# Why does the temperature rise in the boundary layer after a shock wave?

Is it because the BL thickens after the shock so the drag increases, causing a increased friction? Here M=0.82 far from the airfoil, the picture represents the static temperature in Kelvins around the symetrical airfoil at angle of attack 0°.

This is the total temperature for a supposedly everywhere adiabatic flow.

This is not the static temperature but the Mach number, for $$M_{\infty}=0.70$$

Mach number for $$M_{\infty}=0.80$$

Mach number for $$M_{\infty}=0.99$$

Mach number for $$M_{\infty}=1.20$$

I understand that as velocity decreases, temperature rises, but I have another file that shows the evolution of static pressure and it does not increase as static temperature does, after the shock in the BL.

Thank you

• What is this picture representing? The scale/color seems odd and counterintuitive, whatever it is. There is no unit given. – Jpe61 Apr 30 '20 at 22:31
• Can you get the same picture at Mach 0.3, 0.5, 0.7, 0.9, 1.2. Huge info here. Temp rise could be compression or friction. But striking is the bow shock wave changing the shape of the low pressure areas, allowing encroachment of higher pressure from behind the wing (similar to stalling). This "encroachment" may be forming a "tidal bore" on the curved surface. Is this the "shock wave"? Also, is the low pressure area saying "this is what my wing wants to look like at Mach 0.82"? – Robert DiGiovanni Apr 30 '20 at 23:50
• @Jpe61 the first picture is representing the stagnation (static?) temperature around a symetrical airfoil. What would you have set for the color scale? The unit is K. These images have been provided to all students by the teachers because of the current quarantine. – StrangeDoc May 1 '20 at 11:38
• @Robert Digiovanni I edited my post with the pictures of the Mach number for 0.7, 0.8, 0.9 and 1.2 . I will come back with the pictures of the stagnation temp instead of the Mach number as soon as I run the files in Paraview, however I only have the files from 0.7 to 2. Thank you both for your answers! I'll dig deeper with what you gave me :) – StrangeDoc May 1 '20 at 11:41
• @JZYL It is adiabatic. What if it was isothermal? – StrangeDoc May 1 '20 at 14:32

Due to no-slip condition, the flow immediately adjacent to the wall has no velocity relative to the wall. If the flow is adiabatic everywhere and there is no significant viscous stresses (more on that later), then the total enthalpy ($$h_0$$) is constant; if we assume that the gas is calorically perfect (good assumption for air at low supersonic speeds) and that the heat capacity is constant (good assumption for small temperature variation), then:
$$h_0=c_pT_0=c_pT+\frac{1}{2}V^2$$
where $$T$$ is the static temperature, $$T_0$$ is the total temperature, $$V$$ is flow speed.