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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=

+=

+=

fv gv

fgf

fgf

fgf

vxvv

uxuu

hxhh

×+=

×+=

×+=

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

F90T

o

lb/ft7.467v 3g =

1269.sb07.58h

07.58u

f

f

f

===

008.2s

2.1140h

7.110u

g

g

g

=

=

=

shu

( )vvf −

( )fg vv −

/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

hxhh

BTU/lb 686.7u982.2.6458.07u

uxuu

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(64%) .64.0161467.7

.0161300x

vvvv xQuality,

o

fgf

fgf

fgf

gff

fg

f

=

×+=

×+=

=×+=

×+=

=×+=

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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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11

31

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.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

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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

o*

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*

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=

=

×=×=

=

=

=

==

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

( )

( )

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

2

OO

O3

O

3O

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2

OO

32

o

3O

O

33

OO

33

==

==

==

=

==

=

( )

( )

/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.