BioMed Research International

Research Article

Biomethane Production as an Alternative Bioenergy Source from Codigesters Treating Municipal Sludge and Organic Fraction of Municipal Solid Wastes

Table 1

Summary of the mass balance for the proposed integrated biomethanization plant (PE = 100,000; X_PS = 61 g TS/PE·day; ) = 50 g TS/PE·day) (values in brackets show the case for gr/PE·day).


Waste streams	TS (%)	VS/TS (%)	Density (kg/m³)	Total solids (t/day)	Flowrate (m³/day)

Pulper

Inlet (PS)	4.0	80	1010	6	150 (150)
Inlet (SS-OFMSW)	12 (15)	90 (85)	1015–1020	5 (10)	41 (66)
Supernatant phase	15	—	1015	~0.50 (~1.0)	3⁽¹⁾ (5.5)
Bottom phase	12	—	1010	~0.50 (~1.0)	4⁽²⁾ (8)
Outlet (PS+SS-OFMSW)	5.5⁽³⁾ (7)	90 (85)	—	10⁽⁴⁾ (14.2)	184⁽⁵⁾ (202.5)

Codigester

Inlet (PS + SS-OFMSW)	5.5⁽³⁾ (7)	90 (85)	—	10⁽⁴⁾ (14.2)	184⁽⁵⁾ (202.5)
Inert solids	—	10 (15)	—	1.0 (2.13)	—
Excess sludge (P_x)	—	—	—	0.34^(6,7) (0.45)	—
Outlet (digested sludge)	—	83⁽⁸⁾ (75)	—	5.84⁽⁸⁾ (8.615) 4.84^(7,8) (6.485)	—
Solids converted into CH₄	—	—	—	4.16^(7,9) (5.585)	~2220⁽¹⁰⁾ (~2980)

⁽¹⁾If 10% of VS of OFMSW is wasted from the pulper with supernatant, then 0.10 × [(41 × 10³ × 0.12 × 0.90)/(0.15 × 1015)] = 3.
⁽²⁾If 10% of inorganic materials in OFMSW is wasted from the bottom of the pulper, then 0.10 × [(41 × 10³ × 0.12)/(0.12 × 1010)] = 4.
⁽³⁾0.055 = [(150 × 0.04 + 41 × 0.12) − (3 × 0.15 + 4 × 0.12)/184].
⁽⁴⁾10 = (184 × 0.055).
⁽⁵⁾184 = [150 + 41−(3 + 4)].
⁽⁶⁾If VS/TS = 90%, = 0.05, VS Removal Y_obs = 50%, and 1 g VS = 1.5 g COD, then P_x = [(10 × 0.90) × 0.50 × 1.5 × 0.05] = 0.34.
⁽⁷⁾As volatile solids.
⁽⁸⁾5.84 = [(10 × 0.10) + 9 × (1−0.50) + 0.34]; 4.84 = (9 × 0.50 + 0.34); 4.84/5.84 = 0.83.
⁽⁹⁾4.16 = [(10 × 0.90)−4.84].
⁽¹⁰⁾Net CH₄ recovery = 90%; 1 g VS = 1.5 g COD; CH₄ Production/1 kg COD_dest. = 0.395 L (at 35°C), then Q_CH4 = (4160 × 0.90 × 1.5 × 0.395) ≅ 2220.