I am trying to find the pressure drop of air flowing through a circular pipe. The equation provided by the engineers edge website is (P_1^2 - P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.
The link is: http://www.engineersedge.com/fluid_f...ssure_drop.htm
The problem with this equation is that the units don't match up. The flow speed is squared which gives something over seconds squared but there is nothing else in this equation that cancels out the seconds squared. The web page doesnt tell you what units you should be using. Can anyone provide a solution? Thank you.
Nothing wrong with the dimensions
- P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.\
P_2^2)/(2P_1) dimension of pressure
Friction coefficient * L/D dimensionless
density * w^2/2 pressure lb/ft^3/Ft/sec^2*(ft/sec)^2 ....................... lb/ft^2 or pressure
So the left half and right haalf of the equation have the dimension of pressure.
I guess I did not realize that density for imperial units is in slugs/ft^3!