Question

It is proposed to absorb acetone from air using water as a solvent. Operation is at 10 atm and is isothermal at 20°C. The total flow rate of entering gas is 10 kmol /h. The entering gas is 1.2 mol% acetone. Pure water is used as the solvent. The water flow rate is 15 kmol/h. The desired outlet gas concentration should be 0.1 mol % acetone. For this system, Henry's law holds and Ye = 1.5 X where Ye is the mol fraction of acetone in the vapour in equilibrium with a mol fraction X in the liquid.
KGa = 0.4 kmol*m^-3*s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.

Answer 1

Answer:

The meole fraction of acetone in the outlet liquid is $x_1 = 0.0072$

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate $G_s = 10(1-0.012) =9.88 kmol/hr$

Let the entering mole be :$y_1 = 1.2$ % = 0.012

$y_1 =(\dfrac{y_1}{1-y_1})$

$y_1 =(\dfrac{0.012}{1-0.012})$

$y_1 =0.012$

Let the outlet gas concentration be $y_2$ = 0.1% = 0.001

$y_2 = 0.001$

Thus; the mole fraction of acetone in the outlet liquid is:

$G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1$

$9.88(0.012-0.001)=15*x_1$

$9.88(0.011) = 15x_1$

$x_1 = \dfrac{0.10868}{15}$

$x_1 = 0.0072$

The mole fraction of acetone in the outlet liquid is $x_1 = 0.0072$