Question

A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight ethanol and a molar flow rate of a feed is

Answer 1

The answer is
`[(LX_1)/(46) + (L(1-x))/(18)]*0.454` mol/hr

`[(Vy_1)/(46)+(V(1-y))/(18)]*0.454`mol/hr

Step-by-step explanation:

For flash distillation

F = V+L

`(V)/(F) + (L)/(F) = 1`

`(F)/(V) -(L)/(V) = 1`

Fz = Vy+Lx

Y =
`(F)/(V)* Z - (L)/(V)* X` let,
`(V)/(F) = F`

`y = (Z)/(F) -[ (1)/(F) -1]* X`

Highlighted reading

F = 299;
`(V)/(F)` = 0.85 ; z = 0.36

y =
`(0.36)/(0.85) - (-0.15)* X`

= 0.423 + 0.15x ------------(i)

`y^(*)` = -43.99713
`x^(6)` + 148.27274
`x^(5)` - 195.46
`x^(4)`+127.99
`x^(3)`-43.3
`x^(2)`+ 7.469
`x^{}`+ 0.02011

At equilibrium,
`y^(*)` = y

0.423+0.15
`x^{}` =
`y^(*)`

-43.99713
`x^(6)`+ 148.27274
`x^(5)` - 195.46
`x^(4)`+127.99
`x^(3)`-43.3
`x^(2)`+ 7.319
`x^{}`-0.403

F(x) for Newton's Law

Let
`x_(0) = 0`

`x_(1)` =
`\frac{0-[{-0.403}]}{7.319}`

= 0.055

`x_(2)` =
`\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^(') (0.055)}`

=
`\frac{{(0.055)}-(-0.11)}{3.59}`

= 0.085

`x^(3)` =
`\frac{{0.085}-(0.024)}{2.289}`

= 0.095

`x^(4)` =
`\frac{{0.095}-(-0.0353)}{-1.410}`

= 0.07

From This x and y are found from equation (i) and L and V are obtained from
`(V)/(F)` and F values

`[(LX_1)/(46) + (L(1-x))/(18)]*0.454` mol/hr

`[(Vy_1)/(46)+(V(1-y))/(18)]*0.454`mol/hr