnot cycle) is:

ideal =1-(300K)/(273K+1300K)=0.81.Then, the second law efficiency becomes

5.0 II = = 62.081.0

LHV of C6H6=40141kJ/kg=3131MJ/kmol (liquid) (Boiling point of C6H6 is 80 C )

Fuel per available energy is: 1kg/(40140kJ*0.81)=1kmol/2536MJ 1

Fuel per work produced without sequestration: 1kg/20070kJ=1kmol/1565MJ.

(2)

Compression of air (assumptions: T1 =300K and P =1atm)1

From P 2 = T2

k /( k1)

P1 T1 and T2 = T1 + (T T ) / 2 s 1 c

T 2 =832K and P2 =25atm

In the combustion process (assumption: No CO and H2 generation)

C 6 H 6 + 5.7 ( O + 76.3 N 2 ) 6CO + 3 H 2O + 5.776.3 N 2 2 2

From H R (T = 832 K ) = H p (T ) , we get T3 = 2878 K 2 3

Note that

o T 2 T 2 T 2 H R =

h C f 6 H 6 + T o c C p 6 H 6 dT + 5.7 T o c O p 2 dT + 76.3 T o c N p 2 dT

, , , ,

1 From Gibbs free energy change, we get a different value of the maximum work:w = G = h + sT =3131MJ/kmol+300(6 sCO 2 +3 s H 2O -7.5 sO 2 - s C 6 H 6 )max

=3131MJ/kmol+300(6moles*0.214MJ/kmol+3moles*0.189MJ/kmol-7.5moles*0.205MJ/kmol -0.173MJ/kmol)=3173MJ/kmol.

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and

T 3 T 3

o

H = 6 h CO f 2 + T

T

o

3

c CO p 2 dT

+ 3 h

oO H f c O H p dT

+ 5.776.3 To c N p 2 dT . p , , ,2, + T o , 2

While computing the above two expressions, keep in mind that c p is a function of

temperature.

In the mixing process

6CO 2 + 3 O H + 5.776.3 N 2 + O X 2 + 76.3 N 2 ) 6CO + 3 O H + XO2 + 76.3 ( 5.7 + N X( )2 2 2 2

From H R (T ) + H (T ) = H (T =1573 K ) where3 R 2 P 4T 3 T 3

H R (T ) = 6 h o

T 3c CO p 2 dT

+ 3 h O H f c O H p dT + 5.776.3 To c N p 2 dT,3 CO f 2 , , 2 ,2 + T o ,

o + T o T 2 T 2

H (T ) = X T o c O p 2 dT + 76.3 To c N p 2 dT

R 2 , ,

andT 4 T

T

o

4 T 4 o H P (T ) = 6 h , c CO p 2 dT

+ 3 h O H f + c O H p dT + X T o c dT +4 CO f 2 , 2 , 2 O p 2, + T o ,

o

76.3 ( 5.7 + X ) T 4

c N p 2 dT,Towe get X =16.78.

(3)

Separation work (Assume H 2O is all liquid)

T=300K,

16.78O26CO2+24.28*3.76N2+16.78O2

P=1atm

24.28*3.76N2+

6CO2 (gas)

T=300K,P=1atm

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( ( W = nCO 2 ((hCO 2,2 hCO 1,2 ) sT CO 2,2 sCO 1,2 )) + n N 2 ((h N 2,2 h N 1,2 ) sT N 2,2 s N 1,2 )) ( + nO 2 ((hO 2,2 hO 1,2 ) sT O 2,2 sO 1,2 ))

1 1(

X= n RT CO 2 log( ) + (n N 2 + nO 2 ) log( ))

CO 1,2 1 X CO 1,2 = 6.58 MJ

Required work for separation: 58.6MJ/kmol of C6H6 (3.7% of the work produced inthe cycle)

)

T2=300K,

6CO2 (gas) 6CO2 (liquid

T1=300K,P1=1atm P2=100atm

( w = ((hCO 2,2 hCO 1,2 ) sT CO 2,2 sCO 1,2 ))= (560 kJ / kg 810 kJ / kg ) 300 K * ( 3.3 kJ / kgK 9.4 kJ / kgK )= 230 kJ / COofkg 2

Since 6kmol(264kg) of CO 2 is generated by 1kmol of C 6H6, the required work forliquefaction is 60.7MJ/kmol of C6H6 (3.9% of the work produced in the cycle)

Total sequestration work is 119.3MJ/kmol of C6H6 (7.6% of the work produced inthe cycle)

work Ideal(4) = II work Actual

,

work Actual = work Ideal / = 25.1 work Ideal II

Hence, 25% increase of the sequestration work (149.1MJ/kmol of C6H6 and 9.5% ofthe work produced in the cycle.)

