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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
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I
• 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
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I
States, equilibrium, process, path, cycles
• quasi-equilibrium process - allstates in process vary onlyinfinitesimally fromequilibrium states
• cycle - when initial and finalstate are identical
ME 321 Chapters 1-8, Review of Thermo I
Special processes
• Isothermal– constant temperature
• isobaric– constant pressure
• isometric (isochoric)– constant specific volume
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I
Liquid-vapor phase transition
• Subcooled liquid/compressedliquid
• Saturated liquid (about tovaporize)
• Saturated liquid-vapor mixture• Saturated vapor• Superheated vapor
ME 321 Chapters 1-8, Review of Thermo I11
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)
ME 321 Chapters 1-8, Review of Thermo I12
The P-v Diagram Showing Phase Boundaries
ME 321 Chapters 1-8, Review of Thermo I
The T-v Diagram
ME 321 Chapters 1-8, Review of Thermo I14
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”
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I17
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
ME 321 Chapters 1-8, Review of Thermo I18
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
ME 321 Chapters 1-8, Review of Thermo I
• 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)
ME 321 Chapters 1-8, Review of Thermo I20
Superheated Vapors
• For superheated vaporsthe properties areindexed by bothtemperature andpressure
• Double-interpolationmight be needed to getmore accurate values
ME 321 Chapters 1-8, Review of Thermo I
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)
ME 321 Chapters 1-8, Review of Thermo I
• 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
ME 321 Chapters 1-8, Review of Thermo I
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)
ME 321 Chapters 1-8, Review of Thermo I
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)
ME 321 Chapters 1-8, Review of Thermo I
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?
ME 321 Chapters 1-8, Review of Thermo I
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
ME 321 Chapters 1-8, Review of Thermo I27
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
ME 321 Chapters 1-8, Review of Thermo I
...)()()()(5432 +++++=
vTd
vTc
vTb
vTa
vRTP
Other Equations of State
• van der Waals
• Virial
( )Pa
vv b RT+
− =2
ME 321 Chapters 1-8, Review of Thermo I29
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
ME 321 Chapters 1-8, Review of Thermo I30
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
ME 321 Chapters 1-8, Review of Thermo I31
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
ME 321 Chapters 1-8, Review of Thermo I32
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
ME 321 Chapters 1-8, Review of Thermo I33
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
∞
ME 321 Chapters 1-8, Review of Thermo I34
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
ME 321 Chapters 1-8, Review of Thermo I35
• 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
ME 321 Chapters 1-8, Review of Thermo I36
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=
∂∂
ME 321 Chapters 1-8, Review of Thermo I37
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
ME 321 Chapters 1-8, Review of Thermo I38
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.
ME 321 Chapters 1-8, Review of Thermo I39
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:
ME 321 Chapters 1-8, Review of Thermo I40
• 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!!
ME 321 Chapters 1-8, Review of Thermo I41
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 !!!!
ME 321 Chapters 1-8, Review of Thermo I42
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
!!!!
. ..
.
.
ME 321 Chapters 1-8, Review of Thermo I43
Steady-Flow Devices OperateSteadily for Long Periods
Steady-Flow Devices OperateSteadily for Long Periods
ME 321 Chapters 1-8, Review of Thermo I44
++−
++=−∑ 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
ME 321 Chapters 1-8, Review of Thermo I45
++−
++=− 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
Small device so neglectpotential energy changes
0
( )
−+−=
22
22
,ei
eioutturbVVhhmW !! ( )
−+−=
22
22
,ie
ieincompVVhhmW !!
turbW!out
compW!in
ME 321 Chapters 1-8, Review of Thermo I46
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
Small device so neglectpotential energy changes
00
No work is doneby or on
throttling device
ME 321 Chapters 1-8, Review of Thermo I47
( )
( )
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
ME 321 Chapters 1-8, Review of Thermo I48
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
ME 321 Chapters 1-8, Review of Thermo I49
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
ME 321 Chapters 1-8, Review of Thermo I50
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
ME 321 Chapters 1-8, Review of Thermo I51
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
W−== 1,η
%4040.0 ==thη
ME 321 Chapters 1-8, Review of Thermo I52
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
ME 321 Chapters 1-8, Review of Thermo I53
Basic Components of a RefrigerationSystem in Typical Conditions
Basic Components of a RefrigerationSystem in Typical Conditions
ME 321 Chapters 1-8, Review of Thermo I54
• 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
ME 321 Chapters 1-8, Review of Thermo I55
• 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
ME 321 Chapters 1-8, Review of Thermo I56
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.
