# Pressure Drop in Circular Pipe

• 03-03-2011, 01:31 PM
Ian
Pressure Drop in Circular Pipe
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: [url]http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm[/url]

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.

Ian
• 03-03-2011, 03:25 PM
PinkertonD
[QUOTE=Ian;83]The web page doesnt tell you what units you should be using. Can anyone provide a solution? [/QUOTE]

More than one way to skin a Sloth!
calculation pressure drop in pipe
• 03-04-2011, 02:38 PM
Kelly Bramble
[QUOTE=Ian;83]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: [url]http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm[/url]

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.

Ian[/QUOTE]

Yes, I think the equations need units and clarification, as well as an associated calcuator - on the list..
• 03-05-2011, 02:35 PM
zeke
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.
• 03-05-2011, 03:44 PM
Kelly Bramble
Good see you posting around Zekeman? ...

[QUOTE=zeke;99]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.[/QUOTE]
• 03-07-2011, 10:29 AM
Ian
Thanks Zeke!

I guess I did not realize that density for imperial units is in slugs/ft^3!

Ian