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Example 15.4 A domestic wastewater with a flow of 20,000 m /d is to be denitrified. The effluent from the nitrification tank contains 30 mg/L of NO3-N, 2.0 mg/L of dissolved oxygen, and 0.5 mg/L of NO2-N. Assume that an effluent total nitrogen not to exceed 3.0 mg/L is required by the permitting agency. (a) Calculate the quantity of domestic sewage to be provided to satisfy the carbon requirement in mmoles per liter. (b) What is the corresponding BOD5 to be provided?

Solution:

, 2 m v TT/m-^ mmol cells r = — = 0.0625 mmol/L Yc = 3.7( 10 )---

32 mmol sewage fp" = 0.93 from Example 5.2

Ydn = 011-3 = 0.112 p" = 30/40 = 2.14 mmol/L Ydc = 0.05 mg VSS/mg BOD5 (from Table 15.3)

Ydc = 7.08( 10-4 1 = 5.28( 10-5) t' = 05 = 0.0357 mmol/L V0.67y 14

Therefore, r SFW 1 - 2( 0:0625) + [4(0.112)+ 1](0:93)(2.14)

(b) [ BOD5 ie„i(]mg - 400f bod5 Assume f BOD - 0.67

Therefore,

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