DICKY DERMAWAN dickydermawan@gmail.com ITK-233 Termodinamika Teknik Kimia I 3 SKS 2 - PVT Behavior of Fluid, Equation of State.

DICKY DERMAWANwww.dickydermawan.net78 .net

d ickydermawan@gmai l .com

ITK-233Termodinamika Teknik

Kimia I

3 SKS

2 - PVT Behavior of Fluid, Equation of State

P – T Diagram P – v Diagram

T-v Diagram

For the regions of the diagram where a single phase exist:

This means an equation of state exist relating P, V, T. An EOS may be solved for any one of the three quantities as a function of the other two, viz.: V = V(T,P)

Volume expansivity:

Isothermal compressibility:

Equation of State

0)T,V,P(f

dPP

VdT

T

VdV

TP

PT

V

V

1

TP

V

V

1

dPdTV

dV

Example

For liquid acetone at 20oC & 1 bar:Β = 1.487 x 10-3 oC-1 κ = 62 x x 10-6 bar-1 V=

1.287 cm3 g-1

For aceton, find:

a. The value of b. The pressure generated by heating at constant V from 20oC & 1 bar to 30oCc. The change in volume for a change from 20oC & 1 bar to 0oC & 10 bar.

VT

P

Problem 3.1

Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial derivatives.

The isothermal compressibility coefficient () of water at 50oC and 1 bar is 44.18 x 10-6 bar-1. To what pressure must water be compressed at 50oC to change its density by 1%? Assume that is independent of P.

Problem 3.2 & 3.3

Generally, volume expansivity β and isothermal compressibility κ depend on T and P.Prove that

The Tait equation for liquids is written for an isotherm as:

where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.

PT TP

PB

AP1VV 0

Problem 3.4

For liquid water the isothermal compressibility is given by:

where c & b are functions of temperature only. If 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bar at 60oC, how much work is required? At 60oC, b=2700 bar and c = 0.125 cm3 g-1

)bP(V

c

Problem 3.5

Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temperature of 32oF from 1 atm to 3000 atm. The isothermal compressibility of mercury at 32oF is:

κ/atm-1 = 3.9 x 10-6 – 0.1 x 10-9 P (atm)

Problem 3.6

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0oC to 20oC. Determine ΔVt, W, Q, and ΔUt

The properties for liquid carbon tetrachloride at 1 bar & ooC may be assumed independent of temperature:

β = 1.2 x 10-3 K-1

Cp = 0.84 kJ kg-1 K-1

ρ = 1590 kg m-3

Problem 3.7

A substance for which κ is a constant undergo a mechanically reversible process from initial state (P1,V1) to final state (P2,V2), where V is molar volume.

a. Starting with the definition of κ, show that the path of the process is described by

b. Determine an exact expression which gives the isothermal work done on 1 mol of this constant-κ substance

)Pexp()T(AV

P – v Diagram

Virial EOS & Ideal Gas

Virial expansion:

Alternative form:

Compressibility factor

For ideal gas: Z = 1, thus

...P'DP'CP'B1(RTPV 32

...V

D

V

C

V

B1(RTPV 32

RT

PVZ

RTPV

The Ideal Gas

The internal energy of a real gas is a function of pressure and temperature.

This pressure dependency is the result of forces between the molecules.

In an ideal gas, such forces does not exist. No energy would be required to alter the average intermolecular distance, and therefore no energy would be required to bring about volume & pressure changes in an ideal gas at constant temperature. In other word, the internal energy of an ideal gas is a function of temperature only. )T(UU

Implied Property Relations for an Ideal Gas

)T(CdT

dU

T

UC V

VV

RT)T(UPVUH

Thus H is also a funcion of temperature only

)T(CdT

dH

T

HC P

PP

R)T(C)T(CdT

dHC vPP

Isothermal ProcessClosed System, Ideal Gas, Mechanically Reversible

0U

0H

2

1

1

2

P

PlnRT

V

VlnRTdV

V

RTPdVW

WQ

PV = constant

Isobaric ProcessClosed System, Ideal Gas, Mechanically Reversible

dTCU V

)TT(R)VV(PdVPPdVW 1212

dTCH P

HQ

TV-1 = constant

Isochoric (Constant V) ProcessClosed System, Ideal Gas, Mechanically Reversible

dTCU V

0PdVW

dTCH P

UQ

TP-1 = constant

Adiabatic ProcessClosed System, Ideal Gas, Mechanically Reversible

dTCU V

dV

V

RTdTCdVP0dTC

dTCdU

dVPdW

0dQ

vv

v

dTCH P

0Q

PC

R

1

2

1

2

P

P

T

T

vC

R

2

1

1

2

V

V

T

T

P

RTdPdTCv

v

P

C

C

2

1

1

2

V

V

P

P

PdVW

ttanconPV

ttanconsPT

ttanconsVT1

1

V

P

C

C

Polytropic ProcessClosed System, Ideal Gas, Mechanically Reversible

Polytropic process can be considered as general form of process.

