5 replies to this topic

[trevortnt]

Hi I have a question regarding the energy loss in pipes due to friction. Also before the question, am i right in thinking that pressure drop, delta P, is entirely due to friction between pipe walls?

Question
1 Water flows at a rate of 0.008m3/s (volumetric flow) through a pipe 100m long and 10cm diameter. If the friction factor is 0.005 calculate:
i) the energy loss per unit mass due to friction

I'm not sure which equation to use and if indeed it involves consideration of the kinetic energy which I'm very unsure of in its realtion to the energy loss in pipe. Is it lost as heat? and if so is there an equation for this?

please help
thanks

Edited by christo7, 22 August 2009 - 03:34 PM.

[breizh]

Hi ,
Check your textbook on fluid dynamic! Bernouilli and Darcy may help you .
Regards
Breizh

[sheiko]

Hi,

Pressure has the same unit as energy by unit of volume (e.g. Pascal = Joule per cubic meter)...

Edited by sheiko, 23 August 2009 - 04:39 PM.

[latexman]

christo7,

If you can afford it, I highly recommend purchasing a copy of Crane's Technical Paper 410. In my opinion it is one of the best fluid flow references there is. It is concise and thorough, AND it has lots of worked out examples. It's available for sale on the internet at http://www.flowoffluids.com/tp410.htm There is an English or Metric version. It may be in your school's library.

[demank]

U can use a simple equation derived from ISO 13703

pressure drop (kPa/100m) = 6270 x 10⁶ x f x q_L² x d_L / d_i⁵
where qL in m3/h, dL is spesific gravity, di in mm.

Then insert your value, It will give the energy loss per unit mass is:
0.00416 J/kg

[katmar]

The Darcy equation is the correct one to use for calculating the pressure drop. The Darcy equation is used in two different forms. If the last term of the Darcy equation is density x velocity²/2 then it is the Pressure Drop form and if the last term is velocity²/(2 x gravity) then it in terms of head of liquid (since Pressure = density x gravity x height).

As Sheiko has pointed out, the pressure form is equivalent to energy per unit volume. A common form of the Bernoulli equation has each term divided by the density. If you divide pressure by density and noting Sheiko's comment then each term would be in units of energy per mass.

In real life problems you would never specify a pressure drop in these units. You would use either pressure (eg kPa or psi) or head (eg metre or foot). But it is useful to know how all these units relate to each other.