Compressed Air Energy Storage at the North Campus of Durham College and the University of Ontario Institute of Technology-An Engineering Approach

Final Capstone Report
Compressed Air Energy Storage at the
North Campus of Durham College and the
University of Ontario Institute of Technology
-An Engineering Approach
April 11th
, 2016
Faculty of Energy Systems and Nuclear Science
Prepared by:
Submitted to:
Matthew Crummey
100480101
Mr. Sharman Perera
SupervisingProfessor

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Abstract
Work wascontinuedfromthe previoussemesterona CompressedAirEnergyStorage System.This
sectionof the reportwas concernedwithanalyzingthe impactof pairingthe systemwiththe north
campusof DC/UOIT.This wasaccomplishedbyusingthe historical demanddataavailable inconjunction
withthe previouslydesignedSimulinkmodel.Fromthe previoussemestersworkthere were some
designchallengesthatwere identifiedinthe Simulinkmodel thatwere tobe fixedthissemester.
Overall,one of the designchallengeswascompletely fixed,where the otherone,the lackof pressure
and heatlosseswithinthe system,wasfoundtodependonthe physical layoutof all the components,
and thuswas notable to be completed.The resultsof pairingthe Simulinkmodel withthe historical
demandshowedthatforthe base periodof 2013-2014, if the systemwere tobe operatedduring
specifictimeseachday,asavingsof $157 K in electricitycostscouldbe obtained.
Acknowledgements
Doug Crossman forprovidinginformationregardingthe electricity usage of the Durham
College/Universityof OntarioInstitute of Technology campus
StephenCassarforprovidingme withthe electricityusage informationof the campusandanswering
questionsregardingthatinformation
MarilynCrummeyforher proofreading,review, andsupport
SharmanPererafor beingasupportive andhelpfuladvisor
SamanthaSpencer,Jake Tod,and NicholasVantfoortforbeingthe mostawesome capstone groupever

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Tableof Contents
1 Project Identification...................................................................................................................1
1.1 Project Title.........................................................................................................................1
1.2 Problem Statement..............................................................................................................1
2 Literature Review........................................................................................................................1
2.1 Electricity Costs ...................................................................................................................1
2.1.1 Global Adjustment........................................................................................................1
2.2 Electricity Modelling............................................................................................................2
2.2.1 Historical Use...............................................................................................................3
2.2.2 Empirical Survey...........................................................................................................3
2.3 DC/UOIT Electricity Use Data................................................................................................4
3 Design Requirements...................................................................................................................4
4 Preliminary Analysis ....................................................................................................................4
4.1 Wholesale Electricity Price....................................................................................................4
4.2 Global Adjustment...............................................................................................................5
4.2.1 2013-2014 Base Period .................................................................................................5
4.2.2 2014-2015 Base Period .................................................................................................7
4.2.3 2015-2016 Base Period .................................................................................................8
4.3 Total Electricity Costs.........................................................................................................10
4.4 Motivation for Project........................................................................................................10
4.5 Control Method Decision....................................................................................................11
4.5.1 Explanation of the Weighting Factors...........................................................................12
5 Simulink Model Improvement....................................................................................................12
5.1 Thermal Subsystems..........................................................................................................15
5.2 Heat and Pressure Losses...................................................................................................19
5.3 Compressor and Turbine Control Subsystems......................................................................19
5.4 Pressure Calculation...........................................................................................................23
5.5 Summary of Model Changes...............................................................................................24
6 MATLAB™ and Simulink Analysis ................................................................................................27
6.1 Historical Use Data.............................................................................................................27
6.2 SimulinkAnalysis Using Representative Load Curves............................................................28
6.2.1 2013-2014 Representative Load Curve.........................................................................28

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6.2.2 2013-2015 Representative Load Curve.........................................................................30
6.3 SimulinkAnalysis Using Rolling Average ..............................................................................31
6.4 SimulinkAnalysis Using Selected Time of Use......................................................................33
6.5 Summary of Control Methods.............................................................................................35
7 Conclusion and Next Steps.........................................................................................................36
8 References................................................................................................................................37
9 Appendix 1: Sample Calculations................................................................................................38
9.1 Pressure and Heat Losses on the Compressor Side...............................................................38
9.1.1 Constants...................................................................................................................38
9.1.2 States........................................................................................................................38
9.1.3 Compressor 1.............................................................................................................38
9.1.4 Compressor 2.............................................................................................................39
9.1.5 Mass Flow Rate ..........................................................................................................40
9.1.6 Heat Losses................................................................................................................40
9.1.7 Pressure Losses ..........................................................................................................56
9.2 Pressure and Heat Losses on the Turbine Side.....................................................................58
9.2.1 Constants...................................................................................................................58
9.2.2 States........................................................................................................................58
9.2.3 Turbine 1 ...................................................................................................................58
9.2.4 Turbine 2 ...................................................................................................................60
9.2.5 Mass Flow Rate ..........................................................................................................61
9.2.6 Heat Losses................................................................................................................62
9.2.7 Pressure Losses ..........................................................................................................75
9.3 Thermal Subsystem Changes ..............................................................................................76
9.3.1 Glycerin Temperature Out...........................................................................................76
10 Appendix 2: MATLAB Codes.......................................................................................................77
10.1 Importing Time-Use Data and Signal Building ......................................................................77
10.2 Averaging Time-Use Data...................................................................................................77
10.3 RLC Creation......................................................................................................................78
10.4 Averaged RLC Creation.......................................................................................................80
10.5 Selected Time of Use..........................................................................................................82

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Tableof Figures
Figure 1: Total Cost for Electricity during 2013-2014 Base Period.........................................................10
Figure 2: Base Model from Previous Semester...................................................................................13
Figure 3: I/O View of Original Turbine Thermal Subsystem..................................................................15
Figure 4: I/O View of Updated Turbine Thermal Subsystem.................................................................15
Figure 5: Subsystem View of Updated Turbine Thermal Subsystem......................................................17
Figure 6: Subsystem View of Original Turbine Thermal Subsystem.......................................................18
Figure 7: I/O View of the Compressor Controller.................................................................................19
Figure 8: I/O View of the Turbine Controller.......................................................................................19
Figure 9: Subsystem View of Compressor Controller Subsystem..........................................................21
Figure 10: SubsystemView of Turbine Controller Subsystem...............................................................22
Figure 11: Old Pressure Calculation Algorithm....................................................................................23
Figure 12: I/O View of NewPressure Calculation Subsystem...............................................................23
Figure 13: SubsystemView of New Pressure Calculation Subsystem....................................................24
Figure 14: Redesign CAES Model........................................................................................................25
Figure 15: 2013-2014 RLC..................................................................................................................29
Figure 16: 2013-2015 RLC..................................................................................................................30
Figure 17: 7-Day Rolling Average Blocks.............................................................................................31
Figure 18: Discrete FIR Filter set-up...................................................................................................32
Figure 19: Histogram of 2010-2012 Coincident Peak Hours .................................................................33
Figure 20: Histogram of 2010-2015 Coincident Peak Hours .................................................................34
Tableof Tables
Table 1: Base Periods and Adjustment Periods .....................................................................................2
Table 2: Wholesale Electricity Pricesfor 2013-2014 Base Period............................................................5
Table 3: Wholesale Electricity Pricesfor 2014-2015 Base Period............................................................5
Table 4: Coincident Peaks for 2013-2014 Base Period [4] ......................................................................5
Table 5: DC/UOIT Electrical Demand during 2013-2014 CoincidentPeaks...............................................6
Table 6: Peak Demand Factor for DC/UOIT during 2013-2014 Base Period .............................................6
Table 7: GA Costs during 2013-2014 Adjustment Period [10].................................................................6
Table 8: DC/UOIT GA Costs during 2013-2014 AdjustmentPeriod..........................................................7
Table 9: Coincident Peaks for 2014-2015 Base Period [4] ......................................................................7

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Table 10: DC/UOIT Electrical Demand during 2014-2015 Coincident Peaks.............................................8
Table 11: Peak Demand Factor for DC/UOIT during 2014-2015 Base Period ...........................................8
Table 12: Probable Coincident Peaks for 2015-2016 Base Period [4]......................................................8
Table 13: DC/UOIT Electrical Demandfor Probable 2015-2016 Coincident Peaks....................................9
Table 14: Probable Peak Demand Factor for DC/UOIT...........................................................................9
Table 15: DC/UOIT Electrical DemandduringBestCase ScenarioPeakReductionforthe 2013-2014 Base
Period..............................................................................................................................................11
Table 16: Decision Table for Simulink Control Method........................................................................11
Table 17: Compressor Subsystem IO Description...............................................................................20
Table 18: Sample of 2013-2014 base period data for DC/UOIT ............................................................27
Table 19: Sample of Date and Time Conversion of Historical Use Data .................................................28
Table 20: Results of 2013-2014 RLC on Demand and Demand Power Factor.........................................29
Table 21: Results of 2013-2015 RLC on Demand and Demand Power Factor.........................................30
Table 22: Results of 7-Day Rolling Average on Demand and Demand Power Factor ..............................32
Table 23: Hour of Coincident Peaks from 2010 to 2015 base periods...................................................33
Table 24: Results of Selected Time of Use on Demand and Demand Power Factor................................35
Table 25: Summary Table of Control Methods Effectiveness................................................................35
Tableof Abbreviations and Acronyms
AQEW AllocatedQuantityof EnergyWithdrawn
CAES CompressedAirEnergyStorage
DPF DemandPowerFactor
GA Global Adjustment
HOEP HourlyOntarioElectricityPrice
IESO IndependentElectricitySystemOperator
LDC Local DistributionCompany
OPG OntarioPowerGeneration
RLC Representative LoadCurves

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1 Project Identification
1.1 Project Title
Simulation of aCompressedAirEnergyStorage atthe NorthCampusof Durham College andthe
Universityof OntarioInstitute of Technology-AnEngineeringApproach
1.2 ProblemStatement
A CompressedAirEnergyStorage (CAES) systemwasdesignedinthe previoussemesterandwas
simulatedinbothMATLAB™ andSimulink. The real worldthermodynamicefficiencyand economic
impactof the systemneedstobe demonstrated.The thermodynamicefficiency canbe demonstrated in
a digital realmbyusingreal worldelectricityusage data,andthendeterminingif the system will make
an impacton future electricityusage.The economicimpactwillbe determinedbycomparing the change
inelectricityusage withhistorical electricityusage data.
2 LiteratureReview
2.1 Electricity Costs
There are several differentclassesof customersthatthe IndependentElectricitySystemOperator(IESO)
considers andeachof these classeshave differentelectricitycosts.The primaryclass isthe Small
Consumerclasswhichisbilledforits’ electricityusage throughalocal distributioncompany(LDC) such
as Hydro One or TorontoHydro [1]. The LDCs are chargedthe HourlyOntarioEnergyPrice (HOEP) by the
IESO andincorporate the Global Adjustment(GA) intotheirprices [1]. The secondaryclassisthe Large
Consumerclasswhichisanysingle customerthatusesmore than 250 MWh of electricityperyear [1].
Large Consumers are chargedthe commodityprice,whichiscomprisedof the Wholesale Price and
Global Adjustment.The wholesalepricesare determinedinone of three ways [1]:
1. It isset to the HOEP
2. It isset basedonthe consumptionpatternof the LDC
3. It isset byenteringintoafixed-price contractwithanenergyretailer suchasSunwave or
SummitEnergy [2]
2.1.1 Global Adjustment
The Global Adjustment(GA) isacorrectionfactorthat is usedforelectricitysoldto the Large Consumer
classor those inthe Small Consumerclassona contract [3]. It coversthe costs of providingthe required
electricity generationcapacityandpayingforelectricityconservation programs,suchasthe Demand
Response program[3].The Global Adjustmentaccountsforthe difference betweenthe HOEPandthe
ratesfrom variouscontractsfor electricity,suchasBruce Powernuclear,OntarioPowerGeneration
(OPG) nuclearandhydro,and renewable powerpurchasingagreements [3].The Global Adjustmentis
chargedmonthly andvariesmonthly,andthe amountchargeddependsonthe type of customer[3]:
ClassA or ClassB.
2.1.1.1 Class A
ClassA customersare those that have a higherthan5MW hourlypeak demandand are chargedthe GA
basedon howmuchelectricitytheyusedduringthe fivehighestpeakhours (alsoknownascoincident
peaks) duringthe previousbase period [3].

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A base periodisa 12 monthperiodthatruns fromMay 1st
of one yearto April 30th
of the next[4].It is
duringthistime thatthe coincidentpeaksare establishedandthe PeakDemandFactor(PDF) is
determinedforeachcustomer[4].The PDF isthe proportionof the demandthata customeris
responsible forduringthe coincidentpeaks [4].The adjustment,orbilling,periodoccursfromJuly1st
of
the year inwhichthe base periodendstoJune 30th
of the nextyear.Duringthe billingperiod,the PDF
fromthe previousbase periodisusedtodetermine whatproportionof the GA for that monthwill be
chargedto the customer[4].
Table 1: Base Periods and Adjustment Periods
Base Period(Peak-settingPeriod) AdjustmentPeriod(BillingPeriod)
May 1, 2013 to April 30, 2014 July1, 2014 to June 30, 2015
For example,if aClassA customerwasresponsiblefor1% of electricitydemand onaverage duringthe
five peakhoursof the base period,May 1, 2013 to April 30, 2014, theywouldbe charged1% of the total
GA cost setaside forClassA, inthe adjustmentperiod July1,2014 toJune 30, 2015 [4]. ClassA
customersare alwayschargedthe actual GA [3].
2.1.1.2 Class B
ClassB consumers have a peakhourlydemandbetween50kW and 5 MW. The GA ischarged as a rate
perMWh of electricityusedduringthe previous base period[3].ThisGA rate isdeterminedthreetimes,
twoestimates (firstandsecond) andone actual rate. The 1st
Estimate,publishedonthe lastbusinessday
before the givenmonthbegins,iscomprisedof three components:anestimate of the GA basedon the
previousmonth,anestimate of the Ontariodemandforthe givenmonth,andacorrectionfor the
difference betweenthe previousmonth’s1st
Estimate andActual Rate [5]. The 2nd
Estimate,published
on the lastbusinessdayof the givenmonth,isalsocomprisedof three components:actual GA costs for
the Month-to-date,anestimate of the GA and Ontariodemandforthe remainderof the month,anda
correctionforthe difference betweenthe previousmonth’s2nd
Estimate andActual Rate [5]. The Actual
Rate,is basedonthe actual OntarioelectricitydemandandGA costs of the previousmonth,andis
publishedtenbusinessdaysafterthe endof the givenmonth [5].
If the ClassB consumerischargedthrougha LDC, the GA rate theyare charged dependsontheirbilling
cycle.If a ClassB consumerischargeddirectlyfromthe IESO,theirGA rate will be the actual GA rate for
the month[3].
2.2 Electricity Modelling
Modellingelectricitydemandis basedonone of twomethods:takinghistorical use dataand
extrapolatingtothe future,orperformingempirical surveysinwhichactivitiesand/orbehavioursare
recordedandelectricityuse isextrapolatedfromthose. Eachmodellingbase hasitsadvantagesand
disadvantageswhichwill be discussedalongwiththe specifictypesof modellinginthe following
sections.

