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

∂∂

=

∂∂

P=1 atm

Q

liquid

vapor

sat

sat

sat

sat

Given T and PSuper Heat Region if, p<p @T T>T @p

CompressedLiquid Region if, p>p @ T T<T @ p

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

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

superheated

Tp=constant

T

C 6kpa 362.23

kPa 500C 15.71

subcooled

p=constant

v v

Two Phase Real Gas Properties

( )

fg

f

fgf

gf

g

ggff

gf

vvvx

vxvv

xvvx1v

Qualitymm

x

vmvmmv

VVV

−=

×+=

+−=

=

+=

+=

fv gv

fgf

fgf

fgf

vxvv

uxuu

hxhh

×+=

×+=

×+=

/lb.ft 300 vF,90at water of properties theFind 3o =

F90T

o

lb/ft7.467v 3g =

f

f

f

u =58.07h =58.07s =.11165

g

g

g

u =1040.2

h =1107.7

s =2.0083shu

( )vvf −

( )fg vv −

/lbft 300.v 3=lb/ft0161.v 3

f =

f

g f

f f g

f fg

f fg

f fg

o

v vQuality, xv v

300 .0161x .64 (64%)467.7 .0161

u u x (u u )

u u x u

u 58.07 .64 982.2u 686.7 BTU/lb

h h x h

h 58.07 .64 1042.7h 725.4 BTU/lb

s s x s

s .11165 .64 1.8966s 1.3254 BTU/lb R

−=

−

−= =

−

= + × −

= + ×

= + ×=

= + ×

= + ×=

= + ×

= + ×

=

TemperatureTable

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

==

Solve

WindowsEquations

WindowsArrays

WindowsPlot Window

f fg

f fg

10 MPa saturated steam has an enthalpy of 2010 kJ/kg. What is its internal energy?

h h x h

2010 kJ/kg 1407.56 x 1317.1x .4574u u x u

u 1393.04 .4574 1151.4u 1919.69 kJ/kg

EES Programx quality(stea

= +

= +== +

= + ×=

= m,P 10000.,h 2010)x .4576u intenergy(steam,P 10000.,h 2010)u 1919.65 kJ/kg

= === = ==

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

Linear Interpolation with 2 Variables

p 25 mPa p 30 MPaT 450 h 2949.7 h 2821.4

27 25For the pressure table entry, 30 25

= =

= = == = = =

−=

−

h@(T 450 C,p 27 MPa) Steam Superheat TableTable Values

p 27 MPa h 2898.4

.4

The desired pressure, 27 kPa, is 40 % of the differencebetween table values. All the other properties at 27 kPa must be at the same difference.

EES h enthalpy(steam,T 450.,p 27000)h 2901.7 kJ/kg= = == .22% difference, table and interpolation

SPREAD SHEET WORLDEXCELL ADD-IN40 fluids4 sets of units

SPREADSHEET WORLD – THERMAL FLUIDS PROPERTIES

123456789

A B C D E F G H I J

T 200 P 15

CALL TTProps("AIR","EE_F", "P", $C$2,"T", $C$1)

P = v = T = u = s = h = X = STATE = ERROR =RETURN 15 16.29669 200 180.5582 1.085135 225.7937 1 erheated va 0

psia ft^3/lbm deg F BTU/lbm BTU/lbm-R BTU/lbm nd

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)(1.694.6.001043v

vxvv99.61T

.6 totalkg 5 vaporkg 3 xkPa, 100at

1

fgf1

31

1

fgf1

=×+=

×+==

−×+=

×+==

==

100 kpa3 kg2 kgQ

3.88

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 5V

v 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

hhmQ∆HQ

Vp∆EQW∆EQ

3-2 constant,p

−×−=

=−×=

−==

∆+=+=

=

( )

kJ 05.4713Q) 1670.62-(2613.235Q

uumQ∆U∆EQ

0WW∆EQ

2-1 constant,v

12

=×=

−===

=+=

=

p

3.88

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

=

==

==

p

1

23

water 2vapor 3Properties at 1 given p1 and x1x1 0.6p1 100,000

P = v = T = u = s = h = X = STATE =100000 1.016819 99.62524 1670.481 4.936567 1772.163 0.6 Mixed regioPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd

