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Exercise TKP 4105 (Distillation)
Problem 1 (Distillation column for heptane-ethyl benzene) Design 1: Do Problem 11.4-2 from the book (Geankopolis) and in addition: (i) De- rive a general equation for Vbtm as a function of F , R = L/D and q for the case of constant molar flows. (ii) Find the energy consumption Q (heat to the reboiler) given that the heat of vaporization at the normal boiling point (1 atm) is 31.8 kJ/mol for heptane and 35.6 kJ/mol for ethyl benzene. (iii) Compute the minimum number of stages and the minimum reflux ratio and boilup (Rmin, Vmin). (iv) Design 2: To save energy con- sider another design where the reflux ratio is reduced to R = 1. 2 Rmin. Show that this corresponds approximately to V = 1. 1 Vmin in this case. How many stages are required and what is the energy consumption in this case? Compare and discuss the two designs. Note: There are several ways to ”start” the staircase when drawing the McCabe-Thiele diagram. In this case, where both product compositions given and you are to design the column, you are recommended to start at the top and go down to where the operating lines cross and then start from the bottom and go up to where the operating lines cross. You will then get a non-integer number of stages in each section.
Problem 2 (Enriching column for benzene-toluene) Do Problem 11.4- from the book. In addition: What is the distillate composition if the feed composition drops to xF = 0.30 (benzene) with the same reflux ratio and same number of stages? Comment: Note that a batch distillation column (like in the lab) behaves like an enriching column if we assume small holdups so that we have steady-state in the column part. The ”feed” is the vapor that leaves the boiler and the bottoms is returned to the boiler – but note that in a batch column the fraction of light component in the ”feed” drops as time goes because we remove light component in the distillate (top product).
fig.
TKP 4105 Distillation (solution)
Problem 1 (Problem 11.4-2). Heptane-ethyl benzene distillation
Overall balance F = D + B
Component balance xF F = xDD + xB B
Insert total balance into component balance:
xF F = xDD + XB (F − D)
- 42 · 200 = 0. 97 · D + 0. 011 · (200 − D)
Solving for D gives D = 85.297mol/h B = 114.703mol/h
(^00) 0.2 0.4 0.6 0.8 1
1 McCabe−Thiele^ diagram: Heptane^ −^ Ethylbenzene^ (760 mmHg)
Liquid mole fraction Heptane
Ntop = 5.
Nbottom = 4.
XD = 0. XF = 0. XW = 0. Rmin = 1. R = 2.31 * Rmin q = 1
Figure 1: Design 1 with R = L/D = 2.
Next draw the McCabe-Thiele diagram:
- Use the experimental data to draw the equilibrium line.
- From a material balance over the top over the column, assuming constant molar flows, the upper operating line is
y =
L
Vtop
x +
DxD Vtop
L
L + D
x +
DxD L + D
R
R + 1
x +
xD R + 1
The energy that goes to the reboiler is then (we use the heat of vaporization for ethyl benzene since the bottom product is almost pure ethyl benzene)
Q = Vbtm∆Hvap^ = 298.539mol/h · 35 .6kJ/h = 10628kJ/h = 2.95kW
Comment: This is a very small heat duty so this must be a lab column.... An industrial column would typically have numbers that are about 1000 times larger (so the feed would be 200 kmol/h rather than 200 mol/h).
(^00) 0.2 0.4 0.6 0.8 1
1
Rectifier: Heptane − Ethylbenzene (760 mmHg)
Liquid mole fraction Heptane
Vapor mole fraction Heptane
Ntop = 3.
Nbottom = 3.
XD = 0. XF = 0. XW = 0. Rmin = 1. R = 9999 * Rmin q = 1
Figure 2: Minimum number of stages
(iii) The minimum number of stages is found by making a staircase with the diagonal (y = x) as the operating line (corresponding to infinite reflux). We find Nmin = 3.81 + 3.12 = 6.93 (including the reboiler). For this ideal mixture, the minimum reflux is when the operating lines cross on the equilibrium line, that is, at the crossing (x′, y′) between the feed line (q-line) and the equilibrium line. We then get an infinite number of stages on both sides of the feed. We find ( L V
min
xD − y′ xD − x′^
where y′^ = 0.685 is in equilibrium with x′^ = 0.42. Since (L/V ) = L/(L + F ) = R/(R + 1) we get R = (L/V )/(1 − (L/V )) so
Rmin = 0. 518 /(1 − 0 .518) = 1. 08
The corresponding minimum boilup (energy) is: Vmin = 2. 08 · 85 .297 = 177.42[mol/h].
(iv) Design 2: We consider the case with R = 1. 2 · 1 .08 = 1.296. We find V = Vbtm = (1 + 1.296) · 85 .297 = 195.8[mol/h] and we have V /Vmin =
- 8 / 177 .42 = 1.10. The reboiler duty is
Q = Vbtm∆Hvap^ = 6970kJ/h = 1. 94 kW
From the McCage-Thiele diagram the required number of stages is Ntop = 9.15 and Nbottom = 7.30 (including the reboiler). Thus, the total number of theoretical stages (including the reboiler) is
N = 9.15 + 7.30 = 16. 45
In summary, we have for the two cases
R/Rmin V /Vmin = Q/Qmin N/Nmin Design 1 2. 31 1. 68 1. 48 Design 2 1. 20 1. 10 2. 37
Thus, by increasing the number of theoretical stages by 60% (from 10.25 in design 1 to 16.45 in design 2) we have reduced the energy usage by 65% (from 10628 in design 1 to 6979 kJ/h in design 2). Note that for design 2 the energy usage is only 10% higher than the minimum (with an infinite number of stages). Which design is the best? This depends on a trade-off between capital costs (the column) and operating costs (energy). In design 2 the column is about 50% taller, but note that the vapor flow is smaller so we may reduce the column diameter and also the size of the heat exchangers in the top (condenser) and bottom (reboiler)!! So, capital costs are probably similar!! Thus, most likely design 2 is the better one (it has N = 2. 37 Nmin which is in the range 2- recommended by Sigurd!).
(^00) 0.2 0.4 0.6 0.8 1
1 McCabe−Thiele^ diagram: Benzene −^ Toluene (760 mmHg)
Liquid mole fraction Benzene
Ntop = 4.
XD = 0. XF = 0.
Rmin = 2. R = 1.46 * Rmin q = 0
Figure 4: Enriching tower for benzene-toluene: Determining number of stages
Change in feed composition. The feed concentration changes from xF = 0.4 to to xF = 0.3. (Note that it is not possible to maintain the top composition with this feed composition as the feed line (q-line) happens to cross the equilibrium line at this composition, so an infinite number of stages would be required.) The column has N = 4.68 stages, which means the concentrations in the distillate and bottom streams must change. What you need to do is to adjust the operating line (same slope as before because R remains at 4) until you are able to fit in 4.68 stages from the crossing of the operating line with the feed line (which is horizontal at y = 0.3) to the crossing with the diagonal (y = x) at the top of the column. Trial and error gives xD = 0.785 and xW = 0.18. The mass balances then gives D = 19.355 and W = 80.645 kmol/h. Comment: As expected, the top product gets less pure (drops from 0.9 to 0.785). This is similar to what happens in a batch distillation column as the light component is depleted from the boiler.
1 McCabe−Thiele diagram: Benzene − Toluene (760 mmHg)
Figure 5: Enriching tower for benzene-toluene: Change in feed composition
- 00 0.2 0.4 0.6 0.8
- Liquid mole fraction Benzene
- Ntop = 4.
- XD = 0.
- XF = 0.
- Rmin = 3.
- q = R = 1.212 * Rmin
