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Ponchon-Savarit Method
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1000 kg/hr of a mixture containing 42 mole percent heptane and 58 mole percent ethyl benzene is to be fractionated to adistillate containing 97 mole percent heptane and a residue containing 99 mole percent ethyl benzene using a total condenserand feed at its saturated liquid condition. The enthalpy-concentration data for the heptane-ethyl benzene at 1 atm pressureare as follows:
xheptane0 0.08 0.18 0.25 0.49 0.65 0.79 0.91 1.0
yheptane 0 0.28 0.43 0.51 0.73 0.83 0.90 0.96 1.0
Hl (kJ/kmol) x 10-3 24.3 24.1 23.2 22.8 22.05 21.75 21.7 21.6 21.4
Hv (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:
Minimum reflux ratioa.Minimum number of stages at total refluxb.Number of stages at reflux ratio of 2.5c.Condenser dutyd.Reboiler dutye.
Calculations:
Heptane = C7H16
Ethyl benzene = C6H5C2H5
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:
Ponchon-Savarit Method - Mass Transfer Solved Problems - msubbu http://www.che.iitm.ac.in/~ch04d017/sp/mt/Ponchon-Savarit.htm
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zF = 0.42
HF = 22.30098 x 103 kJ/kmol (from graph)
xD = 0.97
HD = 21.53695 x 103 kJ/kmol (from graph)
xW = 0.01
HW = 24.27584 x 103 kJ/kmol (from graph)
At minimum reflux ratio, the tie-line passing through F determines Q' and Q"
Q' = 98.4391 x 103 kJ/kmol (from graph)
Q" = -34.4537 x 103 kJ/kmol (from graph)
HG1 = 53.70453 x 103 kJ/kmol (from graph)
HL0 = HD = 21.53695 x 103 kJ/kmol
Ponchon-Savarit Method - Mass Transfer Solved Problems - msubbu http://www.che.iitm.ac.in/~ch04d017/sp/mt/Ponchon-Savarit.htm
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Reflux ratio, R = (Q' - HG1) / (HG1 - HL0) = (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 103 kJ/kmol
F = 9.9937 kmol/hr
Material balance equations:
F = D + W
F zF = D xD + W xW
Ponchon-Savarit Method - Mass Transfer Solved Problems - msubbu http://www.che.iitm.ac.in/~ch04d017/sp/mt/Ponchon-Savarit.htm
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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 HF = D Q' + W Q"
Substituting for the known quantities,
9.9937 x 22.30098 x 103 = 4.2681 x 134.1235 x 103 + 5.7256 x Q"
Q" = -61.0562 x 103 kJ/kmol
Q' = HD + QC / D
Q" = HW - QB / W
Substituting for the known quantities in the above equations,
134.1235 x 103 = 21.53695 x 103 + QC / 4.2681
QC = 480.53 x 103 kJ/hr = 133.48 kW
-61.0562 x 103 = 24.27584 x 103 - QB / 5.7256
QB = 488.58 x 103 kJ/hr = 135.72 kW
Condenser duty = QC = 133.48 kW
Reboiler duty = QB = 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).
Ponchon-Savarit Method - Mass Transfer Solved Problems - msubbu http://www.che.iitm.ac.in/~ch04d017/sp/mt/Ponchon-Savarit.htm
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Feed is to be introduced at the 7th plate, counting from the top.
For constructing tie-lines in H-x-y diagram, x-y digram is also used.
Ponchon-Savarit Method - Mass Transfer Solved Problems - msubbu http://www.che.iitm.ac.in/~ch04d017/sp/mt/Ponchon-Savarit.htm
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