Chapter 12.6, Problem 98RP

a)

To determine

The enthalpy change per unit mass.

The entropy change per unit mass.

Answer to Problem 98RP

The enthalpy change per unit mass is $765.9\text{Btu}/\text{lbm}$.

The entropy change per unit mass is $0.4617\text{Btu}/\text{lbm}\cdot \text{R}$.

Explanation of Solution

Refer the table A-2Ec, “Ideal gas specific heats of various common gases”, obtain the empirical correlation for the methane as ${\overline{c}}_P=a+bT+c{T}^2+d{T}^3$.

Write the enthalpy change equation $\left({\overline{h}}_2-{\overline{h}}_1\right)$.

$\begin{array}{c}\left({\overline{h}}_2-{\overline{h}}_1\right)={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{c}_{P}dT}\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(a+bT+c{T}^2+d{T}^3\right)dT}\\ ={\left[aT+\frac{b}{2}{T}^2+\frac{c}{3}{T}^3+\frac{d}{4}{T}^4\right]}_{T_1}^{T_2}\\ =a\left(T_2-T_1\right)+\frac{b}{2}\left(T_2^2-T_1^2\right)+\frac{c}{3}\left(T_2^3-T_1^3\right)+\frac{d}{4}\left(T_2^4-T_1^4\right)\end{array}$ (I)

Here, specific heat capacity at constant pressure is $c_P$, initial temperature is $T_1$, final temperature is $T_2$, function of temperatures are $a$, $b$, $c$, and $d$.

Write the work input to the compressor $\left(w_{in}\right)$.

$w_{in}=\frac{{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}}{M}$ (II)

Here, molar mass of $M$.

Write the entropy change equation ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{Ideal}$.

$\begin{array}{c}{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{Ideal}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{c_P}{T}dT}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ =={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(\frac{a}{T}+b+cT+d{T}^2\right)dT}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ ={\left[a\ln \left(T\right)+bT+\frac{c}{2}{T}^2+\frac{d}{3}{T}^3\right]}_{T_1}^{T_2}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ =a\ln \left(\frac{T_2}{T_1}\right)+b\left(T_2^{}-T_1^{}\right)+\frac{c}{2}\left(T_2^2-T_1^2\right)+\frac{d}{3}\left(T_2^3-T_1^3\right)-R_u\ln \left(\frac{P_2}{P_1}\right)\end{array}$ (III)

Here, universal gas constant is $R_u$, initial pressure is $P_2$, and final pressure is $P_2$.

Write the entropy change equation ${\left(s_2-s_1\right)}_{Ideal}$.

${\left(s_2-s_1\right)}_{Ideal}=\frac{{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{Ideal}}{M}$ . (IV)

Conclusion:

Refer table A-1E, “Molar mass, gas constant and critical properties table”, obtain the molar mass, critical temperature, critical pressure, specific heat capacity at constant pressure, and gas constant of methane as $16.043\text{lbm}/\text{lbmol}$, $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$, $343.9\text{ R}$, $673\text{psia}$, and $0.1238\text{Btu}/\text{lbmol}\cdot \text{R}$.

Substitute $4.750$ for $a$, $1560\text{R}$ for $T_2$, $560\text{R}$ for $T_1$, $0.6666\times {10}^{-2}$ for $b$, $0.09352\times {10}^{-5}$ for $c$, and $-0.4510\times {10}^{-9}$ for $d$ in Equation (I).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}=\left\{\begin{array}{l}4.750\left(1560\text{R}-560\text{R}\right)+\frac{0.6666\times {10}^{-2}}{2}\left({\left(1560\text{R}\right)}^2-{\left(560\text{R}\right)}^2\right)+\\ \frac{0.09352\times {10}^{-5}}{3}\left({\left(1560\text{R}\right)}^3-{\left(560\text{R}\right)}^3\right)+\frac{-0.4510\times {10}^{-9}}{4}\left({\left(1560\text{R}\right)}^4-{\left(560\text{R}\right)}^4\right)\end{array}\right\}\\ =12288\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Substitute $12288\text{Btu}/\text{lbmol}\cdot \text{R}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}$ and $16.043\text{lbm}/\text{lbmol}$ for $M$ in Equation (II)