(5) Since no entropy should be generated in the process to minimize the requiredwork, the temperature of the device should be regulated to be the same as that ofenvironment (Isothermal compression). Note that work is supplied for separation andhence should be removed in the form of heat to the environment to keep thetemperature constant. To reject heat while maintaining the temperature requires this

process to occur very slowly.

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(6) Instead of using isothermal compression, one can consider isentropic compressionand heat rejection in a real situation. As we learned from problem II in Homework I,one can reduce the required work using intercooling. If one uses a large number ofintercoolings and compressors, one can bring these compression and heat rejection

processes closer to the ideal isothermal process. However, this will increase thecapital cost of the plant. Using 3 compressors as shown below, the required work is72.6MJ/kmol of C6H6 (20% increase over the isothermal work)

State 1T1=300K, P1=1atm, h1=810kJ/kg

State 2 (Isentropic compression 1)P2=6atm, S2=S1, h2=920kJ/kg

State 3 (heat rejection)T3=300K, P3=6atm, h3=805kJ/kg

State 4 (Isentropic compression 2) P4=30atm, S4=S3, h4=910kJ/kg


State 6 (Isentropic compression 3)P6=100atm,S6=S5, h6=840kJ/kg


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6

II.

In combustion process, this has only O 2 and C 6H6 in the reactant:

HC 6 + XO2 6CO 2 + 3 O H + ( X 5.7 )O2 2

From H R (T = 832 K ) = H p (T =1573 K ) , we get X = .56.1102 3

Hence, in the initial separation of N2

T=300K,110.56O2

110.56*3.76N2+110.56O2

P=1atm

110.56*3.76N2

T=300K,P=1atm

Required work for separation: 674.8MJ/kmol of C6H6 (Significant increase in workdue to the increase in the number of moles being separated)

However, if we do not throw out O 2 after the turbine and circulate it, we only use7.5moles of O2 instead of 110.56moles per 1 mole of C6H6. Hence, the requiredwork is674.8MJ/kmol*7.5moles/110.56moles=45.8MJ/kmol of C6H6

One problem of circulating O2 is that we need to separate O2 from CO 2 in the product as follow:

T=300K,

6CO26CO2+103.06O2

P=1atm

103.06O2

T=300K,P=1atm

The required work of this process is 58.0MJ/kmol of C6H6.

Total required work for separation is 45.8+58=103.8MJ/kmol of C6H6.

There is no change in the liquefaction work of CO 2.

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III.

Combustion (adiabatic, p=1atm) HC 6 + 6(O + 76.3 N 2 ) 6CO + 3 H 2 + 76.3*6 N 2 6 2 2

From H R (T = 300 K ) = H p (T ) , we get T 2 = 2342K 1 2

Separation (Const T)

6CO2

6CO2+6*3.76N2+

3H2T=2342K,P=1atm

6*3.76N2+3H2

T=2342K,P=1atm

Required work for separation: 299MJ/kmol of C6H6 (Significant increase due toincrease in T)

For the liquefaction work, if we assume inlet temperature is 300K, then we get the samerequired work (60.7MJ/kmol of C6H6). However, note that we may be able to useextracted heat from CO 2 since the temperature of the CO 2 after the separator is 2342K.The availability of CO 2 can be calculated in the following process, assuming environmenttemperature is 300K

6CO2 6CO2T=2342K T=300KP=1atm P=1atm

W = H 1 H 2 T (S S 2 ) = 6moles * (112 MJ / kmol 300 K 319.0* MJ / kmol )max o 1= 8.97 MJ

Hence, the total required work for sequestration is(299-97.8+60.7) MJ/kmol of C6H6=262 MJ/kmol of C6H6

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SPRING 04HOMEWORK IIIDUE DATE, March 1, 2004

Most fuel cells run on pure hydrogen, or a mixture of inert gases and hydrogen, as afuel; and pure oxygen, or air as an oxidizer. Storing hydrogen requires either ahigh-pressure tank, if kept in the gaseous phase, or a cryogenic tank, if maintainedin the liquid phase. Instead. One can start with another, easier to store fuel, e.g.,methane CH4, reform that fuel to H2 and CO, and use this mixture in the cell. Hightemperature solid oxide fuel cells are particularly suitable for processing suchmixture. In the homework, you will use a thermochemical properties code, EQUILto estimate the reforming requirements, and the cell efficiency.

The reforming process proposed here is a combined steam reforming and gas shift in onestep, that is:

CH 4 + 2 H 2O CO 2 + 4 H 2 .

However, as we saw in class, this is a reversible reaction and at any state, the mixture, ingeneral, is composed on CH4, H2O, CO, CO2, H2. The mixture composition dependsstrongly on the temperature.

(1) Use EQUIL to determine the composition of the gas mixture as it exits areformer at T = 1100 K. Gaseous methane and liquid water enter the reformerat 300 K. The reformer operates at atmospheric pressure.

(2) Calculate the heat transfer to the reformer per unit mass of the mixture.

