Chapter 12.6, Problem 70P

a)

To determine

The change in enthalpy and entropy of water vapor using departure charts.

Answer to Problem 70P

The change in enthalpy and entropy of water vapor using property table are $\underset{\_}{200\text{Btu}/\text{lbm}}$ and $\underset{\_}{0.191\text{Btu}/\text{lbm}\cdot \text{R}}$.

Explanation of Solution

Write the mean change in enthalpy ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{ideal}$ of ideal gas.

${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{ideal}={\left({\overline{h}}_2\right)}_{ideal}-{\left({\overline{h}}_1\right)}_{ideal}$ (I)

Here, enthalpy of ideal gas at temperature of 280 K is ${\left({\overline{h}}_2\right)}_{ideal}$ and enthalpy of ideal gas at temperature of 250 K is ${\left({\overline{h}}_1\right)}_{ideal}$

Write the mean change in entropy $\left({\left({\overline{s}}_2-{\overline{s}}_1\right)}_{ideal}\right)$ for ideal gas.

${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{ideal}={\overline{s}}^{\circ }{}_2-{\overline{s}}^{\circ }{}_1-R\ln \frac{P_2}{P_1}$ (II)

Here, gas constant is R, initial pressure is $P_1$, final pressure is $P_2$, entropy at final state is ${\overline{s}}^{\circ }{}_2$ and entropy at initial state is ${\overline{s}}^{\circ }{}_1$.

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

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

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

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

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

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

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

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

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

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

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

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

Write the change in enthalpy $\left({\overline{h}}_2-{\overline{h}}_1\right)$ of water vapor using generalized chart relation.

${\overline{h}}_2-{\overline{h}}_1={\left({\overline{h}}_2-{\overline{h}}_1\right)}_{ideal}-R{T}_{cr}\left(Z_{h1}-Z_{h2}\right)$ (VII)

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

Write the change in entropy $\left(s_2-s_1\right)$ of water vapor using generalized chart relation.

${\overline{s}}_2-{\overline{s}}_1={\left({\overline{s}}_2-{\overline{s}}_1\right)}_{ideal}-R\left(Z_{s1}-Z_{s2}\right)$ (VIII)

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

Write the change in enthalpy $\left(h_2-h_1\right)$ of water vapor during a change of state.

$\left(h_2-h_1\right)=\frac{\left({\overline{h}}_2-{\overline{h}}_1\right)}{M_{H_{2}O}}$ (IX)

Here, molar mass of water vapor is $M_{H_{2}O}$.

Write the change in entropy $\left(s_2-s_1\right)$ of water vapor during a change of state.

$\left(s_2-s_1\right)=\frac{\left({\overline{s}}_2-{\overline{s}}_1\right)}{M_{H_{2}O}}$ (X)

Here, molar mass of water vapor is $M_{H_{2}O}$.

Conclusion:

Refer to Table A-4E, “saturated water temperature table”.

Obtain the value of initial pressure $\left(P_1\right)$ at the initial temperature $\left(T_1\right)$ of $400°\text{F}$ as $247.26\text{ psia}$.

From the ideal gas properties of water vapor table A-23E, select the enthalpy of water vapor at temperature of 860 R and 1260 R as $6895.6\text{Btu}/\text{lbmol}$ and $10354.9\text{Btu}/\text{lbmol}$.

$\begin{array}{l}{\left({\overline{h}}_2\right)}_{ideal}=6,895.6\text{Btu}/\text{lbmol}\\ {\left({\overline{h}}_1\right)}_{ideal}=10,354.9\text{Btu}/\text{lbmol}\end{array}$

Substitute $10354.9\text{Btu}/\text{lbmol}$ for ${\left(h_2\right)}_{ideal}$ and $6,895.6\text{Btu}/\text{lbmol}$ for ${\left(h_1\right)}_{ideal}$ in Equation (I).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{ideal}=10,354.9\text{Btu}/\text{lbmol}-6,895.6\text{Btu}/\text{lbmol}\\ =3,459.3\text{Btu}/\text{lbmol}\end{array}$

From the ideal gas specific heats at various common gases table A-23E, select the gas constant of steam as $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$.

