Question

A gas stream contains 15 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole percent in the gas stream leaving the condenser is 4. Liquid hexane condensate is recovered at a rate of 1.0 L/min. You may take the specific gravity of liquid hexane to be 0.6548.

What is the flow rate of the gas stream leaving the condenser in mol/min? mol/min What percentage of hexane entering the condenser is removed as a liquid? %

Answer 1

we know that for the condenser,

using the overall balance,

F = L + V

L = 1L/min = 0.001 m3/min = 0.001*0.6548*1000 kg/min = 0.6548 kg/min

therefore V = F - 0.6548 ............1

writing the material balance over the nitrogen we get,

F.xf = V.xv

where xf = mole fraction of nitrogen in feed = 0.75

xv = mole fraction of nitrogen in gas = 0.96

hence the equation becomes

F*0.75 = V* 0.96 ..........2

solving equation 1 & 2 we get ,

V = flow rate of gas stream = 2.338 kg/min = 0.0327 mol/min

% of hexane leaving as liquid = 0.6548 / 2.992 * 100 =21.88%