Thermodynamic Thermodynamic PropertiesProperties

Property TableProperty Table -- from direct measurement Equation of StateEquation of State --

any equations that relates P,v, and T of a substance

Ideal -Gas Equation of StateIdeal -Gas Equation of State Any relation among the pressure,

temperature, and specific volume of a substance is called an equation of state. The simplest and best-known equation of state is the ideal-gas equation of state, given as

where R is the gas constant. Caution should be exercised in using this relation since an ideal gas is a fictitious substance. Real gases exhibit ideal-gas behavior at relatively low pressures and high temperatures.

Universal Gas ConstantUniversal Gas ConstantUniversal gas constant is given on

Ru = 8.31434 kJ/kmol-K

= 8.31434 kPa-m3/kmol-k

= 0.0831434 bar-m3/kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft3/lbmol-R

ExampleExampleDetermine the particular gas constant for air and hydrogen.

Kkg

kJ0.287

kmolkg

28.97

KkmolkJ

8.1417

MRR u

air

Kkg

kJ4.124

kmolkg

2.016

KkmolkJ

8.1417

Rhydrogen

2

22

1

11

u

u

T

VP

T

VPV/N v T,Rv P

TNRPVRTPv

mRTPV

Ideal Gas “Law” is a simple Ideal Gas “Law” is a simple Equation of StateEquation of State

Percent error for applying ideal gas Percent error for applying ideal gas equation of state to steamequation of state to steam

Question …...Question …...

Under what conditions is it appropriate to apply the ideal gas equation of state?

Ideal Gas LawIdeal Gas Law Good approximation for P-v-T

behaviors of real gases at low densities (low pressure and high temperature).

Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error).

Compressibility FactorCompressibility FactorCompressibility FactorCompressibility Factor The deviation from ideal-gas behavior

can be properly accounted for by using the compressibility factor Z, defined as

Z represents the volume ratio or compressibility.

Ideal Gas

Z=1

Real Gases

Z > 1 or Z < 1

Real GasesReal Gases Pv = ZRT or Pv = ZRuT, where v is volume

per unit mole.

Z is known as the compressibility factor. Real gases, Z < 1 or Z > 1.

Compressibility factorCompressibility factor What is it really doing? It accounts mainly for two things

• Molecular structureMolecular structure

• Intermolecular attractive forcesIntermolecular attractive forces

Principle of Principle of corresponding statescorresponding states

Principle of Principle of corresponding statescorresponding states

The compressibility factor Z is approximately the same for all gases at the same reduced temperature and reduced pressure.

Z = Z(PR,TR) for all gases

Reduced Pressure Reduced Pressure and Temperatureand Temperature

crR

crR T

TT ;

P

PP

where:

PR and TR are reduced values. Pcr and Tcr are critical properties.

Compressibility factor for ten substances Compressibility factor for ten substances (applicable for all gases Table A-3)(applicable for all gases Table A-3)

Where do you find critical-point properties? Where do you find critical-point properties? Table A-7Table A-7

Mol

(kg-Mol)

R

(J/kg.K)

Tcrit

(K)Pcrit

(MPa)

Ar 28,97 287,0 (---) (---)

O2 32,00 259,8 154,8 5,08

H2 2,016 4124,2 33,3 1,30

H2O 18,016 461,5 647,1 22,09

CO2 44,01 188,9 304,2 7,39

Reduced PropertiesReduced Properties This works great if you are given a

gas, a P and a T and asked to find the v.

However, if you are given P and v and asked to find T (or T and v and asked to find P), trouble lies ahead.

Use pseudo-reduced specific volume.

Pseudo-Reduced Pseudo-Reduced Specific VolumeSpecific VolumePseudo-Reduced Pseudo-Reduced Specific VolumeSpecific Volume

When either P or T is unknown, Z can be determined from the compressibility chart with the help of the pseudo-reduced specific volume, defined as

not vcr !

Ideal-Gas ApproximationIdeal-Gas Approximation The compressibility chart shows

the conditions for which Z = 1 and the gas behaves as an ideal gas:

(a) PR < 0.1 and TR > 1

Exercise 3-21 Exercise 3-21 Steam at 600 oC & 1 MPa. Evaluate the specific volume using the steam table and ideal gas law

Tsat = 180oC therefore is superheated steam (Table A-3)

Volume = 0,4011 m3/kg

TTooCC1MPa1MPa

vv

374374600600

180180

TcrTcr

Solution - page 1 Solution - page 1 Part (b) ideal gas law

Rvapor = 461 J/kgK

v = RT/P , v = 461x873/106 = 0.403 m3/kg

Steam is clearly an ideal gas at this state. Error is less than 0.5%

Check: PR = 1/22.09 = 0.05 & TR = 873/647 = 1.35

TEAMPLAYTEAMPLAY

Find the compressibility factor to determine the error in treating oxygen gas at 160 K and 3 MPa as an ideal gas.

