# Aerospace Oblique shock reflection on solid boundary

1. Jan 12, 2010

### alchemist

Hi guys, i need some help on this topic.

Basically, i need to calculate the angle of reflection of a oblique shock on a solid boundary.

Here is the description of the question i have:

an incident supersonic flow reaches a compression corner, with a deflection angle of 15 degrees. Mach number of the inlet flow is 2.6, and static temperature is 288k.

above the compression corner, there is an upper horizontal wall, where the oblique shock would hit and then be reflected. Calculate the angle of reflection of the shock by the upper wall.

What i know:
1. i could calculate the mach number after the flow passes the first shock, as well as the new direction of the flow (15 degrees to floor)

2. I know that the final direction of the supersonic flow would have to be parallel to the upper wall, which means it must be at 0 degrees.

apart from this, i have no idea how the shock would be reflected.

Is there any relationship between the deflected angle and the angle of the oblique shock? or between the change in normal velocity component of the flow to the deflected angle/shock angle?

thank you very much guys!

regards

2. Jan 12, 2010

### minger

It seems as though this a standard homework problem. Before I assist, I will tell you that in like 2 days you'll be doing shock interaction from different families. I know this because I looked at my notes, and that's on the next page. So, ask your professor for some help because it's going to get more confusing. I have an example that has an incoming flow of Ma=2.0 with a half-cone angle of 10°. The upper wall is straight.

From the angle and the incoming mach, you can calculate the shock angle. Important note that you may be missing or forgot:
So, calculate the normal mach number through the shock using the incoming mach number and the shock angle. You can then use this "normal" mach number and the normal mach tables to find the normal mach number after the shock.

Keeping the fact that you should measure all angles with respect to the incoming flow direction, you can then find the mach number (total) after the first shock as:
$$M_2 = \frac{M_{2_n}}{\sin(\beta_1 - \theta_1)}$$
Your flow is now tangential to the cone angle, the upper wall is now 10° to the flow. Rinse and repeat.

I have the numbers for my example if you want them. We will need to see some sort of try before giving you any answers though.

3. Jan 18, 2010

### alchemist

Thanks dude, i've figured out the answer to this question. i just didnt know how to continue with the question after finding out the first shock angle and the mach number of the flow after that.

"rinse and repeat" help me sorted it out. lol

cheers!

4. Jan 18, 2010

### minger

Excellent, glad I could be of help.

5. Feb 1, 2011

### Maurice M ben

Hello
in order to find the angle of the reflected oblique shoch wave.
First you need to do some calculation , based on the given data. you have M1=2.6; and the and wall deflection is theta=15 degrees. temperature is given T=288 degrees Kelvin, but there is a missing data the pressure or the static pressure priciesly.
1-Start using the the oblique shock wave shart.
from the shart we can find the oblique shock wave angle (or the angle of the incident shock wave)
1- for M1=2.6 and theta=15degrees , the shock wave will be Betta=35.3 to 35.6 degrees you have to check this in the shart that are disponible for you.
2- you need to calculate the M2. in order to find M2
first you need to calculate Mn1 the mach number which is perpendicular to the incident wave.For this you can use this equation Mn1=M1*sin(Betta); This is the Mach number normal to the shoch wave in region 1;
Second step , you need to find the downstream Mach number Mn2. and this can be calculated using the following equation;
Mn2^2=[2+(gamma-1)*Mn1^2]/[2*gamma*Mn1^2-(gamma-1)];
Where gamma=1.4, for air and Mn1 is been calculated from previouse steps.
Mn2^2=means Mn2 to power 2.
One Mn2 is found you need to calculate M2 through the following eqaution
1-Since the deflection angle =15 degrees , betta is found, from the oblique shock wave shart you can use this equation to find , since
Mn2=M2*sin(Betta-theta) ===> M2=Mn2/sin(betta-theta);
*step three is very important, here also you need to find Betta(2) or betta' from the oblique shock wave shart,after M2 is found. Once betta(2), is found you find the angle of the reflected shock wave by simply substructing the deflection angle of the wall from betta (2).
For example if we let Phi be the angle between the reflected shock wave and the upper wall
this[ Phi=betta(2)- theta];
Here i would like to clarrify something very important : Notice that the angle betta (2) is not the angle of the reflected shock with respect to the upper wall , betta(2) is the angle between the reflected shock wave and the flow direction in region(2).
For the flow properties behind ech shock wave you need to apply the equation or you can use the table for a normal shock wave.
If you still have a problems or any question about compressible flow here is my e-mail{ london_maurice@hotmail.com]

6. Feb 1, 2011

### Maurice M ben

Hello
A cone is a three dimensional flow is not a two dimensional.
Yes there is a straight oblique shock which emanates from the tip, just as in the case of the wedge (in the case of a supersonioc flow). But the there are no further similarities between this two conditions.
in the case of the wedge the flow is considered 2D flow for a cone the flow is a 3D flow.Also note that the equation for a a supersonic over the cone are based on spherical coordinates. See chapter 13 Jhon D Anderson Modern compressible flow

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