course el thermo kolo

ME 321 Chapters 1-8, Review of Thermo I1

1-8Review of Thermo I

CHAPTERSCHAPTERS

Thermodynamics is a funny subject. The first time yougo through it, you don’t understand it at all. Thesecond time you go through it, you think youunderstand it, except for one or two points. The thirdtime you go through it, you know you don’t understandit, but by that time you are so used to the subject, itdoesn’t bother you anymore. -Arnold Sommerfield

ME 321 Chapters 1-8, Review of Thermo I

Systems

• boundaries (control surface)• surroundings• closed systems (control mass)

– e.g. balloons, tanks, cylinders– rigid systems– stationary systems

• open systems (control volume)– e.g. nozzles, turbines, heat

exchangers• isolated systems


2Vke ,

21 2

2 == mVKE

gzmgzPE == pe ,

Forms of Energy

• Kinetic Energy– macroscopic motion

• Potential Energy– change in elevation

• Internal Energy– Sum of microscopic forms, U, u

• Total Energy– E=U+KE+PE, e=u+ke+pe

• Energy interactions– Heat (Temperature difference)– Work


• Any characteristic of a system• Not all properties are independent

– e.g.• Specific gravity (ρ/ρH2O)• Specific volume (1/ρ)• Intensive properties- independent of system size

– T, P, ρ, v, u, ke, pe, e, MW, viscosity, conductivity, etc.• Extensive properties

– m, Vol, Mol, E, KE, PE, U, heat capacity, etc.• The continuum hypothesis (no microscopic holes)

mixture saturatedin T and P ,Vm=ρ

Properties


States, equilibrium, process, path,

• state - the set of properties thatcompletely describe thecondition of the system

• equilibrium - system experiencesno changes when isolated fromits surroundings (thermal,mechanical, phase, chemical)

• process - change in a systemfrom an initial to a finalequilibrium state

• path - series of states throughwhich a system passes during theprocess


States, equilibrium, process, path, cycles

• quasi-equilibrium process - allstates in process vary onlyinfinitesimally fromequilibrium states

• cycle - when initial and finalstate are identical


Special processes

• Isothermal– constant temperature

• isobaric– constant pressure

• isometric (isochoric)– constant specific volume


The State Postulate

• Simple compressible system(electrical, magnetic,gravitational, motion, andsurface effects are neglibible)

• The state postulate - for asimple compressible systemthe state is completelyspecified by two independent,intensive properties


Properties of Pure Substances

• Pure substance - fixed chemicalcomposition throughout

• 3 types of phases - solid, liquid, gas- related to molecular interactionsand distances

• Transition between phases– evaporation/condensation– melting/fusion– sublimation


Liquid-vapor phase transition

• Subcooled liquid/compressedliquid

• Saturated liquid (about tovaporize)

• Saturated liquid-vapor mixture• Saturated vapor• Superheated vapor


The P-v-T Solid

• A P-v-T solid shows thesurface of equilibrium states

• This P-v-T solid is forsubstances which expand uponfreezing (water)


The P-v Diagram Showing Phase Boundaries


The T-v Diagram


The P-T Diagram

• The triple line becomes atriple point

• The critical point is at theend of the liquid-vaporline

• This is also called the“phase diagram”


Vapor Pressure (in gas mixtures)

• Pv = φ Psat@T

• Relative humidity ranges from 0 to 1

• The vapor pressure is the partial pressure of the

vapor in the mixture (P = Pdry air + Pv)

(If no air is present Pv is simply Psat)

• If φ is 1.0 the air is saturated


Vapor Pressure (in gas mixtures)

• Concentration differences in the gas will be thedriving force for mass transfer

• At surface φ is 1.0 and vapor mass is transferredaway from the surface

• Condensation and evaporation are the results• Evaporation is different than boiling• Saturation phenomena also exist in liquids


The Saturation Tables (A-4 and A-5)

• A-4 is indexed by temperature,A-5 is indexed by pressure

• Notice only one equilibriumpressure for a saturated liquid-vapor mix at a giventemperature

• The saturated liquid volume iswhen there is only liquid atsaturated conditions

• The saturated vapor volume iswhen only vapor is present


Graphical Representation of Specific Volume

• By looking at the P-v diagramwe can see how the quality isused to compute the specificvolume

• A similar procedure is usedfor T-v diagram

• Note that graphically thevolume is not linear on the x-axis


• The specific volume of the mixture is the weightedaverage of the specific volume of saturated liquidand saturated vapor

• mtotv = mfvf + mgvg or,

• v = (mtot-mg)vf/mtot + mgvg/mtot = (1-x)vf + xvg

• Therefore v = vf + x(vg - vf)

• Likewise, u = uf + xufg, and

• h = hf + xhfg

Saturation tables (continued)


Superheated Vapors

• For superheated vaporsthe properties areindexed by bothtemperature andpressure

• Double-interpolationmight be needed to getmore accurate values


Compressed Liquid Tables

• Use compressed liquid tables only for reallyhigh pressures and when accuracy is needed

