VxM sw p ss

In other words, the fraction of the system MLSS in the fictitious bioreactor, fNMi, is proportional to the ratio of the average net specific growth rate in the process (l/W, + bH) relative to the required specific growth rate in the fictitious bioreactor, p.H1, as given by Eq. 10.28. For the case under consideration here, the MLSS concentration in the fictitious bioreactor will be the same as the concentration throughout the process. Therefore, the fraction of the total system volume in which the readily biodegradable substrate is removed, t\ ss, is given by:

VL=/l/(-),+b„\/ Ss<) \ V, V Phi- /\Sm , + Xsu/

where V, is the total system volume.

The fraction given by Eq. 10.31 represents the smallest possible fraction of the system volume within which the readily biodegradable substrate concentration could be reduced to Ssi. The possibility exists that the fraction will be larger because hydrolysis of slowly biodegradable substrate will be occurring, a contribution that was not considered in the derivation of Eq. 10.29. The exact contribution of hydrolysis is difficult to estimate without using a model like ASM No. 1, which would defeat the purpose for which these equations are given. However, the largest possible fraction of the system within which the readily biodegradable substrate might be removed can be calculated by assuming that all of the slowly biodegradable is hy-drolyzed and contributes to substrate removal in the fictitious bioreactor. When that assumption is made, the right parenthetical term, Ss() /(SSo + Xs„), goes to one. Thus, the largest possible fraction can be calculated with Eq. 10.31 with that term set equal to 1.0. The two fractions give the designer bounds within which to apportion the oxygen consumption associated with biomass synthesis from readily biodegradable substrate.

In using Eq. 10.28 to calculate |A,U for substitution into Eq. 10.31, the choice of SS| is very important. It must be large enough to represent the rapid rate of removal of readily biodegradable substrate, but low enough for the assumptions to be valid. Generally, a value of 10% of Ssi, should be adequate.

The total oxygen requirement for nitrification can be calculated with Eq. 10.16. It can be partitioned into two components, synthesis and decay, just as the heterotrophic oxygen requirement was partitioned. However, as shown in Figure 10.17. the effluent ammonia-N concentration from a tanks-in-series system will be less than the concentration from a CMAS system. This will make the oxygen requirement for nitrification slightly larger. Because the effluent ammonia-N concentration cannot be easily predicted without simulation, the effluent ammonia-N concentration should be assumed to be zero. This will provide a slightly conservative estimate. The oxygen requirement associated with synthesis of the autotrophic biomass can be calculated from:

The oxygen requirement associated with decay of the autotrophic biomass can be calculated with an equation like Eq. 10.25:

If the wastewater undergoing treatment is a domestic wastewater, the contribution of autotrophic decay to the total oxygen requirement will typically be negligible. However, this may not be the case for industrial wastewaters, and thus both autotrophic oxygen requirements should be calculated explicitly.

The distribution of the oxygen requirement associated with synthesis of the nitrifiers within a tanks-in-series system is very straightforward since nitrification can be assumed to behave as a zero-order reaction in the first part of such a system. The maximum mass nitrification rate can be estimated from a modified form of Eq. 5.7 that is analogous to Eq. 10.15 for the maximum autotrophic oxygen requirement:

in which X|iAI-V is calculated with Eq. 10.19. Because nitrification behaves as a zero-order reaction over most of the system, the fraction of the system volume over which autotrophic nitrification will occur, fVA, is just given by:

If the value of fV A is greater than 1.0, then nitrification will not be complete, which for the tanks-in-series configuration suggests that the SRT is too short and washout will occur (recall Figure 10.17). Actually, this never should occur since the value of X».A.T-V is required to get RNA mjx. If the SRT were too low, that value could not be calculated. Thus, a value of fVA greater than 1.0 suggests that an error has been made somewhere in the computations. Once the value of fVA is known, the oxygen requirement for nitrification, ROA, is apportioned proportionally over the fraction of the system within which nitrification occurs, just as the oxygen requirement for biomass synthesis from slowly biodegradable substrate was apportioned.

As was done for the oxygen requirement associated with decay of heterotrophs, that associated with decay of the autotrophs should be distributed evenly throughout the entire system.

Example 10.3.4.1

Consider the activated sludge system that was the subject of all of the examples in Section 10.3.3. Instead of a CMAS system, however, the system is to be configured as a CAS system with hydraulic characteristics equivalent to four tanks-in-series. Distribute the steady-state oxygen requirement associated with the summer equalized loading.

a. How should the oxygen requirement associated with biomass synthesis from readily biodegradable substrate be distributed?

