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EngineeringGREAT Solutions

Balancing & Control

Hydronic Engineering Handbook

Balancing of Control Loops

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© 2015 TA Hydronics SA, Eysins

www.imi-hydronics.com

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Balancing of Control Loops > 1. Why balance?

Contents

1. Why balance? ........................................................................................................................4

2. Control loops with two-way control valves ..........................................................................6

2.1 Variableprimaryandsecondaryflows ...........................................................................6

2.2 Variableprimaryflowandconstantsecondaryflow ....................................................13

2.3 Constantprimaryflowandvariablesecondaryflow ....................................................18

2.4 Constantprimaryandsecondaryflow ........................................................................20

3. Control loops with three-way control valves .....................................................................21

3.1 Variableprimaryflowandconstantsecondaryflow ....................................................21

3.2 Variableprimaryandsecondaryflows .........................................................................25

3.3 Constantprimaryandsecondaryflows .......................................................................26

3.4 Constantprimaryflowandvariablesecondaryflow ....................................................28

4. Control loops compared .....................................................................................................30

4.1 Activeprimarynetwork ...............................................................................................31

4.2 Passiveprimarynetwork .............................................................................................35

Appendix AThe authority of two-way control valves ................................................................................36

A.1 Theincompletedefinitionofvalveauthority .................................................................36

A.2 Thecorrectdefinitionofvalveauthorityβ’ ...................................................................38

A.3 Sizingofcontrolvalves ................................................................................................40

Appendix BThe authority of three-way control valves .............................................................................44

B.1 Inmixingfunction ........................................................................................................44

B.2 Indivertingfunction .....................................................................................................45

Appendix CHow to set the BPV to ensure the minimum pump flow ........................................................47

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1. Why balance?

Manypropertymanagersspendfortunesdealingwithcomplaintsabouttheindoorclimate.Thismaybethecaseeveninnewbuildingsusingthemostrecentcontroltechnology.Theseproblemsare widespread:

• Someroomsneverreachthedesiredtemperatures,particularlyafterloadchanges.

• Roomtemperatureskeepswinging,particularlyatlowandmediumloads,eventhoughtheterminalshavesophisticatedcontrollers.

• Althoughtheratedpoweroftheproductionunitsmaybesufficient,designpowercan’tbetransmitted,particularlyduringstartupafterweekendornightsetback.

Theseproblemsfrequentlyoccurbecauseincorrectflowskeepcontrollersfromdoingtheirjob.Controllerscancontrolefficientlyonlyifdesignflowsprevailintheplantwhenoperatingunderdesignconditions.

Theonlywaytogetdesignflowsistobalancetheplant.Balancingmeansadjustingtheflowbymeansofbalancingvalves.Thishastobedoneinthreerespects:

1. Theproductionunitsmustbebalancedtoobtaindesignflowineachboilerorchiller.Furthermoreinmostcases,theflowineachunithastobekeptconstantFluctuationsreducetheproductionefficiency,shortenthelifeoftheproductionunitsandmakeeffectivecontroldifficult.

2. Thedistributionsystemmustbebalancedtomakesureallterminalscanreceiveatleastdesignflow,regardlessofthetotalloadontheplant.

3. Thecontrolloopsmustbebalancedtobringabouttheproperworkingconditionsforthecontrolvalvesandtomakeprimaryandsecondaryflowscompatible.

4. Thishandbookdealswiththebalancingofcontrolloops.Ittellsyouhowtobalance23 commoncontrolloopsusing2-wayand3-waycontrolvalves.Forinformationaboutthebalancingofdistributionsystems,pleaseseeHandbookNo.2.Handbook3concernsBalancingofradiatorsystemswhilehandbook4examinesthestabilisationofthedifferentialpressure.

5

Balancing of Control Loops > 1. Why balance?

Heating Cooling%45

35

25

15

5

20 21 22 23˚C

35

25

15

5

20 21 22 23˚C

%45

Whyistheaveragetemperaturehigherinaplantnotbalanced?Duringcoldweatheritwouldbetoohotclosetotheboilerandtoocoldonthetopfloors.Peoplewouldincreasethesupplytemperatureinthebuilding.Peopleonthetopfloorswouldstopcomplainingandpeopleclosetotheboilerwouldopenthewindows.Duringhotweatherthesameapplies.Itisjustthatitwouldbetoocoldclosetothechiller,andtoohotonthetopfloors.

Onedegreemoreorlessinasingleroomrarelymakesanydifferencetohumancomfortortoenergycosts.Butwhentheaveragetemperatureinthebuildingiswrong,itbecomescostly.

Onedegreeabove20degreesCincreasesheatingcostsbyatleast8percentinmidEurope(12%inthesouthofEurope).Onedegreebelow23degreesCincreasescoolingcostsby15percentinEurope.

Percentage increase in energy costs for every degree C too high, or too low, relative to average building temperature.

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2. Control loops with two-way control valves

2.1 Variable primary and secondary flows

STAD

∆H

C

V

q

ts

tr

∆pV

Fig. 1. Control of a variable flow terminal unit

InFig1thetwo-waycontrolvalvecontrolsthecoiloutputbyadaptingthewaterflow.

Theauthorityofthecontrolvalveβ’=∆pV/∆H.Theterm“authority”isexplainedindetailinAppendixA and B.

Thetwo-waycontrolvalveisselectedtocreate,fullyopenanddesignflow,apressuredrop∆pV=∆H-∆pC-3(kPa)

Moreoverthisvalue∆pVmustbehigherthan0.25x∆Hmax.

Balancing procedure fig 1

1. Openallcontrolvalvesfully.

2. AdjusttodesignflowwithSTAD.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeTAHydronicsHandbookNo2).

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Balancing of Control Loops > 2. Control loops with two-way control valves

ts q

tr

∆pV

STAD

∆H

C

BPV

Fig. 2. A modulating relief valve reduces the differential pressure by a constant value regardless of the flow.

Whenthecontrolvalveisoversized,forinstanceduetothelimitedchoiceofKvvalues,theprimarydifferentialpressurecanbeindirectlyreducedbymeansofaBPVmodulatingreliefvalve.TheBPVreducesthedifferentialpressurebyaconstantvalueregardlessoftheflow.

Thecontrolvalveauthorityβ’=∆pV/(∆H-∆pBPV).


1. Openallcontrolvalvesfully.MakesurealltheBPVsareopen(minimumsetpoint).

2. AdjusttodesignflowwithSTAD.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2)andbeforeyouproceedtostep3.

3. FindoutwhichhandwheelsettingofSTADthatwillcreateapressurelossofatleast3kPainSTADfordesignflow.UsetheTA-SCOPEoraTAnomogramtofindthecorrectsetting.

4. ReadjustSTADaccordingtostep3.TheflowinSTADshouldnowbehigherthandesignvalue.

5. AdjustthesetpointoftheBPVuntilyougetbacktodesignflowinSTAD.MeasuretheflowinSTADasyouadjusttheBPV.

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STAD

∆HC

V

STAP

Fig. 3. A ∆p controller keeps constant the differential pressure a cross the control valve.

Dependingonthedesignoftheplant,thedifferentialpressureavailableonsomecircuitscanvarydramaticallywiththeload.Inthiscase,toobtainandmaintainthecorrectcontrolvalvescharacteristic,thedifferentialpressureacrossthecontrolvalvescanbemaintainedpracticallyconstantwitha∆pcontrollerasrepresentedonfigure 3.

Thedifferentialpressureacrossthecontrolvalve”V”isdetectedononesidebyconnectingthecapillarydownstreamthemeasuringvalveSTAD.ThepressureontheothersideisconnecteddirectlytotheactingmembranebyaninternalconnectionintheSTAP.

Whenthedifferentialpressureacrossthecontrolvalveincreases,theSTAPclosesproportionallytocompensate.

Thecontrolvalve”V”isneveroversizedasthedesignflowisalwaysobtainedforthevalvefullyopenanditsauthorityisandremainsclosetoone.

