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P=1 atm
Q
liquid
vapor
sat
sat
sat
sat
TT PP
if,Region LiquidCompressed
TT PP
if,Region Heat Super P and TGiven
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STEAM PRESUE AND TEMPERATURE TABLES
f
f
f
suh
SaturatedLiquidLine
g
g
g
s
u
h
SaturatedVaporLineT
fv gv
Three TablesTemperature Table
at spaced T’sPressure Table
at spaced P’sSuperheat Table
at spaced T and P6 PropertiesTemperaturePressureVolumeInternal EnergyEnthalpyEntropy
Solid-Liquid-GasPhase Diagram
Saturation liquid internal energy at 0 C 0. Table BaseSaturated liquid enthalpy at 25 C 104.89 kJ/kgSaturated vapor entropy at 25 C 8.558 kJ/kg KEnthalpy at 20 C, 300 kPa
assume saturated liquid enthalpy at 20 C 83.96 kJ/kg Temperature of saturated vapor with an
internal energy of 2396.1 kJ/kg 15 C Enthalpy of vaporization at 10 C 2477.7 kJ/kg
Table A-4 Metric, TTable A-5 Metric, pTable A-4E English, TTable A-5E English, pTEMPERATURE TABLEPRESSURE TABLE
Table A-6, Metric, T,pTable A-6E, English, T,pSUPERHEAT TABLE
Table A-7 Metric T,pTable A-7E English T,p
Two Phase Real Gas Properties
( )
fg
f
fgf
gf
g
ggff
gf
vvvx
vxvv
xvvx1v
Qualitymm
x
vmvmmv
VVV
−=
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=
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+=
fv gv
fgf
fgf
fgf
vxvv
uxuu
hxhh
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/lb.ft 300 vF,90at water of properties theFind 3o =
F90T
o
lb/ft7.467v 3g =
1269.sb07.58h
07.58u
f
f
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008.2s
2.1140h
7.110u
g
g
g
=
=
=
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/lbft 300.v 3=lb/ft0161.v 3
f =RBTU/lb 1.315s
1.8526.64.1296s
sxss
BTU/lb 725.4h1042.7.6458.07h
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BTU/lb 686.7u982.2.6458.07u
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.0161300x
vvvv xQuality,
o
fgf
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fgf
gff
fg
f
=
×+=
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=×+=
×+=
=×+=
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−×+=
=−−
=
−−
= TemperatureTable A-4Epage 880
kJ/kg 1919.65h2010)h10000.,Psteam,intenergy(u
.4576x2010)h10000.,Peam,quality(stx
Program EES
kJ/kg 69.1919u4.1151.45741393.04u
ux uu.4574x
1317.1x 1407.56kJ/kg 2010
hx hh
energy? internal its is What kJ/kg. 2010 ofenthalpy an hassteamsaturated MPa 10
fgf
fgf
====
====
=×+=
+==
+=
+=
Steam at 20 C has an enthalpy of 1800 kJ/kg. What is the internal energy?
kJ/kg 1707.25u2319.0.783.95u
ux uu.7x
2454.1x 83.96kJ/kg 1800
hx hh
fgf
fgf
=×+=
+==
+=
+=
STEAM SUPERHEAT TABLE
SUPERHEAT TABLE
Enthalpy at 700 C and .10 Mpa 3928.2 kJ/kgTemperature at entropy of 8.8642 and .05 Mpa 400 CEnthalpy at .05 MPa and entropy of 10.6662 kJ/kg C 5147.7 kJ/kg
Steam initially at a temperature of 1100 C and a pressure of .10 MPa undergoes a process during which its entropy remains constant to a pressure of .01 MPa. What is the enthalpy and temperature of the steam at the end of the process?
