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THERMODYNAMICS REVIEW
Thermodynamics required for compressible flow:
First Law Energy Equation
Second LawAdiabatic, isentropic, constant entropy process Entropy Calculation
Equation of StateIdeal Gas Law
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First Law of ThermodynamicsExamples:Observations:
wheela brakingbearing wheelain friction togetherhandsyou rubbing
∑ ∑∝
=
WQheat more friction more
heat into ed transferrbecan work
δδ
Experiments:
( ) 0δWδQ
δWδQ
Cycle afor LawFirst m kg .427calorie 1
lbft 778BTU 1
δWCδQ
WCQ
f
=−
=
==
=
=
∫∫∫
∫∫∑∑
W
The First Law is a Fundamental observationof nature, an axiom, which can not be provedbut has never been found to be violated
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First Law of Thermodynamics
( )( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
( ) WQδWδQEE
U(T).PEKE forms, allin energy as E Define property. a therefore
andpath oft independen is δWδQ
δWδQδWδQ
0δWδQδWδQ C and B Processes
0δWδQδWδQ CA Cycle
0δWδQδWδQ BA Cycle
0δWδQ
Cycle afor LawFirst δW δQ
1
221
CB
CB
CA
BA
−=−=−
++
−
−=−
=−−−
=−−−⇒
=−−−⇒
=−
=
∫
∫∫∫
∫∫∫∫∫∫
∫∫∫
AB
C
δWdEδQW∆EQ
Processes afor LawFirst
+=+=
1
2
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( )
shaft
2
shaft1
2
1111
2
111
22211shaftout flowin flowshaft
1111final 1initial 11in flow
Wρgh)2
Vpv(umQ
W)ρgh2
Vvpm(u)ρgh2
Vvpm(uQ
ρgh2
Vu(T)PEKEu(T)E
vp mvp mWWWWW
vp mVpVVppdVW
LawFirstW∆EQ
++++∆=
++++−+++=
++=++=
−+=++=
==−==
+=
∫
Steady Flow Energy EquationOpen, steady flow thermodynamic system - a region in space
QshaftW
p1p2
v2
v1 h2
V
V
h1
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Steady Flow Energy EquationOpen, steady flow thermodynamic system -a region in space
2Vh
2VhQ
flow icnonadiabatan for 2
Vh2
Vh
flow adiabatican for hpvu since
forcesbody without 0,Q 0,W nozzleor diffuser channel, adiabatican for
Wh) g ρ2
V vp(umQ
22
2
21
1
22
2
21
1
shaft
2
+=++
+=+
=+
==
++++∆=
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IDEAL GAS CALCULATION
( )
( )
3
22
ooo
o
o
air
atmospheregage
oo3
3
3o3o
O
atmospheregage
.5047ftV/inft 144psia 514
R459.69F124Rlbmlbf/ ft 53.336lbm 1.2p
T R mV
R lbm
lbfft 53.33628.97
lbmole / R lbm / lbf 1545.15R
psia 514psia 14.7psia 500ppp
kg 9.28K273.16C24/kgm kPa .259813
m 1.2kPa 597RTpVm
/kgm kPa .25981332
K /kmolem kPaor K kJ/kmole 8.314R
kPa 597kPa 97kPa 500ppp
2
=
×+××
==
==
=+=+=
=+×
×==
==
=+=+=
psia. 14.7 is pressure cAtmospheripsia. 500 of pressure gage a and F124at air of lbm 1.2 of volume the isWhat
kPa 97 is pressure cAtmospherikPa. 500 of pressure gage a and C24at oxygen of m 1.2 of mass the isWhat
o
o3
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Ideal Gas Law
TnRpV
TRpv
WeightMolecular nm WeightMolecular molesmass
Kkmole
m kPaorKkmole
kJ8.314R
lbmoleRlbf/lbm 1545.15 R
weightmolecular RR
KR, re, temperatuabsolute - T kPapsia, pressure, absolute - p
mRTpV RTpv
LAW GAS (PERFECT) IDEAL
*
*
o
3
o*
o*
*
oo
=
=
×=×=
=
=
=
==
2
1
2
1
2
1
2
1
2211
23
TT
pp
TT
vv
LAW CHARLES
vpvp LAW BOLYES
)0 and atm (1 STP at gas of /molemolecules 106.023
liters. 22.4 gasany of mole (1) One
LAW SAVOGADRO'
=
=
×=×
×
=
Co
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tCoefficien JT=
∂∂
hpT
h=constant
% Error in assuming water is an ideal gas
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PRINCIPAL OF CORRERSPONDING STATESCOMPRESSIBILITY FACTOR Z
P
criticalR
criticalR
TTT
ppP
=
=
Z is about the same forall gases at the same reduced temperatureand the same reduced pressure where:
mRTpVZ
RTpvZ
=
=
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SPECIFIC HEAT
( ) ( )
v
p
vp
vp
vp
vp
constvv
constpp
cR1
1γγRc
ONLYGASIDEALFORunitssamewithRcc
RdTdTcdTcRdTdudh
RTuhRTpv
pvuh
dTc∆udTc∆hGasIdeal
dTTc∆udTTc∆h
Tuc
Thc
=−γ
−=
=−
+=+=
+==+=
==
==
∂∂
=
∂∂
=
∫ ∫∫∫
==
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( )( )dTTcu
dTTch
v
p
∫∫
=
=
IDEAL GAS IMPROVEMENTS
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Kelvin Planck Statement of the Second LawIt is impossible to construct an engine which, operating in a cycle, will produce no
other effect than the extraction of heat from a single reservoir and the performance of an equivalent amount of work.
