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Substituting r - 4r' and q - 8rTc/(5 - 2Yc), we have

2/5(5 - 2Yc) - 2(r + q)/25r is the number of mmol/L of sewage needed per mmol/L of dissolved oxygen. Therefore, the total mmol/L of sewage needed during the heterotrophic side reaction is milligram moles per liter of sewage needed _ 2 ^ (1558) r' milligram moles per liter of oxygen used 5 (5-2 Yc)

From Equation (15.44), the number of moles of sewage needed per mole of nitrate nitrogen is s + p

Substituting p - 5p' and s - 20p' Ydn, we have s_+p - 20p'Ydn + 5p' - 20Fdn + 5 - (1__9)

From this equation, the total mmol/L of sewage needed during the normal anoxic denitrification is milligram moles per liter of sewage needed _ 4Fd„ + 1 , /-'milligram moles per liter of nitrate nitrogen destroyed 10 p

10 Jp p

The relationship of t (in mmol/L) and t was obtained as follows:

This t may be substituted into

5-2 Yic such that

From Equation (15.45), the number of moles of sewage needed per mole of nitrite nitrogen is t + u

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