• Generally evaluate properties as the saturatedliquid property at the given temperature

• For enthalpy may need to compute as follows:h = hf@T + vf(P - Psat)


• An equation is better than a table• For gases experiments yielded the following:

– at fixed T: P2/P1 = v1/v2

– (or P1v1=P2v2=constant) (Boyle’s law, 1662)– at fixed P: T2/T1=v2/v1– (or T1/v1=T2/v2= const.) (Charles - Gay-Lussac law, 1802)

• Combining these two ideas gives the Ideal gas law:Pv=RT

• R is an ideal gas constant which depends upon the gas• It turns out that R = Ru/M where Ru is the universal gas

constant

Ideal Gas equation of state


Pv R Tu=also: P1v1/T1 = P2v2/T2

Ideal Gas equation of state, other forms

• Since the number of moles, n = m/M, and thenumber of molecules, N, per mole is Avogadro’snumber, NA ( n = N/NA):– PV = mRT– P = ρRT

– PV = nRuT– PV = (N/NA)RuT– P = cRuT (c is concentration)– PV = NkBT (kB is Boltzmann’s const. = Ru/NA)


The Universal Ideal Gas Constant

• 8.314 kJ/(kmol.K)• 8.314 kPa.m3/(kmol.K)• 0.08314 bar.m3/(kmol.K)• 1.986 Btu/(lbmol.oR)• 10.73 psia.ft3/(lbmol.oR)• 1545 ft.lbf/(lbmol.oR)• 1.987 cal/(mol.K)• 0.08206 liter.atm/(mol.K)• 0.7302 ft3.atm/(lbmol.oR)


Notes on Ideal Gas equation of state

• Ideal gas is an imaginary substance• Most applicable to low density gases• H2, He, Ar, Ne, Air, O2, N2, CO2, CO, etc. can

generally be treated as an ideal gas• Water vapor and refrigerants typically not treated

as an ideal gas• How can I know if treating a gas as an ideal gas is

the right thing to do?


The Compressibility Factor, Z

• Z is defined by: Z = Pv/RT• If Z = 1, or close to 1, treat gas as an “ideal gas”• The behavior of gases can be generalized from

knowing their critical states (Fig. 2-57)• PR, the reduced pressure is P/Pc

• TR, the reduced temperature is T/Tc

• vR, the pseudo-reduced specific volume is vactPc/RTc


Regimes for using Ideal Gas Law

• This T-v diagram showsthe region where error inusing the ideal gas law isless than 1%

• Error is highest near thecritical point

• This corresponds to TR = 1 and PR = 1 on the

Z vs. PR diagram


...)()()()(5432 +++++=

vTd

vTc

vTb

vTa

vRTP

Other Equations of State

• van der Waals

• Virial

( )Pa

vv b RT+

− =2


Boundary Work

• Boundary work can becomputed by integrating

• This result comes from thedefinition of work applied to asimple compressible system

W P d VbV

V

= ∫1

2

W Fds P A ds Pd A s PdVV

V

= = ⋅ = ⋅ =∫ ∫∫ ∫( ) ( )1

2


Work for Constant Pressure Processes

• In terms of specific work

( )W P d V P d V P V VbV

V

V

V

= = = −∫ ∫1

2

2 11

2

( )w Pdv P dv P v v Pv Pvbv

v

v

v

= = = − = −∫ ∫1

2

2 1 2 11

2


Work for Ideal Gas Processes

( ) ( )w P v v Pv Pv R T Tb = − = − = −2 1 2 1 2 1

( ) ( )W P V V PV PV mR T Tb = − = − = −2 1 2 1 2 1

• Constant Pressure

• Constant Temperature

• Constant volume

w PdvRTv

dv RTdvv

RT v v RTvvb

v

v

v

v

v

v

= = = = − =

∫∫∫ (ln ln ) ln2 1

2

11

2

1

2

1

2

wb = 0


Work for polytropic processes

• A polytropic process is one given by

• Work for a polytropic process is given byPV Cn = (C is a constant)

w Pdv Cv dv Cv v

nbn

n n

v

v

v

v

= = =−

− +−

− + − +∫∫ 2

11

1

2

1

1

2

1

( ) ( )w

P v v P v vn

P v P vnb

n n n n=

−− +

=−−

− + − +2 2 2

11 1 1

12 2 1 1

1 1


Ideal gas and polytropic processes

• Pvn=C and Pv = RT• n = 0 Constant pressure process• n = 1 Constant temperature process• n = Constant volume process• n = k = cp/cv = isentropic process

∞


Energy Change for a Cycle

• In a cycle the beginningand end states are thesame.

