Pro/II Assignment – Ethanol DistillationRafaela Lima Santos de Souza – ID 1497995

The market now requests a product (ethanol + water) with 85% ethanol fraction. The method of achieving this is by distillation of a dilute feed with 10%mol ethanol, 90%mol water, that originates from a fermentation process. The definition of distillation is: unit operation in which, by the application and removal of heat, a liquid or vapour mixture of two or more compounds is separated into its component fractions of desired purity.

The available distillation column delivers 80% ethanol fraction at the tops stream, with feed rate of 100kmol/h . The bottoms draw-off rate is 90kmol/h and reflux ratio of 3, there are 14 trays and feed arrives in the 8th tray. In order to achieve the new specifications, a new column must be optimised based on the existing one.

The most important specification is achieving the 85% ethanol fraction value. Other desirable features included:

increasing the ethanol recovery to 13kmol/h; to not let any fraction of entering the column ethanol to go to the effluent; to make the ethanol concentration in the effluent become smaller than 100

ppm; and to not reduce the heat efficiency from the current value.

The primary condition can be achieved by replacing the following parameters:

Change the reflux ratio to 5.73, the draw-off rate to 92kmol/h, and the feed tray to 13. The others will be discussed later in this text. Products flowrate will be 8kmol/h with 85%mol ethanol.

The existing column looked like:

Figure 1 – Existing column – 80% ethanol products

The column now looks like this:

Figure 1 - Final solution, with feed arriving at tray 13.

Its heat efficiency is 11.24kmol/h(MW), fraction of ethanol wastage is 0.32, the concentration of ethanol in effluent is 34787 ppm and ethanol fraction in the tops product exactly 0.85. The heat duty 0.717MW is and the condenser duty is -0.507MW.

In order to achieve this, the parameters were slowly altered in case studies in PRO/II.

First of all, the tray number was altered maintaining all other parameters the same. The reason one should prefer the feed to arrive at a lower part of the column is that this way, there is more mixing between the rich in ethanol vapours and the feed, providing a higher ethanol fraction in the final distillate.

Table 1: Tray number x Ethanol fractionSTEP 1 2 3 4 5* 6* 7*TRAY_FEED 8 9 10 11 12 13 14.ETOH_FRAC 0.808 0.815 0.821 0.825 0.758 0.931 0.876

Note: steps with an * have not converged.

Then the reflux ratio was raised to 4.0. The reflux ratio means the quantity of distillate that goes back to the column for further distillation over the quantity of tops final product (L/D usually, in literature). Increasing the reflux ratio implies in a purer product, but also in bigger heat and condenser duties.

Table 2: Tray number x Ethanol fraction (Reflux ratio = 4)STEP 1 2 3* 4* 5* 6* 7*TRAY_FEED 8 9 10 11 12 13 14ETOH_FRAC 0.817 0.825 0.831 0.903 0.723 0.880 0.937

Note: steps with an * have not converged.

As the 10th tray and forward did not converge, the bottoms flowrate was changed to 91kmol/h. This was done because with a greater drawoff rate, more of the heavy product (water) is taken off the column. This causes less vapour to go above the column, so the vapour gets more concentrated on the most volatile product (ethanol), and so it is easier for the simulation to converge.

Table 3: Tray number x Ethanol fraction (Reflux ratio = 4, Bottoms flowrate = 91kmol/h)STEP 1 2 3 4 5 6* 7*TRAY_FEED 8 9 10 11 12 13 14ETOH_FRAC 0.817 0.823 0.829 0.834 0.839 0.808 0.882

Note: steps with an * have not converged.

In order to get a further increase in the ethanol fraction, and as a guess for the 13th tray to converge, the bottoms flowrate was raised to 92.

Table 4: Tray number x Ethanol fraction (Reflux ratio = 4, Bottoms flowrate = 92kmol/h)STEP 1 2 3 4 5 6 7*TRAY_ FEED 8 9 10 11 12 13 14ETOH_FRAC 0.817 0.824 0.830 0.833 0.837 0.841 0.857

Note: steps with an * have not converged.

The 13th tray seems to be getting closer to the desired solution now. The reflux ratio was then raised to 5, in order to have a richer in ethanol tops product.

Table 5: Tray number x Ethanol fraction (Reflux ratio = 5, Bottoms flowrate = 92kmol/h)STEP 1 2 3 4 5 6 7*TRAY_ FEED 8 9 10 11 12 13 14ETOH_FRAC 0.820 0.830 0.833 0.840 0.843 0.846 0.875

Note: steps with an * have not converged.

Then another case study was set, now having the reflux ratio as the variable and the 13th tray as the feed tray.