Isobaric process : δ = 0Isothermal process : δ = 1Adiabatic process : δ = γIsochoric process : δ = ∞

ttanconPV

ttanconsPT

ttanconsVT1

1

Example 3.2

Air is compressed from an initial condition of 1 bar & 25oC to a final state of 5 bar & 25oC by three different mechanically reversible processes in a closed system:

a. Heating at constant volume followed by cooling at constant pressure

b. Isothermal compressionc. Adiabatic compression followed by cooling at constant

volume.Assume air to be an ideal gas with the constant heat

capacities: CV = 5/2 R, CP = 7/2 R.

Sketch the process in a PV diagram & calculate the work required, heat transferred, and the changes in internal energy & entalphy of the air for each processes

Example 3.3

An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system:

a. From an initial state of 70oC & 1 bar, it is compressed adiabatically to 150oC

b. It is then cooled from 150oC to 70oC at constant pressure

c. Finally, it is expanded isothermally to its original state

Sketch the process in a PV diagram & calculate W, Q, DU, and DH for each of the three processes and for the entire cycle.

Take CV = 3/2 R and CP = 5/2R

Problem 3.8

One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 8 bar & T1 = 600 K to P2 = 1 bar by each of the following path:

(a) Constant volume(b) Contant temperature(c) AdiabaticallyAssuming mechanical reversibility, calculate

W, Q, DU, and DH for each of the three processes.

Sketch each path in a single PV diagram

Problem 3.9

An ideal gas initially at 600 K & 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 1-2 pressure decreases isothermally to 3 bar; in step 2-3 pressure decreases at constant volume to 2 bar; in step 3-4 volume decreases at constant pressure; and in step 4-1 the gas returns adiabatically to its initial step.

a. Sketch the cycle on a PV diagramb. Determine (where unknown) both T & P for states

1, 2, 3, and 4c. Calculate W, Q, DU, and DH for each step of the

cycle

Data: CV = 5/2 R, CP = 7/2 R

Problem 3.10

An ideal gas, CV = 3/2 R & CP = 5/2 R is changed from P1 = 1 bar & Vt

1 = 12 m3 to P2 = 12 bar & Vt2 = 1 m3 by the

following mechanically reversible processes:a. Isothermal compressionb. Adiabatic compression followed by cooling at constant

pressurec. Adiabatic compression followed by cooling at constant

volumed. Heating at constant volume followed by cooling at

constant pressuree. Cooling at constant pressure followed by heating at

constant volume

Calculate W, Q, DUt, and DHt for each of these processes, and sketch the paths of all processes on a single PV diagram

Example 3.4

A 400 gram mass of nitrogen at 27oC is held in a vertical cylinder by a frictionless piston. The weight of the piston makes the pressure of the nitrogen 0.35 bar higher than that of the surroundings atmosphere, which is at 1 bar & 27oC. Thus the nitrogen is initially at a pressure of 1.35 bar, and is in mechanical & thermal equilibrium with its surroundings. Consider the following sequence of process:

a. The apparatus is immersed in an ice/water bath and is allowed to come to equilibrium

b. A variable force is slowly applied to the piston so that the nitrogen is compressed reversibly at the constant temperature of 0oC until the gas volume reaches one-half the value at the end of step a. At this point the piston is held in place by latches.

Example 3.4 (cont’)

c. The apparatus is removed from ice/water bath and comes to thermal equilibrium with the surrounding atmosphere at 27oC

d. The latches are removed, and the apparatus is allowed to return to complete equilibrium with its surroundings

Sketch the entire cycle on a PV diagram, and calculate Q, W, DU & DH for the nitrogen for each step of the cycle. Nitrogen may be considered an ideal gas for which CV = 5/2 R and CP = 7/2 R

Cubic Equation of State:van der Waals Equation (1873)

2V

a

bV

RTP

C

2C

2

P

TR

64

27a

C

C

P

RT

8

1b

C

CC P

RT

8

3V

8

3

TR

VPZ

C

CCC

Theorem of Corresponding State; Accentric Factor

All fluids, when compared at the same reduced temperature & reduced pressure, have approximately the same compressibility factor, and all deviate from ideal gas behavior to about the sam degree

Cr T

TT

Cr P

PP

Accentric Factor(Pitzer)

7.0Tsatr rPlog0.1

Critical Constants & Accentric Factors:Paraffins

Tc/K Pc/bar Vc/10-6m3.mol-1 Zc

Critical Constants & Accentric Factors:

Olefin & Miscellaneous Organics

Tc/K Pc/bar Vc/10-6m3.mol-1 Zc

Critical Constants & Accentric Factors:

Miscellaneous Organic Compounds

Tc/K Pc/bar Vc/10-6m3.mol-1 Zc

Critical Constants & Accentric Factors:

Elementary Gases

Tc/K Pc/bar Vc/10-6m3.mol-1 Zc

Critical Constants & Accentric Factors:

