Properties of Pure

Chapter 2

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Properties of PureSubstances

OBJECTIVES

At the end of this chapter students should be able to:

Define the relationship between severalthermodynamics properties.

Draw diagram related to various processes.

Define the phase changing process for pure

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Define the phase changing process for puresubstances.

Use the thermodynamics property table effectively.

Define ideal gases.

Use equations related to ideal gases.

Pure Substance

A substance that has a fixed chemical compositionthroughout.

Examples are water, nitrogen, helium, carbondioxide.

Does not have to be a single chemical composition.

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Does not have to be a single chemical composition.

A mixture of various chemical elements orcompounds also qualifies as pure substances, i.e air.

A mixture of two or more phases of pure substance istill a pure substance as long as the chemicalcomposition of all phases is the same, i.e. water existas ice, liquid water and steam.

Phases of Pure Substances• Three principal phases-solid, liquid and gas

• The arrangement of atoms in different phases:(a) molecules are at relatively fixed positions in asolid,(b) groups of molecules move about each other inthe liquid phase, and(c) molecules move about at random in the gas

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(c) molecules move about at random in the gasphase.

Solid• Three dimensional pattern

• Large attractive forces between atoms ormolecules

• The atoms or molecules are in constant motion –they vibrate in place

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• The higher the temperature – the more vibration

• At sufficiently high temperatures, the velocity(and thus the momentum) of the molecules mayreach a point where the intermolecular forces arepartially overcome and groups of moleculesbreak away

Liquid

• The molecular spacing is not much different fromthat of the solid phase, except the molecules areno longer at fixed positions and they can rotateand translate freely.

• In a liquid, the intermolecular forces are weaker

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• In a liquid, the intermolecular forces are weakerrelative to solids, but still relatively strongcompared to gases.

• The distances between molecules generallyexperience a slight increase as a solid turnsliquid.

Gas

• Molecules are far apart

• High kinetic energy

• Molecules are collide to each others

• Must release a large amount of its energy

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• Must release a large amount of its energybefore it can condense or freeze.

Solid to Liquid to Gas

• On a molecular level, the differencebetween the phases is really a matter oftemperature different

• We identify melting points and

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• We identify melting points andvaporization points based on changes inproperties

– e.g. – big change in specific volume

Consider what happens whenwe heat water at constant

pressure

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Piston cylinderdevice –maintainsconstantpressure

Liquid Water

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T

v

1

2

5

3 4

PHASES CHANGE PROCESSES OF PURE SUBSTANCES:

Compressed liquid (subcooled liquid) meaning liquid that is notabout to vaporize.

Saturated liquid is liquid that is about to vaporize.

Saturated liquid-vapor mixture (saturated mixture) when theliquid and vapor phases coexist in equilibrium.

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P = 1 atmT =20oC

Fig. 2.6At 1 atm and 20oC,

water exists in liquidphase (compressed

liquid)

Heat

P = 1 atmT =100oC

Fig. 2.7At 1 atm and 100oC,water exists as a liquidthat is ready to vaporize(saturated liquid)

Heat

P = 1 atmT =100oC

Fig. 2.8As more heat istransferred, part of thesaturated liquidvaporizes (saturatedliquid-vapor mixture)

Heat

PHASES CHANGE PROCESSES OF PURE SUBSTANCES:

Saturated vapor is vapor that is about to condense.

Superheated vapor is vapor that not about to condense.

The amount of energy absorbed or released during a phase-change process is called the latent heat.

P = 1 atmT =100oC

P = 1 atmT =300oC

5SaturatedMixture

SuperheatedVapor

P = 1 atm

P

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T =100 C

Fig. 2.9At 1 atm pressure,the temperature

remains constant at100oCC until the last

drop of liquid isvaporized (saturated

vapor)

Heat

Fig. 2.10As more heat istransferred, the

temperature of thevapor starts to rise

(superheated vapor)

Heat 1

2 3 4CompressedLiquid

Fig. 2-11: T-v diagram for the heating process of water at constantpressure

v

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Compressed

SuperheatedGas

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Two PhaseRegion

CompressedLiquid

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Critical Point

Critical Point is the point at which the saturated liquidand saturated vapor states are identical.

