Regarding the conversion of units

This post was last edited by ylb913 on 2017-2-24 07:29. There is an issue with the relationship between density and pressure in the table; the data provides a gauge pressure of 0.6 MPa (G represents gauge pressure, while a represents absolute pressure). The calculated density is 7.85, but if 0.6 is considered as gauge pressure, the calculated density becomes 9.16. PV = nRT; therefore, V = nRT/P (pressure must be in absolute terms). n = W/M, so W = nM (where W is mass and M is the molecular weight). ρ = W/V, so ρ = nMP/nRT = MP/RT. Another way to calculate it: since M/22.4 = ρ0 (the basic unit for molecular weight is g/mol; 1 mole corresponds to 22.4 liters, and dividing gives grams per liter, which can be converted to kg/Nm3), with P0 = 1 and T0 = 273, we have ρ/ρ0 = P*T0/T. Thus, ρ = ρ0 * P*T0/T = (M/22.4)*P*273/T = M*P*273/(22.4*T). The density under standard conditions is 32/22.4 = 1.428. At 25 degrees Celsius and a pressure of 6 kg of absolute pressure, the density is 1.428*6*273/298 = 7.85. At 7 kg of absolute pressure, it is 1.428*7*273/298 = 9.16. Of course, there is also the issue of converting between megapascals and standard atmospheres, but this has been ignored here. By multiplying the square of the value by the density given for that value: 2438.3*1.428=3482; this figure is roughly equal to the mass listed in the table. It’s difficult to achieve exact equality due to the precision of the data used.

A G after pressure indicates gauge pressure, while an A indicates absolute pressure. Gauge pressure = Absolute pressure – Atmospheric pressure

It’s very simple to perform calculations with these values: kmol/h * 32 = kg/h; kmol/h * 22.4 = Nm3/h. The density is a bit more difficult to calculate, but the method is the same as that used for spiders