Question

A mixture of hydrocarbons is separated with two distillation columns in series. This mixture, containing 20 mass% ethane, 40 mass% propane and 40 mass% butane, is fed to the first column a rate of 100 moles per hour. The overhead product from the first column contains 95 mole% ethane, 4 mole% propane and 1 mole% butane. The bottoms product from the first column is sent to the second column for further separation. The composition of the overhead product from the second column is 99 mole% propane and 1 mole% butane. The composition of the bottoms product from the second column is 8.4 mole% propane and 91.6 mole% butane. Determine: 1. (5 points) Draw a process flow diagram and clearly label your streams. 2. (5 points) Calculate the flow rate of ethane in the overhead product of the first column. 3. (10 points) Calculate the flow rate of the overhead product leaving the second column. 4. (10 points) Determine the composition of the feed to the second column.

Answer 1

1. Process flow diagram:

(2): Let the moles of ethane(C2H6) be 'y' mol/hr and the moles of propane(C3H8) be 'z' mol/hr.

Hence moles of butane(C4H10) = (100 - y- z) mol/hr

Hence mass of ethane(C2H6) = y mol x 30.0 g/mol = 30y g/hr

mass of propane(C3H8) = z mol x 44.0 g/mol = 44z g/hr

mass of butane(C4H10) = (100 - y- z) mol x 58.0 g/mol = (100 - y- z)x58 g/hr

Hence total mass/hr = 30y g/hr + 44z g/hr + (100 - y- z)x58 g/hr = (5800 - 28y - 14z) g/hr

Given mass % of ethane = 20%

=> 30y g /  (5800 - 28y - 14z) g = 0.2

=> 35.6y = 1160 - 2.8z ----- (1)

Given mass % of propane = 40%

=> 44z g /  (5800 - 28y - 14z) g = 0.4

=> 49.6z = 2320 - 11.2y ----- (2)

By solving eqn(1) and (2) we get

y = moles of ethane = 29.54 mol/hr

z = moles of propane = 38.73 mol/hr

Hence moles of butane = 100 - y - z = 31.73 mol/hr

Also from the process flow diagram it is clear that the second column doesn't contain ethane is liberated in the overhead product of the first column.

Hence  flow rate of ethane in the overhead product of the first column = y mol/hr = 29.54 mol/hr (answer)

(4):

Molar flow rate of ethane in the overhead product of first column = 95 % mol/hr

=> 29.54 mol / n1 = 0.95

=> n1 = 31.09 mol/hr

Hence molar flow rate of feed to the second column = 100 - n1 = 68.91 mol/hr

molar flow rate of propane in the overhead product of first column = 4 % mol/hr

= 31.09 x (4/100) mol/hr = 1.244 mol/hr

Hence composition of propane in the feed to the second column = (z - 1.244)

= (38.73 - 1.244) mol/hr = 37.49 mol propane/hr

molar flow rate of butane in the overhead product of first column = 1 % mol/hr

= 31.09 x (1/100) mol/hr = 0.03109 mol/hr

Hence composition of butane in the feed to the second column = (100 - y - z - 0.03109) mol/hr

= (100 - 29.54 - 38.73 - 1.244) mol/hr = 31.70 mol butane/hr

Composition of the feed to the second column in terms of mole percent is

composition of propane = [37.49 / (37.49 + 31.70)]x100 = 54 mol% propane (answer)

composition of butane = 100 - 54 = 46 mol% butane (answer)

(3): By material balane of propane and butane of the second column

0.99n2 + 0.084n3 = 37.49 mol ----- (3)

0.01n2 = 0.916n2 = 31.70 mol -----(4)

By solving eqn(3) and (4) we get n2 = 34.95 and n3 = 34.44 mol

Hence  flow rate of the overhead product leaving the second column = n2 = 34.95 mol/hr (answer)