A fermentation tank is initially filled with 08 L of broth c

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.

Solution

solution:

1)initial mass in tank

m1=c*V1=100*.8=80 g

2) mass added yrough broth

Q2=.2 L/h

M2\'=C*Q2=20

m2=M2*t=20t

3) mass consume in cell

M3=25

m3=M3*t=25t

4)resultant mass=m=m1+m2-m3=80-5t

mass of glucose is

m=80-5t

5) mass of glucose at t=6 hr

m=80-5*6=30g

6) as concentration equation is

c=m/V=80+20t/.8+.2t=100

hence concentration will not vary if mass absorb in cell is neglected and it is constant to 100 g/L