Hydraulic detention time, Mechanical Engineering

a.The closest point to discharge treated effluent is a lake, but the pipeline is extended so it discharges into the river draining the lake. This allows a conventional secondary treatment level (Table 11-6) and not be required to also reduce NH3, TKN and P, as per Table 11-7. Explain why the difference in receiving waters affects the treatment required.

b. Influent wastewater SS or TSS = 220 mg/l. Primary sedimentation removes 60% of it. The sludge has 6% solids. Calculate the daily volume (liters) and dry solids weight (kg) of primary sludge.

c. The hydraulic detention time F/M= 4 hours, the overflow or surface settling rate is v0 =500 gpd/ft2. Compute the volume and depth of the sedimentation tank.

d. Say the organic SS particles have specific gravity Gs = 1.1. The inorganic SS particles have Gs= 2.65. For which type of material are the removed particles of larger diameter? Illustrate by formula.

e. Activated sludge treatment is used. 2.0 mgd inflow to the aerobic reactor has 100 mg/l BOD5 (So). How many pounds of air at 20% O2 must be transferred into the wastewater each day?

f. Using same F = Q x So, it's desired to keep F/M= 0.2 in the 40,000 ft3tank. Find the desired X (VSS concentration) that must be maintained. Sketch the system used to keep reactor aerated, mixed, and at X.

g. The microorganisms doing the treatment are endogenous, chemheterotrophs Explain what each of these descriptors of metabolism means.

h. The COHNS substrate includes both VS and FS content. What are the 3 possible fates? Yield = 40%. If its 100% degraded (just VS), how many moles of CO2 are produced per mole substrate?

i. Primary & secondary sludge are blended and anaerobically digested in a closed tank for θ = 20 days to reduce VSS. How does this help the plant to be energy self-sufficient?

Posted Date: 3/2/2013 6:25:23 AM | Location : United States

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