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Substituting these newfound values of t and u into 3(t + u)/50 t, we have

From this equation, the total mmol/L of sewage needed during the nitrite-reduction stage of the denitrification is milligram moles per liter of sewage needed _ 3 ^ (15 62) t 'milligram moles per liter of nitrite nitrogen destroyed 10 (5 -2Ydc)

Adding Eqs. (15.58), (15.60), and (15.62) gives the overall milligram moles per liter of sewage, [SEWdenit ]mmol, needed in the denitrification process:

r SEW i _ 2,r' , (4Y dn + 1')fp"p// + 3t' (1563)

[SEWdenit ]mmol may be converted into units in terms of BOD5 in mg/L. From Eqs. (15.53) and (15.54), the ratio of milligrams BOD5 to millimole of sewage is fBODs ((1/4)O2)/(1/50) - 400/BOd. Therefore, the corresponding concentration of [SEWdenit]mmol in terms of BOD5 in mg/L, [BODsdenJmg, is

LBOD5denit]mg - 400/bOD5LSEWdenit]mmol

- 400fBOD,

Example 15.3 After cutting off the aeration to a nitrification plant, the dissolved oxygen concentration is 2.0 mg/L. How much sewage is needed for the carbonaceous reaction in this last stage of aerobic reaction before denitrification sets in? How much NH4-N is produced from this carbonaceous reaction?

Solution:

milligram moles per liter of sewage needed _ 2 , r' milligram moles per liter of oxygen used 5 (5-2 Yc )

' 2 nn^oc i/r v i -v/m^x mmol cells r - -)- - 0.0625 mmol/L Yc - 3.7(10 )-

32 c mmol sewage

Therefore, milligram moles per liter of sewage needed - _2_

r'milligram moles per liter of oxygen used 5[5 _ 2(3.7)(10-4)]

milligram moles per liter of ammonia nitrogen produced - 2(1 - FJ , r' milligram moles per liter of oxygen used 5 (5 - 2 Yc )

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