Question

It is desired that the concentration of i-pentane in the feed to a natural gasoline plant should not be more than 10% by mole. This is prepared by passing a fraction of a hydrocarbon-feed stream made up of n-pentane and i-pentane in the mole ratio 4 : 1 through an i-pentane recovery unit as a sidestream. The stream leaving this tower is free of i-pentane. It is mixed with the fraction of the original mixture that is not passed through the recovery tower to get the feed stock to the natural gasoline plant. What percent of the original hydrocarbon mixture should be treated for removing the isopentane? (Answer: 55.56%)

Answer 1

Solution The process can be described using the below Flow dicgrom amor i-pentane Hydrocarbon ferd Stream Ihr si Recovery Lupendone = 10 n pentane = 80% i pentane - 204 Feed = 100 mol/hr (100-x) mol/ho Basis of calculction = 100 mol/hr of Hydrocarbon feed stream Amount of fresh feed sent to i-pentana recovery = x mollhr i Amount of fresh feed by passed = (100-x)mol/hr. :. Am Performing a component balance for n-pentane : Amount of a pentane in the fred = 0.8x100 = gomols/hr Amount of n-pentane entering and leaving the recovery unit- (0.8.9) mollhv. Amount of n-pentane in the bypassed stream = 0.8(100-x) moll' ? Given that the combined stream as 10% iso-pentane and rest; ve 901 n-pentare. Therefore we can write: 10.8x) + 0.8(100-x) = 0.9 (0.81 +100-2) → 80 - 072* 90 - 0.18% =) 0:181 = 10 = x = 55 56 mola/hvi

Since fresh fresh feed of 0 55:56 feed flowrate is hydrocarbon to be 100 mols / hu; amount of sent to Recovery unit = 55 56 moloche sent to Recovery unit = 55.56% fresh feed Ans Percentage of