Other Thermodynamic Properties:Isobaric (c. pressure) Coefficient

PTv

vv

TT

PP

0Tv

v1

P

Other Thermodynamic Properties:Isothermal (c. temp) Coefficient

TPv

vv

PP

TT

0Pv

v1

T

Other Thermodynamic Properties:

vdPvdTdPPv

dTTv

dvTP

We can think of the volume as being a function We can think of the volume as being a function of pressure and temperature, v = v(P,T).of pressure and temperature, v = v(P,T).Hence infinitesimal differences in volume are Hence infinitesimal differences in volume are expressed as infinitesimal differences in P and expressed as infinitesimal differences in P and T, using T, using and and coefficients coefficients

If If and and are constant, we can integrate for v: are constant, we can integrate for v:

000

PPTTvv

Ln

Other Thermodynamic Properties:Internal Energy, Enthalpy and Entropy

PvuP,Thh

v,Tuu

v,uss

Other Thermodynamic Properties:Specific Heat at Const. Volume

vTu

uu

TT

vv0

Tu

Cv

v

Other Thermodynamic Properties:Specific Heat at Const. Pressure

PTh

hh

TT

PP0

Th

CP

P

Other Thermodynamic Properties:Ratio of Specific Heat

v

p

C

C

Other Thermodynamic Properties:Temperature

ss

uu

vv0

us

T1

v

TT

11

Ideal Gases: u = u(T)Ideal Gases: u = u(T)

Therefore,

dvv

u dT

T

udu

Tv

0

dT)T(C dTT

udu v

v

We can start with du and We can start with du and integrate to get the change in u:integrate to get the change in u:

dT (T)C uuu

and dT,Cdu2

1

T

T v12

v

Note that Cv does change with temperature and cannot be automatically pulled from the integral.

Let’s look at enthalpy Let’s look at enthalpy for an ideal gas:for an ideal gas:

h = u + Pv where Pv can be replaced by RT because Pv = RT.

Therefore, h = u + RT => since u is only a function of T, R is a constant, then h is also only a function of T

so h = h(T)

Similarly, for a change in Similarly, for a change in enthalpy for ideal gases:enthalpy for ideal gases:

(T)dTChhh

and dT,Cdh

p

T

T 12

p

2

1

0P

h & )T(CC pp

Summary: Ideal GasesSummary: Ideal GasesSummary: Ideal GasesSummary: Ideal Gases

For ideal gases u, h, Cv, and Cp are functions of temperature alone. For ideal gases, Cv and Cp are written in terms of ordinary differentials as

gas idealp

gas idealv dT

dhC ;

dT

duC

For an ideal gas,For an ideal gas,

h = u + Pv = u + RT

R dT

du

dT

dh

Kkg

kJ Cp = Cv + R

RatioRatio of specific heats of specific heats is given the symbol, is given the symbol,

1CR

1C

C

vv

p

(T)(T)C

(T)C

C

C

v

p

v

p

Other relations with the Other relations with the ratio of specific heats which ratio of specific heats which

can be easily developed:can be easily developed:

1-R

C and 1-

RC pv

For monatomic gases,For monatomic gases,

constants. are both and

R2

3R , C

2

5C vp

Argon, Helium, and Neon

For all other gases,For all other gases, Cp is a function of temperature and it may be

calculated from equations such as those in Table A-5(c) in the Appendice

Cv may be calculated from Cp=Cv+R.

Next figure shows the temperature behavior …. specific heats go up with temperature.

Specific Heats for Some GasesSpecific Heats for Some Gases Specific Heats for Some GasesSpecific Heats for Some Gases

Cp = Cp(T) a

function of

temperature

Three Ways to Calculate Three Ways to Calculate Δu and ΔhΔu and Δh

Three Ways to Calculate Three Ways to Calculate Δu and ΔhΔu and Δh

Δu = u2 - u1 (table)

Δu =

Δu = Cv,av ΔT

(T) dT C2

1 v

Δh = h2 - h1 (table)

Δh =

Δh = Cp,av ΔT

(T) dT C2

1 p

Isothermal ProcessIsothermal ProcessIsothermal ProcessIsothermal Process Ideal gas: PV = mRT

For ideal gas, PV = mRT For ideal gas, PV = mRT We substitute into the We substitute into the

integralintegral

1

22

1 V

VnmRT

V

dV mRTW

b

dVV

mRT PdV W

b 2

1

2

1

Collecting terms and integrating yields:

Polytropic ProcessPolytropic Process Polytropic ProcessPolytropic Process

PVn = C

Ideal Gas Adiabatic Process and Reversible Work

What is the path for process with expand or contract without heat flux? How P,v and T behavior when Q = 0?

