Chapter 13.3, Problem 63P

To determine

The work required to pump the gas.

Answer to Problem 63P

The work input is $1.86\times {10}^8\text{Btu}$.

Explanation of Solution

Write the expression to obtain the mole number of ${\text{CH}}_4$ $\left(N_{{\text{CH}}_4}\right)$.

$N_{{\text{CH}}_4}=\frac{m_{{\text{CH}}_4}}{M_{{\text{CH}}_4}}$ (I)

Here, molar mass of ${\text{CH}}_4$ is $M_{{\text{CH}}_4}$ and mass of ${\text{CH}}_4$ is $m_{{\text{CH}}_4}$.

Write the expression to obtain the mole number of ${\text{C}}_2{\text{H}}_6$ $\left(N_{{\text{C}}_2{\text{H}}_6}\right)$.

$N_{{\text{C}}_2{\text{H}}_6}=\frac{m_{{\text{C}}_2{\text{H}}_6}}{M_{{\text{C}}_2{\text{H}}_6}}$ (II)

Here, molar mass of ${\text{C}}_2{\text{H}}_6$ is $M_{{\text{C}}_2{\text{H}}_6}$ and mass of ${\text{C}}_2{\text{H}}_6$ is $m_{{\text{C}}_2{\text{H}}_6}$.

Write the expression to obtain the total number of moles $\left(N_m\right)$.

$N_m=N_{{\text{CH}}_{\text{4}}}+N_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ (III)

Write the expression to obtain the mole fraction of ${\text{CH}}_4$$\left(y_{{\text{CH}}_4}\right)$.

$y_{{\text{CH}}_4}=\frac{N_{{\text{CH}}_4}}{N_m}$ (IV)

Write the expression to obtain the mole fraction of ${\text{C}}_2{\text{H}}_6$$\left(y_{{\text{C}}_2{\text{H}}_6}\right)$.

$y_{{\text{C}}_2{\text{H}}_6}=\frac{N_{{\text{C}}_2{\text{H}}_6}}{N_m}$ (V)

Write the expression to obtain the apparent molecular weight of the mixture $\left(M_m\right)$.

$M_m=\frac{m_m}{N_m}$ (VI)

Write the expression to obtain the constant pressure specific heat of the mixture $\left(c_p\right)$.

$c_p={\text{mf}}_{C{H}_4}{c}_p{}_{,C{H}_4}+{\text{mf}}_{C_2{H}_6}{c}_p{}_{,C_2{H}_6}$ (VII)

Here, mole fraction of ${\text{CH}}_{\text{4}}\text{,}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is ${\text{mf}}_{{\text{CH}}_{\text{4}}},\text{and}{\text{mf}}_{{\text{C}}_2{\text{H}}_6}$, specific heat of ${\text{CH}}_4$ is $c_p{}_{,C{H}_4}$, and specific heat of ${\text{C}}_2{\text{H}}_6$ is $c_p{}_{,C_2{H}_6}$.

Write the expression to obtain the apparent gas constant of the mixture $\left(R\right)$.

$R=\frac{R_u}{M_m}$ (VIII)

Here, universal gas constant is $R_u$.

Write the expression to obtain the pseudo-critical temperature of the mixture $\left({T^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{T}_{\text{cr},i}}\\ =y_{{\text{CH}}_{\text{4}}}{T}_{{\text{cr,CH}}_{\text{4}}}+y_{{\text{C}}_2{\text{H}}_6}{T}_{{\text{cr,C}}_2{\text{H}}_6}\end{array}$ (IX)

Here, critical temperature of ${\text{CH}}_4,\text{and}{\text{C}}_2{\text{H}}_6$ are $T_{{\text{cr,CH}}_4},and{T}_{{\text{cr,C}}_2{\text{H}}_6}$.

Write the expression to obtain the pseudo-critical pressure of the mixture $\left({P^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{P}_{\text{cr},i}}\\ =y_{{\text{CH}}_{\text{4}}}{P}_{{\text{cr,CH}}_{\text{4}}}+y_{{\text{C}}_2{\text{H}}_6}{P}_{{\text{cr,C}}_2{\text{H}}_6}\end{array}$ (X)

Here, critical pressure of ${\text{CH}}_4,\text{and}{\text{C}}_2{\text{H}}_6$ are $P_{{\text{cr,CH}}_4},and{P}_{{\text{cr,C}}_2{\text{H}}_6}$.

Write the expression to obtain the reduced temperature $\left(T_R\right)$.

$T_R=\frac{T_m}{{T^{\prime }}_{\text{cr},m}}$ (XI)

Write the expression to obtain the reduced pressure $\left(P_R\right)$.

$P_R=\frac{P_m}{{P^{\prime }}_{\text{cr},m}}$ (XII)

Write the expression to obtain the mass of the mixture $\left(m\right)$.

$m=\frac{V}{\frac{Z_{m}RT}{P}}$ (XIII)

Write the expression to obtain the initial reduced temperature $\left(T_{R,1}\right)$.

