CHAPTER 3Volumetric Properties of

Pure Fluids

Miss. Rahimah Bt. Othman

Email: rahimah@unimap.edu.my

ERT 206/4ERT 206/4ThermodynamicsThermodynamics

COURSE OUTCOME 1 CO1)COURSE OUTCOME 1 CO1)

1.Chapter 1: Introduction to Thermodynamics

2.Chapter 2: The First Law and Other Basic Concepts

2.Chapter 3: Volumetric properties of pure fluids

DESCRIBE and EXPLAIN PVT behavior of pure substances, Virial Equation of State, Ideal Gas, Virial Equation- APPLICATION, Cubic Equation of State, Generalized Correlations for gases and liquids.

4. Chapter 4: Heat effects

5. Chapter 5: Second law of thermodynamics

6. Chapter 6: Thermodynamics properties of fluids

A substance that has a fixed chemical composition throughout is called a Pure Substance.

Pure Substance: - N2, O2, gaseous Air -A mixture of liquid and gaseous

water is a pure substance, but a mixture of liquid and gaseous air is not.

A substance that has a fixed chemical composition throughout is called a Pure Substance.

Pure Substance: - N2, O2, gaseous Air -A mixture of liquid and gaseous

water is a pure substance, but a mixture of liquid and gaseous air is not.

The molecules in a SOLID are kept at their positions by the large spring like inter-molecular forces.

In a solid, the attractive and

repulsive forces between the

molecules tend to maintain them at

relatively constant distances from

each other.

The arrangement of atoms 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.

3. Phase-Change Processes of 3. Phase-Change Processes of Pure SubstancePure Substance

3. Phase-Change Processes of 3. Phase-Change Processes of Pure SubstancePure Substance

Compressed liquid or a subcooled liquid: A liquid that is not about to vaporize.

Saturated liquid: A liquid that is about to vaporize.

Saturated vapor: A vapor that is about to condense.

Saturated liquid-vapor mixture: the liquid and vapor phases coexist in equilibrium.

Superheated vapor: A vapor that is not about to condense

Compressed liquid or a subcooled liquid: A liquid that is not about to vaporize.

Saturated liquid: A liquid that is about to vaporize.

Saturated vapor: A vapor that is about to condense.

Saturated liquid-vapor mixture: the liquid and vapor phases coexist in equilibrium.

Superheated vapor: A vapor that is not about to condense

3. Phase-Change Processes of Pure 3. Phase-Change Processes of Pure SubstanceSubstance

3. Phase-Change Processes of Pure 3. Phase-Change Processes of Pure SubstanceSubstance

Saturated temperature, Tsat: At a given pressure, the temperature at which a pure substance changes phase.

Saturated pressure, Psat: At a given temperature, the pressure at which a pure substance changes phase.

Latent heat: the amount of energy absorbed or released during a phase-change process.

Latent heat of fusion: the amount of energy absorbed during melting.

Latent heat of vaporization: the amount of energy absorbed during vaporization.

Saturated temperature, Tsat: At a given pressure, the temperature at which a pure substance changes phase.

Saturated pressure, Psat: At a given temperature, the pressure at which a pure substance changes phase.

Latent heat: the amount of energy absorbed or released during a phase-change process.

Latent heat of fusion: the amount of energy absorbed during melting.

Latent heat of vaporization: the amount of energy absorbed during vaporization.

2-1

T-v diagram for the heating process of water at constant pressure.

PVT BEHAVIOR OF PURE SUBSTANCES/ FLUIDSPVT BEHAVIOR OF PURE SUBSTANCES/ FLUIDS

The 1-2 line, also known as sublimation curve is where solid-vapor is in equilibrium

The 2-C line, also known as vaporization curve is where liquid-vapor is in equilibrium

The 2-3 line, also known as fusion curve is where solid-liquid is in equilibrium

Triple point, three phases exist in equilibrium (F=0)

Critical point – highest combination of pressure and temperature where the fluid exist in liq-vap equilibrium

3. 3. Property Diagrams for Phase-Change ProcessesProperty Diagrams for Phase-Change Processes3. 3. Property Diagrams for Phase-Change ProcessesProperty Diagrams for Phase-Change Processes

3. 3. Property Diagrams for Phase-Change ProcessesProperty Diagrams for Phase-Change Processes3. 3. Property Diagrams for Phase-Change ProcessesProperty Diagrams for Phase-Change Processes

4. 4. Property TablesProperty Tables4. 4. Property TablesProperty Tables

For most substances, the relationships among thermodynamics properties are too complex to expressed by simple equations. Therefore properties are presented in the form of tables (Property Tables).[ Smith et al. (2005); Cengel & Boles (2002)]

Property tables are given in the Appendix in both SI and English units.

