Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
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Chapter 3, Problem 3.45P

(a)

Interpretation Introduction

Interpretation:

The Zand V value for ethane should be calculated at 50C temperature and 15bar pressure by the truncated virial equation with given virial coefficients.

Concept Introduction:

The value of Z is easily calculated by the given truncated virial equation and value of V is find out by PV=ZnRT .

(a)

Expert Solution
Check Mark

Answer to Problem 3.45P

The Zand V value for ethylene are V=1600.455cm3mol and Z=0.905 respectively.

Explanation of Solution

Given Information:

The virial equation is given as

  Z=PVRT=1+BV+CV2

  gas constant R in CGS unit is 82.05746cm3barmolK

  P=15bar

  T=500C+273.15=323.15K

For ethane, the second and third virial coefficients are: B=-156.7cm3.mol-1 and C=9650cm6.mol-2 .

From given virial equation

  Z=PVRT=1+BV+CV2

  PVRT=1+BV+CV2

Put the given values

  15bar×V82.05746 cm3barmolK×323.15K=1+156.7 cm3molV+9650 cm6 mol2V25.66×104Vcm3mol=V2156.7 cm3molV+9650 cm6 mol2V25.66×104V3cm3mol=V2156.7cm3molV+9650cm6mol2

On solving the equation

  V=1600.455cm3mol

So,

  Z=PVRT=15bar×1600.455 cm3mol82.05746 cm3barmolK×323.15KZ=0.905

(b)

Interpretation Introduction

Interpretation:

The Zand V value for ethane must calculate at 50C temperature and 15bar pressure by the truncated virial equation with a value of B virial coefficient from generalized Pitzer correlation.

Concept Introduction:

In order to find value of Z from given truncated virial equation, first we calculate B virial coefficient from generalized Pitzer correlation and then value of V followed by value of Z find out by PV=ZnRT .

(b)

Expert Solution
Check Mark

Answer to Problem 3.45P

The Zand V value for ethane are V=1634.05cm3mol and Z=0.912 respectively.

Explanation of Solution

Given Information:

The virial equation in reduced conditions is given as

  Z=1+B0+ωB1TrPrOr,V=RTP+RTCPCB0+ωB1

Here B0=0.0830.422Tr1.6 and B1=0.1390.172Tr4.2

For ethane at given temperature and pressure, the critical conditions are written from table B.1 in appendix B,

  TC=305.3Kand PC=48.72bar, ω=0.1

  gas constant R in CGS unit is 83.144cm3barKmol

  P=15bar

  T=500C+273.15=323.15K

For calculation of B virial coefficient from generalized Pitzer correlation,

  Tr=TTcTr=323.15K305.3K=1.059

And

  Pr=PPcPr=15bar48.72bar=0.308

So,

  B0=0.0830.422Tr1.6B0=0.0830.4221.0591.6B0=0.302

and

  B1=0.1390.172Tr4.2B1=0.1390.1721.0594.2B1=3.8035×103

Hence, value of V is

  V=RTP+RTCPCB0+ωB1V=83.144 cm3barKmol×323.15K15bar+83.144 cm3barKmol×305.3K48.72bar0.302+0.1×3.8×103V=1634.05cm3mol

And value of compressibility factor,

  Z=PVRT=15bar×1634.05 cm3mol83.144 cm3barmolK×323.15KZ=0.912

(c)

Interpretation Introduction

Interpretation:

The Zand V value for ethane must calculate at 50C temperature and 15bar pressure by the Redlich/Kwong equation.

Concept Introduction:

The Redlich/Kwong equations is an iterative procedure. So, we will use hit and trial procedure and guess some values of Z to achieve convergent value.

According to Redlich/Kwong equations, the molar volume of saturated vapor can be found using formula:

  V=ZRTP

(c)

Expert Solution
Check Mark

Answer to Problem 3.45P

From Redlich/Kwong equations, the molar volume of ethane is:

V=1622.83cm3mol and Z=0.906 respectively.

Explanation of Solution

Given Information:

The Redlich/Kwong equation is

Z=1+βqβZβ(Z+εβ)(Z+σβ)

                 ....(1)

Here

β=ΩPrTr and q=φα(Tr)ΩTr

From table 3.1 in the example based on Redlich/Kwong equation given in book, the values used to calculate the terms in equation (1) are:

For Redlich/Kwong:

  ε=0,σ=1,ψ=0.42748,Ω=0.08664andα(Tr)=Tr1/2

gas constant R in CGS unit is 83.144cm3barKmol, P=15bar

  T=500C+273.15=323.15K

  Tr=TTcTr=323.15K305.3K=1.059

And

  Pr=PPcPr=15bar48.72bar=0.308

So,

  β=ΩPrTr=0.08664×0.3081.059β=0.0252

  α(Tr)=Tr1/2=1.0590.5=0.972

  q=ψα(Tr)ΩTr=0.42748×0.9720.08664×1.059q=4.527

For ethane, put values in equation (1)

  Z=1+βqβZβ(Z+εβ)(Z+σβ)

  Z=1+0.02524.527×0.0252×Z0.0252(Z+0×0.0252)(Z+1×0.0252)

Using hit and trial method and compare both side of equation, the calculated value is:

  Z=0.906

Hence,

  V=ZRTP=0.906×83.144 cm3barKmol×323.15K15barV=1622.83cm3mol

(d)

Interpretation Introduction

Interpretation:

The Zand V value for ethane must calculate at 50C temperature and 15bar pressure by the Soave/Redlich/Kwong equation.

