1 Thermodynamic Properties are Measurements p,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 Tables Calculation Modules NIST, EES, HYSYM interactive, callable Property Data v T T p v s l L ¶ ¶ = l L ¶ ¶
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1
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
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