An equimolar liquid mixture of n-pentane and n-hexane at 80 degree C and 5 atm i

Question

An equimolar liquid mixture of n-pentane and n-hexane at 80 degree C and 5 atm is fed into a flash evaporation column at a rate of 100 mol/s. When the feed is exposed to the reduced pressure in the column, some amount of the liquid is vaporized. The vapor and liquid phases, which are in equilibrium with each other, are separated and discharged as separate streams at 65 degree C. The liquid product stream contains 41 mole% pentane. Determine the pressure at the product streams, the compositions of the liquid product and the molar flow rates of the liquid and vapor products.

Explanation / Answer

Vapour pressure of Pure n-hexane at 65 C = 672.97 mmHg = 0.885 atm

Vapour pressure of Pure n-pentane at 65 C = 1855.67 mmHg = 2.442 atm

Pressure of n-hexane in the product,

P1 = (1-0.41)*0.885 = 0.522 atm

Pressure of n-pentane in the product stream,

P2 = 0.41*2.442 = 1.001 atm

So, total pressure at the product stream is,

P = P1+P2

= 0.522+1.001 = 1.523 atm

The composition at the product stream contains 0.59 moles of n-hexane at a pressure of 0.522 atm and 0.41 moles of n-pentane at a pressure of 0.1001 atm.

Molar flow rate is given by,

For equimolar solution, molar flow rate was 100 mol/s.

that is, For 0.5 moles, molar fow rate = 100 mol/s

So, for 0.41 moles of n-pentane,

Molar flow rate = 100*0.41/0.5 = 82 mol/s

and for 0.59 moles of n-hexane,

Molar flow rate = 100*0.59/0.5 = 118 mol/s

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