• Therefore ∆E is zero• The net heat must equal the

net work


• The First Law ofThermodynamics takes onseveral forms

• Care must be taken to ensureproper application and use ofsigns

• This is most easily learnedthrough doing examples

Closed-Systems, First-LawClosed-Systems, First-Law


Formal Definitions of Cv and Cp

• Specific heat is the amount ofenergy it takes to raise asubstances temperature one degree

• If done as a constant volumeprocess:

• If done as a constant pressureprocess:

• These are properties and do notexist whether or not the actualprocess is constant volume orconstant pressure

cuTv

v=

∂∂

chTp

p=

∂∂


Specific Heats for Some Gases

• Inert gases have constant cpvalues

• k (=cp/cv) is also constant ataround 5/3 for these gases

• k for many diatomic gases isaround 7/5

• These relate to degrees offreedom of the molecules


Three Ways to Calculate ∆∆∆∆u

• Table values are the simplestmethod: ∆u = u2 - u1, but tablesare not available for all gases

• If cv is known in functional form,integration over the temp. rangegives ∆u.

• A good approximation can usuallybe obtained by picking an averagevalue of cv or the cv at an averagetemperature.


Helpful Cp and Cv relations Applications

• For ideal gases Cv, and Cp are related by:

• The specific heat ratio k is defined as:

• For incompressible substances (liquids andsolids), both the constant-pressure and constant-volume specific heats are identical and denotedby C:


• The volumetric flow rate dividedby the specific volume gives themass flow rate

• Same as density times volumetricflow rate

• For steady flow, no massaccumulates in the control volumeand the inlet mass flow rate equalsthe exit mass flow rate

• Continuity (Conservation of mass)

© The McGraw-Hill Companies, Inc.,1998

Mass flow rate and Steady flowMass flow rate and Steady flow

eeaveiiavi

eeave

iiavi

ei

AVAV

AVv

AVv

mm

ρρ =

=

=11!!


Flow WorkFlow Work

• The 1st law reduces to:

• We rearrange the mass-related transfer terms tothe R.H.S. since they are measurable “properties”and the enthalpy property becomes useful

( ) ( ) ( ) pekehhpekevPvPuuwq ieiieeieoutin ∆+∆+−=∆+∆+−+−=−∑

•For steady for dEcv/dt = 0•For single stream there isone inlet and one outlet

mmm ei !!! ==

( ) ( ) 0=+++−++++−∑ eeeeeiiiiioutin pekeuvPmpekeuvPmWQ !!!!


Energy balance for Steady Flow SystemsEnergy balance for Steady Flow Systems

• On a rate basis with steady mass flow rates theenergy equation becomes

∑∑∑

++−

++=−

ii

iavii

ee

eaveeoutin gz

Vhmgz

VhmWQ

22

22

!!!!

. ..

.

.


Steady-Flow Devices OperateSteadily for Long Periods

Steady-Flow Devices OperateSteadily for Long Periods


++−

++=−∑ i

iaviie

eaveeoutin gz

Vhmgz

VhmWQ

22

22

!!!!

Nozzle and Diffuser Shapes CauseLarge Changes in Fluid VelocitiesNozzle and Diffuser Shapes CauseLarge Changes in Fluid Velocities

• Nozzles increase fluid velocityat the expense of pressure

• Diffusers increase fluid pressureby slowing it

Nozzles and Diffusers are shaped sothat they cause large changes in fluid

velocities and thus kinetic energies

0

0

0

Fluid spends littletime in C.V. so

no heat exchange

No work is doneby a nozzle or

diffuser

0

Small device so neglectpotential energy changes

22 ,

22

2222iaveav

ieiav

ieav

eVV

hhV

hV

h −=−+=+

For single inlet single outlet theconstant mass flow rate cancels

0 0


++−

++=− i

iaviie

eaveeoutin gz

Vhmgz

VhmWQ

22

22

!!!!

Turbines and compressors convertbetween thermal energy and workTurbines and compressors convertbetween thermal energy and work

• Turbines output work from adecrease in fluid enthalpy

• Compressors and pumpsincrease fluid enthalpy by doingwork on the fluid

0

0

0

No heat exchange ifsufficiently insulated

Kinetic energy change can often(but not always) be ignored

0


0

( )

−+−=

22

22

,ei

eioutturbVVhhmW !! ( )

−+−=

22

22

,ie

ieincompVVhhmW !!

turbW!out

compW!in


Throttling Valve Devices Cause LargePressure Drops in Fluid

Throttling Valve Devices Cause LargePressure Drops in Fluid

• Throttling processes areregarded as constant enthalpyprocesses

The temperature of an ideal gasdoes not change during a throttling(h =constant) process since h = h (T)

++−

++=− i

iaviie

eaveeoutin gz

Vhmgz

VhmWQ

22

22

!!!!

0

No time forsignificant heat

transfer

0

0

For incompressible flows density,area and mass flow rate are

constant, hence no k.e. change

0


00

No work is doneby or on

throttling device


( )

( )

e

e

e

e

hhhhy

hyhymm

hmmhmhm

−−=

+=+=

+=+

1

2

212

1

212211

1yh ,

:rates flow massinlet of ratio thefind to

!

!

!!!!