Entropy at 1100 C, .1 MPa 10.1659kJ/kg KEnthalpy at .01 MPa, entropy 10.1659 3705.4 kJ/kgTemperature at .01 MPa, entropy 10.1659 600 C
T.01 MPa
s
Engineering Equation Solver - EESFluid Property Information - 69 fluids availableThermophysical Functions - 25 properties calculatedEquations Window
h=enthalpy(steam, T=200.,P=200) superheated vaporh=enthalpy(steam,T=200.,X=1) saturated vaporu=intenergy(steam,T=200.,X=0.) saturated liquidp=pressure(steam,T=200.,X=0.) saturation pressure
Thermophysical Functionsentropyintenergypressurequalitydensityenthalpyisidealgastemperaturevolume
Function ArgumentsH specific enthalpyP pressureS specific entropyT temperatureU specific internal energyV specific volumeX quality
EES
FLUIDS
FUNCTIONS
Table EES Program
2414.29u
1.)X4.,Psteam,intenergy(u 121.3kJ/kgu
0)X4.,Psteam,intenergy(u
g
g
l
l
=
====
===
g2415.2kJ/ku
g121.45kJ/kul
=
=
g
Pressure TableSaturation internalenergy at 4 kPa
Superheat Table
/kg.001004mv
500.)p30.,Tam,volume(stevg125.67kJ/kh
500.)p30.,Tteam,enthalpy(sh
3l
l
l
l
=
====
===
/kg.001004mv
Cliquid@30 saturated vv
kJ/kg 125.75hC30 @ liquid saturatedh h
3l
ol
l
ol
=
=
==Enthalpy and
volume of water at 150 kPa and 30 C
Temperature TableInternal energy of water at 20 MPa uand 300 C
kJ/kg 1333.446u20000.)p300.,Tsteam,intenergy(u
l
l
====
kJ/kg 1332.0C300 @ liquid saturatedu u
l
ol
==
Linear Interpolation with 2 Variables
ioninterpolat and table,difference .22% kJ/kg 2901.7 h27000)p450.,Tteam,enthalpy(sh
EES
.difference same at the bemust kPa 27at propertiesother theAll values.blebetween ta
difference theof % 40 is kPa, 27 pressure, desired The
4.25302527 entry, tablepressure For the
2821.4h 2949.7h 450TMPa 30p mPa 25p
====
=−−
=======
==
2898.4h MPa 27p
TableSuperheat Steam MPa)27pC, 450h@(T
Linear Interpolation with 2 Variables
( )( ) ( )
ioninterpolat and table,difference .22% kJ/kg 2901.7 h27000)p450.,Tteam,enthalpy(sh
EES
kJ/kg 2898.4h2947.751.32h
7.2947252725302949.7-2821.4h
h@(p25)B CAb xmh
2821.4h 2949.7h 450TMPa 30p mPa 25p
TableSuperheat Steam MPa) 27pC, 450h@(T
====
=+−=
+−−
=
+
=+=
=======
==
2898.4h MPa 27p h(30)=2821.4
h(25)=2949.7
25 27 30
B
p
h@(p=27)
T=450 C
A
C
?
3 kg vapor and 1 kg liquid R-134a is contained in a rigid tank at 20 C.What is the volume of the tank? If the tank is heated until the pressure reaches .6 MPa? What is the quality, and enthalpy of the mixture of liquid and vapor?
kJ/kg 211.27179.71.78779.84h
hxhh
).787(78.7%.0008196.0341
.00008196.027x
12-A Table , 843 page MPa .6 @ vxMPa .6 @ vvvconstantvconstantm constant,V
heating,After
/kgm .027kg 4
m .1082mVv
m .1082V
.0358kg 3.0008157kg 1V
11TableA 842, page C 20 @ vmC 20 @vmV
VVV
2
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2
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33
11
31
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=×+=
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−+×=
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T
v
.6 MPa
20 C.5716 MPa
1
2
Q
3 kg of vapor and 2 kg of liquid R-134a is contained in a piston cylinder device.The volume of the vapor is .1074 cubic meters. What is the temperature andpressure? If the cylinder and its contents are heated until volume is .15 cubicmeters what is the quality?
fgf
fg
f
fg
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g ff3
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.0008157.03v
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11TableA 842, page MPa .5716 C, 20 @ /kgm .0358v
/kgm .0358kg 3
m .1074mV
v
×+=
−=
=−−
=−
=
×+==
===
=
−=
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20 C
T
v
1 2
Q
1. Problem StatementCarbon dioxide is contained in a cylinder
with a piston. The carbon dioxide is compressedwith heat removal from T1,p1 to T2,p2. The gasis then heated from T2, p2 to T3, p3 at constant volume and then expanded without heat transfer to the original state point.
