Chapter3notes ME311 PJF F14

8/11/2019 Chapter3notes ME311 PJF F14

1/65

Chapter 3

PROPERTIES OF PURE SUBSTANCES

Florio

Cengel 14F23

.


2/65

Objectives

Introduce the concept of a pure substance.

Discuss the physics of phase-change processes.

Illustrate the P-v, T-v, and P-T property diagrams and P-v-T

surfaces of pure substances.

Demonstrate the procedures for determining thermodynamic

properties of pure substances from tables of property data.

Describe the hypotheticalsubstance ideal gas and the

ideal-gasPvT equation of state.

Apply the ideal-gas equation of state in the solution of typical

problems.

Introduce the compressibility factor, which accounts for the

deviation of real gases from ideal-gas behavior.

Properties independent properties are measured and

property relations are used to obtain the dependentro erties

9/27/2014 Dr. Florio


3/65

Continuum -PURE SUBSTANCE

Two independent, intrinsic, intensive properties determine the state for a scs, i.e. dependent

,one of the properties cannot be varied while holding the other constant. In other words,

when one changes, so does the other one.

Pure substance: If during a process, a substance has a fixed and

homogeneous chemical composition throughout it is classified as a puresubstance.

Air is a mixture of several gases, but if composition is fixed andhomogeneous it is considered to be a pure substance.


Nitrogen and gaseous air are

pure substances.

A mixture of liquid and gaseous

water is a pure substance, but a

mixture of liquid and gaseous air

is not.


4/65

PHASES OF A PURE SUBSTANCE-molecular arrangement


The molecules

in a solid are

kept at their

positions by the

large spring-like

inter-molecular

forces. In a solid, the attractive andrepulsive forces between

the molecules tend to

maintain them at relatively

constant distances from

each other. Add thermal

energy, results in increaseof momentum eventually

sufficient momentum so a

group of molecules break

free.The arrangement of molecules in different phases: (a) molecules are at relatively

fixed positions in a solid, (b) groups of molecules move about each other in the

liquid phase, and (c) molecules move about at random in the gas phase.

Phase is a homogeneous part of a

system- Homogeneous in state as

well as composition andbounded by a well defined

surface. Phase of pure

substance with Negligible effect

of shear distortion, surface

tension effects , motion or field

effects.


5/65

P=P(T, v) A surface in P-T-v space

Molecular arrangement

No curvature for mixture

surface a ruled surface, but

curvature exists for single

phase , red, regions



6/65

PHASE-CHANGE PROCESSES OF PURE SUBSTANCES-examplefor

constant P process State 1. Compressed liquid (subcooled liquid):A liquid substance that it is not

about to vaporize with a small addition of heat. Water at 1 atm, 20 C, A-4 and 5

State 2. Saturated liquid: A liquid that a portion could be vaporized with asmall addition of heat with no change in T or P Pressure.


State 1-At 1 atm and

20C, water exists in

the liquid phase

(compressed l iqu id).

State 2-At 1 atm

pressure and 100C,

water exists as a liquid

that is ready to vaporize

(saturated l iqu id)-

terminology similar to

chemistry.

Measure

T, Vol,

Mass

fcl


7/65

State 3-Saturated liquidvapor mixture:The state in which the liquid and vapor phasescoexist in mutual equilibrium. A 4&5

State 4-Saturated vapor: A vapor for which a portion of it could condense with a smallamt of cooling.A 4&5

State 5-Superheated vapor: A vapor that is not about to condense (i.e., not a saturatedvapor). A-6


As more heat is transferred, part of the

saturated liquid vaporizes (saturated

l iquidvapor mixture). No change in

intensive state of each phasebut a

change in the extensive state ,

At 1 atm pressure, the

temperature remains constant

at 100C until the all of the

liquid is vaporized(saturated

vapor) or vapor for which a

Infinitesimal q from willcondense.

T>Tsat--As more heat

is transferred, the

temperature of the

vapor starts to

increase

(superheated vapor).

g

f

g

sh


8/65

T-v diagram for the

heating process of

water at constant

pressure.