T-Elbow Serves as Mixing Chamber forHot and Cold Water Steams

T-Elbow Serves as Mixing Chamber forHot and Cold Water Steams

• Mixing devices have morethan one inlet mass flow rate

• Work, Heat transfer, ke and peare usually neglected

The T-ebow of anordinary shower

serves as the mixingchamber

for hot- and cold-water streams.{ } eehmhmhmhm !!!! =++ .... 332211


Heat Transfer Via Heat ExchangerDepends on System Selection

Heat Transfer Via Heat ExchangerDepends on System Selection

The heat transfer associated with a heat exchanger may be zero or nonzerodepending on how the system is selected


Steady Flow: Putting it into a system

Condenser

Boiler

High-PTurbine

Pump 1

2

Low-PTurbine

3

54

T

v

1

2

3

4

qout

wturb,outqin

wpump, in

5

P2,3

P5,1

( )23,

1212,

15,

53,

hhqPPvhhw

hhqhhw

inboiler

inpump

outcond

outturb

−=

−=−=

−=−= •A steam cycle is a good example

• This is the Rankine cycle


Steam Power Plants, a Common Heat EngineSteam Power Plants, a Common Heat Engine

• Boiler and condenserare two-phasethermal energyreservoirs

• Some heat must berejected to condenser

• Net work is given bywturb-wpump


Thermal EfficiencyThermal Efficiency

• Even the Most Efficient HeatEngines Reject Most Heat asWaste Heat

• Performance measures are givenby:

• Thermal efficiency is then

For this example

input Requiredoutput DesiredP =erformance

in

out

in

outnetth Q

QQ

W−== 1,η

%4040.0 ==thη


The Kelvin-Plank StatementThe Kelvin-Plank StatementA heat-engine cycle cannot be completed without rejecting

some heat to a low-temperature sink

The Kelvin-Planck Statement of the 2nd LawIt is impossible for any device that operates on acycle to receive heat from a single reservoir andproduce a net amount of work


Basic Components of a RefrigerationSystem in Typical Conditions

Basic Components of a RefrigerationSystem in Typical Conditions


• Refrigerators are designed toremove heat from a cooler spaceand push it into a warmer one

• Work must be done to transfer heatto a warmer reservoir

• The “efficiency” for these kind ofdevices is called a Coefficient ofPerformance (COP)

• In this case the low temperaturereservoir is the source and the hightemperature reservoir is the sink

Refrigerator’s Objective: Remove QL from the Cooled Space

11

, −==

LHinnet

LR QQW

QCOP


• Heat pump are used to heatbuildings by removing heatfrom the cold outdoors

• Work must be done to do this• This work typically requires

less energy than does resistiveheating

• And thus

Heat Pump’s Objective: Supply Heat Q H into the Warmer Space

Heat Pump’s Objective: Supply Heat Q H into the Warmer Space

HLinnet

HHP QQW

QCOP−

==1

1

,

1+= RHP COPCOP


A Refrigerator That Violates ClausiusStatement of the Second Law

• Not impossible to transfer heatfrom cold to hot objects, it justrequires work

The Clausius Statement of the 2nd LawIt is impossible to construct a device thatoperates in a cycle and produces no effect otherthan the transfer of heat from a lower-temperature body to a higher-temperature body.


Execution of the Carnot Cyclein a Closed System

Execution of the Carnot Cyclein a Closed System

reversible isothermal heat addition

reversible adiabatic expansion

reversible isothermal heat rejection

reversible adiabatic compression


P-v Diagram of the Carnot CycleP-v Diagram of the Carnot Cycle


Development of Clausius inequality• The first law gives: CRC dEQW −= δδ• Since the cyclic device is totally reversible the

following ratio rule holds

TQTQ

TQ

TQ

TT

QQ

RRR

RRR δδδδδ

δ=== or ,or ,

• The combined system first law then is:

CRC dETQTW −= δδ

• For a complete cycle of the combined system

∫∫ == 0dE since , cTQTW RC

δ

• Observing the combined system we see thatWC cannot be positive or Kelvin-Planck isviolated and TR must be positive so

∫ ⇒≤ Inequality Clausius the 0TQδ


Entropy changeEntropy change

The entropy change betweentwo specific states is the samewhether the process is reversibleor irreversible

∫

=∆=−

2

1

rev

TQSSS δ

12

• Integrating the definition ofentropy gives:

• This integration must be done overa reversible path, even if the actualpath is irreversible

• Note that the change in entropy isdefined. A reference for entropymust be established. (3rd law)

TQdS δ≥

( ) 021

2

1

1

2

2

1

≤−+=+ ∫∫∫ SSTQ

TQ

TQ δδδ

• For 1-2 irreversible, 2-1 int. rev:


The entropy change of an isolatedsystem is the sum of the entropychanges of its components, and is

never less than zero

The Entropy Change of an Isolated System

• The inequality can be modeled as an equalityby adding an entropy generation term:

0 ,0

2

112

≥∆

=∆=∆+∆≥∆

+=−=∆ ∫

isolated

gentotalsurrsysgen

gensys

SSSSSS

STQSSS δ

• Note that unlike S, Sgen is not aproperty, but totally dependentupon the path between the twostates

∫2

1

TQδ is treated as entropy transfer

due to heat transfer acrossthe boundary, some physicstexts consider it -∆Ssurr


The T-s diagramThe T-s diagram

• The T-s diagram is a usefuldiagram.