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2.2.1 Historical Use
Thistype of modelling useshistorical electrical demanddatatocreate a model throughthe use of
computational algorithms.One suchalgorithmisthe Model forAnalysisof EnergyandElectricity
Demand(MEAD) [6]. Anothermethod whichuseshistorical datause isconstructingRepresentative Load
Curves(RLC) [7].Historical Use modellingisverygoodformodellingfuturedemandandrepresenting
large systemsata highlevel.Historical Use modellingisnotwell suitedforsystemswhoseelectricityuse
are influx,orsufferfromexcessive temporal variation.
2.2.1.1 MEAD
The MEAD iscomprisedof two modules:MEAD_D whichisusedto predictfuture annual demand,and
MEAD_El is usedto convertthe predictedfuture annual demandintohourlyelectricdemand [6].This
type of model iswell suitedforlarge scale analysisasthe modulesestimate usage forfoureconomic
sectors:industry,transport,households,andservices [6].Variousmodulationfactorsare used inthe
MEAD_El moduleswhichare usedtocalculate hourlydemandfromthe annual demand [6].
2.2.1.2 RepresentativeLoad Curves
A RLC is a dailyloadcurve whichisrepresentative of agroupof historical loadcurvesthatdisplaysimilar
patterns [7]. RLCs are designed tocapture structural (i.e alarge office building constructed ona
previouslyvacantlot) andtemporal (i.e time of day) variations thatoccurfor a givensystem [7].
Furthermore,RLCsare not restrictedtoone system, suchas a large scale powergrid,or a single building,
as the analysisisdone usinga statistical analysis [7].
2.2.2 Empirical Survey
Thistype of modellingusessurveysgeneratedfromthe base usersof electricity(families,workers,
studentsetc.) andapplycertainparameterstothe responses. The resultsfromthe surveyscaneitherbe
useddirectlytocalculate the electricitydemand asinTime-Use Modelling[8],orindirectlyinthe case of
Agent-basedModelling[9].Empirical Surveymodelingisagood type of modellingtouse whenalow
level understandingof the electricityuse of the system needstobe understood.Empirical Survey
modellingisnotwell suitedforlarge systemwithasubstantial numberof base users,asthe time and
cost requiredtocollectandanalyze the datais prohibitive.
2.2.2.1 Time-Use Modelling
Time-Use Modelling(TUM) isusedto create a model of electricityusage fromtime-use data[8].The
basisof thistype of modellingis collectingtime-use data,where userskeepalogof theirelectricity
usage by activityandlength,andthen researchers assigneachof those activitiesabasicelectricityusage
rate [8].These electricityusage rates wouldthenbe compared withthe actual electricityuse of those
users.The mainsource of error withthistype of modelling isthatthe usersdo notkeepan accurate log
of theiractivities [8].Thistype of misrepresentationerroriscommoninalmostall typesof self-reporting
studies. Evenwitherrors,thistype of modellinghasbeenfoundtocreate realisticdemandcurvesthat
are well fittothe actual data [8].
2.2.2.2 Agent-based Modelling
Agent-basedModelling(ABM) isverysimilartoTime-Use Modelling(TUM),inwhichendusers(agents
for ABM) of electricityare examined andloadcurvesare determinedfromtheiruse [9].Agent-based
modelingisacomputational type of modelling,where agentsare notconsideredtobe real people but

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constructsthat have setstatesand rulesthat governtheirbehaviour [9].Anagents’behaviouris
normallybasedonreal life behaviourthatisdeterminedthroughempirical surveys[9]. ABMallowsfor
a wide range of behaviourstobe consideredandthushave beenusedtomodel societiesand
organizationswithahighlevel of accuracy [9].
2.3 DC/UOITElectricity Use Data
Accessto the electricityuse dataforDC/UOIT forthe entire northcampuswasprovidedbyStephan
Cassar and DougCrossmanthroughthe website ‘Utilismart’.Dataisavailable fromMay1st
, 2013 to
presentanddata isaddedfor the previousdatashortlyaftermidnight. The electricityuse isrecorded
every5 minutes,alongwiththe HOEP. Due to the vastamount of data, the raw data will notbe included
inthisreport.
3 Design Requirements
The systemshall be able torespondwithin5minutes
The systemshall be able tostore or release electricityata rate of 1 kW to 500 kW
The systemshall be able torelease 1 MWh of electricity
The systemshall have a positive impacton the electrical costsof the campus
4 Preliminary Analysis
Basedon the literature review,andthe electricityusage dataprovidedbyDC/UOIT,itis clearthat
DC/UOIT wouldbe consideredaLarge Consumer,ClassA.ThismeansthatDC/UOIT paysthe Hourly
OntarioEnergyPrice upfrontandthenpays a certainpercentage of the Global Adjustmentbasedon
theiruse duringthe coincidentpeaks.Thisprovides anincentive tostudyhow aCompressed AirEnergy
Storage systemcouldaffectthe pricesDC/UOITpay forelectricity.Byaccuratelypredictingandreducing
the DC/UOIT demandduringfuture coincidentpeaks,DC/UOITwouldrealize aneconomicbenefit from
the reducedGA charges.The firststep indeterminingthis economicbenefitwouldbe toestablisha
baseline-thefinancial impactof these coincidentpeaksonDC/UOITat the presenttime,utilizingthe
wholesaleprice forelectricity,whichisthe HOEP,and the GA costsfor those peaks.
Giventhe data available regardingDC/UOIT electricityuse,the schedule of the GA,andthe current time
of year, total electricitycosts canonlybe calculatedforone base period,2013-2014. Analysisfortotal
electricitycostswill be done forthe base periodsof 2014-2015, and 2015-2016, up to the information
available.
4.1 Wholesale Electricity Price
The wholesale electricityprice forDC/UOIT,asstatedabove,isthe HOEP. Thusthe costof electricityper
hour (neglectingGA) issimplythe amountof electricityusedduringthathourmultipliedbythe HOEP.
The electricitycostspermonthduringthe 2013-2014 base periodcan be seenin Table 2.

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Table 2: Wholesale ElectricityPrices for 2013-2014 Base Period
The total electricitycostsfromawholesale perspectiveforthe 2013-2014 base periodis$1.563 million.
The electricitycostspermonthduringthe 2014-2015 base periodcan be seenin Table 3.
Table 3: Wholesale ElectricityPrices for 2014-2015 Base Period
The total electricitycostsfromawhole sale perspectiveforthe 2014-2015 base periodis $879.4
thousand.
4.2 Global Adjustment
The global adjustmentforDC/UOITisdependentonthe coincidentpeaksandpeakdemandfactor.For
the 2013-2014 base period,the GA costs have beenfullypaidnow,andforthe 2014-2015 base period,
the coincidentpeaksandPDF have beenfullyestablishedandasper the schedule discussedinthe
literature review,DC/UOITare currentlypay the GA basedonthose peaks.The coincidentpeaksfor the
2015-2016 base period have yetto be fullyestablished,butwinterpeaksare veryuncommoninOntario,
so the analysisforthat base periodwill be performedwiththe establishedsummerpeaks.
4.2.1 2013-2014 Base Period
The coincidentpeaksforthe 2013-2014 base period (May2013-April 2014), whichwere usedtosettle
the GA from July2014 to June 2015, are as follows:
Table 4: Coincident Peaks for 2013-2014 Base Period [4]
May June July August September October
Total HOEP (K$) 90.215 83.94 111.32 88.79 82.142 73.473
November December January February March April
Total HOEP (K$) 52.094 102.936 237.8 254.072 281.174 105.538
May June July August September October
Total HOEP (K$) 60.630 101.744 87.676 86.242 60.552 25.725
November December January February March April
Total HOEP (K$) 51.931 57.301 94.248 115.886 84.009 53.501
Date Hour Ending Demand (MW)
July17, 2013 17 24689
July16, 2013 17 24009
July18, 2013 17 24070
July19, 2013 14 24207
July15, 2013 17 23596

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The DC/UOIT demandduringthose five peakhourscanbe seenin Table 4Table 5. Asall five peakhours
occurredon sequential daysduringthe summerof the base period,the campusdemandsduringthat
periodwere all roughlythe same aswell.
Table 5: DC/UOIT Electrical Demand during 2013-2014 Coincident Peaks
By combiningTable 4and Table 5, the PeakDemandFactor(PDF) for DC/UOITcan be determinedfor
the 2013-2014 base period, asseenin Table 6.
Table 6: Peak Demand Factor for DC/UOIT during 2013-2014 Base Period
The average PDF forDC/UOIT duringthe 2013-2014 Base Periodwas 0.00022144. The meansthat
DC/UOIT contributed0.02%to the total electrical demandduringthe five highestpeaksinthe 2013-
2014 base period.
The nextstepinthis analysisistouse the PDF to determine the economiccostsfromthe GA incurredby
the campus. This involves examiningthe total GA costsduringthe adjustmentperiodforthe base period
of 2013-2014, whichisJuly2014-June 2015, and thenusingthe PDF todetermine DC/UOITshare of
those costs. The total global adjustment costinmillionsof dollars duringthe adjustmentperiod canbe
seeninTable 7.
Table 7: GA Costs during 2013-2014 Adjustment Period [10]
Date Hour Ending UOIT Demand (MW)
July17, 2013 17 5.687
July16, 2013 17 5.067
July18, 2013 17 5.641
July19, 2013 14 5.288
July15, 2013 17 5.026
Date Hour Ending
Ontario Demand
(MW)
UOIT Demand
(MW)
Peak Demand
Factor
July17, 2013 17 24689 5.687 0.000230345
July16, 2013 17 24009 5.067 0.000211049
July18, 2013 17 24070 5.641 0.000234355
July19, 2013 14 24207 5.288 0.000218448
July15, 2013 17 23596 5.026 0.000213003
2014 July August September October November December
Total GA (M$) 672.3 717.6 784.5 987.5 870.2 843.5
2015 January February March April May June
Total GA (M$) 628.1 458.7 706.8 928.6 939.1 943.1

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DC/UOIT’sshare of the above GA costs basedonthe PeakDemandFactor of 0.00022144 can be seenin
Table 8 inthousandsof dollars.
Table 8: DC/UOIT GA Costs during 2013-2014 Adjustment Period
The total GA coststhat DC/UOIT incurredfromJuly2014-June 2015 forthe base periodof May 2013-
May 2014 as seenin Table 8 were $2.099 million.
Thismeansthat on topof the wholesale energypricesthatwere paid basedonthe HOEP duringthe
base period,anadditional $2.099 millionwere chargedtothe school fortheirusage overthe
adjustmentperiod.
4.2.2 2014-2015 Base Period
As previouslystated, notall of the requiredinformationtodetermine the total electricitycostsforthis
base periodhave beendetermineyet,asthe adjustmentperiodforthisbase perioddoesnotenduntil
June 2016. However,the coincidentpeakscanbe determinedandGA costsfrom the secondhalf of 2015
can alsobe determined.
Thus,the coincidentpeaksfor2014-2015 whichare currentlybeingusedtosettle the GA fromJuly2015
to June 2016, can be seenin Table 9.
Table 9: Coincident Peaks for 2014-2015 Base Period [4]
January7, 2015 19 21610
February19, 2015 20 21416
August26, 2014 17 21650
February23, 2015 20 21402
September5,2014 17 21716
It isof interesttonote,thatthree of the peaksforthisbase periodoccurredduringthe winter,whichis
uncommonforOntario.
The DC/UOIT demandduringthose five peakhourscanbe seenin Table 10.
2014 July August September October November December
Total GA (K$) 148.87 158.91 173.72 218.67 192.70 186.78
2015 January February March April May June
Total GA (K$) 139.09 101.57 156.51 205.63 207.95 208.84

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Table 10: DC/UOIT Electrical Demand during 2014-2015 Coincident Peaks
By combiningTable 9and Table 10, the PeakDemandFactor (PDF) can be determinedforthe 2014-2015
base period.
Table 11: Peak Demand Factor for DC/UOIT during 2014-2015 Base Period
The average PeakDemandFactor for DC/UOIT,as seenin Table 11 duringthe 2014-2015 Base Period
was 0.00021658. This meansthatDC/UOIT contributed0.02% to the total electrical demandduringthe
five highestpeaksinthe 2014-2015 Base Period.
4.2.3 2015-2016 Base Period
As of the time of writingthisreport, the 2015-2016 base periodhasnot concluded,andthusthe five
coincidentpeaksseenin Table 12are the current probablypeaks.
Table 12: Probable Coincident Peaks for 2015-2016 Base Period [4]
January7, 2015 19 3.665
February19, 2015 20 4.969
August26, 2014 17 5.564
February23, 2015 20 3.404
September5,2014 17 5.760
Date Hour Ending
Ontario Demand
(MW)
UOIT Demand
(MW)
Peak Demand
Factor
January7, 2015 19 21610.14 3.665 0.000169596
February19, 2015 20 21416.295 4.969 0.00023202
August26, 2014 17 21650.025 5.564 0.000256997
February23, 2015 20 21402.372 3.404 0.000159048
September5,2014 17 21715.628 5.760 0.000265247
July28, 2015 17 22016
July29, 2015 17 21900
August17, 2015 17 21882
July27, 2015 18 21562
September3,2015 14 21429