Process is reversible- If the entire process between state 1 and 5described in the figure is reversed by slowly cooling the water whilemaintaining the pressure at the same value, the water can be returnedto state 1, retracing the same identical path, and in so doing, the

amount of heat released will exactly match the amount of heat addedduring the heating process.

Saturation P and T-

are P,T for which twophase coexist in

mutual equilibrium



9/65

Phases of a pure substance

Solid phase-Molecules have no rotational or translational motion, only vibrational

energy. Attractive force is large, group of molecules vibrate in their lattice

structure. As energy is added to the solid, the spacing usually change as the

molecules reposition in the solid and the vibrational energy increases. Eventually,

a group of molecules break free- liquid. As energy is added, the rotational energy

of that group increases , and additional groups break free. Liquid phase- the attractive force is less than a solid, and decrease as the spacing

between the molecules increases. The distances are greater than in solid and

increase as energy is added. Additional energy goes into rotational motion and

vibrational motion of the group. The rotational energy becomes sufficiently great

enough to overcome the intermolecular attractive force. The molecule breaks free

from other members of the liquid group and the new phase is called a vapor .

The conditions are favorable for this at a surface. Gas translational ke, rotational

ke and potential energy. Close repulsive, farther attractive



10/65

Phase Plots[measures relative to saturation-EX-Constant P eachpoint in a plane represents a state]

9/27/2014

v

gas

TP pressure lowest

P for liquid

Dr. Florio


11/65

9/27/2014

3 Dimensional- Each surface point

represents a unique state of the substance

Projections-view- state plots-

each point on line are

projections of mixture regions

Points off line are unique states

P=P(v,T) a 3D surface

Intersections with planes of const

T

Dr. Florio


12/65

Double interpolation

interpolation 1 property is fixed given P and v

V=v(T,P) v=vl+v/T]PdT+ v/P]TdP

9/27/2014

P1-

table

P2-Table

T

v

vx

px

vx

Dr. Florio


13/65

Know- Useful (Good to Know) Approximations (water)

Slope of the fusion line is -1F/(1000psi)= -1E-03 F/psi= -0.08K/1000kPa

Compressed Liquid (CL) tables not as extensive

Approximation, PPcand T>Tccall fluid a super critical fluid

IF PTccall fluid a gas

If gas and P


14/65

2 phases- Saturation Temperature and Saturation Pressure

The temperature at which water starts boiling depends on the pressure; therefore, if the

pressure is fixed, so is the boiling temperature. Psat =function (Tsat)

Water boils at 100C at 1 atm pressure.

Saturation temperature Tsat: The temperature at which a pure substance changes phase at agiven pressure. [Temperature at which two phases can coexist in mutual equilibrium.] Tsat

function2 (Psat)

Saturation pressure Psat: The pressureat which a pure substance changes phase at a given

temperature.


The liquid

vapor

saturationcurve of a

pure

substance

(numerical

values are for

water).

Psat=P(Tsat)

lnPsat=A-B/Tsat

Vaporization

line


15/65

Latent heat: The amount of energy absorbed or

released during a phase-change process, unit

mass sat phase 1 to sat phase 2 .

Latent heat of fusion: SolidLiquidThe amount

of energy absorbed during melting a unit of

mass . It is equivalent to the amount of energy

released during freezing.

Latent heat of vaporization: Liquid-Vapor The

amount of energy absorbed during

vaporization and it is equivalent to the energyreleased during condensation.

The magnitudes of the latent heats depend on the

temperature or pressure at which the phase change

occurs and are slowly varying functions of state.

At 1 atm pressure, the latent heat of fusion of water is

333.7 kJ/kg and the latent heat of vaporization is2256.5 kJ/kg.



16/65

PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES

The variations of properties during phase-change processes are best studiedand understood with the help of property diagrams such as the T-v, P-v, and

P-T diagrams for pure substances. Critical 374 C, 22 MPa


T-v diagram of

constant-pressurephase-change

processes of a pure

substance at various

pressures

(numerical values

are for water).