Properties at 2 given p2 and v2=v1p2 200,000

P = v = T = u = s = h = X = STATE =200000 1.016819 173.5216 2613.078 7.389259 2816.442 1 erheated vaPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd

V2=m x v2 5.084094

V3= 1.2 * V2 6.100913v3=V3/m 1.220183p3 200,000

Properties at 3 given p3 and v3P = v = T = u = s = h = X = STATE =

200000 1.220183 259.0808 2744.746 7.742318 2988.782 1 erheated vaPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd

Q 2-3=m*(h3-h2) 861.7029W 2-3 = Q 2-3-m*(u3-u2) 203.363Q 1-2=m*(u2-u1) 4712.986

v

Spread Sheet World module solution

p

v

1

23

100 kpa3 kg2 kgQ

EES Solution

?

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

fgf2

2

fgf12

33

11

31

1

gg ff1

gf1

=×+=

×+=

=−−

=

×+===⇒==

===

=

×+×=

−+×=

+=

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

f22

g ff3

2

322

32

3g

33

g1

g1g1

property)(xproperty)(property

(property)property)(propertyx

(83.4%) .834.0008157.0358

.0008157.03v

vvx

C 20 @vxC 20 @ v/kgm .03v

/kgm .035

.15mVv

,.15mVat constant. is pressure theprocess heating theDuring

11TableA 842, page MPa .5716 C, 20 @ /kgm .0358v

/kgm .0358kg 3

m .1074mV

v

×+=

−=

=−−

=−

=

×+==

===

=

−=

===

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=?

Ideal Gas Law

o o

*

IDEAL (PERFECT) GAS LAW

pv RT pV mRT p - absolute pressure, psia,kPa T - absolute temperature, R, K

RRmolecular weight

==

=

o*

3*

o o

*

*

ft lbf R R 1545.15 lbmole

kJ kPa mR 8.314 orkmole K kmole K

mass moles Molecular Weight

m n Molecular Weight

pv R T pV nR T

=

=

= ×= ×

=

=

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

heat specificconstant RTpv

Gas Ideal=

water

2

3 o

gage atmosphere

o 3 o

O

What is the mass of 1.2 m of oxygen at 24 Cand a gage pressure of 500 kPa.Atmospheric pressure is 97 kPa

p p p 500 kPa 97 kPa 597 kPa

8.314 kJ/kmole K or kPa m /kmole KR .2598132

= + = + =

= =

( )

3

3

3 o o

o

3 kPa m /kg alsoTableA 1

pV 597 kPa 1.2 mm 9.28 kgRT .259813 kPa m /kg 24 C 273.16 K

What is the volume mass of 1.2 lbm of air at 124 Fand a gage pressure of 500 psia.Atmospheric pressure is 14.

−

×= = =

× +

gage atmosphere

air o

o o

7 psia.

p p p 500 psia 14.7 psia 514 psia

1545.15 ft lbf / lbmole R ft lbfR 53.336 also Table A 1E in BTU and psi units28.97 lbm R

1.2 lbm 53.336 ft lbf/ lbm R 124 F 459.6m R TVp

= + = + =

= = −

× × += =

( )o

2 2

3

9 R514 psia 144 ft /in

V .5047ft×

=

IDEAL GAS EQUATION FORMS - For Air

P v = m R T3

3 OO

3 33 O

O o

32

kPa mkPa m kmole 8.314 K kmole K

kPa m kPa m kPa m kg .287 K R 8.314 / 28.96kg K kmole K

lb ft ft

=

= =

O

O

3 O2 O O

3

ft lbf lbmole 1545.15 R lbmole R

lb ft lbf ft lbf ft lbm 53.35 R R 1545.15 /28.96ft lbm R lbmole R

psi ft lbm

=

= =

= OO O

3 O2 O O

psi lbf ft lbf .3704 R R 1545.15 /144/28.96lbm R lbmole R

lb BTU ft lbf ft lbm .06855 R R 1545.15 /28.96/ft lbm R lbmole R

=

= = 778

( )

( )

/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.