# Number of plates at Total Reflux

A mixture of benzene and toluene containing 40 mole % of benzene is to be separated to give a product of 90 mole % of benzene at the top and bottom product with not more than 10 mole % of benzene. Using an average value of 2.4 for the volatility of benzene relative to toluene, calculate the number of theoretical plates required at total reflux. Also calculate the minimum reflux ratio, if the feed is liquid and at its boiling point.

Calculations:

The equilibrium curve relationship is given by

y = a*x*/(1 + *x*(a - 1)) = 2.4*x*/(1 + 1.4*x*)

At total reflux, operating line coincides with *y* = *x* line.

The above equation can be written as,

1 + 1.4*x* = 2.4*x*/*y*

1/*x* + 1.4 = 2.4/*y*

1/*x* = 2.4/*y* - 1.4 = (2.4 - 1.4*y*)/*y*

i.e.,

*x* = *y*/(2.4 - 1.4*y*)

We can find analytically, the minimum number of theoretical plates using the above equations as given below:

Starting from the conditions at the top of the distillation column, *y*_{D} = 0.9

For the tray 1:

*y*_{1} = 0.9 and *x*_{1} = 0.9/(2.4 - 1.4 x 0.9) = 0.7895

For the tray 2:

*y*_{2} = 0.7895 and *x*_{2} = 0.7895/(2.4 - 1.4 x 0.7895) = 0.6098

For the tray 3:

*y*_{3} = 0.6098 and *x*_{3} = 0.6098/(2.4 - 1.4 x 0.6098) = 0.3944

For the tray 4:

*y*_{4} = 0.3944 and *x*_{4} = 0.3944/(2.4 - 1.4 x 0.3944) = 0.2134

For the tray 5:

*y*_{5} = 0.2134 and *x*_{5} = 0.2134/(2.4 - 1.4 x 0.2134) = 0.1016

For the tray 6:

*y*_{6} = 0.1016 and *x*_{6} = 0.1016/(2.4 - 1.4 x 0.1016) = 0.045

Since *x*_{W} is only 0.1, we are having fractional tray here. It is estimated as below:

Fractional tray = (0.1016 - 0.1) / (0.1016 - 0.045) = 0.03

Number of theoretical plates including reboiler = 5 + 0.03 = 5.03

The number of theoretical plates at total reflux can also be estimated from Fenske's equation, which is given as,

N_{m} + 1 = ( log (*x*_{D }(1 - *x*_{W}) ) / ( (1 - *x*_{D}) *x*_{W}) ) / log a_{avg}

Where N_{m} + 1 is the minimum number of theoretical stages including the reboiler.

In our problem, *x*_{D} = 0.9; *x*_{W} = 0.1; a_{avg} = 2.4

N_{m} + 1 = 5.02

Calculation of Minimum Reflux ratio:

For the saturated liquid feed, at minimum reflux, 'q' line and enriching operating line cuts at *x* = z_{F} = 0.4 and *y* = 2.4z_{F}/(1 + 1.4z_{F}) =

2.4 x 0.4/(1 + 1.4 x 0.4) = 0.6154

Therefore, the enriching operating line passes through the points (0.9,0.9) and (0.4, 0.6154).

The intersection point of the enriching operating line at the *y* axis (which is equal to *x*_{D}/(R_{m} + 1), say *y*_{I} ) is calculated as follows:

Slope = (0.9 - 0.6154)/(0.9 - 0.4) = 0.2846/0.5 = 0.5692

This is the slope of the line connecting (0.4, 0.6154) and (0, *y*_{I})

(0.6154 - y_{I}) / (0.4 - 0) = 0.5692

*y*_{I} = 0.3877

i.e.,

*x*_{D}/(R_{m} + 1) = 0.3877

0.9/(R_{m} + 1) = 0.3877

R_{m} = 1.321

Minimum reflux ratio = 1.321