Chapter 17.7, Problem 80P

Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma = 2.5. If the pressure and temperature of air are 10.0 psia and 440.5 R, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions.

To determine

The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock and for helium undergoing a normal shock under the same conditions.

Answer to Problem 80P

The actual temperature of air after the normal shock is $942\text{R}$.

The actual pressure of air after the normal shock is $71.25\text{psia}$.

The stagnation pressure of air after the normal shock is $85.3\text{psia}$.

The Mach number value of air after the normal shock is $0.513$.

The velocity of air after the normal shock is $772\text{ft}/\text{s}$.

The Mach number of helium gas after the normal shock is $0.553$.

The actual temperature of helium after the normal shock is $1233\text{R}$.

The actual pressure of helium after the normal shock is $75.6\text{psia}$.

The stagnation pressure of air after the normal shock is $115.46\text{psia}$.

The velocity of air after the normal shock is $2794\text{ft}/\text{s}$.

Explanation of Solution

Refer Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with $k=1.4$”, write the following expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.5.

$\frac{T_2}{T_1}=2.1375$ (I)

$\frac{P_2}{P_1}=7.125$ (II)

$\frac{P_{02}}{P_{01}}=8.5262$ (III)

${\text{Ma}}_2=0.513$

Here, actual temperature after the shock is $T_2$, actual temperature before the shock is $T_1$, actual pressure after the shock is $P_2$, actual pressure before the shock is $P_1$, stagnation pressure  after the shock is $P_{02}$, stagnation pressure before the shock is $P_{01}$, and Mach number after the shock is ${\text{Ma}}_2$.

Write the expression to calculate the velocity of sound after the normal shock.

$c_2=\sqrt{kR{T}_2}$ (IV)

Here, velocity of sound after the shock is $c_2$, and gas constant of air is R.

Write the expression to calculate the velocity of air after the normal shock.

$V_2={\text{Ma}}_2{c}_2$ (V)

The value from the table is not considered for Mach number $\left({\text{Ma}}_2\right)$, because the value for ratio of specific heats $k$ is not 1.4 for helium.

Write the expression to calculate the Mach number for helium after the normal shock.

${\text{Ma}}_2={\left(\frac{{\text{Ma}}_1^2+2/\left(k-1\right)}{{\text{2Ma}}_1^{2}k/\left(k-1\right)-1}\right)}^{1/2}$ (VI)

Here, Mach number of helium before the normal shock is ${\text{Ma}}_1$, Mach number of helium before the normal shock is ${\text{Ma}}_2$, and ratio of specific heats for helium is k.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

$\frac{P_2}{P_1}=\frac{1+k{\text{Ma}}_1^2}{1+k{\text{Ma}}_2^2}$ (VII)

Here, actual pressure of helium after the shock is $P_2$, and actual pressure of helium before the shock is $P_1$.

Write the expression to calculate the actual temperature of helium gas after the normal shock.

$\frac{T_2}{T_1}=\frac{1+{\text{Ma}}_1^2\left(k-1\right)/2}{1+{\text{Ma}}_2^2\left(k-1\right)/2}$ (VIII)

Here, actual temperature of helium after the shock is $T_2$, and actual temperature of helium before the shock is $T_1$.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

$\frac{P_{02}}{P_{01}}=\frac{1+k{\text{Ma}}_1^2}{1+k{\text{Ma}}_2^2}{\left(1+\frac{\left(k-1\right){\text{Ma}}_2^2}{2}\right)}^{k/\left(k-1\right)}$ (IX)

Here, stagnation pressure of helium after the shock is $P_{02}$, and stagnation pressure of helium before the shock is $P_{01}$.

Write the expression to calculate the velocity of sound after the normal shock for helium.

$c_{2,helium}=\sqrt{kR{T}_2}$ (X)

Here, velocity of sound after the shock for helium is $c_{2,helium}$, and gas constant of helium is R.

Write the expression to calculate the velocity of helium after the normal shock $V_{2,helium}$.

$V_{2,helium}={\text{Ma}}_2{c}_{2,helium}$ (XI)

Conclusion:

For air:

Refer Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following properties for air at room temperature.

$\begin{array}{l}R=0.06855\text{}\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.4\end{array}$

Substitute $440.5\text{R}$ for $T_1$ in Equation (I).

$\frac{T_2}{440.5\text{R}}=2.1375$

$\begin{array}{c}{T}_2=440.5\text{R}\times 2.1375\\ =941.56\text{R}\\ =942\text{R}\end{array}$

Thus, the actual temperature of air after the normal shock is $942\text{R}$.

Substitute $10\text{psia}$ for $P_1$ in Equation (II).

$\begin{array}{l}\frac{P_2}{10\text{psia}}=7.125\\ P_2=10\text{psia}\times 7.125\\ P_2=71.25\text{psia}\end{array}$

Thus, the actual pressure of air after the normal shock is $71.25\text{psia}$.

The actual pressure before the normal shock $\left(P_1\right)$ is the equal as the stagnation pressure before the normal shock $\left(P_{01}\right)$, since the flow through the nozzle is isentropic,

Substitute $10\text{psia}$ for $P_{01}$ in Equation (III).

$\frac{P_{02}}{10\text{psia}}=8.5262$

$\begin{array}{c}{P}_2=10\text{psia}\times 8.5262\\ =85.262\text{psia}\\ =85.3\text{psia}\end{array}$

Thus, the stagnation pressure of air after the normal shock is $85.3\text{psia}$.

From Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with $k=1.4$”, write the following expressions of Mach number after the shock for a Mach number of 2.5.

