DIAGNÓSTICO ACTUAL
4.7. DETERMINACIÓN DEL FACTOR DE EFICIENCIA
This section analyses the influence of the sludge age on selected important process parameters of the activated sludge system. All values have been calculated applying the model presented in this chapter.
To broaden the results, the values are given for three different combinations of the parameters Y and Kd, selected to reflect conditions of lower biomass production
898 Activated sludge
(smaller Y and larger Kd) and of larger biomass production (larger Y and smaller Kd).
The main relations presented in Table 31.8 and in Figure 31.4 are (von Sperling, 1996d):
• Production of suspended solids (SS) per unit of BOD5(Sr) removed. Used for the estimation of the production of secondary excess sludge
• Volatile suspended solids (VSS) to total suspended solids (SS) ratio. Used in several design stages
• Oxygen consumption (O2) needed to satisfy the carbonaceous demand per unit of BOD5(Sr) removed. Used for the design of the aeration system
• Mass of mixed liquor volatile suspended solids required (Xv·V) per unit of BOD5(Sr) removed. With the product Xv·V, for a given adopted value of the mixed liquor volatile suspended solids (Xv), the required volume for the reactor (V) may be determined
The following comments can be made with respect to Table 31.8 and Figure 31.4:
• The PSS/Srand VSS/SS ratios were presented in Section 31.9.1.
• The O2/Srratio was presented in Section 31.10.2.
• The influence of the consideration of the influent solids to the reactor and of the presence of primary settling on the production of secondary excess sludge and on the VSS/SS ratio in the reactor can be seen clearly.
• The relations O2/Srand Xv·V/Srare not affected by the presence of primary settling or solids in the influent. Obviously, in a system with primary settling the BOD load to the reactor will be lower, but the values of O2and Xv·V per unit of BOD removed in the reactor will be the same.
• The VSS/SS ratio is little affected by the values of the coefficients Y and Kd.
• The relations O2/Srand Xv·V/Srare highly influenced by the values of the coefficients Y and Kd.
When using the data from Table 31.8 for a quick design, the following points should be taken into consideration (further details are given in Chapter 34):
• If nitrification is desired to be included in the computation of the average oxygen consumption (which is always advisable), the values of the O2/Sr
ratio in the above table starting from the sludge age of 4 days (in warm-climate regions) should be increased by around 50 to 60% (for typical values of influent TKN and assuming full nitrification, oxygen savings through the removal of nitrogen with the excess sludge and absence of intentional denitrification).
• To estimate the oxygenation capacity to be added to the system, the average oxygen consumption needs to be multiplied by a factor, such as the ratio between the maximum flow and the average flow (approximately 1.5 in medium to large plants, and 2.0 in smaller plants). This is the value of the oxygen demand in the field.
Table31.8.Functionalrelationsintheactivatedsludgesystemasafunctionofthepresenceofsolidsintheinfluent,existenceof primarysettling,coefficientsYandKdandthesludgeage Sludgeage(day) RatioSSinthePrimaryCoefficients ItemandunitinfluentsettlingY(g/g)Kd(d−1)261014182226 0.50.090.500.420.370.330.310.290.28 NoNo0.60.080.600.510.450.410.380.360.34 0.70.070.710.610.550.500.470.440.42 ProductionofSS/Sr0.50.090.830.750.700.670.650.630.63 solids(kgSS/YesYes0.60.080.960.870.810.780.750.730.71 kgBOD5rem)0.70.071.040.950.880.840.800.780.76 0.50.091.081.000.950.920.900.880.88 YesNo0.60.081.231.141.091.051.021.000.98 0.70.071.291.201.131.081.061.031.01 VSS/SSratioVSS/SSNoNo0.5–0.70.07–0.090.890.870.850.840.830.820.81 inreactor(g/g)YesYes0.5–0.70.07–0.090.790.760.750.730.720.710.71 YesNo0.5–0.70.07–0.090.750.730.710.700.690.690.68 CarbonaceousO2/Sr––0.50.090.840.951.021.071.101.131.14 oxygen(kgO2/––0.60.080.700.830.910.971.011.051.07 demandkgBOD5rem)––0.70.070.570.700.800.860.910.950.98 VolumeXv·V/Sr––0.50.090.882.163.113.884.555.155.71 ofthe[kgVSS/––0.60.081.072.673.874.855.706.477.17 reactor(kgBOD5/d)]––0.70.071.263.214.695.926.987.938.80 Notes: •Highlightedvalues:Moreusualvaluesinactivatedsludgeplantswithtypicalflowsheets •Percapitacontributions:BOD5=50g/inhabitant·d;SS=60g/inhabitant·day •Removalefficienciesintheprimarysedimentationtank:BOD=30%;SS=60% •Sr:RemovedBOD5load(kgBOD5/d)
900 Activated sludge
Sludge production (Pss) per BOD removed (Sr) 1.4
with solids, without primary settler
without influent solids
Sludge age (d)
Pss/Sr (kg/kg)VSS/SS (kg/kg)
with solids, with primary settler
(in each range: lower curve: Y = 0.5 and Kd= 0.09 d−1; intermediate curve:
Y = 0.6 and Kd= 0.08 d−1; upper curve: Y = 0.7 and Kd= 0.07 d−1)
Conventional activated sludge systems (qc£ 10 d) usually include primary settling, while extended aeration systems (qc≥ 18 d) do not usually include primary settling
0.6
with solids, without primary settler with solids, with primary settler
without influent solids
Sludge age (d) VSS/SS ratio i n the reactor
(variation of the coefficients Y and Kd: small influence; not considered)
Conventional activated sludge systems (qc£ 10 d) usually include primary settling, while extended aeration systems (qc≥ 18 d) do not usually include primary settling
Figure 31.4. Functional relations of process variables with the sludge age
• If one wants to express the oxygen demand under standard conditions (20◦C, clean water, sea level), the field demand has to be divided by a factor between 0.55 and 0.65.
