I want to convert 1.3138 kg/Nm^3 to lbm/ft^3. I've searched a lot of posts and can't find exactly what I want. I know that Nm^3 stands for "normal cubic meter" which means it is at standard temperature and pressure, but I don't know what to do with information. If I am talking about Dry Air, how would I convert that to an actual density value?
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Converting Density from kg/Nm^3 1
- Thread starter sloth4z
- Start date
To convert kg/m3 to lbm/ft3 is straight forward. Convert kg to lbm, then m3 to ft3.
The normal part you leave as is. So, you now should have the kg/Nm3 --> lbm/Nft3.
If you want to change normal condition to your actual process condition, you can do so using any of the equation of states, graphs, correlations, etc. for dry air.
"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
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The normal part you leave as is. So, you now should have the kg/Nm3 --> lbm/Nft3.
If you want to change normal condition to your actual process condition, you can do so using any of the equation of states, graphs, correlations, etc. for dry air.
"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?
kg/Nm[sup]3[/sup] is already a density. One should specify the pressure and temperature, which in this case would probably be 0[sup]o[/sup]C and 1 atm. Conversion to lb/cf at the same normal (or standard) conditions involves applying factors such as 1 ft[sup]3[/sup] = 0.02832 m[sup]3[/sup], and 1 lb = 0.4536 kg.
For changing the selected standard conditions, refer to thread798-106556 and the links therein.
In order to understand how to do the conversions go to Milton Beychoks site at
To do the actual conversions you can download a free calculator that does it all automatically. Click on the link in my signature below and follow the links to Uconeer. Once you have installed and run Uconeer click on the fan icon in the toolbar and the conversion calculator will open up. But make sure you understand what you are doing before you use the calculator. GIGO.
Katmar Software
Engineering & Risk Analysis Software
To do the actual conversions you can download a free calculator that does it all automatically. Click on the link in my signature below and follow the links to Uconeer. Once you have installed and run Uconeer click on the fan icon in the toolbar and the conversion calculator will open up. But make sure you understand what you are doing before you use the calculator. GIGO.
Katmar Software
Engineering & Risk Analysis Software
-
1
- #5
Assuming that ideal gas behavior applies, and that the Normal cubic meter was defined as being at 0 °C and 1 atmosphere of pressure, you can divide kg/Nm[sup]3[/sup] by 16.928 and you will obtain the density as lb/SCF where the SCF is defined as being at 60 °F and 1 atmosphere:
lb/scf = (kg/Nm[sup]3[/sup]) [÷] 16.928
If you want your density at some other temperature and pressure:
lb/cf @ T and P = (lb/scf) [×] (520 [÷] T) [×] (P [÷] 14.696)
where:
T is the temperature in °R = 460 + °F
P is the absolute pressure in psia (pounds per square inch absolute)
If you want to leave your density in the metric units of kg/Nm[sup]3[/sup] but want to convert it to another temperature and pressure:
kg/Nm[sup]3[/sup] @ T and P = (kg/Nm[sup]3[/sup]) [×] (273.15 [÷] T) [×] (P [÷] 101.325)
where:
T is the temperature in °K = 273.15 + °C
P is the absolute pressure in kPa (kiloPascals absolute)
Milton Beychok
(Visit me at www.air-dispersion.com)
.
lb/scf = (kg/Nm[sup]3[/sup]) [÷] 16.928
If you want your density at some other temperature and pressure:
lb/cf @ T and P = (lb/scf) [×] (520 [÷] T) [×] (P [÷] 14.696)
where:
T is the temperature in °R = 460 + °F
P is the absolute pressure in psia (pounds per square inch absolute)
If you want to leave your density in the metric units of kg/Nm[sup]3[/sup] but want to convert it to another temperature and pressure:
kg/Nm[sup]3[/sup] @ T and P = (kg/Nm[sup]3[/sup]) [×] (273.15 [÷] T) [×] (P [÷] 101.325)
where:
T is the temperature in °K = 273.15 + °C
P is the absolute pressure in kPa (kiloPascals absolute)
Milton Beychok
(Visit me at www.air-dispersion.com)
.
- Thread starter
- #6
sethoflagos
Chemical
Are you guy's sure about this?
Nm3 is not a unit of volume, it's 1/22.4 of a kmol of gas (which just happens to occupy 1 m3 at ntp). Hence 1.3138 kg/Nm3 is equivalent to saying Molecular Weight = 29.43. Isn't it?
Nm3 is not a unit of volume, it's 1/22.4 of a kmol of gas (which just happens to occupy 1 m3 at ntp). Hence 1.3138 kg/Nm3 is equivalent to saying Molecular Weight = 29.43. Isn't it?
Nm3 was defined to be Normal meter cubed in the OP. Hence, it is a unit of volume.
"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?
"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?
sethoflagos
Chemical
So what's the volume of 1 Nm3? We don't know - we need more information.
But we do know how many molecules are there.
Beg to differ Ashereng, the units are kmols.
But we do know how many molecules are there.
Beg to differ Ashereng, the units are kmols.
I agree with sethoflagos that the units kg/Nm[sup]3[/sup] are equivalent to Molecular Weight. Nm[sup]3[/sup] defines an amount of matter, which is the number of kmols. If you divide the number of kmols into the total mass you get MW.
One small disagreement. Since the definition of Normal conditions was redefined by IUPAC to be based on 100 kPa and not 1 atm, a Nm[sup]3[/sup] is now 1/22.681 of a kmol of gas.
Katmar Software
Engineering & Risk Analysis Software
One small disagreement. Since the definition of Normal conditions was redefined by IUPAC to be based on 100 kPa and not 1 atm, a Nm[sup]3[/sup] is now 1/22.681 of a kmol of gas.