4. Property Diagramstate points - processes - cycle
T1,p1
T3,p3
T2,p2
p
Thermodynamic Problem Solving Technique
QW
QW
2. Schematicv
5. Property Determination
T2,p2
1 2 3 T pvuhs
3. Select Thermodynamic Systemopen - closed - control volume
a closed thermodynamic systemcomposed to the mass of carbondioxide in the cylinder
p
v
2CO
6. Laws of ThermodynamicsQ=? W=? E=? material flows=?
Linear Interpolation with 3 Variables
ioninterpolat and table,difference .47% kJ/kg, 3003.95 h27000.)p470.,Tteam,enthalpy(sh
EES
4.5020
450500450470eTemperatur
4.52
25302527pressure
C. 500 and C 450between MPa 27pat einterpolatThen 27).p500, and 450(T h@
get to30P and 25pbetween C 500at and C 450at first eInterpolat
3081.1h 4.3161h 500T
2821.4h 2949.7h 450TMPa 30p MPa 25p
====
==−−
==−−
=====
======
=======
==
3129.88h2991.0h 470T2898.4h
MPa 27p
TableSuperheat Steam MPa)27pC,470h@(T
Linear Interpolation with 3 Variables
( )( ) ( )
ioninterpolat and table,difference .47% kJ/kg, 3003.95 h27000.)p470.,Tteam,enthalpy(sh
EES
kJ/kg .2991h
2898.4450-470504500
2898.43129.88h
27)p 450,h(TB CAb xmh
C. 500 and C 450between MPa 27pat einterpolatThen 27).p500, and 450(T h@
get to30P and 25pbetween C 500 and C 450at eInterpolat
3081.1h 4.3161h 500T
2821.4h 2949.7h 450TMPa 30p MPa 25p
TableSuperheat Steam MPa) 27pC, 470h@(T
====
=
+−−
=
==+
=+=
=====
======
=======
==
3129.88h2991.0h 470T2898.4h
MPa 27p
h(T=450)2898.4
450 470 500
B
T
h(470,27)
P=27 MPa h(T=500)3129.88
A
C
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
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*
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=
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=
=
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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
heat specificconstant RTpv
Gas Ideal=
water
( )
( )
3
22
ooo
o
o
air
atmospheregage
o
oo3
3
3o3o
O
atmospheregage
o3
.5047ftV/inft 144psia 514
R459.69F124Rlbmlbf/ ft 53.336lbm 1.2p
T R mV
unitsmolar in 1EA Table also R lbm
lbfft 53.33628.97
lbmole / R lbm / lbf 1545.15R
psia 514psia 14.7psia 500ppp
psia. 14.7 is pressure cAtmospheripsia. 500 of pressure gage a and
F124at air of lbm 1.2 of mass volume theisWhat
kg 9.28K273.16C24/kgm kPa .259813
m 1.2kPa 597RTpVm
1alsoTableA /kgm kPa .25981332
K /kmolem kPaor K ole8.314kJ/kmR
kPa 597kPa 97kPa 500ppp
kPa 97 is pressure cAtmospherikPa. 500 of pressure gage a and
C24at oxygen of m 1.2 of mass theisWhat
2
=
×+××
==
−==
=+=+=
=+×
×==
−==
=+=+=
IDEAL GAS EQUATION FORMS - For Air
P v = m R T
8 /28.96/77R lbm
lbfft 1545.15R R R lbm
BTU .06855 lbm ft ftlb
/144R lbm
lbfft 1545.15R R R lbmole
lbf psi 10.73 lbmole ft psi
96.28 /R lbm
lbfft 1545.15R R R lbm
lbfft 53.35 lbm ft ftlb
R R lbm
lbfft 1545.15 lbmole ft ftlb
96.28/Kmole kg
m kPa8.314RK Kmole kg
m kPa .287 kg m kPa
K Kmole kg
m kPa8.314 mole kg m kPa
OO
O3
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32
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O
33
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33
==
==
==
=
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=
( )
( )
/kgm .8417vkPa 101.325
F24K273.15/kgm kPa .287p
RTv
96.28/Kmole kg
m kPa8.314R
kPa 101.325 and C24at air of volumespecific The
/lbft 13.476v/ftin 144lbf/in 14.7
F75R459.69R lbf/lbmft 53.35p
RTv
R lbf/lbmft 35.53 /28.961545.15Rpsia 14.7 and F75at air of volumespecific The
3
2
oo3
O
3
o
3
222
ooair
o
=
+×==
=
=
×+×
==
==
( )
kPa39.117pm 23m 12 kPa 225p
VVp
V T RV p T Rp
VpT R
VpT Rm
m 23kPa 224T273.15.286
VpRTm
constantT constant,mass
pressure? final theisWhat .m 23 of volumea toemperatureconstant t aat expandskPa 225 of pressure a and m 12 of volumeaat initially Air
2
3
3
2
2
11
21
1122
22
2
11
1
31
11
1
3
3
=
=
==
==
×+×
==
==
pm=constT=const
3m12 33m 2v
SPECIFIC HEATSFOR GASSES
SPECIFIC HEAT C
( ) ( )
vvp
v
p
vp
vp
vp
vp
vp
constvv
constpp
cR1k Rcc ONLYGASIDEALFOR
cc
k andunitssamewith
Rcc
RdTdTcdTcdu anddh for ngsubsistutiRdTdudh atingdifferenti
RTuh RTpv ngsubstituti
pvuh definitionBy
dTcdu dTcdh
dTc∆udTc∆h
constant assumed are heats specific asgidealan For
dTTc∆udTTc∆h
TuC
Th
=−=−
=
+=
+=+=
+==
+=
==
==
==
∂∂
=
∂∂
=
∫ ∫
∫∫==
IDEAL GAS IMPROVEMENTS
Enthalpy, h, internalenergy, u, and entropy, s. are not absolute but
Differences from a base.
