DIAGRAMA DE MAPEO

All the calculations for the design of the reactor, or for the determination of the effluent BOD, were made by assuming that S was the effluent soluble BOD, that is to say, the biochemical oxygen demand caused by the organic matter dissolved in the liquid medium. This BOD could be considered the total effluent BOD from the system, if the final sedimentation tank were capable of removing 100% of the suspended solids flowing into it. However, it is worth remembering that the concentration of these solids that will reach the secondary sedimentation tank is in the order of 3,000 to 5,000 mg/L. Thus, it is expected that they will not be entirely removed, and that a residual fraction will leave with the final effluent. As these solids have a large fraction of organic matter (mainly represented by the biomass), they will still cause an oxygen demand when they reach the receiving body. This demand is named suspended BOD or particulate BOD. Thus, in the final effluent of an activated sludge plant, there are the following fractions:

Total effluent BOD5= Soluble effluent BOD5+ Particulate effluent BOD5

(31.10) The soluble BOD can be estimated using Equation 31.7 or 31.8. For the estima-tion of the particulate BOD, some consideraestima-tions should be made. The solids that generate oxygen demand are only the biodegradable solids, since the inorganic and the inert solids are not an organic substrate that can be assimilated by the bacteria and generate oxygen consumption. By using Equation 31.2, and know-ing the process parameters, one can determine the parameter fb, that is, estimate which fraction of the VSS present in the plant effluent is biodegradable and will, therefore, represent the BOD of the suspended solids. Once this biodegradable fraction is known, the oxygen consumption required to stabilise this fraction can be estimated. For this, Equation 31.11, relative to the stabilisation of the cellular material represented by the formula C5H7NO2, can be used:

C5H7NO2+ 5O2→ 5CO2+ NH3+ 2H2+ Energy (31.11)

MW=113 MW=160 (X_b)

866 Activated sludge

Thus, according to the stoichiometric relationship between the molecular weights (MW), 160 g of oxygen are required for the stabilisation of 113 g of biodegradable solids. Hence, this relationship is:

O2/Xb= 160/113 = 1.42 gO2/g biodegradable solids (31.12) The ultimate biochemical oxygen demand (BODu) of the biodegradable solids is equal to this O2consumption. Thus, expressed in other terms:

BODuof the biodegradable solids= 1.42 mgBODu/mgXb (31.13) In typical domestic sewage, the relationship between BOD5 and BODu is approximately constant, and the ratio BODu/BOD5 is usually adopted as 1.46.

Thus, the ratio BOD5/BODu is the same as the reciprocal of 1.46, that is, BOD5/BODu = 1/1.46 = 0.68 mgBOD5/mgBODu. This means that when reach-ing the fifth day of the BOD test, 68% of the organic matter originally present has been stabilised, or else 68% of the total oxygen consumption takes place by the fifth day. Hence, Equation 31.13 can be expressed as:

BOD5of the SSbiodeg = 0.68 mgBOD5/mgBODu× 1.42 mgBODu/mgXb

BOD5of the SSbiodeg ≈ 1.0 mgBOD5/mgXb (31.14) To express this oxygen demand in terms of the volatile suspended solids, Equa-tion 31.14 needs to be multiplied by fb(= Xb/Xv). The fbvalues can be obtained using Equation 31.2. Hence:

BOD5of VSS≈ 1.0 (mgBOD5/mgXb)× fb(mgX_b/mgVSS)

BOD5of VSS≈ fb(mgBOD5/mgVSS) (31.15)

To make this equation more realistic and yet practical, it is interesting to ex-press the effluent solids not as volatile suspended solids, but as total suspended solids. This is because, in the operational control routine and in the determination of the performance of the treatment system, the usual procedure is to measure the performance of the secondary sedimentation tank based on the effluent total suspended solids concentration. In Section 31.3, the values of the VSS/TSS ratio were presented, and in Section 9.5.8 it was shown how to calculate the value of the ratio. For conventional activated sludge systems, VSS/TSS varies from 0.70 to 0.85, while for extended aeration systems, VSS/TSS varies from 0.60 to 0.75. The BOD5of the total suspended solids will then be:

BOD5of the effluent SS (mgBOD₅/mgSS) = (VSS/TSS)·fb (31.16) Based on the fbvalues resulting from the application of Equation 31.2 and on the typical values of the relationship VSS/TSS described in the paragraph above, and

by applying Equation 31.16, the following ranges of typical values of particulate BOD are obtained:

• conventional activated sludge: 0.45 to 0.65 mgBOD5/mgTSS

• extended aeration: 0.25 to 0.50 mgBOD5/mgTSS

Experimental studies by von Sperling (1990) and Fr´oes (1996) with two ex-tended aeration systems led to a ratio in the range of 0.21 to 0.24 mgBOD5 for each mgSS, close to the lower limit of the theoretical range.

