This is not a homework problem. I am looking for some direction:

A normal shockwave occurs in a gas with an unknown specific heat ratio $\gamma$. The static pressure ratio across the shockwave is $10.6$. The Mach number downstream of the shockwave is $0.495$. I am required to find $\gamma$, where I tried using the normal shockwave equation:

Firstly, I looked into the isentropic tables for the supersonic $M_1$ which at a subsonic $M_2$ downstream of $0.495$ equals $1.71302$ upstream as the thickness of the shockwave is $\approx0$. However after rearranging for $\gamma$ and using $M_1=1.71302$, I am getting a negative $\gamma$, which is not right as $1<\gamma<2$. The area ratio at the found Mach numbers is, $$\frac{A}{A*}=1.35$$Any suggestions on how to approach this problem? Thank you.

  • $\begingroup$ I don't understand what is given and what you have calculated by yourself. The given values do not seem logical to me. Additionally as you are calculating an area ratio it seems to be a normal shock in a Laval nozzle: aere.iastate.edu/~huhui/teaching/2009Sx/AerE343L-AerE311L/… Could you clarify that? Furthermore are you sure there isn't a typo in your first formula for the normal shock (last term should have a different sign)? grc.nasa.gov/www/k-12/airplane/normal.html $\endgroup$ – 2b-t Jan 4 '20 at 11:55
  • $\begingroup$ Thank you, there was indeed a typo in the formula which I have corrected now. Basically, I have been given the pressure ratio between the upstream and downstream of a convergent-divergent nozzle as $10.6$ and the Mach number of the downstream as $0.495$. To calculate the specific heat ratio $\gamma$, I tried manipulating the stated formula to isloate $\gamma$. However, using this approach yields a negative $\gamma$, which is not right. I then looked up the upstream supersonic Mach number $(1.71302)$ at the given downstream subsonic Mach number $(0.495)$. Both were at an area ratio of $1.35$ $\endgroup$ – Mughees Asif Jan 4 '20 at 15:36
  • $\begingroup$ Using the supersonic Mach number, I again tried using the formula but still, $\gamma$ was being calculated as negative. $\endgroup$ – Mughees Asif Jan 4 '20 at 15:38

Let us start with a few known relationships for neutral fluid shock waves: $$ \begin{align} \frac{ P_{2} }{ P_{1} } & = \frac{ 2 \ \gamma }{ \gamma + 1 } M_{1}^{2} - \frac{ \gamma - 1 }{ \gamma + 1 } \tag{0a} \\ \frac{ \rho_{2} }{ \rho_{1} } & = \frac{ U_{1} }{ U_{2} } = \frac{ \left( \gamma + 1 \right) M_{1}^{2} }{ \left( \gamma + 1 \right) + \left( \gamma - 1 \right) \left( M_{1}^{2} - 1 \right) } \tag{0b} \\ C_{s j}^{2} & = \frac{ \gamma \ P_{j} }{ \rho_{j} } \tag{0c} \\ M_{j} & = \frac{ U_{j} }{ C_{s j} } \tag{0d} \end{align} $$ where $P_{j}$ is the scalar pressure in the jth region ($j = 1$ for upstream, $j = 2$ for downstream), $\rho_{j}$ is the mass density in the jth region, $U_{j}$ is the bulk flow speed along the shock normal in the shock rest frame in the jth region, $C_{s j}$ is the speed of sound in the jth region, $M_{j}$ is the Mach number in the jth region, and $\gamma$ is the ratio of specific heats or polytropic index. Equation 0c derives from the assumption that the change across the shock ramp is fast enough that an adiabatic compression can be assumed. Equation 0b comes from the Rankine-Hugoniot relations.

First, to simplify things let $\delta \equiv \tfrac{ \rho_{2} }{ \rho_{1} }$ and $\alpha \equiv \tfrac{ P_{2} }{ P_{1} }$, then we can define the following after a little algebra: $$ M_{1}^{2} = \delta \ \alpha \ M_{2}^{2} \tag{1} $$ which we can use to replace $M_{1}$ in Equations 0a and 0b. We solve the altered version of Equation 0a for $\delta$ to find: $$ \delta = \frac{ \left( \gamma - 1 \right) + \left( \gamma + 1 \right) \alpha }{ 2 \ \gamma \ \alpha \ M_{2}^{2} } \tag{2} $$ Next we set Equation 2 equal to the altered version of Equation 0b and solve for $\gamma$. There are two solutions but only one of them is physically meaningful for a shock, i.e., a compressive sound wave where $\delta$ > 1. That solution is given by: $$ \gamma = \frac{ 1 - \delta }{ 2 \ \delta \ M_{2}^{2} - \left( 1 + \delta \right) } \tag{3} $$

If I use $\delta$ = 10.6 and $M_{2}$ = 0.495, then I get $\gamma$ ~ 1.499 or nearly 3/2. The typical monatomic gas approximation is $\gamma$ = 5/3 ~ 1.67 and for diatomic it goes as $\gamma$ = 7/5 ~ 1.4 (i.e., typical assumption for Earth's atmosphere).

You can also look at some other variations on these expressions at https://physics.stackexchange.com/a/349724/59023 and https://physics.stackexchange.com/a/302879/59023.


I simply used the correlations between the Mach numbers before and after the normal shock

$$ Ma_2^2 = \frac{(\gamma - 1) Ma_1^2 + 2}{2 \, \gamma \, Ma_1^2 - (\gamma - 1)} $$

as well as the correlation between the pressure ratio and the Mach number before the shock

$$ \frac{p_2}{p_1} = \frac{2 \, \gamma \, Ma_1^2 - (\gamma -1)}{\gamma + 1}$$

and solved it numerically with Wolfram Alpha (but you can do the same thing analytically by rearranging the first equation for $Ma_1$ and popping it into the second equation). My results for $Ma_2 = 0.495$ and $\frac{p_2}{p_1} = 10.6$ are $\gamma \approx 1.5$ and $Ma_2 \approx 3$. Which seem reasonable as in this example for $\frac{p_2}{p_1} \approx 6.67$ and air $\gamma = 1.4$ the Mach numbers before and after the shock are $Ma_1 = 1.9$ and $Ma_2 \approx 0.6$ respectively.

I think your mistake was using an isentropic table which assumes a certain gas and thus a certain heat capacity ratio $\gamma$. Most likely the values for dry air at room temperature $\gamma^{(air)} \approx 1.4$. Furthermore as your pressure ratio is quite high this results in a large deviation between the isentropic relation and the actual shock condition.

  • $\begingroup$ Thank you so much! $\endgroup$ – Mughees Asif Jan 4 '20 at 20:33

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