17/65

saturated liquid line

saturated vapor line

compressed liquid region

superheated vapor region saturated liquidvapor mixture

region (wet region)

Critical isotherm Pt of inflection -

2nd=0


At supercritical

pressures (P >Pcr),

there is no distinct

phase-change

(boiling) process.

T-v diagram of a pure substance.The difference between

properties of saturated liq and

vapor decrease until zero at ,

Critical point: The point at which

the saturated liquid and saturatedvapor states are identical.

gasSuper-criticalfluid

P


18/65

P-v diagram of a pure substance. The pressure in a pistoncylinder

device can be reduced by

reducing the weight of the piston.



19/65

Super

critical

Sat solid

Sat-liq



20/65

Extending the Diagrams

to Include

the Solid Phase


P-v diagram of a substance that

contracts on freezing. Vsolid> thus

density lessP-v diagram of a substance that

expands on freezing (such as water).

At triple-point pressureand temperature (a point

on a P-T plot), a

substance exists in three

phases in equilibrium.

For water,

Ttp= 0.01C

Ptp= 0.6117 kPa

min P for

liq

Sat

-liq


21/65

Sublimation: Passing from

the solid phase directly into

the vapor phase.

At low pressures (belowthe triple-point value),

solids evaporate without

melting first (sublimation).

P-T diagram of pure substances.

Vapor- Gas-compress at constant T

or cool at constant P Compare P and T

Phase Diagram- lines are

saturation lines

Gas

PTcr

Super-

critical

fluid


Liquid P>Psat(T)

T


22/65

P-v-T surface of a substance that

contracts on freezing like R(134 a).

P-v-T surface of a substance that

expands on freezing (like water). Or

vsatliq


23/65

PROPERTY TABLES For most substances, the relationships among thermodynamic properties are too

complex to be expressed by simple equations.

Therefore, properties are frequently presented in the form of tables.

Some thermodynamic properties can be measured easily, but u,h,s cannot and are

calculated by using the relations between them and measurable properties.

The results of these measurements and calculations are presented in tables in a

convenient format.


Enthalpy

A Combination Property (SI units)

The

combinationu +Pv is

frequently

encountered

in the analysis

of control

volumes.


24/65

Saturated Liquid and Saturated Vapor States

Table A-4: Saturation properties of water under temperature.

Table A-5: Saturation properties of water under pressure.


A partial list of Table A-4. Computer generated.

Enthalpy of vaporization, hfg(Latent

heat of vaporization):The amount of

energy needed to vaporize a unit mass

of saturated liquid at a giventemperature or pressure. T= 92.5 C

Psat=77.396;

vf= 0.001038

v=vl+v/T]PdT


25/65

Examples:Saturated liquid

and saturated

vapor states ofwater on T-vand

P-vdiagrams.



26/65

Saturated LiquidVapor Mixture


Quality, x : The ratio of the mass of vapor to the total mass of the mixture.

Quality is between 0 and 1 0: sat. liquid, 1: sat. vapor.

The properties of the saturated liquid are the same whether it exists alone or in

a mixture with saturated vapor.

The relative

amounts ofliquid and

vapor phases

in a saturated

mixture are

specified by

the qual ity x.

A two-phase system can be

treated as a homogeneous

mixture for convenience, wellmixed .

Temperature and

pressure are dependent

properties for a mixture.


27/65

Quality is relatedto the horizontal

distances on P-v

and T-v

diagrams.

The v value of a

saturated liquidvapor mixture lies

between the vfand vg

values at the

specified T or P.

y as- v, u, h or s.



28/65

Examples: Saturated liquid-vapormixture states on T-vand P-vdiagrams.

V=0.0617

m3/kg

X=0.4966


S h t d


29/65

Superheated VaporTEFFECTS GREATER THAN


In the region to the right of the

saturated vapor line and at

pressures below the critical

pressure, a substance exists as

gas, if the Temperature is less

than critical a superheatedvapor.

Bar=100kPa,Mpa=1000kPa

In this region, temperature and

pressure are independent

properties.

