Question

4. A supersonic engine inlet is shown below-with a spike centerbody. Suppose the flight Mach number M1 = 2.5, and the pressure is pı = 50,000 N/m². The half-angle of the spike centerbody is 10°, as shown. For a particular mass flow through the engine, it happens that there is an oblique shock at a, and a normal shock wave standing at b. Before entering the second shock wave, the fluid expands through a Prandtl-Meyer turn, as the skech indicates. Determine the Mach number and the pressure M = 2.5 10° just after the normal shock wave.

Answer 1

My 2.086 mach no exet of oblaue shock P2 = Pix 1 + (x Min 2) I+ (2xM20²) 50,000 x [ 1 + (1.4*1.31926 [1+ Clih x 0.7753662) - ] Pe 3 93.30 x 10 N/m2 Expansion fan 2-3 Ma 2.086 Co. = We prandt - mager relation- VCMO - Sr+1) tart (6-12 (M21) - tant | Me-1 For ain Y = 1.4 у см) 6 tant CM2-1) 6 tools CM2-1) 12.0862" - tars' (2.086² - 1 Jo tan veM) : 2 VCM₂) = 28.172

3 V CM₂) = V CM₂) to (for expansion tan VCM3] = 28.72 + 10 = 38.72 V (Mg) = 38.72° 6 x tant (M12 - 1) - tant [M?-D Od solving this we get Mach no M3 Mg = 2.48283 apel Pa = Pax [*+ (1+1)x M, 2] 2 + (N-1) + Mg 2 rapol 3.5 93.3 x103 x Q+00.4* 2.0862) 2t (0.4x 2.48283² P3 50. 1 1868 a 10 x 103 N/m² qaoss normal shock (3-6) Mg = 2.48283 } MA " 2 + (0-1) *M₂ ² 28 xM₃? - (1-1)

6) М. : , 24 (o - х 2.4e283° ) (2.8 x 248 2832) – а.о. Mr = 0.5(4662 Per 37 837 Koncerty e, Mo" МРДАД 1868 xtuT 6 х 2 ки") a tok xo. Ax) Р : в Гя м, 15) Гt + Стиму) 11 мі 1863 Гr (ext') Гін бух - 16ѕѕ* ) PA = 3 кг 0 331. 26 Кра Result exct machno My 2 0.514652 pressure Ph P4 - 351 26 Кра