Chapter 3.8, Problem 92P

(a)

To determine

The final temperature using the ideal gas equation.

Answer to Problem 92P

The final temperature using the ideal gas equation is $\underset{\_}{1720\text{R}}$.

Explanation of Solution

Determine the final temperature using the ideal gas equation.

$T_2=\left(T_1\right)\times \left(\frac{v_2}{v_1}\right)$ (I)

Here, the initial specific volume is $v_1$, the final specific volume is $v_2$, and the initial temperature is $T_1$.

Conclusion:

Refer to Table A-1E to find the gas constant, the critical temperature, and the critical pressure of water as $0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$, $1164.8\text{R}$, and $3200\text{psia}$.

Substitute $400°\text{F}$ for $T_1$, $2v_1$ for $v_1$ in Equation (I).

$\begin{array}{c}{T}_2=\left(400°\text{F}\right)\times \left(\frac{2v_1}{v_1}\right)\\ =\left(400+460\text{R}\right)\left(2\right)\\ =1720\text{R}\end{array}$

Thus, the final temperature using the ideal gas equation is $\underset{\_}{1720\text{R}}$.

(b)

To determine

The final temperature using the compressibility chart.

Answer to Problem 92P

The final temperature using the compressibility chart is $\underset{\_}{1571\text{R}}$.

Explanation of Solution

Determine the reduced pressure at final state.

$P_{R_2}=\frac{P_2}{P_{\text{cr}}}$ (II)

Here, the critical pressure is $P_{\text{cr}}$.

Determine the reduced specific volume at the final state.

$v_{R_2}=\frac{v_2}{R{T}_{\text{cr}}/P_{\text{cr}}}$ (III)

Here, the final state specific volume is $v_2$.

Determine the final temperature using the compressibility chart.

$T_2=\frac{P_2{v}_2}{Z_{2}R}$ (IV)

Conclusion:

Refer Table A-4E to obtain the value of initial pressure and specific volume at the $400°\text{F}$ temperature as $247.26\text{psia}$ and $1.8639{\text{ft}}^{\text{3}}/\text{lbm}$.

Substitute $247.26\text{psia}$ for $P_2$ and $3200\text{psia}$ for $P_{\text{cr}}$ in Equation (II).

$\begin{array}{c}{P}_{R_2}=\frac{\left(247.26\text{psia}\right)}{\left(3200\text{psia}\right)}\\ =0.0773\end{array}$

Substitute $2v_1$ for $v_2$, $1.8639{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_1$, $0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R$, $1164.8\text{R}$ for $T_{\text{cr}}$, and $3200\text{psia}$ for $P_{\text{cr}}$ in Equation (V).

$\begin{array}{c}{v}_{R_2}=\frac{2v_1}{\left(0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(1164.8\text{R}\right)/\left(3200\text{psia}\right)}\\ =\frac{2\times \left(1.8639{\text{ft}}^{\text{3}}/\text{lbm}\right)}{\left(0.216798{\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =\frac{\left(3.7278{\text{ft}}^{\text{3}}/\text{lbm}\right)}{\left(0.216798{\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =17.19\end{array}$

From the Figure A-15, “Nelson-Obert generalized compressibility chart” to obtain the value of compressibility factor at the final state at final reduced pressure and volume of 1.0773 and 17.19 as 0.985.

Substitute $247.26\text{psia}$ for $P_2$, $0.985$ for $Z_2$, $0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R$, $2v_1$ for $v_2$, and $1.8639{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_1$ in Equation (VI).

$\begin{array}{c}{T}_2=\frac{\left(247.26\text{psia}\right)\left(2v_1\right)}{\left(0.985\right)\left(0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =\frac{\left(247.26\text{psia}\right)\left(2\times \left(1.8639{\text{ft}}^{\text{3}}/\text{lbm}\right)\right)}{\left(0.985\right)\left(0.5956\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =\frac{921.7358\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}}{0.58666\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}}\\ =1571\text{R}\end{array}$

Thus, the final temperature using the compressibility chart is $\underset{\_}{1571\text{R}}$.

(c)

To determine

The final temperature using the superheated steam table.

Answer to Problem 92P

The final temperature using the superheated steam table is $\underset{\_}{1578\text{R}}$.

Explanation of Solution

Refer to Table A-6E, “Superheated water”, obtain the below properties at the final specific volume $3.7278{\text{ft}}^{\text{3}}/\text{lbm}$ using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VII)

Here, the variables denote by x and y are temperature and final specific volume.

Show the temperature at $1000°\text{F}$ and $1200°\text{F}$ as in Table (1).

S. No	final specific volume ${\text{ft}}^{\text{3}}/\text{lbm}$ $\left(x\right)$	Temperature,  F $\left(y\right)$
1	3.4403	1000
2	3.7278	$y_2=?$
3	3.9295	1200

Calculate final temperature at final specific volume $3.7278{\text{ft}}^{\text{3}}/\text{lbm}$ for liquid phase using interpolation method.

Substitute 3.4403 for $x_1$, 3.7278 for $x_2$, 3.9295 for $x_3$, $1000°\text{F}$ for $y_1$, and $1200°\text{F}$ for $y_3$ in Equation (VII).

$\begin{array}{c}{y}_2=\frac{\left(3.7278{\text{ft}}^{\text{3}}/\text{lbm}-3.4403{\text{ft}}^{\text{3}}/\text{lbm}\right)\left(1200°\text{F}-1000°\text{F}\right)}{\left(3.9295{\text{ft}}^{\text{3}}/\text{lbm}-3.4403{\text{ft}}^{\text{3}}/\text{lbm}\right)}+1000°\text{F}\\ =1117.5°\text{F}\\ \cong 1118°\text{F}\end{array}$

From above calculation the final temperature of $1118°\text{F}$ at specific volume is $3.7278{\text{ft}}^{\text{3}}/\text{lbm}$.

Unit conversion of temperature from $°\text{F}$ to R.

$\begin{array}{c}{T}_2=1118°\text{F}\\ =1118+\text{460}\text{R}\\ =1578\text{R}\end{array}$

Thus, the final temperature using the superheated steam table is $\underset{\_}{1578\text{R}}$.