## XMTVssniy ijiu sSo x

where p.,u has been calculated with Eq. 10.28. In other words, as long as the fraction of MLSS in the contact tank, fXM (, is greater than or equal to the right side of Eq. 10.45, the effluent quality will be acceptable. Since the system contains only two tanks, the remainder of the biomass is in the stabilization basin. The derivation of Eq. 10.44 is based on the assumption that no slowly biodegradable substrate is used in the contact tank. Some hydrolysis will occur, however, allowing a portion of the slowly biodegradable substrate to be used. The effect of slowly biodegradable substrate utilization in the contact tank is to increase the required fraction of biomass in that tank, with the right parenthetical term in Eq. 10.45 approaching one as all of the slowly biodegradable substrate is used there, an event that is unlikely to occur. Because of the uncertainty associated with the fraction of the slowly biodegradable substrate that will be used in the contact tank, a designer should calculate the required minimum value of fXM.( with Eq. 10.45 twice, once as written and once with the right parenthetical term set to one. A design value for fVM< between those two extremes should then be chosen.

The fraction of the biomass in the contact tank is determined solely by the system hydraulics. If both growth and wastage are neglected, as they were for the SFAS system, then:

1 + a — v where a is the biomass recycle ratio and u is the fraction of the system volume in the contact tank:

where Vs is the volume of the stabilization basin and V, is the total system volume. Many combinations of a and -u can result in the same value of fXM<, as shown in Figure 10.20, where Eq. 10.46 is plotted in two ways. Figure 10.20a shows that for any value of u the fraction of MLSS in the contact tank will increase as the recycle ratio is increased. Thus, the selection of u should be made for the smallest anticipated recycle ratio. Because of the criterion given in Eq. 10.45, the needed effluent quality will be obtained for any larger value of the recycle ratio.

### Example 10.3.5.2

Consider the wastewater that has been the subject of all of the examples in this chapter. Consideration is being given to using a CSAS process with an SRT of 3 days. The effluent quality objective is 10 mg/L as COD of readily biodegradable-organic matter. What fraction of the biomass and what fraction of the system

Figure 10.20 Effect of the biomass recycle ratio on. (a) the fraction of biomass in the contact tank, and (b) the fraction of the system volume in the contact tank, for a CSAS system, as calculated with Eq. 10.46.

Recycle Ratio, a

Figure 10.20 Effect of the biomass recycle ratio on. (a) the fraction of biomass in the contact tank, and (b) the fraction of the system volume in the contact tank, for a CSAS system, as calculated with Eq. 10.46.

volume should be in the contact tank if the smallest recycle ratio that will be used is 0.307

a. What specific growth rate will be required in the contact tank? Calculate the specific growth rate in the contact tank, jjl,, (. using Eq. 10.28 and the kinetic parameters for winter conditions, since they will control. The desired substrate concentration is 10 mg/L. Using the kinetic parameters from

Table E10.2 gives:

b. What is the smallest fraction of the MLSS that can be in the contact tank? Use of (jliu = 1.37 in Eq. 10.45 gives the smallest fraction of the MLSS that can be in the contact tank.

Thus, assuming that no utilization of slowly biodegradable substrate occurs in the contact tank, as long as 15% or more of the MLSS is in that tank, the effluent quality goal can be met. Recalculation of the fraction with the right parenthetical term set to one shows that at least 35% of the MLSS would have to be in the contact tank if all of the slowly biodegradable substrate was used there. This is unlikely to occur, but to provide a factor of safety, choose a fraction of 25% for design.

What fraction of the system tank volume should be in the contact tank with a recycle ratio of 30% ?

The required fraction of the system volume in the contact tank, u can be calculated with a rearranged form of Eq. 10.46:

This value could also have been read from Figure 10.20b. Thus, 59% of the system volume must be in the contact tank to ensure that 25% of the system biomass is there. This large fraction is due to the short SRT of the system. Generally, the larger the SRT, the smaller the contact tank as a fraction of the total system.

After the fraction of biomass and the fraction of total system volume in the contact tank have been selected, the oxygen requirement must be distributed between the two reactors. This must be done for summer and winter conditions, and for the maximum and minimum anticipated recycle ratios. The minimum oxygen requirement will occur in the winter, and it will be used to determine the upper limit on the total system volume, just as with the other systems. Likewise, the summer conditions will give the maximum oxygen requirement, which determines the lower limit on the system volume, as discussed previously. Increasing the recycle ratio shifts some of the oxygen requirement from the stabilization basin to the contact tank, and also increases the oxygen requirement due to nitrification, as shown in Figures 7.24 and 7.25. Consequently, the extreme conditions must be examined so that the controlling situation can be identified.