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The DC/UOIT demandduring those five probablecoincidentare asfollows:
Table 13: DC/UOIT Electrical Demand for Probable 2015-2016 Coincident Peaks
By combiningTable 12 and Table 13, the probable PDFfor DC/UOIT can be determined forthe 2015-
2016 base period.
Table 14: Probable Peak Demand Factor for DC/UOIT
Date Hour Ending Ontario Demand
(MW)
UOIT Demand
(MW)
Peak Demand
Factor
July28, 2015 17 22016 4.851 0.00022034
July29, 2015 17 21900 4.689 0.00021411
August17, 2015 17 21882 5.489 0.000250845
July27, 2015 18 21562 4.061 0.000188341
September3,2015 14 21429 5.005 0.000233562
July28, 2015 17 4.851
July29, 2015 17 4.689
August17, 2015 17 5.489
July27, 2015 18 4.061
September3,2015 14 5.005

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4.3 Total Electricity Costs
While the GA is not chargedduringthe same monthas the wholesale price,combiningthemprovidesan
overviewof the total chargesincurredforthe electricityuse.Thus,the GA paidduringJuly2014 is
assumedtohave beenchargedfor the electricityuse duringMay2013, and soon. Thiscan be seenin
Figure 1.
Figure 1: Total Cost for Electricityduring 2013-2014 Base Period
The total cost of electricityforthe 2013-2014 base periodisapproximately3.662milliondollars,with
the GA representing57%of these costsand the wholesaleprice representingthe other43% of these
costs.
4.4 Motivation for Project
As seeninSection0,the GA comprisesa majorityof the electricitycoststhatDC/UOITincurs.Thus it
shouldbe the focusof anycost reductionorconservationprogram. Thisiswhere an electricity storage
systemandfurthermore Compressed AirEnergyStorage canpotentiallyhave alarge economicreturn.
Giventhe designrequirementswherethe maximumpoweris500 kW and maximumenergystoredis1
MWh, thissystem,if the coincidentpeakswere accuratelypredicted,wouldreduce DC/UOIT’susage
duringthose peaksby0.5 MW. Furthermore,there wouldbe twopossible hoursinwhichthe usage
couldbe reduced.
Under the bestcase scenario,the CAESsystemwouldreduce eachof the coincidentpeakusage of
DC/UOIT by0.5 MW. As all the informationisavailable forthe 2013-2014 base period,the bestcase
scenariowill be appliedtothisbase period.The reduceddemandandPDFcanbe seenin Table 15.
0.000
500.000
1000.000
1500.000
2000.000
2500.000
3000.000
3500.000
4000.000
0.000
50.000
100.000
150.000
200.000
250.000
300.000
Costs(ThousandsofDollars)
Cost(ThousandsofDollars)
Wholesale and GA Prices for 2013-2014 Base
Period
Total HOEP (K$) Total GA (K$) Total Costs (K$)

11
Table 15: DC/UOIT Electrical Demand during Best Case Scenario Peak Reduction for the 2013-2014 Base Period
The average PDFfor the reducedpeaksis0.0002, approximately10% lessthanthe base case PDF of
0.00022. This consequentlyreducesthe total GA costs byapproximately10%,from$2.099 millionto
$1.902 million,asavingsof $196.6 thousand.Thisisa significantamountof moneysavedoverone year,
and presentsastrongcase for buildinganenergystorage system,evenif acapital investmentis
requiredinitially.
Througha preliminaryanalysisof the electricitycostschargedtoDC/UOIT, itcan be seenthat there is
sufficienteconomicmotivationtodetermine if the systemwill operateasdesigned,insimulatedspace,
and whattank capacitywill be neededtoreturn1 MWh of electricity.
4.5 Control Method Decision
Basedon the literature reviewperformedthere are fourmainmethodsthatcan be usedto model
electricityuse.Creatingamodel of electricityuse wouldbe ideal tocontrol the Simulinkmodel asthe
compressorsystemscouldbe runwhenusage isprojectedtobe low and the turbine couldbe runwhen
usage ispredictedtobe high.The weightingasseenin Table 16 was usedto determine the best
modellingmethodforuse withthe Simulinkmodel.
Table 16: Decision Table for Simulink Control Method
Applicability
Ease of Access
of Data
Ease of
Implementation Total
MEAD 5 8 5 18
Representative LoadCurves 8 9 7 24
Time-Use Modelling 7 4 6 17
Agent-BasedModelling 6 4 6 16
It isclear fromthe above table,thatthe RLC modellingmethodisthe bestwaytogo.This isdue to the
fact that an RLC can be appliedtoanydemanddata,whereasMEAD isusedforprovince andstate level
analysis,andTime-Use andAgent-BasedModelingare normallyrestrictedtogroupsof approximately
1000 people orlessdue tothe time requiredtoadministerandanalysisthe surveys.Furthermore,for
RLCs, the data iseasilyobtainablethroughthe Utilismartwebsite asprovidedbyStephenCassarand
DouglasCross.Thisinformationcouldalsobe usedfora MEAD, but surveyswouldneedtobe
administeredforTime-Use orAgent-BasedModelling.Foranyof the modellingmethods,some
manipulationof the final resultswill be requiredtobe able touse the data in the Simulinkmodel.
However,asthe name suggestsandRLC is a curve and thus one of the easierformatsto importinto
Simulink.
Date
Hour
Ending
Total Demand
(MW)
Base Demand
(MW)
Reduced
Demand (MW)
NewDemand
Power Factor
July17, 2013 17 24689 5.687 5.187 0.000210093
July16, 2013 17 24009 5.067 4.567 0.000190223
July18, 2013 17 24070 5.641 5.141 0.000213582
July19, 2013 14 24207 5.288 4.788 0.000197793
July15, 2013 17 23596 5.026 4.526 0.000191813

12
4.5.1 Explanation of the Weighting Factors
Applicabilityisthe measureof howwell scaledthe modellingmethodistothe systembeingexamined,
where 10 meansthat the methodisverywell scaledforthe currentsystem.
Ease of Accessof Data isthe measure of how easythe informationtocomplete the model isobtained,
where 10 meansthe data is veryeasyto access.
Ease of Implementationisthe measure of how easyitwill be touse the resultsinthe Simulinkmodel,
where 10 meansit will be veryeasytouse.
5 Simulink Model Improvement
In thissectionof the report,the various improvementsandchangesthatwere made tothe Simulink
model will be outlinedanddiscussed. Therewere twomainnoteddeficienciesatthe endof workinFall
2016 withthe model:1) The thermal storage loopwasnot followingpropermassconservationand 2)
the model didnotinclude pressure orheatlossesfrompipe components. Thusthe firstchangestothe
model shouldbe done toaddressthese issues. The model thatwascreatedduringthe previous
semestercanbe seenin Figure 2. Thisisthe model thatwill be improvedandexpandedtoprovide a
more realisticsimulationof the CAESsystem.

13
Figure 2: Base Model from Previous Semester

14
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15
5.1 Thermal Subsystems
The notedissue inthe thermal storage issue withregardstomassconservation,wascreatedfromthe
fact that the temperaturesonbothsidesof the heatexchangers(airand glycerin) weresetandthenthe
requiredmassflowrate of glycerinwascalculated basedonthose temperaturesandthe massflowrate
of air.Thisledto a situation inwhich,due tolimitationsplacedonairreheattemperaturesbythe need
to establishthermal gradientsforeffective heattransfertooccur, the massflow rate of glycerinforthe
turbine side wouldbe consistently lessthanthe compressorside. Thiswouldinturnestablisha pattern
where the hottanks withinthe systemwouldnotbe empty bythe endof the turbine cycle,leadingtoa
buildupof hotfluidinthe hottanksand eventuallynofluidinthe coldtankinwhich to cool the airon
the compressorside.While thislimitationwasnotprogrammedintothe model,thatisthe real world
implicationof thissituation. The I/Oview andthe subsystemview of the originalsubsystemcanbe seen
inFigure 3 andFigure 6 respectively.
Figure 3: I/O View of Original Turbine Thermal Subsystem
Thus,it wasdecidedthatto ensure thathot tankswouldbe emptybythe endof the turbine cycle (when
the air tank pressure wasbackdownto 2000 kPa),it wasdecidedthatthe controllingvariableforthe
glycerinmassflowrate wouldbe changedfromthe airmassflow rate to the air tank pressure.However,
withthischange,the outlettemperature of the glycerinonthe turbine side couldnolongerbe setata
constantvalue,andthuswouldfluctuate dependingonthe airand glycerinflow rates.Furthermore,this
wouldlead toa scenarioinwhichglycerinmightbe leavingthe heatexchangersnotcompletelycooled
to the cold tanktemperature,andthusheatrejectionwouldbe required. The I/Oview andthe
subsystemview of the updatedsubsystemcanbe seen inFigure 4 and Figure 5 respectively.
Figure 4: I/O View of Updated Turbine Thermal Subsystem

16

17
The main change betweenthe above subsystemandthe subsystem onthe reverse isthe inclusionof the GlyTemperatureOut and GlyMass
Flowsubsystems.The GlyMassFlowsubsystemutilizesanegative feedbackloopthatusesthe storage pressure inthe airtankto self-correct
the mass flow.The subsystemusesthe airtankpressure andindexesthe glycerinmassflowrate tothat,so that whenthe storage pressure isat
itsminimumthe hottank iscompletelyempty. Inthe GlyTemperature Out subsystemthe standardheattransferequation (𝑄̇ = 𝑚𝑐 𝑃Δ𝑇)is
usedto determine the glycerintemperature outof the systemusingthe heatlosttothe air,the mass flowrate of the glycerinandthe standard
heatcapacity of glycerin.
Figure 5: Subsystem View of Updated Turbine Thermal Subsystem

18
Figure 6: Subsystem View of Original Turbine Thermal Subsystem

19
5.2 Heat and Pressure Losses
The other identifieddesignchallengefromthe previoussemester’swork wasthe that there are no
pressure orheatlosstakenintoaccount withinthe model.Whilethesespressureandheatlosseswere
calculatedona per unitlengthbasis,inAppendix 1: Pressure andHeatLossesonthe CompressorSide,
and Appendix1:Pressure andHeat Lossesonthe Turbine Side,the accurate additiontothe model
cannot be done until aphysical 3D layoutof the full scale systemhasbeendetermined.The calculated
lossesandthe lossesforpipe directionchangesare heavilydependentonthe lengthbetween
componentsandthe orientationinwhichbothworkingfluidsenterandexitthe components.Thus,this
identifiedchallenge isone thatwill continue ontofuture workandwill be addressedonce aCAD model
of the systemhasbeendeveloped.
5.3 Compressor and Turbine Control Subsystems
An additional change thatwasnotan IdentifiedDesignChallengeisthe inclusionof compressorand
turbine control subsystems.The purpose of these control subsystemsistodeterminethe time during
the simulationinwhichthe compressor orthe turbine shouldbe operating.Thisisachange from the
previoussemestermodelinthatthe compressor andturbine were controlledbyasimple stepfunction
that wasonlydependentonthe pressure withinthe airstorage tank,asit was assumedthe maximum
powerinputwasavailable. The I/Oview of the compressorcontrol subsystem view andthe turbine
control subsystem canbe seenin Figure 7 and Figure 8.
Figure 7: I/O View of the Compressor Controller
Figure 8: I/O View of the Turbine Controller
Both subsystemsrequire five inputsandprovide twooutputs.Anexplanationof the input/outputsof
the subsystemscanbe seenin Table 17.

20
Table 17: Compressor Subsystem IO Description
Input/Output Desription
PowerIn The inputfor the electricitiyuse data
Pressure inTank The pressure inthe air storage tank
Energyof Compressor1/Energyof Turbine 1 The per unitmassenergy workrequirement of
the component.Compressor1isconstant while
Turbine 1 is pressure dependent
Energyof Compressor2/Energyof Turbine 2 The per unitmassenergy workrequirementof
the component.Compressor2ispressure
dependentwhileTurbine 2isconstant.
Setpoint The powersetpointof the entire system. Setpoint
was determinedusingeitherthe 24 hour rolling
average or the Representative LoadCurves.
RequiredMassFlow The calculatedmassflow rate of air throughthe
associatedsystem
SystemOn Binaryoutputif the associated systemisturned
on or off.
The outputRequired Mass Flowis calculatedina two-stepprocess.Firstthe PowerInis comparedto
the Setpoint,anda decisionwill be made basedonwhatcomponentisbeingcontrolled.Forexample,if
the compressoristhe systembeingcontrolledand 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛 < 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡,thenthere ispowerthat is
available tobe usedbythe compressor.If 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛 > 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡, thenthe powerrequiredbythe
campusis greaterthan the setpointandrunning the compressorwouldfurtherincrease thatdemand.
Once it has beendeterminedif there issufficientpowertorunthe compressorsystems,the amountof
powermustbe determined.If 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡− 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛 ≥ 500 𝑘𝑊 thanthe compressorsystemswill be
able to run at full power(500 kW),as the powerdemandfromthe campusis at least500 kW lessthan
the setpoint.If 500 𝑘𝑊 > [ 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡 − 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛] > 0 𝑘𝑊,thenthe poweravailable tothe
compressorsystemwouldbe 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡 − 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛,as the poweravailable islessthan500 kW below
the set point.The massflowrate of the air isthendeterminedbydividingthe available powerbythe
total per unitmasswork of both compressors.
The outputSystem On is calculatedbycomparingthe Setpointto the Power In as for the Required Mass
Flowout, and bycomparingthe Pressure inTank to the maximumorminimumpressure,forthe
compressoror turbine respectively.Forthe compressor,if itisfoundthatthere ispoweravailable and
the pressure inthe tank isbelow8 MPa, thenthe compressorsystemcanbe turnedon. Likewise,forthe
turbine system,if itisfoundthatthe needforpower(𝑃𝑜𝑤𝑒𝑟 𝐼𝑛 < 𝑆𝑒𝑡𝑝𝑜𝑖𝑛𝑡,and the pressure intankis
not below2MPa, thenthe turbine systemscanbe turnedon.A Logictable can be seenin
The subsystemviewof the compressorcontrollersubsystemcanbe seenin Figure 9and the subsystem
viewof the turbine controllersubsystemcanbe seenin Figure 10.

