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Trivial question on head loss conversion.

Trivial question on head loss conversion.

Trivial question on head loss conversion.

(OP)
I am trying to convert head loss of feet into corresponding pressure drop in psi.

If I multiply the height in feet by density which is in units of lbm/ft^3, then my result will not be pounds force per square foot, but pounds mass per square foot (this is not pressure!!!).

Am I missing something here?

  Thanks

RE: Trivial question on head loss conversion.

The ackward King George's units!  Thinking back to my university day's, the reason lbm = lbf is g/gC = 1.  I use

ΔP = hLρ/144

ΔP in psi
hL in feet
ρ in lbm/ft3

Good luck,
Latexman

RE: Trivial question on head loss conversion.

Why not use LBS FORCE/ft3 (62.4 lbf/ft3 for water)?

Or you could use 1.9394 slugs/ft3 (that's mass) then multiply by the gravitational constant to get force ( x 32.174 fps = 62.4 lbs force/ft3

 

 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Trivial question on head loss conversion.

ft3, not ft3

Sorry

Good luck,
Latexman

RE: Trivial question on head loss conversion.

(OP)
So will the correct answer then be from the correlation:

P = rho * g * h,

where,

rho = lbm / ft^3 density
g = 32.2
h = head loss in feet

Latexman, your formula
ΔP = hρ/144

ΔP in psi
hL in feet
ρ in lbm/ft

does not contain any conversion from lbm to lbf? Am I just confused? :(

Can I just use rho * g * h?


P.S. This is not for water, we are using nitrogen.

RE: Trivial question on head loss conversion.

32.174 lbm = 1 slug

yes just use ρgh

 

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Trivial question on head loss conversion.

You can also work it as a gradient which is just ρ/144 or 0.433 psi/ft for 62.4 lbm/ft^3 water (the "g" term really does work out to "1" when you use lbm instead of slugs).

So you can just multiply your Hl times 0.433.

David

RE: Trivial question on head loss conversion.

This is a great example of why we should all be working in SI units. But as long as you use any set of consistent units you will get the right answer.  The problem is that the US Customary set of units is NOT a consistent set of units.

At least your physics is correct and for a column of fluid
 pressure = rho * g * h   (in consistent units!)

To answer your question of where does the conversion from lbm to lbf occur consider that the mass part of the density multiplied by acceleration (i.e. gravity) gives you force. And the height (length) cancels part of the volume (length3) portion of density to give area (length2). Thus you have force per area, which is pressure.

If you use the units lbm/ft3 for density, ft/sec2 for gravity, and ft for height you will obtain the pressure in poundals per square foot. Divide this by 144 (as per Latexman) and you have poundals per square inch. Divide again by 32.17 and you have lbf per square inch (=psi).

Or you could express your density in slug/inch3 and your height in inches to get the answer directly in psi.

Or you could move into the 21st century and express density in kg/m3, gravity in m/s2 and height in m to give the answer directly in Pascal.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Trivial question on head loss conversion.

But my gage reads Barg.  So much for SI. smile

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Trivial question on head loss conversion.

Mine reads kg/cm^2.  Is kg(f) any smarter than lbf?  When a unit doesn't have a gestalt with people, they will use units that do.  kPa seems to be a real bust in casual usage or there wouldn't be so many gauges in Bar(g) or kg/cm^2.

David

RE: Trivial question on head loss conversion.

(OP)
Thank you all for the great explanations!!

I would love to move to SI units but my customer isn't that merciful!!


  :)

RE: Trivial question on head loss conversion.

(OP)
Katmar,

Latexman wrote the following,

ΔP = hρ/144

ΔP in psi
hL in feet
ρ in lbm/ft

g does not show up because as stated, we multiply by g and then divide by g and divide by 144 to obtain lbf/in2.

I hope I got it right.

RE: Trivial question on head loss conversion.

Latexman's formula will give you the right answer if you stick to his units. You are multiplying by g (the acceleration of gravity) and dividing by gc (the conversion factor from poundals to pounds force). It just happens that in this set of units g and gc are numerically the same.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Trivial question on head loss conversion.

I use a value of 2.2 feet per 1 p.s.i. It depends on how accurate you need to be.

RE: Trivial question on head loss conversion.

If it happens to be water, that's fine.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Trivial question on head loss conversion.

(OP)
Hello again.

 In the equation P = hL*rho / 144.

hL = v^2 / 2   and NOT  v^2 / 2g correct?

For example, nitrogen moving through a Tee junction at 30 psi and 30 ft/s (flow through run) will decrease in static pressure at the branch by the following amount: this is the way I did it.

h = v^2 / 2g = 30^2 / (2*32.2) = 14 ft static head loss

converted to pressure loss using latexmans equation
ΔP = hρ/144 = 14 * (0.14) / 144 = 0.0136 psi.

Is that right?

Thanks,



 

RE: Trivial question on head loss conversion.

(OP)
In my last post I used h = v^2 / 2g to get h in feet. That is the only way those units work out.  

RE: Trivial question on head loss conversion.

N2 moving through a tee will not decrease pressure by the 30^2 unless the fluid comes to a dead stop.  It changes direction as it moves through out the branch, it doesn't stop.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Trivial question on head loss conversion.

(OP)
BigInch I see what you mean. Assuming I am measuring the static pressure with a probe mounted transverse to flow or similar.


 Are my methods for calculating the head loss still dimensionally consistent? I am using h = v^2 / 2g then
P = rho*h/144.

RE: Trivial question on head loss conversion.

Maybe this is obvious to everyone else, but it helps me to remember that the reason the 'customary pound' units are the way they are is this:  normally (ie, here on the surface of the earth) a pound mass weighs a pound (force).  The problem is, if you apply a force of 1 lbf to a mass of 1 lbm, it does not accelerate at 1 ft/sec2; in fact it accelerates at 32.2 ft/sec2.

That's really the source of all (my) confusion, and the reasons why the units are set up to confuse me.  Like I said, understanding that helps me put the 32.2 in where it goes.
 

RE: Trivial question on head loss conversion.

Actually that depends in which direction the force is applied and acceleration would be 33.2 ft/sec, if the force was applied downward not 32.2.

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

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