Equation of temperature increase by shock wave

In summary, the equation of temperature increase by shock wave is complex and dependent on factors such as Mach number, piston speed, and geometry. The speed of the shock wave can be determined by solving the Navier-Stokes equations, but turbulence may have little effect on the outcome. The normal and oblique shock equations are valid for both laminar and turbulent flows, except near boundaries.
  • #1
ada_ada_2002
16
0
【HELP】Equation of temperature increase by shock wave

Hi there!

What is the equation of the temperature increase when the ideal gas swept by a planar shock wave (Mach number, M)?

Thank you!
 
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  • #2


Another question:
If the shock wave is triggered by a piston, why it can propagate at a speed larger than that of the piston? what determine the speed of a shock?
 
  • #3


I don't have my book handy to tell you the equation, but as for the second question:

The shock wave is a discontinuity in air pressure. Keep in mind that, while the shock wave is caused by your piston, the actual cause is that the air around it is moving in a certain way.

The propagation velocities are in the order of magnitude of the speed of sound. Your piston is probably moving quite slower than that. You get a shock wave because there is a discontinuity in pressure at some point around the piston.

This is the tricky part:

The piston itself is not moving faster than sound, the air at some point(s) reaches those velocities.
 
  • #4


Thank you meldraft!

So, how to calculate the speed of a shock? is there any equations? I just what to know what exactly influence the shock speed.

BTW, I have found the temperature equation
 
  • #5


You would do well to look up normal shock relations. They will have all the relations you are asking for and more. If you intend to do this a lot, it may be a good idea to pick up a book on compressible flow and learn where it all comes from.
 
  • #6


Not intended to do this subject alot, just curious.

Anyway, cannot find relations between initial piston speed, and the consequent shock speed. Just know that shock propagate ahead of the piston. By how much exactly?
 
  • #7


The phenomenon you are asking is very complex because, in contrast to an aircraft, the piston is moving back an forth, which means that the air flow is fully turbulent.

That being said, I imagine that the discontinuity in pressure would be caused by sudden changes in the direction of the airflow, caused by the alternating pressure between upstroke and downstroke.

For an aircraft, these are the equations:

http://www.grc.nasa.gov/WWW/k-12/airplane/normal.html

http://www.grc.nasa.gov/WWW/k-12/airplane/oblique.html

For a piston, unless you can find something in a book that I am not aware of, you would have to solve the Navier-Stokes with turbulence to see when the shock waves are formed.

A rough approximation is to consider a plane that is moving in one direction in a container where the air is moving in the other direction. If you solve the equations, you will get when the Mach number becomes greater than 1, as a function of the displacement of the piston. This value will depend on the relative velocity between the air and the piston, as well as the geometry you assume.
 
  • #8


Your answer is quite clear! Thank you SO MUCH, Meldraft!
 
  • #9


meldraft said:
The phenomenon you are asking is very complex because, in contrast to an aircraft, the piston is moving back an forth, which means that the air flow is fully turbulent.

That being said, I imagine that the discontinuity in pressure would be caused by sudden changes in the direction of the airflow, caused by the alternating pressure between upstroke and downstroke.

For an aircraft, these are the equations:

http://www.grc.nasa.gov/WWW/k-12/airplane/normal.html

http://www.grc.nasa.gov/WWW/k-12/airplane/oblique.html

For a piston, unless you can find something in a book that I am not aware of, you would have to solve the Navier-Stokes with turbulence to see when the shock waves are formed.

A rough approximation is to consider a plane that is moving in one direction in a container where the air is moving in the other direction. If you solve the equations, you will get when the Mach number becomes greater than 1, as a function of the displacement of the piston. This value will depend on the relative velocity between the air and the piston, as well as the geometry you assume.

Turbulence has absolutely no effect in this case. The only time turbulence would have an effect is in the case of a boundary layer, and then only because you change the shape of the boundary layer. The normal and oblique shock equations are equally valid for laminar and turbulent flows except near boundaries since the normal and oblique shock relations assume inviscid flow. Then again, that means that they aren't really valid near boundaries for laminar flow either.

In general, in compressible fluid mechanics like this, the inviscid approximation is very good because (a) the flows generally involve very high Reynolds numbers, so the viscous terms are small in comparison to the rest of the terms and (b) the boundary layer is typically very thin until you start stepping into the hypersonic regime, where it can actually get very thick.