To develop an expression to the adiabatic process is necessary employ:

1. Reversible work mode: dW = PdV2. Adiabatic hypothesis: dQ =03. Ideal Gas Law: Pv=RT4. Specific Heat Relationships5. First Law Thermodynamics: dQ-dW=dU

Ideal Gas Adiabatic Process and Reversible Work (cont)

First Law:

Using P = MRT/V

Integrating from

(1) to (2)

dTvMCPdV0

dUdWdQ

TdT

RC

VdV

11

V

1

1

2

1

2VV

TT

Ideal Gas Adiabatic Process and Reversible Work (cont)

2

1

1

2VV

PP

1

2

1

1

2VV

TT

Using the gas law :

Pv=RT, other

relationship amid

T, V and P are

developed

accordingly:

1

1

2

1

2PP

TT

Ideal Gas Adiabatic Process and Reversible Work (cont)

An expression for

work is developed

using PV =

constant.

i and f represent

the initial and final

states

1

PVPV

VV1

PVW

V

dVPVPdVW

if

1i

1f

i

i

Ideal Gas Adiabatic Process and Reversible Work (cont)

The path representation are lines where Pv = constant.

For most of the gases, 1.4

The adiabatic lines are always at the righ of the isothermal lines.

The former is Pv = constant (the exponent is unity)

PP

vv

Q = 0

Q = 0T=const.

T=const.

ii

ffff

Polytropic ProcessPolytropic ProcessA frequently encountered process for gases is the polytropic process:

PVn = c = constant

Since this expression relates P & V, we can calculate the work for this path.

2

1

V

Vb PdVW

Polytropic ProcessPolytropic Process

Constant pressure 0 Constant volume Isothermal & ideal gas 1 Adiabatic & ideal gas k=Cp/Cv

Process Exponent n

Boundary work for a gas which Boundary work for a gas which obeys the polytropic equationobeys the polytropic equation

111

11

12

1 2

1

2

1

2

1

n n

VV

-n

Vv

v c

V

dV cPdV W

nn-n

nb

We can simplify it We can simplify it furtherfurther

The constant c = P1V1n = P2V2

n

11

1

1122

1111

1222

, nn

VPVP-n

VVP)(VVP Wnnnn

b

Polytropic ProcessPolytropic Process

1

11

1

2

1122

2

1

2

1

, nV

VnPV

, nn

VPVP

dVV

c PdVW

n

b

Exercise 3-11Exercise 3-11A bucket containing 2 liters of water at 100oC is

heated by an electric resistance.

a) Identify the energy interactions if the system boundary is i) the water, ii) the electric coil

b) If heat is supplied at 1 KW, then how much time is needed to boil off all the water to steam? (latent heat of vap at 1 atm is 2258 kJ/kg)

c) If the water is at 25oC, how long will take to boil off all the water (Cp = 4.18 J/kgºC)

Solution - page 1Solution - page 1Part a)

If the water is the system then Q > 0 and W = 0. There is a temperature difference between the electric coil and the water.

If the electric coil is the system then Q < 0 and W < 0. It converts 100% of the electric work to heat!

Solution - page 2Solution - page 2

The mass of water is of 2 kg. It comes from 2liters times the specific volume of 0.001 m3/kg

To boil off all the water is necessary to supply all the vaporization energy:

Evap = 2285*2=4570 KJ

The power is the energy rate. The time necessary to supply 4570 KJ at 1KJ/sec is therefore:

Time = 4570/1 = 4570 seconds or 1.27 hour

Part b)

Solution - page 3Solution - page 3The heat to boil off all the water initially at 25oC is the sum of : (1) the sensible heat to increase the temperature from 25oC to 100oC, (2) the vap. heat of part (b)

The sensible heat to increase from 25oC to 100oC is determined using the specific heat (CP = 4.18 kJ/kgoC)

Heat = 4.18*2*(100-25)=627 KJ

The time necessary to supply (4570+627)KJ at 1KJ/sec is :

Time = 5197/1 = 5197 seconds or 1.44 hour

Part c)

Exercise 3-12Exercise 3-12A bucket containing 2 liters of R-12 is left outside

in the atmosphere (0.1 MPa)

a) What is the R-12 temperature assuming it is in the saturated state.

b) the surrounding transfer heat at the rate of 1KW to the liquid. How long will take for all R-12 vaporize?