Clausius Statement of the Second Law
It is impossible to have a system operating in a cycle which transfers heat from a cooler to a hotter body without work being done on the system by the surroundings
Reversible Heat Engine Reversible Refrigerator
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Carnot Power Cycle
14 process3,2 process expansion, adiabatic Reversible43 process2,1 process
fer,heat trans emperatureconstant t Reversible
→→
→→
inQW
InputRequiredEffectDesiredEfficiency ==
H
LHCYCLE Q
QQη −=
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Carnot Principles
1. No engine operating between two heat reservoirs, each having a fixed temperature, can be more efficient than a reversible engine operating between the same reservoirs.
Carnotactual ηη ≤
2. All reversible engines operating between two heat reservoirs, each having its own fixed temperature, have the same efficiency.
3. The efficiency of any reversible engine operating between two reservoirs is independent of the nature of the working fluid and depends only on the temperature of the reservoirs.
4. An absolute temperature scale can be defined in a manner independent of the thermometric material.
2
1
2
1
TT
=
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5-15
FIGURE 5-47Proof of the firstCarnot principle.
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Thermodynamic Temperature Scale
h
lhl
l
h
l
h
313221
3
2
2
1
3
1
313
132
3
221
2
1
313
1
1
3
31
TTQQand
TT
if,onlysatisfiedbecanequationthis)T,f(T)T,f(T)T,f(T
ng,substitutiQQ
identity,by
)T,f(TQQ)T,f(T
QQ)T,f(T
schematicsenginefrom
)T,function(TQQ
QQ1η
)T,function(Tη
==
×=
=
===
=
−=
=
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0
0
=
=
∫
∑
processesreversible T
dQTQδ
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Clasius Inequality
0TδQ
0TQ
TQ
TT1
QQ1
TTT
QQQ
Cycle Reversible afor
l
l
h
h
h
l
h
l
h
lh
h
lh
=
=−
−=−
−=
−
∫
∫ ≤ 0TQδ
0TδQ
0TQ
TQ
TT1
QQ1
TTT
QQQ
CycleleIrreversiban for
l
l
h
h
h
l
h
l
h
lh
h
lh
⟨
⟨−
−⟨−
−⟨
−
∫
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Entropy Definition and Change
∫
∫∫
∫
∫
∫
∫∫∫
∫
=
=≥
≥
≤+
≤−+
≤+=
≤
TdsδQTδQds
TδQds
TδQds
0dsTδQ
0SSTδQ
ENTROPY S PROPERTYA DEFINE
0TδQ
TδQ
TδQ
process leirreversiban andreversible a of composed cycle afor
InequalityClasius0TδQ
revirrev
1
irrev2
1
irrev2
12
1
irrev2
2
rev1
1
irrev21
2reversible
irreversible
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Entropy Change of an Ideal Gas
−
=−
−=
=
=
−=
=+−−=
+=−−=++=
+=
1
2
1
2p12 p
plnR TT lncss
pdpR
TdTcds
pdpRdp
Tv
TdTc
Tdh , gas idealan For
dpTv
Tdhds
vdp-dhTds pdvvdppdvdhTds
pdvduTds into ngSubstitutivdppdvdhdu vdppdvdudh
pvuh
p
p
+
=−
+=
=
=
+=
+==
+=
1
2
1
2v12 v
vlnR TT lncss
Law Second Tdsδq LawFirstδwduδq
vdvR
TdTcds
vR
Tp
dTcdu :gas idealan For T
pdvTduds
pdvduTds
v
v
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( )
( )
( )
( )
entropyconstant also is constant pv process, adiabatic
0cc
cc
cc
cc
cpplnss
cc1cpplnss
pplncc
ppln1css
pplncc
pplncss
ppRln
TTlncss
pp
vv
TT
RTpv from substitutevv
pp
vpvpconstantpv
vp
v
p
v
v
v
p
p1
212
vpp1
212
1
2vp
1
2p12
1
2vp
1
1
2p12
1
2
1
2p12
1
1
2
1
2
1
1
2
2
1
1
2
2211
=
=
−−−
=−
−−
γ−γ
=−
−−
γ−γ
=−
−−
=−
−
=−
=
=
=
=
=
=
γ
γ−γ
γ−γ
−γ
γ
γγ
γ
Isentropic Adiabatic Process
IdealGas0∆s,Isentropic
Adiabaticconstantpvconstantplnvln
gintegratin
0p
dp1
1_v
dv11
1
0vdpRcpdv1
Rc
0pdvvdpRcpdv
Rc
Rvdp
RpdvdT,
RpvTgasidealanfor
0pdvdTc0δWdU
0δQprocess AdiabaticLawFirstδWdUδQ
vv
vv
v
==
=+γ
=
−γ
+
+
−γ
=+
+
=++
+==
=+=+
=+=
γ
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ADIABATIC PROCESSQ=0, 0s constant,pv =∆=γ
γ−γ
−γ
γ
γγ
γ
=
=
=
=
=
=
1
1
2
1
2
1
1
2
2
1
1
2
2122
pp
vv
TT
RTpv substitutevv
pp
vpvpconstantpv