4 first law analysis (open systems)

8/13/2019 4 First Law Analysis (Open Systems)

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Chapter#4FirstLawAnalysis(opensystems)

Controlvolumeanalysis

Massbalanceequation

Energybalanceequation

Steadystateflowprocesses

Charginganddischargingofcylinders

ControlVolumeAnalysis Focus on a specified region in space through which material

flows continuously in and out

(e.g. turbines, compressors, pumps, heat exchangers)

Heat and work interaction across control surface

Control volume need not be constant in volume and shapeand may undergo configuration change.

Most often, we deal with control volumes that are of fixedshape and size.

To analyze continuous flow processes


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Massanalysisforopensystems

Let us consider a control volume,

let dmi= mass entering in control volume during time intervaldt

dme= mass leaving in time dt

m(t) = mass inside control volume at time t

m(t + dt) = mass inside control volume at time (t + dt)

Writin law of conservation of mass

m (t + dt) = m(t) + dmidme

m(t+dt) m(t)=dmi dme

Dividebydt,

[m(t+dt) m(t)]/dt =dmi/dt dme /dt

In imit t0;dm/dt =dmi/dt dme /dt


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Ifcontrolvolumehasmanyinletportandexitports,

Rateofmassaccumulation=

Netrateofmassin Netrateofmassout

(Equationofcontinuity)

Particularly,forsteadystateprocess,

dm/dt =0,

m=constant

andtotalmassrate(in)=Totalmassrate(out)

Energyanalysisforopensystems Considerfollowinganopensystem(e.g.steamturbine),


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Let m(t) = total mass inside control volume at time t

m(t +dt) = total mass inside at time (t + dt)

e(t) = u + V2/2 + gz = specific energy of matter inside the

control volume at time t

e(t + dt) = specific energy of matter at time (t + dt)

Piand Pe= pressure at inlet and exit ports

Viand Ve= velocity at inlet and exit ports

viand ve= specific volume at inlet and exit ports

eiand ee= specific energy at inlet and exit ports

Writingmassbalance,

Workdonebysurroundingoncontrolvolume,bycompressing

massAincontrolvolume=

Workdoneonsurroundingoncontrolvolume,releasingmass

Bfromcontrolvolume=

Energyofsystemattimet=

Energyofsystemattime(t+dt)=


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Applyingfirstlawofthermodynamicsfortimeintervaldt,

E2 E1=Q W

Rearranginganddividebydt,

Uselimitdt0 ute=u+V2/2+ z andh=u+Pv

Rearranging,

Inwords,

Rateofenergyaccumulation=Rateofenergyinflow

Rateofenergyoutflow

Note:Aboveequationisusefulforanalysisofunsteadyflow

processes.


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SteadyStateFlowProcess In any steady state flow process,

= constant

and dm/dt = 0, m = constant (no accumulation)

The state of matter at the inlet, exit and at any given point

inside control volume does not change with time.

Therefore, dE/dt = 0, E = constant,

,

First law for steady state open system (SFE),

ApplicationsofsteadystateflowprocessesTurbine

High pressure steam, expanded to a lower pressure with help

of nozzles

Pressure energy converted into kinetic energy

High velocity fluid impacts on set of blades

Delivering of shaft work


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Writing first law for steady flow process,

Ignoring heat losses, small change in kinetic and potential

energy through a turbine,

Compressor

Operatinginreverseofturbine

,

Ignoringheatlosses,kineticandpotentialenergychange,

Here,he>hi


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Nozzle

Primarilyusedtoincreaseflowvelocity

,

Neglectingpotentialenergyforanadiabaticnozzle,

Ifinletvelocityisnegligible(Vi0),

Ve={2(hi he)}

Foridealgas,puthi he=CP(Ti Te)above,

V ={2C (T T )}

Ve={2CPTi(1 Te/Ti)}

Particularlyforadiabaticexpansionthroughnozzle,

PutthisinVeabove,


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Diffuser

Canbethoughtasreversetonozzle

Firstlawanalysisforanadiabaticdiffuser,

HeatExchanger/Condenser

Transferofenergyasheatfromonefluidtoanother


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First law for a heat exchanger can be written as,

As, shaft work = 0, heat losses = 0 and change in

kinetic and potential energy = small

Therefore,

ThrottlingValves

Flowrestricting devices that cause a significant

pressure drop in the fluid Pressure drop in the fluid is often accompanied by a

large drop in temperature (or, sometimes, the

temperature rise)


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Small devices, flow may be assumed to be adiabatic(Q = 0), since neither sufficient time nor largeenough area for any effective heat transfer to take

place

Writing first law for steady state device,

No shaft work (Wsh = 0), small potential energychange and insignificant increase in kinetic energy,So

hi= he (Isenthalpic device)

or ui+ Pivi= ue+ Peve

MixingChambersWritingfirstlawforsteadyflow,

Ignoringheattransfer,shaftwork,potentialandkinetic

energychanges,


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TransientFlowProcesses Unsteadyflowprocesses

Differentmassflowrates(inandout)

Propertiesincontrolvolumevarywithtime

Charginganddischargingofagascylinder

Chargingofacylinder

Considercylinderascontrolvolumeandwritingfirst

lawforunsteadyflowsystems,


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Here,

Ignoringchangeinpotentialandkineticenergy

Equationreducesto,

Where,m=massofgasincontrolvolumeattimet

and u = s ecific internal ener in control volume at

timet.

Integrateaboveeq.fromstartofchargingoperation

(t=0)toendofchargingoperation(t=t),

Writingmassbalanceforcontrolvolume,

Integratingfromtimet=0totimet,

Putvalueinequation(1),


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Dischargingofacylinder

Writing first law for flow process across control

volume

Here,

Ignoringpotentialenergychange

Integratingfromtimet=0totimet,

Writingmassbalancefordischarging,

Integrating,


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Putthisvalueinequation(2),

Example(Dischargingofacylinder):A 750 L rigid tank, shown in figure 3, initially contains water at250C and 50% liquid and 50% vapour by volume. A valve at the

bottom of the tank is opened and liquid is slowly withdrawn.

Heat transfer takes place such that the temperature remains

cons an . n e amoun o ea rans er requ re o e s a e

where half the initial mass is withdrawn.

Solution:

Writing equation for discharging a cylinder with neglecting

,

Usesat.temperaturetableat250C,

vf=0.001251m3/kg,vg=0.0500037m

3/kg


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So,massofliquid=mf=0.375/0.001251=299.76kg

andmassofvapour =mg=0.375/0.050037=7.49kg

Totalinitialinternalenergyintankm0u0=mfuf+mgug=mf(hf Psatvf)+mg(hg Psatvg)

= 299.76 1085.8 3977.6*0.001251 + 7.49 2800.43977.6*0.050037)=343464kJ

Initialtotalmass=299.76+7.49=307.25kg

Finalmassintank=(307.25)/2=153.63kg

Infinalstate(250C),specificvolumevf=0.75/153.63

=0.004882m3/kg

Atfinalstate,usev=x.vg+(1 x)vf0.004882=xfinal(0.050037)+(1 xfinal)0.001251

xf=0.074

Now,ufinal =xfinal.ug+(1 xfinal)uf

=xfinal(hg Psatvg)+(1 xfinal)(hf Psatvf)=0.074(2800.4 3977.6*0.050037)+(1 0.074)(1085.8

3977.6*0.001251)

=193.45kJ/kg (bysteamtable)

=1151.87

Also,he=hf=enthalpyofliquidwithdrawing

=1085.34kJ/kg

Puttingallvalueinenergyequationabove,Q=6450kJ

4 first law analysis (open systems)

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