21
The if1 subsystematthe topof the subsystemisusedtodetermineif the currentelectrical usage islessthe setpoint.If itis,thatmeansthat the
system’selectricityuse iscurrentlybelowthe averagepowerusage (setpoint) forthattime periodandthusthere iselectricitythatcanbe stored
for lateruse. The If Action Subsystem2is usedtodetermine thatprovidedthereisexcesselectricity,howmuchpowerthe compressorshave to
use.If the poweravailableislessthe maximum ratedpowerof the compressorsystem,thanthe available powerwill be equal tothe available
power. If the poweravailable isgreaterthanthe maximumratedpowerof the compressorsystem,the poweravailablewill be settothe
maximumratedcompressorpower. The ifsubsystematthe bottomof this subsystemisusedtodetermineif the compressorsystems shouldbe
turnedon.This done byonce again comparingthe currentelectrical usage tothe average,andalsocheckingtoensure the currentair tank
pressure islessthanthe maximumpressure.The requiredmassflowrate iscalculatedbydividingthe available powerbythe total energyper
unitmass of the compressors.Thisrequiredmassflowrate will onlybe outputfromthe subsystemif the signal entering If2isa 1 indicatingthe
compressorsystemistobe turnedon.
Figure 9: Subsystem View of Compressor Controller Subsystem

22
Figure 10: Subsystem View of Turbine Controller Subsystem
The basisof the turbine controllersubsystemisthe same asthe compressorcontrollersubsystemandthusthe subsystemappearsroughly
similaraswell.The If1subsystemisusedtocompare the input powertothe setpointand then If ActionSubsystem2and 4 are usedto
determine the amountof powerthe turbinesshouldoutput.The Ifsubsystemisusedtodetermineif the turbine systemsshouldbe turnedonor
not.Once againthisisdone by comparingthe inputpowertothe setpointandalsothe tankpressure tothe minimumpressure todetermineif
there isair available togeneratedpowerwith.The requiredmassflowisdeterminedbydividingthe requiredpowerbythe energyperunitmass
available fromthe turbines. Thisrequiredmassflowrate will onlybe outputfromthe subsystemif the signal entering If2isa 1 indicatingthe
compressorsystemistobe turnedon.

23
5.4 Pressure Calculation
Whenthe modificationsweremade toaddressthe mass conservationinthe glycerinloop,itwasnoted
that pressure calculationsinthe secondcompressorandfirstturbine were notcorrect,althoughthe
overall pressure calculationswere correct.Thischange wasimplementedtoensure thatnomatterthe
outputselected inthe model,the signalforpressure wouldbe the same.
The old pressure calculationalgorithmwhichwasfoundinbothCompressor2and Turbine 1 of Figure 2,
can be seenin Figure 11.
Figure 11: Old Pressure Calculation Algorithm
As seeninthe algorithm,anintegratorisusedtoadd the currentlydeterminedpressure
increase/decreasetothe pastdeterminedpressure.Thiscreatedasituationwhere the signal goingto
Pressure Outcontinuallyincreasesoverthe durationof the simulation,anddoesnotstaywithinthe
8000 kPa maximumsetpoint. However,eventhoughthisdeviationwascorrectedbyasimilaralgorithm
inTurbine 1, it isdesiredthatat all pointsof the simulationthe physical parametersstaywithinthe
limitsasdetermine bythe designrequirementsandthusanew subsystemwas createdtocalculate the
pressure irrespectiveof locationwithinthe model. The I/Oview of thisnew subsystemcanbe seenin
Figure 12.
Figure 12: I/O View of New Pressure Calculation Subsystem
Thisnewsubsystemusesthe compressorandturbine massflow ratestodeterminethe airtankstorage
pressure.Thiscalculationcanbe seenin Figure 13.

24
Figure 13: Subsystem View of New Pressure Calculation Subsystem
By comparingthe compressorandturbine flow rate inthe same subsystem, eventhoughtheywould
neverprovide anon-zeronumberatthe same time,the integratorusedwillproduce anaccurate
pressure increase anddecrease tothe outputwhichcanthenbe routedto Compressor2, Thermal
Subsystem1,Thermal Subsystem2,Turbine2,and bothcontrollersubsystems.Thisallowsthe
eliminationof the previouslyusedloopsinthe Compressor2and Turbine 1 subsystemsandmaintains
congruencyof parametersthroughoutthe model.
5.5 Summary of Model Changes
Overall,one identifieddesignchallengewascompletelyaddressedandthe appropriate systemwas
redesigned.The otheridentifieddesignchallenge wasnotable tobe completelyaddressasthe
dimensionsof the systemwere notknownandthusthe determinedpressureandheatlossesonaper
unitlengthbasiswere notable tobe appliedintothe system.The twootherdesignchangesthatwere
made was to include acompressorandturbine controllersubsystemthatisable touse a setpointto
determine whenthe compressorandturbine systemsshouldbe onor not.Thisadditionwasnecessary
for the nextstepinthe project.The additionof a global pressure calculationsubsystemwastoensure
that throughoutthe systemthe pressure wasbeingroutedatitstrue value,notinflatedatone point
and correctedat another. Thiswasmostlyan aestheticchange asthose false pressure valueswere not
usedto determine anyoperatingcharacteristicsbefore beingcorrected. The redesignedsystemcanbe
seeninFigure 14. Several GotoandFrom blockswere usedtocleanupthe model andmake iteasierto
understandwhatsignal were goingwhere.However,due tothe complexitiesof the turbine systems,the
signal routingforthose subsystemsisstilldifficulttounderstand

25
Figure 14: Redesign CAES Model

26

27
6 MATLAB™ and Simulink Analysis
The primarypurpose of thiscapstone projectwasto determine if the inclusionof aCAESsystemat the
north campusof DC/UOIT couldpotentiallyhave areal worldimpactonelectricitycosts.The firststep
was to create a time basedsignal inSimulinkusingthe historical use data.The nextstepwasto
implementacontrol algorithmthatwoulddetermine the setpointforusingwithinthe controller
subsystems.Three differenttypesof control were used:Representative LoadCurves(asdetermined
fromthe decisionstable),a7-dayrollingaverage,andaselectedtime of use algorithm.
6.1 Historical UseData
As mentionedabove,electrical usage datafromDC/UOITis available fromMay1st
, 2013 to the
beginningof the presentday.Due tothe constraintsplacedbothbythe limitedhistorical use data
available fromDC/UOITandthe fact that the GA is charged16 monthsafterthe start of its associated
base period,onlyelectrical use dataduringthe periodof May 2013 to April 2014 will be analyzed. A
sample of thisdata whichcoversthe firsthourof the 2013-2014 base periodcanbe seenin Table 18.
Table 18: Sample of 2013-2014 base period data for DC/UOIT
DATE TIME kW
5/1/2013 0:05 4823.3
5/1/2013 0:10 4840.9
5/1/2013 0:15 4780.4
5/1/2013 0:20 4808.2
5/1/2013 0:25 4767.8
5/1/2013 0:30 4815.8
5/1/2013 0:35 4795.6
5/1/2013 0:40 4820.8
5/1/2013 0:45 4742.7
5/1/2013 0:50 4868.6
5/1/2013 0:55 4778
5/1/2013 1:00 4755.2
5/1/2013 1:05 4793
For thisdata to be usable ina Simulinkmodel,itfirsthastobe convertedfromdate andtime to a
numericformthat representedthe numberof secondsthatdemandwasrecordedat.Asa full yearof
data was to be used,andthe time stepbetweeneachelectrical demandentrywas5 seconds,eachtime
stepina numericsequencewouldbe 300 from0 to 31536000. Also,as the data couldonlybe
downloadedbytwomonthsata time,six datasetscoveringMay 2013 to April 2014 were downloaded
and combinedintoone spreadsheet.Fromthere,afunctioninExcel wascreatedthatconvertedthe
informationfromthe DateandTime intoa secondtime value.Anexampleof thiscalculation,alongwith
the formulausedcan be seenin Table 19.

28
Table 19: Sample of Date and Time Conversion of Historical Use Data
A B C D E
1 DATE TIME kW =((A2-A$A2)+B2)+(3600*24))
2 2013-05-01 0:05 4823.3 300 4823.3
3 2013-05-01 0:10 4840.9 600 4840.9
4 2013-05-01 0:15 4780.4 900 4780.4
5 2013-05-01 0:20 4808.2 1200 4808.2
6 2013-05-01 0:25 4767.8 1500 4767.8
7 2013-05-01 0:30 4815.8 1800 4815.8
8 2013-05-01 0:35 4795.6 2100 4795.6
9 2013-05-01 0:40 4820.8 2400 4820.8
10 2013-05-01 0:45 4742.7 2700 4742.7
11 2013-05-01 0:50 4868.6 3000 4868.6
12 2013-05-01 0:55 4778 3300 4778
13 2013-05-01 1:00 4755.2 3600 4755.2
14 2013-05-01 1:05 4793 3900 4793
The formulawas appliedtothe entire Dcolumn,andthenColumnD andE were movedtoanother
sheetsothat the data can be importedintoMATLAB™ usingthe function xlsread. Once the data was in
MATLAB, the data was usedtocreate a signal inSIMULINKusingthe signalbuilderfunction.The code
usedcan be seenin Appendix 2:MATLAB Codes,ImportingTime-Use DataandSignal Building.The signal
createdfromthat code was usedas the basisfor the analysisof the CAESsystem.
6.2 Simulink Analysis Using Representative Load Curves
Representative LoadCurves,asmentionedabove,are createdfromananalysisperformedonthe data
to whatthe average electricityusage duringaperiodis.RLCsare normallycreatedfroma
multidiscriminate analysis,butdue totheircomplexityanduse of machine learningtocreate,a
simplifiedRLCwascreatedby determiningthe meanandstandarddeviationforeachperiod. Asthe data
was providedin5minute intervals,the firststeptocalculatingthe RLCwouldbe average the usage for
each hourof use. The MATLAB™ code usedforthiscan be seeninAppendix 2:MATLABCodes,
AveragingTime-Use Data.
The nextstepwouldbe to determiningthe actual RLC,whichcan alsobe done usinga MATLAB™ code.
Thiscode seenin Appendix 2:MATLAB Codes, RLCCreation,while beingsetuptofindRLC for the 2013-
2014 base period,caneasilybe adaptedtoanotherbase period.Thiscode sortsthe hour averageddata
intotheirrespective monthsthanfindsthe average andstandarddeviationforeachhourduringthat
month,graphsit and thencreatesa Simulinksignalforuse inthe Simulinkmodel. The resultsfromusing
RLCs for control can be seenbelowinthe proceedingsections.
6.2.1 2013-2014 RepresentativeLoad Curve
The representativeloadcurve forthe 2013-2014 Base Periodcan be seenin Figure 15. It isof interestto
note that the monthwiththe highestdemandoverall wasApril 2014 andthe monthwiththe lost
average demandwasDecember2013.

29
Figure 15: 2013-2014 RLC
The resultsof implementingthe above RLCintoSimulinkcanbe seenbelow in Table 20.
Table 20: Results of 2013-2014 RLC on Demand and Demand Power Factor
Date
Hour
Ending
Total Demand
(MW)
Base Demand
(MW)
Modeled
Demand (MW)
New Demand
Power Factor
July 17, 2013 17 24689 5.687 5.630 0.000228
July 16, 2013 17 24009 5.067 5.068 0.000211
July 18, 2013 17 24070 5.641 5.630 0.000234
July 19, 2013 14 24207 5.288 5.745 0.000237
July 15, 2013 17 23596 5.026 5.027 0.000213
Examiningthe datainthe above table,itisclearthat utilizingthe 2013-2014 RLC for model control is
ineffectiveatreducingthe DemandPowerFactorsignificantly,and actuallyincreasesit,asthe demand
duringthe peakon July19 is increasedby457 kW. Thisis furthercompoundedbythe factthat the
turbine isonlyrunningatverylowpower,<60 kW, duringthe twoof the otherpeaks(17 and 18), and
thusnot providing ameaningful decrease indemandduringanyof the peaks.
The 2013-2014 RLC increasedthe overall DPFfrom 221.440 ∗ 10−6 to224.675 ∗ 10−6, resultinginan
increase inGA costsfrom $2099.25 𝐾 to $2129.92 𝐾.

30
6.2.2 2013-2015 RepresentativeLoad Curve
The constructionof the 2013-2015 RLC is done byfirstrunningthe AveragingTime-UseDataandRLC
Creationcodesforthe 2014-2015 BPdata. Thenthe AveragedRLCCreation code in Appendix2:MATLAB
Codesisusedto create the averagedRLC. The summarygraph fromthiscode can be seenin Figure 16.
Figure 16: 2013-2015 RLC
The resultsof implementingthe above RLCintothe Simulinkmodelcanbe seenin Table 21.
Table 21: Results of 2013-2015 RLC on Demand and Demand Power Factor
Date
Hour
Ending
Total Demand
(MW)
Base Demand
(MW)
Modeled
Demand (MW)
New Demand
Power Factor
July 17, 2013 17 24689 5.687 5.614 0.000227
July 16, 2013 17 24009 5.067 5.068 0.000211
July 18, 2013 17 24070 5.641 5.614 0.000233
July 19, 2013 14 24207 5.288 5.778 0.000239
July 15, 2013 17 23596 5.026 5.027 0.000213
Examiningthe datainthe above table,itisclearthat utilizingthe 2013-2015 RLC for model control is
ineffectiveatreducingthe DemandPowerFactorsignificantly,andactuallyincreasesit,asthe demand
duringthe peakon July19 isincreasedby 490 kW. Thisis furthercompoundedbythe factthat the

31
turbine isonlyrunningatverylowpower,<75 kW, duringthe twoof the otherpeaks(17 and 18), and
thusnot providingameaningful decrease indemandduringanyof the peaks.
The 2013-2015 RLC increasedthe overall DPFfrom 221.440 ∗ 10−6 to224.680 ∗ 10−6, resultinginan
increase inGA costsfrom $2099.25 𝐾 to $2129.97 𝐾.
6.3 Simulink Analysis Using Rolling Average
From the resultsabove itisclearthat usinga RLC for control doesnot provide the requiredPDFandGA
reductions.Thus,anewmethodof control mustbe designed.Itwasdeterminedthata 7-dayrolling
average wouldbe optimal asto capture the seasonal changesbutnotbe affectedbyanyabnormal
peaks.ThiswasaccomplishedinSimulinkbyusingthe setof blocksasseenin Figure 17.
Figure 17: 7-Day Rolling Average Blocks
In the above seriesthe demanddatafromthe 2013-2014 base periodisfedintothe first Rate Transition
blockand the setpointfor the compressorandturbine controllersisproducedbythe second Rate
Transition block.The Rate Transition blocksare usedto convertthe sample time fromthe 30 seconds
usedthroughoutthe analysisto3600 seconds(1 hour) so that the Discrete FIR Filter can average the
valuesenteringiteveryhourof the simulation.The coefficientforthe filterissettobe an array 168
entrieslongwitheachentrybeingequal to 1/168,where 168 isthe numberof hoursin7 days,the
desireddurationof the rollingaverage.The initial state(s) of the filterwasalsosetto be 5000, an
arbitraryyet reasonable demand,which are usedtocalculate the average until after7daysintothe
simulation.The set-upof the filtercanbe seenin Figure 18.

