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
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.
ΔP = hLρ/144
ΔP in psi
hL in feet
ρ in lbm/ft3
Good luck,
Latexman
RE: Trivial question on head loss conversion.
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.
Sorry
Good luck,
Latexman
RE: Trivial question on head loss conversion.
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.
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.
So you can just multiply your Hl times 0.433.
David
RE: Trivial question on head loss conversion.
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.
**********************
"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.
David
RE: Trivial question on head loss conversion.
I would love to move to SI units but my customer isn't that merciful!!
:)
RE: Trivial question on head loss conversion.
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.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Trivial question on head loss conversion.
RE: Trivial question on head loss conversion.
**********************
"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.
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.
RE: Trivial question on head loss conversion.
**********************
"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.
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.
with units in MKS and US
attached,
**********************
"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.
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.
**********************
"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/