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Pressure drop in pipe sections

Pressure drop in pipe sections

Pressure drop in pipe sections

(OP)

When calculating pressure drop in long pipes I've used the following formula.

dp = 7.57*(q^1.85)*(L)*(10^4)) / ((d^5)*p) (1)

where

dp = pressure drop (kg/cm2)

q = air volume flow at atmospheric conditions (FAD) (m3/min)

L = length of pipe (m)

d = inside diameter of pipe (mm)

It's probably something silly buy I've just noticed it doesn't appear the units balance.... what's up with that?

Thanks,

Slader99


RE: Pressure drop in pipe sections

What does the 7.57 coefficient represent? You'll likely answer your question when you answer that question.

RE: Pressure drop in pipe sections

You haven't said what you're using for p in the denominator, but this appears to be initial (presumably start pressure) in absolute units. Didn't take long to google this...

This came from the engineering toolbox website

The pressure drop in compressed air lines can be calculated by using the formula


dp = 7.57 * q^1.85 * L * 10^4 / (d^5 * p) (1)

where

dp = pressure drop (kg/cm2)

q = air volume flow at atmospheric conditions (FAD) (m3/min)

L = length of pipe (m)

d = inside diameter of pipe (mm)

p = initial abs. pressure (kg/cm2)

They even gicve you a handy calculation box - http://www.engineeringtoolbox.com/pressure-drop-co...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

RE: Pressure drop in pipe sections

How in the world do you expect volume^1.87 to ever cancel with anything else? This is an empirical equation that fell out of somebody's statistical analysis of a data set (or maybe several data sets). It only works with the units that it requires, no conversion of the equation to use other units will ever give you meaningful answers.

Any equation with kg/cm^2 must have a cg term buried in there to get the bastardized force units back to a mass. Somehow the developer deleted the friction term (or assumed it as a value that can be rolled into the constant). For "long" pipes I wouldn't use any equation without an explicit friction term. But I wouldn't use any equation that I didn't know the underlying assumptions and could verify that my current case satisfied those assumptions. I use empirical equations all the time, but I always make sure that I'm not extrapolating outside the original dataset. They call that "Engineering".

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

RE: Pressure drop in pipe sections

(OP)
Thanks for the responses. As David mentioned it was the volume^1.87 that was throwing me for a loop. As Rconner suggested the 7.57 coefficent likely includes the cg term and would allow for my unit balance.

RE: Pressure drop in pipe sections

The page on the toolbox site further down does say that pressure drops greater than 1 kg/cm2 (14 psi) are not relevant , i.e. valid, probably due to the expansion effect.

I agree, we all use some rough and ready calces, but "long" pipes with compresisble fluids need to be analysed properly. A simple fixed formula like this doesn't work.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

RE: Pressure drop in pipe sections

Looks like some kind of Manning formula adapted for gas flow. Good luck with that.

Independent events are seldomly independent.

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