Kg/hr to Nm3/hr

[chmcl]

How to convert 45kgs/hr of hydrocarbon @ 0.01067bar and 88 Deg C (M.Wt 160) to Nm3/hr and M3/hr?

 

[MortenA]

Nm3 is just another way of saying molar flow.

So if you have a combined MW of 160 - divide your mass flow by this and use the factor 44.618 mol/Nm3 to find the flow in Nm3.

Sound like a school assignment because half the numbers are just smokescreen.

Best regards

Morten

 

[zdas04]

I think it is an alien problem. There must be a planet somewhere with a hydrocarbon gas with a molecular weight of 160. n-Decane has a MW of 142.385, and its boiling point is 174C. That is the heaviest hydrocarbon in my list of hydrocarbon gases. Radon is a gas at room temperature and its molar mass is 222, so maybe you have a "hydrocarbon" that is 95% radon?

MortonA's constant, assumes that he knows the "normal" or "standard" conditions. That is good if you do, but those conditions change from jurisdiction to jurisdiction and contract to contract. I find it more reliable to work with density than MW.

Anyway. volume flow rate = (mass flow rate)/[ρ]. So if you have mass flow rate, you only need to calculate the density of the gas as thoughh it were at your "normal" (or "standard") conditions to get Nm^3/hr. I'm not even going to guess what M3/hr might be.

David

 

[25362]

1. The given MW is probably that of a mixture of hydrocarbons.
2. MortenA gave you the answer for standard TP conditions taken at 1 ata (= 1.01325 bar) and zero Celsius (= 273.15 K).
3. If M3/hr means "actual" m³/h, then you can find it with the previous result assuming ideality using:

(PV/T)_actual = (PV/T)_normal​

4. zdas04 is right about the definition of normal conditions, which might affect a bit the results. For example, if NTP refer to 273.15 K and 1 bar, then the mol/m³ factor changes by 1.325%.
5. I also think it sounds like a school assignment.

 

[tickle]

It might not be a gas and the OP is after standard liquid flow. I will not comment further on the assumption that this is a homework question.

 

[cloa]

If you think its a homework question, red flag it. Anyway at that tiny pressure 0.01 bar, its almost certainly a gas.

 

[MortenA]

David when we are splitte hair:

When dealing whith "normal volume flow" then you dont have to have a gas - since its a molar reference. Therefore you could have a "normal flow" of a multiphase fluid with an average molecular weight of 160 g/mol of say 6.3 Nm3/hr at the given conditions. Not very "normal" but not against convention. Now another thing strikes me as strange. The mass unit kgs/hr what is "kgs" - is that because the real mass flow unit is kg/sec?

However, you are right: chmcl you should be careful to check that your definition of "normal" or "standard" conditions are similar to the one used by the person who assigned the task to you.

cloa: well as david points out n-decane has a boiling point of 174ºC at atm conditions and a MW of 142 - and the temp here is just 88º.

Best regrads Morten

 

[zdas04]

MortenA,
I often see people use "kgs" to mean "KiloGramS". It is nearly always confusing, but people (especially students) do insist on doing it.

Also, I can't remember ever seeing liquid volumes converted to standard conditions. I often see a temperature adjustment in liquid sales contracts that are based on volume, but I don't think I've ever seen one that bothered with a pressure adjustment. That is why I assumed that when he wanted to convert mass flow rate to volume flow rate at "normal" conditions I assumed gas. He's never been back to see if he got an answer so I think we can assume anything we want.

David

 

[Spoonful]

Hi all,

here are some realistic numbers, here I have got compressed natural gas @ operating condition: P:6.895 kpag. T:80 degree C, mole weight: 17.6, Compressibility: 0.0134.

I have converted from 11000 Sm^3/h to 178 m^3/hr using (PV/T)op = (PV/T)std.

I just want to double check my calculation is correct for this. and by using this (PV/T)op = (PV/T)std what assumption do I have to make?

Thanks in advance

 

[katmar]

Two things:

(1) You would probably get more response if you started a new thread rather than tacking on to the end of an old one.

(2) A compressibility factor of 0.0134 is impossibly low. If your natural gas is mostly methane your Z should be very close to 1 under the given conditions. Where did you get this compressibility factor from?

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[zdas04]

Somewhere that gives him the complement of compressibility.

David

 

[Spoonful]

hummm that Z value dosen't looks so right, i might got it wrong, i will check it tmr when I back to the office. Its 10:35pm now in Australia

 

[Spoonful]

I was reading the wrong line Z=0.909, where the0.0134 is the gas viscosity.