How is the volume fraction of a gas mixture converted to a mass ratio?
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C2: 0.1%, Propane: 14.2%, Propylene: 46.8%, C4: 38.8%, C5: 0.1%. How do I convert the volume percentages of this mixture into mass percentages?
The volume percentage of each component is multiplied by its respective molecular weight to obtain its mass, and the masses of all components are added together to get the total mass. The mass of each component, compared to the total mass, is its respective mass percentage. You don’t have detailed carbon-hydrogen ratios for C2, C4, and C5; you can assume one, such as treating them all as alkanes, or……
I calculated it in the same way as well: if 100 t is fed into the gas-phase unit, and the mass fraction of propylene is 20%, meaning 20 t of feed, then the actual product yield should be 20 t divided by that amount. So why is it still less than 80%?
Project C2: Propane, Propylene, C4, C5 – Total; v%: 0.1, 14.2, 46.8, 38.8; 0.1, 100. Molecular weights: 30, 44, 42, 58, 72. Masses: 0.03, 6.248, 19.656, 22.504, 0.072; total mass: 48.51. m%|: 0.06, 12.88, 40.52, 46.39, 0.15
I did the same, but it’s not because the deviation calculated in that way is too large. Let me explain: assuming the gas separation unit operates at a rate of 100 t/h, with propylene accounting for 40% of the mass, that means 40 t/h of propylene enters the system. The amount of propylene that is output as a product from the unit is Xt/h. Why is X/40 actually less than 80%, meaning that less than 80% of the propylene is separated and included in the product? This pattern has been observed in the data collected over several months; I’m not sure why Is the analyzed data not representative? Is the deviation still large between this calculation and the actual value? What do you, fellow sailors, think caused this?
Please analyze the propylene content in other products such as ethane and propane; it is possible that the recovery rate of propylene is low, and errors in the measurement data and analysis data also need to be taken into account.
The original poster must be making a mistake by using the method of simply multiplying by the molecular weight. Because C4 and C5 are not just one type of compound – for example, C5 includes n-pentane, isobutane, and olefins – such a calculation is certainly not accurate. The best approach is to conduct sample analysis before making calculations.
The volume fraction ratio is the same as the molar ratio; by multiplying the molar fraction by the molecular weight of each component, the corresponding mass value is obtained. The remaining step is to add these values together and calculate the percentage