32
Figure 18: Discrete FIR Filter set-up
The resultsof the implementationof the rollingaverage canbe seenin Table 22.
Table 22: Results of 7-Day Rolling Average on Demand and Demand Power Factor
Examiningthe datainthe above table,itisclearthat utilizingthe 7-dayrollingaverage formodel control
isineffective atreducingthe DemandPowerFactorsignificantly,andactually slightly increasesit,asthe
demandduringthe peakonJuly 15 is increasedby 3 kW. Thisisfurthercompoundedbythe factthat the
turbine neverrunsduringanyof the peaksand thusdoesnot provide ameaningful decrease indemand
duringany of the peaks.
The 7-day rollingaverage increasedthe overall DPFfrom 221.440 ∗ 10−6 to221.465 ∗ 10−6, resulting
inan increase inGA costs from $2099.25 𝐾 to $2099.49 𝐾.
Date
Hour
Ending
Total Demand
(MW)
Base Demand
(MW)
Modeled
Demand (MW)
New Demand
Power Factor
July 17, 2013 17 24689 5.687 5.687 0.000230
July 16, 2013 17 24009 5.067 5.067 0.000211
July 18, 2013 17 24070 5.641 5.641 0.000234
July 19, 2013 14 24207 5.288 5.288 0.000218
July 15, 2013 17 23596 5.026 5.029 0.000213

33
6.4 Simulink Analysis Using Selected Time of Use
While the 7-dayrollingaverage didnotgreatlyincrease GA costs,itisnot an effectivecontrol method
and thusanothernewmethodmustbe found.Thisnew methodisbasedonexaminingwhenthe
previous coincidentpeakswereandbasingthe model’soperationof the compressorandturbine
systemsaroundthose times.The hourendingeachcoincidentpeakfromMay1st
, 2010 to April 10th
,
2016 can be seenin Table 23 withdata takenfrom [4].
Table 23: Hour of Coincident Peaks from 2010 to 2015 base periods
Peak 2010 2011 2012 2013 2014 2015
1 16 16 16 17 19 17
2 15 17 17 17 20 17
3 16 16 16 17 17 17
4 16 17 14 14 20 18
5 16 12 16 17 17 14
Usinga histogramtoanalyze the data above,the mostcommoncoincidentpeakhourscanbe found.A
histogramforthe three yearsbefore the 2013-2014 base periodcanbe seenin Figure 19 and a
histogramof all the data can be seenin Figure 20.
Figure 19: Histogram of 2010-2012 Coincident Peak Hours

34
Figure 20: Histogram of 2010-2015 Coincident Peak Hours
As seeninbothof the above figures,the mostcommonhoursfor the coincidentpeakstoendare 16:00
and 17:00 (4:00pm and 5:00pm). Thus,from a historical perspective,from15:01 to 17:00 isthe most
optimal time torun the turbine systems.The optimaltime torunthe compressorsystemtocharge the
systemwouldbe whenthere isnochance of a coincidentpeak,whichisfromhistoricaldatawouldbe
any time from24:00 to 06:00. Appropriately,the compressorsystemsshouldbe runatmidnightuntil
the systemischarged,and thenthe turbine systemsshouldbe runfrom 15:01 to 17:00 to ensure that
the majorityof the historical coincidentpeaksare covered.
Thiscontrol methodwasimplementedinSimulinkbycreatingasignal that from 24:00 to 15:00 has a
value of 10,000 whichissignificantlyhigherthanmaximumdemandof 9,371.9. Thiswill ensure thatno
matterthe demandduringthisperiod,the compressorwillbe the preferredsystem.However,another
part of the controllersubsystemisthatthe pressure hasto be at or below 8000 kPa. Subsequently,once
the systemreachedthatmaximumpressure,nomatterthe setpoint, the compressorsystemswillnot
function.
A similarapproachwastakenfor the turbine system, inthatfrom15:01 to 23:59, the createdsignal has
a value of 0 to ensure thatthe turbine,nomatterwhat the currentdemandiswill alwaysrun.Analogous
to the compressorcontrollersubsystem, the turbinecontrollersubsystemwillonlyrunwhenthere is
more than 2000 kPa inthe air storage tank.Subsequently,whenthe systemreachesthe minimum
pressure,nomatterthe setpoint,the turbine systemswill notfunction.
The code usedtocreate the Simulinksignalcanbe seenin Appendix2:MATLAB Codes, SelectedTimeof
Use.The resultsfromusingthiscreatedsignal inthe model canbe seenin Table 24.

35
Table 24: Results of Selected Time of Use on Demand and Demand Power Factor
Date
Hour
Ending
Total Demand
(MW)
Base Demand
(MW)
Modeled
Demand (MW)
New Demand
Power Factor
July 17, 2013 17 24689 5.687 5.188 0.000210
July 16, 2013 17 24009 5.067 4.568 0.000190
July 18, 2013 17 24070 5.641 5.142 0.000214
July 19, 2013 14 24207 5.288 5.288 0.000218
July 15, 2013 17 23596 5.026 4.527 0.000192
Examiningthe datainthe above table,itisclearthat utilizingthe selectedtime of use formodel control
isextremelyeffective atreducingthe DemandPowerFactorsignificantly.The demandduringall of the
1700 endingpeakswasreducedby500 kW, as wasplannedfromthe signal used.The peakduringthe
1400 endpeakwas notaffected,asthe systemwasdesignedtoonlyincrease demandduringthe night.
The selectedtime of use control methoddecreasedthe overallDPFfrom 221.440 ∗ 10−6 to 204857 ∗
10−6, resultinginadecrease inGA costsfrom $2099.25 𝐾 to $1942.04 𝐾.
6.5 Summary of Control Methods
Four differentcontrol methodswere usedtoexamine their effectiveness atreducingGA costs.The first
two,whichutilizedRepresentativeLoadCurves,were selectedoutof a groupof four possible electricity
modellingmethod.TheywereultimatelyineffectiveatreducingGA costs andactuallyincreasedthem.
Due to the failure of the RLCs,twonewmethodsof control were developed,one whichwasagainbased
on the demand,andanotherwhichwasnot. The one basedon the demandwasa 7-day rollingaverage,
whichalsowas ineffective atreducingthe GA costs,but didnot increase itasmuch as the RLCs did.The
fourthmethod,termedSelectedTimeof Use,wascreatedby examiningduringwhathoursthe
coincidentpeaksoccurredhistorical andrunningthe turbinesystemsduringthe toptwohoursevery
day.The compressorsystemsunderthismethodwouldbe runningaftermidnighteachdayuntil the air
tank wasfull.Thismethodultimatelyprovedtoprovide asignificantdecreaseinGA costs,of
approximately $157.21 𝐾.A summarytable of the differentmethods,the effectonthe DPFand GA can
be seenin Table 25.
Table 25: Summary Table of Control Methods Effectiveness
Demand
Power Factor
GA Costs
(K$)
Percentage of
Base
Monetary
Change (K$)
Base Case 0.000221440 2099.25 0 0
RLC 2013-2014 0.000224675 2129.92 101.461% 30.672343
RLC 2013-2015 0.000224680 2129.97 101.463% 30.716383
RollingAverage 0.000221465 2099.49 100.011% 0.2410592
SelectedTime ofUse 0.000204857 1942.04 92.511% -157.207

36
7 Conclusion and Next Steps
The two mainobjectivesof this capstone projectwastoaddressthe identifieddesignchallengesfrom
the previoussemestersworkandtodetermine the effectivenessof aCAESsystemif coupledwith
DC/UOIT. Onlyone of the identifieddesignchallengeswasable tobe completelyaddress,thatisthe
mass conservationinthe glycerinloop.The lackof pressure andheatlosseswere notable tobe
completelyaddressedasthe physical dimensionsof the systemwasnotknown,andas suchthe
calculationsperformedinAppendix1:Sample Calculationswere notable tobe appliedtothe model.
It was determinedthat couplingthe designedCAESsystemtothe NorthCampusof DC/UOIT wouldonly
be effective if therewere selectedtimesduringthe daythatthe systemwouldbe storingandwouldbe
releasingenergy.Assuch,withthe compressorrunningfrom24:00 to approximately02:30,and the
turbine runningfrom13:01 to 17:00 everyday,a savingsinGlobal Adjustmentcostsof $157.2 𝐾 could
be obtained.
It shouldbe notedthatevenwithoutthe pressure andheatlossesappliedinthe system, the GA savings
wouldlikelyremainthe same,asthe systemwouldalwaysbe designedtorelease1MWh of electricity
at a rate of 500 kW. Includingthe pressure andheatlosscalculationswouldcause the compressor
poweranddurationof operationtoincrease,the turbine sizestoincrease,andwouldalsocause the air
storage tank volume toincrease aswell.The compressorpowerwouldincreaseasthe compressor
selectedwouldhave aslightlyhigherpressure ratiothancurrent(8.894) as to overcome asleastpartly,
the pressure andheatlossesinthe compressorside totry and ensure that8000 kPais the final pressure
inthe airstorage tank.The volume of the air storage tankwouldneedtobe increasedtoensure that
evenwithpressure andheatlossesonthe turbine side,the same powerandtotal energywouldbe
released.Thisisalsothe reasoningforthe largerturbines,aswithareducedpressure entering them,
theywill produce lessworkperunitmass,thusneedtobe largerto accommodate a largermass flow
rate.
The overall nextstepwouldbe todeterminea3D physical layoutof the system, mostlikelyinaCAD
environment,todetermine the pipe andcomponentdimensionssothatan accurate determinationof
the pressure andheatlossescan occur. A secondnextstep wouldbe todesigna glycerinheatrejection
system,because asthe systemwasredesigned,the glycerinisleavingthe turbinethermal systemswith
temperature still greatlyabove ambientconditions.A thirdnextstepwouldbe torevisitthe RLCsand
examine if there isawayto make theiruse as a control methodmore effective.A possible startingpoint
on thiswouldbe tocombine SelectedTime of Use withthe RLC,as to ensure thatthe turbine isalways
runningduringthe mostprobablycoincidentpeakhours,butcanbe run at othertimes.

37
8 References
[1] “IESO ElectricityPricinginOntario.”[Online].Available:http://www.ieso.ca/Pages/Ontario’s-Power-
System/Electricity-Pricing-in-Ontario/default.aspx.[Accessed:08-Feb-2016].
[2] “List of RetailersandMarketers|OEB.” [Online].Available:
http://www.ontarioenergyboard.ca/oeb/Consumers/Energy%20Contracts/List%20of%20Retailers%
20and%20Marketers.[Accessed:15-Feb-2016].
[3] “IESO Global Adjustment.”[Online].Available:http://www.ieso.ca/Pages/Ontario%27s-Power-
System/Electricity-Pricing-in-Ontario/Global-Adjustment.aspx.[Accessed:08-Feb-2016].
[4] “IESO Global AdjustmentforClassA.”[Online].Available:
http://www.ieso.ca/Pages/Participate/Settlements/Global-Adjustment-for-Class-A.aspx.[Accessed:
09-Feb-2016].
[5] “IESO Global AdjustmentforClassB.”[Online].Available:
http://www.ieso.ca/Pages/Participate/Settlements/Global-Adjustment-for-Class-B.aspx.[Accessed:
11-Feb-2016].
[6] A. Hainoun, “Constructionof the hourlyloadcurvesanddetectingthe annual peakloadof future
Syrianelectricpowerdemandusingbottom-upapproach,”Int.J.Electr. PowerEnergy Syst.,vol.31,
no.1, pp.1–12, Jan.2009.
[7] P. Balachandraand V.Chandru,“Modellingelectricitydemandwithrepresentativeloadcurves,”
Energy,vol.24, no. 3, pp. 219–230, Mar. 1999.
[8] J. Widén,M.Lundh,I. Vassileva,E.Dahlquist,K.Ellegård,andE.Wäckelgård,“Constructingload
profilesforhouseholdelectricityandhotwater fromtime-use data—Modellingapproachand
validation,”Energy Build.,vol.41,no. 7, pp. 753–768, Jul.2009.
[9] T. Zhang,P.-O.Siebers,andU.Aickelin,“Modellingelectricityconsumptioninoffice buildings:An
agentbasedapproach,”Energy Build.,vol.43, no.10, pp.2882–2892, Oct. 2011.
[10] “IESO Global Adjustment - Archive.”[Online].Available:
http://www.ieso.ca/Pages/Participate/Settlements/Global-Adjustment-Archive.aspx.[Accessed:12-
Feb-2016].
[11] Y. A. Çengel andA.J.Ghajar, Heat and masstransfer:fundamentals&applications,5thedition.New
York, NY: McGraw-Hill Education,2015.
[12] E. Barbour, D. Mignard,Y. Ding,and Y. Li, “AdiabaticCompressedAirEnergyStorage withpacked
bedthermal energystorage,”Appl.Energy,vol.155, pp. 804–815, 2015.

