Chapter 3: Evaluating Properties

Property Relations in Engineering Thermodynamics

Lava flowing into the Pacific Ocean in Hawaii. Photo courtesy of Mike Benson.

ENGINEERING CONTEXT

To apply the energy balance to a system of interest requires knowledge of the properties of the system and how the

properties are related.

The objective of this chapter is to introduce property relations relevant to engineering thermodynamics. As part of the

presentation, several examples are provided that illustrate the use of the closed system energy balance introduced in Chap. 2 together with the property relations considered in this chapter.

Question: Where do properties such as internal energy come from?

In every problem thus far, we have been given the thermodynamic properties:

T, P, v, u, (h, s …)

Know how to measure T, P, v.

The properties fix the STATE.

But, where do they come from? What about u, etc?

State of Matter

• State of a system at equilibrium is its condition, described by its thermodynamic properties

• Most of the thermodynamic properties are dependent on each other

State of Matter

• Energy is a property of matter (E=U+PE+KE)– Properties are Point functions, they depend

only on the State (Exact differentials)

• Work and Heat are not properties of matter. They are actions performed by environment on system, or vice versa (Inexact differentials)

• State describes the condition of a system as described by its list of properties

Examples

Liquid H2O, vapor H2O, mixture of liquid/vapor (such as steam)-All H2O, but all at different states

Boiling Point, Freezing Point- different for each substance, but clearly same phenomena

Simple Compressible Substance

Water, Air, and most other common engineering fluids are called:

“Simple Compressible substances”

Definition: The only reversible way to perform work on these fluids is by expansion or compression

2

1

V

BySystemV

W PdV= ∫

Models Other Than Simple Compressible Substance

Must have other means to perform reversible work

Examples:

Simple elastic system

Simple magnetic system

Simple Compressible Substance“Simple” means only one way system energy can be significantly altered by work

Empirical evidence suggests one independent property for each independent way system energy can be changed (one for Q and one for W)

For Simple Compressible substances:– We require two independent thermodynamic properties to

uniquely establish the state of the matter (this is the State Principle)

– Which two properties would you choose?– How about the ones we can measure?

(T,v)(P,v)(P,T) because sometimes not independent

Equations of StateSuppose we pick (T,v) as two independent

intensive properties. Then, the rest of the intensive properties can be expressed as functions of these two

P=P(T,v)u=u(T,v)

Other properties we haven’t discussed yeth=h(T,v) [enthalpy]s=s(T,v) [entropy]

The relationship between state properties is referred to as the Equation of State

Properties would be determined from experimental data

Certain mathematical relationships between properties are required by thermodynamics

Equations of State

Forms of “Equation” of State• Equations: e.g. Ideal gas, Incompressible Liquid• Tables of measured (and computed) properties• Plots or graphical representations of the tables

Understand and tabulate behavior of measurable properties

P-v-T

Find relationship between measurable properties and non-measurable properties

u, h, s

Example Equation of State:

Perfect Gas

PV = mRT

R is gas constant

Pv = RT

P= RT/v = P(T, v) This says that P is a function of T and v

The P-v and T-v diagramsPv RT=

1CPv

=P

v

T

v

2PT v C vR

= =

T P

An Experiment with WaterConstant Pressure Heating in Piston-Cylinder System

Q

PTAll LiquidM, V1, v1

Liquid & VaporM, V2, v2

All VaporM, V1, v1

Q

PT

W

Q

PT

W

W

v

T

Liqu

id

Mix

ture

Vap

or

An Experiment with WaterWeight keeps tank at constant pressure

What’s going on with M, P, V, v and T?

Q

PT

W

P P

Phase Change and P-v-T Surface• Constant Pressure Heating in Piston-Cylinder

at P = 1 atm

Note three parts: all liquid expansion, mixed phase expansion and all vapor expansion

Not all expansions will occur at 1 atm

What happens if we increase the pressure by adding more weight?

Phase Change and P-v-T Surface

• Constant Pressure Heating in Piston-Cylinder at P = 2 atm

P-v-T Surface• Constant Pressure Heating in Piston-Cylinder

at P = 2 atmP

v

T

P = 1 atm

P = 2 atm

Note that a 3-Dimensional surface is created. This is a form of Equation of State (not in equation form).

