Chapter 17.7, Problem 96P

To determine

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

Answer to Problem 96P

The Mach number value of air after the normal shock through the nozzle is $0.5039$.

The actual temperature of air after the normal shock through the nozzle

is $604.34\text{K}$.

The actual pressure of air after the normal shock through the nozzle is $447.76\text{kPa}$.

The stagnation pressure of air after the normal shock though the nozzle

is $532.5\text{kPa}$.

The velocity of air after the normal shock through the nozzle is $248.31\text{m}/\text{s}$.

Thus, the Mach number of helium gas after the normal shock through the nozzle

 is $0.5455$.

Thus, the actual temperature of helium after the normal shock through the nozzle is $799.362\text{K}$.

Thus, the actual pressure of helium after the normal shock through the nozzle

 is $475.7\text{kPa}$.

Thus, the stagnation pressure of helium after the normal shock though the nozzle

is $602.5\text{kPa}$.

Thus, the velocity of helium after the normal shock through the nozzle is $907.5\text{m}/\text{s}$.

Comparison between the obtained results of air and helium is shown in below Table:

Parameters/ Conditions	Air	Helium
Mach number value	$0.5039$	$0.5455$
Actual temperature of air	$604.34\text{K}$	$799.362\text{K}$
Actual pressure of air	$447.76\text{kPa}$	$475.7\text{kPa}$
Stagnation pressure	$532.5\text{kPa}$	$602.5\text{kPa}$
Velocity	$248.31\text{m}/\text{s}$	$907.5\text{m}/\text{s}$

Explanation of Solution

Write the expression for the velocity of sound after the normal shock.

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

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

Write the expression for the velocity of airafter the normal shock.

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

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

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

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 for the actual pressure of helium gas after the normal shock.

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

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 for the actual temperature of helium gas after the normal shock.

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

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 for the actual pressure of helium gas after the normal shock.

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

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 for the velocity of sound after the normal shock.

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

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

Write the expression for the velocity of helium after the normal shock.

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

Conclusion:

Refer to Table A-33, “One-dimensional normal-shock functions for an ideal gas with k 5 1.4”, obtain the expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.6 before the shock.

${\text{Ma}}_2=0.5039$

Thus, the Mach number value of air after the normal shock through the nozzle is $0.5039$.

$\frac{T_2}{T_1}=2.2383$ (IX)

$\frac{P_2}{P_1}=7.7200$ (X)

$\frac{P_{02}}{P_{01}}=9.1813$ (XI)

Here, 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 ${\text{Ma}}_2$, actual temperature after the shock is $T_2$ and actual temperature before the shock is $T_1$.

Substitute $270\text{K}$ for $T_1$ in Equation (IX).

$\begin{array}{l}\frac{T_2}{270\text{K}}=2.2383\\ T_2=270\text{K}\times 2.2383\\ T_2=604.34\text{K}\end{array}$

Thus, the actual temperature of air after the normal shock through the nozzle

is $604.34\text{K}$.

Substitute $58\text{kPa}$ for $P_1$ in Equation (X).

$\begin{array}{l}\frac{P_2}{58\text{kPa}}=7.7200\\ P_2=58\text{kPa}\times 7.7200\\ P_2=44.76\text{kPa}\end{array}$

Thus, the actual pressure of air after the normal shock through the nozzle is $447.76\text{kPa}$.

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

Substitute $58\text{kPa}$ for $P_{01}$ in Equation (XI).

$\begin{array}{l}\frac{P_{02}}{58\text{kPa}}=9.1813\\ P_2=58\text{kPa}\times 9.1813\\ P_2=532.5\text{kPa}\end{array}$

Thus, the stagnation pressure of air after the normal shock though the nozzle

is $532.5\text{kPa}$.

Refer to thermodynamics properties table and interpret the value of k, and R for a temperature of $270\text{K}$, and a pressure of $58\text{kPa}$ for air as 1.4, and $0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute 1.4 for k, $0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $604.34\text{K}$ for $T_2$ Equation (I).

$\begin{array}{c}{c}_2=\sqrt{1.4\times 0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}\times 604.34\text{K}}\\ =492.27\text{m}/\text{s}\end{array}$

Substitute 0.5039for ${\text{Ma}}_2$, and $492.27\text{m}/\text{s}$ for $c_2$ in Equation (II).

