Question

If a 100 kg mol mixture of n-hexane and n-octane containing 50 mole% n-hexane is to be separated by flash distillation @ 1 atm to obtain 40 kg mol in distillated (ie. V=40 kg mol) a) Whatare the compositions of the liquid and vapor phases when V = 40 kg mol? b) What is the flash drum temperature?

Answer 1

(a)
`x_1=0.34\\x_2=0.66\\y_1=0.74\\y_2=0.24`

(b)
`95.24^0C`

Step-by-step explanation:

Hello,

The Rachford-Rice equation should be used in this case:

`(z_1(K_1-1))/(1+V/F(K_1-1)) +(z_2(K_2-1))/(1+V/F(K_2-1)) =0\\(0.5((P_(1,sat))/(P)-1))/(1+V/F((P_(1,sat))/(P)-1)) +(z_2((P_(2,sat))/(P)-1))/(1+V/F((P_(2,sat))/(P)-1)) =0`

The distribution coefficients could be approximated via Raoult's law and the Antoine equation for the vapor pressure:

`K_i=(P_(i,sat))/(P)=\frac{10^{A-(B)/(C+T) }}{P}`

For n-hexane: A=6,9895, B=1216,92 and C=227,451

For n-octane: A=7,14462, B=1498,96 and C=225,874

*Those values and the Antoine law are used in °C and mmHg.

Replacing each value in the Rachford-Rice equation and solving for T, one gets the flash drum temperature as 95.24°C (attached picture), thus, the compositions in both the liquid and the vapor phases are:

`x_1=(z_1)/(1+V/F(K_1-1))=(0.5)/(1+0.4*(2.175-1)) =0.34\\x_2=1-x_1=0.66\\y1=K_1*x_1=2.175*0.34=0.74\\y_2=1-y_1=0.24`

Best regards.