Example 147

Problem: A rotating biological contactor receives a flow of 2.2 MGD with a BOD content of 170 mg/L and suspended solids concentration of 140 mg/L. If the K value is 0.7, how many pounds of soluble BOD enter the RBC daily?

Solution:

Total BOD = Particulate BOD + Soluble BOD 170 mg/L = (140 mg/L x 0.7) + x mg/L 170 mg/L = 98 mg/L + x mg/L 170 mg/L - 98 mg/L = x x = 72 mg/L soluble BOD

Now, lb/day soluble BOD may be determined:

Soluble BOD = Soluble BOD (mg/L) x Flow (MGD) x 8.34 lb/gal = 72 mg/L x 2.2 MGD x 8.34 lb/gal = 1321 lb/day

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