Pump head formula 4

[w1999]

Hello all,

I am new in engineering field, I was loking for a formula to find a pump head, I found one: Head=total pipe length*50% for elbows and fitting * 0.04. The last term I did not know what represent.
I know the formula for head is:V sq/ 2g.

Please, does any one can tell me what the 0.04 stand for.
Regards,

 

[pmover]

there are two book resources, in addition to your fluid textbook, that i highly recommend. . .

1) Cameron Hydraulic Textbook (now offered from Flowserve) and 2) Crane Technical Paper No. 410.

Both are available online and provide the reader excellent knowledge regarding hydraulics, etc.

prudent decision in that you are questioning formulas . . . highly recommend learning the fundamentals, principles, and how to apply them in your work.

oh, v^2/2g is referred to velocity head, not just head . . .

good luck!
-pmover

 

[jcfoley]

I agree with pmover on the Crane document. This is a must have in your technical library.

 

[LHA]

It looks like you just have the velocity head equation, hv. The "50% for elbows and fitting * 0.04" is trying to account for minor losses, hm. 0.04 is probably a weighted K. You need a few more things to get TDH, and P required of the pump.

delta z, ft + hv, ft = v^2/2g + hf, ft + hm, ft = Sum K * hv = TDH, ft
delta P, psi = TDH, ft/ 2.31

Using Hazen-Williams equation, hf = 10.44 * L *gpm^1.85 / (C^1.85 * Dia^4.8655)

Sum K = Specific K for each fitting * quantity of that fitting


Remember: The Chinese ideogram for “crisis” is comprised of the characters for “danger” and “opportunity.”
-Steve

 

[25362]

He is another possibility for the reason of the factor. Since you are speaking of rules of thumb and roughly approximate figures, for water in pipes, after taking the equivalent length at 150% of the piping length in ft o/a of fittings, you multiply by 4 ft of water/100 ft of pipe or 0.04 to obtain the total pressure drop, when assuming the following flow rates:

for 3in. Sch 40 pipe, 30 m³/h
for 4in. Sch 40 pipe, 60 m³/h
for 6in. Sch 40 pipe, 180 m³/h

Flow rates that would result in about the same friction drop of 4 ft/100 ft of pipe, or 0.04. Just a thought.

 

[w1999]

Thank you all for these inputs and help.

 

[lilliput1]

You are refering to friction head. Pump head is the capacity of the pump such that:

bhp = (GPM x Ft. wg pump head)/(3960 x pump efficiency in decimals)

Average efficiency is 0.65 with larger pumps better (to .80)while smaller pumps lower(to .40)

pump head in ft. wg. should be larger than sum of pipe friction + static head if open circuit + velocity head + equipment pressure drop + valve pressure drop + safety allowance for pipe aging + allowance for balancing valve.

 

[w1999]

Thank you lilliput1
for this formula, and all variable details.