Answer / hermann

1Nm3 (normal m3) of air is a volume of 1m3 of air under
normal conditions of pressure (1 atm = 101.330 kPa) and
temperature (0 deg C = 273 K) (Europe);

Since air can be assumed to behave as an ideal gas under
those conditions, its compressibility factor is 1; so in
the relation:PV=ZmRT; Z=1;R for air= 0.287 kPa.m3/(kg.K);
V=1m3; therefore 101.330*1=1*m*0.287*273; so m = 1.295 kg.

Conversion factor: 1 Nm3/hr = 1.295 kg/hr

Answer / kamal joshi

1 Nm3/hr = 1.295 kg/hr

Answer / neelakanth pattar

As per ideal gas law
PV = mRT
Data : P-pressure-101.330Kpa at atm
T-273K std
R-0.287 Kpa/Kr.k
V-1 m3

So. 101.330 x 1 = m x 0.287 x 273
m = 1.295 Kg
i.e , 1 Nm3/hr = 1.295 Kg/hr
20 Nm3/hr = 20x1.2929 kg/hr

Answer / 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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