(3) Since the reformed gas contains, besides H2, CO, we will use it in a SolidOxide fuel cell, which itself is maintained at 1100 K, and is supplied, besidesthe reformed gases, with air at 1100 K. Calculate the maximum electricalwork transfer from the fuel cell, per unit mass of the reformer mixture, and theheat transfer to the cell from an environment maintained at 1100 K.

(4) Calculate the maximum EMF for this cell (open circuit voltage), knowing thatfor each molecule of H2 or CO consumed, the anode produces two electrons.

The curve in the next page shows the impact of the temperature at the reformer exit onthe composition of the reformed gases. It was obtained using EQUIL, starting with onemole of CH4 and two moles of water in the liquid form.

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HW 3 SOLUTION

1. Q

Reformer

reforming

H2H1

CH4+2H2O(L) CO,CO2,H2,H2O,CH4300K T=1100K

Using Equil one can calculate the mole fractions of the final stage and enthalpies ofthe initial and final stages. In the "Gas Chemistry input" (chem.inp) file in Equil, youneed to specify CO, CO2, H2, H2O and CH4. Also, you need to supply "Surfacechemistry input" for H2O in liquid form. 1 An example of input files is as follow:

Gas Chemistry Input ELEMENTS C H O ENDSPECIES CO H2 CO2 CH4 H2O END

Surface Chemistry InputBULKH2O(L)END

Application InputREAC CH4 1REAC H2O(L) 1TPPRES 1TEMP 300CNTNENDREAC CH4 1REAC H2O(L) 1TPPRES 1TEMP 1100 END

Using the above input files, you will get the following answer:

1 You should not put H2O(L) in the "Gas Chemistry input". The gas chemistry input is for gases only.

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P (atm) 1.0000E+00 1.0000E+00T (K) 3.0000E+02 3.0000E+02V (cm3/gm) 4.7205E+02 4.8916E+02H (erg/gm) -1.2410E+11 -1.2380E+11U (erg/gm) -1.2458E+11 -1.2429E+11S (erg/gm-K) 6.2814E+07 6.3889E+07W (gm/mol) 1.7358E+01 1.7358E+01

Mol FractionsH2 0.0000E+00 1.3448E-05CO2 0.0000E+00 3.3621E-06CH4 3.3333E-01 3.3333E-01H2O 0.0000E+00 1.2061E-02H2O(L) 6.6667E-01 6.5459E-01

GAS PHASEMols 1.0000E+00 1.0362E+00W (gm/mol) 1.6043E+01 1.6112E+01V (cm3/gm) 1.5345E+03 1.5279E+03

Mol FractionsCO 0.0000E+00 1.1463E-13H2 0.0000E+00 3.8935E-05CO2 0.0000E+00 9.7337E-06CH4 1.0000E+00 9.6503E-01H2O 0.0000E+00 3 .4919E-02

BULK PHASE: BULK1Mols 2.0000E+00 1.9638E+00

Mol FractionsH2O(L) 1.0000E+00 1.0000E+00

**** CONTINUING TO NEW PROBLEM*******

KEYWORD INPUT

REAC CH4 1REAC H2O(L) 2TPPRES 1TEMP 1100Reached end of input ...

Constant temperature and pressure problem.

MIXTURE: INITIAL STATE: EQUILIBRIUM STATE:

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P (atm)T (K)V (cm3/gm)H (erg/gm) U (erg/gm) S (erg/gm-K)W (gm/mol) Mol FractionsCO H2CO2CH4H2OH2O(L)

GAS PHASEMolsW (gm/mol) V (cm3/gm)Mol FractionsCOH2CO2CH4H2O

1.0000E+001.1000E+031.7327E+03-5.6216E+09-7.3773E+094.8881E+071.7358E+01

0.0000E+000.0000E+000.0000E+003.3333E-010.0000E+006.6667E-01

1.0000E+001.6043E+015.6263E+03

0.0000E+000.0000E+00

0.0000E+001.0000E+000.0000E+00

BULK PHASE: BULK1Mols 2.0000E+00Mol FractionsH2O(L) 1.0000E+00

Total CPUtime: 1 (seconds)

Mole fractions after reformer

1.0000E+001.1000E+038.6517E+03 -4.4884E+10-5.3651E+101.9509E+081.0432E+01

1.5953E-016.3843E-013.9961E-028.4767E-041.6123E-010.0000E+00

4.9915E+001.0432E+018.6522E+03

1.5953E-016.3843E-013.9961E-028.4767E-041.6123E-01

0.0000E+00

0.0000E+00

CH4 CO CO2 H2 H2O

Mole fractions 8.48e-4 0.16 0.04 0.638 0.161 Number of moles/(per kg mixture ofCH4 and 2H2O)

0.08 15.4 3.8 61.3 15.5

2.