Table 6: Reflux ratio x Ethanol fraction (Feed tray = 13, Bottoms = 92kmol/h) STEP 1 2 3 4 5 6 7 8 9

REF_RAT 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

ETOH_FRAC

0.846 0.846 0.848 0.847 0.849 0.849 0.850 0.849 0.849

It seems that the solution is somewhere between the reflux ratio of 5.5 and 5.8. Using a smaller step:

Table 7: Reflux ratio x Ethanol fraction (Feed tray = 13, Bottoms = 92kmol/h) STEP 1 2 3 4 5 6 7 8 9 10 11REF_RAT 5.60 5.61 5.62 5.63 5.64 5.65 5.66 5.67 5.68 5.69 5.70ETOH_FRAC 0.849 0.849 0.848 0.849 0.850 0.849 0.848 0.849 0.850 0.850 0.849

STEP 12 13 14 15 16 17 18 19 20 21 22 23REF_RAT 5.71 5.72 5.73 5.74 5.75 5.76 5.77 5.78 5.79 5.800 5.81 5.82

ETOH_FRAC 0.849 0.850 0.850 0.849 0.849 0.850 0.850 0.849 0.849 0.850 0.850 0.849 STEP 24 25 26 27 28 29 30

REF_RAT 5.83 5.84 5.85 5.86 5.87 5.88 5.89ETOH_FRAC 0.849 0.850 0.850 0.850 0.850 0.850 0.850

Then some values have got the desired fraction of 0.85 ethanol fraction. However, when approached as single runs, the values below 5.72 with the feed tray 13 and bottoms flowrate of 92kmol/h gave an ethanol tops fraction smaller than 0.849. Running with reflux ratio of 5.73 brings at first the wanted solution of 0.85. Values higher than 5.73 are not interesting because they imply in bigger heat and condenser duties, larger costs for achieving the same purity.

The secondary wish list is not possible to accomplish under the given conditions. The feed comes from a fermentation process upstream, so its composition is fixed. By a simple mass balance, and knowing that a distillation column operates under steady-state regimen, one should see that it’s impossible for 13kmol/h of ethanol to come out of the column if only 10 kmol/h of ethanol are entering.

Mass balance for ethanol:In: 10% x 100 kmol/h = 10 kmol/hOut of the new column: 85% x 8kmol/h + 3.5% x 92kmol/h = 10 kmol/h = maximum ethanol possibly going out, as an steady-state operation (no accumulation).

It is impossible that no ethanol in the feed goes into the effluent. That happens because if a more ethanol-rich product is desired, and altering the reflux ratio only can’t do that, it is imperative to remove water from the tops fraction. But by increasing the drawoff rate to any value above 90kmol/h (or even less, because no perfect separation could occur between ethanol and water, which form azeotrope) one excludes the possibility of all alcohol that goes into the column (10kmol/h) exiting from the tops. This way, there is always going to exist ethanol wastage. Its value has changed from 0.19 to 0.32.

The concentration of ethanol in effluent is directly linked to the ethanol wastage, and it’s not possible to keep it under 100 ppm if it is wanted an 85% ethanol fraction at tops, for the same reasons, therefore it has increased from 21232 ppm to 34787 ppm.

The heat efficiency could not be maintained the same. In order to have a higher fraction ethanol product, the reboiler must have a larger duty because of the larger reflux ratio number (more liquid goes back to the column for further distillation, so more liquid needs to be vaporised). Also, there is less product getting out of this new column. The formula for heat efficiency is total product flowrate / reboiler duty, and the numerator got smaller and the denominator got bigger. This way, the heat efficiency obtained (11.24 kmol/hr(MW)) has to be smaller than that of the starter column (17.74kmol/hr(MW)).

To achieve this kind of optimisation solution, the variable costs must be balanced. There is always going to exist a trade-off between the quality (purity) of the product, the quantity of ethanol obtained and the running costs of the operation. One could have the same fraction of ethanol in the tops but with a higher reflux ratio and a higher drawoff rate at the bottoms, therefore having smaller heat efficiency and bigger ethanol wastage. Achieving just the required fraction of ethanol is what makes it possible to lower the costs while still inside the major specification. The resolution presented in this study is only one of the possible solutions to the market demands, but it is certainly a very efficient one. References:

M.T. Tham(1997). Distillation, an introduction. http://lorien.ncl.ac.uk/ming/distil/distil0.htm - Visited 25/11/2014McCabe, W. and Smith, J. (1976). Unit operations of chemical engineering. New York: McGraw-Hill.

Appendix

Figure 1: Tray number casestudy

Figure 2: Reflux ratio first casestudy

Figure 3: Reflux ratio second casestudy

Figure 4: Tray and flowrate specifications

Figure 5: Performance specifications