Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇...

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Volumetric Properties of Pure Fluid Application of The Virial Equation Prepared By : Agung Ari Wibowo ST., M.Sc Thermodynamics State Polytechnic of Malang 2017

Transcript of Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇...

Page 1: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Volumetric Properties of Pure Fluid

Application of

The Virial Equation

Prepared By : Agung Ari Wibowo ST., M.Sc

Thermodynamics

State Polytechnic of Malang 2017

Page 2: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Ideal Gas and Real Gas

Ideal Gas Real Gas

V1V2

What are the differences?

Ideal Gas

β€’ No molecular interactionβ€’ Exist at Low Pressureβ€’ PV = RT

Real Gas

β€’ Molecular interactionβ€’ Compressibility Factorβ€’ PV = ZRT

Page 3: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Compressibility Factor

RT

BP

RT

PVZ 1

21

V

C

V

B

RT

PVZ

πœ•π‘

πœ•π‘ƒπ‘‡

= 𝐡′ + 2𝐢′𝑃 + 3𝐷′𝑃2 +β‹―

𝑃 ~1

𝑉

Z of Gas Methane at Various Temperature

Virial Equation :

Page 4: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Virial Equation

21.

V

C

V

B

RT

PVZc

Calculate V and Z for Isopropanol gas at 200 C and 10 Bar. By the following equation : a. Ideal Gas

RT

BP

RT

PVZb 1. B = -388 cm3/mol

C = -26.000 cm6/mol2

Page 5: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Virial Equation

a. Ideal GasPV = RT Z = 1V = RT/P

=83,14 cm3bar/mol Kβˆ— 473,15 K

10 π‘π‘Žπ‘Ÿ

= 3934 cm3/mol

V = 3934 -388= 3546 cm3/mol

RT

BP

RT

PVZb 1.

𝑉 =𝑅𝑇

𝑃+ 𝐡 Lihat perbedaannya !!

V1 > V2 , disebabkan oleh ZZ2 = 1- 0,099 = 0,901

V1

V2

Page 6: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Virial Equation

21.

V

C

V

B

RT

PVZc

𝑉 =𝑅𝑇

𝑃1 +

𝐡

𝑉+

𝐢

𝑉2

𝑉 = 3934 1 +𝐡

𝑉+

𝐢

𝑉2

Perlu dilakukan iterasiTebak nilai V awal = V gas idealV0 = 3934 cm3/mol

Masukan ke persamaan = 3934 1 +𝐡

𝑉+

𝐢

𝑉2

V1 trial 1 = 3934 1 +βˆ’388

3934+

βˆ’26000

39342= 3539 cm3/mol

Check apakah sama dengan V0 , ternyata tidak sama

V2 trial 2 = 3934 1 +βˆ’388

3539+

βˆ’26000

35392= 3495 cm3/mol

Diulang terus sampe Vn = Vn-1

V = 3488 cm3/mol

Page 7: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Cubic Equation of State

Page 8: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Van Der Waals EOS

Proposed by J.D van der Waals 1873

𝑃 =𝑅𝑇

𝑉 βˆ’ π‘βˆ’

π‘Ž

𝑉2

β€œa and b are positive constant”

Given values of a and b for a particular fluid, one can calculate P as a function of V for various values of T c

c

c

c

P

RTb;

P

TRa

864

27 22

Page 9: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

A Generic Cubic EOS

𝑃 =𝑅𝑇

𝑉 βˆ’ π‘βˆ’

π‘Ž(𝑇)

(𝑉 + πœ–π‘)(𝑉 + πœŽπ‘)

β€’ πœ– π‘Žπ‘›π‘‘ 𝜎 have same value vor all substance

β€’ π‘Ž 𝑇 are temperature dependent, and different

for each EOSβ€’ b also dependent parameter and different for

each EOSEOS :1. Van der Waals2. Redlich/Kwong (RK)3. Soave/Redlich/Kwong (SRK)4. Peng/Robinson (PR)

Page 10: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

A Generic Cubic EOS

𝑃 =𝑅𝑇

𝑉 βˆ’ π‘βˆ’

π‘Ž(𝑇)