AlladditionaldifferentialpressureisappliedtotheSTAP.Thecontrolofthedifferentialpressureisquiteeasyincomparisonwithatemperaturecontrolandasufficientproportionalbandcanbeusedtoavoidhunting.

Astheflowsarecorrectateachterminal,nootherbalancingprocedureisrequired.IfallcontrolvalvesarecombinedwithSTAP,thenbalancingvalvesinbranchesandrisersarenotnecessarybutfordiagnosispurposes.


1. Openfullythecontrolvalve”V”.

2. PresettheSTADtoobtainatleast3kPafordesignflow.

3. Adjustthesetpoint∆pLofthedifferentialpressurecontrollerSTAPtoobtaininSTADthedesignflow.

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Balancing valve on each terminal

qp

STAP

STAD

STADor

TBV

VV

C C

∆H

Fig. 4. A differential pressure controller STAP stabilises the differential pressure across a set of terminal units.

Whenseveralsmallterminalunits”C”areclosetoeachother,itcanbesufficienttostabilisethedifferentialpressureonthewholesetaccordingtofigure4.

ThesupplypressureistransmittedtotheSTAPbymeansofacapillarymountedonthebalancingvalveinthesupplypipe.

Whenthedifferentialpressure∆Hincreases,thevalveSTAPshutstocompensate.Eachcontrolvalves”V”ischosentocreate,fullyopenandatdesignflow,approximatelythesamepressuredropasitscoilunit.


1. OpentheSTAPfully. ThecontrolvalvesVarefullyopen.

2. BalancetheterminalsofthebranchaccordingtotheTABalancemethod,whichdoesnotdependonthedifferentialpressure∆Havailable.STADserveaspartnervalves.

3. AdjustthesetpointoftheSTAPtoobtaindesigntotalflowqpthroughtheSTAD.


Balancing valve on each terminal

10

tp

qb

qp qstr

∆H

STAD-1 STAD-2

BPV

ts

B

AD

V

Fig.5. A secondary pump creates a sufficient differential pressure. The secondary water temperature is necessarily different from that of the primary.

Ifthedifferentialpressure∆Histoolowtogiveareasonableauthorityforthecoilcontrolvalves,asecondarypumpcancreateasufficientdifferentialpressure.

ThesolutioninFig5canalsobeusedwhentheprimarydifferentialpressureistoohigh.

Thesecondarywatertemperaturetscanbeconstantorvariable,butisnecessarilydifferentfromtheprimarytemperaturetp.Inheatingts<tp,whileincoolingts>tp.

Atlowloads,thedifferentialpressureacrossthesecondarytendstoincrease.Whenthispressureexceedsacertainvalue,theBPVopenstoallowaminimumflowtoprotectthepump.Thisflowalsolimitsthetemperaturedropinthepipessothatthenecessarywatertemperatureisobtainedthroughoutthesecondarynetwork.


The secondary

1. Openallcontrolvalvesfully.ClosetheBPV.

2. BalancethecoilsinthesecondarysystemwithSTAD-2asthePartnervalve (seeHandbookNo2).

3. SettheBPVonthemaximumallowed∆pforthecoilcontrolvalves.

4. Closethecoilcontrolvalves.

5. SettheBPVtoobtaintheminimumpumpflow(seeAppendixC).

The primary

1. OpenthecontrolvalveV.

2. Iftheprimaryflowisunknown,calculateitusingtheformulaonpage15.

3. AdjusttheprimarydesignflowwithSTAD-1.Dothisaspartofthebalancing

4. procedurefortheentireprimarysystem(seeHandbookNo2).

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tp

qb

qp

qs

tr

∆H

STAD-1

STAD-2

BPV

ts

B

AD

∆T

t1t2VC

Fig.6. A differential temperature controller maintains a minimum flow qb in the bypass, such that ts = tp.

Ifthesecondarywatertemperaturemustbeequaltothatoftheprimary,thecircuitinFig6(heatingonly)orthecircuitinFig7(bothheatingandcooling)maybeused.

Toobtaints=tp,theflowqbthroughthebypassmustbegreaterthanzero.A∆TcontrolleractsontheprimarycontrolvalveVtoensureaminimumflowqbintherightdirection.The∆Tcontrollerkeepst2slightlyhigherthant1.Normally,thesetpointofthe∆Tcontrollerisbetween1and2degrees.


The secondary

1. Openallcontrolvalves.ClosetheBPV.

2. BalancethecoilsinthesecondarysystemwithSTAD-2asthePartnervalve (seeHandbookNo2).

3. SettheBPVonthemaximumallowed∆pforthecoilcontrolvalves.

4. Closethecoilcontrolvalves.


The primary


2. Iftheprimaryflowisunknown,calculateitusingtheformulabelow.

3. AdjusttoprimarydesignflowwithSTAD-1.Dothisaspartofthebalancing procedurefortheentireprimarysystem(seeHandbookNo2).

qp=1.05qs


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tp

qb

qp

qs

tr

∆H

STAD-1

ts = tp

B

AD

C

STAP

STAD-2

Fig. 7. A differential pressure controller keeps the flow constant in the bypass, ensuring a constant differential pressure across this bypass.

ThecircuitinFig7maybeusedincoolingplantswhere∆Histoolowtogiveasufficientauthorityforthecoilcontrolvalves,andwhere∆Hvariesgreatly.

ThecontrolvalveSTAPmaintainsasmallandconstantflowinthebypass,regardlessofvariationsin∆H.ThissmallflowismeasuredbymeansofSTAD-2.When∆Hincreases,theSTAPclosescorrespondingly,ensuringaconstantdifferentialpressureacrossthebalancingvalveSTAD-2.


1. Openallcontrolvalves.

2. SetSTAD-2tocreate,for5%ofdesignflowqs,apressurelosscorrespondingtotheselectedsetpointofthe∆pcontroller.UsetheTA-SCOPE,oraTAnomogram,tofindthecorrectsettingforSTAD-2.

3. BalancethesecondarycircuitwhereSTAD-1isthePartnervalve (seeHandbookNo.2).

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2.2 Variable primary flow and constant secondary flow

qb

tpts

tr

qp

qs = ct

∆H

STAD-2STAD-1

V

B

A

C

tr

Fig. 8. Control of coil output for a coil supplied at constant flow.

Thiscircuitisfrequentlyusedbothinheatingandincooling.Thecoilsupplytemperaturetsisadaptedtothepowerneedthroughcontroloftheprimaryflow.

If,underdesignconditions,tsmustbeequaltotp,themaximumflowqpintheprimarymustbeequaltoorgreaterthanthesecondaryflowqs.Otherwise,theinstalledpowercannotbetransmittedtothesecondarysincethedesignvaluetsccannotbeobtained.Primaryandsecondaryflowsmustbecompatible.TheseflowsareadjustedbythebalancingvalvesSTAD-2andSTAD-1.

A floor heating example:Assumethattsc=50°C,whichiswellbelowthetp=80°C.Thecontrolvalvemustthenbeselectedforarelativelysmallflow.Forareturntemperaturetrc=45°C,theformulabelowshowsthattheprimaryflowwillbeonly14%ofthesecondaryflow.Ifthecontrolvalveisselectedforthisflow,itcanoperateoveritsentirerange.Thelimitof50°Cforthecircuitsupplytemperaturewillnotbeexceededatthemaximumvalveopening.Ifthesecondarypumpfails,theprimaryflowpassesthroughthebypass,preventingoverheatinginthecircuit.


1. Opencontrolvalvesfully.

2. AdjusttosecondarydesignflowwithSTAD-2.


4. AdjusttheprimaryflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2).

qp=qs =qs =0.14qs

ts-tr 50 - 45

tp-tr 80 - 45


14

qb

tpts

tr

qp

qs = ct

∆H

STAD-2STAD-1

V

B

A

C

tr

Fig. 9. A check valve in the bypass allows a certain waterflow through the coil C even if the secondary pump fails.