The heterotrophic oxygen requirement can be distributed by a technique similar to that used to distribute the oxygen requirement in a CAS system, with special consideration of the unique characteristics of the CSAS system. The oxygen requirement for synthesis of biomass from readily biodegradable substrate can be calculated with Eq. 10.23. All of it will occur in the contact tank. The oxygen requirement for synthesis of biomass from slowly biodegradable substrate can be calculated with Eq. 10.24. Because slowly biodegradable substrate is entrapped in the MLSS where it undergoes hydrolysis, which is a slow reaction, it is likely to occur in both tanks. Although the relative amount of slowly biodegradable substrate utilization occurring in each tank can only be determined by simulation with a model like ASM No. 1 or 2, in the absence of such simulations it seems reasonable to distribute the oxygen requirement to the two tanks in proportion to the mass of MLSS in them. The oxygen requirement due to heterotrophic biomass decay can be calculated with Eq. 10.25. It, too, should be distributed in proportion to the mass of MLSS in each tank. Because of the effects of the recycle ratio on the distribution of MLSS in the system, its extreme values should be considered. The maximum value will maximize the heterotrophic oxygen requirement in the contact tank and minimize it in the stabilization basin, whereas the minimum recycle ratio will have just the opposite effect.

Distribution of the oxygen requirement associated with nitrification is more difficult and cannot be done precisely with analytical equations. However, it is possible to approximate it for those situations in which substantial nitrification will occur, i.e., when the system SRT is well in excess of the minimum SRT for nitrification. The rationale used to derive the equations is similar to that used to derive Eqs. 10.44 and 10.45, as well as 10.29 and 10.30, with additional criteria. First, nitrification will occur in the stabilization basin, making the concentration of am-monia-N entering the contact tank in the RAS flow approach zero. Second, when stable partial nitrification is occurring in the contact tank, it cannot be assumed that the mass of nitrogen oxidized in the stabilization tank is much less than the mass oxidized in the contact tank. In fact, it may be more. Furthermore, neither can it be assumed that the change in ammonia-N concentration across the contact tank is approximately equal to change in concentration across for an equivalent CMAS system, since more nitrification will occur in the latter. Rederiving Eq. 10.30 with these changes, and with the assumption that all forms of nitrogen are equally available in both tanks, gives the following approximate expression for the fraction of autotrophs in the contact tank, fXA (:

where p. vt is the specific growth rate of the autotrophs in the contact tank as calculated with Eq. 10.28 using SNH<, which is the ammonia-N concentration in that tank. The fraction of autotrophs in the contact tank is the same as the fraction of the MLSS in the contact tank, fXM because the mixed liquor is homogeneous throughout the system. Substitution of Eq. 10.28 into Eq. 10.48 and rearrangement gives a quadratic equation allowing the effluent ammonia-N concentration, SNn<, to be calculated for any given biomass recycle ratio, a, with its associated value of fXM,, as determined by Eq. 10.46:

where SN j is the available nitrogen concentration, as given by Eq. 10.17. The value obtained from Eq. 10.49 is only approximate because of the approximate nature of Eq. 10.48. Use of the highest anticipated value of a in the solution of Eq. 10.49 will minimize the ammonia-N concentration in the contact tank, which will maximize the oxygen requirement associated with nitrification in that tank, while minimizing the oxygen requirement in the stabilization basin. Use of the lowest anticipated recycle ratio will have just the opposite effect.

Once the ammonia-N concentration in the contact tank has been estimated, the oxygen requirement due to nitrification can be apportioned to the two vessels. Because particulate organic nitrogen will have a different fate than soluble nitrogen, a

distinction must be made between the two, even though that was not done in the derivation of Eq. 10.48. Let Sn s be the available soluble nitrogen:

Then the oxygen requirement due to synthesis of autotrophic biomass from soluble nitrogen in the contact tank can be calculated from a modified form of Eq. 10.32:

ROa.sn, = F(Sn.;, - Smk )(4.57 - Y/VI i() xh.t) (10.51)

The oxygen requirement in the stabilization tank associated with synthesis of autotrophic bacteria from soluble nitrogen is then:

The value of a used in Eq. 10.52 should be consistent with the value used to solve Eq. 10.49. The total oxygen requirement associated with synthesis of autotrophic biomass from particulate organic nitrogen can be estimated as:

It is apportioned to each of the vessels in proportion to the fraction of biomass in each. That apportionment will depend on the recycle ratio, as discussed previously. The oxygen requirement due to decay of autotrophs can be ignored because its contribution is less than the uncertainty associated with the other determinations.

Transient-state oxygen requirements can be estimated by combining the techniques presented in Section 10.3.2 with those presented above. As with the other multitank activated sludge systems, there will be considerable uncertainty associated with the estimates because of the need to distribute the transient oxygen requirement spatially. Consequently, simulation with a model like ASM No. 1 is a much more certain way of estimating such requirements.

