ScottDombroskyZackMeyerMichaelCzajkowskiDominicJosephsSpauldingRobSwahl(unofficialmember)

PrinciplesofenergyEngineeringEGEE302

Group14FinalProjectFall2013

A600MWpulverizedcoalfiredpowerplantisoperatinginCarbonCounty,Wyoming.Thisplantusescoalfromalocalcoalminewiththefollowingcompositionwith18%excessair.Proximate:Moisture 12.0%VM 38.4%FC 43.6%Ash 6.0%HeatingValue(BTU/lb)Asreceived 11,030Ultimate(wt.%):Carbon 62.6Hydrogen 4.5Nitrogen 1.2Sulfur 0.7Ash 6.0Oxygen 13.0Thepowerplantoverallthermalefficiencyis33.0%.Theplantisrequiredtocapture99%ofthesulfurdioxideemissions.Theplantalsousesaparticulatecollectiondevice(ESP)withanefficiencyof99.9%.Theaverageambienttemperatureis80Fand60%RH.TheForcedairdraftfanandInducedairdraftfanbalancethefurnacedraft.Ashanalysisshowedthattheunburntfuellossis0.2%.Assumethatfuelentersatambienttemperature.ThereisaselectivecatalyticreductionsysteminstalledtoreducetheNOxby80%usingNH3.Assumethat40%ofthefuelnitrogenformsNOthatreactswithNH3toreducetoN2overacatalyst.Theparticulatecollectiondeviceisabaghouse.Assumethat5%ofthetotalashflowsasbottomashfromthecombustionchamberandtherestiscapturedbytheESP.Thecleangasisgoingthroughthestack.Thestackneedstobemaintained80Fabovethedewpointofthegas.Thesteamtohighpressureturbineflowsat1000Fand2520psig.Theflowrateofhighpressuresteamis4.08millionlb/h.Ofthissteam3.77millionlbsafterexpansionto653psigwillbegoingbacktotheheatertoheatitto1000Fand587psig.

Prepareaflowchartforthisplantandlabeltheflowsquantities,pressuresandtemperaturesofeachflowaroundallthecomponents.Performmaterialandenergybalancesbycalculatingthefollowing:Coalandairfeedratesintons/hCalculatetheairflowrateinACFHthroughtheFDfan,Fluegascompositiononwetanddrybasis,FluegasflowrateinSCFHthroughtheIDfan,Fluegasproducedperlboffuel,Fluegasmolecularweight,Fluegasmolecularweight(dry)AshflowratefromtheESPintons/h.Ammoniarequirementintons/hWhatpercentoftheheatinputiscapturedbythehighpressureturbineandintermediatepressureturbinesteam?Heatlossthroughthestack.Theadiabatictemperaturethatcanbeachievedbythiscoalinthecombustionchamber.Recordsofmeetings,howtheresponsibilitiestocarryouttheprojectweredivided,whoperformedwhichpartandsincethereareseveralvaluesthathavetobetransmittedfromoneparttoanother,documenttheproblemsencounteredandhowthegrouphasovercomethosemustbedocumentedandsubmittedwiththeprojectreportsubmission.Thereportmustbelookingprofessionalandmustbeinwordandallcalculationsmustbedoneusingexcel.Ifthereareanyquestionsoraftertheflowchartispreparedyoucanconsultmetomakesurethatyouontherighttrackifnecessary.Youmustclearlywriteyourassumptionsandreasonsforthoseassumptions.

Group14EGEE302Date:22November10December,2013

Allgroupmemberindividuallylookedovertheprojectinattemptstotrytofigureouttheschematicandstateadateforthemeetinglocation.Atthistimeasetupofthegroupexcelspreadsheetwasmadetoinputquestionsorotherinquiriesthatthegroupasawholehad.Date:11December,2013

GroupMeeting:

Allfourofthegroupmembersmetandworkedontheprojecttogether.WewerejoinedbyoneofourclassmatesRobertSwahlwhohadasimilarprojectasoursandcontributedtoourproject.Whatourmainfocuswastodoeachproblemtogethersoastoeliminateconfusionandkeeptheworkschedulewasontrack.Thefirstproblemwehadwasactuallytryingtounderstandthequestionsthattheprojectwasasking,suchasdecidingwhattheschematiclookedlikeandwhatthecorrectprocesseswere.WeeventuallyagreedontheschematicusedinthespreadsheetaftergoingtoofficehoursandthestudysessionthatwasprovidedonWednesdayDecember11,2013.

