An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The sludge is aerated for 6 hr. The food-to-microorganism ratio is 0.30. The recirculation ratio is 0.2. The surface loading rate of the secondary clarifier is 800 gal/day-ft2. The final effluent has a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the activated sludge treatment processes (secondary BOD removal), (b) aeration tank volume, (c) MLSS, and (d) secondary clarifier surface area? If 0.5 pound of oxygen is required for each pound of BOD entering the aeration tank and the density of air is approximately 0.075 lb/ft3, and the air is 20.9% oxygen by volume, calculate air requirements per day.

(a) BOD removal efficiency = 89.51%

(b) Aeration tank volume = 4732m³

(c) MLSS = 1706.669 mg/L

(d) Secondary clarifier surface area 6250ft²

(e) Air requirement =  12930.284 lb

Explanation:

See the attached file for explanation.  This is continuation from page 3 of the attached file.

Air required = 172403.792 ft3

since, density of air = 0.075lb/ft3

air required = 0.075*172403.792 lb

                    = 12930.284 lb

answer:

the answer is d

explanation:

answer:

explanation:

assault rifle is a rapid-fire, magazine-fed rifle designed for military use. it is a shoulder-fired weapon that allows the shooter to select between semi-automatic (requiring you pull the trigger for each shot), fully automatic (hold the trigger and the gun continuously fires) or three-shot-burst modes.