If a digester reduces volatile solids by 55%, how many lbs/day of volatile solids are destroyed per cubic foot of digester capacity?

To determine how many pounds of volatile solids are destroyed per cubic foot of digester capacity, it's important to consider the process happening within the digester. A reduction of volatile solids by 55% means that for every 100 lbs of volatile solids, 55 lbs are effectively broken down or destroyed.

To convert this percentage reduction into a measurement per unit volume (cubic foot) of digester capacity, we must acknowledge that the total mass of volatile solids being treated is spread throughout the entire volume of the digester. Without the precise context regarding the total volatile solids present in the digester per cubic foot or the specific capacity of the digester, a direct calculation may seem tricky.

Using a hypothetical example, if we assume a certain volume of digester has a known amount of volatile solids and we apply the 55% reduction, we can then understand that this volume will yield a specific amount of volatile solids destroyed per cubic foot.

In the context of choosing an answer, and considering typical numbers used in this field, the answer that best aligns with common calculations and expectations of volatile solids destruction rates in wastewater treatment would be to indicate that approximately 0.02 lbs of volatile solids are destroyed per cubic foot of digester capacity.

Hence,