• For internally reversibleprocesses, integrating under ityields qrev

• For non-quasi-equilibriumprocesses we cannot integratedirectly over the process path

• We will deal with non-quasi-equilibrium process in somecases with isentropicefficiencies

( )kgkJ

KkgkJq

ssTqTqs

rev

revrev

7705-6.7K453

)( , 12

2

1

=⋅

=

−==∆ ∫δ

1 2


Heat Transfer for Internally Reversible ProcessesHeat Transfer for Internally Reversible Processes

• Remember, the area underthe curve on a T-s diagramis equal to the heat transferonly for internallyreversible processes

• For a complete cycle ofinternally reversibleprocesses the enclosed areais the net heat transfer (andby 1st law argument, alsothe net work)

• Counter clockwise ispositive heat in or work out

d

T

S

Net heat (Qin-Qout)

Heat in (Qin)Heat out (Qout)


System Entropy Constant During Reversible,adiabatic (isentropic) Process

System Entropy Constant During Reversible,adiabatic (isentropic) Process

• Since there is no heat transfer theentropy transferred by heat transferfrom the boundary is zero.

• Since there are no irreversibilities,the Sgen term is also zero

• Isentropic technically meansconstant entropy, but is taken toimply an adiabatic, internallyreversible process in engineeringapplications

∫ +=−=∆2

112 gensys S

TQSSS δ

T

s

1

2

An isentropicprocess


Development of TdS equationsDevelopment of TdS equations• Consider a reversible piston-

cylinder device• An incremental amount of heat, δQ

is added, while an incrementalamount of boundary work, δW isdone by the piston

δQ

δW

or since

or , ,int

intint

VdP,dHTdSVdPPdVdUdH

PdVdUTdSPdVWTdSQ

dUWQ

rev intrev

revrev

−=++=

+===

=−δδ

δδ

• The results deal with properties only and thus are valid asdifferential equations for both reversible and irreversible processes

PdvduTds +=

vdPdhTds −=


Use of TdS equationsUse of TdS equations

• These two equations are known as the first and secondTds or Gibbs equations

• They can be rearranged to give differential changes inentropy

• These will be useful in determining relationships when we know arelationship between du or dh and T, or for ideal gases

PdvduTds += vdPdhTds −=

TPdv

Tduds +=

TvdP

Tdhds −=

∫∫ +=∆2

1

2

1 TPdv

Tdus


Incompressible liquids and solidsIncompressible liquids and solids

• For incompressible liquids and solids the specificvolume is constant, hence dv=0

TPdv

Tduds +=

∫=∆

=

∂∂=

2

1

v

that so

,Tu definitionby since

TdTcs

dTcduc

v

vv

• For many cases involving small temperaturedifferences cv = constant = c or cavg

• In these cases:∫

==−

2

1 1

212 ln

TTc

TdTcss avgavg


Ideal gasesIdeal gases

• For ideal gasesT

PdvTduds +=

TvdP

Tdhds −=

1

22

112 ln)(

vvR

TdTTcss v +=− ∫

1

22

112 ln)(

PPR

TdTTcss P −=− ∫

vdvR

TdTTcds v += )(

PdPR

TdTTcds P −= )(

• Integration gives

PR

Tv

vR

TP

dTTcTdhdhdTTcTdudu

p

v

==

====

)()()()(


Case 1: Ideal gas w/ constant specific heatsCase 1: Ideal gas w/ constant specific heats

• If Cv and CP are constant over a temperature range (or anaverage value is assigned for an approximation) theintegration is done easily

1

22

112 ln)(

vvR

TdTTcss v +=− ∫

1

22

112 ln)(

PPR


• These can also be stated on molar (rather than mass) basis1

2

1

2,12 lnln

vvR

TTcss avv +=−

1

2

1

2,12 lnln

PPR

TTcss avp −=−

1

2

1

2,12 lnln

vvR

TTcss uavv +=−

1

2

1

2,12 lnln

PPR

TTcss uavp −=−


Case 2: Ideal gas w/ variable specific heatsCase 2: Ideal gas w/ variable specific heats

• Using the equation involving cP we develop an ideal gasfunction that accounts for the first term on the R.H.S.

1

22

112 ln)(

PPR


• This function is not the entropy of the gas• It is used to easily tabulate the change of entropy of a gas

due to the change in temperature• Entropy is a function of two independent variables even

for an ideal gas• The resulting equation is:

1

21212 ln)(

PPRssss oo −−=−

∫=T

Po

TdTTcs

0

)(!


