This p may be substituted into s = 4pYdn such that s = 20p'Ydn.
Substituting these newfound values of p and s into 2p - 3s/20p, we have
2p-3s = 1-6Ydn (1548) -----2---0----p------ = --------1---0------d--- n (15.48)
(1 - 6Ydn)/10 = (2p - 3s)/20p is the number of mmol/L of ammonia nitrogen produced per mmol/L of nitrate nitrogen destroyed. The p' mmol/L of nitrate nitrogen constitutes a fraction of the total concentration of nitrate nitrogen produced during the nitrification process. Let this concentration be p" mmol/L and let fp« be the fraction of p" destroyed. Then, p' = fp« p". Therefore, the mmol/L of ammonia nitrogen produced during the normal anoxic denitrification is milligram moles per liter of ammonia nitrogen produced p 'milligram moles per liter of nitrate nitrogen destroyed 1 - 6 Y 1 - 6 Y
The equation to calculate the amount of ammonia nitrogen produced from the nitrite-reduction side reaction may be derived in a similar manner. We will no longer go through the steps but simply write the answer at once:
milligram moles per I Her of ammonia nitrogen produced = 3(1-i^) ^ (1550) t' milligram moles per liter of nitrite nitrogen destroyed 10( 5-2 Ydc)
t is the mmol/L of nitrite nitrogen that corresponds to the t meq/L of nitrite nitrogen destroyed or used during the nitrite-reduction stage of the denitrification process. It is assumed that all of the nitrites that appeared after the aeration is shut off are being converted to the nitrogen gas.
Adding Eqs. (15.47), (15.49), and (15.50) gives the concentration of NH4-N in milligram moles per liter, [NH4-N]mmol, that will appear in the effluent of the nitrification-denitrification process:
[NH4-N]- = ^^ r' + ^f p"p" + ^^ ' (15.51)
The total nitrogen concentration that will appear at the effluent of the nitrification-denitrification process as a result of the consumption of the residual oxygen from nitrification, destruction of nitrate, and destruction of nitrite in the denitrification step is the sum of the ammonia nitrogen in the above equation plus the nitrate nitrogen not destroyed in the denitrification step. Let [IN ]mmol be the milligram moles per liter of total nitrogen in the effluent. Thus,
= 2(1 -r' + L^f "p" + 3(1t' + (1- f ")p" 5 (5 - 2 Yc) 10 Jpp 10 (5 - 2Ydc) U J p >p (15.52)
Again, t' = mmol/L of nitrite nitrogen appearing after the nitrification step, assumed totally destroyed; p" = mmol/L of nitrate nitrogen appearing after the nitrification step, with fp" fraction destroyed; and r' = mmol/L of dissolved oxygen remaining right after aeration is cut off, assumed totally consumed.
Note that the production of effluent nitrogen depends upon the values of the cell yields. For Yc equal to or greater than 1, no effluent ammonia nitrogen is produced from the heterotrophic side of the denitrification; for 6Ydn equal to or greater than 1, no effluent ammonia nitrogen is produced from the normal anoxic denitrification;
and for Ydc equal to or greater than 1, no effluent ammonia nitrogen is produced from the nitrite-reduction side of the denitrification. These facts should be considered in the calculation by setting the values equal to zero when the respective conditions are met.
Three types of cell yields are used in the previous derivations: Yc, Ydn, and Ydc. The units of Yc and Ydc are in terms of moles of organisms per unit mole of sewage. The units of Ydn are in terms of moles of organism per unit mole of nitrate nitrogen. The units normally used in practice for Yc and Ydc, however, are either in terms of mass of organisms or cells (approximated by the volatile suspended solids value, VSS) per unit mass of BOD5 or mass of organisms or cells per unit mass of COD. For the case of Ydn, the units used in practice are in terms of mass of the organisms or cells per unit of mass of the nitrate nitrogen. Unlike the conversion of Yc and Ydc from mass basis to mole basis which is harder, the conversion of Ydn from mass basis to mole basis is very straightforward; thus, we will address the conversion of the former.
Let the cells yielding Yc and Ydc be designated collectively as YBOD when expressed in terms of mass cells per mass BOD5 and let them be designated collectively as YCOD when expressed in terms of mass cells per mass COD. To make the conversion from YBOD or YCOD to Yc or Ydc, the electrons released by sewage as the electron donor must be assumed to be all taken up by the oxygen electron acceptor. This is because we want the conversion of one to the other—partial taking up of the electrons does not make the conversion. The two half reactions are reproduced below for convenience.
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