The temperature, pressure and specific volume atcritical point are called, respectively, criticaltemperature T , critical pressure P and critical

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temperature Tcr, critical pressure Pcr and criticalspecific volume vcr.

For water Pcr=22.06 MPa, Tcr=373.95oC,vcr=0.003106 m3/kg

Above the critical point there is not a distinct phasechange process.

P-v diagram of a pure substance.

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The pressure in a piston–cylinderdevice can be reduced by

reducing the weight of the piston.

P-v Diagram of a Substance that Contracts onFreezing

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P-v Diagram of a Substance that Expands onFreezing

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Phase Diagrams

• So far we have sketched

– T – v diagram

– P – v diagram

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– What about the P – T diagram?

- This diagram is often called the phasediagram since all three phases areseparated from each other by three lines.

P-T Diagram of a Pure Substance

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Contracts on Freezing Expands on Freezing

Combine all three

• We can put all three properties – P,V & T on the samediagram

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The P-v-T Surface• The state of a simple compressible substance is fixed

by any two independent, intensive properties.

• Once the two appropriate properties are fixed, all theother properties become dependent properties. HereT and v may be viewed as the independent variables(the base) and P as the dependent variable (the

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(the base) and P as the dependent variable (theheight).

• All the points on the surface represent equilibriumstates.

• All states along the path of a quasi-equilibriumprocess lie on the P-v-T surface since such a processmust pass through equilibrium states.

Property Tables

• P - pressure

• T - temperature

• v – specific volume

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• v – specific volume

• u – specific internal energy

• h – specific enthalpy h = u + Pv

• s – specific entropy (define in Chapter 6)

Enthalpy – A Combination Property

• In the analysis of certain types of processes,particularly in power generation and refrigeration, wefrequently encounter the combination of properties

u + Pv.

• For the sake of simplicity and convenience, thiscombination is defined as a new property, enthalpy,and given the symbol H:

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and given the symbol H:

H = U + PV (kJ)

or per unit mass,

h = u + Pv (kJ/kg)

The combination u + Pv is frequentlyencountered in the analysis of control volumes.

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Saturated Liquid and Saturated Vapor States

Saturation Properties

• Saturation Pressure is the pressure at whichthe liquid and vapor phases are in equilibriumat a given temperature.

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• Saturation Temperature is the temperature atwhich the liquid and vapor phases are inequilibrium at a given pressure.

Table A-4 and A-5

• A-4 - pg 916

– Saturated water temperature table

• A-5 - pg 918

– Saturated water pressure table

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– Saturated water pressure table

u u u

h h h

s s s

fg g f

fg g f

fg g f

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g stands for gas

f stands for fluid

fg stands for the difference between gas and fluid

Enthalpy of vaporization, hfg (Latent heat of vaporization):The amount of energy needed to vaporize a unit mass ofsaturated liquid at a given temperature or pressure.

Examples: Saturated liquid and saturated vapor states of wateron T-v and P-v diagrams.

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Example 2-1 Pressure of Saturated Liquid in a Tank

A rigid tank contains 50 kg of saturated liquid water at 90ºC.Determine the pressure in the tank and the volume of the tank.

Solution

Since saturation conditions exist in the tank, the pressure must bethe saturation pressure at 90ºC:

P = Psat@90ºC = 70.14 kPa (Table A-4)

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The specific volume of the saturated liquid at 90ºC is

v = vf@90ºC = 0.001036 m3/kg (Table A-4)

The total volume of the tank is

v = mv = (50 kg)(0.001036 m3/kg) = 0.0518 m3

Saturated Liquid-Vapour Mixture

Saturated Mixture

During a vaporization process, a substance exists aspart liquid and part vapor.

