Chapter 7.7, Problem 25P

(A)

Interpretation Introduction

Interpretation:

The rate at which the work is done

Concept Introduction:

Write the molar entropy equation for a reversible, steady-state, adiabatic turbine.

${\underset{\_}{S}}_2-{\underset{\_}{S}}_1=0$

Here, final molar entropy of the fluid leaving the turbine is ${\underset{\_}{S}}_2$, and initial molar entropy of the fluid entering the turbine is ${\underset{\_}{S}}_1$.

Write the expression to calculate the Lee Kesler correlation residuals entropy.

${\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2-{\underset{\_}{S}}_2^{ig}\right)+\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)$

Here, final molar entropy for an ideal gas state is ${\underset{\_}{S}}_2^{ig}$, and initial molar enthalpy for an ideal gas state is ${\underset{\_}{S}}_1^{ig}$.

Write the expression to calculate the reduced temperature $\left(T_{r1}\right)$ for initial condition.

$T_{r1}=\frac{T}{T_c}$

Here, critical temperature is $T_c$, and temperature is T.

Write the expression to calculate the reduced pressure $\left(P_{r1}\right)$ for initial condition.

$P_{r1}=\frac{P}{P_c}$

Here, critical pressure is $P_c$, and pressure is P.

Calculate parameter ${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}$.

$\begin{array}{c}{\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}=\frac{C_V^{\ast }}{T}dT+\frac{R}{\underset{\_}{V}}d\underset{\_}{V}\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_V^{\ast }}{T}dT+}R\ln \frac{{\underset{\_}{V}}_2}{{\underset{\_}{V}}_1}\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P^{*}-R}{T}dT+}R\ln \left(\frac{R{T}_2/P_2}{R{T}_1/P_1}\right)\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P^{*}-R}{T}dT+}R\ln \left(\frac{P_1{T}_2}{P_2{T}_1}\right)\end{array}$

Here, constant volume heat capacity on a molar basis for ideal gas is $C_V^{*}$, change in temperature is $dT$, molar volume is $\underset{\_}{V}$, final and initial molar volume is ${\underset{\_}{V}}_2\text{and}{\underset{\_}{V}}_1$ respectively, final temperature is $T_2$, initial temperature is $T_1$, and change in molar volume is $d\underset{\_}{V}$.

Write the energy balance for a reversible turbine.

$\frac{{\dot{W}}_{S,rev}}{\dot{\eta }}={\underset{\_}{H}}_2-{\underset{\_}{H}}_1$

Here, rate of shaft work for reversible turbine is ${\dot{W}}_{S,rev}$, net efficiency is $\dot{\eta }$, and final and initial molar enthalpies are ${\underset{\_}{H}}_2\text{and}{\underset{\_}{H}}_1$ respectively.

Write the expression to calculate the Lee Kesler correlation residuals enthalpy.

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2-{\underset{\_}{H}}_2^{ig}\right)+\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$

Here, final molar enthalpy is ${\underset{\_}{H}}_2$, initial molar enthalpy is ${\underset{\_}{H}}_1$, final molar enthalpy for an ideal gas state is ${\underset{\_}{H}}_2^{ig}$, and initial molar enthalpy for an ideal gas state is ${\underset{\_}{H}}_1^{ig}$.

Explanation of Solution

Given information:

n-butane.

Flow rate five moles per second.

Inlet pressure is 15 bar and temperature is 500 K.

Final pressure is 1 bar.

Isentropic efficiency is 80%.

Refer the Appendix table C 1, “Physical properties”, obtain the critical properties of butane.

Critical temperature, $T_c=425.12\text{K}$

Critical pressure, $P_c=37.96\text{bar}$

Acentric factor, $\omega =0.2$

Calculate the reduced temperature $\left(T_{r1}\right)$ for initial condition.

$T_{r1}=\frac{T}{T_c}$        (1)

Substitute $\text{500}\text{K}$ for T and $\text{425}\text{.12}\text{K}$ for $T_c$ in Equation (1).

$\begin{array}{c}{T}_{r1}=\frac{\text{500}\text{K}}{\text{425}\text{.12}\text{K}}\\ =1.18\end{array}$

Calculate the reduced pressure $\left(P_{r1}\right)$ for initial condition.

$P_{r1}=\frac{P}{P_c}$        (2)

Substitute $\text{15}\text{bar}$ for T and $37.96\text{bar}$ for $P_c$ in Equation (2).

