A fermentation tank is initially filled with 0.8 L of broth containing glucose a
ID: 1057197 • Letter: A
Question
A fermentation tank is initially filled with 0.8 L of broth containing glucose at a concentration of 100g/L, as well as some antibiotic-producing cells. At time t=0, additional broth (containing 100 g/L glucose) is added to the tank at a rate of 0.2 L/h. The cells consume the glucose at a rate of 25g/h. What is the mass of glucose in the tank as a function of fermentation time? What is the mass of glucose in the tank after 6 hours of fermentation? What is the glucose concentration in the tank (g/L) as a function of time? The total liquid volume in the lank can be considered as the sum of the volume of added broth and the volume of the initial tank contents. You may ignore mass losses due to evaporation and cell metabolism.Explanation / Answer
answer:(a)
initially the mass of the glucose in the tank with 0.8 L = 80 gm ( t =0)
rate of addition of broth is 0.2L/h.
after one hour weight of additional glucose would be = 20gm
so, the total mass of the glucose after one hour (t =1) = 100 gm but the cells consume the glucose at a rate of 25g/h.
therefore, amount of glucose left after one hour ( t=1) would be = 75 gm is the function of of fermentation times
total volume broth added is =1.2 L during 6 hours of addition, hence total amount of glucose addded = 120 gm and amount of glucose consumed by the cells during 6 hour = 150 gm and after 6 hour of fermentation the total amount of glucose = 75 + 120 = 195 gm
Therefore, mass of glucose in the tank after 6 hours of fermentation = 195 - 150 = 45 gm
(b) now, the total volume of liquid = 0.8 + 0.2 + 1.2 = 2.2 L and amount of glucose is 45 gm after 7 hours
therefore, glucose concentration would be 20.45 g/L after 7 hours and the this value whould be 2.92 g/L per unit hour time.
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