( )
( )
( ) ( )
22E-A 22,-A to17-A 17, -A Tables
Tcc,Tcc
dTTcu
dTTch
vvpp
T
StateBaseTable
v
T
StateBaseTable
p
==
=
=
∫
∫
IDEAL GAS WITH VARIBLE SPECIFIC HEAT
2-104
( )( ) ( )( )
( )( ) ( )( )
( )
( )
g449.46kJ/k∆h 101.325)p600.,Titrogen,enthalpy(n101.325)p1000.,Titrogen,enthalpy(n∆h
e)EESkJ/kg 448.583kg/kgmolemole/28.0112566kJ/kg∆h
kJ/kgmole 12566kJ/kgmole 17563kJ/kgmole 30,129Kh@600Kh@1000h∆ 18ETableA d)
kJ/kg 415.66001000 kJ/kg 1.039448∆h
kJ/kg 1.039448.5K@800c heat, specific re temperatuc)Room
448.5kJ/kg6001000K kJ/kg 1.121∆Tc∆h
kJ/kgK 1.121K@800c range, re temperatuover theheat specific b)AveragekJ/kg 447.8 8.013kJ/kmole/2 2544∆h
kJ/kmole 125446001000102.873808141600100010.8081
31
6001000.0001571.5600100028.9h∆
dTcTbTac dT,Tch∆ a)
EES. e) 18E,A Table d) 2a,-A Table re temperaturoomat heat specific c) 2b,-A Table re temperatuaverage at theheat specific b)
, 2cA Tableequation heat specific empirical a) :using F)1340 C,(726K 1000 K to600 from heated isit as kJ/kgin nitrogen of ∆h, change,enthalpy theDetermine
OO
Op
p
Op
449335
12
32pp
OOOO
===−===
===−=−=
−=−=
=
=−==
=
==
=−×−−×+
−+−=
+++==
−
−
−−
∫
PRINCIPAL OF CORRERSPONDING STATESCOMPRESSIBILITY FACTOR Z
P
criticalR
criticalR
TTT
)202(p
pP
=
−=
Z is about the same forall gasses at the same reduced temperatureand the same reduced pressure where:
mRTpVZ
RTpvZ
=
=
VAN DER WAALS EQUATION OF STATE - 1873
( )
critical
critical
critical
2critical
2
T2
2
T
2criticalcritical
critical
2
p 8TRb
p 64T R 27a
0dv
p0dvp
va
bvRTp
molecules gas of volumeb
forcesular intermolecva
criticalcritical
==
=
∂=
∂
−−
=
−
−
=−
+ 22)-(2 RTbv
vap 2
critical point
0vp
v
0vp
T
T
=
∂∂
∂∂
=
∂∂
p
v
THERMODYNAMIC PROPERTY MEASURMENT
Thermodynamic properties are independent of path or process and are exact differentials.
Heat and Work are not exact differentials but are dependent on process or path.
gas real afor valueagas, idealan for 0pT
tCoefficienTompson JoulepT
cTu
cTh
equations 60 time,aat 2 properites amic thermodyn6
dvvuds
sudu
v)f(s,u
h
h
vv
pp
sv
=
∂∂
=
∂∂
=
∂∂
=
∂∂
∂∂
+
∂∂
=
=
vT Tp
vs
∂∂
=
∂∂
LawFirst with theused
tCoefficien JT=
∂∂
hpT
h=constant
Course Property Sources
1) Ideal Gas Law with constantspecific heats
2) TablesSteamRefrigerantAir
3) EES CDNIST
for home work convenience
EQUATION OF STATE ERRORS
nitrogen
NIST Webbook Propertiesfttp://webbook.nist/gov/chemistry/fluidTemperature Table for Water in .1 degree incrementsfrom 40 to 40 degrees.
Select Units
Select Table Type
Select fluid
Set low and high temperatureand temperature increment.