### Example 10.3.5.3

Continue considering the CSAS system whose analysis was begun in Example 10.3.5.2. What would the summer time steady-state oxygen requirement be if 59% of the system volume was in the contact tank and the system was being operated with a recycle ratio of 0.30? As determined in Example 10.3.5.2, the contact tank of such a system would contain 25% of the MLSS.

a. What is the heterotrophic oxygen requirement in each tank?

The oxygen requirement for heterotrophic synthesis from readily biodegradable substrate can be determined with Eq. 10.23. Since the SRT is the same as in all previous examples, this computation is the same as the one made in Part a of Example 10.3.4.1. Thus, ROM SS = 1,840 kg O.Vday. All of this will occur in the contact tank.

The oxygen requirement for heterotrophic synthesis from slowly biodegradable substrate can be determined with Eq. 10.24. This computation is the same as the one made in Part b of Example 10.3.4.1. Thus, ROMAS = 2,400 kg 0:/day. Since the contact tank contains 25% of the biomass, 25% of this oxygen requirement will occur in it and 75% in the stabilization tank. Thus,

The oxygen requirement associated with heterotrophic decay can he estimated with Eq. 10.25. This was done in Part c of Example 10.3.4.1. Thus, ROh.i, = 2,020 kg 0:/day. It should also be distributed to the two vessels in proportion to the fraction of the MLSS that each contains. Thus,

Summing these values gives the heterotrophic oxygen requirement in each tank:

b. What is the autotrophic oxygen requirement in each tank?

The first task is to estimate the concentration of ammonia-N leaving the contact tank using Eq. 10.49. In that equation, a = 0.30, fXM , = 0.25, ^ = 1.3 day ', ("X = 3 days, bA = 0.12 day KM, = 1.9 mg N/L, and Sv„ = 30.5 mg N/L. Setting up the quadratic equation gives:

Solution of this equation reveals that SSM(, the ammonia-N concentration leaving the contact tank, is 16.6 mg ,'L as N. This compares to a concentration of 2.1 mg/L from the CMAS system (see Example 10.3.3.2). This difference is consistent with the differences between the two systems shown in Figure 7.20.

The oxygen requirement from oxidation of soluble nitrogen in the contact tank can be calculated with Eq. 10.51 after Ss ., has been estimated with Eq. 10.50 using the value of SVi, of 30.5 calculated in Example 10.3.3.2:

RO.vsm- = (40,000X22.0 - 16.6)[4.57 - (0.2)( 1.20)] = 935,000 g 0;/day = 935 kg O./day

The oxygen requirement associated with oxidation of soluble nitrogen in the stabilization basin can be calculated with Eq. 10.52 using a value of 16.6 mg/L as the ammonia-N concentration entering the basin through the RAS flow, which is 30% of the system influent flow:

ROvsss = (0.30)(40,000)( 16.6)[4.57 - (0.2)(1.20)] = 862,000 g 0:/day = 862 kg 0_,/day

The oxygen requirement from oxidation of particulate organic nitrogen can be calculated with Eq. 10.53:

Since 25% of the biomass is in the contact tank, 25% of this oxygen requirement will be exerted there, and 75%- in the stabilization basin. Consequently, the oxygen requirement from oxidation of particulate organic nitrogen in the two vessels is:

ROa nss< = (0.25)( 1,470) = 368 kg O./day RO vxnvs = (0.75)(1,470) = 1,102 kg O./day

Summing these values gives the autotrophic oxygen requirement in each tank:

ROv, = 935 + 368 = 1,303 kg OVday ROa.s = 862 + 1,102 = 1,964 kg 0:/day.

c. What is the total oxygen requirement in each tank?

The total oxygen requirement in each tank is obtained by summing the heterotrophic and autotrophic requirements:

RO, = 2,945 + 1,303 = 4,248 kg O./day ROs = 3,315 + 1,964 = 5,279 kg O./day

The total oxygen requirement in the whole system is 9,530 kg 0:/day, which is less than that required by the CMAS system as calculated in Example 10.3.3.2. The difference is due to the lower degree of nitrification in the CSAS system.

The contact tank comprises 59% of the system volume and experiences 45% of the oxygen requirement, whereas the stabilization basin contains 41% of the system volume and experiences 55% of the oxygen requirement. These percentages are specific to the operating conditions imposed and depend on the SRT, the recycle ratio, and the fraction of system volume in each tank, as illustrated in Figures 7.25 and 7.27.

