Question

it is anticipated that 265 lbs/day ss will be pumped from the primary clarifier of a new plant. if the primary clarifier sludge has a solids content of 4.4%, how many gallons per day of sludge will be pumped from the clarifier?

715 gpd
722 gpd
745 gpd
738 gpd

Explanation:

Calculate total daily mass of wet sludge

$$ \text{Total Sludge Mass} = \frac{265\text{ lbs/day}}{0.044} \approx 6022.73\text{ lbs/day} $$

Convert wet sludge mass to volume in gallons

$$ \text{Sludge Volume} = \frac{6022.73\text{ lbs/day}}{8.34\text{ lbs/gal}} \approx 722.15\text{ gpd} $$

Select the closest matching option

$$ 722.15\text{ gpd} \approx 722\text{ gpd} $$

Answer:

• (A) 715 gpd
• (B) 722 gpd (Correct answer)
• (C) 745 gpd
• (D) 738 gpd