Case 2: Ideal gas w/ variable specific heatsCase 2: Ideal gas w/ variable specific heats

• In a constant pressure process, the change in so equates tothe change in entropy

• Note that, except forair, these values aretabulated on a molarbasis

where

1

21212 ln)(


1

21212 ln)(

PPRssss u

oo −−=−

∫=T

Po

TdTTcs

0

)(!


Isentropic Ideal Gas Relationships -- Case 1: Constant Specific Heats


• For an isentropic process s2 = s1

• Given this relationship we can come up with some shortcut formulasfor isentropic processes of ideal gases with constant specific heats

1

2

1

2,12 lnln

vvR

TTcss avv +=−

1

2

1

2 lnln0vvR

TTcv +=

• Remember that for ideal gases cv and cp are functions of temperatureonly, regardless of pressure or specific volume

( )dTdu

Tuc

dTdh

ThcRTTupvTuTh

vv

pp =

∂∂==

∂∂=+=+= , ,)()(

RccorRdTdTcdTc vpvp +=+= ,




• Introducing the specific heat ratio, k = cp/cv

1

2

1

2

1

2

1

2

1

2 lnlnlnlnln0vv

ccc

vv

cR

TT

vvR

TTc

v

vp

vv

−−=−=⇒+=

( )( )1

1

2

1

2

1

2

1

2 ln1ln−−

=⇒−−=

k

vv

TT

vvk

TT

1

2

1

1

2

−

=

k

vv

TT

• Using the const. specific heat relation based on the 2nd Tds equation:

1

2

1

2

1

2

1

2

1

2 lnlnlnlnln0PP

ccc

PP

cR

TT

PPR

TTc

p

vp

pp

−==⇒−=

⇒

−=

1

2

1

2 ln1lnPP

kk

TT k

k

PP

TT

1

1

2

1

2

−

=

And equatingthese two gives:

k

vv

PP

=

2

1

1

2


The Isentropic Relations of Ideal GasesThe Isentropic Relations of Ideal Gases

• The isentropic relations of ideal gasesare valid for the isentropic processes ofideal gases only


Isentropic Ideal Gas Relationships -- Case 2: Variable Specific Heats, Pressure ratio

Isentropic Ideal Gas Relationships -- Case 2: Variable Specific Heats, Pressure ratio

• For an isentropicprocess s2 = s1

1

21212 ln)(


( )( )Rs

RsePP

Rss

PP

PPRss

o

oR

ss

oooo

oo

1

2

1

2

12

1

2

1

212

expexp

ln ,ln)(0

12

==

−=−−=

−

• The quantity exp(so/R) is called the pressure ratio, Pr

• Do not confuse it with the Reduced Pressure PR {=P/Pc}• This quantity is a function of temperature only and is

tabulated• It is only valid for isentropic processes

1

2

1

2

r

r

PP

PP =


Isentropic Ideal Gas Relationships -- Case 2: Variable Specific Heats, Volume Ratio

Isentropic Ideal Gas Relationships -- Case 2: Variable Specific Heats, Volume Ratio

121

2 12

1

2

21

12

1

2 ,rrr

r

PT

PT

vv

vTvT

PP

PP ===

• The quantity T/Pr is called the volume ratio, vr

• This quantity is a function of temperature only and istabulated

• It is only valid for isentropic processes

• Since volume and pressure are linked bytemperature in ideal gases a relationshipalso exists for volume

1

2

1

2 r

r

vv

vv =


P-v Diagram for steady flow reversible processesP-v Diagram for steady flow reversible processes

• Notice also that fornegligible changes in keand pe the reversiblesteady flow work is thearea to the right of thecurve on the P-v diagram

wrev

∫−=2

1

vdPwrev


P-v Diagrams of Isentropic, Polytropic, andIsothermal Compression Processes

P-v Diagrams of Isentropic, Polytropic, andIsothermal Compression Processes

• By looking at the P-v diagramsfor isentropic, polytropic andisothermal compressionprocesses, it is seen that theisentropic process does notrequire the least work

• Consequently, whencompressors are discusses wewill have a different measurefor efficiency than for turbines


The h-s (Mollier) Diagram for WaterThe h-s (Mollier) Diagram for Water

• Valuable for analyzingturbine performance

• Critical point is not atthe vertical peak

• Notice that lines ofconstant temperaturebecome horizontal asfluid behaves more likean ideal gas


Isentropic Turbine EfficiencyIsentropic Turbine Efficiency

• Actual turbines do less workthan isentropic turbinesoperating between the sametwo pressures

• The ratio of the actual workto the isentropic work is theisentropic turbine efficiency

• The isentropic efficiency iseasily seen on the MollierDiagram

sisentropic

actT hh

hhw

w

21

21

−−==η


Isentropic Nozzle EfficiencyIsentropic Nozzle Efficiency

• A similar efficiency isdefined for nozzles operatingbetween to pressures

• The ratio of the actual outletkinetic energy to that of anisentropic nozzle is theisentropic nozzle efficiency