• If one wants to express the relations in terms of the BOD load applied to the reactor, instead of the removed load, the corresponding values must be multiplied by 0.93 to 0.98, which correspond to the typical BOD removal efficiencies [(So–S)/So].
0
To obtain the required reactor volume, simply divide the value of Xv.V (kg) by the adopted value of Xv (kg/m3)
Oxygen consumption (O2) per BOD removed (Sr)
(in the range: lower curve: Y = 0.7 and Kd= 0.07 d−1; intermediate curve:
Y = 0.6 and Kd= 0.08 d−1; upper curve: Y = 0.5 and Kd= 0.09 d−1)
This consumption is for the average carbonaceous demand. To obtain the total demand, the consumption for nitrification and the provision of oxygen for the maximum flow must be considered
Figure 31.4 (Continued )
To facilitate the implementation of a quick automated design tool in the com-puter, thus avoiding the need to refer to Table 31.8 and allowing a continuous solution for any sludge age within the range (not only those given in Table 31.8), von Sperling (1996d) made regression analyses which correlated the various vari-ables and ratios in Table 31.8 with the sludge age. The structure adopted for the regression equation was the multiplicative (y= a·xb). The results are included in Table 31.9.
In all the regressions the fitting was excellent (coefficients of determination R2greater than 0.98). The utilisation of the equations is as follows. For example,
902 Activated sludge
Table 31.9. Regression analysis comparing several relations included in Table 31.8 and the sludge age (influent solids considered)
Coefficients Equation y= a·(θc)b Relation Solids in Primary
(y) the influent settling Y (g/g) Kd(d−1) a b
SS/Sr Yes Yes 0.5 0.09 0.900 −0.110
(kgSS/ 0.6 0.08 1.053 −0.115
kgBOD5rem) 0.7 0.07 1.158 −0.126
Yes No 0.5 0.09 1.145 −0.081
0.6 0.08 1.318 −0.087
0.7 0.07 1.401 −0.098
VSS/SS (g/g) Yes Yes 0.5–0.7 0.07–0.09 0.817 −0.043
Yes No 0.5–0.7 0.07–0.09 0.774 −0.038
O2/Sr – – 0.5 0.09 0.777 0.118
the oxygen consumption per unit of BOD removed (O2/Sr) for the sludge age of 8 days and the intermediate coefficient values (Y= 0.6 and Kd= 0.08 d−1) will be (from Table 31.9) a= 0.630 and b = 0.161. The equation is: O2/Sr = 0.630 · (θc)0.161= 0.630 × (8)0.161= 0.88 kgO2/kgBOD5removed. This value is consistent with Table 31.8, after interpolating between the sludge ages of 6 and 10 days.
The detailed design sequence of the activated sludge, using the various process formulae introduced in this chapter, is presented in Chapter 34.
Example 31.11
Undertake a quick design of the biological reactor, based on the data included in Table 31.8 and the associated remarks. Determine the volume of the reactor, the oxygen consumption, the power of the aerators and the production and removal of the excess sludge.Consider an extended aeration plant, with a sludge age of 25 days and a MLVSS concentration of 3,500 mg/L. Use the influent load of 3,350 kgBOD5/d (estimated in the example in Chapter 2 and also adopted in the detailed design in Chapter 34). Take into account the solids in the influent and assume that the system will not have primary settling.
