CHE425: Problem set #4
1)Run program MCCABE. You need to copy the folder CHE425 into the H: drive or your flash drive. Open the folder CHE425 and double click on DOSBox.exe. When the prompt “C:\>” appears, type MCCABE and press “ENTER” to run the program. Copy and report the score and performance number from the program. Type “e” or “exit” and press “ENTER” to exit the DOSBox program.
2) A 1000 lb-mole per hour mixture of 5% methane, 20% ethane, 25% propane, 30% n-butane, and 20% n-pentane at 47 psia and 90F is to be separated by a flash drum. Write a Matlab program using the data in Table 2 to determine the fraction of the feed vaporized and the compositions of the vapor and liquid streams leaving the flash drum. Compare the results with those from Provision.
Table 2 Equilibrium K values for light hydrocarbon systems
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(1) ln K = A/T2 + BC ln(P) + D/P2
(2) ln K = A/T2 + BC ln(P) + D/P
CompoundABCDForm
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Methane2928608.2445.895159.8465(1)
Ethane687248.27.90694.86649.02654(1)
Propane970688.67.15059.769846.90224(2)
n-Butane12805577.94986.964550(1)
n-Pentane15248917.33129.891430(1)
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Ans: Flash temperature, T(R) = 550.00, V/F = 0.9771
Species: Methane Ethane Propane n-Butane n-Pentane
Feed z = 0.0500 0.2000 0.2500 0.3000 0.2000
x = 0.0011 0.0200 0.0825 0.2992 0.5973
y = 0.0511 0.2042 0.2539 0.3000 0.1907
Ki = 47.3323 10.2064 3.0789 1.0029 0.3192
3) The equilibrium data for the acetone (1)-water (2) system are given at 25oC as follows:
x1 / 0.0194 / 0.0289 / 0.0449 / 0.0556 / 0.0939y1 / 0.5234 / 0.6212 / 0.7168 / 0.7591 / 0.8351
P(mmHg) / 50.1 / 61.8 / 81.3 / 91.9 / 126.1
At 25oC the vapor pressures are: P1sat = 230.05 mmHg and P2sat = 23.76 mmHg. Use these data to calculate the following for the air-water-acetone system.
(a) The vapor mol fraction of acetone in equilibrium with a liquid containing 0.0939 mol fraction acetone at 25oC and a pressure of 2 atm.
(b) The vapor mol fraction of water in equilibrium with the liquid described in part (a).
Ans:a) 0.0693b) 0.0137
4.[1]A distillation column is fed 100 kmol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs pressure. The feed is liquid at the boiling point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52:1. Calculate the kmol/h distillate, kmol/h bottoms, and the number of theoretical trays needed using the McCabe-Thiele method. Equilibrium data for benzene-toluene system at 101.32 kPa are given below, where x and y are mole fraction of benzene.
x / 0 / 0.130 / 0.258 / 0.411 / 0.581 / 0.780 / 1.000y / 0 / 0.261 / 0.456 / 0.632 / 0.777 / 0.900 / 1.000
Ans: D = 50 kmol/h, B = 50 kmol/h
We have about 6 theoretical stages or 5 theoretical trays plus a reboiler.
5.[2]A saturated liquid feed of 200 mol/h at the boiling point containing 42 mol % heptane and 58% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol % heptanes and a bottoms containing 1.1 mol % heptanes. The reflux ratio used is 2.5:1. Calculate the mol/h distillate, mol/h bottoms, the number of theoretical trays needed, and the feed tray number. Equilibrium data for heptane-ethyl benzene system at 101.32 kPa are given below, where x and y are mole fraction of heptane.
x / 0 / 0.080 / 0.250 / 0.485 / 0.790 / 1.000y / 0 / 0.230 / 0.514 / 0.730 / 0.904 / 1.000
Ans: D = 85.3 kmol/h, B = 114.7 kmol/h
We have about 11theoretical stages or 10 theoretical trays plus a reboiler. Feed is on tray 6 from top.
6.[3]A total feed of 200 mol/h at the boiling point containing 42 mol % heptane and 58% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol % heptanes and a bottoms containing 1.1 mol % heptanes. The feed enters the tower is a mixture of saturated liquid and vapor with 40 mol % liquid. The reflux ratio used is 2.5:1. Equilibrium data for heptane-ethyl benzene system at 101.32 kPa are given below, where x and y are mole fraction of heptane.
x / 0 / 0.080 / 0.250 / 0.485 / 0.790 / 1.000y / 0 / 0.230 / 0.514 / 0.730 / 0.904 / 1.000
Calculate the following.
(a)Mol/h distillate, mol/h bottoms.
(b)Minimum reflux ratio.
(c)Minimum theoretical trays at total reflux
(d)Theoretical number of trays required for an operating reflux ratio of 2.5:1.
Ans: D = 85.3 kmol/h, B = 114.7 kmol/h
Rmin = 1.68
6.6 theoretical trays plus reboiler
11.2 theoretical trays plus reboiler
7.[4]A distillation column is fed 100 kmol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs pressure. The feed is saturated vapor at the dew point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52:1. Equilibrium data for benzene-toluene system at 101.32 kPa are given below, where x and y are mole fraction of benzene.
x / 0 / 0.130 / 0.258 / 0.411 / 0.581 / 0.780 / 1.000y / 0 / 0.261 / 0.456 / 0.632 / 0.777 / 0.900 / 1.000
Calculate the following.
(a)Minimum reflux ratio Rm.
(b)Minimum theoretical trays at total reflux
(c)Theoretical number of trays required for an operating reflux ratio of 1.5*Rm.
Ans: Rmin = 1.92506
(b) # of equilibrium stages (including reboiler) = 4.93446
3.94 theoretical trays plus reboiler at total reflux
(c) # of equilibrium stages (including reboiler) = 7.65925
6.7 theoretical trays plus a reboiler
8.[5]An enriching tower is fed 100 kmol/h of a saturated vapor feed containing 40 mol % benzene and 60 mol % toluene at 101.32 kPa abs pressure. Figure 8 depicts an enriching tower where the feed stream is the saturated vapor and the bottoms stream is the saturated liquid. There is no reboiler in an enriching tower. The distillate is to contain 90 mol % benzene and the reflux ratio is 4.0:1. Equilibrium data for benzene-toluene system at 101.32 kPa are given below, where x and y are mole fraction of benzene.
x / 0 / 0.130 / 0.258 / 0.411 / 0.581 / 0.780 / 1.000y / 0 / 0.261 / 0.456 / 0.632 / 0.777 / 0.900 / 1.000
Calculate the kmol/h distillate D and bottoms B and their compositions. Also, calculate the number of theoretical plates required.
Figure 8 Enriching tower with feed F = V and bottoms B = L.
Ans: D = 20 kmol/h, B = 80 kmol/h
# of equilibrium stages = 4.67952
[1] Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 753
[2] Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 754
[3] Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 754
[4] Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 754
[5] Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 754