Chemical Forums

Specialty Chemistry Forums => Chemical Engineering Forum => Topic started by: gera19 on April 14, 2006, 02:30:50 AM

Title: Friction factor Q
Post by: gera19 on April 14, 2006, 02:30:50 AM
See Q 1b
i used the formula for change in pressure=2Cf X density X velocity^2 X L/d but there are simply too many unknowns. Pls help.
Title: Re: Friction factor Q
Post by: Donaldson Tan on April 15, 2006, 07:42:34 AM
Can you type the question out instead posting the PDF?

It will be easier for all of us to view the question.
Title: Re: Friction factor Q
Post by: Donaldson Tan on April 18, 2006, 03:25:19 AM
I assume you are using the Fanning Friction Factor because this is a chemical engineering problem.

However, this is not required to solve the problem. You only need to find the frictional loss (in energy).

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fmath%2F8%2F2%2Fc%2F82cf4ee7e0214e063ea7d24787380f8a.png&hash=55c75098d4e49e95920b48c9734bd3ec)
The original bernoulli equation (see above) does not take in account of energy input and friction losses.

I have modified it to answer the question. It is still the same principle being used to solve the problem, ie. the first law of thermodynamics: conservation of energy.

energy in = energy out
uin2/2 + ghin + Pin/? + Wpump = uout2/2 + ghout + Pout/? + Floss
uin2/2  + Pin/? + Wpump = uout2/2 + g(hout - hin) + Pout/? + Floss

where Floss is the frictional loss and Wpump is energy supplied by pump to soybean oil

Since the exit and entrance pipes are of the same diameter, mass flow rate is constant and the soybean oil is incompressible, then the velocity of the oil must remain constant, ie. uin = uout = u

This means the above equation reduces to:
Pin/? + Wpump = Pout/? + Floss + g(hout - hin)
Floss = Wpump - g(hout - hin) - (Pout - Pin)/?