Question 3. In many fermentations, the maximum amount of cell mass must be obtai

Question

Question 3. In many fermentations, the maximum amount of cell mass must be obtained. However, the amount of mass that can be made is limited by the cell volume. Cells occupy a finite volume and have a rigid shape so they cannot be packed beyond a certain limit. There will always be some water remaining in the space between cells which represents the void volume that as best can be as low as 40% of the fermenter volume. Calculate the maximum cell mass on a dry basis per liter of the fermenter that can be obtained if the wet cell density is 1.1 g/cm3 . Note the cells 2 themselves consist of about 75% water and 25% solids and the cell mass is reported as dry weight in the fermenter industry.

Explanation / Answer

Assume a fermetor volume of 100 cm3

Assuming that 40% of this volume is void space occupied by water and the rest 60% is occupied by cells.

So total cell volume = 60 cm3

Now these cells are wet as they have water inside them

Wet cell density = 1.1 g/cm3

So total mass of these wet cells = 1.1*60 = 66 g

Now this mass contains 75% water mass and 25% dry mass.

So the actual dry cell mass = 0.25*66 = 16.5 g

Now we get 16.5 g dry cell mass when fermentor volume is 100 cm3 ( = 0.1 L), so on a per lier basis, the maximum dry cell mass that can be obtained is = 16.5/0.1 = 165 g

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