The tables in English units carry the same number as the corresponding tables in SI, followed by an identifier E.

For example: Tables A-6 and A6-E list properties of superheated water vapor.[Tables F-2 in Smith et al.]

Table A-4 until A-8: show internal energy (u), enthalpy (h) and entropy (s).

For most substances, the relationships among thermodynamics properties are too complex to expressed by simple equations. Therefore properties are presented in the form of tables (Property Tables).[ Smith et al. (2005); Cengel & Boles (2002)]

Property tables are given in the Appendix in both SI and English units.

The tables in English units carry the same number as the corresponding tables in SI, followed by an identifier E.

For example: Tables A-6 and A6-E list properties of superheated water vapor.[Tables F-2 in Smith et al.]

Table A-4 until A-8: show internal energy (u), enthalpy (h) and entropy (s).

4(a). Saturated Liquid and Saturated Vapor States 4(a). Saturated Liquid and Saturated Vapor States 4(a). Saturated Liquid and Saturated Vapor States 4(a). Saturated Liquid and Saturated Vapor States

The properties of saturated liquid and saturated vapor for water are listed in Tables A-4 and A-5.

Table A-4 properties is listed under temperature, and Table A-5 under pressure.

Subscript:f = properties of saturated liquidg = properties of saturated vaporfg = difference between properties of saturated vapor and liquid

For example:vf = specific volume of saturated liquidvg = specific volume of saturated vaporvfg = vg - vf

Saturated pressure (Psat): - The pressure when liquid and vapor phases are in equilibrium at a given temperature value.

Saturated temperature (Tsat): - The temperature when liquid and vapor phases are in equilibrium at a

given pressure value.

The properties of saturated liquid and saturated vapor for water are listed in Tables A-4 and A-5.

Table A-4 properties is listed under temperature, and Table A-5 under pressure.

Subscript:f = properties of saturated liquidg = properties of saturated vaporfg = difference between properties of saturated vapor and liquid

For example:vf = specific volume of saturated liquidvg = specific volume of saturated vaporvfg = vg - vf

Saturated pressure (Psat): - The pressure when liquid and vapor phases are in equilibrium at a given temperature value.

Saturated temperature (Tsat): - The temperature when liquid and vapor phases are in equilibrium at a

given pressure value.

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

Solution

The state of the saturated liquid water is shown on a T-v diagram. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90 oC.

P = Psat@90oC = 70.14 kPa (Table A-4)

The specific volume of the saturated liquid at 90 oC is

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

Then the total volume of the tank is

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

A piston-cylinder device contains 2 ft3 of saturated water vapor at 50-psia pressure. Determine the temperature and the mass of the vapor inside the cylinder.

Solution

The state of the saturated water vapor is shown on a P-v diagram. Since the cylinder contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this pressure.

T = Tsat@90oC = 281.03 oF (Table A-5E)

The specific volume of the saturated vapor at 50 psia is;

v = vf@50 psia = 8.518 ft3/lbm (Table A-5E)

Then the mass of water vapor inside the cylinder becomes;

lbm 0.235/lbmft 8.518

ft 23

3

v

Vm

A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine (a) the volume change and (b) the amount of energy added to the water. Solution (a) The process described is illustrated in a P-v diagram. The volume change per unit mass during a vaporization process is vfg, which is the

difference between vg and vf.

Reading these values from Table A-5 at 100 kPa and substituting yield

vfg = vg - vf = 1.6940 – 0.001043 = 1.6930 m3/kg

Thus, ∆V = mvfg = (0.2 kg) (1.6930 m3/kg) = 0.3386 m3

(b) The amount of energy needed to vaporize a unit mass of a substance at a given pressure is the enthalpy of vaporization at that pressure, which is;

hfg = 2258.0 kJ/kg for water at 100 kPa.mhfg = (0.2 kg)(2258 kJ/kg) = 451.6 kJ.

4(b). Saturated Liquid-Vapor Mixture4(b). Saturated Liquid-Vapor Mixture4(b). Saturated Liquid-Vapor Mixture4(b). Saturated Liquid-Vapor Mixture

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

To analyze this mixture, we need to know the proportions of the liquid and vapor phases in the mixture.

This to be done by defining a new property called the quality, x as the ratio of the mass of vapor to the total mass of the mixture:

Consider a tank that contains a saturated liquid-vapor mixture. The volume occupied by saturated liquid is Vf, and the volume occupied by saturated vapor is Vg. The total volume V is the sum of the two;

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

To analyze this mixture, we need to know the proportions of the liquid and vapor phases in the mixture.