Someotherproblemswehadweredecidingifthechoicetouseambientorstandardtemperaturewereappropriateforcertainsituations,butmostoftheconflictarosefromdifferentassumptionseachteammemberwasmakingduetothedifficultyoftheproblem.Thequestionsweresolvedwithcollaborativeeffortandbyapplyingcommonsensetojustifythatouranswerswerereasonable.

Theproblemwearecurrentlystuckonisheatlossthroughthestackbutwecantcurrentlysolvethequestionbecausethereisnotenoughinformationtounderstandwhattoactuallysolve.Currentlywemadealistofquestionfortomorrow'sclass(12December,2013)inordertofinishoutthelastthreebulletpointsoftheproject.Date:12December,2013Classnotes:

Duringclassthegroupaskedquestionstotheunsolvedproblemsofyesterday.Theresultswere.Heatlossthroughstack:DeltaH=integral(StandardTemp.toAdiabaticflametemp.)Heatlosstotheturbines:(energyintotheturbine/energyfromtheboiler)Adiabaticflametemperature:energyfromcoal&inputairenergyofproducts=0

GroupMeeting:AllfourmemberplusRobertSwahlmetupontheprojectaftertheEGEE302class.As

ofthebeginningofthemeeting,weattemptedtofinishoffthelast3problemsintheproject.Thehardestproblemthatweworkedonwastocalculatetheadiabatictemperatureinthecombustionchamber.

Asagroup,wewereabletoresolveallthequestionswehadandcreatedthefollowingsolutionsbelow.

AlltheanswerstotheproblemscorrespondtothetabAnswers/BulletedPointsintheexcelspreadsheet.

Forourschematicwecreatedthediagramusingthegeneralinformationgiveninthetemplateoftheproject.ToconfirmthatthediagramwascorrectourgroupconsultedwithDr.Pisupati.1. Coalandairfeedratesintons/h.

Tosolveforthecoalfeedrateweusedthetotalpowerofthecoalfireplantdividedbythe33%thermalefficiencysothatwewouldgettherequiredenteringenergy.TogetfromMWtoBtu/hrweusedaconversionfactorthendividedbythe11030(Btu/lb)coalHHVtofindthelbcoal/hrandthenconvertedto(toncoal/hr).

Answer:281.22toncoal/h

Solvingfortheairfeedrate.Theactualairsent977.70(lbair/h)dividedbyacoalfeedrateassumption(100lbcoal/h)togettheairfeedrate9.78(lbair/lbcoal)(withrespecttothefuelfeedrate)andusedthecoalfeedrate281.22(toncoal/h)tofind(tonair/h).

Answer:2753.402tonair/h

2. CalculatetheairflowrateinACFHthroughtheFDfan.

BecauseourRT/Pvalueisgreaterthan20L/mole,weassumedwewereabletousetheidealgaslawinordertofindthedensityoftheairflowrate.Thevaluethatwascalculatedwas.0000363(ton/ft^3).Weneededtofindtheairflowratethroughthefanbydividingthe(ton/ft^3)bytherequiredairfeedrate2753.40(tonair/h)toget75,839,908.68(SCFH),actualtemperature539.67wasfoundstandardconditionsat492Rand1atmwereassumedtocalculatethe(ACFH).

Answer:83,188,055.93ACFH

3. Fluegascompositiononwetanddrybasis.

Usingtheinitialconditionsthatweregivenintheprojectfromtheultimateanalysis,webrokethecoalintoeachofitscomponentsanddeterminedtheirproductswhencombusted.Fromtheseproductswefoundthemolefractionsbelow.Answer: molefraction(wet) molefraction(dry)

CO2 0.15075 0.16452H2O 0.08369 NO 4.95E05 5.41E05

SO2 6.32E05 6.90E05O2 3.02E05 3.30E05N2 0.76541 0.83532

4. FluegasflowrateinSCFHthroughtheIDfan.

Solvingforthewetfluegasmolarflowratetotal34.609(lbmole/h)wemultiplythemolecularweightofthegas29.582(lb/lbmole)wewouldgettheamountofwetfluegasenteringthestacks1,023.60(lbwetfluegas/h)andusingthefactor359SCF/lbmole,theSCFHcanbecalculated.

