The wastewater treatment plant operates anaerobic digesters that have a total volume of 824,000 gallons. The average flow rate of sludges into the digester is 11.3 gpm(gallons per minute). The total solids concentration of the sludge is about 2%, and of that 70% is volatile solids.
Calculate the volatile solids loading rate to the digester and express your answer in units of
loading rate = kg VS/(m^3 volume *day)
The Attempt at a Solution
824000 gal=3119.179 m^3
I first converted the rate of flow to m^3/day and got 61.59
(61.59 m^3/day)(x g VS/ 100 g sludge)(1 g/ 1 mL)(1000 mL/ L)(1000 L/ m^3)(1 kg/ 1000 g) = x kg/day
I know I have to account for the volatile solids, but (0.02)(0.7)= 0.14 g VS/100 g sludge
When plugged in to the loading rate formula, I get:
86.226 kg/ 3119.179 m^3-day
which equals 0.028, clearly not in the 2 to 5 range we were given.