$\begin{array}{c}{P}_{r1}=\frac{\text{15}\text{bar}}{37.96\text{bar}}\\ =0.40\end{array}$

Refer the Figure 7-18, “Molar entropy residual function for a compound with $\omega =0$, as determined using the Lee-Kesler equation”, obtain the value of ${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^0$ corresponding to $P_r$.

${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^0=-0.2$

Refer the Figure 7-19, “Correction to molar entropy residual function for a compound with $\omega =1$, as determined using the Lee-Kesler equation”, obtain the value of ${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^1$ corresponding to $P_r$.

${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^1=-0.1$

Calculate parameter ${\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}$.

${\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}=R\left[{\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^0+\omega {\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^1\right]$        (3)

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $-0.2$ for ${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^0$, $-0.1$ for ${\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^1$, and $0.2$ for $\omega$ in Equation (3).

$\begin{array}{c}{\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}=R\left[{\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^0+\omega {\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)}^1\right]\\ =\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\left[\left(-0.2\right)+\left(0.2\right)\left(-0.1\right)\right]\\ =-1.8\text{J}/\text{mol}\cdot \text{K}\end{array}$

Calculate the Lee Kesler correlation residuals entropy.

${\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2-{\underset{\_}{S}}_2^{ig}\right)+\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)$        (4)

Re-write the Equation (4).

${\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)$        (5)

Substitute 0 for ${\underset{\_}{S}}_2-{\underset{\_}{S}}_1$ and $-1.8\text{J}/\text{mol}\cdot \text{K}$ for ${\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}$ in Equation (5)

$\begin{array}{l}{\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)\\ 0=\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left(-1.8\text{J}/\text{mol}\cdot \text{K}\right)\\ {\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}=-1.8\text{J}/\text{mol}\cdot \text{K}\end{array}$

Refer the Appendix D, “Heat capacity”; obtain the ideal gas heat capacity of a compound as a function of temperature.

$\begin{array}{l}\frac{C_P^{*}}{R}=A+BT+C{T}^2+D{T}^3+E{T}^4\\ C_P^{*}=R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]\end{array}$

Here, constants are A, B, C, D, and E.

Calculate parameter ${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}$.

${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P^{*}-R}{T}dT+}R\ln \left(\frac{P_1{T}_2}{P_2{T}_1}\right)$        (6)

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]$ for $C_P^{*}$ in Equation (6).

$\begin{array}{l}{\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{\left(A-1\right)+BT+C{T}^2+D{T}^3+E{T}^4}{T}dT+}R\ln \left(\frac{P_1{T}_2}{P_2{T}_1}\right)\\ =\left\{R\left[\begin{array}{l}\left(A-1\right)\ln \frac{T_2}{T_1}+B\left(T_2-T_1\right)+\frac{C\left(T_2^2-T_1^2\right)}{2}\\ +\frac{D\left(T_2^3-T_1^3\right)}{3}+\frac{E\left(T_2^4-T_1^4\right)}{4}\end{array}\right]+R\ln \left(\frac{P_1{T}_2}{P_2{T}_1}\right)\right\}\end{array}$        (7)

Refer the appendix table D.1, “Ideal Gas Heat Capacity”, obtain the constants for the butane.

Name	Formula	A	${\text{B×10}}^{\text{3}}$	${\text{C×10}}^{\text{5}}$	${\text{D×10}}^{\text{8}}$	${\text{E×10}}^{\text{11}}$
Butane	${\text{C}}_4{\text{H}}_{10}$	5.547	5.536	8.057	–10.571	4.134

substitute $-1.8\text{J}/\text{mol}\cdot \text{K}$ for ${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, $\text{500}\text{K}$ for $T_1$, 5.547 for A, $-5.536\times {10}^{-3}$ for B, $8.057\times {10}^{-5}$ for C, $-10.571\times {10}^{-8}$ for D, and $4.134\times {10}^{-11}$ for E, $\text{15}\text{bar}$ for $P_1$, and $\text{1}\text{bar}$ for $P_2$ in Equation (7).

