Answer:
Explanation:
From the information given:
Feed F = 150.0 kmol/hr
The saturated liquid mixture of the distillation column is [tex]X_F[/tex] = 30%
Reflux ration = 2.0%
methanol distillate mole fraction [tex]X_D[/tex] = 0.990
recovery of methanol in the distillate = 97.0%
The distillate flow rate D can be determined by using the formula;
[tex]D = 0.97 (F* X_F)[/tex]
D = 0.97 × 150 × 0.3
D = 43.65 kmol/h
The bottom flow rate Balance B on the column is:
F = D + B
150 = 43.65 + B
B = ( 150 - 43.65 )kmol/h
B = 106.35 kmol/h
The methanol mole fraction in the bottom [tex]x_M[/tex] can be computed by using the formula:
[tex]F*X_F = DX_D + BX_B[/tex]
150(0.3) = 43.65(0.999) + 106.3([tex]X_B[/tex])
45 = 43.60635 + 106.3([tex]X_B[/tex])
45 - 43.60635 = 106.3([tex]X_B[/tex])
1.39365 = 106.3([tex]X_B[/tex])
[tex]X_B[/tex] = 1.39365 / 106.3
[tex]X_B[/tex] = 0.013
the fractional recovery of water in the bottoms f is calculated as:
[tex]f = \dfrac{B(1-X_B)}{F_X_F}[/tex]
[tex]f = \dfrac{106.35(1-0.013)}{150\times 0.7}[/tex]
[tex]f = \dfrac{106.35(0.987)}{150\times 0.7}[/tex]
f = 0.99969