A partial

listing of

Table A-6.

At a specified

P, superheated

vapor exists at

a higher h than

the saturatedvapor.

Compared to saturated vapor,

superheated vapor is characterized by

T

Psat(T)

Tsat(P)


30/65

What is the Phase description? One yes then stop

Navigation of Tables - Questions to answer.

1. Is P> Psat (T) or T< Tsat (P) if yes CL

2. Is P < Psat (T) or T> Tsat (P) if yes SHV

3. Is P= Psat (T) if yes mixture

4. Is T= Tsat (P) if yes Mixture

y= u, h, s, v

5. Is y> yg (at T or P) if yes SH

6 Y


31/65

Compressed Liquid-T effects less than


Compressed liquid is characterized by

y v, u, or s

Example uuf(T)

A more accurate relation for h

A compressed liquid

may be approximated

as a saturated liquid at

the given temperature.

At a given P

and T, a puresubstance will

exist as a

compressed

liquid if

The compressed liquid properties

depend on temperature much more

strongly than they do on pressure.

u(T,v)

T=


32/65

The specific volume of water at a specified state is to be determined

using the incompressible liquid approximation and it is to be compared

to the more accurate value. U,s,v

Analys isThe state of water is compressed liquid. From the steam

tables, h =hf+vf(P-Psat (T) ) If P< 5 mPa use this method OR 500 psia

7)-A(TableC100

MPa5/kgm0.001041 3

v

T

P

Based upon the incompressible liquid approximation,

@ 100 Cv v (Table A-4)100 C fSat liquidT 30.001043 m /kgThe error involved is

0.19%

100001041.0

001041.0001043.0ErrorPercent

which is acceptable in most engineering calculations.



33/65

3-26


T, C P, kPa h, kJ / kg x Phase

description

200 0.7

140 1800

950 0.0

80 500 - - -

800 3162.2 - - -

Complete the following Table for water- keep the digits .


34/65

Example 1

9/27/2014 (stop) Dr. Florio

T, C P, kPa h, kJ / kg x Phase description

120.21 200 2045.8 0.7 Saturated mix ture

140 361.53 1800 0.565 Saturated mix ture

177.66 950 752.74 0.0 Saturated li quid

80 500 335.37 - - - Compressed l iqu id

350.0 800 3162.2 - - - Superheated vapor

Completed table for H2O:


35/65

Example


6E)-A(Table

/lbmft29.2

psia02513

1

1F550

T

P

v

H2O

250 psia

1 lbm

2.29 ft3

P

v

2

1

A rigid container that is filled with 1 lbm of water at 250 psia, 2.29 ft3 is cooled

to 100 F. The initial temperature and final pressure are to be determined.

AnalysisSince v= 2.29/1 since v > vgat 250 psia initial

state is superheated vapor. The temperature isdetermined to be

This is a constant volume cooling process (v= V/m= constant).

Since at the final state vf


36/65

1 sat@98 C 94.39 kPabtP P

C99.0 [email protected] TT

40 cm

5 cm

g p p p g

The boiling temperature in a 40-cm deep pan in the same atm is to be determined.

Assumpt ionsBoth pans are full of water and . density of liquid water

is approximately = 1000 kg/m3.AnalysisThe pressure at the bottom where it is boiling of the 5-cm pan is the

saturation pressure

corresponding to the boiling temperature of 98C:Table A-4)

The pressure difference between the bottoms of two

pans is

Then the pressure at the bottom of the 40-cm deep pan is

P= 94.39 + 3.43 = 97.82 kPa

Then the boiling temperature becomes

(Table A-4-5)

1 1 2 2 1 1

1 1

2 2

94.39 0.49035 93.9

( ) 93.9 .4*9.807 97.8

bt atm bt atm atm bt

atm bt

bt

P P gh P P gh P P gh

P P gh kPa

P Patm g h kPa

( )( ) / ( )l h l l h l P P P P T T T T

( )( ) / ( )l h l l h l

T T T T P P P P



37/65

atm atm

2

2 2

given 100(20 kg)(9.81 m/s ) 1 kPa

(100 kPa)0.01 m 1000 kg/m s

119.61 kPa

mg

P P P kPaA

C104.7 kPa119.61@satTT

W = mg

Patm

P

51A vertical piston-cylinder device is filled with water and covered with a 20-kg piston

that serves as the lid. The boiling temperature of water is to be determined.

