Question

A sludge pump pumps at a rate of 40 GPM. The raw sludge density is 4.5 percent solids. How many minutes per hour should the pump operate if the inflow to the clarifier is 1.8 MGD, the influent suspended solids are 210 mg/L, and the effluent suspended solids are 60 mg/L

Answer 1

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.