Q reforming =H 2-H 1

Q reforming =7.910MJ/(kg mixture of CH4 and 2H2O)

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3.

Fuel cell

Qcell

Wmax

CO,CO2,H2,H2O,CH4T=1100K

H2O+CO2+N2T=1100K

Air,T=1100K

Required O2 in the cell per one mole of mixture after reformer can be calculated as

O2Required

CH4 8.48e-4 8.48e-4*2

CO 0.16 0.16/2H2 0.638 0.638/2

Total 0.4

Hence, 0.4 mole of O2 and 0.4*3.76=1.504 mole of N2 is required in the cell per onemole of mixture. In terms of mass, 5.25kg of air is required per kg of (CH4+2H2O)

To calculate G , we use Equil again and an example of input files is as follow:

Gas Chemistry Input ELEMENTS C H O N ENDSPECIES CO CO2 CH4 O2 H2 H2O N2 END

Application InputREAC CH4 8.48E-4REAC CO 0.16 REAC CO2 0.04REAC H2 0.638REAC H2O 0.161REAC O2 0.4 REAC N2 1.504PT PRES 1TEMP 1100 END

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Solutions are as follow:

Constant temperature and pressure problem.Calling SUBROUTINE EQUIL

WORKING SPACE REQUIREMENTSPROVIDED REQUIRED

INTEGER 341 341REAL 1196 1196CHAR 7 7

MIXTURE: INITIAL STATE: EQUILIBRIUM STATE:

P (atm) 1.0000E+00 1.0000E+00T (K) 1.1000E+03 1.1000E+03V (cm3/gm) 4.0094E+03 3.4593E+03H (erg/gm) 1.3874E+08 -3.1069E+10U (erg/gm) -3.9238E+09 -3.4574E+10S (erg/gm-K) 1.0306E+08 9.3398E+07W (gm/mol) 2.2513E+01 2.6092E+01

Mol FractionsCO 5.5099E-02 1.1214E-04CO2 1.3775E-02 8.0049E-02CH4 2.9203E-04 1.0086E-16O2 1.3775E-01 8.7922E-13H2 2.1971E-01 4.4343E-04H2O 5.5444E-02 3.1913E-01

N2 5.1793E-01 6.0027E-01

Total CPUtime: 2 (seconds)

W = G = ( H TS 2 ) + ( H TS 1 )max 2 1=-(-3.1069E+10-1100*9.3398E+07)+(1.3874E+08-1100* 1.0306E+08)= 2.06x10 10erg/gm=2.06MJ/(kg mixture)

We can get W max per unit mass of the reformer mixture by

W max =2.06MJ/(kg mixture)*(1kg+5.25kg)/ (kg mixture of CH4 and 2H2O)=12.9MJ/(kg mixture of CH4 and 2H2O),

Qcell can be calculate from energy balance as

Qcell W = H 2 H 1 =-6.66MJ/(kg mixture of CH4 and 2H2O)max

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4.

Mole fractions after reformer CH4 CO CO2 H2 H2O

Mole fractions 8.48e-4 0.16 0.04 0.638 0.161

Number of moles/(per kg mixture ofCH4 and 2H2O)

0.08 15.4 3.8 61.3 15.5

If we represent W max per mole of H2 and CO used,12.9MJ/(kg mixture of CH4 and 2H2O)* (kg mixture of CO and2H2)/(61.3+15.5 moles

of H2 and CO)=168J/mol

Voltage= 168*1000J/mol /(2*96485C/mol)=0.87Volt

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SPRING 04HOMEWORK IV

DUE DATE, March 10, 2004

1. For an overall reaction ofH2O H + + OH -

(a) Use the Table 4.1 in the Lecture Note Electrochemical Energy Conversion andStorage to define two half cell reactions for the above reaction and calculate theelectrochemical potential of the reaction under standard conditions.

(b) Calculate the molalities of H + and OH - in deionized water at 298K from the Gibbsfree energy of reaction. Show calculation step by step.

2. An exhaust gas analyzer uses an electrochemical solid state cell (sensor) to measureoxygen concentrations in the exhaust. This cell consists of a solid-state electrolytezirconium dioxide (ZrO 2) with yttria (Y 2O3) that provides oxygen ion conduction in thecell, and Pt electrodes used for anode and cathode, depicted as,

2 2 2 3 2( , ) ( , )+eg air Pt O exhaust p ZrO Y O O air p Pt

The electrode reaction that occurs at high temperature at both the anode-electrolyte andcathode-electrolyte interfaces is

O2 + 4e 2O2-

(a) Please derive the governing equation for the cell potential as a function of pressures,Peg and P air .

(b) What would the cell voltage be when the partial pressure of oxygen in the exhaust gasis 0.02atm?