$\begin{array}{c}{V}_2=0.5039\times 492.27\text{m}/\text{s}\\ =248.31\text{m}/\text{s}\end{array}$

Thus, the velocity of air after the normal shock through the nozzle is $248.31\text{m}/\text{s}$.

Substitute 2.6for ${\text{Ma}}_1$, and 1.667 for k in Equation (III).

$\begin{array}{c}{\text{Ma}}_2={\left(\frac{{2.6}^2+2/\left(1.667-1\right)}{\text{2}\times {2.6}^2\times 1.667/\left(1.667-1\right)-1}\right)}^{1/2}\\ =0.5455\end{array}$

Thus, the Mach number of helium gas after the normal shock through the nozzle

 is $0.5455$.

Substitute 1.667 for k, 2.6for ${\text{Ma}}_1$, and 0.5455 for ${\text{Ma}}_2$. in Equation (IV).

$\frac{P_2}{P_1}=\frac{1+1.667\times {\left(2.6\right)}^2}{1+1.667\times {\left(0.5455\right)}^2}$

$\frac{P_2}{P_1}=8.2009$ (XII)

Substitute 1.667 for k, 2.6for ${\text{Ma}}_1$, and 0.5455 for ${\text{Ma}}_2$. in Equation (V).

$\frac{T_2}{T_1}=\frac{1+{2.6}^2\left(1.667-1\right)/2}{1+{0.5455}^2\left(1.667-1\right)/2}$

$\frac{T_2}{T_1}=2.9606$ (XIII)

Substitute 1.667 for k, 2.6for ${\text{Ma}}_1$, and 0.5455 for ${\text{Ma}}_2$. in Equation (VI).

$\frac{P_{02}}{P_{01}}=\frac{1+1.667\times {2.6}^2}{1+1.667\times {0.5455}^2}{\left(1+\frac{\left(1.667-1\right){0.5455}^2}{2}\right)}^{1.667/\left(1.667-1\right)}$

$\frac{P_{02}}{P_{01}}=10.389$ (XIV)

Substitute $270\text{K}$ for $T_1$ in Equation (XIII).

$\begin{array}{l}\frac{T_2}{270\text{K}}=2.9606\\ T_2=270\text{K}\times 2.9606\\ T_2=799.362\text{K}\end{array}$

Thus, the actual temperature of helium after the normal shock through the nozzle is $799.362\text{K}$.

Substitute $58\text{kPa}$ for $P_1$ in Equation (XII).

$\begin{array}{l}\frac{P_2}{58\text{kPa}}=8.2009\\ P_2=58\text{kPa}\times 8.2009\\ P_2=475.7\text{kPa}\end{array}$

Thus, the actual pressure of helium after the normal shock through the nozzle

 is $475.7\text{kPa}$.

Since, $\left(P_1=P_{01}\right)$

Substitute $58\text{kPa}$ for $P_{01}$ in Equation (XIV).

$\begin{array}{l}\frac{P_{02}}{58\text{kPa}}=10.389\\ P_2=58\text{kPa}\times 10.389\\ P_2=602.5\text{kPa}\end{array}$

Thus, the stagnation pressure of air after the normal shock though the nozzle

is $602.5\text{kPa}$.

Refer Table A–1, “Molar mass, gas constant, and critical2point properties”, obtain

the value of k, and R for a temperature of $270\text{K}$, and a pressure of $58\text{kPa}$ for helium as 1.667, and $2.0769\text{}\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Substitute 1.667 for k, $2.0769\text{}\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $799.362\text{K}$ for $T_2$ Equation (VII).

$\begin{array}{c}{c}_2=\sqrt{1.667\times 2.0769\text{}\text{kJ}/\text{kg}\cdot \text{K}\times 799.362\text{K}}\\ =1663.59\text{m}/\text{s}\end{array}$

Substitute 0.5455 for ${\text{Ma}}_2$, and $1663.59\text{m}/\text{s}$ for $c_2$ in Equation (VIII).

$\begin{array}{c}{V}_2=0.5455\times 1663.59\text{m}/\text{s}\\ =907.5\text{m}/\text{s}\end{array}$

Thus, the velocity of air after the normal shock through the nozzle is $907.5\text{m}/\text{s}$.