From the ideal gas properties of water vapor table A-23E, select the entropy of water vapor at temperature of 860 R and 1260 R as $48.916\text{Btu}/\text{lbmol}\cdot \text{R}$ and $52.212\text{Btu}/\text{lbmol}\cdot \text{R}$.

$\begin{array}{l}{\overline{s}}^{\circ }{}_2=52.212\text{Btu}/\text{lbmol}\cdot \text{R}\\ {\overline{s}}^{\circ }{}_1=48.916\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Substitute $52.212\text{Btu}/\text{lbmol}\cdot \text{R}$ for ${\overline{s}}^{\circ }{}_2$, $48.916\text{Btu}/\text{lbmol}\cdot \text{R}$ for ${\overline{s}}^{\circ }{}_1$, $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for R, $247.2\text{psi}$ for $P_2$ and $247.2\text{psi}$ for $P_1$ in Equation (II).

$\begin{array}{c}{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{ideal}=\left(52.212-48.916\right)\text{Btu}/\text{lbmol}\cdot \text{R}-\left(1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\right)\ln \left(\frac{247.2\text{psi}}{247.2\text{psi}}\right)\\ =3.296\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

From the Nelson-Obert generalized compressibility chart A-15, select the initial state of compressibility factor $Z_{s_1}$ and $Z_{h_1}$ at reduced pressure and temperature of 0.07727 and 0.7380 as 0.1171 and 0.1342.

$\begin{array}{c}{Z}_{s_1}=0.1171\\ Z_{h_1}=0.1342\end{array}$

Substitute 860 R for $T_1$ and 1,164.8 R for $T_{cr}$ in Equation (III).

$\begin{array}{c}{T}_{R1}=\frac{860\text{R}}{1,164.8\text{R}}\\ =0.7380\end{array}$

Substitute $247.26\text{psi}$ for $P_1$ and $3200\text{psi}$ for $P_{cr}$ in Equation (IV).

$\begin{array}{c}{P}_{R1}=\frac{247.26\text{psi}}{3,200\text{psi}}\\ =0.07727\end{array}$

Substitute 1,260 R for $T_2$ and 1,164.8 R for $T_{cr}$ in Equation (V).

$\begin{array}{c}{T}_{R2}=\frac{1,260\text{R}}{1,164.8\text{R}}\\ =1.081\end{array}$

Substitute $680.56\text{psi}$ for $P_2$ and $3,200\text{psi}$ for $P_{cr}$ in Equation (VI).

$\begin{array}{c}{P}_{R2}=\frac{680.56\text{psi}}{3,200\text{psi}}\\ =0.07727\end{array}$

From the Nelson-Obert generalized compressibility chart A-15, select the initial state of compressibility factor $Z_{s_2}$ and $Z_{h_2}$ at reduced pressure and temperature of 0.07727 and 1.081 as 0.04595 and 0.07213.

$\begin{array}{c}{Z}_{s_2}=0.04595\\ Z_{h_2}=0.07213\end{array}$

Substitute 0.07213 for $Z_{h2}$, 0.1342 for $Z_{h1}$, $3459.3\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{ideal}$, $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$ and $1164.8\text{R}$ for $T_{cr}$ in Equation (VII).

$\begin{array}{c}{\overline{h}}_2-{\overline{h}}_1=\left[\begin{array}{l}3,459.3\text{Btu}/\text{lbmol}-\left(1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\right)\\ \left(1,164.8\text{R}\right)\left(0.07213-0.1342\right)\end{array}\right]\\ =3,602.9\text{Btu}/\text{lbmol}\end{array}$

Substitute $3.296\text{Btu}/\text{lbmol}\cdot \text{R}$ for ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{ideal}$, 0.04595 for $Z_{s2}$, 0.1171 for $Z_{s1}$ and $1.9858\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$ in Equation (VIII).

$\begin{array}{c}{\overline{s}}_2-{\overline{s}}_1=3.296\text{Btu}/\text{lbmol}\cdot \text{R}-\left(1.9858\text{Btu}/\text{lbmol}\cdot \text{R}\right)\left(0.04595-0.1171\right)\\ =3.4373\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

From the molar mass properties table A-1, select the molar mass $M_{H_{2}O}$ of water vapor as $18.015\text{lbm}/\text{lbmol}$.