38
9 Appendix 1: Sample Calculations
9.1 Pressure and Heat Losses on the Compressor Side
9.1.1 Constants
𝑘 = 1.4
𝜂 𝑝𝑜𝑙 = 0.85
𝑐 𝑝,𝑎𝑖 𝑟 = 1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
𝑐 𝑝,𝑔𝑙𝑦𝑐𝑒𝑟𝑖𝑛 = 3.716
𝑘𝐽
𝑘𝑔 ∗ 𝐾
9.1.2 States
𝑇1 = 293 𝐾, 𝑃1 = 100 𝑘𝑃𝑎, 𝑃2 = 894.48 𝑘𝑃𝑎, 𝑃4 = 8000 𝑘𝑃𝑎, 𝑇3 = 303 𝐾
𝑇5 = 303 𝐾, 𝑊̇ 𝑡𝑜𝑡𝑎𝑙 = 500 𝑘𝑊, 𝑇∞ = 293 𝐾, 𝑇𝑔𝑙𝑦,1 = 293 𝐾, 𝑇𝑔𝑙𝑦,2 = 550 𝐾
𝑇𝑔𝑙𝑦,3 = 520 𝐾
9.1.3 Compressor 1
9.1.3.1 Work
The work requiredona perunitmass basisof a compressorcan be determinedusingthe polytropic
efficiency,whichaccountsfor irreversibilities andheattransfer,andthe ratioof specificheatcapacities,
usingthe followingequation [11].
𝑤 𝑐𝑜𝑚𝑝,1 = 𝑐 𝑝 𝑇1 ((
𝑃2
𝑃1
)
𝑘−1
𝜂 𝑝𝑜𝑙∗𝑘
− 1)
A polytropicefficiencyof 85%is used,andwasdeterminedfromthe currentlyoperatingUSA McIntosh
CAES facility,andaspecificheatcapacityratio of 1.4 is standardfor air[11].
𝑤 𝑐𝑜𝑚𝑝,1 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(293 𝐾)((
894.48 𝑘𝑃𝑎
100 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,1 = 322.15
𝑘𝐽
𝑘𝑔
9.1.3.2 Temperature
The temperature atthe outletof a compressorcanbe determinedusingthe polytropicefficiency,which
accounts forirreversibilities andheattransfer,andthe ratioof specificheatcapacities,usingthe
followingequation[11].
𝑇2 = 𝑇1 (
𝑃2
𝑃1
)
𝑘−1

39
𝑇2 = 293 𝐾 (
894.48 𝑘𝑃𝑎
100 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
𝑇2 = 611.96 𝐾
9.1.4 Compressor 2
9.1.4.1 Work
Due to the variable pressurenature of the secondcompressor,the workwill be determinedatthree
pressure stagesandthenaveraged.
𝑤 𝑐𝑜𝑚𝑝,2,𝑃1 = 𝑐 𝑝 𝑇3 ((
𝑃4,1
𝑃2
)
𝑘−1
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃1 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
2000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃1 = 95.05
𝑘𝐽
𝑘𝑔
𝑤 𝑐𝑜𝑚𝑝,2,𝑃2 = 𝑐 𝑝 𝑇3 ((
𝑃4,2
𝑃2
)
𝑘−1
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃2 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
5000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃2 = 239.72
𝑘𝐽
𝑘𝑔
𝑤 𝑐𝑜𝑚𝑝,2,𝑃3 = 𝑐 𝑝 𝑇3 ((
𝑃4,3
𝑃2
)
𝑘−1
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃3 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
8000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑃3 = 333.1
𝑘𝐽
𝑘𝑔
𝑤 𝑐𝑜𝑚𝑝,2=
𝑤 𝑐𝑜𝑚𝑝,2,𝑃1 + 𝑤 𝑐𝑜𝑚𝑝,2,𝑃2 + 𝑤 𝑐𝑜𝑚𝑝,2,𝑃3
3
𝑤 𝑐𝑜𝑚𝑝,2=
95.05
𝑘𝐽
𝑘𝑔
+ 239.72
𝑘𝐽
𝑘𝑔
+ 333.1
𝑘𝐽
𝑘𝑔
3

40
𝑤 𝑐𝑜𝑚𝑝,2 = 222.62
𝑘𝐽
𝑘𝑔
9.1.4.2 Temperature
𝑇4,1 = 𝑇3 (
𝑃4,1
𝑃3
)
𝑘−1
𝑇4,1 = 303𝐾 (
2000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
𝑇4,1 = 397.11 𝐾
𝑇4,2 = 𝑇3 (
𝑃4,2
𝑃3
)
𝑘−1
𝑇4,2 = 303𝐾 (
5000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
𝑇4,2 = 540.34 𝐾
𝑇4,3 = 𝑇3 (
𝑃4,3
𝑃3
)
𝑘−1
𝑇4,3 = 303𝐾 (
8000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
0.85∗1.4
𝑇4,3 = 632.82 𝐾
𝑇4 =
𝑇4,1 + 𝑇4,2 + 𝑇4,3
3
𝑇4 =
397.11 𝐾 + 540.34 𝐾 + 632.82 𝐾
3
𝑇4 = 523.42 𝐾
9.1.5 Mass Flow Rate
𝑚̇ 𝑐𝑜𝑚𝑝 =
𝑃
( 𝑤 𝑐𝑜𝑚𝑝,1 + 𝑤 𝑐𝑜𝑚𝑝,2)
=
500 𝑘𝑊
322.15
𝑘𝐽
𝑘𝑔
+ 222.62
𝑘𝐽
𝑘𝑔
=
500 𝑘𝑊
544.77
𝑘𝐽
𝑘𝑔
⇒ 𝑚̇ 𝑐𝑜𝑚𝑝 = 0.92
𝑘𝑔
𝑠
9.1.6 Heat Losses
9.1.6.1 Compressor 1
The heat lossfromthe compressorcan be determine bycomparingthe workdone whenthe polytropic
efficiencyis100%, whenthere are nolossesdue toirreversibilities,tothe actual work done withthe
realisticpolytropicefficiencyof 85%.

41
𝑤 𝑐𝑜𝑚𝑝,1,𝑎𝑐𝑡 = 322.15
𝑘𝐽
𝑘𝑔
𝑤 𝑐𝑜𝑚𝑝,1,𝑡ℎ𝑒𝑜 = 𝑐 𝑝 𝑇1 ((
𝑃2
𝑃1
)
𝑘−1
− 1)
𝑤 𝑐𝑜𝑚𝑝,1,𝑡ℎ𝑒𝑜 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(293 𝐾)((
894.48 𝑘𝑃𝑎
100 𝑘𝑃𝑎
)
1.4−1
1∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,1,𝑡ℎ𝑒𝑜 = 257.51
𝑘𝐽
𝑘𝑔
𝑞 𝑐𝑜𝑚𝑝,1 = ∆𝑤 𝑐𝑜𝑚𝑝,1 = 𝑤 𝑐𝑜𝑚𝑝,1,𝑎𝑐𝑡 − 𝑤 𝑐𝑜𝑚𝑝,1,𝑡ℎ𝑒𝑜 = 322.15
𝑘𝐽
𝑘𝑔
− 257.51
𝑘𝐽
𝑘𝑔
⇒
𝑞 𝑐𝑜𝑚𝑝,1 = 64.65
𝑘𝐽
𝑘𝑔
By multiplyingthe perunitmassheat lossbythe maximummassflow rate,the heatrate lossfrom
Compressor1 can be determined:
𝑄̇ 𝑐𝑜𝑚𝑝,1 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑞 𝑐𝑜𝑚𝑝,1 = (0.92
𝑘𝑔
𝑠
) (64.95
𝑘𝐽
𝑘𝑔
) ⇒
𝑄̇ 𝑐𝑜𝑚𝑝,1 = 59.75 𝑘𝑊
9.1.6.2 Heat Exchanger 1
The total heattransferredfromthe airin the heatexchangeris:
𝑄̇ 𝐻𝑋 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝,𝑎𝑖 𝑟∆𝑇 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝,𝑎𝑖𝑟( 𝑇1 − 𝑇3) = (0.92
𝑘𝑔
𝑠
) (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(611 𝐾 − 303 𝐾) ⇒
𝑄̇ 𝐻𝑋 = 286.2 𝑘𝑊
The requiredmassflowrate of glycerinrequiredtoacceptthe heatis:
𝑄̇ 𝐻𝑋 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝,𝑔𝑙𝑦∆𝑇 ⇒ 𝑚̇ 𝑔𝑙𝑦 =
𝑄̇ 𝐻𝑋
𝑐 𝑝,𝑔𝑙𝑦( 𝑇𝑔𝑙𝑦,2 − 𝑇𝑔𝑙𝑦,1)
=
286.2 𝑘𝑊
(3.716
𝑘𝐽
𝑘𝑔 ∗ 𝐾
) (550 𝐾 − 293 𝐾)
⇒
𝑚̇ 𝑔𝑙𝑦 = 0.3
𝑘𝑔
𝑠
9.1.6.3 Compressor 2
The heat lossfromthe secondcompressorcan be determinedbycomparingthe workdone whenthe
polytropicefficiencyis100%,whenthere are no lossesdue toirreversibilities,tothe actual workdone
withthe realisticpolytropicefficiencyof 85%. The workwill be calculatedatthe three pressures,as
done above.
𝑤 𝑐𝑜𝑚𝑝,1,𝑎𝑐𝑡 = 222.62
𝑘𝐽
𝑘𝑔

42
𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃1 = 𝑐 𝑝 𝑇3 ((
𝑃4,1
𝑃2
)
𝑘−1
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃1 = (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
2000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
1∗1.4
− 1)
𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃1 = 79.10
𝑘𝐽
𝑘𝑔
𝑃4,2
𝑃2
)
𝑘−1
− 1)
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
5000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
1∗1.4
− 1)
𝑘𝐽
𝑘𝑔
𝑃4,3
𝑃2
)
𝑘−1
− 1)
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(303 𝐾)((
8000 𝑘𝑃𝑎
894.48 𝑘𝑃𝑎
)
1.4−1
1∗1.4
− 1)
𝑘𝐽
𝑘𝑔
𝑤 𝑐𝑜𝑚𝑝,𝑡ℎ𝑒𝑜,2 =
𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃1 + 𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃2 + 𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜,𝑃3
3
𝑤 𝑐𝑜𝑚𝑝,𝑡ℎ𝑒𝑜,2 =
79.10
𝑘𝐽
𝑘𝑔
+ 194.36
𝑘𝐽
𝑘𝑔
+ 266.27
𝑘𝐽
𝑘𝑔
3
𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜 = 179.91
𝑘𝐽
𝑘𝑔
𝑞 𝑐𝑜𝑚𝑝,1 = ∆𝑤 𝑐𝑜𝑚𝑝,1 = 𝑤 𝑐𝑜𝑚𝑝,2,𝑎𝑐𝑡 − 𝑤 𝑐𝑜𝑚𝑝,2,𝑡ℎ𝑒𝑜 = 222.62
𝑘𝐽
𝑘𝑔
− 179.91
𝑘𝐽
𝑘𝑔
⇒
𝑞 𝑐𝑜𝑚𝑝,1 = 42.71
𝑘𝐽
𝑘𝑔
By multiplyingthe perunitmassheatlossperunitmassby the maximummass flow rate,the heatrate
lossfromthe compressor1 can be determined:
𝑄̇ 𝑐𝑜𝑚𝑝,1 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑞 𝑐𝑜𝑚𝑝,1 = (0.92
𝑘𝑔
𝑠
) (42.71
𝑘𝐽
𝑘𝑔
) ⇒

43
𝑄̇ 𝑐𝑜𝑚𝑝,1 = 39.29 𝑘𝑊
9.1.6.4 Heat Exchanger 2
The total heattransferredfromthe airin the heatexchangeris:
𝑄̇ 𝐻𝑋 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝,𝑎𝑖𝑟∆𝑇 = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝,𝑎𝑖𝑟( 𝑇1 − 𝑇3) = (0.92
𝑘𝑔
𝑠
) (1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(523.42 𝐾 − 303 𝐾) ⇒
𝑄̇ 𝐻𝑋 = 204.8 𝑘𝑊
The requiredmassflowrate of glycerinrequiredtoacceptthe heatis:
𝑄̇ 𝐻𝑋 = 𝑚̇ 𝑔𝑙𝑦 𝑐 𝑝,𝑔𝑙𝑦∆𝑇 ⇒ 𝑚̇ 𝑔𝑙𝑦 =
𝑄̇ 𝐻𝑋
𝑐 𝑝,𝑔𝑙𝑦( 𝑇𝑔𝑙𝑦,2 − 𝑇𝑔𝑙𝑦,1)
=
204.8 𝑘𝑊
(3.716
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(500 𝐾 − 293 𝐾)
⇒
𝑚̇ 𝑔𝑙𝑦 = 0.27
𝑘𝑔
𝑠

44
9.1.6.5 Between Compressor 1 and HX 1
To determine the heatlossbetweenthe firstcompressorandthe firstheatexchangeronaper length
basis,three heattransferquantitiesneedtobe determined:
1. Internal forcedconvectionbetweenthe airandthe pipe
2. Heat conductionthroughthe pipe
3. External natural convectionbetweenthe pipe andthe air
It isassumedthat Schedule 4010-inch stainlesssteel pipesare used.The internal diameterof thispipe is
25.45 cm and the outerdiameteris27.305 cm.The heatconductionthroughthe pipe can be easily
modelledthroughthe use of the thermal conductivityof the pipe,which,asastainlesssteel,isaround
15.6 𝑊 𝑚 ∗ 𝐾⁄ [12, p.915].
9.1.6.5.1 Internal Forced Convection
To determine the heattransfercoefficientafterthe compressorinlet,the Reynoldsnumbermustbe
foundfirst.
𝜌( 𝑇, 𝑃) = 𝜌(611 𝐾, 894.48 𝑘𝑃𝑎) ⇒ 𝜌 = 5.082
𝑘𝑔
𝑚3
𝑉̇ =
𝑚̇
𝜌
=
0.92
𝑘𝑔
𝑠
5.082
𝑘𝑔
𝑚3
⇒ 𝑉̇ = 0.181
𝑚3
𝑠
𝜐( 𝑇, 𝑃) = 𝜐(611 𝐾,894.48 𝑘𝑃𝑎) ⇒ 𝜐 = 6.84 ∗ 10−7
𝑚2
𝑠
𝑅𝑒 =
𝑉̇ 𝐷ℎ
𝜐𝐴
=
𝑉̇ 𝐷
𝜐
𝜋
4
𝐷2
=
𝑉̇
𝜐
𝜋
4
𝐷
=
(0.020
𝑚3
𝑠
)
(6.84 ∗ 10−7 𝑚2
𝑠
)(
𝜋
4
(0.2545 𝑚))
⇒ 𝑅𝑒 = 1.474 ∗ 105
The Reynoldsnumbermustnowbe usedtodetermine the frictionfactorof the stainlesssteelpipe.Fora
Reynoldsnumberof 1.474 ∗ 105 anda material of stainlesssteel (𝜀 = 0.0002 𝑐𝑚) [12,p.500], the
relative roughnessis 7.86 ∗ 10−6.
𝑓 (𝑅𝑒,
𝜀
𝐷
) = 𝑓(1.474 ∗ 105, 7.86 ∗ 10−6) ⇒ 𝑓 = 0.016 𝑓𝑟𝑜𝑚 𝑀𝑜𝑜𝑑𝑦 𝐶ℎ𝑎𝑟𝑡
To determine the correctNusseltnumbercorrelationtobe used,the Prandltnumberof the airneedsto
be found.
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(611 𝐾, 894.48 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.7051
Givena Reynoldsnumberof 1.474 ∗ 105 anda Prandltnumberof 0.7051, the secondPetukhov
equationcanbe used [12, p.497].
𝑁𝑢 =
( 𝑓 8⁄ ) 𝑅𝑒𝑃𝑟
1.07 + 12.7( 𝑓 8⁄ )0.5(𝑃𝑟
2
3⁄
− 1)
=
(0.016/8)(1.474 ∗ 105)(0.7051)
1.07 + 12.7(0.016 8⁄ )0.5((0.7051)
2
3⁄
− 1)
⇒
𝑁𝑢 = 218.4
The heat transfercoefficientcannowbe determinedusingthe flow correlation.

