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 kmolm^-3s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-06-05T05:24:43+02:00

Answer:

The meole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]

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 [tex]G_s = 10(1-0.012) =9.88 kmol/hr[/tex]

Let the entering mole be :[tex]y_1 = 1.2[/tex] % = 0.012

[tex]y_1 =(\dfrac{y_1}{1-y_1})[/tex]

[tex]y_1 =(\dfrac{0.012}{1-0.012})[/tex]

[tex]y_1 =0.012[/tex]

Let the outlet gas concentration be [tex]y_2[/tex] = 0.1% = 0.001

[tex]y_2 = 0.001[/tex]

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

[tex]G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1[/tex]

[tex]9.88(0.012-0.001)=15*x_1[/tex]

[tex]9.88(0.011) = 15x_1[/tex]

[tex]x_1 = \dfrac{0.10868}{15}[/tex]

[tex]x_1 = 0.0072[/tex]

The mole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]