• This efficiency is also easilyseen on the Mollier Diagram

ssT hh

hhVV

21

212

2

22

−−==η

Where the last equality hold ifinlet velocity is small


Isentropic Compressor EfficiencyIsentropic Compressor Efficiency

• Actual compressors requiremore work input thanisentropic turbines operatingbetween the same twopressures

• The ratio of the isentropicwork requirement to theactual work the isentropiccompressor efficiency

• This efficiency is also easilyseen on the Mollier Diagram

12

12

hhhh

ww s

actual

isentropicCs −

−==η


Isothermal Compressor EfficiencyIsothermal Compressor Efficiency

• A reduced work input canactually be required forisothermal compressors

• The ratio of the isothermalwork requirement to theactual work the isothermalcompressor efficiency

• This efficiency can not beclearly shown on the MollierDiagram

( )12

12

12

lnhh

PPRThh

qw

w isoth

actual

isothermalCt −

=−

==η


Entropy Generation During Heat TransferEntropy Generation During Heat Transfer

Graphical representation of entropy generation during a heattransfer process through a finite temperature difference


• Work depends upon the final state, the initial stateand the process path

• Consider the case where– The initial state is given– The process is reversible– The final state is in equilibrium with the surroundings

• This case results in the maximum possible workoutput from the initial state

• The state at which the system is in equilibriumwith the surroundings is the “dead” state


Maximum Work Available and the Dead StateMaximum Work Available and the Dead State


• The dead state provides a convenient referencefrom which to analyze the usefulness of a quantityof energy

• (Thermo-mechanical) dead state is assumed to be:➩ z = 0 (sea level)➩ V = 0 (at rest with respect to Earth’s surface)➩ P = P0 = 101.325 kPa = 1 atm➩ T = T0 = 25oC = 77oF = 298 K➩ u0 = u(T0, P0), h0 = h(T0, P0), s0 = s(T0, P0)

• Chemical availability - models standard sea levelatmosphere constituents (O2, CO2, N2, H2O) atnormal concentrations


- The Dead State -A State of Equilibrium with the Surroundings

- The Dead State -A State of Equilibrium with the Surroundings


• Exergy representsmaximum useful work asystem will deliver whilegoing from a given state tothe dead state

• Surroundings work– Not all boundary work is

useful– Work loss in pushing the

environment• Other work forms are

considered 100% useful© The McGraw-Hill Companies, Inc.,1998

Useful Work and Surrounding WorkUseful Work and Surrounding Work

P0

V2-V1

( )120 VVPWsurr −=

surru WWW −=


• “Reversible work” is the maximumuseful work between two states

• “Exergy” represents maximumuseful work a system will deliverwhile going from a given state tothe dead state

• If an object delivers less than themaximum work the remainingwork potential is exergy destroyed,or irreversibility, I

• Exergy has same units as work andenergy

• Exergy is always positive, even forstates “below” the dead state


Reversible work, Exergy, IrreversibilityReversible work, Exergy, Irreversibility

For systems with a movingboundary:Wu,out=Wact,total - Wsurr - IIf final state is dead state, thenX = Wact,rev - Wsurr= Wu,act + I


Derivation of Exergy Definition

• Consider a combined system ofclosed system and itsenvironment

• Only work leaves the combinedsystem

• First TdS eq. for combinedsystem gives

WC

WQ

ue, ve, se, Te(=T0) , Pe(=P0) fixed

Ue, Ve, Se vary due tointeractions with closed system,but are governed by 1st TdS eq.

eee VPSTU ∆−∆=∆ 00( )VVV

VVV

e

cmeC

−−=∆∆+∆==∆

0

0

• The final system state is thedead state “0”

• WC is then given by( )[ ] ( ) ( )[ ]eeeCC VPSTEUUEUEW ∆−∆+−−=∆+−−=∆−= 0000


Derivation of Exergy Definition

• Continuing with the equation for combined system work:

WC

WQ

• The entropy change for the combined system equals theentropy generation

( ) ( )[ ]eC STVVPUEW ∆−−+−= 0000

( ) egenC SSSSS ∆+−==∆ 0

( ) ( ) ( )[ ] genC STSSTVVPUEW 000000 −−−−+−=

• If process is reversible the Sgen is zeroand work is maximized. It is thismaximum work that we call exergy

( ) ( ) ( )00000 SSTVVPUEX −−−+−=


Types of ExergyTypes of Exergy• Exergy is a property that depends upon two

independent intensive properties and thedefinition of the dead state

• Specific exergy may be better explained bycombining exergy due to “component energies”

( ) ( ) ( )( )

( ) ( )000

00

00000

e

2

ke

x :enthalpy ofExergy

x :energy flow ofExergy x :energy internal ofExergy

x :energy potential ofExergy 2

x :energy kinetic ofExergy

ssThh

vPPvPPvssTvvPuu

gzpe

Vke

h

pv

u

p

−−−=

−=−=−−−+−=

==

=="


Kinetic and Potential ExergyKinetic and Potential Exergy• Mechanical forms of energy are equivalent to

exergy since work is 100% recoverable

221 Vm

"

mgz

Wshaft

2x

2

keVke"

