Air is
passing through 4-in ID pipe at 14.7 psi and 70
degrees Fahrenheit. A manometer
connected to the pitot tube
measures a pressure drop of one inch of mercury. What is the velocity of air? Use pitot
tube constant as 1.0.
Solution:
For
manometers, pressure drop (delta P) is directly related to delta h. If manometer fluid density is rho_m then, delta P is equal to delta h times rho_m minus rho times g over gc. If height is measured in meter, density in kilogram
per meter cubed, then delta P values is in pascals.
Density of
mercury is 13.56 gram per cc or 1.356 times 104 kilogram per cubic
meter. Density of air is 1.196 kilogram
per cubic meter. Pressure drop is
calculated as 3.378 times 103 pascals.
If we
consider that air is incompressible, then velocity of the fluid passing though
the pipe is given by C time square root of 2 gc delta P over rho, where
C is the ‘Pitot Tube Constant”.
Outlet
pressure is found as 9.795 times 104 pascals. For
gases with adiabatic flow conditions, velocity is given as 74.69 meters per
second.
Note that Pi
is inlet pressure and
If the gas
velocity is found greater than 200 feet per second, treat case as flow of
compressible fluid.