Question

Air ows over a wall at a supersonic speed. The wall turns towards the ow generating an oblique shock wave. This wave is found to be at an angle of 400 to the initial ow direction. A scratch on the wall upstream of the shock wave is found to generate a very weak wave that is at an angle of 300 to the ow. Find the angle through

Answer 1

10.65°

Step-by-step explanation:

To calculate the initial Mach Number

`Sin \beta _(weak) = (1)/(Ma_1)`

`Ma_1 = (1)/(sin \beta_weak)`

`Ma_1 = (1)/(sin 30)`

`Ma_1 = 2`

Now, to calculate the angle turned by the wall; we have the following expression.

`tan \theta = (2 cot \beta_i (Ma_1^2 sin^2 \beta_i -1 ))/(Ma_1^2(k+cos2 \beta_i)+2)`

where:

`\beta_i` = the angle between the initial flow direction = 40°

K is the ratio of specific heats o fair = 1.4

`tan \theta = (2 (cot 40) (2^2 (sin 40)^2 -1 ))/(2^2(1.4+cos2 (40))+2)`

`tan \theta = (1.56)/(8.295)`

`tan \theta = 0.188`

`\theta = tan ^(-1)( 0.188)`

`\theta = 10.65^0`