I am having some difficulty in determining a consistent dimensionless form of the Euler equations. In the form I'm presently using, the ICs are set by:

rho_inf = 1.0 u_inf = M_inf * cos(alpha) v_inf = M_inf * sin(alpha) p_inf = 1/(gamma*M_inf^2) Et_inf = (p_inf/(gamma-1)) + 0.5*rho_inf*(u^2 + v^2)

The problem is, I get the proper wave angle but ratios of quantities across the shock are incorrect. Any suggestions?

Thanks, jvn

Given Mach number M and angle of attach α

Let unprimed denote dimensional quantities and primed denote non-dimensional.

q = free-stream speed = sqrt(u² + v²)

u' = u/q

v' = v/q

ρ' = ρ/ρ_∞

The pressure and temperature is determined from M.

So at free-stream

q' = 1

u' = cos(&alpha

v' = sin(&alpha

ρ' = 1

p' = 1/(γ M²), using definition of Mach number

Energy/volume

E' = p'/(γ -1) + ρ' q'²/2

I think your anomaly arises from the inconsistent nondimensionalization that you appear to be using. You use the speed of sound as a reference speed when nondimensionalizing the velocities, but you use the freestream speed as a reference speed when nondimensionalizing the pressure and energy. Praveen listed the initial / freestream dimensionless quantities where he consistently uses the freestream speed as a reference speed. Similarly you can write down the corresponding quantities if you consistently used the speed of sound as the reference speed. Just work it out both ways yourself to gain confidence in your understanding of nondimensionalization. Keep in mind the dimensionless form of the Euler equations used in the code. One usually chooses reference quantities so that the freestream Mach number does not explicitly enter the dimensionless form of the Euler equations in the code.

Thank you, Praveen and Ananda. I decided to use the dimensionless formulation because the "dimensional" one, using units that is, kept on crashing due to sqrt or power functions going out of range. I assumed that these problems were due to dealing with numbers with several orders of magnitude difference between them. Thanks again! jvn