Trivial question on head loss conversion.

[dbecker]

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

 

[Latexman]

The ackward King George's units! Thinking back to my university day's, the reason lb[sub]m[/sub] = lb[sub]f[/sub] is g/g[sub]C[/sub] = 1. I use

[Δ]P = h[sub]L[/sub][ρ]/144

[Δ]P in psi
h[sub]L[/sub] in feet
[ρ] in lb[sub]m[/sub]/ft[sub]3[/sub]

Good luck,
Latexman

 

[BigInch]

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/

 

[Latexman]

ft[sup]3[/sup], not ft[sub]3[/sub]

Sorry

Good luck,
Latexman

 

[dbecker]

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
?P = h?/144

?P in psi
hL in feet
? 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.

 

[BigInch]

32.174 lbm = 1 slug

yes just use [ρ]gh



**********************
"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/

 

[zdas04]

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 H[sub]l[/sub] times 0.433.

David

 

[katmar]

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 (length[sup]3[/sup]) portion of density to give area (length[sup]2[/sup]). Thus you have force per area, which is pressure.

If you use the units lbm/ft[sup]3[/sup] for density, ft/sec[sup]2[/sup] 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/inch[sup]3[/sup] 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/m[sup]3[/sup], gravity in m/s[sup]2[/sup] and height in m to give the answer directly in Pascal.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

But my gage reads Barg. So much for SI.

**********************
"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/

 

[zdas04]

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

 

[dbecker]

Thank you all for the great explanations!!

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

 

[dbecker]

Katmar,

Latexman wrote the following,

?P = h?/144

?P in psi
hL in feet
? 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.

 

[katmar]

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 g[sub]c[/sub] (the conversion factor from poundals to pounds force). It just happens that in this set of units g and g[sub]c[/sub] are numerically the same.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Dharmax]

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

 

[BigInch]

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/

 

[dbecker]

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
?P = h?/144 = 14 * (0.14) / 144 = 0.0136 psi.

Is that right?

Thanks,

 

[dbecker]

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

 

[BigInch]

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/

 

[dbecker]

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.

 

[BigInch]

Velocity Head derived from K.E.
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/