ThisisessentiallythesamecircuitasinFig8.However,acheckvalveisaddedtopreventcirculationindirectionBAinthebypass.

Ifthecircuitisusedindistrictheatingandtheprimarycontrolvalveisoversized,thecheckvalvepreventsheatingofthereturnwater.Ifthecircuitisusedforaheatingcoilincontactwithoutsideair,thecheckvalveeliminatestheriskoffreezingduetosecondarypumpfailure.

Notethatit’simpossibletoobtainaprimaryflowgreaterthanthesecondaryflow.


tsc equal to tp:

1. ClosethecontrolvalveV.

2. AdjusttosecondarydesignflowqscwithSTAD-2.


4. AdjusttheprimaryflowtothesameflowqscwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2).

tsc not equal to tp:



3. Ifprimaryflowisunknown,calculateitusingtheformulabelow.

4. Openthecontrolvalve.

5. AdjusttheprimarydesignflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2).

(ts-tr)

(tp-tr)

qp=qs

15

qb

tp

tr

qp

qs = ct

∆H

STAD-2STAD-1

V

B

A

tr

C C

ts

Fig. 10. A primary constant flow distribution converted to primary variable flow.

It’scommontoconvertexistingconstantwaterflowdistributionsystemsintovariableflowinlargeplants.Therearethreereasons:1)Thesupplywatertemperaturecanthenbekeptconstantwithouthavingtokeepinserviceallproductionunitsatallloads.2)Avariabledistributionflowmeansreducedpumpingcosts.3)Theplantcanbedesignedwithadiversityfactor.

Normally,thesecondarysidecontinuestoworkwithconstantflow.

Afterconversion,wecannotworkwithts=tp.WhenthevalveViscompletelyopen,wecangetts=tpwithflowreversalinthebypass.Sincethedemandissatisfiedinthissituation,thereisnosignaltomakethetwo-wayvalveclose.Itremainsopenandwearebacktoaconstantflowdistributionsystem.Toavoidthis,tshastobeadjustedsothatts<tpinheatingandts>tp in cooling.

P

1 + (tsc -trc)

(tp -trc)

P

100( -1)

%qp=

Theprimaryflowwillvaryasafunctionoftheload:

Pistheloadinpercentofdesignpower.

Nowassumethattp=6°C,tsc=8°Candtrc=12°C.ForP=50%,wegetqp=75%.Thustheflowdemandis75%forapowerdemandof50%.

Beforeconvertingtheconstantflowsystemintoavariableflowsystem,theflowdemandwas100%forapowerdemandof50%.

Thisconversiondoesnotchangereallytheprimaryintoatruevariabledistributionastheflowin%remainshigherthanthepowerin%.


1. Balancethethree-wayvalvecircuits(seeHandbookNo2).STAD-2isthePartnervalve.

2. Iftheprimaryflowqpisunknown,calculateitusingtheformulabelow.


4. AdjusttheprimaryflowqpwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2).


16

(ts -tr)

(tp -tr)

qp=qs

Fig 11a Fig 11b

≈ 0

qp

STAD-3

qb

STAD-2

ts

tr

qs

STAD-1V

CB

A

F+E-

≈ 0

BPV

qb

STAD-2

STAD-1V

C

F+E-

tptp

Fig. 11. Secondary pumps induce flow through the distribution network.

Ifthedistributionnetworkisalowpressurelosspassivecircuit,circulationmaybeinducedbysecondarypumps.

ThebalancingvalveSTAD-3createsacertaindifferentialpressurebetweenFandE.ThispressuregeneratestheprimaryflowqpinthecontrolvalveV,throughFCBandAE.Thedifferentialpressure∆pcEFisobtainedforqb=qsc-qpc.Thisimpliesthatqschastobegreaterthanqpc.WhenthecontrolvalveVisclosed,thebypassflowqb=qscand∆pEFisatitsmaximum.ThisisalsothedifferentialpressureappliedacrosstheclosedcontrolvalveV.Toobtainagoodauthorityforthisvalve,it’simportanttoavoidbigchangesin∆pEF.Thismeansthatqpchastobeassmallaspossiblecomparedtoqsc.Consequently,thissystemcanonlybeconsideredifthereisalargedifferencebetweents and tp,asforexampleinfloorheating.

Theflowthroughthebypassisgivenbythisformula:

(tp -ts)

(tp -tr)

qb=qs

Assumingthatthesecondaryflowqsismoreorlessconstant,thecontrolvalveauthorityβ’ = ∆pV/∆pEFmax.

Example: Floorheatingwithtp=80°C,ts=50°C,tr=45°Candqs=100.Atfullload,qb=100(80-50)/(80-45)=85.7.Atthisflow,thebalancingvalveSTAD-3inthebypassmustcreateadifferentialpressurethatcompensatesforthepressurelossofthetwo-wayvalve(8kPaforinstance)andoftheprimarycircuit(5kPa),atotalof13kPa.Whenthetwo-wayvalveisclosed,atzeroload,theflowqb changesto100(assumingthattheincreaseinthepressurelossEFhaslittleeffectontheflowqs)

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andthepressurelossintheSTAD-3becomes∆pEFmax=13x(100/85.7)2=18 kPa.

Thecontrolvalveauthorityisthereforeβ’=8/18=0.44.

TheSTAD-3canbereplacedbyamodulatingreliefvalveBPV(Fig11b)whichkeepsaconstantdifferentialpressureEF.Inthefloorheatingexample,thisimprovescontrolvalveauthorityfrom0.44to0.61.


STAD-3 in the bypass (Fig 11a):


2. SettheSTAD-3tocreateapressureloss∆pEF=∆pcV+pressuredropintheprimarycircuit(8+5=13kPainourexample)foraflowrateqb=(qsc-qpcinthebypass.UsetheTA-SCOPE,ortheTAnomogram,tofindthecorrecthandwheelsettingforSTAD-3.

3. SettheSTAD-1tocreateapressurelossof3kPaforprimarydesignflow.Usethe TA-SCOPE,oraTAnomogram,tofindthecorrecthandwheelsettingforSTAD-1

4. ClosethecontrolvalveV.AdjusttodesignflowwithSTAD-2.

5. Iftheprimaryflowqpcisunknown,calculateitusingtheformulabelow.

6. OpenthecontrolvalveV.ReadjustSTAD-3toobtainqp=qpcmeasuredinSTAD-1.

(tsc -trc)

(tp -trc)

qp=qs

BPV in the bypass (Fig 11b):


8. SettheSTAD-1tocreateapressurelossof3kPaforqp=qpc.UsetheTA-SCOPE,oraTAnomogram,tofindthecorrecthandwheelsettingforSTAD-1.

9. OpenSTAD-2.AdjusttheBPVtoobtaindesignflowinSTAD-1.

10. AdjustSTAD-2toobtaindesignflowinthesecondary.


18

2.3 Constant primary flow and variable secondary flow

tp

qb

qp

qs

trSTAD

tp

BPV

A

B

∆H

Fig. 12. A modulating relief valve BPV stabilizes the differential pressure applied to small units.

Whentheavailabledifferentialpressureontheprimaryistoohighforthesecondary,thecircuitinFig12maybeused.

ThesetpointoftheBPVcanbeselectedwithinarangefrom8to60kPa.Thismakesitpossibletoensuregoodworkingconditionsforthecoilcontrolvalves(goodauthority)regardlessofvariationsinthedifferentialpressure∆H.TheBPVensuresaconstantdifferentialpressurebetweenAandB.STADcreatesapressurelossof(∆H-∆pBPV).

Balancingprocedurefig12

1. Openallcontrolvalves.CloseallBPVs.

2. Balancethecoilsagainsteachother,thebranchagainstotherbranchesandtheriseragainstotherrisers(seeHandbookNo2).Dothisbeforeyouproceedtostep3.