$\begin{array}{c}-1.8\text{J}/\text{mol}\cdot \text{K}=\left\{R\left[\begin{array}{l}\left(A-1\right)\ln \frac{T_2}{T_1}+B\left(T_2-T_1\right)+\frac{C\left(T_2^2-T_1^2\right)}{2}\\ +\frac{D\left(T_2^3-T_1^3\right)}{3}+\frac{E\left(T_2^4-T_1^4\right)}{4}\end{array}\right]+R\ln \left(\frac{P_1{T}_2}{P_2{T}_1}\right)\right\}\\ =\left\{\begin{array}{l}8.314\left[\begin{array}{l}\left(5.547-1\right)\ln \frac{500}{T_1}+\left(-5.536\times {10}^{-3}\right)\left(T_2-500\right)\\ +\frac{\left(8.057\times {10}^{-5}\right)\left(T_2^2-{500}^2\right)}{2}+\\ \frac{\left(-10.571\times {10}^{-8}\right)\left(T_2^3-{500}^3\right)}{3}\\ +\frac{\left(4.134\times {10}^{-11}\right)\left(T_2^4-{500}^4\right)}{4}\end{array}\right]\\ +\left(8.314\right)\ln \left[\frac{\left(15\right)T_2}{\left(1\right)\left(500\right)}\right]\end{array}\right\}\end{array}$        (8)

Solve Equation (8) to obtain the final temperature:

$T_2=419.5\text{K}$

Calculate the Lee Kesler correlation residuals enthalpy.

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2-{\underset{\_}{H}}_2^{ig}\right)+\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$        (9)

Re-write the Equation (8).

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$        (10)

Refer the Figure 7-16, “Molar enthalpy residual function for a compound with $\omega =0$, as determined using the Lee-Kesler equation”, obtain the value of ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0$ corresponding to $T_{r1}\text{and}{P}_{r1}$.

${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0=-0.3$

Refer the Figure 7-17, “Correction to the molar enthalpy residual function for a compound with $\omega =1$, as determined using the Lee-Kesler equation”, obtain the value of ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1$ corresponding to $T_{r1}\text{and}{P}_{r1}$.

${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1=-0.1$

Calculate parameter $\frac{{\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}}{R{T}_c}$.

${\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}=R{T}_c\left\{{\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0+\omega {\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1\right\}$        (11)

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $\text{425}\text{.12}\text{K}$ for $T_c$, $-0.3$ for ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0$, $-0.1$ for ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1$, and $0.2$ for $\omega$ in Equation (11).

$\begin{array}{c}{\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}=R{T}_c\left\{{\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0+\omega {\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1\right\}\\ =\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{425}\text{.12}\text{K}\right)\left\{\left(-0.3\right)+0.2\left(-0.1\right)\right\}\\ =-1,131\text{J}/\text{mol}\end{array}$

Calculate parameter ${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}$.

${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}={\displaystyle {\int }_{T_1}^{T_2}{C}_P^{\ast }}dT$        (12)

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]$ for $C_P^{*}$ in Equation (12).

$\begin{array}{c}{\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}={\displaystyle {\int }_{T_1}^{T_2}R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]}dT\\ =R{\displaystyle {\int }_{T_1}^{T_2}\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]}dT\end{array}$        (13)

substitute $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, $\text{500}\text{K}$ for $T_1$, $\text{419}\text{.5}\text{K}$ for $T_2$, 5.547 for A, $-5.536\times {10}^{-3}$ for B, $8.057\times {10}^{-5}$ for C, $-10.571\times {10}^{-8}$ for D, and $4.134\times {10}^{-11}$ for E, in Equation (13).

$\begin{array}{c}{\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}=8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\left[\begin{array}{l}5.547\left[\left(419.5-500\right)\text{K}\right]+\\ \frac{-5.536\times {10}^{-3}{\left[\left(419.5-500\right)\text{K}\right]}^2}{2}+\\ \frac{8.057\times {10}^{-5}{\left[\left(419.5-500\right)\text{K}\right]}^3}{3}+\\ \frac{-10.571\times {10}^{-8}{\left[\left(419.5-500\right)\text{K}\right]}^4}{4}+\\ \frac{4.134\times {10}^{-11}{\left[\left(419.5-500\right)\text{K}\right]}^5}{5}-\left[\left(419.5-500\right)\text{K}\right]\end{array}\right]\\ {\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}=-11,170\text{J}/\text{mol}\end{array}$

Substitute $-11,170\text{J}/\text{mol}$ for ${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}$, and $-1,131\text{J}/\text{mol}$ for ${\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}$ in Equation (10).

$\begin{array}{c}{\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)\\ =\left(-11,170\text{J}/\text{mol}\right)-\left(-1,131\text{J}/\text{mol}\right)\\ =-10,039\text{J}/\text{mol}\left(0.001\text{kJ}/\text{mol}/1\text{J}/\text{mol}\right)\\ =-10.039\text{kJ}/\text{mol}\end{array}$

Calculate the actual work is done $\left({\dot{W}}_s\right)$

${\dot{W}}_s=\eta \left({\underset{\_}{H}}_2-{\underset{\_}{H}}_1\right)$        (14)

Substitute 80% for $\eta$ and $-10.039\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_2-{\underset{\_}{H}}_1$ in Equation (14).