Concept Introduction:

The Soave/Redlich/Kwong equations is an iterative procedure. So, we will use hit and trial procedure and guess some values of Z to achieve convergent value.

According to Redlich/Kwong equations, the molar volume of ethylene can be found using formula:

  V=ZRTP

(d)

Expert Solution
Check Mark

Answer to Problem 3.45P

From Soave/Redlich/Kwong equations, the molar volume of ethane is:

V=1624.62cm3mol and Z=0.907 respectively.

Explanation of Solution

Given Information:

The Soave/Redlich/Kwong equation is

Z=1+βqβZβ(Z+εβ)(Z+σβ)

                 ....(1)

Here

β=ΩPrTr and q=φα(Tr)ΩTr

From table 3.1 in the example based on Soave/Redlich/Kwong equation given in book, the values used to calculate the terms in equation (1) are:

For Soave/Redlich/Kwong:

  ε=0,σ=1,ψ=0.42748,Ω=0.08664andα(Tr)=[1+(0.480+1.574ω0.176ω2)(1Tr1/2)]2

gas constant R in CGS unit is 83.144cm3barKmol, P=15bar

  T=500C+273.15=323.15K

The values of reduced temperature and pressure are same as found in subpart (c), From reference subpart (c),

Tr=1.059 and Pr=0.308

So,

  β=ΩPrTr=0.08664×0.3081.059β=0.0252

  α(Tr)=[1+(0.480+1.574ω0.176ω2)(1Tr1/2)]2α(Tr)=[1+(0.480+1.574×0.10.176×0.12)(11.0591/2)]2α(Tr)=0.9634

  q=φα(Tr)ΩTr=0.42748×0.96340.08664×1.059q=4.489

For ethane, put values in equation (1)

  Z=1+βqβZβ(Z+εβ)(Z+σβ)

  Z=1+0.02524.489×0.0252×Z0.0252(Z+0×0.0252)(Z+1×0.0252)

Using hit and trial method and compare both side of equation, the calculated value is:

  Z=0.907

Hence,

  V=ZRTP=0.907×83.144 cm3barKmol×323.15K15barV=1624.62cm3mol

(e)

Interpretation Introduction

Interpretation:

The Zand V value for ethane must calculate at 50C temperature and 15bar pressure by Peng/Robinson equation.

Concept Introduction:

The Peng/Robinson equation is an iterative procedure. So, we will use hit and trial procedure and guess some values of Z to achieve convergent value.

According to Peng/Robinson equation, the molar volume of ethylene can be found using formula:

  V=ZRTP

(e)

Expert Solution
Check Mark

Answer to Problem 3.45P

From Soave/Redlich/Kwong equations, the molar volume of ethylene is:

V=1604.91cm3mol and Z=0.896 respectively.

Explanation of Solution

Given Information:

The Peng/Robinson equation is

Z=1+βqβZβ(Z+εβ)(Z+σβ)

                 ....(1)

Here

β=ΩPrTr and q=φα(Tr)ΩTr

From table 3.1 in the example based on Peng/Robinson equation given in book, the values used to calculate the terms in equation (1) are:

For Peng/Robinson equation:

  ε=12,σ=1+2,ψ=0.45724,Ω=0.07779andZC=0.30740α(Tr)=[1+(0.37464+1.54226ω0.26992ω2)(1Tr1/2)]2

gas constant R in CGS unit is 83.144cm3barKmol, P=15bar

  T=500C+273.15=323.15K

The values of reduced temperature and pressure are same as found in subpart (c), From reference subpart (c),

Tr=1.059 and Pr=0.308

So,

  β=ΩPrTr=0.07779×0.3081.059β=0.0226

  α(Tr)=[1+(0.37464+1.54226ω0.26992ω2)(1Tr1/2)]2α(Tr)=[1+(0.37464+1.54226×0.10.26992×0.12)(11.0591/2)]2α(Tr)=0.97

  q=ψα(Tr)ΩTr=0.45724×0.970.07779×1.059q=5.384

For ethylene, put values in equation (1)

  Z=1+βqβZβ(Z+εβ)(Z+σβ)

  Z=1+0.02265.384×0.0226×Z0.0226(Z+(12)×0.0226)(Z+(1+2)×0.0226)

Using hit and trial method and compare both side of equation, the calculated value is:

  Z=0.896

Hence,

  V=ZRTP=0.896×83.144 cm3barKmol×323.15K15barV=1604.91cm3mol

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Chapter 3 Solutions

Introduction to Chemical Engineering Thermodynamics

Ch. 3 - Prob. 3.11PCh. 3 - Prob. 3.12PCh. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27PCh. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - Prob. 3.30PCh. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - Prob. 3.33PCh. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - Prob. 3.40PCh. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Prob. 3.43PCh. 3 - Prob. 3.44PCh. 3 - Prob. 3.45PCh. 3 - Prob. 3.46PCh. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Prob. 3.75PCh. 3 - Prob. 3.76PCh. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95P
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