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12345 [234]
3 years ago
9

A 300-gallon anaerobic digester will be loaded daily with a feedstock that contains two parts dairy manure and one-part water by

volume. The feedstock contains 6% volatile solids (VS) by weight and has a density of 37.5 lb/ft'. What volume of feedstock would be required each day to maintain an organic loading rate (OLR) of 2.0 kg VS/m/day? What is the hydraulic retention time (HRT) in the anaerobic digester tank for a loading rate of 2.0 kg VS/m/day?
Chemistry
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

The volume of feedstock needed to mantain an organic load rate of 2 kgVS/day is 0.055 m3/day of feedstock.

The HRT is 20.6 days.

Explanation:

First, we calculate how many kg is 1 m3 of feedstock. We know the density, so we can calculate the mass:

M=\rho*V=37.5\frac{lb}{ft^3}*1m^3*(\frac{3.281ft}{1m})  ^3=1324.5lb=600.7 kg

If the VS are 6% in weight,

M_{vs}=0.06*M=0.06*600.7\,kg/m^3=36,0kgVS/m3

The volume per day needed to feed 2 kg of VS/day is:

V=\frac{2kg}{36kg/m^3}= 0.055m3/day=5.5litres/day

The HRT depends on the volume of the tank and the flow. Its equation is

HRT=\frac{V}{Q}=\frac{300gal}{0.055 m^3/day}*\frac{1m^3}{264.172gal}\\   \\HRT=20.6\,days

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