$\begin{array}{c}{\dot{W}}_s=80\%\left(-10.039\text{kJ}/\text{mol}\right)\\ =0.8\left(-10.039\text{kJ}/\text{mol}\right)\\ =-8.0312\text{kJ}/\text{mol}\end{array}$

Thus, the rate at which work is done is $-8.0312\text{kJ}/\text{mol}$.

(B)

Interpretation Introduction

Interpretation:

The rate at which the work is done

Concept Introduction:

Write the reduced temperature $\left(T_r\right)$.

$T_r=\frac{T}{T_c}$

Here, critical temperature is $T_c$ and temperature is T.

Write the reduced pressure $\left(P_r\right)$.

$P_r=\frac{P}{P_c}$

Here, pressure is $P$.

Write the acentric factor in a manner analogous to Soave’s $m$.

$\kappa =0.37464+1.54226\omega -0.26993{\omega }^2$

Write the $\alpha$ as a function of the reduced temperature.

$\alpha ={\left[1+\kappa \left(1-T_r^{0.5}\right)\right]}^2$

Write the Peng-Robinson parameter a at the critical point.

$a_c=\frac{0.45724R^2{T}_c^2}{P_c}$

Write the van der Waals parameter $a$.

$a=a_c\alpha$

Write the van der Waals parameter $b$.

$b=\frac{0.07780R{T}_c}{P_c}$

Write the Peng-Robinson equation.

$P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{\underset{\_}{V}\left(\underset{\_}{V}+b\right)+b\left(\underset{\_}{V}-b\right)}$

Here, molar volume is $\underset{\_}{V}$, parameters of Robinson equation are a, b, gas constant is R, temperature and pressure is T and P respectively.

Write the compressibility factor.

$Z=\frac{P\underset{\_}{V}}{RT}$

Write the residual properties of A.

$A=\frac{aP}{R^2{T}^2}$

Write the residual properties of B.

$B=\frac{bP}{RT}$

Write the compressibility factor $\left(Z\right)$.

$Z=\frac{\underset{\_}{V}}{\underset{\_}{V}-b}-\frac{a}{RT}\left(\frac{1}{\underset{\_}{V}+b}\right)$

Write the expression to calculate the Lee Kesler correlation residuals enthalpy.

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2-{\underset{\_}{H}}_2^{ig}\right)+\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$

Here, final molar enthalpy is ${\underset{\_}{H}}_2$, initial molar enthalpy is ${\underset{\_}{H}}_1$, final molar enthalpy for an ideal gas state is ${\underset{\_}{H}}_2^{ig}$, and initial molar enthalpy for an ideal gas state is ${\underset{\_}{H}}_1^{ig}$.

Explanation of Solution

Calculate accentric factor in a manner analogous to Soave’s $m$.

$\kappa =0.37464+1.54226\omega -0.26993{\omega }^2$        (15)

Substitute $0.2$ for $\omega$ in Equation (15).

$\begin{array}{c}\kappa =0.37464+1.54226\left(0.2\right)-0.26993{\left(0.2\right)}^2\\ =0.6722\end{array}$

Calculate the value of $\alpha$ expressed as a function of the reduced temperature.

$\alpha ={\left[1+\kappa \left(1-T_r^{0.5}\right)\right]}^2$        (16)

Substitute $0.6722$ for $\kappa$ and $1.18$ for $T_r$ in Equation (16).

$\begin{array}{c}\alpha ={\left[1+0.6722\left(1-{1.18}^{0.5}\right)\right]}^2\\ =0.88\end{array}$

Calculate the value of parameter a at the critical point.

$a_c=\frac{0.45724R^2{T}_c^2}{P_c}$        (17)

Substitute $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for $R$, $\text{425}\text{.12}\text{K}$ for $T_c$, and $\text{37}\text{.96}\text{bar}$ for $P_c$ in Equation (17).

$\begin{array}{c}{a}_c=\frac{0.45724{\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)}^2{\left(\text{425}\text{.12}\text{K}\right)}^2}{\text{37}\text{.96}\text{bar}}\\ =1.504\times {10}^{-5}{\text{m}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}\left(\frac{{10}^{12}{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}{1{\text{m}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}\right)\\ =1.504\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}\end{array}$

Calculate the van der Waals parameter $a$.

