This question is incomplete, the complete question is;
Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.
Note that standard sea level density and pressure are 1.23 kg/m3 (0.002377 slug/ft3) and 1.01 x 105 N/m2 (2116lb/ft3), respectively.
Answer:
the velocity of the airplane is 267.2 ft/s
Step-by-step explanation:
Given the data in the question;
throat-to-inlet area ratio A₂/A₁ = 0.75
density of air ρ = 0.002377 slug/ft³
the pressure at inlet p₁ = 2116 lb/ft³
the pressure at the throat p₂ = 2050 lb/ft³
Now, for a venturi duct, the velocity of the airplane V is given as;
V = √[ (2( p₁ - p₂ )) / (ρ( [A₁/A₂]² - 1 )) ]
so we substitute in our values
V = √[ (2( 2116 - 2050 )) / (0.002377 ( [1/0.75]² - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 ( 1.7778 - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 × 0.7778 ) ]
V = √[ 132 / 0.0018488 ]
V = √[ 71397.663349 ]
V = 267.2 ft/s
Therefore, the velocity of the airplane is 267.2 ft/s