Solution:
(a) Estimation of the removed BOD load
The removed BOD load can be estimated as 95% of the applied BOD. Thus:
Sr = 0.95 × 3,350 kgBOD5/d = 3,183 kgBOD5/d
Example 31.11 (Continued )
(b) Estimation of the VSS/SS ratio in the reactor and the resulting MLSS concentration
According to Table 31.8 (considering the influent solids and not using primary settling), after a linear interpolation between the sludge ages of 22 and 26 days for the sludge age of 25 days, one gets: VSS/SS= 0.68.
For Xv = 3,500 mgMLVSS/L, the resulting MLSS concentration is:
MLSS= 3,500/0.68 = 5,147 mg/L = 5.147 kg/m3 (c) Estimation of the reactor volume
According to Table 31.8, by adopting the intermediate coefficient values (Y= 0.6; Kd = 0.08 d−1) and interpolating between the sludge ages of 22 and 26 days, one gets: Xv·V/Sr = 7.0 d−1.
For Xv= 3.5 kg/m3 (= 3,500 mgMLVSS/L) and Sr= 3,183 kgBOD5/d, the resulting reactor volume is:
V= 7.0 × 3,183/3.5 = 6,366 m3
(d) Estimation of the production and removal of excess sludge
According to Table 31.8, by considering the influent solids and not adopting primary settling, one gets: Px/Sr= 0.98 (interpolating between the sludge ages of 22 and 26 days).
For Sr= 3,183 kgBOD5/d, the sludge production is calculated as:
Px= 0.98 × 3,183 = 3,119 kgSS/d
If the sludge is removed directly from the reactor, its concentration is the same as MLSS (X). Thus, the excess sludge flow is (disregarding the loss of solids in the final effluent) is:
Qexreactor= Px/X = 3,119/5.147 = 606 m3/d
Note that this value is different from the value V/θc (= 6,366/25 = 255 m3/d), usually adopted when controlling the system by the sludge age (hydraulic control), disregarding the influent solids.
If the sludge is removed from the return sludge line, its concentration is the same as RASS (Xr). For a recirculation ratio R (=Qr/Q) equal to 1.0 (adopted), one has:
Xr = X·(R + 1)/R = 5.147 × (1 + 1)/1 = 10.294 kg/m3(=10,294 mg/L)
Qexreturn sludge line= Px/Xr= 3,119/10.294 = 303 m3/d
904 Activated sludge
Example 31.11 (Continued )
The excess sludge flow removed from the return sludge line (303 m3/d) is half of the flow removed from the reactor (606 m3/d), due to the fact that the solids concentration in the return sludge line (10,294 mg/L) is twice the concentration in the reactor (5,147 mg/L).
(e) Calculation of the oxygen consumption and aerator power requirements
According to Table 31.8, O2/Sr = 1.06 kgO2/kgBOD5(interpolating between the sludge ages of 22 and 26 days).
For Sr= 3,183 kgBOD5/d:
O2carbonaceous= 1.06 × 3,183
= 3,374 kgO2/d (average carbonaceous demand) To take into account the nitrification in the total O2 consumption, the car-bonaceous demand value must be increased by 50 to 60%. By adopting a value of 55%, the total average demand (disregarding denitrification) is:
O2total= 1.55 × 3,374 = 5,230 kgO2/d
To take into account the demand under maximum load conditions, the aver-age oxygen demand must be multiplied by a correction factor. This factor may be adopted varying between 1.5 and 2.0 (≈Qmax/Qaverage), depending on the size of the plant. Adopting a factor of 2.0, one has:
Total maximum O2= 2.0 × 5,230
= 10,460 kgO2/d (in the field, under operational conditions) To express it in standard conditions, the field value should be divided by a factor that varies between 0.55 and 0.65. By adopting the value of 0.60, one has:
O2standard= 10,460/0.60 = 17,433 kgO2/d = 726 kgO2/hour By adopting a standard oxygenation efficiency of 1.8 kgO2/kW·hour for mechanical aeration (see Chapter 11), the power requirement is:
Power required= 726/1.8 = 403 kW = 537 HP (f) Comments
• If the detailed design sequence presented in Chapter 34 had been followed, it could be verified that the values found for the volume of the reactor,
Example 31.11 (Continued )
production of excess sludge and oxygen requirements are very similar to those obtained in the present quick design (differences of less than 2.5%).
• This example could have been also undertaken based on the equations presented in Table 31.9. The results obtained should be very similar.
• The design did not foresee the intentional denitrification in the reactor.
Although still little used in most developing countries, its implementation in a more systematic way should be encouraged, especially in warm-climate countries.
• To complete the plant, the designs of the secondary sedimentation tanks and the preliminary treatment units (screen and grit chamber) and sludge processing units (thickening and dewatering) are still needed. The design of these units is simpler than the design of the reactor and associated variables.