$a=a_c\alpha$        (18)

Substitute $0.88$ for $\alpha$, and $1.504\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for $a_c$ in Equation (18).

$\begin{array}{c}a=\left(1.504\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}\right)\left(0.88\right)\\ =1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}\end{array}$

Calculate the van der Waals parameter $b$.

$b=\frac{0.07780R{T}_c}{P_c}$        (19)

Substitute $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for $R$, $\text{425}\text{.12}\text{K}$ for $T_c$, and $\text{37}\text{.96}\text{bar}$ for $P_c$ in Equation (19).

$\begin{array}{c}b=\frac{0.07780\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)\left(\text{425}\text{.12}\text{K}\right)}{\text{37}\text{.96}\text{bar}}\\ =7.24\times {10}^{-5}{\text{m}}^{\text{3}}/\text{mol}\left(\frac{{10}^6{\text{cm}}^3/\text{mol}}{1{\text{m}}^3/\text{mol}}\right)\\ =72.4{\text{cm}}^3/\text{mol}\end{array}$

Calculate constant A.

$A=\frac{aP}{R^2{T}^2}$        (20)

Substitute $1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for $a$, $\text{15}\text{bar}$ for $P$, $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for $R$, and $\text{500}\text{K}$ for $T$ in Equation (20).

$\begin{array}{c}A=\frac{\left(1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}\right)\left(\text{15}\text{bar}\right)}{{\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)}^2{\left(\text{500}\text{K}\right)}^2}\\ =0.116\end{array}$

Calculate constant B.

$B=\frac{bP}{RT}$        (21)

Substitute $72.4{\text{cm}}^3/\text{mol}$ for $b$, $\text{15}\text{bar}$ for $P$, $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for $R$, and $\text{500}\text{K}$ for $T$ in Equation (21).

$\begin{array}{c}B=\frac{\left(72.4{\text{cm}}^3/\text{mol}\right)\left(\text{15}\text{bar}\right)}{\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)\left(\text{500}\text{K}\right)}\\ =0.026\end{array}$

Calculate the molar volume $\left(\underset{\_}{V}\right)$.

$P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{\underset{\_}{V}\left(\underset{\_}{V}+b\right)+b\left(\underset{\_}{V}-b\right)}$        (22)

Substitute $\text{15}\text{bar}$ for $P$, $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for R, $\text{500}\text{K}$ for T, $1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for a, and $72.4{\text{cm}}^3/\text{mol}$ for b in Equation (22).

$\begin{array}{l}\text{15}\text{bar}=\frac{\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)\left(\text{500}\text{K}\right)}{\underset{\_}{V}-72.4{\text{cm}}^3/\text{mol}}-\frac{1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}{\left[\begin{array}{l}\underset{\_}{V}\left(\underset{\_}{V}+72.4{\text{cm}}^3/\text{mol}\right)+\\ 72.4{\text{cm}}^3/\text{mol}\\ \left(\underset{\_}{V}-72.4{\text{cm}}^3/\text{mol}\right)\end{array}\right]}\\ \underset{\_}{V}=73.52{\text{m}}^{\text{3}}/\text{kmol}\end{array}$

Calculate the compressibility factor $\left(Z\right)$.

$Z=\frac{\underset{\_}{V}}{\underset{\_}{V}-b}-\frac{a}{RT}\left(\frac{1}{\underset{\_}{V}+b}\right)$        (23)

Substitute $73.52{\text{m}}^{\text{3}}/\text{kmol}$ for $\underset{\_}{V}$, $8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}$ for R, $\text{500}\text{K}$ for T, $1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for a, and $72.4{\text{cm}}^3/\text{mol}$ for b in Equation (23).

$\begin{array}{c}Z=\left\{\begin{array}{c}\frac{73.52{\text{m}}^{\text{3}}/\text{kmol}}{73.52{\text{m}}^{\text{3}}/\text{kmol}-72.4{\text{cm}}^3/\text{mol}}-\frac{1.34\times {10}^7{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}{\left(8.314\times {10}^{-5}{\text{m}}^{\text{3}}\cdot \text{bar}/\text{mol}\cdot \text{K}\right)\left(\text{500}\text{K}\right)}\\ \left(\frac{1}{73.52{\text{m}}^{\text{3}}/\text{kmol}+72.4{\text{cm}}^3/\text{mol}}\right)\end{array}\right\}\\ =0.901\end{array}$

Calculate parameter ${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}$.

${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}={\displaystyle {\int }_{T_1}^{T_2}{C}_P^{\ast }}dT$        (24)

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]$ for $C_P^{*}$ in Equation (24).