3. Closethecontrolvalvesofthisbranch.

4. ReducetheBPVsetpointslowlyuntilyougetbackto2/3ofthedesignflowinSTAD.

5. (Seealsohandbook4-appendix5.5forcomplementaryexplanations).

19

tp

qb

qp

qs

tr

∆H

STAD-1 STAD-2

BPV

ts = tp

Fig. 13. Reducing or increasing the differential pressure applied across coils by means of a secondary pump.

Whentheprimarydifferentialpressureistoohighortoosmallforthesecondary,thecircuitinFig13presentsapossiblesolution.Inthiscircuit,areliefvalveisusedtoobtainaminimumflowtoprotectthepump.STAD-1isessentialtoavoidshort-circuitingtheprimarysystem.

Balancingprocedurefig13


2. BalancethecoilsagainsteachotherwithSTAD-2asthePartnervalve(seeHandbookNo2).

3. SettheBPVonthemaximumpermitteddifferentialpressureforthecoilcontrolvalves.

4. Closethecoilcontrolvalvesofthisbranch.

5. Ifnecessary,reducetheBPVsetpointuntilyouobtaintheminimumpumpflow(seeAppendixC).

6. AdjusttoprimarydesignflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo.2).


20

2.4 Constant primary and secondary flow

tp

qb

qp

qstr

STAD-1

tp

BPV

A

B

∆H

D

C

qv

ts

C

STAD-2

V

Fig. 14. Constant primary and secondary flows.

Acoilissuppliedatconstantflow.Thesupplywatertemperatureismodifiedbythetwo-waycontrolvalveV.Thistemperaturehastobeadjustedsothatts<tpinheatingandts>tp in cooling. The BPV keepsthedifferentialpressureCDconstant.ThisisthedesignpressurelossforthecontrolvalveV,whichhasanauthoritycloseto1afterbalancing.




3. AdjusttheprimaryflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2)andbeforeyouproceedtostep4.


5. MeasuretheflowinSTAD-1.ReducetheBPVsetpointslowlyuntilyougetbackto2/3ofthedesignflowinSTAD-1.

6. AdjustthesecondarydesignflowwithSTAD-2.

(tsc -trc)

(tp -trc)

qp=qs

(Seealsohandbook4-appendix5.5forcomplementaryexplanations).

21

3. Control loops with three-way control valves

3.1 Variable primary flow and constant secondary flow

Passive primary network

Apassiveprimarynetworkisadistributionnetworkwithoutapump.Thesecondarypumppressurizesboththeprimaryandthesecondary.

tg ts

tr qs

E C

DSTAD

G

qg

HV

L

qb

tr

Fig. 15. Mixing circuit associated with a production unit.

Fig 15 shows a circuit controlled by a three-way mixing valve. The primary circuitconsists of an exchanger, a bypass line or a boiler that can either accept zero flowor be equipped with a bypass pump that generates a minimum flow. The three-wayvalve should be selected for a pressure loss at least equal to that in G, and at least 3 kPa.


1. Openthethree-wayvalvecompletely.

2. AdjusttodesignflowwithSTAD.

Balancing of Control Loops > 3. Control loops with three-way control valves

22

tg ts

tr qs

E

STAD-2

G

qg

HV

L

qb2

tr

C

D

qb1

B

A

tp

STAD-3

STAD-1

Fig. 16. Mixing circuit with intermediate bypass.

Whentheflowqsinthecircuitisgreaterthanthedesignflowthroughtheproductionunit,abypassABinsurescompatibilitybetweentheflows.

ThepressuredropcreatedbytheSTAD-3,forawaterflowqb1=qsc -qgc,isthenecessarydifferentialpressuretocompensatethepressuredropsintheSTAD-1+G+the3-wayvalve.

Thepressuredropcreatedbythethree-waycontrolvalveforthedesignflowqgcmustbeequalorhigherthanthedesignpressuredropinGandaccessorieswithaminimumof3kPa.


1. Openthethree-waycontrolvalve”V”.

2. Calculatethedesignflowqb1requiredintheSTAD-3andtheflowqgcinSTAD-1withformulabelow.

3. STAD-3andSTAD-1arebalancedaccordingtotheTABalancemethod(SeeHandbook2).

4. AdjusttheflowqswiththeSTAD-2.

(tsc-trc)

(tg -trc)

qgc =qsc qb1 =qsc -qgc

23

Active primary network

Anactiveprimarynetworkisadistributionsystemwithitsownpump.

Theprimarypumpcreatesadifferentialpressurewhichforcesthewaterflowthroughthesecondarycircuits.

L

tr

qs

qb C

∆H

STAD-2

V

E

tsqp tp

D

F

-

+H

tr

CSTAD-1

Fig. 17. Mixing valve with differential primary pressure and compensation balancing valve.

Thethree-wayvalveinFig17issuppliedbyaprimarydifferentialpressure∆H.Thispressuremaydisturbthefunctionofthethree-wayvalve.Thewaterflowqbinthebypassmayreverseandcancelthemixingfunctionofthecontrolvalve.

Topreventthis,thebalancingvalveSTAD-1hasbeeninstalled.ThepressurelossinSTAD-1shouldbe∆Hfordesignflowqpc.

Thedesignpressurelossacrossthethree-wayvalvemustbeatleastequalto∆Htogiveanauthorityof0.5.Thispressurelosshastobecoveredbythesecondarypump.


1. Closethethree-wayvalve.

2. AdjusttosecondarydesignflowwithSTAD-2

3. Openthethree-wayvalve.

4. ContinuetomeasuretheflowwithSTAD-2.AdjustSTAD-1toobtainthesameflowasinstep2.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2).


24

qb

qs

tp

tr

ts

qp

C

V2

∆p = min

STADV1

∆H

Fig. 18. Elimination of the primary differential pressure by means of a pressure controller.

Insomeplantsthethree-wayvalvesdonotworksatisfactorilybecauseofatoohighprimarydifferentialpressure.Sometimesadifferentialpressurecontrollerisinstalledtoeliminateorreducethispressuretoareasonablevalue,asshowninFig18.

Thisisanexpensivesolution.However,itmaybeconsideredifthedifferentialpressurecontrollerisusedtosupplyseveral3-wayvalvesandwhenavariableflowsystemisrequiredinthedistribution.Ifaconstantprimaryflowisaccepted,thedesignoffigure20isbetter.



2. AdjusttosecondarydesignflowwithSTAD.

3. Adjustthesetpointofthedifferentialpressurecontrollertoavalueaslowaspossible.

25

3.2 Variable primary and secondary flows

Passive primary network

qb

qstp

tr

ts

qp

∆p≈ 0

+

BPV

STAD

-V

Fig. 19. The three-way valve prepares the water temperature in the distribution system.

Thethree-wayvalvecontrolsthesecondarywatertemperature.Thetwo-waycontrolvalvesmakethefinetuningoftheenergysupplybyadaptingtheflowtothedemands.

Thethree-wayvalvehasanauthoritycloseto1.Atlowloads,themodulatingreliefvalveBPVensuresaminimumpumpflow,andalsoreducesthetemperaturedropinthepipeline.

Note:Belowacertainflow,athree-wayvalvewillworkwithlaminarratherthanturbulentflow.Thenthethree-wayvalvetemporarilylosesitsbasiccharacteristicsandthecontrolloopbecomesdifficulttostabilize.Thus,theminimumflowcontrolledbytheBPVmustbehighenoughtocreateapressurelossofatleast1kPainthethreewayvalve.


1. Openallcontrolvalves.ClosetheBPV.

2. Balancethesecondarysystem(seeHandbookNo2)withSTADasthePartnervalve.

3. Closealltwo-waycontrolvalves.



26

3.3 Constant primary and secondary flows

Active primary network

qs

trSTAD-2

A

∆H

B D

qp

tp

qb

ts

STAD-1

V

C

Fig. 20. The balancing valve STAD-1 and the bypass AB, eliminate the primary differential pressure across the three-way valve.