Calculate the oxygen requirement with Eq. 10.23. The readily biodegradable substrate concentration is 115 mg/I. (= 1 15 g/m") and the flow rate is 40,000 m '/day.

RO,(ss = (40,()()())(115)| 1 -- (0.50)( 1.20)] = 1,840.000 g 0../day = 1,840 kg 0:/day

Calculate the specific growth rate in a fictitious bioreactor capable of removing 90'/'i of the readily biodegradable substrate. In that case, Ss, = 1 1.5 mg/L. Utilization of Eq. 10.28 gives:

Use it in Eq. 10.31 to calculate the smallest possible fraction of the total system volume within which the readily biodegradable substrate is removed. The SRT of the CAS system is 3 days.

The largest possible fraction within which the readily biodegradable substrate is removed can be calculated by setting the right parenthetical term equal to 1.0, giving a value of 0.17. Thus, the readily biodegradable substrate will be removed in 7.5 to 17 percent of the system volume. Since the CAS system behaves as lour equal tanks-in-series, 25% of the system volume is in each lank. Since the readily biodegradable substrate will be removed in less than 25'.'$ of the volume, all of the oxygen requirement associated with it should be apportioned to tank No. 1.

b. How should the oxygen requirement associated with biomass synthesis from slowly biodegradable substrate be distributed?

Calculate the oxygen requirement with Eq. 10.24. The slowly biodegradable substrate concentration is 150 mg/L (= 150 g/m') and the How rate is 40,000 m '/day.

R01I V, = (40,000)(150)[1 - (0.50)( 1.20)] = 2,400,000 g 0../day = 2,400 kg O./day

Use Eq. 10.27 to estimate the fraction of the system volume within which biomass growth on slowly biodegradable substrate occurs. Since the total system SRT is 3 days, all of the substrate must be fairly easy to degrade. Therefore, it seems reasonable to assume that (-), Ns has a value of 2 days.

Each tank in the system contains 25% of the system volume. Since 679i of the system volume has 100% of the oxygen requirement, 25% of the volume will have 37% of the oxygen requirement, i.e., 0.25/0.67 = 0.37. Therefore, tanks No. 1 and No. 2 will both receive 37% of the oxygen requirement and tank No. 3 will receive the remainder, or 26%. Therefore,

RO„as., = (0.37)(2,400) = 888 kg 0;/day RO|| xs: = (0.37)(2,400) = 888 kg O./day ROn xs , = (0.26)(2,400) = 624 kg 0:/day

Tank No. 4 will have no oxygen requirement associated with use of slowly biodegradable substrate.

c. How should the oxygen requirement associated with heterotrophic biomass decay be distributed?

The total oxygen requirement for heterotrophic decay can be calculated with Eq. 10.25:

The oxygen requirement associated with decay should be distributed equally to all reactors. Since each reactor contains 25% of the system volume, each will have 25% of the oxygen requirement. Thus, each tank will require 505 kg O./day for biomass decay.

d. How should the oxygen requirement associated with synthesis of autotrophic biomass be distributed?

Calculate the oxygen requirement for nitrification with Eq. 10.32. The value of SN , is obtained from Eq. 10.17, and can be assumed to be the same as for the CMAS system, or 30.5 mg/L as N:

RO,vsv = (40,000)(30.5)[4.57 - (0.2)(1.20)] = 5,280,000 g 0;/day = 5,280 kg CK/day

The fraction of the system volume over which nitrification will occur can be calculated with Eq. 10.35, which requires knowledge of the maximum mass nitrification rate from Eq. 10.34. That equation, however, requires the mass of autotrophic biomass in the system, which can be calculated with Eq. 10.19. Just as the oxygen requirement was slightly higher in the CAS system because of the lower effluent ammonia-N concentration, so will the mass of autotrophic biomass be higher. In this case it will be:

[1 + (0.20)(0.12)(3.0)](0.20)(30.5) 1 + (0.12)(3.0)

= 577,000 g autotrophic MLSS

It can now be substituted into Eq. 10.34. Since the minimum DO concentration in the system is to be 1.5 mg/L, this gives:

= <577.0«)) (() 75'^ , 5) = 2,500,000 g N/day

This can now be used with Eq. 10.35 to find the fraction of the system volume over which nitrification occurs:

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