Can Construct P-v-T surface for Water

Plane at P = 2 atm

Plane at P = 1 atm

Looking down at T-v projection

P

v

T

Water Expands During Freezing (Unusual)Critical Point: Tc, Pc, vc

Saturated liquid and vapor are no longer distinguishable

P < Pc P = Pc P > Pc

Q Q Q

Heating @ Constant V & M: Constant v

“Bomb”

22.1 MPa

Phases

Critical Point

Triple Point

T

v

Critical Pressure line

Liquid

Liquid-Vapor Mixture

Superheated Vapor

Subcritical Pressure line

Saturation Line

The Liquid, Sub-cooled or Compressed Liquid Region

T

v

Liquid

Saturation Linef g

Phigh

Plow

P & T are independent

- and are therefore enough to specify state

Could other pairs ofproperties be used?

Which ones?

Two-Phase, Liquid-Vapor Mixture“Under the Dome”

T

v

f g

Phigh

Plow

P & T under dome notenough to specify state

P & T are not independent

Note that v could be used

Quality Definition and Use“Under the Dome”

T

v

f g

Phigh

Plow

Quality = x

vapor

liquid vapor

mxm m

=+

0 < x < 1

X = 0 X = 1

“Quality” of steam for steam engine

vapormx

m=

QualityFor use in Tables A-2 and A-3

• For Saturated Mixture (Liquid-Vapor) Region– Quality; x; an intensive

property– x gives fraction that is

vapor (gas)– (1-x) gives Moisture

Content

gf

g

mmm

x+

≡

0 ≤ x ≤ 1; x = 0 → Saturated Liquid (subscript ‘f’) x = 1 → Saturated Vapor (subscript ‘g’)

‘fg’ → ‘g’-’f’

Superheated Vapor

P & T are independent

- and are therefore enough to specify state

Could other pairs ofproperties be used?

Which ones?

T

v

f g

Phigh

Plow

PhasesCould do same for P-v Diagram

Note: Far enoughaway from mixed phase (dome) region,Perfect Gas is goodmodel for superheatedvapor

Perfect Gas Model

Retrieving Thermodynamic PropertiesIndex to Tables in SI UnitsIndex to Tables in English Units

SI:Table A-5: Properties of Compressed Liquid Water

English:Table A-5E: Properties of Compressed Liquid Water

Saturated (mixed phase/dome), Superheated, Compressed

Water, Refrigerant 22, Refrigerant 134a, Ammonia, Propane

Ideal Gas

Tabulations of Water Properties

T

v

Liquid

Liquid-Vapor Mixture

Superheated Vapor

Saturation Line

Table A5, A5E

Table A2, A3 (A2E, A3E)

Table A4, A4E

f g

Sat Liquid

Sat vapor

Two Saturated Tables for Each

One for Saturated Temperature

One for Saturated Pressure

Sometimes we know T, sometimes P

Compressed or Sub-cooled

LiquidTable A-5, A-5E

Note:sub-cooled tables are sparse because it is accurate to use incompressible liquid model

T

v

f g

Saturated Water

Table A-2, A3, A2E, A3E

T

v

f g

Superheated Vapor

Table A-4, A-4E

T

v

f g293 F

500 F

Computing Properties Under the Dome

1

(1 )

liquid vapor

liq vap

liq f vap g

vap

liq

f g

V V V

V VVvm m mm v m v

vm mm

xm

mx

mv x v xv

= +

= = +

= +

=

= −

∴ = − +

It is not enough to know T, P in order to establish state under dome

Need T or P, and one other property

(x= Quality)

( ) ( )1 f g f g f f fgv x v xv v x v v v xv= − + = + − = +

How to Locate States in Tables

Use Saturation Tables and T-v Sketches

Note isobars

How to Locate States in Tables

Note Isotherms

Use Saturation Tables and P-v Sketches

Example 1For Water

What is the phase for T = 100 C, P = 25 bar ?

What is the phase for T = 100 C, P = 0.7 bar ?

Need to define Psat for T = 100 C in both cases.

Then compare P to Psat on appropriate diagram.

How to Locate States in Tables

T = 100C

What is Psat?

Table A-2 Psat = 1.014 bar

1.014 bar

What is phase for T=100 C, P=25 bar ?