AnalysisThe pressure in the cylinder is determined from a force balance on the piston,

PA = PatmA + W

or,

The boiling temperature is the saturation temperature corresponding to this pressure,

(Table A-5)


43


38/65

43


/kgm12322.0kg100m322.12 3

3

1 mVv

R-134a

200 kPa100 kg

12.322 m3

P

v

2 1

A piston-cylinder device that is filled with 100 kg of R-134a at 200 kP

and a volume of 12.322 m3is cooled at constant pressure to half its

original volume.

The final temperature and the change of total internal energy are to

be determined.

AnalysisThe initial specificvolume is

The initial state( v > vg ) is superheated and the internal energy at

this state is

13)-A(TablekJ/kg08.263/kgm12322.0

kPa20013

1

1

u

P

v

The final specific volume is

/kgm06161.02

/m12322.0

2

33

12

kgvv

This is a constant pressure process. The final state is determined to be saturated mixture whose temperature is

12)-A(TablekPa200@sat2 C10.09 TT

The internal energy at the final state is (Table A-12)

kJ/kg61.152)21.186)(6140.0(28.38

6140.0/kgm)0007533.0099867.0(

/kgm)0007533.006161.0(

22

3

32

2

fgf

fg

f

uxuu

xv

vv

Hence, the change in the internal energy is

2 1 152.61 263.08u u u

110.47 kJ/kg

h = -110.47-12.322 = -122.79 kJ /kg

Vf


39/65

3-47


2 2

evap

evap

evap

evap

( / 4) [ (0.25 m) / 4](0.10 m)4.704 kg

0.001043

Assuming steady rate

4.704 kg0.001742 kg/s

45 60 s

f f

D Lm

mm

t

V

v v H2O

1 atm

Water is boiled at 1 atm pressure in a 25 cm pan placed on an electric burner. The liquid

water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water

is to be determined.

Propert ies Neglecting hydrostatic Pressure change, and assuming constant rate The

properties of water at 1 atm and thus at a saturation temperature of Tsat= 100C are hfg=2256.5 kJ/kg and vf = 0.001043 m

3/kg (Table A-4).

Analysis From Conservation of mass, the constant rate of evaporation of water is the

rate of loss of liquid

Since the amount of heat required to change the phase of a

unit mass from saturated liquid to saturated vapor at the

same temperature is hfg the rate of heat transfer to water

becomes thus

cv

evap

Q-W= dU/dt +mh:

is a constant and m for liquid system -is related to dM/dt = -m

m where = h

(0.001742 kg/s)(2256.5 kJ

g

f f out g

out f out

f f out f f

fg

CV is the liquidQ PVOL U m h

PVOL U h h

Q m h

/kg)3.93 kW

evap fgQ m h


40/65

59


2 sat@1 MPaT T 179.88 CH2O

300C

1 MPa

T

v

2

1

0.8 kg of Superheated steam at 300 C, 1MPa is contained in a piston-cylinder

device is cooled at constant pressure until half of the mass condenses. i.e.

Moisture is 0.5 . The final temperature and the volume change are to be

determined, and the process should be shown on a T-vdiagram.

