Answer:
M∞ = 0.53
M∞ = 1.5
M∞ = 3.1
Explanation:
Find: For each case the free stream Mach number.
-Pitot pressure=1.22×10^5N/m2 , static pressure=1.01 × 10^5N/m2 .
Solution:
- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:
P = 1.01 × 10^5 .. static pressure
Po = 1.22×10^5 ... pitot pressure ( hydrodynamic )
- Take the ratio:
P / Po = (1.01 × 10^5) / (1.22×10^5) = 0.8264.
- Use Table A.13 and look up the ratio P/Po = 0.8264 for Mach number M∞.
M∞ = 0.53
Find:
-Pitot pressure=7222 lb/ft^2 , static pressure=2116 lb/ft^2
Solution:
- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:
P = 2116 .. static pressure
Po = 7222 ... pitot pressure ( hydrodynamic )
- Take the ratio:
P / Po = (2116) / (7222) = 0.2930.
- However, since this is supersonic, a normal shock sits in front of the Pitot tube. Hence, Po is now the total pressure behind a normal shock wave. Thus, we have to use Table A.14.
P1 = 2116 .. static pressure
Po2 = 7222 ... pitot pressure ( hydrodynamic )
- Take the ratio:
Po2 / P1 = (7222) / (2116) = 3.412.
- Use Table A.14 and look up the ratio Po2/P1 = 3.412 for Mach number M∞.
M∞ = 1.5
Find:
-Pitot pressure=13197 lb/f^t2 , static pressure=1020 lb/ft^2
Solution:
- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:
P = 1020 .. static pressure
Po = 13197 ... pitot pressure ( hydrodynamic )
- Take the ratio:
P / Po = (1020) / (13197) = 0.0772.
- Again, since this is supersonic, a normal shock sits in front of the Pitot tube. Hence, Po is now the total pressure behind a normal shock wave. Thus, we have to use Table A.14.
P1 = 1020 .. static pressure
Po2 = 13197 ... pitot pressure ( hydrodynamic )
- Take the ratio:
Po2 / P1 = (13197) / (1020) = 12.85.
- Use Table A.14 and look up the ratio Po2/P1 = 12.85 for Mach number M∞.
M∞ = 3.1