Miscellaneous Inorganic Compounds

Tc/K Pc/bar Vc/10-6m3.mol-1 Zc

Cubic Equation of State:Redlich/Kwong Equation (1949)

)bV(V

)T(a

bV

RTP

C

2C

2

rC

2C

2

r P

TRT42748.0

P

TR)T()T(a 2

1

C

C

C

C

P

RT08664.0

P

RTb

3

1

TR

VPZ

C

CCC

Cubic Equation of State:Soave/Redlich/Kwong Equation (1972)

)bV(V

)T(a

bV

RTP

C

2C

2

rC

2C

2

r P

TR),T(42748.0

P

TR),T()T(a

C

C

C

C

P

RT08664.0

P

RTb

3

1

TR

VPZ

C

CCC

22/1r

2r )T1()176.0574.1480.0(1),T(

Cubic Equation of State:Peng - Robinson Equation (1976)

)bV(V

)T(a

bV

RTP

C

2C

2

rC

2C

2

r P

TR),T(45724.0

P

TR),T()T(a

C

C

C

C

P

RT07779.0

P

RTb

30740.0TR

VPZ

C

CCC

22/1r

2r )T1()26992.054226.137464.0(1),T(

Example 3.8

Given that the vapor pressure of n-butane at 350 K & 9.4573 bar, find the molar volumes of saturated vapor and saturated liquid n-butane at these conditions as given by:

a. Van der Waalsb. Redlich/Kwongc. Soave/Redlich/Kwongd. Peng-Robinson

Application of The Virial Equations

At low to moderate pressure, it is common to use truncated virial equation:

At pressure above the range of applicability of the above eqn, the appropriate form is:

Benedict/Webb/Rubin equation:

and its modifications are inspired by volume expansion virial eqn and are used in the petroleum & natural gas industries for light hydrocarbons.

RT

BP1

RT

PVZ

2V

C

V

B1

RT

PVZ

2223632

2000

Vexp

V1

TV

c

V

a

V

abRT

V

T/CARTB

V

RTP

Example 3.7

Reported values for the virial coefficients of isopropanol vapor at 200oC are:B = -388 cm3 mol-1 C = -26000 cm6 mol-2

Calculate V and Z for isopropanol at 200oC & 10 bar by:

a. The ideal gas equationb. Two term truncated pressure expansion virial

equationc. Three term truncated volume expansion virial

equation

Pitzer Correlation for the Second Virial Coefficient

r

r

C

C

T

P

RT

BP1

RT

BP1Z

10

C

C BBRT

BP

6.1r

0

T

422.0083.0B

2.4r

1

T

172.0139.0B

10 ZZZ

Z0

Pr

Tr

Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler

Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler 10 ZZZ

Z0

Pr

Tr

Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler 10 ZZZ

Z1

Pr

Tr

Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler 10 ZZZ

Z1

Pr

Tr

Real Gas, EOS

Calculate Z and V for steam at 250oC and 1,800 kPa by the following:

a. Ideal gas equationb. Truncated virial equation with the following

experimental values of virial coefficient:

B = -152.5 cm3 mol-1 C = -5,800 cm6

mol-2

c. Pitzer correlation for virial equationd. Pitzer-type correlation of Lee – Kesslere. Steam tablef. van der Waals equation g. Redlich/Kwong equationh. Soave/Redlich/Kwong equationi. Peng-Robinson Equation

Correlation for Liquids: Rackett Equation

2857.0r )T1(

CCsat ZVV

Correlation for Liquids:Lydersen, Greenkorn & Hougen

Problem 3.45

A 30 m3 tank contains 14 m3 of liquid n-butane in equilibrium with its vapor at 25oC. Estimate the total mass of n-butane in the tank. The vapor pressure if n-butane at the given temperature is 2.43 bar.

Ideal Gas

A cylinder pressure vessel having an inside diameter of 50 cm and height of 1.25 cm. Amount of the ammonia gas is confined in that a pressure vessel. Measured pressure and temperature of gas are 30 bars and 200oC. What are the specific volume and amount of the ammonia gas in pressure vessel (in SI unit), assuming ammonia is an ideal gas?

Real Gas, EOS

Calculate Z and for steam at 250oC and 1,800 kPa by the following:

The ideal gas equationThe truncated virial equation with the following

experimental values of virial coefficient:B = -152.5 cm3 mol-1 C = -5,800

cm6mol-2The van der Waals equation The Redlich/Kwong equationThe steam tableData: critical point of steam, TC = 647.1 K; PC =

220.55 bar; = 55.9 cm3mol-1

1st law after diagram

1 kmole of ideal gas (Cv = 3R/2) undergoes a three-step mechanically reversible cycle in a closed system.

Gas in initial state at 500 kPa and 27 oC is heated at constant pressure

Followed by adiabatic expansion until the its pressure become 120 kPa

Finally step, gas is isothermally compressed to initial condition

From that process description:Illustrate the cycle process at P-V diagram in T parameter!

b. Calculate H, U, Q and W for each step of process and total!