Solution - page 1Solution - page 1Part a)

From table A-2, at the saturation pressure of 0.1 MPa one finds:

• Tsaturation = - 30oC

• Vliq = 0.000672 m3/kg

• vvap = 0.159375 m3/kg

• hlv = 165KJ/kg (vaporization heat)

Solution - page 2Solution - page 2Part b)

The mass of R-12 is m = Volume/vL, m=0.002/0.000672 = 2.98 kg

The vaporization energy:

Evap = vap energy * mass = 165*2.98 = 492 KJ

Time = Heat/Power = 492 sec or 8.2 min

Exercise 3-17Exercise 3-17

A rigid tank contains saturated steam

(x = 1) at 0.1 MPa. Heat is added to

the steam to increase the pressure to

0.3 MPa. What is the final

temperature?

Solution - page 1Solution - page 1PP

0.1MPa0.1MPa

vv

99.6399.63ooCC

Tcr = 374Tcr = 374ooCC

0.3MPa0.3MPa133.55133.55ooCC

superheatedsuperheated

Process at constant volume, search at the superheated steam table for a temperature corresponding to the 0.3 MPa and v = 1.690 m3/kg -> nearly 820oC.

Exercise 3-30Exercise 3-30 Air is compressed reversibly and adiabatically

from a pressure of 0.1 MPa and a temperature of 20oC to a pressure of 1.0 MPa.

a) Find the air temperature after the compressionb) What is the density ratio (after to before

compression)c) How much work is done in compressing 2 kg of

air?d) How much power is required to compress 2kg

per second of air?

In a reversible and adiabatic process P, T and v follows:

PP

vv

Q = 0

Q = 0T=const.

T=const.

ii

ffff

Solution - page 1Solution - page 1

2

1

1

2VV

PP

1

2

1

1

2VV

TT

1

1

2

1

2PP

TT

Solution - page 2Solution - page 2

Part a) The temperature after compression is

Part b) The density ratio is

)oC293( K566

1.01

293TPP

TT4.14.0

2

1

1

212

179.51.0

1PP

VV 4.11

1

21

1

2

2

1

1

2

Solution - page 3Solution - page 3Part c)

The reversible work:

Part d)

The power is:

KW391

1TTRM

dtdW

P 12

KJ391

4.02935662872

1TTRM

1

PVPVW 1212

REV

Exercícios – Capítulo 3Propriedades das Substâncias

Puras

Exercícios Propostos: 3.6 / 3.9 / 3.12 / 3.16 / 3.21 / 3.22 / 3.26 / 3.30 / 3.32 / 3.34

Team Play: 3.1 / 3.2 / 3.4

Ex. 3.1) Utilizando a Tabela A-1.1 ou A-1.2, determine se os estados da água são de líquido comprimido, líquido-vapor, vapor superaquecido ou se estão nas linhas de líquido saturado ou vapor saturado. a) P=1,0 MPa; T=207 ºC b) P=1,0 MPa; T=107,5 ºC c) P=1,0 MPa; T=179,91 ºC; x=0,0 d) P=1,0 MPa; T=179,91 ºC; x=0,45 e) T=340 ºC; P=21,0 MPa f) T=340 ºC; P=2,1 MPa g) T=340 ºC; P=14,586 MPa; x=1,0 h) T=500 ºC; P=25 MPa i) P=50 MPa; T=25 ºC

a) Vapor superaquecido

b) Líquido comprimido

c) Líquido saturado

d) Líquido-Vapor

e) Líquido comprimido f) Vapor superaquecido

g) Vapor saturado

h) Vapor superaquecido - Fluido

i) Líquido comprimido - Fluido

Ex. 3.2) Encontre o volume específico dos estados “b”, “d” e “h” do exercício anterior.

b) P=1,0 MPa; T=107,5 ºC vvl=0,001050(utilizar referência T=107,5ºC)

d) P=1,0 MPa; T=179,91 ºC; x=0,45 v=0,08812[v=(1-x)vl+x(vv)]

h) T=500 ºC; P=25 MPa v=0,011123 m3/kg(Tabela A-1.3 Vapor Superaquecido)

Ex. 3.4) Amônia a P=150 kPa, T=0 ºC se encontra na região de vapor superaquecido e tem um volume específico e entalpia de 0,8697 m3/kg e 1469,8 kJ/kg, respectivamente. Determine sua energia interna específica neste estado.

u =1339,345 kJ/kg(u = h - P v)