FRICTION LOSS THRU .25" ID TUBING 3

[JCA]

I AM LOOKING FOR FRICTION LOSS DATA ON .25" SS TUBING. THE VISCOSITY OF THE LIQUID IS 17,000 CPS, THE FLOW RATE IS 1 LB/MIN. WE ARE TRYING TO PUMP THRU 4' OF .25" SS TUBING THEN THE CUSTOMER WANTS TO PUMP THRU A .080" ORIFACE. I CANNOT FIND FRICTION LOSS DATA FOR 1/4" TUBING TO VERIFY IF THIS IS EVEN POSSIBLE. CAN ANYONE HELP.

 

[jhelm]

You will need to calculate the pressure drop using a formula such as the Darcy-Weisbach equation. The Cameron Hydraulic Data book published by Ingersoll-Rand has a detailed explanation of the calculations necessary. Here is a short version that may be of some help:

Darcy-Weisbach equation ( LV)2
friction loss= f ----------
D2g

where
L = pipe length - ft
D =average inside diameter of pipe - ft
V =average pipe velocity in ft/sec
g =gravitational constant (32.174 ft/sec squared)
f =friction factor (dimensionless number that depends on the roughness of pipe
ID and the Reynolds number)

With the viscosity you state, I believe you will have laminar flow (Reynold's number below 2000) so the roughness of the pipe ID has no effect and your friction factor would become:
64
f = ----------
R

where
VD
R = Reynold's number = -----------
v

where

V and D as above
v = Kinematic viscosity - ft squared/second

If your Reynold's number is below 2000, you have laminar flow. If it is above 4000, the flow is turbulent and you will need to determine the friction factor. That is more involved an requires yet another calculation using the Colebrook formula, unless you have access to further data about the 1/4 ID tubing. For that, you might try the vendor of the tubing.

More information is also available from the Hydraulic Institute and their Pipe Friction Manual. See their website at

http://www.pumps.org.

Regards and good luck, Jeff Helm

 

[rick4]

The Darcy-Weisbach equation in the previous post for friction loss should not have the "L" term squared.

h = (f L/D) V²/(2g)

In the above form, a loss coefficient (for valves, fittings) can be substituted for (f L/D).

 

[jhelm]

Rick, thanks for your correction. I did not notice the error in the formula when I posted it. I was having trouble working out how to get the formulas to be clear.


Regards, Jeff