Analysis(b) System is a fms. At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the

final pressure, You are given the moisture and Pressure final state

(Table A-5)

(c) The quality at the final state is specified to be 1-x2= 0.5. The specific

volumes at the initial and the final states are

/kgm0.25799C300

MPa1.0 31

1

1

v

T

P (Table A-6)

/kgm.097750

)001127.019436.0(5.0001127.05.0

MPa1.0

3

222

2

fgf xx

Pvvv

Thus,

3

2 1

3

( ) (0.8 kg)(0.09775 0.25799) m /kg

or the decrease in volume is 0.128.m .

m 30.1282 mV v v

Heat is lost from a piston-cylinder device that contains steam at a specified state of 3.5 MPa, 5


41/65

kJ/kg1.2821C6.247

MPa5.31

1

1

h

T

P Steam3.5 MPa Q

Heat is lost from a piston cylinder device that contains steam at a specified state of 3.5 MPa, 5

deg superheat. The heat is lost until the piston hits the stops , at which point the fluid is

saturated liquid. Cooling continuesuntil the temperature reaches 200 C. The initial

temperature, the enthalpy change, and the final pressure and quality are to be determined.

Analysis(a) The saturation temperature of steam at 3.5 MPa is

[email protected] MPa = 242.6C (Table A-5)

Then, Since degrees superheat is T- Tsat,

the initial temperature becomes

T1 = 242.6+5 = 247.6C

(Table A-6)

(b) The properties of steam when the piston

first hits the stops are

22 1 32 21049.7 kJ/kg3.5 MPa

sat liq 0( ) 0.001235 m /kgf

f

h hP Px all liquid v v

(Table A-5)

Then, the enthalpy change of steam becomes

kJ/kg-1771 2821.17.104912 hhh

(c) At the final state (Table A-4 )3

3 23 3

3 3

3 1

0.001235 m /kg200 C saturated

h 852.4 /1968.7 /

v vT P

kJ kg xh h kJ kg

1555 kPa0.0006

(Table A-4 )

The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.

Const P process + constant V process.

T

v

2

1

(TTL)/(THTL)]p

=(h- hL)/(hHhL)p


R f St t d R f V l


42/65

Reference State and Reference Values The values of u, h, and s cannot be measured directly, and they are calculated from measurable

properties using the relations between properties.

However, those relations give the changes in properties, not the values of properties at specifiedstates.

Therefore, we need to choose a convenient reference stateand assign a value of zerofor aconvenient property or properties at that state.

The reference state for water is saturate liquid at 0.01C and differs for other substances or tables.

Some properties may have negative values as a result of the reference state chosen.

Sometimes different tables list different values for some properties at the same state as a result ofusing a different reference state.

However, In thermodynamics we are concerned with the changes in properties, and the reference

state chosen is of no consequence in calculations.



43/65

P-V-T relation for idealgases-modeling equations


T1

T2

T3

p

Tvp

0

[ ]*lim up

PvR AGN Boltzman

T

1. Negligible molecular PE

2. Negligible attractive or repulsiveforces

3. Negligible volume of molecule

4. perfectly elastic collisions

5. Space between molecules is large

relative to the volume of the

molecule

6. State due to molecular kinetics,

translation, rotation and vibration.-

on the molecular-electron level


44/65

The universal gas constant


*

*

8.314 /

1545 /

u

MW AVGN mass of molecule

R R MW

kJ kmol K

ft lbf lbmol R

Mass= MW*NMW = Molecular weight, kg/kmol

or lbm/lbmol.


45/65

THE IDEAL-GAS EQUATION OF STATE Equation of state: An equation that relates thermodynamic equilibrium

properties. Also commonly used relative to an equation that relates the

pressure, temperature, and specific volume of a substance. The simplest and best-known equation of state for substances in the gas phase isthe ideal-gas equation of state. This equation predicts the P-v-T behavior of a gasquite accurately within some properly selected region, but notall gas states.


R: gas constant

M or MW: [molar mass =AVGADOs *

mass molecule](kg/kmol)

Ru: universal gas constant

Ideal gas equation

of state

Different substances have different

gas constants.


46/65

Properties per

unit mole are

denoted with a

bar on the top.

The ideal-gas

relation often is

not applicable to

real gases; thus,care should be

exercised when

using it.