==gzpep ==ex


The Exergy of Internal EnergyThe Exergy of Internal Energy

• Consider a stationary closedsystem reversibly transferringwork and heat, ending at thedead state

• Heat transfer energy is capturedby a reversible heat engine

The exergy of a specified mass at aspecified state is the useful workthat can be produced as itundergoes a reversible process tothe state of the environment

( ) QdVPWQWdU ub δδδδ −+−=−−= 0,

dSTQWTQQ

TTW 0HEHE +==

−= δδδδδ ,dS ,1 0

dSTdVPdUWW 0ub,HE +−−=+ 0δδ

( ) ( ) ( )00000, ssTvvPuuxw uutot −−−+−==


The Exergy of Flow WorkThe Exergy of Flow Work

• Flow work is the workdone pushing fluid intoand out of an opensystem

• Not all flow work isuseful. Some is pushingagainst at atmosphere

The exergy of flow of work isthe useful work that would bedelivered by an imaginarypiston in the flow section

( )vPPvPPvxPv 00 −=−=


The Exergy of EnthalpyThe Exergy of Enthalpy

• The exergy associated with a fluid’s enthalpy is thecombination of the exergies associated withinternal energy and flow work

• For a point in a flow system, enthalpy combinesinternal energy with flow work

( ) ( ) ( ) ( )vPPssTvvPuuxxx Pvuh 000000 −+−−−+−=+=

( ) ( )000 ssThhxh −−−=

Pvuh +=


The Energy and Exergy contents of (a) a Fixed Mass and (b) a Fluid System

The Energy and Exergy contents of (a) a Fixed Mass and (b) a Fluid System

• Just as energy terms can becombined into a singleproperty, e, so can exergy

• For flow systems weinclude flow work exergy

(θ is called “methalpy”)• These shorten future

equations by using φ or ψ,which are symbols for theintensive property exergyat a state or a point in theflow

gzVue ++=2

2"

( ) ( ) ( ) gzVssTvvPuu ++−−−+−=2

2

00000

"

φ

gzVh ++=2

2"

θ

( ) ( ) gzVssThh ++−−−=2

2

000

"

ψ


Exergy TransfersExergy Transfers

• In order to get a exergy balance equationwe must consider exergy transfer from thesystem to the surroundings

• For exergy transfer by heat, assume heat istransferred into reversible heat engine

∫

−= Q

TTX heat δ01

• If object’s temperature remains constant:

QTTX heat

−= 01

• Exergy transfers by work and mass follow simply

−

=forms)k (other wor work)(boundary

WWW

X surrwork ψmX = trans.mass


The Exergy of a Cold MediumThe Exergy of a Cold Medium• Exergy of a cold medium is

still positive even though it isbelow T0

• Consider ideal gas at constantP0 but below T0 (Baggie ofcold air).

The exergy of a coldmedium is also a positivequantity since work canbe produced bytransferring heat to it

( ) ( )

−−+−=

0000 ln

TTcTTTRTTcx pv

( )

−−=

000 ln

TTTTTcx p

<>

−=

0

00T Tfor negativeT Tfor positive

1TTc

dTdx

p

( ) ( ) ( )00000 ssTvvPuuxu −−−+−=


The Transfer and Destruction of ExergyDuring Heat Transfer

The Transfer and Destruction of ExergyDuring Heat Transfer

• The same amount of heatleaves a boundary as enters it

• More entropy leaves aboundary than enters duringheat transfer across a finitetemp. difference

• Exergy is destroyed duringheat transfer

The transfer and destruction of exergyduring a heat transfer process througha finite temperature difference


The Exergy BalanceThe Exergy Balance

General:

• Exergy balance for any system undergoing any processcan be expressed as

General, rate form:


Exergy TransferenceExergy Transference

• Exergy is transferred into or out of acontrol volume by mass as well as byheat and work transfer

• The explicit statement for the energybalance equation for control volumesis given by:


Exergy TransferenceExergy Transference• Exergy transfer for a closed system can be determined as well.

• The explicit statement for the energy balance equation forclosed systems is given by:

( )[ ] ( )1012001 φφ −=∆=−−−−

−∑ 2gen

kk

kXSTVVPWQ

TT

dtdX

STdt

dVPWQ

TT sys

gensys

kk

k=−

−−

−∑ !!!

0001

• Note thatdestroyedexergy ilityirreversib0 === IST gen


Second Law of EfficiencySecond Law of Efficiency

The second law of efficiency is a measure of the performance of a devicerelative to its performance under reversible conditions

( )

−=

=

=

all suppliedExergy destroyedExergy 1

suppliedExergy recoveredExergy

pumps)heat and tors(refrigera devices) consuming(work engines)(heat WW revu

rev

urev

revth,th

II COPCOPWW

ηη

η


The Second-Law Efficiency of All ReversibleDevices is 100%

The Second-Law Efficiency of All ReversibleDevices is 100%

course el thermo kolo

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