3. Design of packed tower for extraction. An inlet water solution of 400kg/h con

Question

3. Design of packed tower for extraction. An inlet water solution of 400kg/h containing 10 wt% acetone is extracted with the pure solvent trichloroethane in a countercurrent packed tower at 25 °C. Water and trichloroethane are essentially immiscible. The exit concentration in the water (raffinate) stream is set at 1.0 wt % acetone. (a) Plot the equilibrium data given in the table in example 12.7-4 and fit the data using a second order polynomial. (b) Using the feed condition, calculate L' (the mass flow rate of the inert water). Calculate L (c) For V-300kg/h (the mass flow rate of the inert trichloroethane), calculate y, (solute concentration in the extract phase at the exit). (Hint: make a mass balance on the solute over the entire tower.) (d) Calculate V. (f) Calculate, and (g) If the height of a transfer unit Ho1m, using the equation (12.7-22), calculate the number of transfer units No and the tower height. (Note that Ho and No are different from Ho ad No derived during the class. The equation we derived was based on the extract phase. A similar equation can be derived based on the raffinate phase, which is the next problem. Either way, you can estimate the tower height-) (h) Using the method introduced in the class, derive (12.7- 22)

Explanation / Answer

(a) The x-y data can be plotted and second order polynomial can be fitter using MS-excel by following steps:

(i) For plot: enter x and y data in two separate columns (e.g. A and B) (ii) Select both the columns and go to insert -->Scatter (chart). This will display a plot of x vs. y.

(ii) For fit: Right click the plot. Select 'Add trendline'. (Format trendline box will pop-up). Select trend type polynomial and order 2.Also tick mark on 'Display equation on chart'. Close. It will result in following:

(b) The total feed rate, F (10% Acetone + 90% water) is 400 kg/h. So the flow rate of inert water in feed (L') is:

L' (kg/h) = Total feed flow rate X wt fraction of water = 400 X (90/100) = 360 kg/h = L'

L1 is referred to what is not clearly mentioned. I am assuming it's the total raffinate (1 wt% acetone in water) flow rate. This can be calculated by making component balance for water. Since water is only in feed and raffinate the following holds good: L' = 99% water X L1

L' = 360 kg/h

L1 = 363.74 kg/h. Please note that theinert water flow rate in raffinate will be 360 kg/h.

(c) The mass flow rate solvent (inert trichloroethane)is given as: V' = 300 kg/h

The overall mass balance on entire tower: Feed in + Solvent in = Extract out + Raffinate out

                                                    F + V' = V2 + L1

So Extract flow rate V2 = 336.26 kg/h (answer for part (d))

To calculate y2making component balance for acetone:

(F x 0.10) + (V' x 0) = (V2 x y2) + ( L1 x 0.01) [ The acetone content in feed = 10/100, in solvent =0, in raffinate = 1/100)

Substitutine F = 400, V' = 300, V2 = 336.26 and L1 =363.74, we can obtain y2

y2 = 0.1081

Solute wt% in extract is 10.81%

Question (f) is not clear. Question (g) and (h) will require more time and can be posted as another problem.

Hope this will help.

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