Determine the daily volume of methane and total gas produced in an anaerobic digester that is operated at 35°C under the following conditions: biosolids flow = 300 m³/d; bsCOD = 5,000 g/m³ . Assume Y = 0.04 g VSS/g COD and 95% bsCOD removal.

Question

Accepted Answer

a. Use a steady-state mass balance to determine the amount of the influent COD converted to methane:
[latex]0 = COD_{in} − COD_{eff} − COD_{vss} − COD_{CH_{4}}[/latex]

b. Determine the individual COD values.
[latex]COD_{in} = (5, 000 g/m^{3})(300 m^{3}/d) = 1.5 × 10^{6} g/d[/latex]
[latex]COD_{eff} = (1 − 0.95)(1.5 × 10^{6} g/d) = 7.50 × 10^{4} g/d[/latex]

Noting from Equation 23-42 in Chapter 23, that the COD of waste activated sludge is 1.42 ([latex]P_{x}[/latex]),
[latex]C_{5}H_{7}NO_{2} + 5O_{2} \rightleftharpoons 5CO_{2} + 2H_{2}O + NH_{3} + energy[/latex] (23-42)
[latex]COD_{vss}[/latex] = (1.42 g COD/g VSS)(0.04 g VSS/g COD)(0.95)(1.5 × 10[latex]^{6}[/latex] g/d)
= 8.09 × 10[latex]^{4}[/latex] g/d

c. Solve the mass balance for
[latex]COD_{CH_{4}}[/latex]

[latex]COD_{CH_{4}} = COD_{in} − COD_{eff} − COD_{vss}[/latex]
[latex]= 1.5 × 10^{6} g/d − 7.50 × 10^{4} g/d − 8.09 × 10^{4} g/d = 1.34 × 10^{6} g/d[/latex]

d. Correct the theoretical methane gas production for the actual anaerobic digester temperature. From Equation 27-22, the theoretical production is 0.35 L of CH[latex]_{4}[/latex]/g COD.

[latex]\frac{22.414 L/mole}{64 g mole} =[/latex] 0.35 L/g COD (27-22)
Actual gas production =[latex](0.35 L of CH_{4}/g COD) \frac{(273.15 K + 35 K)}{273.15 K} [/latex]
[latex]= 0.395 L of CH_{4}/g COD[/latex]

e. The methane production is then
[latex]CH_{4} production = (1.34 × 10^{6} g COD/d)(0.395 L of CH_{4}/g COD)(10^{−3} m^{3}/L) [/latex]
= 531 or 530 m³/d

f. Assuming the methane is 65% of the gas volume, the total gas volume is

Total [latex]\sout{V}_{gas} = \frac{531 m^{3}/d}{0.65} [/latex] = 817 or 820 m³/d

Question 27.5: Determine the daily volume of methane and total gas produced...

Related Answered Questions

Select a CBFP to dewater an anaerobic digester sludge. The sludge is primary plus waste activated sludge that has a solids concentration of 2.5% and a specific gravity of 1.02. The design sludge flow rate is 171.2 m³/d. The plant managers wish to operate on a two-shift basis, that is, 16 h/d to ...

Verified Answer:

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: ...

Verified Answer:

Verified Answer:

Verified Answer:

An anaerobic digester for Omega Three (Examples 27-2, 27-3, and 27-4) is being considered as an alternative to alkaline stabilization or aerobic digestion. Determine the volume of the digester and the flow rate of methane. The following data have been obtained for this analysis: Temperature of ...

Verified Answer:

The Omega Three activated sludge plant generates a peak sludge flow rate of 270 m³/d of a mixed primary and thickened waste activated sludge. The solids concentration is 3.8%. One-third of the sludge is to be pumped 200 m to a stabilization facility during each shift. Select a pipe diameter and ...

Verified Answer:

Verified Answer: