Answer:
The pump should operate at 6.25 minutes per hour
Explanation:
We have to find the minute per day pumping rate first.
Therefore,
Sludge removal lbs/ day = Sludge pumped lbs/day
Formula is given as:
(Suspended solids removed (mg/L) × Clarifier inflow(MGD) × 8.34lbs/gal) ÷ ( solids contents in % ÷ 100) = GPM Sludge pumping rate × minute per day pumping rate × 8.34 lbs/gal
From the question,
Suspended solids removed is calculated as : Influent suspended solids concentration - Effluent suspended solids concentration
= 210 mg/L - 60mg/L
Suspended solids removed = 150mg/L
Clarifier Inflow(MGD) = 1.8 MGD
Sludge density = 4.5% solids
GPM = 40 GPM
Minute per day pumping rate = is unknown so we represent it as y
Hence,
(150 mg/L × 1.8 MGD Flow × 8.34lbs/gallon) ÷ ( 4.5% ÷ 100) = 40 GPM × y minute per day pumping rate × 8.34 lbs/gallon
(150 mg/L × 1.8 MGD Flow × 8.34lbs/gallon) ÷ 0.045 = 40 GPM × y minute per day pumping rate × 8.34 lbs/gallon
50040 = 333.6y minute per day
y = 50040 ÷ 333.6
y = 150 minute per day pumping rate
In other to get the minute per hour pumping rate, we would convert the minute per day to minute per hour,
Therefore,
24 hours = 1 day
So we have,
150 minute per day ÷ 24 hours
= 6.25minute per hour
The pump should operate at 6.25 minutes per hour.