After the oxygen requirements have been estimated for winter and summer and for both the maximum and minimum planned recycle ratios, the next task in the design of a CSAS system is to determine the total bioreactor volume, thereby fixing the volume of each tank, the MLSS concentration in each, and the final settler size. This is done in the same way as for the other activated sludge designs. An extra level of complexity is involved, however, because it is not apparent before the exercise whether the contact tank or the stabilization basin will control the design. For instance, in the example above, the stabilization basin had the greater requirement per unit volume, with 55% of the oxygen requirement in 41% of the volume. Thus, for that operational situation (summer with minimum recycle flow), the stabilization basin will control the minimum system volume. The engineer must consider all of the possible scenarios to determine which actually control the upper and lower limits on the system volume. Once those volumes have been determined, the range of permissible MLSS concentrations is fixed by the mass of MLSS in the system as calculated with Eq. 9.11. Finally, that information is used with the anticipated recycle ratios and projections of solids settling properties to arrive at an economic combination of bioreactor and settler sizes.

10.3.6 Batch Reactors—Sequencing Batch Reactor Activated Sludge

As discussed in Section 7.8, batch processes offer opportunities for flexibility that can be advantageous in some circumstances. For example, when flows are highly variable or when the character of the contaminants changes on a regular and periodic basis, batch processes allow those special needs to be met. Furthermore, by changing the length of the fill period within a cycle, batch processes can be made to behave like continuous flow processes with hydraulic characteristics anywhere between a perfect CSTR and a perfect plug-flow reactor (PFR). Consequently, sequencing batch reactor activated sludge (SBRAS) systems have found increasing popularity in recent years.

Although SBRAS systems are often treated as if special design techniques are required, they actually operate according to the same principles as other activated sludge processes, as discussed in Section 7.8.2, and can be designed according to them. Consequently, the mass of MLSS in the system is defined by Eq. 9. II, the overall daily solids wastage rate is given by Eq. 9.12, the total daily heterotrophic oxygen requirement is given by Eq. 9.13, and the total daily autotrophic oxygen requirement by Eq. 10.16, provided that the effective SRT is used in their computation. The three primary differences between SBRAS and other activated sludge systems are that the oxygen requirement must be distributed to each of the discrete operating cycles and then apportioned within each cycle in a manner consistent with the length of the fill period, the fraction of each operating cycle that is not devoted to biological reaction must be accounted for in the design, and the interaction between the bioreactor and the secondary clarifier must be analyzed differently because biological reaction and sedimentation occur in the same vessel (although at different times in the cycle). The first of these can be accomplished by the same techniques used to apportion steady- and transient-state oxygen requirements in the various continuous-flow activated sludge systems. The second can be accounted for through use of the effective HRT, t,., and the effective SRT, when using the CMAS equations listed above. The third can be considered through simple procedures to be presented below.

The effective HRT and effective SRT account for the portion of the SBRAS cycle that is not utilized for biological reaction, i.e., that portion devoted to settling and decanting (draw), as illustrated in Figure 7.42. In Section 7.8.2, £ is defined as the fraction of the total cycle devoted to fill plus react. Using those definitions, the effective HRT is defined as:

and the effective SRT is defined as:

Selection of the effective SRT is governed by the same considerations as the selection of the SRT for any other activated sludge system. Relatively low values can be used if the removal of biodegradable organic matter is the primary objective. However, SBRAS systems are often used in small wastewater treatment plants where sludge stabilization is also an important consideration. Thus, effective SRTs of 10 days or more are often used.

Selection of the system volume, and hence the effective HRT, must consider not only mixing and oxygen transfer, but also the anticipated settling properties of the MLSS. That is necessary because the bioreactor volume must be sufficient to contain not only the flow added per cycle, Ft, but the volume retained after effluent is decanted to contain the recycled biomass, Vh„ where:

As discussed in Section 7.8.2, N, is the number of cycles per day and a is analogous to the recycle ratio in a continuous flow system. Consequently, when an SBRAS system is being designed as a simple activated sludge system with no nutrient removal capabilities:

The primary factors considered in the selection of both Nc and £ are adding influent wastewater and removing treated effluent, allowing a sufficient time for settling and decanting in the operating cycle, and allowing for peak hydraulic flows through the bioreactor. These factors, in turn, are affected by the total number of bioreactors selected and the desired operational cycle. Values of Nc often range from 4 to 6 cycles per day. Since the time required for settling and decanting is relatively constant, whereas the length of a cycle gets shorter as the number of cycles per day is increased, the value of £ decreases as Nc increases. Nevertheless, values of £ often range from 0.5 to 0.7. Considerable flexibility exists in the selection of both parameters.

The minimum bioreactor volume associated with a given number of cycles per day will result when Vhr is just large enough to contain the MLSS retained for use in the next cycle after the effluent has been decanted. The mass of MLSS in the bioreactor after solids settling and effluent decanting is the same as the mass of MLSS at the end of the react period, which is given by Eq. 9.11. Therefore:

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