$\begin{array}{c}{w}_{in}=\frac{12288\text{Btu}/\text{lbmol}\cdot \text{R}}{16.043\text{lbm}/\text{lbmol}}\\ =765.9\text{Btu}/\text{lbm}\end{array}$

Thus, the enthalpy change per unit mass is $765.9\text{Btu}/\text{lbm}$.

Substitute $4.750$ for $a$, $1560\text{R}$ for $T_2$, $560\text{R}$ for $T_1$, $0.6666\times {10}^{-2}$ for $b$, $0.09352\times {10}^{-5}$ for $c$, $-0.4510\times {10}^{-9}$ for $d$, $500\text{psia}$ for $P_2$, $50\text{psia}$ for $P_1$, and $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R_u$ in Equation (III).

$\begin{array}{c}{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{Ideal}=\left\{\begin{array}{l}4.750\ln \left(\frac{1560\text{R}}{560\text{R}}\right)+0.6666\times {10}^{-2}\left(\left(1560\text{R}\right)-\left(560\text{R}\right)\right)+\frac{0.09352\times {10}^{-5}}{2}\left({\left(1560\text{R}\right)}^2-{\left(560\text{R}\right)}^2\right)+\\ \frac{-0.4510\times {10}^{-9}}{3}\left({\left(1560\text{R}\right)}^3-{\left(560\text{R}\right)}^3\right)-1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\times \ln \left(\frac{500\text{psia}}{50\text{psia}}\right)\end{array}\right\}\\ =7.407\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Substitute $7.407\text{Btu}/\text{lbmol}\cdot \text{R}$ for ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{Ideal}$ and $16.043\text{lbm}/\text{lbmol}$ for $M$ in Equation (IV).

$\begin{array}{c}{\left(s_2-s_1\right)}_{Ideal}=\frac{7.407\text{Btu}/\text{lbmol}\cdot \text{R}}{16.043\text{lbm}/\text{lbmol}}\\ =0.4617\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Thu, the entropy change per unit mass is $0.4617\text{Btu}/\text{lbm}\cdot \text{R}$.

b)

To determine

The change in enthalpy per unit mass.

The change in entropy per unit mass.

Answer to Problem 98RP

The change in enthalpy per unit mass is $767.4\text{Btu}/\text{lbm}$.

The change in entropy per unit mass is $0.4628\text{Btu}/\text{lbmol}\cdot \text{R}$.

Explanation of Solution

Calculate the reduced temperature $\left(T_{R1}\right)$ at initial state.

$T_{R1}=\frac{T_1}{T_{cr}}$ (V)

Here, critical temperature is $T_{cr}$ and initial temperature is $T_1$.

Calculate the reduced pressure $\left(P_{R1}\right)$ at initial state.

$P_{R1}=\frac{P_1}{P_{cr}}$ (VI)

Here, critical pressure is $P_{cr}$ and initial pressure is $P_1$.

Calculate the reduced temperature $\left(T_{R2}\right)$ at final state.

$T_{R2}=\frac{T_2}{T_{cr}}$ (VII)

Here, critical temperature is $T_{cr}$ and final temperature is $T_2$.

Calculate the reduced pressure $\left(P_{R2}\right)$ at final state.

$P_{R2}=\frac{P_2}{P_{cr}}$ (VIII)

Here, critical pressure is $P_{cr}$ and final pressure is $P_1$.

Calculate the change in enthalpy $\left(h_2-h_1\right)$ using generalized chart relation.

$h_2-h_1=R{T}_{cr}\left(Z_{h1}-Z_{h2}\right)+{\left(h_2-h_1\right)}_{ideal}$ (IX)

Here, change in enthalpy of ideal gas is ${\left(h_2-h_1\right)}_{ideal}$ and gas constant is $R$.