1.0 bar

25 bar

T = 100C

The phase is Compressed LiquidUse Table A-5

P

Compressed or Sub-cooled

LiquidTable A-5, A-5E

Use Table A-5 or A-5E to get properties

T

v

f g

What is phase for T=100 C, P=0.7 bar ?

1.0 bar

0.7 barT = 100C

The phase is Superheated VaporUse Table A-4

P

Superheated Vapor

Table A-4, A-4E

T

v

f g293 F

500 F

Processes on DiagramsExamples:

1: Constant volume (V, v) condensation2: Constant temperature condensation

1

2

Example Problem

Text 3.27 (5th Ed; not in 6th Ed)

P1 = 30 Bar1

Substances that Expand When FreezingUnusual Behavior

Water is a common substance

Water Expands During Freezing (Unusual)Solid, Liquid & Vapor regions

Single phase regions: 2 intensive properties fix equilibrium state

Note Two-phase regions:Liquid-vapor, liquid-solid, solid-vaporP & T NOT independent State fixed by T & v or P & v

3 phases can exist in equilibriumalong Triple Line where T & Pfixed for range of v

Water Expands During Freezing (Unusual)Saturated states:

Where a phase change begins or ends:Saturated liquid, saturated vapor, saturated solid

Vapor dome: blue area composed of mixed (liquid-vapor) states

Saturated liquid states/line

Saturated vapor states/line

Water Expands During Freezing (Unusual)

Critical Point: Tcr, Pcr, vcr

Highest pressure whereliquid & vapor can co-exist

Top of vapor dome

Saturated liquid and vapor are no longer distinguishable

Water Expands During Freezing (Unusual)

Critical Point: Tcr, Pcr, v’cr

All substances have CP

See Table A-1 & A-1E

Water Expands During Freezing (Unusual)3 phases can exist in equilibriumalong Triple Line where T & Pfixed for range of v. All 3 exist ONLYat the Triple T & P

Ice (s)

Water (l)

Water (v)

Note: No AIRin this system

AIR is a different system: O2 & N2

T = 273.16 KP = 0.6113 kPaP = 0.00602 atm

Projection of P-v-T Surface: P-TCalled “Phase Diagram”

Expands on Freezing (Water)

Point here represents a line on 3-D surface

Saturation T & P: Tsat, Psat(where phase change occurs)

Tsat

Psat

Melting/Freezing

Boiling (Evaporation)/Condensation

Sublimation

Key Definitions:1. Triple Point2. Critical Point

Projection of P-v-T Surface: P-TCalled “Phase Diagram”Expands on Freezing (Water)

Dry Ice

Triple Point: T = 273.16 K & P = 0.6113 kPa

Melting/Freezing

Projection of P-v-T Surface: P-TCalled “Phase Diagram”Expands on Freezing (Water)

Solid-liquid line slopes to left for expansion upon freezing

Physical Meaning ?

0dPdT

<

Lower freezing/meltingTemperature at higher pressure

Projection of P-v-T Surface: P-vCalled “P-v Diagram”Expands on Freezing (Water)

P

vNote isotherms for T < Tc, T = Tc and T > TcDifferent diagrams useful for different problems

T

v

Projection of P-v-T Surface: T-vCalled “T-v Diagram”Expands on Freezing (Water)

Note isobars for P < Pc, P = Pc and P > PcDifferent diagrams useful for different problems

Substances That Contract When FreezingUsual Behavior

The P-v-T surface for a substance that contracts upon freezing

Compare Substances That Expand and Contract Upon Freezing

Contraction ExpansionBasically The Same

Projection of P-v-T Surface: P-vCalled “P-v Diagram”

Contracts on Freezing

Note isotherms for T < Tc, T = Tc and T > TcDifferent diagrams useful for different problems

The P-v-T surface for a substance that contracts upon freezing

Projection of P-v-T Surface: P-TCalled “Phase Diagram”

Contracts on Freezing

Sublimation Dry Ice

Boiling (Evaporation)/Condensation

Sublimation

Melting/Freezing

Linear Interpolation:Between values in the tables

H L L

H L L

v v v vSlopeT T T T

− −= =

− −

Subscripts:L – Value in table at lower endH – Value in table at upper endNone – value of interest

H L L

H L L

v v v vSlopeT T T T

− −= =

− −

0.2275 0.2060 0.2060240 200 215 200

v− −=

− −

( )215 200 0.2275 0.2060 0.2060240 200

v− − + = −

0.2141 v=

Linear Interpolation:Between values in the tables

Double Interpolation

h at P = 65 psiand T = 425oF ?