Substitute $3,602.9\text{Btu}/\text{lbmol}$ for $\left({\overline{h}}_2-{\overline{h}}_1\right)$ and $18.015\text{lbm}/\text{lbmol}$ for $M_{H_{2}O}$ in Equation (IX).

$\begin{array}{c}\left(h_2-h_1\right)=\frac{3,602.9\text{Btu}/\text{lbmol}}{18.015\text{lbm}/\text{lbmol}}\\ =200\text{Btu}/\text{lbm}\end{array}$

Thus, the change in enthalpy $\left(h_2-h_1\right)$ of water vapor during a change of state is $\underset{\_}{200\text{Btu}/\text{lbm}}$.

Substitute $3.4373\text{Btu}/\text{lbmol}\cdot \text{R}$ for $\left({\overline{s}}_2-{\overline{s}}_1\right)$ and $18.015\text{lbm}/\text{lbmol}$ for $M_{H_{2}O}$ in Equation (X).

$\begin{array}{c}\left(s_2-s_1\right)=\frac{3.4373\text{Btu}/\text{lbmol}\cdot \text{R}}{18.015\text{lbm}/\text{lbmol}}\\ =0.191\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Thus, the change in entropy $\left(s_2-s_1\right)$ of water vapor during a change of state is $\underset{\_}{0.1908\text{Btu}/\text{lbm}\cdot \text{R}}$.

b)

To determine

The change in enthalpy and entropy of water vapor using property table.

Answer to Problem 70P

The change in enthalpy and entropy of water vapor using property table are $\underset{\_}{222\text{Btu}/\text{lbm}}$ and $\underset{\_}{0.214\text{Btu}/\text{lbm}\cdot \text{R}}$.

Explanation of Solution

Write the change in enthalpy $\left(\Delta h\right)$ of water vapor using property table.

$\Delta h=\left(h_2-h_1\right)$ (XI)

Write the change in entropy $\left(\Delta s\right)$ of water vapor using property table.

$\Delta s=\left(s_2-s_2\right)$ (XII)

Refer to Table A-4, “saturated water temperature table”.

Obtain the initial enthalpy $\left(h_1\right)$ of saturated liquid and entropy $\left(s_1\right)$ of saturated liquid at temperature of $400°\text{F}$ as $1201.4\text{Btu}/\text{lbm}$ and $1.5279\text{Btu}/\text{lbm}\cdot \text{R}$ respectively.

$\begin{array}{l}{h}_1=1,201.4\text{Btu}/\text{lbm}\\ s_1=1.5279\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Refer to Table A-4, “saturated water pressure table”.

Obtain the final enthalpy $\left(h_2\right)$ of saturated liquid and entropy $\left(s_2\right)$ of saturated liquid at pressure of 247.26 psi as $1,423.6\text{Btu}/\text{lbm}$ and $1.7418\text{Btu}/\text{lbm}\cdot \text{R}$ respectively.

$\begin{array}{l}{h}_2=1,423.6\text{Btu}/\text{lbm}\\ s_2=1.7418\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Conclusion:

Substitute $1,201.4\text{Btu}/\text{lbm}$ for $h_1$ and $1,423.6\text{Btu}/\text{lbm}$ for $h_2$ in Equation (XI).

$\begin{array}{c}\Delta h=1,423.6\text{Btu}/\text{lbm}-1,201.4\text{Btu}/\text{lbm}\\ =222\text{Btu}/\text{lbm}\end{array}$

Thus, the change in enthalpy $\left(\Delta h\right)$ of water vapor using property table is $\underset{\_}{222\text{Btu}/\text{lbm}}$.

Substitute $1.5279\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_1$ and $1.7418\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_2$ in Equation (XII).

$\begin{array}{c}\Delta s=1.7418\text{Btu}/\text{lbm}\cdot \text{R}-1.5279\text{Btu}/\text{lbm}\cdot \text{R}\\ =0.214\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Thus, the change in entropy $\left(\Delta s\right)$ of water vapor using property table is $\underset{\_}{0.214\text{Btu}/\text{lbm}\cdot \text{R}}$ .