The determination of the BOD5 of the final effluent is, therefore, essentially dependent on the estimation of the suspended solids concentration in the effluent from the secondary sedimentation tank. Unfortunately, there are no widely ac-cepted rational approaches that can be safely used to estimate the effluent solids concentration, since the number of variables involved in the clarification function of secondary sedimentation tanks is very high. There are some empirical criteria that correlate the solids loading rate in the settling tank and other variables with the effluent SS concentration, but these relationships are very site specific.

Designers usually assume a SS concentration to be adopted in the design (equal to or lower than the SS discharge standard), and through this value the particulate BOD5is estimated. Based on a desired value of total BOD5 in the effluent, and with the estimated particulate BOD5, by difference, the required soluble BOD5is obtained (simple rearrangement of Equation 31.10). With this value the biological stage of the treatment plant can be properly designed.

Example 31.4 illustrates the complete calculation of the effluent total BOD5of an activated sludge system.

Example 31.4

For the conventional activated sludge system described in Example 31.1, cal-culate the concentrations of partical-culate, soluble and total BOD in the effluent.

Assume that the design value for the effluent SS concentration is 30 mg/L.

Data already obtained in Examples 31.1 and 31.3:

S= 8 mg/L fb= 0.72 Solution:

(a) Particulate BOD₅in the effluent from the secondary sedimentation tank

Adopt the VSS/SS ratio equal to 0.8 (see above). The particulate BOD5 is calculated using Equation 31.16:

Particulate BOD5= (VSS/SS)·fb= 0.8 × 0.72 = 0.58 mgBOD5/mgSS For 30 mg/L of effluent suspended solids, the effluent particulate BOD5is:

30 mgSS/L× 0.58 mgBOD5/mgSS= 17 mgBOD5/L

868 Activated sludge

Example 31.4 (Continued )

(b) Summary of the effluent BOD₅concentrations Soluble BOD= 8 mg/L (calculated in Example 31.3) Particulate BOD= 17 mg/L

Total BOD= 8 + 17 = 25 mg/L

If, for example, a better effluent quality, with a total effluent BOD5 of 20 mg/L were desired, there would be two possibilities. The first would be to reduce the effluent SS concentration (effluent polishing), to decrease the particulate BOD5. The second would be to allow a maximum value for the soluble BOD5

of 3 mg/L (= 20 – 17 mg/L). In this case, the reactor should be redesigned.

(c) Efficiency of the system in the BOD removal

The efficiency of the system in the BOD removal is given by:

E(%)= BOD5influent− BOD5effluent DBO5influent ·100

The biological removal efficiency (that considers only the soluble BOD in the effluent) is:

E= 100·(300 − 8)/300 = 97%

The overall removal efficiency (considering total BOD in the effluent) is:

E= 100·(300 − 25) = 92%

In the calculation of the reactor volume and of the BOD removal, S is considered as the soluble effluent BOD, and Sois the total influent BOD. This is because the organic suspended solids, which are responsible for the influent particulate BOD, are adsorbed onto the activated sludge flocs, and subsequently undergo suc-cessive transformations into simpler substrate forms, until they become available for synthesis. Only after this transformation to soluble organic solids will they be removed by similar mechanisms to those that acted on the soluble BOD. Thus, the influent particulate BOD will also generate bacterial growth and oxygen demand, but with a time lag compared to soluble BOD. In dynamic models this time lag should be taken into account, but it has no influence in steady-state models. This is the reason why Sois considered as the total influent BOD.

Another aspect to be remembered is that, if the treatment system is provided with primary sedimentation tanks, such as the conventional activated sludge system, part of the influent BOD is removed by sedimentation, corresponding to the settled fraction of the volatile suspended solids. These will undergo subsequent separate digestion processes in the sludge treatment line and will not enter the reactor. The BOD5removal efficiency of primary sedimentation tanks usually ranges from 25%

to 35%, that is, to say, the influent BOD to the reactor (So) is 65% to 75% of the raw sewage BOD.