To analyze this mixture requires the proportions ofliquid and vapor phases in the mixture.

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liquid and vapor phases in the mixture.

A new property known as quality, x, is defined as theratio of mass of vapor to the total mass of themixture

x = (mvapor)/(mtotal)

mtotal = mliquid + mvapor

Consider a tank that contains a saturated liquid-vapour mixture.The volume occupied by saturated liquid is Vf, and the volumeoccupied by saturated vapour is Vg. The total volume V is thesum of the two:

V = Vf + Vg where V = mv

mtvav= mfvf+ mgvg

mt = mf + mg

mtvav= (mt – mg )vf+ mgvg

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Dividing by mt,

vav= (1 – x )vf+ xvg where x= mg/mt

vav= vf+x vfg

x = (vav – vf)/vfg

The analysis can be repeated for internal energy and enthalpy,

uav = uf + xufg hav = hf + xhfg

Summarized in a single equation as

yav = yf + xyfg

where y is v, u, or h.

The values of the average properties of the mixtures are alwaysbetween the values of the saturated liquid and the saturated vapourproperties, yf yav yg

P or TSat. vapor

P or T

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A C

vf vg v

P or T

ABX = ----

AC

vfg

vav - vf

Fig. 2-41 Quality is related to the horizontal distances on P-vand T-v diagrams.

vvf < v < vg

vf vg

Sat. vaporvg

Sat. liquidvf

Fig. 2-41 The v value of a saturated liquid-vapor mixture liesbetween the vf and vg values at the specified T or P.

v

B

Fraction of the material that is gas

x = 0 the material is all saturated liquid

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x = 0 the material is all saturated liquid

x = 1 the material is all saturated vapor

x is not meaningful when you are out of thesaturation region

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X = 0 X = 1

y y x y y

y x y

f g f

f fg

( )01

= yg

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When x = 0 we have all liquid, and y = yf

When x = 1 we have all gas, and y = yf + yfg = yg

Example 2-3 Pressure and Volume of Saturated Mixture

A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water isin liquid form and the rest is in the vapour form, determine

(a) the pressure in the tank and

(b) the volume of the tank

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Example 2-4 Properties of Saturated Liquid-Vapour Mixture

An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of160 kPa. Determine

(a) the temperature of the refrigerant

(b) the quality

(c) the enthalpy of the refrigerant

(d) the volume occupied by the vapour phase

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Superheated Vapor can be characterized by:

Lower Pressures (P<Psat at a given T)

Higher Temperatures (T>Tsat at a given P)

Higher specific volume (v>vg at a given P or T)

Higher internal energy (u>ug at a given P or T)

Superheated Vapour

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g

Higher enthalpies (h>hg at a given P or T)

Example 2-5

Determine the temperature of water at a state of P=0.5 MPa and

h=2890 kJ/kg.

Superheated Water Properties

Table A-6

Page 920

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Compressed Liquid can be characterized by:

Higher Pressures (P>Psat at a given T)

Lower Temperatures (T<Tsat at a given P)

Lower specific volume (v<vf at a given P or T)

Lower internal energy (u<uf at a given P or T)

Lower enthalpies (h<hf at a given P or T)

Compressed Liquid

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Lower enthalpies (h<hf at a given P or T)

In the absence of compressed liquid data, a generalapproximation is to treat compressed liquid as saturated liquidat the given temperature.

Compressed liquid water table – Table A-7 (page 924)

Example 2-5

Determine the internal energy of compressed liquid water at

80oC and 5 MPa

T, oC

5MPa

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80oC

u

5MPa

T=80oCP=5MPa

Solution:

At 80°C, the saturation pressure of water is 47.39 kPa, and since 5MPa > Psat, we obviously have compressed liquid,

(a) From the compressed liquid table (Table A-7)

P = 5 MPa, T = 80°C

u = 333.82 kJ/kg

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(b) From the saturation table (Table A-4),

u uf@80°C = 334.97 kJ/kg

The error involved is

(334.82 – 333.97)/333.82 x 100 = 0.25%

which is less than 1 percent.

