Chapter 4, Problem 89P

To determine

The final temperature using the ideal gas equation, the compressibility chart, and steam table.

Explanation of Solution

Given:

The temperature of the saturated water vapor $\left(T\right)$ is $350°\text{C}$.

The constant pressure process.

The final specific volume $\left(v_2\right)$  is equal to double the initial specific volume $\left(v_1\right)$ .

$v_2=2v_1$.

Calculation:

Refer to Table A-1, “Molar mass, gas constant, and critical-point properties”, to obtain the gas constant $\left(R\right)$, the critical temperature $\left({\text{T}}_{\text{cr}}\right)$, and the critical pressure of water $\left({\text{P}}_{\text{cr}}\right)$ as $0.4615\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$, $647.1\text{ K}$, and $22.06\text{ MPa}$.

Write the expression to obtain the final temperature using the ideal gas equation.

  $\begin{array}{c}{T}_2=\left(T_1\right)\times \left(\frac{v_2}{v_1}\right)\\ T_2=\left(350°\text{C}+273\right)\text{K}\times \left(\frac{2v_1}{v_1}\right)\\ =1246\text{ K}\end{array}$

Thus, the final temperature using the ideal gas equation is $1246\text{ K}$.

Refer Table A-4, “Saturated water: Temperature table”, to obtain the pressure at $350°\text{C}$ temperature and quality of 1 as $16,529\text{ kPa}$.

Write the expression to obtain the initial reduced pressure $\left(P_{R1}\right)$.

  $\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{\text{cr}}}\\ P_{R1}=\frac{16529\text{kPa}\left(\frac{1\text{ MPa}}{{10}^3\text{kPa}}\right)}{22.06\text{MPa}}\\ =0.749\end{array}$

Write the expression to obtain the initial reduced temperature $\left(T_{R1}\right)$.

  $\begin{array}{c}{T}_{R1}=\frac{T_1}{T_{\text{cr}}}\\ T_{R1}=\frac{623\text{K}}{647.1\text{K}}\\ =0.963\end{array}$

Refer Table A-15, “Nelson-Obert generalized compressibility chart”, obtain the value of the initial compressibility factor $\left(Z_1\right)$ and initial pseudo-reduced specific volume $\left(v_{R1}\right)$ corresponding to the initial reduced pressure $\left(P_{R1}\right)$ of $0.749$ and the initial reduced temperature $\left(T_{R1}\right)$ of $0.963$ as $0.593$ and $0.75$ respectively.

Write the expression to obtain the final pseudo-reduced specific volume $\left(v_{R2}\right)$.

  $\begin{array}{c}{v}_{R2}=\left(1.8\right)v_{R1}\\ v_{R2}=\left(2\right)\left(0.75\right)\\ =1.5\end{array}$

Obtain the final reduced pressure $\left(P_{R2}\right)$.

  $P_{R2}=P_{R1}=0.749$

Refer Table A-15, “Nelson-Obert generalized compressibility chart”, obtain the value of the final compressibility factor corresponding to the final reduced pressure $\left(P_{R2}\right)$ of $0.749$ and the final pseudo-reduced specific volume $\left(v_{R2}\right)$ of $1.5$ as $0.88$.

Write the expression to obtain the final temperature $\left(T_2\right)$.

  $\begin{array}{c}{T}_2=\frac{P_2{v}_2}{Z_{2}R}\\ =\frac{P_2}{Z_2}\frac{v_{R2}{T}_{\text{cr}}}{P_{\text{cr}}}\end{array}$

  $\begin{array}{c}{T}_2\text{=}\frac{16529\text{ kPa}\times 1.5\times 647.1\text{ K}}{0.88\times 22060\text{ kPa}}\\ =826\text{ K}\end{array}$

Thus, the final temperature using the compressibility chart is $826\text{ K}$.

Refer Table A-4, “Saturated water: Temperature table”, to obtain the pressure and specific volume at $350°\text{C}$ temperature and quality of 1 as $16,529\text{ kPa}$ and $0.008806{\text{m}}^3/\text{kg}$.

Obtain the final specific volume $\left(v_2\right)$.

  $\begin{array}{c}{v}_2=2v_1\\ =2\left(0.008806{\text{m}}^3/\text{kg}\right)\\ =0.017612{\text{m}}^3/\text{kg}\end{array}$

Refer to Table A-6, “Superheated water”, obtain the final temperature $\left(T_2\right)$ as $477°\text{C}$ corresponding to pressure at $16,529\text{ kPa}$ and specific volume at $0.017612{\text{m}}^3/\text{kg}$.

Convert the temperature from $°\text{C}$ to $\text{K}$.

  $\begin{array}{c}{T}_2=\left(477°\text{C}+273\right)\text{ K}\\ =\text{750 K}\end{array}$

Thus, the final temperature using the superheated steam table is $750\text{ K}$.