Additionally, saying things like "solve the Navier-Stokes equations with turbulence" really has no meaning. Turbulence is already built into the Navier-Stokes equations. It is fully described by them. The only difference in the Navier-Stokes equations in this case and what are traditionally used is that since this is a compressible flow, one cannot drop the bulk viscosity term as is typical. This still has nothing to do with turbulence.
 
  • #10


boneh3ad said:
Turbulence has absolutely no effect in this case. The only time turbulence would have an effect is in the case of a boundary layer, and then only because you change the shape of the boundary layer. The normal and oblique shock equations are equally valid for laminar and turbulent flows except near boundaries since the normal and oblique shock relations assume inviscid flow. Then again, that means that they aren't really valid near boundaries for laminar flow either.

You are saying that you can predict where the shock wave will be shaped and how it will look like, as a time dependent phenomenon, in a small closed container, and, possibly, fuel coming in, by ignoring turbulence. If this sounds right to you, I'm not one to argue :smile: I will just point out that the equations are still valid, but your field does not look like you would expect for laminar flow.

boneh3ad said:
Additionally, saying things like "solve the Navier-Stokes equations with turbulence" really has no meaning. Turbulence is already built into the Navier-Stokes equations.

The person that manages to solve that will get a Nobel prize :biggrin: That is why we have turbulence models, or choose to neglect it.
 
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  • #11


meldraft said:
You are saying that you can predict where the shock wave will be shaped and how it will look like, as a time dependent phenomenon, in a small closed container, and, possibly, fuel coming in, by ignoring turbulence. If this sounds right to you, I'm not one to argue :smile: I will just point out that the equations are still valid, but your field does not look like you would expect for laminar flow.

Without knowing more about his scenario, I can't say for sure, but there are plenty of piston systems that create shocks in laminar flow, and plenty that create them in turbulent flow. It just depends on where in the flow you are looking and a host of other factors. A piston in an open pipe is a pretty canonical problem, and you can look at the pressure created by the piston to determine the shock velocity and location as a function of time.

Consider that the piston compresses the air directly in front of it, but this compression travels at the speed of sound, so it moves faster than the piston. However, after each infinitesimal compression, the speed of sound actually gets a little bit faster behind that wave, so if the piston is moving fast enough that it compresses the air quickly enough, each successive compression will slowly catch up to the previous one until eventually they can coalesce into a shock.

meldraft said:
The person that manages to solve that will get a Nobel prize :biggrin: That is why we have turbulence models, or choose to neglect it.

That isn't really true. People solve them directly all the time. It is called direct numerical simulation, or DNS. The real problem is that due to the incredibly large range of length scaled involved in turbulence, you need a large, incredibly fine mesh in order to do that, so the only real way to get a converged solution is to put the problem on a supercomputer and let it run. Right now, DNS can do a pretty good job of simulating turbulent flows, but only at low Reynolds number and it definitely can't simulate the transition from laminar to turbulent flow given what we know now, so it definitely has its limitations.
 

1. What is the equation for calculating the temperature increase caused by a shock wave?

The equation for calculating the temperature increase caused by a shock wave is: ΔT = (γ - 1)ρu²/Cp, where ΔT is the temperature increase, γ is the specific heat ratio, ρ is the density of the material, u is the shock wave velocity, and Cp is the specific heat capacity at constant pressure.

2. How does the specific heat ratio affect the temperature increase caused by a shock wave?

The specific heat ratio, γ, is a measure of how much energy is required to raise the temperature of a material. A higher γ value means that the material can absorb more heat before its temperature increases significantly. Therefore, a higher γ value will result in a lower temperature increase caused by a shock wave.

3. How does the density of the material affect the temperature increase caused by a shock wave?

The density of the material, ρ, plays a role in the equation for calculating temperature increase by shock wave. A higher density means that there is more material to absorb the energy of the shock wave, resulting in a lower temperature increase. On the other hand, a lower density will result in a higher temperature increase.

4. How does the shock wave velocity affect the temperature increase?

The shock wave velocity, u, is a key factor in determining the temperature increase caused by a shock wave. The higher the velocity, the more energy is transferred to the material, resulting in a higher temperature increase. Additionally, a higher shock wave velocity can also cause a greater amount of compression in the material, which can also contribute to a higher temperature increase.

5. Can the equation for temperature increase by shock wave be used for all materials?

No, the equation is specifically designed for ideal gases. It may provide a rough estimate for other materials, but it will not be accurate. The specific heat ratio, density, and specific heat capacity at constant pressure vary for different materials, so the equation will need to be modified for each material to get an accurate result.

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