This to be done by defining a new property called the quality, x as the ratio of the mass of vapor to the total mass of the mixture:

Consider a tank that contains a saturated liquid-vapor mixture. The volume occupied by saturated liquid is Vf, and the volume occupied by saturated vapor is Vg. The total volume V is the sum of the two;

10 ;

xmm

m

mm

m

m

mx

gf

g

vaporliquid

vapor

total

vapor

gggfff

gf

gf

vmVvmVmvV

mmm

VVV

, ,

4(b). Saturated Liquid-Vapor Mixture-cont’4(b). Saturated Liquid-Vapor Mixture-cont’4(b). Saturated Liquid-Vapor Mixture-cont’4(b). Saturated Liquid-Vapor Mixture-cont’

Therefore, Therefore,

fgf

fgf

fg

f

fgfg

fgf

gf

gf

gf

gg

ggff

ggff

xhhh

xuuu

v

vvx

vvv

vvxvv

xvvxv

xm

mm

m

m

mm

m

m

mx

m

vm

m

vmv

vmvmmv

enthalpy; andenergy internalfor repeated becan analysis This

with

)( or

)1(

1

since

A rigid tank contains 10 kg of water at 90 oC. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine (a) the pressure in the tank and (b) the volume of the tank.

Solution

(a) Since the two phases coexist in equilibrium, we have a saturated mixture and the pressure must be the saturation pressure at the given temperature;

P = Psat@90oC = 70.14 kPa (Table A-4)

(b) At 90 oC, we have vf = 0.001036 m3/kg and vg = 2.361 m3/kg (Table A-4)

One way of finding the volume tank is to determine the volume occupied by each phase and then add them:

3

33

4.73m

/kg)1m(2kg)(2.36/kg)m 36kg)(0.0010 (8

ggffgf vmvmVVV

(b) Another way is to first determine the quality x, then the average specific volume v, and finally the total volume:

33

3

33

m 4.73 /kg)m kg)(0.473 (10

and

/kgm 0.473

/kg]m)001036.0361.2[(/kgm 0.001036

2.0kg 10

kg 2

mv V

xvvv

m

mx

fgf

f

g

A 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa. Determine (a) the temperature of the refrigerant, (b) the quality, the enthalpy of the refrigerant, and (d) the volume occupied by the vapor phase.

Solution

(a) Basically, we do not know whether the refrigerant is in the compressed liquid, superheated vapor, or saturated mixture region.

This can be determined by comparing a suitable property to the saturated liquid and saturated vapor values. From the given information, we can determine the specific volume;

At 160 kPa, we read

/kgm 0.02kg 4

m 0.080 33

m

Vv

12-A Table /kgm 0.1229

/kgm 0.00074353

3

g

f

v

v

Obviously, vf ˂ v ˃ vg, and, the refrigerant is in the saturated mixture region. Thus, the temperature must be the saturation temperature at the specified pressure: T = Tsat@ 160 kPa = -15.62 oC (Table A-12)

(b) Quality can be determined from;

kJ/kg 62.7

kJ/kg) 8.18(0.158)(20kJ/kg 29.78

fgf xhhh

(c) At 160 kPa, we also read from Table A-12 that hf = 29.78 kJ/kg and hfg = 208.18 kJ/kg. Then;

0.1580.0007435-0.1229

0.007435-0.02

fg

f

v

vvx

(d) The mass of the vapor is;

and the volume occupied by the vapor phase is;

The rest of the volume (2.3 L) is occupied by the liquid.

kg 0.632kg) 0.158)(4( tg xmm

L) 77.7(or m 0.0777/kg)m kg)(0.1229 0.632( 33 ggg vmV

4(c). Superheated Vapor 4(c). Superheated Vapor 4(c). Superheated Vapor 4(c). Superheated Vapor

In the region to the right of the saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor.

Superheated vapor is characterized by;

Lower pressures (P < Psat at a given T)Higher temperatures (T > Tsat at a given P)Higher specific volumes (v > vg at a given P or T)Higher internal energies (u > ug at a given P or T)Higher enthalpies (h > hg at a given P or T)

In the region to the right of the saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor.

Superheated vapor is characterized by;

Lower pressures (P < Psat at a given T)Higher temperatures (T > Tsat at a given P)Higher specific volumes (v > vg at a given P or T)Higher internal energies (u > ug at a given P or T)Higher enthalpies (h > hg at a given P or T)

Determine the temperature of water at a state of P = 0.5 MPa and h = 2890 kJ/kg.

