Chapter 13, Problem 1P

To determine

The changes in pressure and entropy.

Explanation of Solution

Given:

Initial  pressure $\left(p_1\right)$ is $120\text{psi}$.

Initial temperature $\left(T_1\right)$ is $40°\text{F}$.

Final temperature $\left(T_2\right)$ is $120°\text{F}$.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio $\left(\frac{p_1}{p_2}\right)$ and calculate final pressure $\left(p_2\right)$.

 $\begin{array}{l}\frac{p_1}{p_2}=\frac{\rho R{T}_1}{\rho R{T}_2}\\ \frac{p_1}{p_2}=\frac{T_1}{T_2}\end{array}$

$\begin{array}{l}\frac{120\text{psi}}{p_2}=\frac{460+40°\text{F}}{460+120°\text{F}}\\ p_2=139.2\text{psi}\end{array}$

Calculate the change in pressure $\left(\Delta p\right)$.

$\Delta p=p_2-p_1$

$\begin{array}{c}\Delta p=\left(139.2-120\right)\text{psi}\\ =19.2\text{psi}\end{array}$

Thus, the change is pressure is $\underset{\_}{19.2\text{psi}}$.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant $\left(R\right)$ as $1,554\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}$ and specific heat ratio $\left(k\right)$ as 1.4.

Write an equation for change in enthalpy $\left(\Delta h\right)$ and calculate specific heat at constant pressure $\left(c_p\right)$.

$\begin{array}{l}\Delta h=\frac{kR}{k-1}\left(T_2-T_1\right)\\ c_p\Delta T=\frac{kR}{k-1}\Delta T\\ c_p=\frac{kR}{k-1}\end{array}$

$\begin{array}{c}{c}_p=\frac{1,554\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R} \times 1.4}{1.4-1}\\ =5439\text{}\text{ ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}\end{array}$

Calculate the change in entropy $\left(\Delta s\right)$.

$\begin{array}{c}\left(\Delta s\right)=c_p\text{ln}\frac{T_2}{T_1}-R\ln \frac{p_2}{p_1}\\ =\left\{\begin{array}{l}\left(5439\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}\right)\left[\ln \left(\frac{580°\text{R}}{500°\text{R}}\right)\right]\\ -\left(1554\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}\right)\left[\ln \left(\frac{139.2\text{psi}}{120\text{psi}}\right)\right]\end{array}\right\}\\ =576.61\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}\end{array}$

Thus, entropy change $\left(\Delta s\right)$ is $\underset{\_}{576.61\text{ft}\cdot \text{lb}/\text{slug}\cdot °\text{R}}$.