Katmar Software
Engineering & Risk Analysis Software
1.3138kg/m^3 is 0.0820 lbm/ft^3.
Sloth4Z, be careful with your units. "N" in metric means newtons which is a measure of force. Saying "normal meters" is meaningless in the metric system; there is no abnormal meter, for example.
Also, the correct measurement for mass in the imperial system is "slugs". Since "kg" is correct for the metric mass measure, I think it proper that you should list the density as "slugs/ft^3". This is not to say that "lbm/ft^3" is wrong, just be aware that the computations involving mass would be out by a factor of gravity acceleration. I prefer the metric system for this reason, then convert the answer to imperial if the need be.
Trick question!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Sloth4Z, be careful with your units. "N" in metric means newtons which is a measure of force. Saying "normal meters" is meaningless in the metric system; there is no abnormal meter, for example.
Also, the correct measurement for mass in the imperial system is "slugs". Since "kg" is correct for the metric mass measure, I think it proper that you should list the density as "slugs/ft^3". This is not to say that "lbm/ft^3" is wrong, just be aware that the computations involving mass would be out by a factor of gravity acceleration. I prefer the metric system for this reason, then convert the answer to imperial if the need be.
Trick question!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
The conversion by Cockroach from kg/m[sup]3[/sup] to lb/ft[sup]3[/sup] is correct only if the P,T conditions (normal or standard) are the same. Mass wouldn't change but volume might. Refer to thread798-106556, and to the above post by mbeychok.
![[ponder]](/data/assets/smilies/ponder.gif "[ponder] [ponder]")
there is something ![[hammer]](/data/assets/smilies/hammer.gif "[hammer] [hammer]")
![[curse]](/data/assets/smilies/curse.gif "[curse] [curse]")
metrics![[rofl]](/data/assets/smilies/rofl.gif "[rofl] [rofl]")
The use of the capital "N" to represent Normal and Newton is confusing. Strictly we should put a point (full stop) between the units if there were such a thing as a Newton metre cubed. It would be written as N.m[sup]3[/sup]
On the other hand, when we talk Normal metre cubed there would be no point involved and it would be written as Nm[sup]3[/sup]. But this convention is so sloppily applied that you can't really rely on it and you have to take the context into account to determine whether the N stands for Normal or Newton.
I'm 100% with Cockroach in being against any system of units that involves the dreaded g[sub]c[/sub]. It is purely by coincidence that g[sub]c[/sub] has the same numerical value as the acceleration of earth's gravity in the US customary system of units. This coincidence leads to untold confusion. But being realistic, we engineers will have to deal with g[sub]c[/sub] for a while yet.
Katmar Software
Engineering & Risk Analysis Software
On the other hand, when we talk Normal metre cubed there would be no point involved and it would be written as Nm[sup]3[/sup]. But this convention is so sloppily applied that you can't really rely on it and you have to take the context into account to determine whether the N stands for Normal or Newton.
I'm 100% with Cockroach in being against any system of units that involves the dreaded g[sub]c[/sub]. It is purely by coincidence that g[sub]c[/sub] has the same numerical value as the acceleration of earth's gravity in the US customary system of units. This coincidence leads to untold confusion. But being realistic, we engineers will have to deal with g[sub]c[/sub] for a while yet.
Katmar Software
Engineering & Risk Analysis Software
Just a thought: it would be easier to understand if mass would be expressed in kg, and weights in N.
A man with a mass of 70 kg on Earth would still have the same mass on the Moon.
His weight (a force=m.g) on the surface of the Earth would be
9.8 m/s[sup]2[/sup][×]70 kg = 686 N.
On the Moon his weight would be 1.6 m/s[sup]2[/sup][×]70 kg = 112 N, and on Mars it would be 3.74 m/s[sup]2[/sup][×]70 kg = 262 N.
Thus a rope supporting a maximum weight of 350 N couldn't support that man on Earth, but it could on the Moon and on Mars.
sethoflagos
Chemical
Good point 25362.
But the sitution is actually worse than having to worry which planet you're on. The gravitational 'constant' gc is tied to sea level at latitude 45 or thereabouts. Now I live within a stones-throw of the equator where gravity's a little (not much, but a little) stronger.
Is my psi the same as your psi?
You know where you stand with N/m2 !
Katmar, thanks for the info on ntp, though I know nothing of this IUPAC of which you speak. Which pub do they hold their meetings in?
But the sitution is actually worse than having to worry which planet you're on. The gravitational 'constant' gc is tied to sea level at latitude 45 or thereabouts. Now I live within a stones-throw of the equator where gravity's a little (not much, but a little) stronger.
Is my psi the same as your psi?
You know where you stand with N/m2 !
Katmar, thanks for the info on ntp, though I know nothing of this IUPAC of which you speak. Which pub do they hold their meetings in?
Sethoflagos,
The gravity at the equator is a little, not much, weaker -not stronger- because of the slight increase in distance from Earth's center.
From the CRC Handbook of Chemistry and Physics: the mean acceleration of gravity is 9.78036 m/s[sup]2[/sup] at equator vs 9.83208 m/s[sup]2]/sup] at poles.
![[surprise]](/data/assets/smilies/surprise.gif "[surprise] [surprise]")
sloth4z said:
Guys, sloth4z clarified that the N stands for normal.
I still think he is confused about what normal is, and how to get it to his working conditions, or to standard.
He is converting kg/m3 to lbm/ft3, and then try to figure out how to deal with the difference in temp and press.
"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?
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