Mass = Molal mass Mole number

Variousexpressions

of ideal gas

equation

Ideal gas equation at two

states for a fixed mass

Real gases

behave as an ideal

gas at low

densities (i.e., low

pressure, high

temperature).

y

Nv mv Volv MW v

y MW



47/65

The ideal gas state surface


I deal Gas State PVT Sur face

p

VT


48/65

Specific heats of the ideal gas


kinetic + inter molecular Potential

/ | [ / | ] (molecular PE)

( )

( ) or h(T)

v v

v

v

p v

p

du u T dT T P T P dv

u u T

u duc

T dT

h u T RT

h dhc c R

T dT

I t l d th l f th


49/65

Internal energy and enthalpy of the

ideal gas


2

1

2

1

2 1

2 1

Remember values depend on table referencefor u,h,s but differences do NOT

Mix tables use- ex. one table with differentrefernce states and values.A change taken from

T

vTT

pT

u T u T c dT

h T h T c dT

3 1

2 1 3 2 3 1

a single tableis independent of reference

h to connect ig ;common, connecting state 2 must be an igand steam table values

[ ] [ ]tablesteam tableig

h h

h h h h h h h

Water Vapor behaving as an Ideal Gas


50/65

Water Vapor behaving as an Ideal Gas

At vapor pressures below 10 kPa,

water vapor can be treated as an ideal

gas, regardless of its temperature,

with negligible error (less than 0.1percent)-

At higher pressures, however, the

ideal gas assumption yields

unacceptable errors, particularly in

the vicinity of the critical point and

the saturated vapor line.

In steam power plant applications,

however, the pressures involved are

usually very high; therefore, ideal-gas

relations should not be used.


Percentage of error ([|vtable- videal|/vtable] 100) involved in

assuming steam to be an ideal gas, and the region where steam

can be treated as an ideal gas with less than 1 percent error.

1-1/Z

3 78


51/65

3-78


kg5.846K)K)(298/kgmkPa(0.287

)mkPa)(1.0(500

kPa200

K)K)(308/kgmkPakg)(0.287(5

3

3

1

1

3

1

11

A

A

B

B

RT

P

m

P

RTm

V

V 3

m2.21

Air

V = 1 m3

T= 25C

P= 500 kPa

Air

m= 5 kg

T= 35C

P= 200 kPa

A B

Two rigid tanks connected by a valve to each other contain air at specified conditions.

Assuming a final equilibrium temperature of 20 C, the volume of the second tank and

the final equilibrium pressure when the valve is opened are to be determined.

Assumpt ions At specified conditions, air behaves as an ideal gas.

Propert ies The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).

AnalysisCalling first and the second tanks A and B. Treating air as an ideal gas, the

volume of the second tank and the mass of air in the first tank are determined to be

Thus,

kg10.8465.05.846

m3.212.211.03

BA

BA

mmm

VVV

Then the final equilibrium pressure becomes

kPa284.1m3.21

K)K)(293/kgmkPakg)(0.287(10.8463

32

2 V

mRTP


52/65

Compressibility factor-real gas

Not all fluids have a complete set of property valuesthat are tabulated for all states.

It was found experimentally that if for two fluids theratio of two independent properties such as P and T

are the same at two states, then the ratio of a thirdproperty such as specific volume were approximatelythe same. This led to the law of corresponding states.And the use of the new properties of reduced P ,

Temperature and pseudo specific volume with thereference state the critical state, the properties beingcalled reduced properties.



53/65

COMPRESSIBILITY FACTORA MEASURE

OF DEVIATION FROM IDEAL-GAS BEHAVIOR


The compressibility factor is

unity for ideal gases.

A New property-

Compressibility factor Z

A factor that accounts forthe deviation of real gases

from ideal-gas behavior at

a given temperature and

pressure.

The farther away Z is from unity, the more

the gas deviates from ideal-gas behavior.

Gases behave as an ideal gas at low densities

(i.e., low pressure, high temperature).

The temperature (or pressure ) of a gas is high

(or low relative) to its critical temperature (or

pressure). If a gas and Pr


54/65

Real Gases- Search for a Universal equation of state-

reduce variable range


P

1

Z

T1

T2

T3

T1< T2< T3I deal Gas

TRvp


55/65


56/65

Law of Corresponding states Ideal gas equation:

1. Molecule volume is negligible.

2. Attractive force negligible.

3. Repulsive force negligible

4. Perfectly elastic collisions.

5. Energy solely associated with kinetics on molecular scale

But- Attractive forces tend to decrease volume, repulsive tends to increase

Z= vreal /v ideal at same P,T= Pr vr/Tr=Z(Pr,Tr , Zcr)

All gases are not the same due to different molecular force fields moleculararrangements.

