thermodynamic properties are measurements
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Thermodynamic Properties are Measurementsp,T,v, u ,h,s - measure directly
-measure by change
Tables
Curve fits Tables
Correlation's, Boyles Law Tables pv=c @ T=c limited hand
calculations
Equations of State, pv=RT TablesCalculation Modules
NIST, EES, HYSYMinteractive, callable
PropertyData
vT Tp
vs
∂∂
=
∂∂
2
P=1 atm
Q
liquid
vapor
kg/m3
3
4
sat
sat
sat
sat
TT PP
if,Region LiquidCompressed
TT PP
if,Region Heat Super P and TGiven
<>
>
<
5
kPa 8587.9C 300 @p C 195.04kpa 1400 @T
C 300 and kpa 1400at water
saturation
saturation=
=kpa 362.23C 6 @p
C 15.71kpa 500 @T C 6 and kPa 500at 134aR
saturation
saturation=
=−
C 195.04kpa 1400kPa 8587.9
C 300
T
v
superheatedC 6
kpa 362.23kPa 500
C 15.71
T
v
subcooled
p=constantp=constant
6
7
Three TablesTemperature Table
at spaced T’sPressure Table
at spaced P’sSuperheat Table
at spaced T and P6 PropertiesTemperaturePressureVolumeInternal EnergyEnthalpyEntropy
Figure A-9E
8
TEMPERATURE TABLESaturation properties only as a function of temperature
9
PRESSURE TABLE saturation properties only as a function of pressure
10
Saturation liquid internal energy at .01 C. 0. kJ/kg Table BaseSaturation vapor internal energy at 15 C. 2395.5 kJ/kg Saturation vapor entropy at 10 C. 8.8999 kJ/kg KEnthalpy at 5 C, 1 bar
approximate saturated liquid enthalpy at 5 C 21.020 kJ/kgTemperature of saturated vapor at 2381.8 kJ/kg
internal energy. 5 CEnthalpy of vaporization at 5 C 2489.1 kJ/kg Volume at 10 C, 1 bar
approximate saturated liquid volume at 6 C .001000 cubic m/kg
11fgf
fgf
fgf
sxssuxuuhxhh
×+=
×+=
×+=
fv
( )
−=
×+=
×+−=
=
+=
+=
fg
f
fgf
gf
g
gglf
gf
vvvx
vxvvvxvx1v
mm
x
vmvmmvVVV
gv
Two Phase Real Gas Properties
12
Steam at 20 Chas an enthalpy of 1800 kJ/kg.What is theinternal energy? ( )
kJ/kg 1706.78u913.833.2402.783.913u
ux uu.7x
2453.5x 83.915kJ/kg 1800
hx hh
fgf
fgf
=−×+=
+==
×+=
+=
13
TEMPERATURE TABLETable A-4, A-4EPRESSURE TABLETable A-5, A-5E
SUPERHEAT TABLETable A-6, A-6E
Water
Table A-7,A-7E
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Engineering Equation Solver - EESFluid Property Information - 69 fluids availableThermophysical Functions - 25 properties calculated
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
EQUATION WINDOW
Thermophysical Functionsentropyintenergypressurequalitydensityenthalpyisidealgastemperaturevolume
Function ArgumentsH specific enthalpyP pressureS specific entropyT temperatureU specific internal energyV specific volumeX quality
Available from Text Student Resources CD
15
EES
FLUIDS
FUNCTIONS
16
17
18
19
Solve
WindowsEquations
20
WindowsArrays
21
WindowsPlot Window
22
23kJ/kg 1931.3ukJ/kg 2204.6.762kJ/kg 251.4u
5-A Table Pressure kPa 20 @uxkPa 20 @uu
uxuuC60.06kPa 20 @ re temperatusaturationT
phase two.762xkJ/kg 2357.5xkJ/kg 251.422047.2
5-A Table Pressure kPa 20 @hxkPa @20hkJ/kg 2047.2EXPANSIONTHEAFTER
kJ/kg 2047.2hkJ/kg 1404.8.5kJ/kg 1344.8h
4-A Table eTemperatur 300 @hxC 300 @hh
hxhhEXPANSIONTHEBEFORE
fgf
fgf
O
fgf
Ofg
Of
fgf
=×+=
×+=
×+===
⇒=×+=
×+=
=×+=
×+=
×+=
T
s
steam? expanded the of energyinternal and phase e,temperatur the is WhatkPa. 20 to enthalpy
constant at expanded is 50% qualityof a withC300 at Steam O
300 C
20 kPa
24
SUPERHEATED TABLEsuperheat properties as a function of temperature and pressure
25
Enthalpy at 600 C and 4.5. MPa 3670.9kJ/kgTemperature at entropy of 6.7714 and 4. MPa 400 CInternl energy at 4.5 MPa and entropy of 7.0323 3096. kJ/kg
26
COMPRESSED LIQUID (SUBCOOLED LIQUID) TABLESubcooled properties as a function of temperature and pressure
27
A saturated mixture of 2 kg water and 3 kg vapor in contained ina piston cylinder device at 100 kpa. Heat is added and the piston,initially resting on stops, begins to move at a pressure of 200 kpa.Heating is stopped when the total volume in increased by 20%. Find:
a) the initial and final temperatures.b) the mass of liquid water when the pressure reaches 200 kPa and the
piston starts to move. c) the work done by the expansion.