25/100 between 400 and 500 atboth P = 60 and P = 80

Then

5/20 between interpolatedh values at P = 60 and P = 80

EnthalpyThe combination of internal energy and the product of pressure, P and volume, V, occurs frequently in Thermodynamics (as we will see).

For convenience we will give this combination of terms a new name: Enthalpy and the symbol, H

Notes: Combinations of Properties are PropertiesUnits must be consistent: Energy, Energy/mass, Energy/mol

H U P Vh u P v

= + ⋅= + ⋅

Extensive form

Intensive form

h u P v= + ⋅ Molar form

Data for specific enthalpy are determined same as for specific volume with diagrams and saturated tables.

Determine if in compressed liquid, saturated or superheatedregion.

If in saturated region, then use quality (x) approach

( ) ( )1 f g f g f f fgu x u xu u x u u u x u= − + = + − = + i

( ) ( )1 f g f g f f fgh x h xh h x h h h x h= − + = + − = + iSpecific enthalpy, h:

Specific internal energy, u:

Saturated Water

Table A-2, A3, A2E, A3E

T

v

f g

Ufg not here, but could be

Example

Problem 3.66

Text Example Problems

Compute h from known T and u.Saturated region, so x determined, then used for h

Evaluate T, v and h for water where P = 0.1 MPaAnd u is known. This u > Usat at this P. Thus, this isin superheated region. Can get h = h(P, u)

Can also get h from h = u + Pv

Reference States and Reference ValuesWhen no chemical reactions are involved, only the difference in energies are important. This is case for this course. Thus, the reference state can be arbitrary.

Combustion is important case where chemical reactions occurAnd special care required for reference states. (Combustion not done in AME230)

Ref1 = 0Ref2 =2

5

8 =c

mgzPEg

Z = 8 - 5 = 3Z = (8-2) – (5-2) = 6-3 = 3

Reference States and Reference Values

Reference states for water, ammonia, propane, Refrigerant 22 and Refrigerant 134a

Water: uref = 0 at saturated liquid at 0.01oC (left side of dome)h = u + Pv can be calculated from this reference state

All other substances: href = 0 at saturated liquid at -40oC (SI) or -40oF (English)u = h - Pv can be calculated from this reference state

Will lead to negative energy values in some cases (OK)

Evaluating Properties of Liquids & Solids

Approximations for Liquid using Saturated Liquid Data

Because u, v vary little with P at fixed T make the following approximation:

( , ) ( )

( , ) ( )f

f

v T P v T

u T P u T

≈

≈

And also: ( , )

( , ) ( ) ( )f f

h T P u pv

h T P u T pv T

= +∴

≈ +

Convince yourself that: f fh u≈ 1f

f

Pvu

and

( ) ( ), fh T P h T≈Thus:

Interpretation on T-v diagram

T

v

Liquid

f gT

v

Saturation state

Actual state

Interpretation on P-v diagram

P

v

Saturation state

Actual state

T

u & h are weak functions of Pressure

u & h are strong functions of Temperature

Approximations for Liquid using Saturated Liquid Data

Why is this important: We can calculate the state in the liquid region using only P, T and the properties of saturated liquid that are tabulated in the saturation liquid tables

It is not necessary to have extensive tabulations of data in the liquid region

Specific Heats: and

Definitions of specific heats

Specific heat at constant volume:

Specific heat at constant pressure:

vv

ucT

∂ = ∂

vc pc

pp

hcT∂ = ∂

Specific Heats: and vc pc

vv

ucT

∂ = ∂

kJ/kg K kJ/kmol K Btu/lb oR Btu/lbmol oR

T

vuv

( ),u u T v=

Surface interpretation

Special case for measuring cv

Q U W U m u= ∆ + = ∆ = ∆

vv

ucT

∂ = ∂

Q m

T

Q

T

u

Rigid

What is the Specific Heat?