$T_{R\text{,1}}=\frac{T_m}{{T^{\prime }}_{\text{cr},m}}$ (XIV)

Here, mixing critical temperature is ${T^{\prime }}_{\text{cr},m}$.

Write the expression to obtain the initial reduced pressure $\left(P_{R,1}\right)$.

$P_{R\text{,1}}=\frac{P_m}{{P^{\prime }}_{\text{cr},m}}$ (XV)

Here, mixing critical pressure is ${P^{\prime }}_{\text{cr},m}$.

Write the expression to obtain the final reduced temperature $\left(T_{R,2}\right)$.

$T_{R\text{,2}}=\frac{T_m}{{T^{\prime }}_{\text{cr},m}}$ (XVI)

Write the expression to obtain the final reduced pressure $\left(P_{R,2}\right)$.

$P_{R\text{,2}}=\frac{P_m}{{P^{\prime }}_{\text{cr},m}}$ (XVII)

Write the expression to obtain the enthalpy change for the ideal gas mixture.

${\left(h_1-h_2\right)}_{\text{ideal}}=c_p\left(T_1-T_2\right)$ (XVIII)

Here, initial state enthalpy is $h_1$ and final state enthalpy is $h_2$.

Write the expression to obtain the enthalpy change with departure factors.

$h_1-h_2={\left(h_1-h_2\right)}_{\text{ideal}}-R{T^{\prime }}_{\text{cr},m}\left(Z_{h1}-Z_{h2}\right)$ (XIX)

Write the expression to obtain the work input $\left(W_{\text{in}}\right)$.

$W_{\text{in}}=m\left(h_1-h_2\right)$ (XX)

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of ${\text{CH}}_{\text{4}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ as $16.0\text{lbm}/\text{kmol}$ and $30.0\text{lbm}/\text{kmol}$.

Substitute $75\text{lbm}$ for $m_{{\text{CH}}_4}$ and $16.0\text{lbm}/\text{mol}$ for $M_{{\text{CH}}_4}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{CH}}_4}=\frac{75\text{lbm}}{16.0\text{lbm}/\text{mol}}\\ =4.6875\text{lbmol}\end{array}$

Substitute $25\text{lbm}$ for $m_{{\text{C}}_2{\text{H}}_6}$ and $30\text{lbm}/\text{mol}$ for $M_{{\text{C}}_2{\text{H}}_6}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{C}}_2{\text{H}}_6}=\frac{25\text{lbm}}{30.0\text{lbm}/\text{mol}}\\ =0.8333\text{lbmol}\end{array}$

Substitute $4.6875\text{lbmol}$ for $N_{{\text{CH}}_4}$, and $0.8333\text{lbmol}$ for $N_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ in Equation (III).

$\begin{array}{c}{N}_m=4.6875\text{lbmol}+0.8333\text{lbmol}\\ =5.5208\text{lbmol}\end{array}$

Substitute $4.6875\text{lbmol}$ for $N_m$ and $5.5208\text{lbmol}$ for $N_{{\text{CH}}_4}$ in Equation (IV).

$\begin{array}{c}{y}_{{\text{CH}}_4}=\frac{4.6875\text{lbmol}}{5.5208\text{lbmol}}\\ =0.8941\end{array}$

Substitute $4.6875\text{lbmol}$ for $N_m$ and $5.5208\text{lbmol}$ for $N_{{\text{C}}_2{\text{H}}_6}$ in Equation (V).

$\begin{array}{c}{y}_{{\text{C}}_2{\text{H}}_6}=\frac{0.8333\text{lbmol}}{5.5208\text{lbmol}}\\ =0.1509\end{array}$

Substitute 5.5208 lbmol for $N_m$ and 100 lbm for $m_m$ in Equation (VI).

$\begin{array}{c}{M}_m=\frac{100\text{kg}}{5.5208\text{lbmol}}\\ =18.11\text{lb}/\text{kmol}\end{array}$

From the Table of ideal gas specific heats of various common gases, write the constant pressure specific heats of ${\text{CH}}_{\text{4}}$, and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ at $300°\text{F}$ as $0.532\text{Btu}/\text{lbm}\cdot \text{R}$ and $0.427\text{Btu}/\text{lbm}\cdot \text{R}$.

Substitute $0.532\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p{}_{,C{H}_4}$, $0.427\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p{}_{,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$, 0.75 for ${\text{mf}}_{{\text{CH}}_{\text{4}}}$, and 0.25 for ${\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ in Equation (VII).

$\begin{array}{c}{c}_p=\left(0.75\right)\left(0.532\text{Btu}/\text{lbm}\cdot \text{R}\right)+\left(0.25\right)\left(0.427\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ =0.506\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R_u$ and $18.11\text{lbm}/\text{lbmol}$ for $M_m$ in Equation (VIII).