kJ/kg 1670.622088.2.6417.40u
uxuu /kgm 1.0168v
.001043)(1.694.6.001043vvxvv
99.61T
.6 totalkg 5 vaporkg 3 xkPa, 100at
1
fgf1
31
1
fgf1
=×+=
×+==
−×+=
×+==
==
100 kpa3 kg2 kgQ
3.88
28
p
v
1
23
( )
( )
( )
( )
( ) kJ 203.2m 5.08m 6.096kPa 2000W
VVp0pdVpdVW
kJ/kg 65.2988hMPa .2P1.2192,v @h h
MPa .2at 3Point
dsuperheate.8857v
1.2192kg 5
m 6.096v
6.096V1.2VkJ/kg 2816.47h
)MPa .2P1.0168,vh@hkJ/kg 2613.23u
MPa .2P1.0161,vu@u6A Table fromion Interpolat
m 5.081.0161kg 5Vv vMPa, 2 .at 2Point
33
232
3
2
2
1
3
33
3
3
23
2
2
2
2
32
21
g
=−×+=
−+=+=
====
⇒=
==
=×==
====
===−
=×=
=
∫∫
( )
( )233-13-1
3-2
23
uumQW,lyalterative
kJ 9.860Q)47.2816(2988.655Q
hhmQΔHQ
VpΔEQWΔEQ
3-2 constant,p
−×−=
=−×=
−==
∆+=+=
=
( )
kJ 05.4713Q) 1670.62-(2613.235Q
uumQΔUΔEQ
0WWΔEQ
2-1 constant,v
12
=×=
−===
=+=
=
3.88
29
kJ/kg 2988.65h
173.1.1989-1.316231.1989-1.2192 vof ratio
3092.1 1.31623 300 1.2192
2971.2 1.1989 250h v T
1.2192) vMPa, .2p ( @enthalpy 6-A TableSuperheat
hfor ion Interpolat
3
3
=
==
==
30
p
v
1
23
100 kpa3 kg2 kgQ
EES Solution
∫×= pdvmW
31
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
3084.8h 9.3165h 500T
2821.h 2950.6h 450TMPa 30p MPa 25p
==−−
==−−
=====
======
=======
==
3133.46h2992.64h 470T
2898.76hMPa 27p
TableSuperheat Steam MPa) 27pC, 470h@(T
Linear Interpolation with 3 Variables
32
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
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
=
=
×=×=
=
=
=
==
Ideal Gas Law
33
T18
8.314pv
T18
1545.15pv
:for water
heat specificconstant RTpv
Model Gas Ideal
×=
×=
=
% Error in assuming water is an ideal gas
34
( )
( )
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
unitsmolar in 1EA Table also 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
1alsoTableA /kgm kPa .25981332
K /kmolem kPaor K ole8.314kJ/kmR
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
35
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
kJ8.314 mole kg m kPa
OO
O3
2
OO
O3
O
3O
O3
2
OO
32
o
3O
O
33
OO
3
==
==
==
=
==
=
P v = m R T
IDEAL GAS EQUATION FORMS - For Air
36
( )
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
2
3
3
2
2
11
21
1122
22
2
11
1
31
11
1
=
=
==
==
×+×
==
==
pressure? final the isWhat .m 23 of volume a to etemperaturconstant aat expandskPa 225 of pressure a and m 12 of volume aat initiallyAir
3
3
p
v
m=constT=const
3m 12 33m 2
37
SPECIFIC HEAT
( ) ( )
v
v
p
vp
vp
vp
vp
constvv
constpp
cR1k
cc
k
29)-(4 Eq ONLYGASIDEALFORunitssamewithRccRdTdTcdTc
RdTdudhRTuh
RTpvpvuh
dTcΔudTcΔhGasIdeal
dTTcΔudTTcΔhTuc
Thc
=−
=
=−+=
+=+=
=+=
==
==
∂∂
=
∂∂
=
∫ ∫∫∫
==
38
( )( )dTTcu
dTTch
v
p
∫∫
=
=
IDEAL GAS IMPROVEMENTS
Figure 4-24
39
AIR TABLEair properties as a function of temperature with variable temperature dependent specfic heat
40
NIST Webbook Propertiesfttp://webbook.nist/gov/chemistry/fluidTemperature Table for Water in .1 degree incrementsfrom 40 to 40 degrees.
Select Units
Select Table Type
41
Select fluid
42
Set low and high temperatureand temperature increment.
43
44
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