Question

The hormone estrogen is produced in the ovaries of women and elsewhere in men and post-menopausel women, and is also given in hormone replacement therapy, a common treatment for women who have had a hysterectomy. unfortunately, it also binds to estrogen receptor in breast tissue and can activate cells to become cancerous. tamifoxen is a drug that aso binds to estrogen receptors but does not activate cells, in effect blocking the receptors from access to estrogen and inhibiting the growth of breast cancer cells.

tamifoxifen is given in tablet form. in the manu. process, fine ground powder contains tamoxifen (tam) abd 2 inactive filers- lactose monohydrate (lac) and corn starch (cs). the powder is mixed with a 2nd stream with water and suspended polyvinylpyrrolidone ((pvp) binders, to keep the tablets from crumbling. the slurry leaving the mixer goes to a dryer where 94.2% of the water is vaporized. the wet powder leaves the dryer containing 8.8%wt tam, 66.8% lac, 21.4%cs, 2% pvp, and 1%water. after additional processing the powder forms into tablets. 17.13kg of wet powder is required for the production.

A) taking a basis of 100,000 tablets produced, draw and label a process flowchart, labeling masses of individual component rather than total masses and component mass fractions.it is unnecessary to label the stream between the mixer and the dryer. carry out a degree of freedom analysis of the overall 2 unit process

B) calculate the masses and compositions of the streams that must enter the mixer to make 100,000 tablets.

C) why was it necessary to label the stream between the mixer and the dryer? under what circumstances would it have been necessary?

D) go back to the flowchart of part A. without using the mass of the wet powder (17.13 kg) or any of the results of part B in your calculations, determine the mass fractions of the stream components of the powder fed to the mixer and verify that they match your solution to part B. (take basis of 100 kg of wet powder)

E) suppose a student does part D before doing part B and relabel the powder feed to the mixer on the flowchart of part A with an unknown total mass m1 and the three now known mole fractions. (sketch the resulting flowchart) the student then does a degree of freedom analysis, counts 4 unknowns (masses of the powder, pvp, and water fed into mixer and mass of water evaporated in the dryer), and 6 equations (5 material balances for the 5 species and the percentage evaporation), for a net of -2 degrees of freedom. since there are more equations that unknowns, it should not be possible to get a unique solution for the four unknown. nevertheless, the student writes 4 equations, solves the 4 unknowns and verifies that all of balance equations are satisfied. there must have been a mistake in the degree of freedom calculation, what was it?

Answer 1

There are four unknowns. They are

1. Tam, lac, CS, PV and water

1. Mass percent of tam in the wet powder
2. Amount of lac in the wet powder
3. Amount og Cs in the wet powder
4. Amount of PV in the wet powder.

Since the mass fractionsof water in the wet powder= 1-( sum of mass fractions of all the other four)

There is no change in the amount of raw material except water, this amount remains the same before and after dryer .This amount is 5.8% of totalraw material entering the dryer and hence water presentbefore the dryer can be calculated.

1. Weight of wet powder leaving the dryer= 17.13 kg

It contains 8.8 wt% tam, amount of tam= 17.13*8.8/100=1.5074

Amount of lac is 66.8 %, amount of lac= 17.13*66.8/100=11.44284kg

Amount of Cs is 21.4%, amount of CS= 17.13*21.4/100=3.66582

Amount of PV is 2%, amount of PV= 17.13*2/100= 0.3426kg

Total of tam+lac+CS+PV= 16.9587 kg

Amount of water= 17.13*1/100= 0.1713 kg

Let x= amount entering the dryer, 94.2% water is removed and rest is sent for further processing into tables= 5.8% of this is leaving the dryer

Therefore x*5.8/100=16.9587. x =16.9587*100/5.8=292.3914 kg

Percentages by mass ; tqam= 100*1.5074/ 16.9587 = 8.89 % tam = 100*11.4424/16.9587=67.474%

CS= 100*3.66582/16.9587=21.62%

PV= 0.3426 kg water= 0.1713+292.3914*0.94=0.1713+275.4327   =275.6057kg

Total= 0.3426+275.6057=275.9483 kg

Mass percentages : PV= 100*0.3426/275.9483 =0.124 and water= 100-0.124= 99.876

If the amount of water needs to be known then only the mixer and dryer needs to be separated.

1. Whether one takes one kg or 100 kg, the mass fractions does not vary and only the masses of substances will vary depending upon the amount of product produced

For 100 kg of wet powder,

Tam= 8.8 kg, lac= 66.8 kg, CS= 21.4 kgPV =2 kg and water= 1 kg

Since water is the only component that is undergoing change, rest of the substances can be considered as inert .

5.8*x= 99, x= 99*100/5.8=1706.897 kg

It contains 94.2% water= 1706.897 kg

Masses : tam= 8.8 kg,lac= 66.8CS= 21.4 , total = 8.8+66.8+21.4= 99 kg

PV= 2 kgand water =1+0.942*1706.897=1608.897 kg

Mass Percentages

Tam= 100*8.8/99=8.89, CS= 100*21.4/99=21.62lac= 100*66.8/99=67.47

                      Total of PV + water= 1608.897+2= 1610.897

                  Mass % water = 100*2/1610.897=0.124 and water =100-0.124=99.876