3. Consider an electrochemical cell with different concentration of aqueous CuCl 2 solutions in each 1/2 cell. 3.5 molal CuCl 2 on one side and 1 molal CuCl 2 solution on theother side separated by an ion conducting membrane permeable only to Cu +2 and Cl -1 ions. The electrodes are Pt coated with a thin film of copper metal.

(a) Initially will the cell produce a current against a resistive load? Explain.(b) What is the maximum amount of work that such a cell would produce?(c) What happens to the concentrations of ions in the cell after a period of time?(d) Is there a way that you could modify this cell to determine the mean ionic activity of

CuCl 2 as a function of concentration? (hint- Consider coupling the cell to a standardhydrogen electrode via a salt bridge.

The molality is the number of moles of solute present in 1000 grams of water whilemolarity is the number of moles of solute present in one liter of water. Mole fraction,

1000/18imolality

X molality

=+

. State all assumptions made in answering questions 1-3.

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Homework IV Solution

1. For an overall reaction ofH2O H + + OH -

(a) Use the Table 4.1 in the Lecture Note Electrochemical Energy Conversion andStorage to define two half cell reactions for the above reaction and calculate theelectrochemical potential of the reaction under standard conditions.

H2O+e - 1/2H 2+OH - -0.828Volt (Anode)1/2H 2 H + + e - 0 volt (Cathode)

H2O H ++OH - -0.828Volt

(b) Calculate the molalities of H + and OH - in deionized water at 298K. Show calculationstep by step and state all assumptions made.

Note that is related to mole fractions of H + and OH - by the following equationsassuming that the activity coefficients of the species are 1:

Eq. 12

( , ) exp( )1

l

o H OH H H OH OH r

p H O

a a m m G K T P

a RT

+ + + = = =

Assuming 1 H OH +

= = ,

( , ) exp( )1

o

H OH r pm m G K T P RT

+ = =

Eq.2 or GnF = .From Eqs. 1 and 2, one can get K w=[H +][OH -]=1.0x10 -14 at 298K. If pH=7, then[H+]=[OH -]=1.0x10 -7. Then, the molalities (m H+ and m OH- ) and molarities ([H +] and [OH -

]) of H + and OH - are identical, which is equal to 1.0x10 -7mole per 1000g water.

The molality is the number of moles of solute present in 1000 grams of water while molarity is thenumber of moles of solute present in one liter of water. Note that 1000 grams is equivalent to oneliter of water.

2. An exhaust gas analyzer uses an electrochemical solid state cell (sensor) to measureoxygen concentrations in the exhaust. This cell consists of a solid-state electrolytezirconium dioxide (ZrO 2) with yttria (Y 2O3) that provides oxygen ion conduction in thecell, and Pt electrodes used for anode and cathode, depicted as,

2 2 2 3 2( , ) ( , )+eg air Pt O exhaust p ZrO Y O O air p Pt The electrode reaction that occurs at high temperature at both the anode-electrolyte andcathode-electrolyte interfaces is

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O2 + 4e 2O 2-

(a) Please write the governing equation for the cell potential as a function of pressures, P eg and P air .

If one assumes that P eg >P air and P eg and P air are the partial pressure of O 2 in theexhaust and the air, respectively, the potential generated by partial pressure differenceis as follow:

)log(air

eg

P

P

nF RT =

Note that Pt electrodes in the air acts as a cathode due to higher partial pressure of O 2.Also, it is assumed that the activity coefficients are 1.

(b) What would the cell voltage be when the partial pressure of oxygen in the exhaust gasis 0.02atm?

State all assumptions made in answering (a) -(b)

If we assume T=300K, P air =0.21atm, then we get)21.0/02.0log()964854/(300314.8 = =15.2mV

3. Consider an electrochemical cell with different concentration of aqueous CuCl 2 solutions in each 1/2 cell. 3.5 molal CuCl 2 on one side and 1 molal CuCl 2 solution on theother side separated by a ion conducting membrane permeable only to Cu +2 and Cl -1 ions.

The electrodes are Pt coated with a thin film of copper metal.

(a) Initially will the cell produce a current against a resistive load? Explain.

A concentration cell produces a voltage as it attempts to reach equilibrium, which willoccur when the concentration in both cells are equal. To reach this point, Cu 2+ ions inthe concentrated solution (3.5molal solution) are reduced while oxidation increasesCu 2+ in the dilute solution. However, oxidation of Cl -1 cannot occur since the standard

potential of Cl - is higher than that of Cu 2+. Hence, released electrons from the anode by Cu 2+ oxidation can produce current passing through a resistive circuit. To satisfythe local electroneutrality, Cl - should migrate from the concentration cell through the

membrane to the dilute cell. These reactions are represented as follow:

Electrode 1 reaction (Cathode, 3.5molal solution)Cu 2+ +2e - -> Cu(s)

2 20.337 /( ) log(1/ )cathode Cu CuV RT nF m + += where activity coefficient is equal to 1.