$\begin{array}{c}R=\frac{1.9858\text{Btu}/\text{lbmol}\cdot \text{R}}{18.11\text{lbm}/\text{lbmol}}\\ =0.1097\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

From the Table of critical properties, write the critical temperature and pressure for ${\text{CH}}_{\text{4}}$ as 343.9 R and 673 psia and critical temperature and pressure for ${\text{C}}_2{\text{H}}_6$ as 549.8 R and 708 psia.

Substitute 0.8491 for $y_{{\text{CH}}_4}$, 343.9 R for $T_{{\text{cr,CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, and 549.8 R for $T_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (IX).

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}=0.8491\left(343.9\text{R}\right)+0.1509\left(549.8\text{R}\right)\\ =375.0\text{R}\end{array}$

Substitute 0.8491 for $y_{{\text{CH}}_4}$, 673 psia for $P_{{\text{cr,CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, and 708 psia for $P_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (X).

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}=0.8491\left(673\text{psia}\right)+0.1509\left(708\text{psia}\right)\\ =678.3\text{psia}\end{array}$

Substitute 760 R for $T_m$ and 375.0 R for ${T^{\prime }}_{\text{cr},m}$ in Equation (XI).

$\begin{array}{c}{T}_R=\frac{760\text{R}}{375.0\text{R}}\\ =2.027\end{array}$

Substitute 1,300 psia for $P_m$ and 678.3 psia for ${P^{\prime }}_{\text{cr},m}$ in Equation (XII).

$\begin{array}{c}{P}_R=\frac{1,300\text{psia}}{678.3\text{psia}}\\ =1.917\end{array}$

From the Table of Nelson-Obert generalized compressibility chart, write the compressibility factor, $Z_m$ as 0.963.

Substitute 0.963 for $Z_m$, $2\times {10}^6{\text{ft}}^{\text{3}}$ for V, $0.5925\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, 760 R for T, and 1,300 psia for P in Equation (XIII).

$\begin{array}{c}m=\frac{2\times {10}^6{\text{ft}}^{\text{3}}}{\frac{0.963\left(0.5925\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)760\text{R}}{1,300\text{psia}}}\\ =5.996\times {10}^6\text{lbm}\end{array}$

Substitute 760 R for $T_m$ and 375.0 R for ${T^{\prime }}_{\text{cr},m}$ in Equation (XIV).

$\begin{array}{c}{T}_{R,1}=\frac{760\text{R}}{375.0\text{R}}\\ =2.027\end{array}$

Substitute 1,300 psia for $P_m$ and 678.3 psia for ${P^{\prime }}_{\text{cr},m}$ in Equation (XV).

$\begin{array}{c}{P}_{R,1}=\frac{1,300\text{psia}}{678.3\text{psia}}\\ =1.917\end{array}$

From the Table of generalized enthalpy departure chart, write the initial enthalpy departure, $Z_{h1}$ as 0.487.

Substitute 660 R for $T_m$ and 375.0 R for ${T^{\prime }}_{\text{cr},m}$ in Equation (XVI).

$\begin{array}{c}{T}_{R,2}=\frac{660\text{R}}{375.0\text{R}}\\ =1.76\end{array}$

Substitute 20 psia for $P_m$ and 678.3 psia for ${P^{\prime }}_{\text{cr},m}$ in Equation (XVII).

$\begin{array}{c}{P}_{R,2}=\frac{20\text{psia}}{678.3\text{psia}}\\ =0.0295\end{array}$

From the Table of generalized enthalpy departure chart, write the final enthalpy departure, $Z_{h2}$ as 0.0112.

Substitute $0.506\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, 760 R for $T_1$, and 660 R for $T_2$ in Equation (XVIII).

$\begin{array}{c}{\left(h_1-h_2\right)}_{\text{ideal}}=0.506\text{Btu}/\text{lbm}\cdot \text{R}\left(760\text{R}-660\text{R}\right)\\ =50.6\text{Btu}/\text{lbm}\end{array}$

Substitute $50.6\text{Btu}/\text{lbm}$ for ${\left(h_1-h_2\right)}_{\text{ideal}}$, 375 R for ${T^{\prime }}_{\text{cr},m}$ $0.1097\text{Btu}/\text{lbm}\cdot \text{R}$ for R, 0.487 for $Z_{h1}$, 0.0112 for $Z_{h2}$ in Equation (XIX).

$\begin{array}{c}{h}_1-h_2=50.6\text{Btu}/\text{lbm}-\left(0.1097\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(375\text{R}\right)\left(0.487-0.0112\right)\\ =31.02\text{Btu}/\text{lbm}\end{array}$

Substitute $31.02\text{Btu}/\text{lbm}$ for $h_1-h_2$ and $5.996\times {10}^6\text{lbm}$ for m in Equation (XX).

$\begin{array}{c}{W}_{\text{in}}=5.996\times {10}^6\text{lbm}\left(31.02\text{Btu}/\text{lbm}\right)\\ =1.86\times {10}^8\text{Btu}\end{array}$

Thus, the work input is $1.86\times {10}^8\text{Btu}$.