(𝑉 + πœ–π‘)(𝑉 + πœŽπ‘)

π‘Ž 𝑇 = πœ“π‘Ž(π‘‡π‘Ÿ)𝑅

2𝑇𝑐2

𝑃𝐢

𝑏 = Ω𝑅𝑇𝐢𝑃𝐢

EOS Calculation :

Value of Tc , Pc, and πœ” are listed in App. B

π‘‡π‘Ÿ =𝑇

𝑇𝐢; π‘ƒπ‘Ÿ =

𝑃

𝑃𝐢

Page 11: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

A Generic Cubic EOS

EOS Calculation Liquid:

PV = ZRT

Page 12: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Generalized Equation for Gases

Page 13: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Pitzer Correlation

The most well known correlation related with EOS, is correlation developed by Pitzer and coworkers :

a. Compressiblity Factor (Z)b. Second Virial Eq (B)

Pitzer and Curl correlation (1955, 1957)

10 ZZZ

Dimana Z0 dan Z1 fungsi (Tr=T/Tc) dan (Pr=P/Pc)

The values can be determined from The Lee/Kesler Gener

alized-correlation Tables (Lee and Kesler, AIChE J., 21, 5

10-527 (1975) provided in App. E, p. 667

Compressibility Factor :

Page 14: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Pitzer Correlation

Second Virial Eq Correlation

𝑍 = 1 +𝐡𝑃

𝑅𝑇

Original Form :

𝑍 = 1 + αΈ‚π‘ƒπ‘Ÿπ‘‡π‘Ÿ

By Correlation:

αΈ‚=𝐡𝑃𝐢

𝑅𝑇𝐢

αΈ‚ =𝐡0 + πœ”π΅1

𝐡0 = 0.083 +0.422

π‘‡π‘Ÿ1.6

𝐡1 = 0139 +0.172

π‘‡π‘Ÿ4.2

Pitzer Correlation:

Page 15: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Generalized Correlation

Third Virial Eq Correlation

𝑍 = 1 + 𝐡𝜌 + 𝐢𝜌2

Original Form :

𝑍 = 1 + αΈ‚π‘ƒπ‘Ÿπ‘‡π‘Ÿπ‘

+ Δˆπ‘ƒπ‘Ÿπ‘‡π‘Ÿπ‘

Ĉ =𝐢0 + πœ”πΆβ€²

𝐢0 = 0.01407 +0.024322

π‘‡π‘Ÿβˆ’

0.00313

π‘‡π‘Ÿ10.5

By Correlation:

𝜌 =1

𝑉

Ĉ =𝐢𝑃𝐢

2

𝑅2𝑇𝐢2

𝐢′ = βˆ’0.02676 +0.05539

π‘‡π‘Ÿ2.7 βˆ’

0.00242

π‘‡π‘Ÿ10.5

Page 16: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Generalized Equation for Liquids

Page 17: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

Generalized Correlation for Liquids

Generalized correlation for liquids

Rackett equation (Racket, J. Chem. Eng. Data, 15 (1970) 514-517:

estimation of molar volume of saturated liquids)

Lyderson, Greenkorn and Hougen:

estimation of liquid molar volume

With accuracy of 1-2%π‘‰π‘†π‘Žπ‘‘ = 𝑉𝐢𝑍𝐢(1βˆ’π‘‡π‘Ÿ)

0.2857

πœŒπ‘Ÿ =𝜌

πœŒπ‘=𝑉𝐢𝑉

𝑉2 = 𝑉1πœŒπ‘Ÿ1πœŒπ‘Ÿ2

If we know T1 we can evaluate πœŒπ‘Ÿ1

Page 18: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

EOS Application

In Calculation of process design, estimation of fluid properties is a must.

To choose the most suitable methods, here is the guide

Page 19: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

EOS Application

Page 20: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak

EOS Application

Page 21: Thermodynamics State Polytechnic of Malang 2017Virial Equation 2. 1 V C V B RT PV c Z 𝑉= 𝑅𝑇 𝑃 1+ 𝑉 + 𝑉 6 𝑉=3934 1+ 𝑉 + 𝑉 6 Perlu dilakukan iterasi Tebak