A
annie
I need help in understanding how they calculated the exit Mach Number in Problem 32. :brickwall:
Thanks
Annie
Thanks
Annie
Annie,
Not positive, but at first look I believe that you can get to the solution by using table for the isentropic flow functions and the normal shock relations. We assume isentropic flow from inlet to standing shock, and from the downstream side of the shock to the exit plane.
For a shock in the diverging section, we have sonic conditions at the throat (M = 1). I'm using x for values upstream of the shock and y for downstream values.
Using Mx = 1.80 and the table of normal shock functions, we can now use the chain rule to find: A_exit/Ay* = (A_exit/Ax*)(Ax*/Ay*) = (A_exit/Ax*)(Poy/Pox) = (3)(0.81268) = 2.438 (substitution possible since we assume no chane in cross-sectional area across the shock wave)
Using the isentropic flow functions, there are two Mach number choices for this ratio (A/A*) and we choose the subsonic value since it is downstream of the standing shock. I got roughly 0.25.
Hope this helps. Please double check the calculations.
Newton
Enter your email address to join: