Question

3. A 18" half-angle wedge is mounted at angle of Back in the section of a supersonic wind tunnel. The tunnel reservoir pressure and temperature and 350 K, respectively. When the tunnel is operating the wave angle of oblique Shockwaves formed from the leading edge of the wedge is measured to be 41.8" (a) Calculate the exit Mach number (5 ps) and ext-to-throat area ratio (5 ps) of the nozzle (b) Calculate the local pressure (5 pts. local temperature 5 local density (5 pts) and local velocity (5 pts) at the exit of the note Reservoir To = 350k Pe5

Answer 1

To =350k Po = 5 atm Throat i B = 41-8° - 180 From oblique shock (A-3) tables, Corresponding to ß= 418, 0=18°, mach Number in 2-4 otherwise, substitute O, B in. tano = 2 cotß [m, sin²B-1] mi? [R+ cos2B] +2 tan 18 - 200t (418) [m, "sin?(418)-1] mi [1:4+ cos2 (419)] +2 0.324 = 2.2368[m 20-444-2] m, 2 [1-511] +2 By solving above Egn, - 0.504 m, ² - 2.8848 m, = 2.41

Oblique mach Number Before The Shock in 2-4- & The Mach number at nozzle erit in 2-4 Forum i sentropic flow (A-1) tables, @M=24, 432.403, +0:0629 -0447 Exit to throat Area ratio = 2.403. estagnation properties are same throughout the igentropic flow, 0.342 atm = 34.65 kPa local pressure (P2) = 580-8684 = local temperature (Iz) = 35080.4647 = 162.645k. local density (2) = P2 RT2 =34-65 0-2878162-645 = 0.742 kg/m3 local velocity (V2) = m₂ x K RT2 = 2-4 XJ1-4x287 X 162-445 -613-53 mly

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