Note that Cl 2(g) (Cl 2(g)+2e - -> 2Cl - ) is not available and hence reduction of Cu 2+ only occurs in the cathode.

Electrode 2 reaction (Anode, 1molal solution)

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Cu(s) -> Cu 2+ +2e -

2 20.337 /( ) log( )anode Cu Cu

V RT nF m + +=

Potential in the resistive load can be calculated as2 2 2

2 2 2

[ ] [ ]log log

[ ] [ ] Dilute DiluteCu Cu Cu

cathode anodeConc ConcCu Cu Cu

m m RT RT nF m nF m

+ + +

+ + +

= = = assuming the

activity coefficients of Cu2+ ions in the dilute and concentrated solutions are the same.

Note that 2 2[ ] [ ]concentrated DiluteCu Cum m+ +> and hence 0> .

(b) What is the maximum amount of work that such a cell would produce?

Note that concentration of Cu 2+ changes from 3.5 to (3.5+1)/2=2.25 in theconcentration cell and from 1 to 2.25 in the dilute cell.

If we define X is the change in the molal values in the concentrated solution andassume that the activity coefficient of Cu 2+ is 1 or does not change within the range

between 1 to 3.5 molal, we get

2

2

log( )Cu Dilute

Cu Conc

m RT nF m

+

+

= =

Since X changes from 0 to 1.25.

Work =- +==

25.1

0

)5.3

1log( dX

X X

RT nF G

Assuming T=300K, we get Work=1.47kJ/(mole of CuCl 2)

(c) What happens to the concentrations of ions in the cell after a period of time?

Over a long period of time the potential will drop to zero when the concentrations ofCu 2+ become 1.25 molal in both cells. Also, Cl -1 ions will move to the dilute solution

to satisfy local electroneutrality (1.25 molal in both cells).

(d) Is there a way that you could modify this cell to determine the mean ionic activityof CuCl 2 as a function of concentration? (hint- Consider coupling the cell to astandard hydrogen electrode via a salt bridge. State all assumptions made inanswering (a) -(d)

If the cathode of the cell is connected to the hydrogen electrode, we get the followingreaction:

Hydrogen electrode (Anode)

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2H ++2e - -> H 2 Electrode 2 reaction (Cathode, 3.5molal solution)Cu 2+ +2e - -> Cu(s)

Potential between two electrodes can calculated as

( )

22

2

28.314 /( ) 300ln( ) 0.337 ln(1/ )2 96485 /

sCu H O

H Cu

a a RT J mol K K V Cu

nF C mol P a

+

+

+ = =

Hence, one can get value by measuring .

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SPRING 04HOMEWORK V

DUE DATE, March 29, 2004

Problem #1In this problem, you will practice working with finite rate reaction equations in thecontext of synthetic fuel production. Review the ozone chemistry example in the note

before working on this example. You will discover how operating conditions must beselected carefully to maximize the yield of the desired chemical while minimizing others.

A catalytic reactor is used for the synthesis of methanol, CH 3OH, starting with water andcarbon monoxide that are produced from coal gasification. The catalyst in the reactor isactive at temperatures between 300 and 450 K. The reactor runs at isothermal conditions.The important elementary reactions for the conversion of carbon monoxide and waterto methanol are:

(1) CO + 2H 2 CH 3OH(2) CO + H 2O CO 2 +H 2(3) CH 3OH H 2 + CH 2O

While these are, in principle, not elementary reactions, for the purpose of this analysisthey can be considered as if they were. These reactions take place in the gas phase, andunder isothermal conditions. The first two reactions are reversible, the third in anirreversible dissociation of methanol that, if possible, should be avoided.

The feed gas into the reactor consists of 7/15 hydrogen, 1/5 carbon monoxide, 1/5 carbondioxide and 2/15 steam, all by volume. The total molar flow rate is 300 mol/s, the inlettemperature of the feed gas ranges from 300 - 400 K, and the reactor pressure rangesfrom 1 160 atm.

The objective of the following analysis is to find the conditions at which we canmaximize the production of methanol within the reactor, including the pressure,temperature and residence time. The residence time is defined as the time the gas spendswithin the reactor to reach the final state. The following data are provided for thereactions:

K 131667 2 m3

2

C 1= ( 001987.0 T )

exp 30620 1 1 mol

R T 298

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103943 K C 2 = 9834 1 1

exp

R T 298

23

5.2exp933.0

18000

31400 1 1

m 1 s k f = 1, R T mol 330 3 1 1

mk f exp636.0 = R 2, T mol 300 s

5.1exp244.0

H o OH CH f 3

28956

R 1 1

1 s k f = 3, T 325

201kJ mol J mol K ( ) / = 44 /( ) = , c,CH P

OH 3

H o OCH f 2 kJ mol 5.35 J K mol ) ( ) = 116 / /(, c = ,CH P O2

The activation energies in these expression are given in cal , and the universal gasconstant is R=1.987 cal/mol K . Assume that the inlet gas and the reactor temperatures arethe same.