Iftheprimaryflowmaybeconstant,it’ssimpletoavoidatoohighdifferentialpressureontheprimaryofamixingthree-wayvalve.It’sjusttoinstallabypassABandtocompensatetheprimarydifferentialpressurewiththebalancingvalveSTAD-1.Theauthorityofthethree-wayvalvewillthenbecloseto1.





4. AdjusttheprimaryflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo.2).

(tsc -trc)

(tp -trc)qp=qs

27

qs

trSTAD-2A

∆H

B

D

F

qp

tp

qbp qbs

ts

STAD-1V

C

Fig. 21. When tsc is not equal to tp, it is better to place the bypass on the secondary side.

Whenthedesigntemperaturetscisnotequaltotp,thecircuitinFig21isoftenpreferabletothatinFig 20.

TheflowinthecontrolvalveislowerinFig21thaninFig20(qpinsteadqs),thusallowingtheuseofasmallerthree-wayvalve.

Theauthorityofthethree-wayvalveiscloseto1.





4. AdjusttheprimaryflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo.2).

(tsc -trc)

(tp -trc)qp=qs


28

3.4 Constant primary flow and variable secondary flow

tr

qs

qbSTAD-3 C

∆H

STAD-1V

tp

qp

tp

Fig. 22. A three-way mixing valve in a diverting circuit.

Thethree-wayvalveusedasamixingvalveinadivertingcircuitcansupplythecoilatvariableflowandconstantwatersupplytemperature,whilekeepingprimaryflowconstant.Thisway,thethree-wayvalveeliminatesinteractivitybetweencircuitsontheprimaryside.

Thethree-wayvalveshouldcreateadesignpressurelossequaltoorgreaterthanthepressurelossincircuitCtoensureanauthorityofatleast0,5.

Note: The most important balancing valve is the STAD-1. STAD-3 can be omitted if∆pC<0.25∆H.


1. Openallthree-wayvalves.

2. AdjusttodesignflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2)andbeforeyouproceedtostep3.


4. MeasuretheflowinSTAD-1.AdjusttodesignflowwithSTAD-3.

29

tr

qs

qbSTAD-3 C

STAD-1V

tp

qp

tpBA

≈ 0

H

Fig. 23. Circuit in diversion on a passive distribution.

Ifthedistributionsystemispassive(noactivedifferentialpressure),thereisaneedforaseparatepump.Thispumpcanbecommontoseveralcircuits.

Note:ThemostimportantbalancingvalveisSTAD-1.STAD-3canbeomittedif∆pC<0.25H.


1. Openallthree-wayvalves.

2. AdjusttodesignflowwithSTAD-1.Dothisaspartofthebalancingprocedurefortheentireprimarysystem(seeHandbookNo2)andbeforeyouproceedtostep3.

3. Closeallthree-wayvalves.

4. MeasuretheflowinSTAD-1.AdjusttodesignflowwithSTAD-3.


30

4. Control loops compared

Variable primary water flow

Variablesecondarywaterflow Constantsecondarywaterflow

2-way 5 2-way 8 - 9 -10

3-way 19 3-way 17 - 18

Constant primary water flow

Variablesecondarywaterflow Constantsecondarywaterflow

2-way 12 - 13 2-way 14

3-way 22 3-way 20 - 21

Same functions obtained with two way or three way control valves.

31

4.1 Active primary network

STAD

��H

C

V

q

ts

tr

��pV

STAD

��HC

V

STAP

ts q

tr

��pV

STAD

��HC

BPV

tp

qb

qp qstr

STAD-1 STAD-2

BPV

ts

B

A

tp

qb

qp

qs

tr

STAD-1

STAD-2

BPV

ts

B

A

��T

t1t2VC

tp

qb

qp

qs

trSTAD-1

ts = tp

B

A

CSTAP

STAD-2

1

7

6

5

3

2

��pV > ��H/2 *

��pSTAD = ��H -��pV - ��pC

ßß' = ��pV / ��H

��pV > (��H - ��pBPV)/2 *

��pSTAD > 3 kPa

��pBPV = ��H - ��pV - ��pC - ��p STAD

ßß' = ��pV / ( ��H - ��pBPV)

��pV > Min STAP set point ≥10 kPa��pSTAD ≥3 kPa

ßß' close to one

qs > qp

��pV > ��H/2 *

��pSTAD-1 = ��H -�pV

ßß' = ��pV / ��H

ts = tp

qs < qp

��pV > ��H/2 *


ßß' = ��pV / ��H

ts = tp

��pV > ��H/2 *

��pSTAD-1 = ��H - ��pV - ��p STAD-2

ßß' = ��pV / ��H

Balancing of Control Loops > 4. Control loops compared

Variable primary and secondary water flows. Variables are represented by their design values -

Values recommended (*).

32

8��pV > ��H/2 *

ßß' = ��pV / ��H

ßß' = ��pV / ( ��pV + ��pH )

qs > qp

��pV > ��H *

��pSTAD-1 = ��H

ßß' = ��pV / ��H

��pV1 > ��H/2 *

��pSTAD-1 = ��H -��pV

ßß'V1 = ��pV1 / (��H -��p )

��pV2 > 3 kPa *

��pSTAD-1 = ��H - ��pV1

qb

tpts

tr

qp

qs = ctSTAD-2STAD-1

V

B

A

C

tr

qb

tpts

tr

qp

qs = ctSTAD-2STAD-1

V

B

A

C

tr

qb

qs

tp

tr

ts

qp

C

V2

�p= min

STAD-2V1

�H STAD-1

L

tr

qs

qb C

��H

STAD-2

V

E

tsqp tp

D

-

+H

tr

CSTAD-1

18

17

9

qs > qp

��pV > ��H/2 *


Variable primary water flow and constant secondary water flow. Variables are represented by their design values - Values recommended (*).

33

12

ßß' = ∆∆pV / ( ∆∆pV + ∆∆pC )

∆∆pSTAD-3 = ∆∆pC

∆∆pSTAD-1 = ∆∆H -∆∆pBPV

∆∆pSTAD-1 = ∆∆H - ∆∆pV - ∆∆pC

ts = tp

∆∆pV > ∆∆pC *

∆∆pSTAD-1 = ∆∆H

tp

qb

qp

qs

trSTAD

tp

BPV

A

B

tp

qb

qp

qs

trSTAD-1 STAD-2

BPV

ts = tp

tr

qs

qbSTAD-3 C

∆H

STAD-1V

tp

qp

tp

22

13

ts = tp

ts = tp


Constant primary water flow and variable secondary water flow. Variables are represented by their design values - Values recommended (*).

34

14

ßß' = ∆∆pV / ∆∆pBPV

∆∆pSTAD-1 = ∆∆H -∆pBPV

∆∆pSTAD-1 = ∆∆H - ∆pV

∆∆pV > 8 kPa

∆∆pSTAD-1 = ∆∆H

21

20

qs > qp

tp

qb

qp

qstr

STAD-1

tp

BPV

A

B

∆H

D

C

qv

ts

C

STAD-2

V

qs

trSTAD-2

A

∆H

B D

qg

tg

qb

ts

STAD-1

V

C

qs

trSTAD-2A

∆H

B

D

qg

tg

qbp qbs

ts

STAD-1V

C

∆∆pV > 3 kPa *

ßß' = 1

∆∆pV > 3 kPa *

ßß' = 1

Constant primary and secondary water flows. Variables are represented by their design values - Values recommended (*).