ME 321 Chapters 1-8, Review of Thermo I57
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
ME 321 Chapters 1-8, Review of Thermo I58
P-v Diagram of the Carnot CycleP-v Diagram of the Carnot Cycle
ME 321 Chapters 1-8, Review of Thermo I59
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
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
ME 321 Chapters 1-8, Review of Thermo I60
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:
ME 321 Chapters 1-8, Review of Thermo I61
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
ME 321 Chapters 1-8, Review of Thermo I62
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
ME 321 Chapters 1-8, Review of Thermo I63
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)
ME 321 Chapters 1-8, Review of Thermo I64
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
ME 321 Chapters 1-8, Review of Thermo I65
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 −=
ME 321 Chapters 1-8, Review of Thermo I66
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
ME 321 Chapters 1-8, Review of Thermo I67
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
ME 321 Chapters 1-8, Review of Thermo I68
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
==
====
)()()()(
ME 321 Chapters 1-8, Review of Thermo I69
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
TdTTcss P −=− ∫
• 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 −=−
ME 321 Chapters 1-8, Review of Thermo I70
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
TdTTcss P −=− ∫
• 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
)(!
ME 321 Chapters 1-8, Review of Thermo I71
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)(
PPRssss oo −−=−
1
21212 ln)(
PPRssss u
oo −−=−
∫=T
Po
TdTTcs
0
)(!
ME 321 Chapters 1-8, Review of Thermo I72
Isentropic Ideal Gas Relationships -- Case 1: Constant Specific Heats
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 +=+= ,
ME 321 Chapters 1-8, Review of Thermo I73
Isentropic Ideal Gas Relationships -- Case 1: Constant Specific Heats
Isentropic Ideal Gas Relationships -- Case 1: Constant Specific Heats
• 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
ME 321 Chapters 1-8, Review of Thermo I74
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
ME 321 Chapters 1-8, Review of Thermo I75
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)(
PPRssss oo −−=−
( )( )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 =
ME 321 Chapters 1-8, Review of Thermo I76
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 =
ME 321 Chapters 1-8, Review of Thermo I77
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
ME 321 Chapters 1-8, Review of Thermo I78
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
ME 321 Chapters 1-8, Review of Thermo I79
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
ME 321 Chapters 1-8, Review of Thermo I80
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
ME 321 Chapters 1-8, Review of Thermo I81
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
ME 321 Chapters 1-8, Review of Thermo I82
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
ME 321 Chapters 1-8, Review of Thermo I83
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
ME 321 Chapters 1-8, Review of Thermo I84
Entropy Generation During Heat TransferEntropy Generation During Heat Transfer
Graphical representation of entropy generation during a heattransfer process through a finite temperature difference
ME 321 Chapters 1-8, Review of Thermo I85
• 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
© The McGraw-Hill Companies, Inc.,1998
Maximum Work Available and the Dead StateMaximum Work Available and the Dead State
ME 321 Chapters 1-8, Review of Thermo I86
• 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 McGraw-Hill Companies, Inc.,1998
- The Dead State -A State of Equilibrium with the Surroundings
- The Dead State -A State of Equilibrium with the Surroundings
ME 321 Chapters 1-8, Review of Thermo I87
• 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 −=
ME 321 Chapters 1-8, Review of Thermo I88
• “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
© The McGraw-Hill Companies, Inc.,1998
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
ME 321 Chapters 1-8, Review of Thermo I89
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
ME 321 Chapters 1-8, Review of Thermo I90
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 −−−+−=
ME 321 Chapters 1-8, Review of Thermo I91
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
−−−=
−=−=−−−+−=
==
=="
ME 321 Chapters 1-8, Review of Thermo I92
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
ME 321 Chapters 1-8, Review of Thermo I93
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 −−−+−==
ME 321 Chapters 1-8, Review of Thermo I94
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 −=−=
ME 321 Chapters 1-8, Review of Thermo I95
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 +=
ME 321 Chapters 1-8, Review of Thermo I96
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
ME 321 Chapters 1-8, Review of Thermo I97
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
ME 321 Chapters 1-8, Review of Thermo I98
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 −−−+−=
ME 321 Chapters 1-8, Review of Thermo I99
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
ME 321 Chapters 1-8, Review of Thermo I100
The Exergy BalanceThe Exergy Balance
General:
• Exergy balance for any system undergoing any processcan be expressed as
General, rate form:
ME 321 Chapters 1-8, Review of Thermo I101
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:
ME 321 Chapters 1-8, Review of Thermo I102
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
ME 321 Chapters 1-8, Review of Thermo I103
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
ME 321 Chapters 1-8, Review of Thermo I104
The Second-Law Efficiency of All ReversibleDevices is 100%
The Second-Law Efficiency of All ReversibleDevices is 100%