1000 kg/hr of a mixture containing 42 mole percent heptane and 58 mole percent ethyl benzene is to be fractionated to a distillate containing 97 mole percent heptane and a residue containing 99 mole percent ethyl benzene using a total condenser and feed at its saturated liquid condition. The enthalpy-concentration data for the heptane-ethyl benzene at 1 atm pressure are as follows:
x_{heptane} |
0 |
0.08 |
0.18 |
0.25 |
0.49 |
0.65 |
0.79 |
0.91 |
1.0 |
y_{heptane} |
0 |
0.28 |
0.43 |
0.51 |
0.73 |
0.83 |
0.90 |
0.96 |
1.0 |
H_{l} (kJ/kmol) x 10^{-3} |
24.3 |
24.1 |
23.2 |
22.8 |
22.05 |
21.75 |
21.7 |
21.6 |
21.4 |
H_{v} (kJ/kmol) x 10^{-3} |
61.2 |
59.6 |
58.5 |
58.1 |
56.5 |
55.2 |
54.4 |
53.8 |
53.3 |
Calculate the following:
Calculations:
Heptane = C_{7}H_{16}
Ethyl benzene = C_{6}H_{5}C_{2}H_{5}
Average molecular weight of feed solution
= 0.42 x 100 + 0.58 x 106 = 103.48
Molal flow rate of feed, F = 1000/103.48 = 9.9937 kmol/hr
The H-x-y diagram is constructed with the above data as given below:
z_{F} = 0.42
H_{F} = 22.30098 x 10^{3 }kJ/kmol (from graph)
x_{D} = 0.97
H_{D} = 21.53695 x 10^{3} kJ/kmol (from graph)
x_{W} = 0.01
H_{W} = 24.27584 x 10^{3} kJ/kmol (from graph)
At minimum reflux ratio, the tie-line passing through F determines Q' and Q"
Q' = 98.4391 x 10^{3} kJ/kmol (from graph)
Q" = -34.4537 x 10^{3} kJ/kmol (from graph)
H_{G1} = 53.70453 x 10^{3} kJ/kmol (from graph)
H_{L0} = H_{D} = 21.53695 x 10^{3} kJ/kmol
Reflux ratio, R = (Q' - H_{G1}) / (H_{G1} - H_{L0}) = (98.4391 - 53.70453) / (53.70453 - 21.53695)
= 1.3907
Minimum Reflux ratio = 1.3907
With the x-y data, the following graph is drawn:
Minimum number of stages at total reflux is found from the x-y diagram and = 6.97
Number of stages at reflux ratio of 2.5:
(Q' - 53.70453) / (53.70453 - 21.53695)= 2.5
Q' = 134.1235 x 10^{3} kJ/kmol
F = 9.9937 kmol/hr
Material balance equations:
F = D + W
F z_{F} = D x_{D} + W x_{W}
i.e.
9.9937 = D + W
9.9937 x 0.42 = 0.97 D + 0.01 W
Solving,
0.96 D = 4.0974
D = 4.2681 kmol/hr
W = 9.9937 - 4.2681 = 5.7256 kmol/hr
Energy balance equation:
F H_{F} = D Q' + W Q"
Substituting for the known quantities,
9.9937 x 22.30098 x 10^{3 }= 4.2681 x 134.1235 x 10^{3} + 5.7256 x Q"
Q" = -61.0562 x 10^{3} kJ/kmol
Q' = H_{D} + Q_{C} / D
Q" = H_{W} - Q_{B} / W
Substituting for the known quantities in the above equations,
134.1235 x 10^{3} = 21.53695 x 10^{3} + Q_{C} / 4.2681
Q_{C} = 480.53 x 10^{3} kJ/hr = 133.48 kW
-61.0562 x 10^{3} = 24.27584 x 10^{3} - Q_{B} / 5.7256
Q_{B} = 488.58 x 10^{3} kJ/hr = 135.72 kW
Condenser duty = Q_{C} = 133.48 kW
Reboiler duty = Q_{B} = 135.72 kW
Number of stages is estimated from Ponchon-Savarit method as shown in the graph, and is equal to 11 (including the reboiler).
Feed is to be introduced at the 7^{th} plate, counting from the top.
For constructing tie-lines in H-x-y diagram, x-y digram is also used.
Last Modified on: 04-Feb-2022
Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in