35

4.2 Passive primary network

∆pV > ∆p1 *

ß' = pV / (∆pV + ∆p1)

qp < q s

∆pSTAD-3 =

∆p1 + ∆pV + ∆pSTAD-1

∆pSTAD-1 ≥ 3 kPa *

∆pV ≥ ∆pSTAD-3 / 2 *

ß' = ∆pV /∆pSTAD-3 max

(11a)

V

qp

t p

qs

t g

STAD-1 STAD-2

STAD-3

L>10 dt r

qs

t s

t p

qb

qp

STAD-1 STAD-2

STAD-3

L>10 dt r

qs

t st p qb

qp

STAD-1 STAD-2

BPV

V

L>10 dt r

qs

t st p

qb

qp

STAD

V

L>10 d

t r

t s

qb2

qg

V

t r

qb1

(11b)

(16)

(15)

∆p= 0

qp < q s

∆pBPV =

∆p1 + ∆pV + ∆pSTAD-1

∆pSTAD-1 ≥ 3 kPa *

∆pV ≥ ∆pSTAD-3 / 2 *

ß' = ∆pV /∆pBPV

- +∆p

qp < q s

∆pV > ∆p1 *

ß' = ∆pV / (∆pV + ∆p1)


Variable primary water flow and constant secondary water flow. Variables are represented by their design values - Values recommended (*).

36

Appendix A

The authority of two-way control valves

A.1 The incomplete definition of valve authority

Thestaticcharacteristicofacontrolvalveisdefinedforaconstantdifferentialpressureacrossthevalve.Butthispressureisrarelyconstantinaplant.Therefore,therealcharacteristicofacontrolvalveisnotthesameasthetheoreticalone.

Whenthecontrolvalveisfullyopen,thedifferentialpressure∆pminisequaltotheavailabledifferentialpressureminuspressurelossesinterminalunit,pipesandaccessories.

Whenthecontrolvalveisclosed,pressurelossesintheotherelementsdisappearsincetheflowiszero.Theentireavailabledifferentialpressure∆Hmax=∆pmaxisthenappliedacrossthecontrolvalve.

Butthecontrolvalveissizedbasedon∆pminsinceitisatthispressurelossthedesignflowistobeobtained(fullyopenvalve).

Whenthevalveisnearitsclosedposition,therealflowishigherthanthetheoreticalsincethedifferentialpressureisgreaterthan∆pmin.Thetheoreticalcharacteristicisdistorted.Thedegreeofthisdistortiondependsontheratio∆pmin/∆pmax.

Thisratioisthecontrolvalveauthority.

β =Dpmin

Dpmax

Coil

STADSupply Return

∆pmax∆pmin

pressures

Thedifferentialpressureappliedtothecontrolvalvedependsonitsdegreeofopening.

37

Distortion of a linear valve characteristic as a function of its authority.

Distortion of the “EQM” characteristic of a valve as a function of its authority.

100

0 10 20 30 40 50 60 70 80 90 100

90

80

70

60

50

40

30

20

10

0h %

q %

β = 0

.1

0 .25

0 .5

1

100

90

80

70

60

50

40

30

20

10

0

q %

0 10 20 30 40 50 60 70 80 90 100

h %

0.25

0.5 1

β = 0

.1

Thelowertheauthority,thebiggerthedistortionofthetheoreticalvalvecharacteristic.

Consideravalvewithalinearcharacteristic,designedtoobtainthedesignflowexactlyatfullopening,butwithafairlylowauthorityof0.1.Ata10%valvelift,theflowinthecircuitisabout30%.

Supposethattheterminalunitisaheatingcoilwithadesigntemperaturedropof10K.Then30%ofthewaterflowproduces80%ofthedesignpower.

Thefinalresultisthatthecoiloutputis80%ofthedesignpoweratacontrolvalveliftofonly10%.Undertheseconditionsthereislittlehopetoobtainstablecontrol.Thesituationwouldbeevenworseif,forthesameauthority,thecontrolvalvewasoversized!

Anauthorityof0.5isacceptablesinceitdoesnotgreatlydeformthevalvecharacteristic.Inotherwords,thepressurelossatdesignflowinafullyopencontrolvalvemustbeequivalenttoatleasthalftheavailabledifferentialpressure.

Notethatthedesignflowdoesnotappearinthedefinitionofvalveauthority.

Thecurvesinthefiguresaboveareplottedassumingdesignflowwhenthecontrolvalveisfullyopen.Butthatisrarelythecaseinpractice,sinceitisdifficulttoavoidacertaindegreeofoversizing.

Whenacontrolvalveisoversized,∆pminisreduced,assumingconstant∆pmax.Thus,thecontrolvalveauthorityisalsoreduced.Thetheoreticalvalvecharacteristicwillbeheavilydistortedandcontrolbecomesdifficultatsmallloads.

However,anoversizedcontrolvalvecanhaveagoodauthority.Ifthedifferentialpressureappliedtoacircuitisdoubled,∆pminand∆pmaxincreaseinthesameproportionandtheauthorityremainsunchanged,althoughthere’snowanoverflowinthecircuit.

What,then,willhappentothevalveauthorityinacircuitexposedtoavariabledifferentialpressure?

Then∆pmaxand∆pminwillvarysimultaneouslyinthesameproportions.Thevalveauthorityβthusremainsconstant.

However,thevalvecharacteristicisdeformed,despitethefactthattheauthorityβisthesame.

Therefore,theauthorityasdefinedabove,doesnotgiveenoughinformationabouttherealdistortionofthevalvecharacteristic.

Balancing of Control Loops > Appendix A. The authority of two-way control valves

38

A.2 The correct definition of valve authority β’

Wegetamorecoherentdefinitionoftheauthorityifwerelatethepressurelossinthecontrolvalvefordesignflowtothemaximumpressurelossinthevalve:

Dpacrossthefullyopencontrolvalveanddesignflow

Dpvalveshutβ’ =

0 10 20 30 40 50

h %

q %

60 70 80 90 100

90

100

110

120

130

140

80

70

60

50

40

30

20

10

0

β = 0.5, β' = 0.5

β = 0.5, β' = 0.32

β = 0.5, β' = 0.22β = 0.5, β' = 0.13

Theoretical

Flow as a function of valve lift when the circuit supply pressure varies with constant authority β.

Thefigureshowsthattheauthorityβ’takesintoaccountthedistortionofthevalvecharacteristic.Thisisnottruefortheauthorityβaccordingtotheconventionaldefinition.

Thetwoauthorityfactorsrelatetoeachotheraccordingtotheformulabelow

(Sqistheoverflowfactor):

β=(Sq)2 . β’

Sq≥1withthevalveopen.Whenthemaximumflowisequaltodesignflow,β= β’.

39

140

120

90 1000

20

40

60

80

100

20 30 40 50 60 70 80100

q : flow in %

b- with STAD

a- theoretical

h: lift in %

c- without STAD

Influence on the control valve characteristic through limiting the maximum flow by a balancing valve.

Can a balancing valve be put in series with a control valve?

AcontrolvalvewithexactlythecalculatedKv-valueisusuallynotavailableonthemarket.Consequently,theinstalledvalveismoreorlessoversized.Duringstartupafternightsetback,whenmostcontrolvalvesareopen,theoverflowinfavouredunitscreatesunderflowsinothers.It’sthereforeessentialthattheflowthroughthecontrolvalveislimitedbyabalancingvalve.

Thefigureaboveshowshowthistypeoflimitationinfluencesthecontrolvalvecharacteristic.Withoutabalancingvalve,theoverflowinthefullyopencontrolvalvewillbe22%anditsauthorityβ = 0.5 accordingtotheconventionaldefinitionofvalveauthority.Butthisisarathermisleadinginformationabouttheauthority,sincetheflowiswrong.

Theauthorityβ’=0.34indicatestherealdistortionofthevalvecharacteristic.

Theauthorityβ’isthesamewithorwithoutthebalancingvalve,anddependsmainlyontheinitialchoiceofcontrolvalve.

Byinstallingabalancingvalve,wecanobtainthecorrectwaterflowunderdesignconditionsandimprove thecontrolvalvecharacteristic.


40

A.3 Sizing of control valves

The Kv factor

Acontrolvalvecreatesacomplementarypressurelossinthecircuittolimitthewaterflowtotherequiredvalue.Thewaterflowdependsonthedifferentialpressureappliedtothevalve:

(F3.10) 1000p

Kvq ⋅ρ

∆ ⋅=

Kvisthevalveflowfactor.

