Loading rate of volatile solids in an anaerobic digester

[musicmar]

Homework Statement

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
kg VS/(m^3-day)

Homework Equations

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
Then:
(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 into 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.

 

[haruspex]

(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

Looks like you are using x to mean two different things.

(0.02)(0.7)= 0.14 g VS/100 g

No, if you want /100g then it's 2% of 70% of 100g.

86.226 kg/ 3119.179 m^3-day

Why divide by the total capacity? (Why is the capacity even montioned? Maybe to test that you only use the relevant info.)