1. Write down the reaction rate equations for the three reactions.2. Write down the formation rates for each species in terms of the time rate of

change of its molar concentration.3. Integrate these equations for p = 1, 10 and 100 atm, at temperatures in the

given range, take T = 320, 340, 360 and 380 K.4. Use these results to determine the impact of the pressure on the concentration

of methanol in the products. Can you explain this trend form the reactionequations?

5. Use the results to determine the optimum operating conditions for maximizingmethanol production in terms of p, T and residence time.

6. How much heat transfer is required in your optimum case to keep thetemperature constant within the reactor?

Integrating the differential equations is most easily done using Matlab ordinarydifferential equations solver, Sungbae will post an example on how to do that.

Problem #2

Consider the electrode reaction O + e = R. Under the conditions that bulk concentrations[O]*=[R]*=1mM, k o= 10 -6cm/s, and =0.3.

1. Calculate the exchange current density, j o=io/A in microamps/cm 2.2. Draw a current density overpotential curve for this reaction for currents up to

600 microamps/cm 2 anodic and cathodic. Neglect mass transfer effects.3. Draw log vs. a curves for the current ranges in question 2.

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Homework V solution

Problem 1.

1. The reaction rate equations for the three reactions are as follow:11,1, / C f b K k k =

22,2, / C f b K k k =

[ ][ ] [ ]OH CH k H COk R b f r 31,221,1, = [ ][ ] [ ][ ]222,22,2, H COk O H COk R b f r = [ ]OH CH k R f r 33,3, =

Note that you should consider backward reactions in the first two reactions.

2. The rates of formation of each species are as follow:[ ]

2,1, r r R Rdt COd

=

[ ]3,2,1,

2 2 r r r R R Rdt H d

++=

[ ]3,1,

3r r R Rdt

OH CH d =

[ ] 2,2 r Rdt O H d

=

[ ]2,

2

r Rdt

COd =

[ ]3,

2r Rdt

OCH d =

3.

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4. Increasing the pressure raises the mole fraction of CH3OH when T 340, but does nothave a significant effect after T 360. It is due to the following reasons:1) In the first reaction, which generates CH 3OH, the forward reaction favors higher

pressure since the number of moles decreases during the forward reaction. Thus, byincreasing the pressure of the reactor, the concentration of CH 3OH increases more

rapidly.2) After T 360, even at low pressure, the first reaction is sufficiently fast. Soincreasing the pressure does not help. Also, note that the concentration of CH 3OHdrops back to almost zero with time due to the third reaction (dissociation ofCH3OH) indicating that the third reaction becomes active after T 360.

5. The maximum mole fraction of CH 3OH one can obtain is ~33% at T=340K and P=10-100atm with a residence time less than 1msec.

6. The mole concentration changes at T=340K, P=10atm and residence time=1msec areas follow:

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HOMEWORK VI Solution

1)

One can get the equilibrium fuel cell voltage from the following equation:

F n

P T G

e

r ),(= Eq.(1)

Note that r G is the function of T and P. Using Equil, one can get)150,353( kPa P K T G r == =230kJ/(mole of H2). Using Eq.(1), we get

Volt

moleC

molekJ 19.1

/96485*2

/230 ==

2)

a ,cathode )/

/ln( ,

Ai

Ai

F n

RT cathodeo

e =

a ,anode )/

/ln(

)1(,

Ai

Ai

F n

RT anodeo

e =

A R A

i Ri ==

= r cell a ,cathode - a ,anode

Assuming =0.5 , we get the following plot

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3)

A H

AW

H

W

r

out

r

out ond

//

sec &&

=

=

)/(1047.1/964852

1/284

21/ 23 mW

A

i

A

i

mol C mol kJ

F A

ih A H

r r

=== &

=out W & current density x cell potential (W/m 2)

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How to use CHEMKIN

1. To open a Chemkin Window

1) logon Athena

2) At the Athena prompt, typeathena % add chemkin athena % chemkin

3) The following Chemkin window will pop up.

Figure 1 Chemkin window

4) Select an application from Chemkin window. The available applications andtheir functions are described below:

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Aurara: well stirred reactor simulation. Creslaf: channel flow simulation. Equil: equilibrium state calculation. Oppdif: diffusion flame between two opposing nozzles. Plug: Plug-flow chemical reactor simulation. Premix: steady, laminar, one-dimensional premixed flamesimulation. Senkin: predicts homogeneous gas-phase chemical kinetics in a

closed system with sensitivity analysis. Shock: predicts the state of the products behind incident andreflected shocks. Spin: simulates one-dimensional rotating-disk. Surftherm: analyzes the thermochemical and kinetic data in gasphaseand surface chemical reaction mechanisms.

