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

∂∂

P=1 atm

liquid

if,Region LiquidCompressed

if,Region Heat Super P and TGiven

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

superheatedC 6

kpa 362.23kPa 500

C 15.71

subcooled

p=constantp=constant

Three TablesTemperature Table

at spaced T’sPressure Table

at spaced P’sSuperheat Table

at spaced T and P6 PropertiesTemperaturePressureVolumeInternal EnergyEnthalpyEntropy

Figure A-9E

TEMPERATURE TABLESaturation properties only as a function of temperature

PRESSURE TABLE saturation properties only as a function of pressure

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

sxssuxuuhxhh

×+−=

vxvvvxvx1v

vmvmmvVVV

Two Phase Real Gas Properties

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

=−×+=

TEMPERATURE TABLETable A-4, A-4EPRESSURE TABLETable A-5, A-5E

SUPERHEAT TABLETable A-6, A-6E

Table A-7,A-7E

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

FLUIDS

FUNCTIONS

WindowsEquations

WindowsArrays

WindowsPlot Window

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

×+===

⇒=×+=

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

20 kPa

SUPERHEATED TABLEsuperheat properties as a function of temperature and pressure

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

COMPRESSED LIQUID (SUBCOOLED LIQUID) TABLESubcooled properties as a function of temperature and pressure

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

−×+=

100 kpa3 kg2 kgQ

( ) 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

=−×+=

−+=+=

===−

∫∫

( )233-13-1

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

−===

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

100 kpa3 kg2 kgQ

EES Solution

∫×= pdvmW

4.5020

450500450470eTemperatur

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

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'

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

×=×=

Ideal Gas Law

8.314pv

1545.15pv

:for water

heat specificconstant RTpv

Model Gas Ideal

% Error in assuming water is an ideal gas

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

×+××

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

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

P v = m R T

IDEAL GAS EQUATION FORMS - For Air

kPa39.117pm 23m 12 kPa 225p

V T RV p T Rp

VpT Rm

m 23kPa 224T273.15.286

constantT constant,mass

pressure? final the isWhat .m 23 of volume a to etemperaturconstant aat expandskPa 225 of pressure a and m 12 of volume aat initiallyAir

m=constT=const

3m 12 33m 2

SPECIFIC HEAT

( ) ( )

constvv

constpp

29)-(4 Eq ONLYGASIDEALFORunitssamewithRccRdTdTcdTc

RdTdudhRTuh

RTpvpvuh

dTcΔudTcΔhGasIdeal

dTTcΔudTTcΔhTuc

=−+=

∂∂

∫ ∫∫∫

( )( )dTTcu

∫∫

IDEAL GAS IMPROVEMENTS

Figure 4-24

AIR TABLEair properties as a function of temperature with variable temperature dependent specfic heat

NIST Webbook Propertiesfttp://webbook.nist/gov/chemistry/fluidTemperature Table for Water in .1 degree incrementsfrom 40 to 40 degrees.

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Set low and high temperatureand temperature increment.

Thermodynamic Properties are Measurements

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