${\text{Ma}}_2=0.513$

Thus, the Mach number value of air after the normal shock is $0.513$.

Substitute 1.4 for k, $0.06855\text{}\text{Btu}/\text{lbm}\cdot \text{R}$ for R, and $942\text{R}$ for $T_2$ in Equation (IV).

$\begin{array}{c}{c}_2=\sqrt{1.4\times 0.06855\text{}\text{Btu}/\text{lbm}\cdot \text{R}\times 942\text{R}}\\ =\sqrt{1.4\times 0.06855\text{}\text{Btu}/\text{lbm}\cdot \text{R}\times 942\text{R}\left(\frac{25,037{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{Btu}/\text{lbm}}\right)}\\ =1504.47\text{ft}/\text{s}\end{array}$

Substitute 0.513 for ${\text{Ma}}_2$, and $1504.47\text{ft}/\text{s}$ for $c_2$ in Equation (V).

$\begin{array}{c}{V}_2=0.513\times 1504.47\text{ft}/\text{s}\\ =771.7\text{ft}/\text{s}\\ =772\text{ft}/\text{s}\end{array}$

Thus, the velocity of air after the normal shock is $772\text{ft}/\text{s}$.

For helium:

Refer Table A-E, “Ideal-gas specific heats of various common gases”, obtain the following properties for helium.

$\begin{array}{l}R=\text{0}\text{.4961}\text{}\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.667\end{array}$

Substitute 2.5 for ${\text{Ma}}_1$, and 1.667 for k in Equation (VI).

$\begin{array}{c}{\text{Ma}}_2={\left(\frac{{2.5}^2+2/\left(1.667-1\right)}{\text{2}\times \text{2}{\text{.5}}^2\times 1.667/\left(1.667-1\right)-1}\right)}^{1/2}\\ ={\left(\frac{9.248}{30.24}\right)}^{1/2}\\ =0.553\end{array}$

Thus, the Mach number of helium gas after the normal shock is $0.553$.

Substitute 1.667 for k, 2.5 for ${\text{Ma}}_1$, and 0.553 for ${\text{Ma}}_2$. in Equation (VII).

$\frac{P_2}{P_1}=\frac{1+1.667\times {\left(2.5\right)}^2}{1+1.667\times {\left(0.553\right)}^2}$

$\frac{P_2}{P_1}=7.5631$ (XII)

Substitute 1.667 for k, 2.5 for ${\text{Ma}}_1$, and 0.553 for ${\text{Ma}}_2$. in Equation (VIII).

$\frac{T_2}{T_1}=\frac{1+{\left(2.5\right)}^2\left(1.667-1\right)/2}{1+{\left(0.553\right)}^2\left(1.667-1\right)/2}$

$\frac{T_2}{T_1}=2.7989$ (XIII)

Substitute 1.667 for k, 2.5 for ${\text{Ma}}_1$, and 0.553 for ${\text{Ma}}_2$. in Equation (IX).

$\frac{P_{02}}{P_{01}}=\frac{1+1.667\times {\left(2.5\right)}^2}{1+1.667\times {\left(0.553\right)}^2}{\left(1+\frac{\left(1.667-1\right){\left(0.553\right)}^2}{2}\right)}^{1.667/\left(1.667-1\right)}$

$\frac{P_{02}}{P_{01}}=11.546$ (XIV)

Substitute $10\text{psia}$ for $P_1$ in Equation (XII).

$\begin{array}{l}\frac{P_2}{10\text{psia}}=7.5632\\ P_2=10\text{psia}\times 7.5632\\ P_2=75.6\text{psia}\end{array}$

Thus, the actual pressure of helium after the normal shock is $75.6\text{psia}$.

Substitute $440.5\text{R}$ for $T_1$ in Equation (XIII).

$\frac{T_2}{440.5\text{R}}=2.7989$

$\begin{array}{c}{T}_2=440.5\text{R}\times 2.7989\\ =1232.91\text{R}\\ =1233\text{R}\end{array}$

Thus, the actual temperature of helium after the normal shock is $1233\text{R}$.

Since the flow through the nozzle is isentropic $\left(P_{01}=P_1\right)$.

Substitute $10\text{psia}$ for $P_{01}$ in Equation (XIV).

$\begin{array}{l}\frac{P_{02}}{10\text{psia}}=11.546\\ P_{02}=10\text{psia}\times 11.546\\ P_{02}=115.46\text{psia}\end{array}$

Thus, the stagnation pressure of air after the normal shock is $115.46\text{psia}$.

Substitute $1.667$ for k, $0.4961\text{}\text{Btu}/\text{lbm}\cdot \text{R}$ for R, and $1233\text{R}$ for $T_2$ in Equation (X).

$\begin{array}{c}{c}_{2,helium}=\sqrt{1.667\times 0.4961\text{}\text{Btu}/\text{lbm}\cdot \text{R}\times 1233\text{R}}\\ =\sqrt{1.667\times 0.4961\text{}\text{Btu}/\text{lbm}\cdot \text{R}\times 1233\text{R}\left(\frac{25,037{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{Btu}/\text{lbm}}\right)}\\ =5052.72\text{ft}/\text{s}\end{array}$

Substitute $0.553$ for ${\text{Ma}}_2$, and $5052.72\text{ft}/\text{s}$ for $c_2$ in Equation (XI).

$\begin{array}{c}{V}_{2,helium}=0.553\times 5052.72\text{ft}/\text{s}\\ =2794.15\text{ft}/\text{s}\\ =2794\text{ft}/\text{s}\end{array}$

Thus, the velocity of air after the normal shock is $2794\text{ft}/\text{s}$.