Thread: Pressure Drop in Circular Pipe

1. 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: 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.

Ian  Reply With Quote

2. Originally Posted by Ian The web page doesnt tell you what units you should be using. Can anyone provide a solution?
More than one way to skin a Sloth!
calculation pressure drop in pipe  Reply With Quote

3. Originally Posted by Ian 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.

Ian
Yes, I think the equations need units and clarification, as well as an associated calcuator - on the list..  Reply With Quote

4. 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.  Reply With Quote

5. Good see you posting around Zekeman? ... Originally Posted by 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.  Reply With Quote

6. Thanks Zeke!

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

Ian  Reply With Quote

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flow, friction factor, pipe, pressure, velocity  Posting Permissions

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