In the next section, we describe how to use Equil. Other applications can be used in asimilar manner. However, there is one distinct difference between Equil and otherapplications. Equil application doesnt use reaction mechanism data, however others use it.

2. How to Use Equil

Equil computes the chemical equilibrium state of an ideal gas or solution mixture.

1) Click the Equil button in the Chemkin window.

2) The window in Figure 2 appears.

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Figure 2 Equil window

3) To calculate equilibrium state, one needs to generate two input files:chem.inp and gas_equil.inp.

4) If you click the edit button in Gas Chemistry, you can view and editchem.inp file as shown in Figure 3. This chem.inp file contains elementsand species data.

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Figure 3 Chem.inp file

5) You can generate your own file and name it ****.inp instead of chem.inpfile, but the file should be in the directory specified in the working directory. Togenerate the ***.inp file, either use the text editor inAthena or use a PC and send the file to Athena through FTP.

6) Next, you need to generate Gas_equil.inp. When you click the edit button

in Equil, you will see Figure 4.

Figure 4 Gas_equil.inp

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REAC in Figure 4 stands for reactant; each line contains a reactantidentified by its chemical symbol and its number of moles in the mixture.The reactants in the Figure 4 example are H 2 , O 2 and N 2 . Their number of moles is: 2, 1 and 3.76,respectively.

HP means constant enthalpy and pressure conditions. Other options are available, e.g., EV, etc.

TEMP specifies the starting temperature (Units-K)TEST specifies an estimate of the equilibrium temperature (Units-K)PRES specifies the starting pressure (Units-atm)CNTN is used to continue the computation in a different initial condition

after END keyword. Therefore, if CNTN is used, one can get more than one solution. In Figure 4, the initial temperature (keyword TEMP) is changedfrom 300 to 400.

END signifies the end of the keywords for a particular equil calculation.

7) Also, you can generate your own file and name it ****.inp instead ofGas_equil.inp file.

8) Once chem..inp and Gas_equil.inp files are ready, you can calculate theequilibrium conditions. Click the run button and the program will start computingthe equilibrium conditions.

9) After the computation is finished, click view the button of equil.out. You will seethe following text.

EQUIL: Chemkin interface for Stanjan-IIICHEMKIN-III Version 3.20, 2000/09/11DOUBLE PRECISIONWORKING SPACE REQUIREMENTSPROVIDED REQUIREDINTEGER 579 579REAL 2052 2052CHARACTER 34 34CKLIB: CHEMKIN-III GAS-PHASE CHEMICAL KINETICS LIBRARY,DOUBLE PRECISION Vers. 5.31 2000/10/02Copyright 1995, Sandia Corporation.The U.S. Government retains a limited license in this software.KEYWORD INPUTREAC H2 2REAC O2 1REAC N2 3.76HPTEMP 300TEST 2000PRES 1CNTNENDConstant pressure and enthalpy problem.EQUIL: Chemkin interface for Stanjan-IIICHEMKIN-III Version 3.20, 2000/09/11DOUBLE PRECISIONWORKING SPACE REQUIREMENTSPROVIDED REQUIRED

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T (K) 5.0000E+02 2.4884E+03

V (cm3/gm) 1.9620E+03 8.4546E+03H (erg/gm) 2.8378E+09 2.8378E+09U (erg/gm) 8.4978E+08 -5.7288E+09S (erg/gm-K) 9.5035E+07 1.1194E+08W (gm/mol) 2.0911E+01 2.4150E+01Mol FractionsH2 2.9586E-01 2.0206E-02H 0.0000E+00 3.3760E-03O2 1.4793E-01 7.6135E-03O 0.0000E+00 1.1825E-03OH 0.0000E+00 1.0962E-02HO2 0.0000E+00 2.2850E-06H2O 0.0000E+00 3.1430E-01

N2 5.5621E-01 6.4236E-01H2O2 0.0000E+00 1.9554E-07

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Solving ODEs in MATLAB

To solve ODEs, one can use ode solvers in the Matlab, e.g., ode23tb. You can get help by typing help ode23tb in the Matlab command window. As an example, consider thefollowing Van der Pol equation:

2221

1 )1( x x xdt

dx= ,

12 x

dt

dx= ,

)0(1 =t x =0 and 25.0)0(2 ==t x .

Use the two m files attached to solve the above ODEs. Put both m files in the samedirectory and execute the example.m file. Then you will get a time plot of x1 and x2 .

Note that the van.m file contains information of the Van der Pol equation and theexample.m file specifies initial conditions, start and end integrations times and callsvan.m file to solve the ODEs.

1. example.m file%%%start timeto=0;

%%%end timetf=30;

%%%time stepstep=0.5;

%%% initial conditionsxo=[0 0.25]';

%%% solve the odes%%% t: time , x: output[t,x]=ode23tb('van',[to:step:tf],xo);

%%% plot solutions plot(t,x)xlabel('time')

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