what is the conversion factor for nm3 to kg/hr in air flow
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Answer Posted / rajesh gajera
The ideal gas law demonstrates that: PV=nRT in which n=m/M
n= mole number
m= mass or mass flow (kg/h)
V= volume or volumetric flow (m3/hr)
M= molecular weight of the gas for example Molecular weight of nitrogen (N2) =28 kg/kgmole
In Normal conditions according to the latest SI definition P=100 kpa and T=0 C or 273.15 K
R = gas constant =8.314 pa.m3/mol.K
So, PV=(m/M)*R*T or m(kg/h) = (P*V*M)/(R*T)
at Normal conditions then we have:
m(kg/h) = M* 100* V (Nm3/h) / (8.314*273.15) or
m(kg/h) = 0.044 M* V (Nm3/h)
example: 20 Nm3/h of Nitrogen is equal to:
m = 0.044* 28 (molecular weight) * 20 (Nm3/h)
m = 7 kg/h
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