Solution

At 0.5 Mpa, the enthalpy of saturated water vapor is hg = 2748.7 kJ/kg. Since h > hg, now we have superheated vapor.

Under 0.5 MPa. In Table A-6 we read;

Obviously, the temperature is between 200 and 250 oC. By linear interpolation it is determined to be:

T = 216.4 oC.

T, oC h, kJ/kg

200 2855.4

250 2960.7

4(d). Compressed Liquid4(d). Compressed Liquid4(d). Compressed Liquid4(d). Compressed Liquid

In general, a compressed liquid is characterized by;

Higher pressures (P > Psat at a given T)Lower temperatures (T < Tsat at a given P)Lower specific volumes (v < vg at a given P or T)Lower internal energies (u < ug at a given P or T)Lower enthalpies (h < hg at a given P or T)

In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature.

This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure.

In general, a compressed liquid is characterized by;

Higher pressures (P > Psat at a given T)Lower temperatures (T < Tsat at a given P)Lower specific volumes (v < vg at a given P or T)Lower internal energies (u < ug at a given P or T)Lower enthalpies (h < hg at a given P or T)

In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature.

This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure.

Determine the internal energy of compressed liquid water at 80 oC and 5 MPa, using (a) data from the compressed liquid table and (b) saturated liquid data. What is the error involved in the second case?

Solution

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

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

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

The error involved is

which is less than 1 percent.

kJ/kg 333.72 C 80

MPa 5o

u

T

P

kJ/kg 334.86C 80@ o

fuu

% 0.34 100 x 72.333

72.33386.334

EXERCISE 1EXERCISE 1EXERCISE 1EXERCISE 1Determine the missing properties and the phase descriptions in the following table of water:

T, oC P, kPa u, kJ/kg x Phase description

(a) 200 0.6

(b) 125 1600

(c) 1000 2950

(d) 75 500

(e) 850 0.0

EXERCISE 1EXERCISE 1EXERCISE 1EXERCISE 1Determine the missing properties and the phase descriptions in the following table of water:

Answer:

T, oC P, kPa u, kJ/kg x Phase description

(a) 120.23 200 1719.49 0.6 Saturated liquid-vapor mixture

(b) 125 232.1 1600 0.535 Saturated liquid-vapor mixture

(c) 395.6 1000 2950 No meaning Superheated vapor

(d) 75 500 313.90 No meaning Compressed liquid

(e) 172.96 850 731.27 0.0 Saturated liquid

• Equation of state: Any 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 is the ideal-gas equation of state. This equation predicts the P-v-T behavior of a gas quite accurately within some properly selected region.

U = U(T) (Ideal gas) (3.15)

6. The Ideal-Gas Equation of State6. The Ideal-Gas Equation of State

7. Compressibility Factor - A Measure of Deviation from Ideal-Gas Behavior

7. Compressibility Factor - A Measure of Deviation from Ideal-Gas Behavior

The ideal-gas equation is very simple and thus very convenient to use. But, gases deviate from ideal-gas behavior significantly at states near the saturation region and the critical point.

This deviation from ideal-gas behavior at a given temperature and pressure can accurately accounted by correction factor called as compressibility factor, Z.

The ideal-gas equation is very simple and thus very convenient to use. But, gases deviate from ideal-gas behavior significantly at states near the saturation region and the critical point.

This deviation from ideal-gas behavior at a given temperature and pressure can accurately accounted by correction factor called as compressibility factor, Z.

ideal

actual

v

v Z

ZRTPv

RT

PvZ

as; expressed be alsocan It

or

7. Compressibility Factor - A Measure of Deviation from Ideal-Gas Behavior

7. Compressibility Factor - A Measure of Deviation from Ideal-Gas Behavior

Gases behave differently at a given temperature and pressure, but they behave very much the same at temperatures and pressures normalized with respect to their critical temperatures and pressures.

The normalization is done as;

where PR = reduced pressure, TR = reduced temperature. When P and v, or T and v, are given instead of P and T, the generalized

compressibility chart can still be used to determine the third property, but it would involve tedious trial and error.

Therefore, it is necessary to define one more reduced property called the pseudo-reduced specific volume , vR as:

The Generalized Compressibility Chart can be used for all gasses. [Figure A-30a until A-30c]

Gases behave differently at a given temperature and pressure, but they behave very much the same at temperatures and pressures normalized with respect to their critical temperatures and pressures.

The normalization is done as;

where PR = reduced pressure, TR = reduced temperature. When P and v, or T and v, are given instead of P and T, the generalized

compressibility chart can still be used to determine the third property, but it would involve tedious trial and error.