Amount of energy required to change the temperature of a unit mass of substance one degree: [ kJ/kg K ]

vucT∆

≅∆

( ),h h T P=

kJ/kg K kJ/kmol K Btu/lb oR Btu/lbmol oR

T

PhP

Surface interpretation

pp

hcT∂ = ∂

Specific Heats: andvc pc

Specific Heat Ratio, kp

v

ck

c=

See Figure 3.9 for T and P variations of cp of water vapor

Vapor phases of other substance exhibit similar behavior

Specific heat data available for:

Solids and liquids (Tables A19, A19E)

Gases (Tables A20, A21, A20E, A21E)

Incompressible Substance Model (Liquid or Solid)

Equation of State: v = constant

Definition of specific heat:

( )vduc TdT

=

vv

ucT

∂ = ∂

For an incompressible substance, cv only depends on T

( ), ( )u u T v u T= =

Change in Internal Energy for Incompressible Substance

Integrating

2

1

2 1

( )vT

vT

du c T dT

u u c dT

=

− = ∫

Or if cv is taken as constant

( ) ( )2 1 2 1vu u c T T− = −

Equality of Specific Heats for Incompressible Substance

p

h duT dT∂ =∂

( ) ( ),h T P u T Pv= +

p vc c c= =

By Definition:

Differentiate equation above wrt T, holding P constant:

Change in Enthalpy for Incompressible Substance

( ) ( ) ( )2 1 2 1 2 1h h u u v P P− = − + −

( ) ( ),h T P u T Pv= +

( ) ( )2

12 1 2 1

T

Th h c T dT v P P− = + −∫

( ) ( )2 1 2 1 2 1h h c T T v P P− = − + −

dh du Pdv vdP du vdP= + + = +0

Change in Enthalpy for Incompressible Substance

( ) ( ) ( )2 2

1 12 1 2 1

T T

T Th h c T dT v P P c T dT− = + − ≈∫ ∫

( ) ( ) ( )2 1 2 1 2 1 2 1h h c T T v P P c T T− = − + − ≈ −

Constant specific heat, c; Negligible pressure contribution

Variable specific heat, c; Negligible pressure contribution

Approximations for Liquids and Solids

• Using Saturated Liquid Data ( @ T )

(‘Compressed Liquid Rule’)

• Using ‘Incompressible Substance Model’

f

f

f

f

v vh hu us s

≈

≈

≈

≈

2 1 2 1

2 1 2 1 2 1

2 1 2 1

( )( ) ( )

( )

p vc c cu u c T Th h c T T v P P

h h c T T

= =

− = −− = − + −

− ≈ −

Note: Internal energy change and Enthalpy change often “same”

Note temperature-dependenceIteration may be necessary

P vc c c= =

Example Problem:

Potato at 20oC immersed in water bath at 5oC

Universal Gas Constant

PvT

P

R0

limP

Pv RT→

=

Same for all gases

v (volume per mole)

8.314 kJ/kmol K1.986 Btu/lbmol R1545 ft lbf/lbmol R{R =

Ideal Gas Model

.....(M is mass)ˆ ......(N is number of moles)

whereˆ Universal gas constantˆ Molecular weight

Pv RTPV MRT

PV NRT

RRM

==

=

= =See Table A-1

ˆ 8.314 kJ/kmol-KR =

Ideal Gas Approximations for Vapor

Approximate the behavior of superheated vapor as though the mean free path between molecules is large

-Billiard ball model of molecular collisions

-When is the approximation valid?Low PressureHigh Temperature(compared to what?)

T

v

f g

Compressibility Factor, Z

Z = 1.0 Implies Perfect GasZ deviation from 1.0 implies departure from PG modelCompressibility factor and chart can be used for non-PG

PvZRT

=

Pv RT= OK when vapor behaves as Perfect Gas

Need a way to decide when Perfect Gas model is OK

Define “Compressibility Factor”, Z, to do this:

T

v

Pc1, Tc1

Pc2, Tc2

Pc3, Tc3

Pc4, Tc4

Reduced Pressure, PR

Reduced Temperature, TR

Rc

PPP

=

Rc

TTT

=

Basis for Generalized Compressibility Concept

Pseudo-Reduced specific volume Rc c

vvRT P

′ =

T

v

“Collapse” of all Curves to a Generalized Curve, Scaled by Critical Properties

Principle of Corresponding States

Compressibility Factor

R TRc c

Pv p TZ pRT p T

= = =

Under what conditions is Z=1?