ρ isthedensity.Forwater,ρ =1000kg/m3at4°Candρ=970at80°C.

qistheliquidflowinm3/h.

∆pisthedifferentialpressureexpressedinbars.

ThemaximumKvvalue(Kvs)isobtainedwhenthevalveisfullyopen.Thisvaluecorrespondstothewaterflowexpressedinm3/h,foradifferentialpressureof1bar.

ThecontrolvalveisselectedsothatitsKvsvaluewillgivedesignflowforthedifferentialpressureavailablewhenthevalveisworkingunderdesignconditions.

ItisnoteasytodeterminetheKvsneededforacontrolvalvesincetheavailabledifferentialpressureforthevalvedependsonmanyfactors:

• Theactualpumphead.

• Pressurelossesinpipesandaccessories.

• Pressurelossesinterminalunits.

Thesepressurelossesalsodependontheaccuracywithwhichbalancingisdone.

Duringthedesignoftheplant,thetheoreticallycorrectvaluesforpressurelossesandflowsforvariouscomponentsarecalculated.Butcomponentswithexactlythespecifiedpropertiesarerarelyavailable.Theinstallermustnormallyselectstandardvaluesforpumps,controlvalvesandterminalunits.

Controlvalves,forinstance,areavailablewithKvsvalueswhichincreaseinageometricprogression,called a Reynard series:

Kvs:1.01.62.54.06.31016.....

Eachvalueisabout60%greaterthanthepreviousvalue.

Itisunusualtofindacontrolvalvethatcreatesexactlythedesiredpressurelossfordesignflow.If,forinstance,acontrolvalvethatcreatesapressurelossof10kPaforthedesignflowisneeded,itmayoccurthatthevalvewiththenearesthigherKvsvaluecreatesapressurelossofonly4kPa,whilethevalvewiththenearestlowerKvsvaluecreatesapressurelossof26kPafordesignflow.

41

∆p (bar), q (m3/h) ∆p (kPa), q (l/s) ∆p (mm WG), q (l/h) ∆p (kPa), q (l/h)

pKvq ∆⋅= p36Kv

q ∆= pKv10q ∆= pKv100q ∆=2

Kvq

p

=∆

2

Kvq

36p

=∆

2

Kvq

0.01p

=∆

2

Kvq

0.1p

=∆

pq

Kv∆

=p

q36Kv

∆=

pq

0.1Kv∆

=p

q0.01Kv

∆=

Some formulas involving the flow, Kv and ∆p (ρ = 1000 kg/m3).

Inaddition,pumpsandterminalunitsareoftenoversizedforthesamereason.Thismeansthatcontrolvalvesfrequentlyhavetoworkneartheirclosedposition,resultinginunstablecontrol.Itisalsopossiblethatthesevalvesperiodicallyopentoamaximum,definitelyduringstartup,creatinganoverflowintheirunitandunderflowsinotherunits.Weshouldthereforeaskthequestion:

What to do if the control valve is oversized?

Wehavealreadyseenthatweusuallycan’tfindexactlythecontrolvalvethatwewant.

Takethecaseofa2000wattcoil,designedforatemperaturedropof20K.Itspressurelossis6kPaforthedesignflowof2000x0.86/20=86l/h.Iftheavailabledifferentialpressureis32kPaandpressurelossesinpipesandaccessoriesare4kPa,thedifferenceis32-6-4=22kPawhichmustbeappliedacrossthecontrolvalve.

TherequiredKvsvalueis0.183.

IfthelowestavailableKvsvalueis0.25,forexample,theflowwillbe104l/hinsteadofthedesired86l/h,anincreaseby21%.

Invariable-flowsystems,differentialpressuresappliedtoterminalunitsarevariablesincepressurelossesinpipesdependontheflow.Controlvalvesareselectedfordesignconditions.Atlowloads,themaximumpotentialflowinallunitsisincreasedandthereisnoriskofcreatingunderflowsinsomeunits.Underdesignconditions,andwhenmaximumloadisrequired,it’sessentialtoavoidoverflows.

a- Flow limitation by means of a balancing valve in series

Iftheflowintheopencontrolvalveunderdesignconditionsisgreaterthantherequiredvalue,abalancingvalveinseriescanbeusedtolimitthisflow.Thisdoesnotchangetherealauthorityofthecontrolvalve,andweevenimproveitscharacteristic(seefigurepage41).Thebalancingvalveisalsoadiagnostictoolandashut-offvalve.


42

STAD

∆H

C

V

q

ts

tr

∆pV

A balancing valve limits the flow in the control valve without changing its authority β’

b - Reduction of the maximum valve lift

Tocompensateforoversizingofacontrolvalve,thevalveopeningmaybelimited.ThissolutioncanbeconsideredforEqualpercentagecharacteristicvalves,sincethemaximumKvvaluecanbesignificantlyreducedwithareasonablereductioninthemaximumopening.Ifthedegreeofopeningisreducedby20%,themaximumKvvalueisreducedby50%.

Inpractice,balancingiscarriedoutbymeansofbalancingvalvesinserieswithfullyopencontrolvalves.Thebalancingvalvesarethenadjustedineachcircuitsuccessivelyinordertogiveapressurelossof3kPaatthedesignflow.

Thecontrolvalveliftisthenlimitedtocreate3kPainthebalancingvalve.Sincetheplantisandremainsbalanced,theflowisthereforeactuallyobtainedunderdesignconditions.

c- Flow reduction using a ∆p control valve in series

Thedifferentialpressureacrossthecontrolvalvecanbestabilizedaccordingtofigurebelow.

STAD

∆HC

V

STAP

A ∆p controller keeps the differential pressure across the control valve constant.

ThesetpointofthedifferentialpressurecontrolvalveSTAPischosentoobtainthedesignflowforthecontrolvalvefullyopen.Inthiscase,thecontrolvalveisneveroversizedanditsauthorityiskeptclosetoone.Balancingprocedureisdescribedonpage10.

43

Some rules of thumb

Whentwo-waycontrolvalvesareusedonterminalunits,mostcontrolvalveswillbeclosedoralmostclosedatlowloads.Sincewaterflowsaresmall,pressurelossesinpipesandaccessoriesarenegligible.Theentirepumpheadisappliedacrossthecontrolvalves,whichmustthenbeabletoresistthispressure.Thisincreaseinthedifferentialpressuremakescontroldifficultatsmallflows,sincetherealvalve authorityβ’isreducedgreatly.

Assumethatacontrolvalveisdesignedforapressurelossof4%ofthepumphead.Iftheplantworksatsmallflow,thedifferentialpressureacrossthiscontrolvalveincreasesfrom4toalmost100%.Thedifferentialpressureisthusmultipliedby25.Forthesamevalveopening,allflowsarethenmultipliedby5(√25=5).

Thevalveisforcedtoworknearitsclosedposition.Thismayresultinnoiseandhunting(underthesenewworkingconditions,thevalveisoversizedfivetimes).

Thisiswhysomeauthorsrecommendthattheplantisdesignedsothatthedesignpressurelossinthecontrolvalvesisatleast25%ofthepumphead.Then,atlowloads,flowoversizingofcontrolvalvesdoesnotexceedafactor2.

It’snotalwayspossibletofindcontrolvalvescapableofresistingsuchlargedifferentialpressurewithoutgeneratingnoise.Itisalsodifficulttofindvalvessmallenoughtosatisfytheabovecriterionwhenlowpowerterminalunitsareused.Then,differentialpressurevariationsintheplantshouldbelimited,forexamplebytheuseofsecondarypumps.

Takingthisadditionalconceptintoconsideration,thesizingofatwo-waycontrolvalvemustsatisfythefollowingconditions:

1. Whentheplantoperatesundernormalconditions,theflowinthefullyopencontrolvalvemustbethecalculatedflow.Iftheflowisgreaterthanthis,abalancingvalveinserieswilllimittheflow.Anauthorityof0.30isthenacceptableforaPItypecontroller.Iftheauthorityislower,thecontrolvalveshouldbereplacedforasmallerone.

