Chapter 12, Problem 41EP

To determine

he pressure at the location when velocity of air and sound are equal.

The temperature at the location when velocity of air and sound are equal.

The ratio of the areas to entrance area.

Answer to Problem 41EP

The pressure at the location when velocity of air and sound are equal to each other is $17.27\text{psi}$.

The temperature at the location when velocity of air and sound are equal to each other is $\text{53}9°\text{R}$.

The ratio of the areas to entrance area is $0.584$.

Explanation of Solution

Given information:

The initial pressure of the air is $30\text{psi}$, the initial temperature of the air is $630°\text{R}$ and the velocity of the air is $450\text{ft}/\text{s}$.

Write the expression for the stagnation temperature.

   $T_o=T_1+\frac{V^2}{2c_p}$     ...... (I)

Here, the initial temperature is $T_1$, the velocity of the air is $V$ and the specific heat of the air is $c_p$.

Write the expression for the stagnation pressure.

   $p_o=p_1+{\left(\frac{T_o}{T_1}\right)}^{k\left(k+1\right)}$     ...... (II)

Here, the initial pressure is $p_1$ and the specific heat ratio is $k$.

Write the expression for the pressure ratio.

   $R_p=\frac{p_2}{p_o}$     ...... (III)

Here, the pressure at exit is $p_2$.

Write the expression for the temperature ratio.

   $R_T=\frac{T_2}{T_o}$     ...... (IV)

Here, the temperature at exit is $T_2$.

Write the expression for the Mach number at inlet.

   $Ma=\frac{V}{c}$     ...... (V)

Here, the velocity of the sound is $c$.

Write the expression for the velocity of sound.

   $c=\sqrt{kRT}$

Substitute $\sqrt{kRT}$ for $c$ in Equation (V).

   $Ma=\frac{V}{\sqrt{kRT}}$     ...... (VI)

Write the expression for the ratio of area.

   $\frac{A^{*}}{A}=\frac{1}{\frac{1}{Ma}{\left[\left(\frac{2}{k+1}\right)\left(1+\frac{k-1}{2}M{a}^2\right)\right]}^{\frac{\left(k+1\right)}{2\left(k-1\right)}}}$     ...... (VII)

Here, the area at inlet is $A$, the specific heat ratio is $k$ and the area at unity Mach number is $A^{*}$.

Calculation:

Consider the air as an ideal gas. Therefore, the list of properties for the ideal gas at room temperature is

The specific heat ratio is $1.4$.

The gas constant is $1716{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}$.

The specific heat at constant pressure is $6019{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}$.

Substitute $450\text{ft}/\text{s}$ for $V$, $630°\text{R}$ for $T_1$ and $6019{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}$ for $c_p$ in Equation (I).

   $\begin{array}{c}{T}_o=630°\text{R}+\frac{{\left(450\text{ft}/\text{s}\right)}^2}{2\left(6019{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}\right)}\\ =630°\text{R}+\frac{\left(202500{\text{ft}}^2/{\text{s}}^2\right)}{2\left(6019{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}\right)}\\ =630°\text{R}+16.82°\text{R}\\ =646.82°\text{R}\end{array}$

Substitute $1.4$ for $k$, $30\text{psi}$ for $p_1$, $630°\text{R}$ for $T_1$ and $646.82°\text{R}$ for $T_0$ in Equation (II).

   $\begin{array}{c}{p}_o=30\text{psi}{\left(\frac{646.82°\text{R}}{630°\text{R}}\right)}^{1.4\left(1.4+1\right)}\\ =30\text{psi}{\left(1.026\right)}^{3.36}\\ =30\text{psi}\left(1.090\right)\\ =32.70\text{psi}\end{array}$

Since, the velocity of the air is equal to the velocity of sound hence the value of Mach number is 1.

Refer to table “One dimensional isentropic compressible flow function for the an ideal gas” to obtain the pressure ratio as $0.5283$ and temperature ratio as $0.8333$ at Mach number $Ma=1$.

Substitute $0.5283$ for $R_p$ and $32.70\text{psi}$ for $p_o$ in Equation (III).

   $\begin{array}{l}0.5283=\frac{p_2}{\left(32.70\text{psi}\right)}\\ p_2=\left(0.5283\right)\left(32.70\text{psi}\right)\\ p_2=17.27\text{psi}\end{array}$

Substitute $0.8333$ for $R_T$ and $646.82°\text{R}$ for $T_o$ in Equation (IV).

   $\begin{array}{l}0.8333=\frac{T_2}{\left(646.82°\text{R}\right)}\\ T_2=\left(0.8333\right)\left(646.82°\text{R}\right)\\ T_2=538.99°\text{R}\\ T_2\approx \text{53}9°\text{R}\end{array}$

Substitute $1.4$ for $k$, $1716{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}$ for $R$, $630°\text{R}$ for $T_1$ and $450\text{ft}/\text{s}$ for $V$ in Equation (VI).

   $\begin{array}{c}Ma=\frac{450\text{ft}/\text{s}}{\sqrt{1.4\left(1716{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{°R}\right)630°\text{R}}}\\ =\frac{450\text{ft}/\text{s}}{\sqrt{\left(1513512{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}}\\ =\frac{450\text{ft}/\text{s}}{1230.24\text{ft}/\text{s}}\\ =0.366\end{array}$

Substitute $1.4$ for $k$ and $0.366$ for $Ma$ in Equation (VII).

   $\begin{array}{c}\frac{A^{*}}{A}=\frac{1}{\frac{1}{0.366}{\left[\left(\frac{2}{1.4+1}\right)\left(1+\frac{1.4-1}{2}{\left(0.366\right)}^2\right)\right]}^{\frac{\left(1.4+1\right)}{2\left(1.4-1\right)}}}\\ =\frac{1}{2.73{\left[\left(0.833\right)\left(1+0.2\left(0.1339\right)\right)\right]}^3}\\ =\frac{1}{2.73\left[0.6257\right]}\\ =0.584\end{array}$

Conclusion:

The pressure at the location when velocity of air and sound are equal is $17.27\text{psi}$.

The temperature at the location when velocity of air and sound are equal is $\text{53}9°\text{R}$.

The ratio of the areas to entrance area is $0.584$.