45
ℎ1 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(611 𝐾,894.48 𝑘𝑃𝑎) ⇒ 𝑘 = 0.04677 𝑊 𝑚 ∗ 𝐾⁄
ℎ1 =
0.04677
𝑊
𝑚 ∗ 𝑘
0.2545 𝑚
218.4 ⇒ ℎ1 = 40.14
𝑊
𝑚2 ∗ 𝐾
9.1.6.5.2 External Natural Convection
To determine the heattransfercoefficientforthe natural convectionthatoccurson the pipe surface,
the Rayleighnumbermustbe found.The correlationforthisnumberisasfollows [12,p.541]:
𝑅𝑎 𝐿 =
𝑔𝛽(𝑇𝑠 − 𝑇∞)𝐿 𝑐
3
𝜐2 𝑃𝑟
𝑤ℎ𝑒𝑟𝑒 𝛽 =
1
𝑇𝑓𝑖𝑙𝑚
The characteristiclength, 𝐿 𝐶,for a circularpipe isequal to the outerdiameter,whichinthiscase is
0.2731 𝑚.
To determine Rayleigh’snumber,the surface temperature of the pipe mustbe known.Thistemperature
couldonlybe determinedafterthe heattransferrate throughthe pipe hasbeendetermined.Thus,a
worstcase scenariowill be assumedwhere the surface temperature of the pipe isequal tothe bulkair
temperature inside the pipe whichis 611 𝐾.
𝛽 =
1
( 𝑇𝑠 + 𝑇∞)
2
=
1
(611 𝐾 + 293 𝐾)
2
=
1
452 𝐾
⇒ 𝛽 = 0.0022 𝐾−1
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟( 452 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.698
𝜐( 𝑇, 𝑃) = 𝜐(452 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝜐 = 3.27 ∗ 10−5
𝑚2
𝑠
𝑅𝑎 𝐿 =
3
𝜐2 𝑃𝑟 =
9.81
𝑚
𝑠2 (0.0022 𝐾−1)(611 𝐾 − 293 𝐾)(0.2731 𝑚)3
(3.27 ∗ 10−5 𝑚2
𝑠
)
2 (0.698) ⇒
𝑅𝑎 𝐿 = 1.313 ∗ 108
Rayleighnumbercanbe connectedtoNusselt’snumberandfurthertothe heattransfercoefficient
usingthe followingcorrelation [12,p.542]:
𝑁𝑢 =
{
0.6 +
0.589𝑅𝑎 𝐿
1
6⁄
[1 + (
0.559
𝑃𝑟
)
9/16
]
8/27
}
2
=
{
0.6 +
0.589(1.313 ∗ 108)1/6
[1 + (
0.559
0.698
)
9/16
]
8/27
}
2
⇒
𝑁𝑢 = 134.8
ℎ2 =
𝑘
𝐷
𝑁𝑢

46
𝑘( 𝑇, 𝑃) = 𝑘(452 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝑘 = 0.03689 𝑊 𝑚 ∗ 𝐾⁄
ℎ2 =
0.04677
𝑊
𝑚 ∗ 𝑘
0.2731 𝑚
19.87 ⇒ ℎ2 = 18.21
𝑊
𝑚2 ∗ 𝐾
9.1.6.5.3 Heat Transfer Determination
The heat transfercan be determinedusingthe followingequation [12,p.163]:
𝑄̇ =
𝑇2 − 𝑇∞
𝑅 𝑡𝑜𝑡𝑎𝑙
𝑤ℎ𝑒𝑟𝑒: 𝑅 𝑡𝑜𝑡𝑎𝑙 = 𝑅 𝑐𝑜𝑛𝑣,1 + 𝑅 𝑐𝑦𝑙 + 𝑅 𝑐𝑜𝑛𝑣,2
The individual resistancescanbe determinedasfollows [12,p.163]:
𝑅 𝑐𝑜𝑛𝑣,1 =
1
ℎ 𝑖 𝐴 𝑖
=
1
ℎ1 (2𝜋
𝐷𝑖
2
)
=
1
40.14
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2545 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,1 = 0.0156
𝑚 ∗ 𝐾
𝑊
𝑅 𝑐𝑦𝑙 =
ln (
𝑟2
𝑟1
)
2𝜋𝐿𝑘
=
ln (
𝐷 𝑜
2
𝐷𝑖
2
)
2𝜋𝑘
=
ln(
0.2731 𝑚
0.2545 𝑚
)
2𝜋 (15.6
𝑊
𝑚 ∗ 𝐾
)
⇒ 𝑅 𝑐𝑦𝑙 = 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
1
ℎ2 𝐴2
=
1
ℎ2 (2𝜋
𝐷 𝑜
2
)
=
1
18.21
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2731 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,2 = 0.0320
𝑚 ∗ 𝐾
𝑊
𝑅 𝑡𝑜𝑡𝑎𝑙 = 𝑅 𝑐𝑜𝑛𝑣,1 + 𝑅 𝑐𝑦𝑙 + 𝑅 𝑐𝑜𝑛𝑣,2 = 0.0156
𝑚 ∗ 𝐾
𝑊
+ 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
+ 0.0320
𝑚 ∗ 𝐾
𝑊
⇒
𝑅 𝑡𝑜𝑡𝑎𝑙 = 0.048
𝑚 ∗ 𝐾
𝑊
The per unitlengthheattransferrate can now be determined
𝑄̇ =
611 𝐾 − 293 𝐾
0.048
𝑚 ∗ 𝐾
𝑊
⇒ 𝑄̇ = 6582
𝑊
𝑚
Furthermore,the temperature decreaseperunitlengthcanalsobe determined.
𝑄̇ = 𝑚̇ 𝑐𝑜𝑚𝑝 𝑐 𝑝Δ𝑇 ⇒ Δ𝑇 =
𝑄̇
𝑚̇ 𝑐 𝑝
=
6.582
𝑘𝑊
𝑚
0.92
𝑘𝑔
𝑠
(1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)
⇒ Δ𝑇 = 7.35
𝐾
𝑚

47
9.1.6.6 Between HX 1 and Compressor 2
The heat lossesbetweenthe firstheatexchangerandthe secondcompressorwillfollow the same
processas above where three heattransferquantitiesneedtobe found:
The same assumptionswill be usedinthiscalculationaswell.
𝜌( 𝑇, 𝑃) = 𝜌(303 𝐾, 894.48 𝑘𝑃𝑎) ⇒ 𝜌 = 10.31
𝑘𝑔
𝑚3
𝑉̇ =
𝑚̇
𝜌
=
0.92
𝑘𝑔
𝑠
10.67
𝑘𝑔
𝑚3
⇒ 𝑉̇ = 0.089
𝑚3
𝑠
𝜐( 𝑇, 𝑃) = 𝜐(303 𝐾,894.48 𝑘𝑃𝑎) ⇒ 𝜐 = 1.82 ∗ 10−6
𝑚2
𝑠
𝑅𝑒 =
𝑉̇ 𝐷ℎ
𝜐𝐴
=
𝑉̇ 𝐷
𝜐
𝜋
4
𝐷2
=
𝑉̇
𝜐
𝜋
4
𝐷
=
(0.089
𝑚3
𝑠
)
(1.82 ∗ 10−6 𝑚2
𝑠
)(
𝜋
4
(0.2545 𝑚))
⇒ 𝑅𝑒 = 2.448 ∗ 105
𝑓 (𝑅𝑒,
𝜀
𝐷
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(303 𝐾, 894.48 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.713
Givena Reynoldsnumberof 2.448 ∗ 105 anda Prandltnumberof 0.713, the secondPetukhovequation
can be used [12, p.497].
𝑁𝑢 =
( 𝑓 8⁄ ) 𝑅𝑒𝑃𝑟
1.07 + 12.7( 𝑓 8⁄ )0.5 (𝑃𝑟
2
3⁄
− 1)
=
(0.015 8⁄ )(2.448 ∗ 105)(0.713)
1.07 + 12.7(0.015 8⁄ )0.5((0.713)
2
3⁄
− 1)
⇒
𝑁𝑢 = 341.3
ℎ2,1 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(303 𝐾,894.48 𝑘𝑃𝑎) ⇒ 𝑘 = 0.02687 𝑊 𝑚 ∗ 𝐾⁄
ℎ2,1 =
0.02687
𝑊
𝑚 ∗ 𝑘
0.2545 𝑚
341.3 ⇒ ℎ1 = 36.03
𝑊
𝑚2 ∗ 𝐾
𝛽 =
1
( 𝑇𝑠 + 𝑇∞)
2
=
1
(303 𝐾 + 293 𝐾)
2
=
1
298 𝐾
⇒ 𝛽 = 0.0034 𝐾−1
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(298 𝐾,100 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.708

48
𝜐( 𝑇, 𝑃) = 𝜐(298 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝜐 = 1.58 ∗ 10−5
𝑚2
𝑠
𝑅𝑎 𝐿 =
3
𝜐2 𝑃𝑟 =
9.81
𝑚
𝑠2 (0.0034 𝐾−1)(303 𝐾 − 293 𝐾)(0.2731 𝑚)3
(1.58 ∗ 10−5 𝑚2
𝑠
)
2 (0.708) ⇒
𝑅𝑎 𝐿 = 2.695 ∗ 107
𝑁𝑢 =
{
0.6 +
0.589𝑅𝑎 𝐿
1
6⁄
[1 + (
0.559
𝑃𝑟
)
9/16
]
8/27
}
2
=
{
0.6 +
0.589(2.695 ∗ 107)1/6
[1 + (
0.559
0.708
)
9/16
]
8/27
}
2
⇒
𝑁𝑢 = 82.16
ℎ2,2 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(298 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝑘 = 0.02624 𝑊 𝑚 ∗ 𝐾⁄
ℎ2,2 =
0.02624
𝑊
𝑚 ∗ 𝑘
0.2731 𝑚
82.16 ⇒ ℎ2,2 = 7.893
𝑊
𝑚2 ∗ 𝐾
𝑄̇ =
𝑇2 − 𝑇∞
1
ℎ 𝑖 𝐴𝑖
=
1
ℎ2,1 (2𝜋
𝐷𝑖
2
)
=
1
36.03
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.245 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,1 = 0.0174
𝑚 ∗ 𝐾
𝑊
𝑅 𝑐𝑦𝑙 =
ln (
𝑟2
𝑟1
)
2𝜋𝐿𝑘
=
ln (
𝐷 𝑜
2
𝐷𝑖
2
)
2𝜋𝑘
=
ln(
0.2731 𝑚
0.2545 𝑚
)
2𝜋 (15.6
𝑊
𝑚 ∗ 𝐾
)
⇒ 𝑅 𝑐𝑦𝑙 = 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
1
ℎ2 𝐴2
=
1
ℎ2 (2𝜋
𝐷 𝑜
2
)
=
1
7.893
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2731 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,2 = 0.0738
𝑚 ∗ 𝐾
𝑊
𝑚 ∗ 𝐾
𝑊
+ 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
+ 0.0738
𝑚 ∗ 𝐾
𝑊
⇒
𝑅 𝑡𝑜𝑡𝑎𝑙 = 0.0919
𝑚 ∗ 𝐾
𝑊

49
𝑄̇ =
303 𝐾 − 293 𝐾
0.0919
𝑚 ∗ 𝐾
𝑊
⇒ 𝑄̇ = 108.8
𝑊
𝑚
𝑄̇
𝑚̇ 𝑐 𝑝
=
0.109
𝑘𝑊
𝑚
0.92
𝑘𝑔
𝑠
(1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)
⇒ Δ𝑇 = 0.117
𝐾
𝑚