2. Thepumpheadshouldbesuchthatthepressurelossinthetwo-waycontrolvalvescanbeselectedtobeatleast25%ofthispumphead.

ForOn-Offcontrollers,theauthorityconceptismeaninglesssincethecontrolvalveiseitheropenorclosed.Itscharacteristicisthereforenotveryimportant.Inthiscasetheflowislimited,withnofundamentalrestriction,byabalancingvalveinseries.


44

Appendix B

The authority of three-way control valves

B.1 In mixing function

Athree-wayvalveusedinmixingcansupplyacircuitatconstantflowandvariableinletwatertemperature.

Theprimarywaterattemperaturetpismixedwiththereturnwaterattemperaturetrinthenecessaryproportiontoobtaintherequiredmixtemperaturets.

tp ts

tr qs

E C

DSTAD-2

G

qp

HV

L

STAD-3qb

tr

Three way valve in mixing function.

WhenportEopens,portLclosesinthesameproportion.Thethirdcommonportremainsopen.WhenportEisclosed,thethree-wayvalveisclosedandnoenergycanbeextractedfromtheprimary.Thetemperaturetsisthenequaltotrwhichgraduallyreachesthemeantemperatureoftheroom.

ThebalancingvalveSTAD-2canadjusttheflowtotherequiredvalue.Inprinciple,ahydraulicresistanceequaltothatofGmustbecreatedinthebypassbymeansofSTAD-3inordertogivethesamewaterflowqsregardlessofwhetherthethree-wayvalveisopenorclosed.Inthiscase,thethree-wayvalveisbalanced.

The authority of the three-way valve

Wewillreplacethethree-wayvalvebytwotwo-wayvalvesworkinginopposition.Wewillthenobtainthesamemixingfunction.

ts

qs

VE

STAD-2

G

qp

H

VL

tr

C

D

qb

tr

tp C qs

A three-way valve can be represented by two two-way valves working in opposition.

45

ThevalveVErepresentsthecontrolport.Itspressuredropforthedesignflow=∆pV.Ifthecircuitflowqsisconstant,thepumppressureheadHisconstant,asarepressurelossesinthecircuit.Theresultisthatthepressuredifference∆pDCisconstant.ThispressuredifferenceisappliedtovalveVEwhenitisclosed.Bydefinition,thevalveauthorityisgivenbytheratio∆p(valveopen)to∆p(valveclosed).Thus:

∆pV ∆pV

∆pDC ∆pV+∆pGβ’ = =

Thisauthorityisequalto0.5ormoreif∆pV>or=∆pG.Thismeansthatthepressurelossacrossthethree-wayvalvemustbeatleastequaltothepressurelossinthevariableflowcircuitG,includingthepipes.

Thecircuitbelowgivesaconstantflowintheproductionunitandthethree-wayvalveauthorityiscloseto1.

ts

qs

E

STAD-2

G

qp

HV

L

tr

STAD-1

B

A

C

D

qb

qg

tr

A bypass AB and a primary pump can give a constant production flow and a three-way valve authority close to 1.

Infact,thethree-wayvalvedrawsfromanddischargesintothebypassABwhichactuallyformsavirtualproductionunitwithnopressureloss.Inthiscasetheauthorityofthethree-wayvalveis:

DpV

DpV+∆pDBAEβ’ =

Since∆pDBAEislow,theauthorityofthecontrolvalveiscloseto1.

B.2 In diverting function

Whenusedindivertingfunction,athree-wayvalvecansupplyacircuitatvariableflowandconstantwater supply temperaturewhile keeping theprimaryflowpracticallyconstant.

Diverting valve mounted in a diverting circuit.

L

tr

qs

qbSTAD-3 C

∆H

STAD-1

V

E ts

qp

ts

Balancing of Control Loops > Appendix B. The authority of three-way control valves

46

TheprimaryflowistransferredthroughportEorbypassedthroughportL.Inprinciple,it’sconstant.ThebalancingvalveSTAD-1locatedintheconstant-flowpipe,limitstheflowbycreatingaconstantpressureloss.

Sincethethree-wayvalveinadistributioncircuitisusedtokeeptheprimaryflowconstanttoavoidinteractivitybetweencircuits,itwouldbelogicaltotakewhateveractionisnecessarytoensurethatthispurposeisactuallysatisfied.

ThisisdonebyplacingabalancingvalveSTAD-3inthebypassinordertocreateapressuredropequivalenttothatofCforthesameflow.Inthisway,theprimaryflowisunchangedifportEorportLisfullyopensincethehydraulicresistancesinserieswiththeseportshavethesamevalue.

ThemostimportantbalancingvalveisthisSTAD-1.STAD-3canbeomittedif∆pCislowerthan0.25times∆H.

Note:

Three-wayvalvesareusuallydesignedtoperformamixingfunction:twoinputsandoneoutput.Theiruseindivertingfunctionwithoneinputandtwooutputswillgeneratewatercirculationthroughthevalveinthedirectionoppositetotheplanned.Forsomevalves,thisreversalmayleadtoasignificantincreaseinthenoiselevelandavalvechatteringphenomenon.

tr

qs

qbSTAD-3 C

∆H

STAD-1V

tp

qp

tp

Diverting circuit using a three way mixing valve

Thisiswhyadivertingfunctionwithathree-waymixingvalveisobtainedbyplacingthevalveinthereturncircuitasshowninthefigure.Thisthenprovidesthesamefunction,whilerespectingthewatercirculationdirectionthroughthevalve.

Inbothcasesthevalveauthorityis:

∆pV

∆pV+∆pCβ’ =

Toobtainanauthorityofatleast0.5,thepressurelossinthethree-wayvalvemustbeequaltoorgreaterthanthepressurelossintheterminalunitC.

47

Appendix C

How to set the BPV to ensure the minimum pump flow

InsomecasesaBPVmodulatingreliefvalveisinstalledtoensureaminimumflowtoprotectthepump,asinthefigure.

Ifthisminimumflowisforinstance10%ofdesignflow,thepressurelossinthebalancingvalveSTAD-2isonly1%ofthepressurelossatdesignflow.Normally,thisisafartoolowvaluetoallowprecisemeasurement.Sohowcanwemeasuresuchasmallflowasqsmin?

Thefollowingmethodcanbeused:

a) FindoutwhichhandwheelsettingofSTAD-2thatwillcreate3kPafortheminimumpumpflowqsmin,forinstance10%ofdesignflow.UsetheTA-SCOPE,oraTAnomogram,tofindthecorrectsetting.

b) AdjustSTAD-2tothissettingtemporarily.Closethetwo-waycontrolvalves.

c) OpentheBPVslowlyuntilyouobtaintheminimumpumpflowqsmininSTAD-2.

d) ReopenSTAD-2toitspresetposition.

Whenthecoilcontrolvalvescloseandtheflowqsgoesbelowthespecifiedminimumflowqsmin,theBPVopens.TheBPVthenbypassesaflowqsminforaslongastheflowqsinthecoilcontrolvalvesremainsbelowqsmin.

tp

qb

qp

qs

tr

∆H

STAD-1 STAD-2

BPV

ts = tp

Thismethodisonlyapplicableiftheflowmeasurementdeviceisofthevariableorificetype,astheSTADbalancingvalve.

Balancing of Control Loops > Appendix C. How to set the BPV to ensure the minimum pump flow

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Balancing of Control LoopsHydronicEngineeringHandbookN°1

“Balancingcontrolloops”isthefirsthandbookinTAHydronicsseriesofpublicationsabouthydronicdesignand balancing. The second handbook dealswithbalancingofdistributionsystems,thethirdwithbalancingofradiatorsystemsandthefourthwithhydronicbalancingwithdifferentialpressurecontrollers.

hydronic engineering handbook engineering great solutions · 2020. 9. 22. · primary system (see...

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