Question

The air in the driven section of a shock tube is initially at the temperature and pressure of 300 k and 0.1 atm.

After the diaphragm is broken, an incident shock wave with a velocity of 1440 m/s is propagating into the driven section where a pitot tube is placed.

The pitot tube is facing toward the oncoming incident shock wave With its axis parallel to the axis of the shock tube.

calculate :

a) the total pressure of the air behind the shock wave ?

b) the pressure pitot measures after the incident shock sweeps?

Answer 1

Ans- Shock waure Shock tube (2. grven P2 = 0.1 alm W= 144om/s sttimang Now het mdke shock WAme 2. Sho ck wae. M2 = JoRTz R= 2875/k Jor Air 1440 J1.4x287x 3v0 M, = 4.15 %3D Normel Sholk teble for My-4-15 from Pol o.12232 Po2 - 19.92625

from Isentropic teble for M= 4155 Pa 0.0654' Poz Po1= Oo12232 X Po, a) %3D D 12232X Pz 6.0054 z 0-12232 Poj = 2.2652 etin P, 19.92625 P2 19.92625xO:l atm P, = P,-1.992625etra