Therefore, it is necessary to define one more reduced property called the pseudo-reduced specific volume , vR as:

The Generalized Compressibility Chart can be used for all gasses. [Figure A-30a until A-30c]

crR

crR T

TT

P

PP and

/ crcr

actualR PRT

vv

Determine the specific volume of refrigerant-134a at 1 MPa and 50 oC, using (a) the ideal-gas equation of state and (b) the generalized compressibility chart. Compare the values obtained to the actual value of 0.02171 m3/kg and determine the error involved in each case.

Solution The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are determine from Table A-1 to be;

(a) The specific volume of refrigerant-134a under the ideal-gas assumption is;

Therefore, treating the R-134a vapor as an ideal gas would result in an error of (0.02632-0.02171)/0.02171=0.212, or 21.2 percent in this case.

K 374.3

MPa 4.067

K /kg.m kPa. 0.0815 3

cr

cr

T

P

R

/kg0.02632m

kPa 1000

K 323 K /kg.m kPa. 0.0815 33

P

RTv

(b) To determine the correction factor Z from the compressibility chart, we can first need to calculate the reduced pressure and temperature:

Thus,

The error in this result is less than 2 percent. Therefore, in the absence of tabulated data, the generalized compressibility chart can be used with confidence.

84.0

0.863K 374.3

K 323

0.246MPa 4.067

MPa 1

Z

T

TT

P

PP

crR

crR

)/kgm 02632.0)(84.0( 3 idealZvv

Determine the pressure of water vapor at 600 oF and 0.514 ft3/lbm, using (a) the steam tables (b) the ideal equation, and (c) the generalized compressibility chart.

Solution

The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are determine from Table A-1E to be;

(a) The pressure at the specified state is determined from Table A-6E to be;

This is the experimentally determined value, and thus it is the most accurate.

R .31651

psia 2043

R /lbm.ft psia. 0.05956 3

cr

cr

T

P

R

psia 1000F 600

/lbmft 0.0514o

3

P

T

v

(b) The pressure of steam under the ideal-gas assumption is determined from the ideal-gas relation to be;

Therefore, treating the steam as an ideal gas would result in an error of (1228-1000)/1000 = 0.228, or 22.8 percent in this case.

(c) To determine the correction factor Z from the compressibility chart (Fig A-30), we first need to calculate the pseudo-reduced specific volume and the reduced temperature.

Thus,

Using the compressibility chart reduced the error from 22.8 to 5.7 percent.

psia

v

RTP 2281

/lbmft 0.514

R 1060 R /lbm.ft psia. 0.59563

3

33.0

0.91R 1165.3

R 1060

373.2R) R)(1165.3 /lbm.psia.ft (0.5956

psia) /lbm)(3204ft (0.514

/ 3

3

R

crR

crcr

actualR

P

T

TT

PRT

vv

.psia 1057psia) 4(0.33)(320 crR PPP

8. Other Equations of State8. Other Equations of State

Van der Waals Equation:

Redlich/Kwong Equation:

Soave/Redlich/Kwong Equation:

Peng/Robinson Equation:

Van der Waals Equation:

Redlich/Kwong Equation:

Soave/Redlich/Kwong Equation:

Peng/Robinson Equation:

cr

cr

cr

cr

P

RTb

P

TRa

bV

a

bV

RTP

8 and

64

27 where

constant,a

22

2

constant )(2/1

a,bbVVT

a

bV

RTP

constant )(

a,bbVV

a

bV

RTP

constant 22

a,bbbVV

a

bV

RTP

8. Other Equations of State –cont’8. Other Equations of State –cont’

Viral Equation: Viral Equation:

...)D'PC'PB'Pa(Z 321

RT

PVZ

...132

V

D

V

C

V

BZ

)13.3(' aRT

BB

)13.3()(

'2

2

bRT

BCC

)13.3(

)(

23'

3

3

cRT

BBCDD

(3.11)

(3.10)

(3.12)

Equations of State ApplicationEquations of State Application

Equation Application

Ideal-gas Pr < 0.05, Tr >1.5 or, P < 2 atm, T > 25 oC

Van der Waals Only for vapor.Lower pressure.

Viral Equation(C = D =…0)

Only for vapor.Tr > 0.436 Pr +0.6

Redlich/Kwong Only for vapor.Lower pressure.

Soave/Redlich/Kwong Saturated vapor and saturated liquid.Higher pressure.

Peng Robingson Saturated vapor and saturated liquid.Higher pressure.Standard to hydrocarbon industry.

Thank youThank youThank youThank you

Prepared by, Prepared by,

MISS RAHIMAH OTHMANMISS RAHIMAH OTHMAN