Law of corresponding states- States of a gas dependent on force fields aroundmolecules, for corresponding states, similar molecular force fields, similarequations of state.

Thus, for two gases their states correspond to one another if two of ratio of theirPressures and Temperatures or specific volumes are the same as thecorresponding ratios at their critical states T2/T1=T2c/T1c


Water Vapor does not always behave as an Ideal Gas.Error< 1 % if Pr800 C


57/65

1atm or P< 66M Pa andT>800 C


Error< 1 % if Pr


58/65

Comparison of Z factors for various gases, experimental.

Gases deviate from the

ideal-gas behavior the

most in the neighborhood

of the critical point.

Reduced

temperature

Reduced

pressure

Pseudo-reduced

specific volume

Zcan also be determined from

a knowledge of PRand vR.

Fig A-15

Z(Pr,Tr , Zcr)



59/65



60/65



61/65

98


)error%4.86(kPa10,000

K)K)(150/kgmkPa(0.29683

/kgm0.004452 3

P

RTv

N2

10 MPa

150 K

The specific volume of nitrogen gas is to be determined using the ideal gas relation and the

compressibility chart. The errors involved in these two approaches are also to be determined.

Propert iesThe gas constant, the critical pressure, and the critical temperature of nitrogen are, from

Table A-1,

R= 0.2968 kPam3/kgK, Tcr= 126.2 K, Pcr= 3.39 MPa

Analysis(a) From Nitrogen tables v= 0.00245 m3/kg b. From the ideal gas equation of state,

(c) From the compressibility chart (Fig. A-15),

54.0

1.19

K126.2

K150

2.95MPa3.39

MPa10

Z

T

TT

P

PP

cr

R

cr

R

Thus,

error)(0.7%/kg)m04452(0.54)(0.0 3ideal /kgm0.002404 3 vv Z

actualR

cr

cr

vv

RT

P

OTHER EQUATIONS


62/65

OTHER EQUATIONS

OF STATE


Several equations have been proposed torepresent the P-v-T behavior of substances

accurately over a larger region with no

limitations.

Van der Waals

Equation of State Critical isothermof a pure

substance has

an inflection

point at the

critical state.

This model includes two effects not considered

in the ideal-gas model:the intermolecular

attraction forces and the volume occupied by the

molecules themselves. The accuracy of the van

der Waals equation of state is often inadequate.

Beattie Bridgeman Equation of State


63/65

Beattie-Bridgeman Equation of State

It is known to be reasonably

accurate for densities up to

about 0.8cr.

Benedict-Webb-Rubin Equation of State

The constants are given in Table 3-4. This equation can handle substances

at densities up to about 2.5 cr.

Virial Equation of State

The coefficients a(T), b(T), c(T), and so on, that are

functions of temperature alone are called virial coefficients.



64/65

Complex equations of

state represent the P-v-

T behavior of gasesmore accurately over a

wider range.Percentage of error involved in various equations of

state for nitrogen

(% error = [(|vtable- vequation|)/vtable] 100).


Summary


65/65

Summary Pure substance Phases of a pure substance

Phase-change processes of pure substances Compressed liquid, Saturated liquid, Saturated vapor,

Superheated vapor Saturation temperature and Saturation pressure

Property diagrams for phase change processes The T-vdiagram, The P-vdiagram, The P-Tdiagram, The P-v-

Tsurface Property tables

Enthalpy Saturated liquid, saturated vapor, Saturated liquid vapor

mixture, Superheated vapor, compressed liquid Reference state and reference values

The ideal gas equation of state Is water vapor an ideal gas?

Compressibility factor Other equations of state

Documents

Chapter3notes ME311 PJF F14