Question

Caffeine is often added to headache medications because it affects the flow of blood in vessels and may enhance the pain relieving qualities of the medication. You have been tasked with evaluating one process that extracts caffeine from a coffee slurry, and you decide to study a small scale version of the process. A coffee slurry containing a mass fraction of 0.01 g caffeine/g, 0.08 g other solids/g, and 0.91 g water/g is fed to a mixer at a rate of 515.0 g/min. Na2CO3 is added to the mixer such that the mass ratio of Na2CO3 to coffee slurry is 0.0300. This aqueous solution proceeds to an extractor where CH2Cl2 is added such that the mass ratio of CH2Cl2 to coffee slurry is 0.380. Two streams leave the extractor. The first is a mostly aqueous stream that leaves the process containing some caffeine, all of the other solids in the original slurry, all of the Na2CO3, all of the water and 2% of the mass of CH2Cl2 entering the extractor. The second stream contains the remaining caffeine and CH2Cl2, and enters an evaporator that completely vaporizes the CH2Cl2. The vaporized CH2Cl2 is condensed and is merged with the fresh CH2Cl2 on the way to the extractor. The dried caffeine crystals exit the evaporator and leave the process. Of the caffeine entering the process a fraction of 0.970 leaves the process in this product stream. Label the process flow diagram below. Label any quantities as "unknown" if you have to perform a calculation to obtain the numeric value. DO NOT LEAVE ANY ANSWER SPACE BLANK.

Answer 1

Mass flow diagram for extractor

Mass of caffeine entering the extractor = (0.01 g caffeine / g) x 515 g = 5.15 g

Mass of other solid entering the extractor = (0.08 g other solid / g) x 515 g = 41.2 g

Mass of water entering the extractor = (0.91 g H2O / g) x 515 g = 468.65 g

Mass of Na2CO3, m2 entering the extractor = 15.45 g / min

m4 = 195.7 g/min

m7 = 5.15 g caffeine x 0.97 = 4.9955 g caffeine / min

2 % of CH2Cl2 removes through aqueous layer

Hence or the first minute, the mass of CH2Cl2 that will remove through the aqueous layer

= (2/100)*195.7 = 3.914 g

and the rest of CH2Cl2 = 195.7 - 3.914 =191.786 will be recycled

But for the rest of the minutes, 2% of 195.7 + 191.786 = 387.486 g will be recycled

Hence 2% of 387.486 g = 7.750 g CH2Cl2

Hence m6 = 98 % of 387.486 g = 379.736 g /min = flow rate of recycled CH2Cl2

m5 = m4+m6 = 195.7 g + 379.736g = 575.44 g/ min

Ratio of flow rate of recycled to the fresh CH2Cl2 = m5/m4 = 575.44g / 195.7g = 2.94