Answer:367,483.017SCFH5. Fluegasproducedperlboffuel,Fluegasmolecularweight,Fluegasmolecularweight(dry).

Thefluegasmolecularweightwascalculatedusingthesumofallthemolecularweightsofitscomponentsmultipliedbytheirmolefractions.Thesameprocedureforcalculatingthedrymolecularweightbyremovingthemoisture.Usingthe1,023.630(lbwetfluegas/hr)anddividingbythe100lb/hrcoalfeedratebasistogetthe(lbfluegas/lbfuel).

Answer(molecularweight):29.577Answer(molecularweightdry):30.632651745658Answer(lbfluegas/lbfuel):10.2362957406556

6. AshflowratefromtheESPintons/h.

TheanalysissaidthattheESPhadanefficiencyof99.9%.Fromthis,weareabletotaketheamountofashfromtheultimateanalysisandmultiplyitby0.1%inordertofindouthowmuchashisallowedthroughtheESP.

Answer:0.00000285ton/h7. Ammoniarequirementintons/h(NO=NitrogenOxide)

Wenotedthat80%oftheproducedNOwasreducedwithNH3tocreateN2andH2O.Usingthiswecreatedabalancedchemicalequation.Fromtheultimateanalysis,wefoundthemolesofNOgoingtotheaminescrubber.Fromthis,andathechemicalequationwewereabletocalculatetherequiredmassflowrateofammoniarequiredtothisamountofNO.

Answer:0.00007(tonNH3/h)8. Whatpercentoftheheatinputiscapturedbythehighpressureturbineandintermediatepressureturbinesteam?WefoundtheenthalpiesofthecoalandtheairgoingtoeachturbinefromthesteamtablesB8(fromthebook).Weadjustedtheenthalpiestoourcalculatedfeedrate,theenthalpiesforthehighpressureturbineandlowpressureturbinewerefoundtobe1457.370(btu/lb)and1,517.590(btu/lb).Whenthenfiguredoutthetotalwetaircomingintotheboiler0.736151797813036(btu/hr).Tofindthetotalboilerenthalpywefirstfoundtheenthalpyofcoalbyusingthe11,030(BTU/lb)HHVmultipliedbythecoalfeedrate562,433.034(lbcoal/hr).Theenthalpyofairiscalculatedbythecoalfeed562,433.034(lbcoal/hr)multipliedbytheamountofwetair0.736151797813036(btu/hr).Thetotalboilerenthaplyis6,603,731,861.921.Togetthehighandlowpressureturbinesenthalpywemultipliedtheamountofmolesgoingthroughthesystemwitheach4,080,000.000lb/hourand3,770,000.000pounds/hwecangetourpercentages.

AnswerHighPressure:90.041%AnswerLowpressure:86.638%

9. Heatlossthroughthestack.

WetableB.9(fromthebook)andtheheatcapacitiestofindthechangeinenthalpyfromstacktemperaturetoambienttemperatureofallthegasesinthestack.Usingthetotalenthalpy805.59(btu/lbmol)ofthegasandthetotalwetflowrateofthegas34.605lbmole/htofindtheheatlossinbtu/hr.web

Answer:27,877.05btu/hr

10. Theadiabatictemperaturethatcanbeachievedbythiscoalinthecombustionchamber.

Wesetthesumoftheenthalpiesequaltozerousingthegivenheatofformationoftheproductsintheflueandtheairconstantsrelatedtothesensibleenthalpy.WefoundtheheatofformationfromB.1andB.2(fromthebook).Tofindtheadiabatictemperaturewecanusetheintegraloftheourreferencetemperature(25C)toouradiabatictemperature.ThenusetheCpvaluestofindthesensibleenthalpycalculatepluggingthatintoexcelandusingtheGoalSeekfunctionweobtainedtheadiabatictemperature.

Answer:2039.7K