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18-Env-B5 Industrial & Hazardous Waste Management · December 2015

Question 6 of 18: Gas Production Rate from an Anaerobic Chicken-Manure Digester

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment, 2nd ed.; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed.; Davis & Cornwell, Introduction to Environmental Engineering, 6th ed.; LaGrega, Buckingham & Evans, Hazardous Waste Management, 2nd ed.; CCME, Guidelines for the Management of Biomedical Waste in Canada (1992); Canadian Environmental Protection Act (CEPA), 1999; Basel Convention on the Control of Transboundary Movements of Hazardous Wastes (1989); provincial hazardous waste regulations (e.g. BC's Environmental Management Act and Hazardous Waste Regulation).

Question 6: Gas Production Rate from an Anaerobic Chicken-Manure Digester (8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantitySymbolValue
Chicken population—200,000
Manure production per chicken—0.00019 m3/d
COD of raw manure feedCt150,000 g/m3
Anaerobic reactor HRT—12 d
Net biomass (VSS) production rate per reactor volumerg600 g/m3·d
Fraction of COD removed—0.79
Methane yield per kg COD destroyed—0.37 m3/kg

Find. The daily methane (biogas) production rate from the anaerobic digester treating the combined manure stream.

Approach. Compute the influent flow and reactor volume from the population and HRT, find the total COD load and the mass actually destroyed, subtract the COD equivalent of the new biomass synthesized (which is not available for methane conversion), then apply the given methane yield to the net COD stabilized.

  1. Manure (feed) flow rate. Total flow contributed by 200,000 chickens: $$Q = 200{,}000 \times 0.00019\ \text{m}^3/\text{d} = 38\ \text{m}^3/\text{d}$$
  2. Reactor volume. From the HRT $\theta = V/Q$: $$V = Q\,\theta = 38\ \text{m}^3/\text{d} \times 12\ \text{d} = 456\ \text{m}^3$$
  3. Total COD loading rate. Mass load $= Q \times C_t$: $$\text{COD load} = 38\ \text{m}^3/\text{d} \times 150{,}000\ \text{g/m}^3 = 5{,}700{,}000\ \text{g/d} = 5700\ \text{kg/d}$$
  4. COD actually removed (destroyed) in the digester. Applying the given removal fraction: $$\text{COD}_{removed} = 0.79 \times 5700 = 4503\ \text{kg/d}$$
  5. Net biomass (VSS) produced. The volumetric growth rate $r_g$ applies over the whole reactor volume: $$P_x = r_g \times V = 600\ \text{g/m}^3\text{d} \times 456\ \text{m}^3 = 273{,}600\ \text{g/d} = 273.6\ \text{kg VSS/d}$$
  6. COD stabilized for energy (methane) production. Not all destroyed COD becomes methane — some is diverted into new cell mass. Each kg of VSS synthesized represents 1.42 kg of COD (the standard oxygen-demand equivalent of biological cell tissue), so the COD available for conversion to methane is: $$\text{COD}_{stabilized} = \text{COD}_{removed} - 1.42\,P_x = 4503 - 1.42(273.6) = 4503 - 388.5 = 4114.5\ \text{kg/d}$$
  7. Methane (gas) production rate. Applying the given methane yield to the net stabilized COD: $$\boxed{V_{CH_4} = 0.37\ \text{m}^3/\text{kg} \times 4114.5\ \text{kg/d} \approx 1522\ \text{m}^3/\text{d}}$$
Final results
QuantityValue
Manure flow, Q38 m3/d
Reactor volume, V456 m3
Total COD load5700 kg/d
COD removed4503 kg/d
Net biomass produced, Px273.6 kg VSS/d
COD stabilized for gas production4114.5 kg/d
Methane production rate≈ 1522 m3/d
Check: assumes the 1.42 g COD/g VSS cell-mass conversion factor (standard for biological solids, Metcalf & Eddy) to net out biomass synthesis from the COD destroyed before applying the methane yield — this is the conventional McCarty-type anaerobic digester gas-production method and is the only way the supplied rg term is used in the calculation.