BMP testingGuide

Theoretical methane potential: the Buswell and Boyle formula with calculation examples

How to calculate the maximum yield of CH₄ and CO₂ from a molecular formula or elemental composition. Three numerical examples, the limitations of the method and comparison with the BMP result.

Published 4 min readFeedstockData Editorial Team

Other language versions: polski.

Kolby Erlenmeyera z zielonożółtą zawiesiną na półce w jasnym świetle.
Photo Oak Ridge National Laboratory / Wikimedia Commons, licence CC BY 2.0, source file; changes: cropping

The theoretical methane potential is the maximum amount of CH₄ that can be obtained from a given organic matter if it were completely converted to methane and carbon dioxide. It is calculated from the Buswell formula and, once nitrogen and sulphur are taken into account, from the Boyle formula. The result is an upper limit, not a forecast: actual BMP is lower, because part of the substrate (e.g. lignin) does not undergo anaerobic degradation and part of the carbon is used to build microbial biomass.

The formula

For matter with the molecular formula CcHhOoNnSs, degradation proceeds according to the equation:

CcHhOoNnSs + (c − h/4 − o/2 + 3n/4 + s/2) H₂O → (c/2 + h/8 − o/4 − 3n/8 − s/4) CH₄ + (c/2 − h/8 + o/4 + 3n/8 + s/4) CO₂ + n NH₃ + s H₂S

The molar volume of an ideal gas at standard conditions (0 °C, 101.325 kPa) is 22.414 l/mol. The yield in NmL CH₄/g is calculated as: (moles of CH₄ × 22,414) / molar mass of the substrate [g/mol]. The CH₄ fraction in the theoretical biogas is CH₄ / (CH₄ + CO₂).

Example 1: glucose (C₆H₁₂O₆, 180.16 g/mol)

CH₄: 6/2 + 12/8 − 6/4 = 3 + 1.5 − 1.5 = 3 mol; CO₂: 3 mol. CH₄ fraction = 50%. Yield = 3 × 22,414 / 180.16 ≈ 373 NmL CH₄/g.

Example 2: model protein (C₅H₇NO₂, 113.12 g/mol)

CH₄: 2.5 + 0.875 − 0.5 − 0.375 = 2.5 mol; CO₂: 2.5 mol; yield = 2.5 × 22,414 / 113.12 ≈ 495 NmL CH₄/g, CH₄ fraction 50%. The nitrogen goes into ammonia (1 mol of NH₃ per mole), which does not appear in the gas but raises the pH and the ammonium nitrogen content in the digester.

Example 3: fat (triolein C₅₇H₁₀₄O₆, 885.4 g/mol)

CH₄: 28.5 + 13 − 1.5 = 40 mol; CO₂: 28.5 − 13 + 1.5 = 17 mol; CH₄ fraction ≈ 70%. Yield = 40 × 22,414 / 885.4 ≈ 1,013 NmL CH₄/g.

Example 4: biomass of known elemental composition

Assumptions (illustrative, per 100 g VS): C 48 g, H 6.5 g, O 40 g, N 2 g; the remainder omitted. Moles: C 4.00; H 6.5; O 2.5; N 0.143. CH₄ = 4/2 + 6.5/8 − 2.5/4 − 3 × 0.143/8 = 2.000 + 0.8125 − 0.625 − 0.054 ≈ 2.13 mol; yield ≈ 2.13 × 22,414 / 100 ≈ 478 NmL CH₄/g VS. CH₄ fraction in the gas: 2.13 / 4.00 ≈ 53%. If the determinations indicate a different composition, the result will change accordingly — which is why a reliable elemental analysis is so important.

Alternative: COD

If the chemical oxygen demand (COD) in g O₂/g VS is known, the following relationship can be used: 1 g COD ≈ 350 NmL CH₄ (about 0.35 l CH₄ per g COD at standard conditions). The method is simple, but it assumes complete conversion of COD into methane, so it gives a ceiling value.

Interpretation: degradability

Degradability index = measured BMP / theoretical BMP. For easily degradable substrates (sugars, fats) it is usually high, for lignocellulosic ones — clearly lower, and it is an important indication of whether pretreatment makes sense. A low ratio may result from:

  • lignin and other recalcitrant fractions,
  • inhibition in the test (too high a substrate dose, acidification),
  • use of part of the carbon for biomass,
  • gas losses or measurement error.

Limitations of the method

Limitation Consequence
Assumes full conversion result overstated relative to practice
Requires elemental composition from the same batch error when using tabulated data
Does not account for kinetics says nothing about the rate
CO₂ partly dissolves in the liquid actual gas composition differs from the calculated one
Mixtures require a combined composition or a weighted average

Checklist for use

  1. Obtain the C, H, O, N, S composition (elemental analyser) and VS from the same sample.
  2. Convert to moles, apply the formula, state the assumptions (standard conditions, molar volume).
  3. Compare with BMP; record the ratio.
  4. If the BMP value is above theory — look for an error.
  5. Remember that the formula does not replace the test: it serves for a plausibility check and for preliminary estimation.

Good practice in a test report is to give both values side by side — this allows the reader to assess what part of the potential has been realised.

Application in planning the mix

For a mixture of substrates the theoretical potential can be calculated as a weighted average by the VS of each component, provided their elemental composition is known. Such a calculation is useful when assessing whether the yield declared for a plant does not exceed physical limits. If the methane production balance of the whole biogas plant, related to the VS fed, approaches the theoretical value, there is an error in the data (e.g. in substrate weighing or gas measurement). In practice, technical yields are clearly below the theoretical ceiling.

Sources

  1. Holliger C. et al. (2016), Towards a standardization of biomethane potential tests, Water Sci. Technol. 74(11), 2515–2522
  2. Angelidaki I. et al. (2009), Defining the biomethane potential (BMP) of solid organic wastes and energy crops, Water Sci. Technol. 59(5), 927–934

Questions and answers

Can the theoretical potential be lower than the BMP result?

It should not be. If the test result exceeds the theoretical ceiling calculated from a reliable elemental composition, this is a sign of error: in gas measurement, in the VS determination, in the elemental composition, or an overstated inoculum contribution.

This question in the FAQ