Calculate the change in entropy $\left(s_2-s_1\right)$ using generalized chart relation.

$\left(s_2-s_1\right)=R{T}_{cr}\left(Z_{s1}-Z_{s2}\right)+{\left(s_2-s_1\right)}_{ideal}$ (X)

Here, change in entropy of ideal gas is ${\left(s_2-s_1\right)}_{ideal}$.

Conclusion:

Substitute 560 R for $T_1$ and 343.9 R for $T_{cr}$ in Equation (V).

$\begin{array}{c}{T}_{R1}=\frac{560\text{R}}{343.9\text{R}}\\ =1.63\end{array}$

Substitute $50\text{psia}$ for $P_1$ and $673\text{psia}$ for $P_{cr}$ in Equation (VI).

$\begin{array}{c}{P}_{R1}=\frac{50\text{psia}}{673\text{psia}}\\ =0.0743\end{array}$

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”, select the compressibility factor $Z_{h1}$ and $Z_{s1}$ at initial state of reduced pressure and temperature of $1.63$ and $0.0743$ as $0.03313$ and $0.01617$

$\begin{array}{l}{Z}_{h1}=0.03313\\ Z_{s1}=0.01617\end{array}$

Substitute 1560 R for $T_2$ and 343.9 R for $T_{cr}$ in Equation (VII).

$\begin{array}{c}{T}_{R2}=\frac{1560\text{ R}}{\text{343}\text{.9 R}}\\ =4.54\end{array}$

Substitute $500\text{psia}$ for $P_2$ and $673\text{psia}$ for $P_{cr}$ in Equation (VIII).

$\begin{array}{c}{P}_{R2}=\frac{500\text{psia}}{673\text{psia}}\\ =0.743\end{array}$

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”, select the compressibility factor $Z_{h2}$ and $Z_{s2}$ at final state of reduced pressure and temperature of 4.54 and $0.743$ as 0 and 0.00695.

$\begin{array}{l}{Z}_{h2}=0\\ Z_{s2}=0.00695\end{array}$

Substitute $c_p\left(T_2-T_1\right)$ for ${\left(h_2-h_1\right)}_{ideal}$.

$h_2-h_1=R{T}_{cr}\left(Z_{h1}-Z_{h2}\right)+c_p\left(T_2-T_1\right)$

Here, specific heat at constant pressure is $c_p$, final temperature is $T_2$, and initial temperature is $T_1$.

Substitute 0 for $Z_{h2}$, 0.03313 for $Z_{h1}$, 1560 R for $T_2$, 560 R for $T_1$, $0.1238\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$, $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $c_p$, and 343.9 R for $T_{cr}$ in Equation (IX).

$\begin{array}{c}{h}_2-h_1=\left\{\begin{array}{l}0.1238\text{Btu}/\text{lbmol}\cdot \text{R}\left(343.9\text{R}\right)\left(0.03313-0\right)+\\ 1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\left(1560\text{ R}-560\text{ R}\right)\end{array}\right\}\\ =767.4\text{Btu}/\text{lbm}\end{array}$

Thus, the change in enthalpy per unit mass is $767.4\text{Btu}/\text{lbm}$.

Substitute 0.00695 for $Z_{s2}$, 0.01617 for $Z_{s1}$, 1560 R for $T_2$, 560 R for $T_1$, $0.1238\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$, $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $c_p$, and 343.9 R for $T_{cr}$ in Equation (X).

$\begin{array}{c}{s}_2-s_1=\left\{\begin{array}{l}0.1238\text{Btu}/\text{lbmol}\cdot \text{R}\left(343.9\text{R}\right)\left(0.01617-0.00695\right)+\\ 1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\left(1560\text{ R}-560\text{ R}\right)\end{array}\right\}\\ =0.4628\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Thus, the change in entropy per unit mass is $0.4628\text{Btu}/\text{lbmol}\cdot \text{R}$.