Question

You wish to remove solute Q from a heavy oil stream by stripping the oil with air. The stripper is expected to operate at 60°C and 2.2 atm pressure. At this temperature and pressure the oil can be considered to be nonvolatile and the air is insoluble in the oil. The inlet liquid flows at 100 kgmole/hr (total flow) and is 20 mole% Q. You wish to reduce the concentration of Q in the oil to 1 mole% Q. The inlet air stream is terribly polluted (probably by your plant) and contains 0.2 mole% Q. Equilibrium data taken in your laboratory (which you trust totally) indicate that at 60°C and 2.2 atm y = 1.5x (where y and x are mole fractions of Q in the air and oil, respectively). What is the minimum flowrate of the entering stream of air (including the Q mole fraction) required to achieve the desired cleanup, corresponding to an infinite number of stages? How many ideal stages will be required if the column operates at 1.5 times the minimum gas flowrate?

Answer 1

I am solving this question on solute free basis. So, I will use solute free liquid and solute free gas flow rate instead of normal liquid and gas. The mole fraction will also be on solute free basis.

Inlet liquid flow rate = 100 kmole/hr

Composition of solute Q in liquid = 0.20

Liquid flow rate on solute free basis = 100 × (1 - 0.2)

L_{​​​​​​s}​​​​​ = 80 kmol/hr

Mole fraction of Q in liquid on solute free basis (initial)= 20/80

X_{​​​​​​o} = 0.25

Inlet air stream flow = G_{​​​​​​s}​​​

Mole fraction of Q in gas (initial) = 0.002 / ( 1 - 0.002)

= 0.002

Equilibrium line can be assumed same on solute free basis

Y = 1.5 X

For infinite number of stages will occur when the operating line touches the equilibrium line. It means the exit stream mole fraction will be satisfied by the equilibrium line. The slope of operating line is given by ( L_{​​​​​​s} / G_{​​​​​​s} )

れ ㄥ

( L_{​​​​​​s} / G_{​​​​​​s} )_max slope line will touch the equilibrium line which will give the infinite number of stages

Material balancing :

so answer is 6

0 002) uli O 3 Ls (0.25-0.0101) = kmol me nintmtt 7.179 kmo'/ mo l m s IS X 77-17 9 0.691