Linear Interpolation

A B

100 5

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150 10

120 X

510

5

100150

100120

x

Example 2-6

The Use of Steam Tables to Determine Properties

Determine the missing properties and the phase descriptions in thefollowing table for water:

T,°C P, kPa u,kJ/kg x PhaseDescription

(a) 200 0.6

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(b) 125 1600

(c) 1000 2950

(d) 75 500

(e) 850 0.0

Ideal Gas Law• Any equation that relates the pressure,

temperature and specific volume of asubstances is called an equation of state.

• The simplest equation of state for substances inthe gas phase is the ideal-gas equation ofstate, Pv = RT also known as ideal-gas law.

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state, Pv = RT also known as ideal-gas law.

• Gas constant, R is different for each gas and isdetermined from R=Ru/M.

- where Ru= universal gas constant (kJ/kmol.K)

M = molar mass (kg/kmol)

-Tabulated in the back of the book ( page 910)

Ideal Gas Law

• Ideal-gas equation of state can be written inseveral forms:- V = mv PV=mRT- mR=(MN)R = NR PV=NR T

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- mR=(MN)R = NRu PV=NRuT- V = Nv Pv=RuTwhere N = mole number,v=molar specific volume

• Pv=RT is the form that always be used.

When does the ideal gaslaw apply?

• PR and TR are called the reduced pressure andreduced temperature respectively.

PR = P/PCR and TR = T/TCR (Critical point –pg 910)

• At very low pressures (PR<<1), gases behave

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• At very low pressures (PR<<1), gases behaveas an ideal gas regardless of temperature.

• At high temperature (TR>2), ideal-gas behaviourcan be assumed with good accuracy regardlessof pressure (except when PR >>1).

• The deviation of a gas from ideal-gas behaviouris greatest in the vicinity of the critical point.

Is water Vapor an Ideal Gas

Air-conditioning=>water vapor in theair can be treated as

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air can be treated asan ideal gas

Steam power plant=>ideal gas relationsshould not be used

Ideal Gas Law

PV PV

• For a fixed mass, the properties of an ideal gasat two different states are related to each otherby

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PV

R T

PV

R T1 1

1

2 2

2

Example 2-7

Determine the mass of the air in a room whose dimensions are 4 mx 5 m x 6 m at 100 kPa and 25°C

Solution:

The gas constant of air is R = 0.287 kPa.m3/kg K (Table A-1)pg 910

T = 25°C + 273 = 298 K

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V = (4 m)(5 m)(6 m) = 120 m3

m = PV/RT = (100 kPa)(120 m3)/(0.287 kPa.m3/kg K)(298 K)

= 140.3 kg

Example 2-8

An automobile tire with a volume of 0.6 m3 is inflated to a gagepressure of 200 kPa. Calculate the mass of air in the tire if thetemperature is 20°C.

Solution:

Patm = 100 kPa

P = 200 + 100 = 300 kPa

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P = 200 + 100 = 300 kPa

T = 20 + 273 = 293 K

m = PV/RT = (300 000 N/m2)(0.6 m3)/(287 Nm/kg K)(293 K)

= 2.14 kg

Example 2.9

The pressure in an automobile tire dependson the temperature of the air in the tire. Whenthe air temperature is 25oC, the pressuregage reads 210 kPa. If the volume of the tireis 0.025 m3, determine the pressure rise in

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is 0.025 m3, determine the pressure rise inthe tire when the air temperature in the tirerises to 50oC. Also, determine the amount ofair that must be bled off to restore pressure toits original value at this temperature. Assumethe atmospheric pressure is 100kPa.

SUMMARY

• Definition for pure substance, compressed liquid,superheated vapour, saturation temperature andsaturation pressure

• Definiton for quality

• Reading properties using thermodynamics tables

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• Reading properties using thermodynamics tables

• Ideal-gas equation of states

END OF CHAPTER 2

THANK YOU

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