50
The heat lossesbetweenthe secondcompressorandsecondheatexchangerwill follow the same
processas above where three heattransferquantitiesneedtobe found:
The same assumptionswill be usedinthiscalculationaswell.The average temperature (523.42 𝐾)as
determinedabove will be used,andanaverage outletpressure of 5000 kPa will alsobe used.
𝜌( 𝑇, 𝑃) = 𝜌(523.42 𝐾,5000 𝑘𝑃𝑎) ⇒ 𝜌 = 32.67
𝑘𝑔
𝑚3
𝑉̇ =
𝑚̇
𝜌
=
0.92
𝑘𝑔
𝑠
10.67
𝑘𝑔
𝑚3
⇒ 𝑉̇ = 0.029
𝑚3
𝑠
𝜐( 𝑇, 𝑃) = 𝜐(523.42 𝐾,5000 𝑘𝑃𝑎) ⇒ 𝜐 = 8.72 ∗ 10−7
𝑚2
𝑠
𝑅𝑒 =
𝑉̇ 𝐷ℎ
𝜐𝐴
=
𝑉̇ 𝐷
𝜐
𝜋
4
𝐷2
=
𝑉̇
𝜐
𝜋
4
𝐷
=
(0.029
𝑚3
𝑠
)
(8.72 ∗ 10−7 𝑚2
𝑠
)(
𝜋
4
(0.2545 𝑚))
⇒ 𝑅𝑒 = 1.617 ∗ 105
𝑓 (𝑅𝑒,
𝜀
𝐷
) = 𝑓(1.617 ∗ 105, 0.000075) ⇒ 𝑓 = 0.016 𝑓𝑟𝑜𝑚 𝑀𝑜𝑜𝑑𝑦 𝐶ℎ𝑎𝑟𝑡
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(523.42 𝐾, 5000 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.7089
Givena Reynoldsnumberof 1.617 ∗ 105anda Prandltnumberof 0.7089,the secondPetukhov
𝑁𝑢 =
( 𝑓 8⁄ ) 𝑅𝑒𝑃𝑟
1.07 + 12.7( 𝑓 8⁄ )0.5(𝑃𝑟
2
3⁄
− 1)
=
(0.016 8⁄ )(1.617 ∗ 105)(0.713)
1.07 + 12.7(0.016 8⁄ )0.5((0.7089)
2
3⁄
− 1)
⇒
𝑁𝑢 = 240.3
ℎ3,1 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(523.42 𝐾,5000 𝑘𝑃𝑎) ⇒ 𝑘 = 0.0424 𝑊 𝑚 ∗ 𝐾⁄
ℎ3,1 =
0.0424
𝑊
𝑚 ∗ 𝑘
0.2545 𝑚
240.3 ⇒ ℎ1 = 40.01
𝑊
𝑚2 ∗ 𝐾
𝛽 =
1
( 𝑇𝑠 + 𝑇∞)
2
=
1
(523.4 𝐾 + 293 𝐾)
2
=
1
408.2 𝐾
⇒ 𝛽 = 0.0025 𝐾−1

51
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(408.2 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.699
𝜐( 𝑇, 𝑃) = 𝜐(408.2 𝐾,100 𝑘𝑃𝑎) ⇒ 𝜐 = 2.74 ∗ 10−5
𝑚2
𝑠
𝑅𝑎 𝐿 =
3
𝜐2 𝑃𝑟 =
9.81
𝑚
𝑠2 (0.0025 𝐾−1)(408.2 𝐾 − 293 𝐾)(0.2731 𝑚)3
(1.58 ∗ 10−5 𝑚2
𝑠
)
2 (0.699) ⇒
𝑅𝑎 𝐿 = 1.499 ∗ 108
𝑁𝑢 =
{
0.6 +
0.589𝑅𝑎 𝐿
1
6⁄
[1 + (
0.559
𝑃𝑟
)
9/16
]
8/27
}
2
=
{
0.6 +
0.589(1.499 ∗ 108)1/6
[1 + (
0.559
0.699
)
9/16
]
8/27
}
2
⇒
𝑁𝑢 = 140.5
ℎ3,2 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(408.2 𝐾,100 𝑘𝑃𝑎) ⇒ 𝑘 = 0.034 𝑊 𝑚 ∗ 𝐾⁄
ℎ2 =
0.034
𝑊
𝑚 ∗ 𝑘
0.2731 𝑚
140.5 ⇒ ℎ2 = 17.5
𝑊
𝑚2 ∗ 𝐾
𝑄̇ =
𝑇3 − 𝑇∞
1
ℎ 𝑖 𝐴𝑖
=
1
ℎ3,1 (2𝜋
𝐷𝑖
2
)
=
1
40.01
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2545 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,1 = 0.0156
𝑚 ∗ 𝐾
𝑊
𝑅 𝑐𝑦𝑙 =
ln (
𝑟2
𝑟1
)
2𝜋𝐿𝑘
=
ln (
𝐷 𝑜
2
𝐷𝑖
2
)
2𝜋𝑘
=
ln(
0.2731 𝑚
0.2545 𝑚
)
2𝜋 (15.6
𝑊
𝑚 ∗ 𝐾
)
⇒ 𝑅 𝑐𝑦𝑙 = 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
1
ℎ2 𝐴2
=
1
ℎ2 (2𝜋
𝐷 𝑜
2
)
=
1
17.5
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2731 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,2 = 0.0333
𝑚 ∗ 𝐾
𝑊
𝑚 ∗ 𝐾
𝑊
+ 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
+ 0.0333
𝑚 ∗ 𝐾
𝑊
⇒
𝑅 𝑡𝑜𝑡𝑎𝑙 = 0.0497
𝑚 ∗ 𝐾
𝑊

52
𝑄̇ =
523.4 𝐾 − 293 𝐾
0.0497
𝑚 ∗ 𝐾
𝑊
⇒ 𝑄̇ = 4641
𝑊
𝑚
𝑄̇
𝑚̇ 𝑐 𝑝
=
4.641
𝑘𝑊
𝑚
0.92
𝑘𝑔
𝑠
(1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)
⇒ Δ𝑇 = 4.99
𝐾
𝑚

53
9.1.6.8 Between HX 2 and Tank
The heat lossesbetweenthe secondheatexchangerandthe tank will follow the same processasabove
where three heattransferquantitiesneedtobe found:
The same assumptionswill be usedinthiscalculationaswell.The average compressor2outletpressure
of 5000 𝑘𝑃𝑎 will be used.
𝜌( 𝑇, 𝑃) = 𝜌(303 𝐾,5000 𝑘𝑃𝑎) ⇒ 𝜌 = 57.94
𝑘𝑔
𝑚3
𝑉̇ =
𝑚̇
𝜌
=
0.92
𝑘𝑔
𝑠
57.94
𝑘𝑔
𝑚3
⇒ 𝑉̇ = 0.015
𝑚3
𝑠
𝜐( 𝑇, 𝑃) = 𝜐(303 𝐾, 5000 𝑘𝑃𝑎) ⇒ 𝜐 = 3.37 ∗ 10−7
𝑚2
𝑠
𝑅𝑒 =
𝑉̇ 𝐷ℎ
𝜐𝐴
=
𝑉̇ 𝐷
𝜐
𝜋
4
𝐷2
=
𝑉̇
𝜐
𝜋
4
𝐷
=
(0.015
𝑚3
𝑠
)
(3.37 ∗ 10−7 𝑚2
𝑠
)(
𝜋
4
(0.2545 𝑚))
⇒ 𝑅𝑒 = 2.354 ∗ 105
𝑓 (𝑅𝑒,
𝜀
𝐷
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(303 𝐾,5000 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.741
Givena Reynoldsnumberof 2.354 ∗ 105 anda Prandltnumberof 0.7148, the secondPetukhov
𝑁𝑢 =
( 𝑓 8⁄ ) 𝑅𝑒𝑃𝑟
1.07 + 12.7( 𝑓 8⁄ )0.5 (𝑃𝑟
2
3⁄
− 1)
=
(0.015 8⁄ )(2.354 ∗ 105)(0.713)
1.07 + 12.7(0.015 8⁄ )0.5((0.741)
2
3⁄
− 1)
⇒
𝑁𝑢 = 82.16
ℎ2,1 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(303 𝐾, 5000 𝑘𝑃𝑎) ⇒ 𝑘 = 0.02858 𝑊 𝑚 ∗ 𝐾⁄
ℎ2,1 =
0.02858
𝑊
𝑚 ∗ 𝑘
0.2545 𝑚
82.16 ⇒ ℎ1 = 40.51
𝑊
𝑚2 ∗ 𝐾
𝛽 =
1
( 𝑇𝑠 + 𝑇∞)
2
=
1
(303 𝐾 + 293 𝐾)
2
=
1
298 𝐾
⇒ 𝛽 = 0.0034 𝐾−1

54
𝑃𝑟( 𝑇, 𝑃) = 𝑃𝑟(298 𝐾,100 𝑘𝑃𝑎) ⇒ 𝑃𝑟 = 0.708
𝜐( 𝑇, 𝑃) = 𝜐(298 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝜐 = 1.58 ∗ 10−5
𝑚2
𝑠
𝑅𝑎 𝐿 =
3
𝜐2 𝑃𝑟 =
9.81
𝑚
𝑠2 (0.0034 𝐾−1)(303 𝐾 − 293 𝐾)(0.2731 𝑚)3
(1.58 ∗ 10−5 𝑚2
𝑠
)
2 (0.708) ⇒
𝑅𝑎 𝐿 = 2.695 ∗ 107
𝑁𝑢 =
{
0.6 +
0.589𝑅𝑎 𝐿
1
6⁄
[1 + (
0.559
𝑃𝑟
)
9/16
]
8/27
}
2
=
{
0.6 +
0.589(2.695 ∗ 107)1/6
[1 + (
0.559
0.708
)
9/16
]
8/27
}
2
⇒
𝑁𝑢 = 82.16
ℎ2,2 =
𝑘
𝐷
𝑁𝑢
𝑘( 𝑇, 𝑃) = 𝑘(298 𝐾, 100 𝑘𝑃𝑎) ⇒ 𝑘 = 0.02624 𝑊 𝑚 ∗ 𝐾⁄
ℎ2,2 =
0.02624
𝑊
𝑚 ∗ 𝑘
0.2731 𝑚
82.16 ⇒ ℎ2,2 = 7.893
𝑊
𝑚2 ∗ 𝐾
𝑄̇ =
𝑇2 − 𝑇∞
1
ℎ 𝑖 𝐴𝑖
=
1
ℎ2,1 (2𝜋
𝐷𝑖
2
)
=
1
40.51
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2545 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,1 = 0.0154
𝑚 ∗ 𝐾
𝑊
𝑅 𝑐𝑦𝑙 =
ln (
𝑟2
𝑟1
)
2𝜋𝐿𝑘
=
ln (
𝐷 𝑜
2
𝐷𝑖
2
)
2𝜋𝑘
=
ln(
0.2731 𝑚
0.2545 𝑚
)
2𝜋 (15.6
𝑊
𝑚 ∗ 𝐾
)
⇒ 𝑅 𝑐𝑦𝑙 = 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
1
ℎ2 𝐴2
=
1
ℎ2 (2𝜋
𝐷 𝑜
2
)
=
1
7.893
𝑊
𝑚2 ∗ 𝐾
(2𝜋(0.2731 𝑚))
⇒ 𝑅 𝑐𝑜𝑛𝑣,2 = 0.0738
𝑚 ∗ 𝐾
𝑊
𝑚 ∗ 𝐾
𝑊
+ 7.19 ∗ 10−4
𝑚 ∗ 𝐾
𝑊
+ 0.0738
𝑚 ∗ 𝐾
𝑊
⇒
𝑅 𝑡𝑜𝑡𝑎𝑙 = 0.090
𝑚 ∗ 𝐾
𝑊

55
𝑄̇ =
303 𝐾 − 293 𝐾
0.0900
𝑚 ∗ 𝐾
𝑊
⇒ 𝑄̇ = 111.1
𝑊
𝑚
𝑄̇
𝑚̇ 𝑐 𝑝
=
0.111
𝑘𝑊
𝑚
0.92
𝑘𝑔
𝑠
(1.01
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)
⇒ Δ𝑇 = 0.119
𝐾
𝑚

56
9.1.7 Pressure Losses
9.1.7.1.1 Pipe Losses
The pressure lossdue tofrictionwithinthe pipe perunitlengthcanbe determinedbythe follow
correlation [12,p. 488]:
∆𝑃 = 𝑓
𝜌𝑉2
2𝐷𝑖
The velocityof the fluidcanbe foundas follows:
𝑉 =
𝑉̇
𝐴 𝑖
=
𝑉̇
𝜋
4
𝐷𝑖
=
0.181
𝑚3
𝑠
𝜋
4
(0.2545 𝑚)2
⇒ 3.56
𝑚
𝑠
The pressure lossdue tofrictionwithinthe pipe perunitlengthis:
∆𝑃 = 0.016
(5.082
𝑘𝑔
𝑚3)(3.56
𝑚
𝑠
)
2
2(0.2545 𝑚)
⇒ ∆𝑃 = 2.022
𝑃𝑎
𝑚
9.1.7.2 Between HX 1 and Compressor 2
The pressure lossdue tofrictionwithinthe pipe perunitlengthcanbe determined bythe follow
∆𝑃 = 𝑓
𝜌𝑉2
2𝐷𝑖
𝑉 =
𝑉̇
𝐴 𝑖
=
𝑉̇
𝜋
4
𝐷𝑖
=
0.089
𝑚3
𝑠
𝜋
4
(0.2545 𝑚)2
⇒ 1.75
𝑚
𝑠
∆𝑃 = 0.015
(10.31
𝑘𝑔
𝑚3)(1.75
𝑚
𝑠
)
2
2(0.2545 𝑚)
⇒ ∆𝑃 = 0.93
𝑃𝑎
𝑚
9.1.7.3 Between Compressor 2 and HX2

57
∆𝑃 = 𝑓
𝜌𝑉2
2𝐷𝑖
𝑉 =
𝑉̇
𝐴 𝑖
=
𝑉̇
𝜋
4
𝐷𝑖
=
0.029
𝑚3
𝑠
𝜋
4
(0.2545 𝑚)2
⇒ 0.57
𝑚
𝑠
∆𝑃 = 0.016
(32.67
𝑘𝑔
𝑚3)(0.57
𝑚
𝑠
)
2
2(0.2545 𝑚)
⇒ ∆𝑃 = 0.33
𝑃𝑎
𝑚
9.1.7.4 Between HX 2 and Tank
∆𝑃 = 𝑓
𝜌𝑉2
2𝐷𝑖
𝑉 =
𝑉̇
𝐴 𝑖
=
𝑉̇
𝜋
4
𝐷𝑖
=
0.015
𝑚3
𝑠
𝜋
4
(0.2545 𝑚)2
⇒ 0.295
𝑚
𝑠
∆𝑃 = 0.015
(57.94
𝑘𝑔
𝑚3)(0.294
𝑚
𝑠
)
2
2(0.2545 𝑚)
⇒ ∆𝑃 = 0.149
𝑃𝑎
𝑚

Compressed Air Energy Storage at the North Campus of Durham College and the University of Ontario Institute of Technology-An Engineering Approach

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