Low pressures

High temperatures

Note data for 10 gases collapsingto Generalized form

Compressibility Factor, Z

R is a constant for a gas with molecular weight M

R is Universal Gas ConstantR

Where: RRM

=

8.314 kJ/kmol K1.986 Btu/lbmol R1545 ft lbf/lbmol R{R =

If PG model not good & vapor tables not available

Z = 1

Implies Perfect Gas

PvZRT

=

Generalized Compressibility ChartIf not Perfect Gas & vapor tables not available

Example Problem:

3.85 Z factor and PG comparison

Ideal Gas Model

First Law Aspects:Calculation of Energy Changes

Internal Energy, Enthalpy and Specific Heats of Ideal/Perfect Gases

( ) ( ),h h T P h T= =

( )PP

dhdh dT c T dTdT

= =

By definition, enthalpy is a function of Temperature ONLY

P T

h hdh dT dPT P∂ ∂ = + ∂ ∂

0

( ) ( ) ( )2

12 1

T

PTh T h T c T dT− = ∫

( ) ( ) ( )2 1 2 1Ph T h T c T T− = −

Variable cp

Constant cp

Relationship Between Specific Heats

( ) ( ) ( )h T u T Pv u T RT= + = +

dh du RdT dT

= +

( ) ( )P vc T c T R= + ( ) ( )P vc T c T R= +

( )( )

1P

v

c Tk

c T= >

( )1P

kRc Tk

=−

( )1vRc Tk

=−

Ratio of Specific Heats, k

(Ideal Gas)

Where: ( )Pdhc TdT

= ( )vduc TdT

=

Variation of Specific Heat of Gases with Temperature (Figure)

Note monatomic Gases

Note: constant value “OK” for small temp range

Temperature Variation of Specific Heats in Tables, Figures and Equations

( ) 2 3 4Pc TT T T T

Rα β γ δ ε= + + + +

Constants in Table A-21

Tabular data available in Table A20(see next slide)

Note: constant value “OK” for small temp range

P Pc cR R=

Handling Variable Specific Heats

1

2

T

v

21-A Table see...1

)()(

)(2

1

12

−=

=−

=− ∫

Rc

Rc

RTcTc

dTTcuu

pv

vp

T

Tv

Handling Variable Specific Heats

{ }

( ) ( )

( ) ( )51

52

41

42

31

32

21

2212

432

12

54

32))(1(

1

1

2

1

2

1

2

1

TTTT

TTTTTT

dTTTTT

dTRc

dTRc

Ruu

T

T

T

T

pT

T

v

−+−

+−+−+−−

=−++++

=

−==−

∫

∫∫

εδ

γβα

εδγβα

Example Problem:

3.95 Energy balance and PG model [ Cp(T) used ]

Polytropic Process of Ideal Gas

1 1 2 2n nPV PV=nPV = constant

Any Gas (Could be Perfect Gas, but doesn’t have to be)

- nLog (P)

Log (V)

P

V

Experimental data define n. Polytropic if fits definition aboveNote what happens for different values of n

2 1

1 2

nP VP V

=

Polytropic Process of Ideal Gas

1 1 2 2n nPV PV= 2 1

1 2

nP VP V

=

2

1

2 2 1 1

1V

V

PV PVPdVn−

=−∫

2

1

21 1

1

lnV

V

VPdV PVV

=∫

( )1 1

2 2 1

1 1 2

n n nT P VT P V

− −

= =

( )2

1

2 1

1V

V

mR T TPdV

n−

=−∫

2

1

2

1

lnV

V

VPdV mRTV

=∫

nPV = constant

( )1n ≠ ( )1n =

Any Gas

Ideal Gas Only

( )1n ≠ ( )1n =

Work:

Work:

Other Equations of State• Van der Waals’

• Redlich-Kwong

• Benedict-Webb-Rubin

• Virial

2va

bvRTP −−

=

( )RK

1/2RK RK

aRTPv b v v b T

= −− +

20

0 0 2 2 3 6 3 2 2CRT 1 (bRT a) a 1 / v

P (B RT A ) c expv T v v v v T v

− α + γ −γ = + − − + + +

2RT B C

P 1 ......v v v

= + + +

Table A-24 has constants

Example Problem:

3.98 Polytropic process

End