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Example 15.1 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. If the total nitrate nitrogen is to be destroyed by 95%, calculate the total ammonia nitrogen produced from the reactions and total nitrogen in the effluent.

Solution:

Referring to Table 15.3, assume:

Yc = 0.35 mg VSS/mg BOD, = 7.08( 10-')( and assume f BOD = 0.67.

Yc = 7.08( 10-4)(lO^ mmo' Cdls = 3.7(10-4) c v \0.67jmmol sewage v y mmol cells mmol sewage 'mmol sewage r' = |f = 0.0625 mmol/L

0 9/113

Ydn = 9/.'43 = 0.112 fp. = 0.95 p" = 30/14 = 2.14 mmol/L Ydc = 0.05 mg VSS/mg BOD5 (from Table 15.3)

Ydc = 7.08( 10-4 )f'°051 = 5.28( 10-5) t' = 05 = 0.0357 mmol/L V 0.67y 4

Therefore,

[ NH4 N]mmol = 5?[^'_2t(3'.7V)(,1^:4)](0.0625 )+ T0 (0.95 )(2.14)

= 5.0 (10-3) + 0.067 + 2.22( 10-3) = 0.074 mmol/L ^ 1.036 mg/L Ans [ TN ]mmol = 0.074 + (1 - 0.95)(2.14) = 0.181 mmol/L ^ 2.53 mg/L Ans

Example 15.2 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. If the total nitrate nitrogen in the effluent is limited by the state agency to 3.0 mg/L, calculate the percent destruction of the nitrate nitrogen and the ammonia nitrogen concentration in the effluent.

Solution:

[tn] , = 2(1-Yc) r' + l~6Yd-f ,,p"+ 3(1-Fdc) t' + (1- f „)p" [ TN ]mmol 5(5-,Fc- r+ 10 fp p + 10(5- 2Yd--1+( 1 fp)P

= 5.0(10 ) + 1 ^fp"(2.14) + 2.22(10 ) + (1 - fp") (2.14)

0.214 = 5.0 (10-3) + ^f-11"'-fp,,( 2.14) + 2.22 (10-3) + ( 1- fp")( 2.14)

= 5.0 (10-3) + 0.070fp" + 2.22 (10-3) + 2.14 - 2.14 fp 2.07 fp" = 1.93 fp" = 0.93 Ans rNH N1 - 2(1 "Y') r' + l~6Y*f v" + ^^

- 5.0(10-3) + 1 -6100112)(0.93)(2.14) + 2.22( 10-3)

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