A mixture of benzene and toluene containing 40 mole % of benzene is to be separated to give a product of 90 mole % benzene at top and a bottom product with not more than 10mole % benzene. Using an average value of 2.4 for relative 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 is at its boiling point
The equilibrium curve relationship is given by y = ax/(1 + x(a - 1)) = 2.4x/(1 + 1.4x)
At total reflux, operating line coincides with y = x line. The above equation can be written as,
1 + 1.4x = 2.4x/y
1/x + 1.4 = 2.4/y
1/x = 2.4/y - 1.4 = (2.4 - 1.4y)/y i.e.,
x = y/(2.4 - 1.4y)
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, yD = 0.9
For the tray 1:
y1 = 0.9 and x1 = 0.9/(2.4 - 1.4 x 0.9) = 0.7895
For the tray 2:
y2 = 0.7895 and x2 = 0.7895/(2.4 - 1.4 x 0.7895) = 0.6098
For the tray 3:
y3 = 0.6098 and x3 = 0.6098/(2.4 - 1.4 x 0.6098) = 0.3944
For the tray 4:
y4 = 0.3944 and x4 = 0.3944/(2.4 - 1.4 x 0.3944) = 0.2134
For the tray 5:
y5 = 0.2134 and x5 = 0.2134/(2.4 - 1.4 x 0.2134) = 0.1016
For the tray 6:
y6 = 0.1016 and x6 = 0.1016/(2.4 - 1.4 x 0.1016) = 0.045
Since xW 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,
Nm + 1 = ( log (xD (1 - xW) ) / ( (1 - xD) xW) ) / log aavg
Where Nm + 1 is the minimum number of theoretical stages including the reboiler.
In our problem, xD = 0.9; xW = 0.1; aavg = 2.4
Nm + 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 = zF = 0.4 and y = 2.4zF/(1 + 1.4zF) = 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 xD/(Rm + 1), say yI ) 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, yI)
(0.6154 - yI) / (0.4 - 0) = 0.5692
yI = 0.3877 i.e.,
xD/(Rm + 1) = 0.3877
0.9/(Rm + 1) = 0.3877
Rm = 1.321
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