$\begin{array}{c}{\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}={\displaystyle {\int }_{T_1}^{T_2}R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]}dT\\ =R{\displaystyle {\int }_{T_1}^{T_2}\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]}dT\end{array}$        (25)

substitute $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, $\text{500}\text{K}$ for $T_1$, $\text{419}\text{.5}\text{K}$ for $T_2$, 5.547 for A, $-5.536\times {10}^{-3}$ for B, $8.057\times {10}^{-5}$ for C, $-10.571\times {10}^{-8}$ for D, and $4.134\times {10}^{-11}$ for E, in Equation (25).

$\begin{array}{c}{\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]dT}\\ 8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\left[\begin{array}{l}5.547\left[\left(419.5-500\right)\text{K}\right]+\\ \frac{-5.536\times {10}^{-3}{\left[\left(419.5-500\right)\text{K}\right]}^2}{2}+\\ \frac{8.057\times {10}^{-5}{\left[\left(419.5-500\right)\text{K}\right]}^3}{3}+\\ \frac{-10.571\times {10}^{-8}{\left[\left(419.5-500\right)\text{K}\right]}^4}{4}+\\ \frac{4.134\times {10}^{-11}{\left[\left(419.5-500\right)\text{K}\right]}^5}{5}-\left[\left(419.5-500\right)\text{K}\right]\end{array}\right]\\ {\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}=-11,170\text{J}/\text{mol}\end{array}$

Refer the Figure 7-16, “Molar enthalpy residual function for a compound with $\omega =0$, as determined using the Lee-Kesler equation”, obtain the value of ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0$ corresponding to $T_{r1}\text{and}{P}_{r1}$.

${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0=-0.3$

Refer the Figure 7-17, “Correction to the molar enthalpy residual function for a compound with $\omega =1$, as determined using the Lee-Kesler equation”, obtain the value of ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1$ corresponding to $T_{r1}\text{and}{P}_{r1}$.

${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1=-0.3$

Calculate the term ${\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}$.

${\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}=R{T}_c\left\{{\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0+\omega {\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1\right\}$        (26)

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $\text{425}\text{.12}\text{K}$ for $T_c$, $-0.3$ for ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0$, $-0.3$ for ${\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1$, and $0.2$ for $\omega$ in Equation (26).

$\begin{array}{c}{\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}=R{T}_c\left\{{\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^0+\omega {\left[\frac{\underset{\_}{H}-{\underset{\_}{H}}^{ig}}{R{T}_c}\right]}^1\right\}\\ =\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{425}\text{.12}\text{K}\right)\left\{\left(-0.3\right)+0.2\left(-0.3\right)\right\}\\ =-1,272.40\text{J}/\text{mol}\end{array}$

Calculate the Lee Kesler correlation residuals enthalpy.

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2-{\underset{\_}{H}}_2^{ig}\right)+\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$        (27)

Re-write the Equation (27).

${\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)$        (28)

Substitute $-11,170\text{J}/\text{mol}$ for ${\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}$, and $-1,272.40\text{J}/\text{mol}$ for ${\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}$ in Equation (28).

$\begin{array}{c}{\underset{\_}{H}}_2-{\underset{\_}{H}}_1=\left({\underset{\_}{H}}_2^{ig}-{\underset{\_}{H}}_1^{ig}\right)-\left({\underset{\_}{H}}_1-{\underset{\_}{H}}_1^{ig}\right)\\ =\left(-11,170\text{J}/\text{mol}\right)-\left(-1,272.40\text{J}/\text{mol}\right)\\ =-9,897.6\text{J}/\text{mol}\left(0.001\text{kJ}/\text{mol}/1\text{J}/\text{mol}\right)\\ =-9.897\text{kJ}/\text{mol}\end{array}$

Calculate the actual work is done $\left({\dot{W}}_s\right)$

${\dot{W}}_s=\eta \left({\underset{\_}{H}}_2-{\underset{\_}{H}}_1\right)$        (29)

Substitute 80% for $\eta$ and $-9.897\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_2-{\underset{\_}{H}}_1$ in Equation (29).

$\begin{array}{c}{\dot{W}}_s=80\%\left(-9.897\text{kJ}/\text{mol}\right)\\ =0.8\left(-9.897\text{kJ}/\text{mol}\right)\\ =-7.9\text{kJ}/\text{mol}\end{array}$

Thus, the rate at which work is done is $-7.9\text{kJ}/\text{mol}$.