the 7.2. The growth of baker\'s yeast (S. cerevisiae) on glucose may be simply d
ID: 700754 • Letter: T
Question
the 7.2. The growth of baker's yeast (S. cerevisiae) on glucose may be simply described by following equation CH,,06+3 02 + 0.48 NH,-> 0.48 C6H,,No, + 4.32 H,O+ 3.12 CO2 yeast In a batch reactor of volume 10 1, the final desired yeast concentration is 50 gdw/1. a. Determine the concentration and total amount of glucose and (NH4),SO4 in the nutrient medium. b. Determine the yield coefficients YxIs (biomass/glucose) and Yro (biomass/oxygen). c. Determine the total amount of oxygen required. d. If the rate of growth at exponential phase is r, 0.7 gdw/l-h, determine the rate of oxygen consumption (g O,/1-h) e. Calculate the heat-removal requirements for the reactor (recall equation 6.26).Explanation / Answer
Solution:-
C6H12O6 + 3O2 + 0.48 NH3 ------> 0.48 C6H10NO3 + 4.32H2O + 3.12CO2
2NH3 + H2SO4 -----> (NH4)2SO4
Molecular weight of yeast (C6H10NO3) = 144 g/mole
Molecular weight of glucose (C6H12O6) = 180 g/mole
Molecular weight of Ammonium Sulfate (NH4)2SO4 = 132.14 g/mole
Batch reactor volume = 100000 liters
Final east concentration = 50 g/l
a) Determine the concentration and total amount of glucose and ammonium sulfate in the nutrient medium?
Therfore total yeast (C6H10NO3) = 100000*50 = 5000000 gram
Mole of yeast = (5000000 / 144) = 34722.22 mole of yeast
0.48 mole of yeast required = 1mole of glucose
34722.22 mole of east required = 34722.22/0.48 = 72337.963 mole of glucose
Total glucose = 72337.963 * 180 (MW of glucose) = 13020833.3 gram
0.48 mole of yeat require = 0.48 mol3 of Ammonia (NH3)
34722.22 mole of yeast required = 34722.22 mole of NH3
2 mole of NH3 = 1 mole of (NH4)2SO4
34722.22 mole of NH3 = (34722.22 / 2) = 17361.11 mole of (NH4)2SO4
Total (NH4)2SO4 = 17361.11 * 132.14 (MW of (NH4)2SO4) = 2294097.22 g
Concentration of glucose = 13020833.3 / 100000 = 130.208 g/l
Concentration of (NH4)2SO4 = 2294097.22 / 100000 = 22.940 g/l
b) Determine the yield coefficients Yx/s (biomass/glucose) and YX/O (biomass/oxygen)
Yx/s (biomass/glucose) = 5000000 / 13020833.3 = 0.384
0.48 mole of yeast required = 3 mole of oxygen
34722.22 mole of yeast = 3 * 34722.22 / 0.48 = 217013.889 mole
Total oxygen = 217013.889 * 16 (MW of oxygen) = 3472222.22 gram
YX/O (biomass/oxygen) = 5000000 / 3472222.22 = 1.44
C) Determine the total amount of oxygen required?
0.48 mole of yeast required = 3 mole of oxygen
34722.22 mole of yeast = 3 * 34722.22 / 0.48 = 217013.889 mole
Total oxygen = 217013.889 * 16 (MW of oxygen) = 3472222.22 gram
d) If the rate of growth at exponential phase is rx = 0.7 gdw/l-h, determine the rate of oxygen consumption (g O2/l-h)?
Yeast rate = 0.7g/l-h = 0.7/144 = 0.00486 mole /l- hr
0.48 mole of yeast required = 3 mole of oxygen
0.00486 mole/l-hr of yeast required = 3 * 0.00486 / 0.48 = 0.03038 mole/l-hr
rate of oxygen consumption = 0.03038 * 16 = 0.486 g/l-hr
e)Calculate the heat removal requirements for the reactor?
Q = Volume * rate of growth * (1/free energy released by glucose)
Volume = 100000 l
rate of growth = 0.7
free energy released by glucose = 0.42 g/Kcal
Q= (100000 * 0.7 * (1 / 0.42) = 166666.667 Kcal/hr needs to be removed from the reactor
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