An aerobic digester for Omega Three (Examples 27-2 and 27-3) is being considered as an alternative to alkaline stabilization. Determine the volume of the digester and the required air flow rate to meet a Class B pathogen density (PSRP). The following data have been obtained for this analysis:

Question

Winter conditions govern
Temperature of digester contents = 15°C
VSS = 0.8 TSS
VSS reduction goal = 40%
Influent BOD = 5,200 mg/L
Fraction of influent BOD consisting of primary solids = 0.35
Digester suspended solids concentration is 70% of the incoming thickened sludge concentration
Diffused air mixing is to be used

Accepted Answer

a. Check the requirements for volatile solids reduction and the SRT to meet PSRP (see Chapter 18). Find that the product of temperature and sludge age from Figure 27-11 at 40% VSS is 500°C ⋅ d. At a temperature of 15°C,
[latex]SRT= \frac{500°C ⋅ d}{15°C} =[/latex] 33.33 d

This is less than the PSRP requirement for an SRT of 60 d at a temperature of 15°C, so the 60 d SRT governs (see Table 18-15).

b. Using Equation 27-14, compute the volume of the digester. Note from Example 27-2 that the influent concentration is 3.8% or 38,000 g/m³ and that X = (38,000) (0.7).
[latex]\sout{V} = \frac{Q(X_{i} + F_{PS})}{X(k_{d}P_{v} + 1/SRT)} [/latex] (27-14)
[latex]\sout{V} = \frac{270 m^{3}/d[38, 000 g/m^{3} + (0.35)(5, 200 g/m^{3})]}{(38, 000 g/m^{3})(0.7)[(0.06 d^{ − 1})(0.80) + 1/60 d]} ][/latex]
[latex] = \frac{1.08 × 10^{7}}{1.72 × 10^{3}} =[/latex] 6.25 × 10³ m³

c. From Examples 27-2 and 27-3, the mass of sludge is
[latex]M_{sl}[/latex]= (3 cycles/d)(3,461 kg/cycle) = 10,383 kg/d

d. The required oxygen is estimated using the design parameter of 2.3 kg O /kg VSS destroyed from Table 27-12.
VSS = (0.80)(TSS) = (0.80)(10, 383 kg/d) = 8, 306.4 kg/d
[latex]VSS_{dest}[/latex] = (0.40)(8, 306.4 kg/d) = 3, 322.56 kg/d
[latex]O_{2} required[/latex] = (3, 322.56 kg/d)(2.3 kg O[latex]_{2}[/latex] /kg VSS) = 7, 641.89 kg/d

e. Assuming a 10% oxygen transfer efficiency, an air density of 1.204 kg/m³ , and a mass fraction of oxygen in air of 0.232, the daily volume of air is estimated as
[latex]\sout{V} _{air} = \frac{7,641.89 kg/d}{(0.10)(1.204 kg/m^{3})(0.232)} [/latex] = 273, 581 or 274,000 m³/d or 190 m³/min

f. Check the air required for oxygen and mixing.
[latex]Q_{air} = \frac{190 m^{3}/ min}{6.25 × 10^{3} m^{3}} [/latex] = 0.030 m³/ min ⋅ m³

This is within the recommended aeration ranges of 0.024–0.030 m³/min ⋅ m³ for mixed primary and activated sludge and 0.02–0.040 m³/min ⋅ m³ for diffused air mixing.

Comments:
1. Two digesters are provided for redundancy.
2. The aeration system is independent of that used for the activated sludge process.

Question 27.4: An aerobic digester for Omega Three (Examples 27-2 and 27-3)...

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