Appendix 2

United Kingdom

Calculation of λ-Shift factor (S_λ)

Calculation

The λ-shift factor (S_λ)^f1 shall be calculated by means of equation (9-1):

S_{λ} = \frac{2}{(1 - \frac{inert%}{100}) (n + \frac{m}{4}) - \frac{O_{*}^{2}}{100}}

(9-1)

Where:

S_λ

λ-shift factor;

inert %

% by volume of inert gases in the fuel (i.e. N₂, CO₂, He, etc.);

O_{*}^{2}

% by volume of original oxygen in the fuel;

n and m

refer to average C_nH_m representing the fuel hydrocarbons, i.e.:

$n = \frac{1 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} %}{100}] + 3 \times [\frac{C_{3} %}{100} + 4 \times [\frac{C_{4} %}{100} + 5 \times [\frac{C_{5} %}{100} + ..]]]}{\frac{1 - diluent%}{100}}$	(9-2)
$m = \frac{4 \times [\frac{{CH}_{4} %}{100}] + 4 \times [\frac{C_{2} H_{4} %}{100}] + 6 \times [\frac{C_{2} H_{6} %}{100} + …8 \times [\frac{C_{3} H_{8} %}{100} + ..]]}{\frac{1 - diluent%}{100}}$	(9- 3)

Where:

CH₄

% by volume of methane in the fuel;

C₂

% by volume of all C₂ hydrocarbons (e.g.: C₂H₆, C₂H₄, etc.) in the fuel;

C₃

% by volume of all C₃ hydrocarbons (e.g.: C₃H₈, C₃H₆, etc.) in the fuel;

C₄

% by volume of all C₄ hydrocarbons (e.g.: C₄H₁₀, C₄H₈, etc.) in the fuel;

C₅

% by volume of all C₅ hydrocarbons (e.g.: C₅H₁₂, C₅H₁₀, etc.) in the fuel;

diluent

% by volume of dilution gases in the fuel (i.e.: O₂*, N₂, CO₂, He, etc.).

Examples for the calculation of the λ-shift factor S_λ:

Example 1: G₂₅: CH₄ = 86 %, N₂ = 14 % (by volume)

n = \frac{1 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} %}{100}] + ..}{\frac{1 - diluent %}{100}} = \frac{1 \times 0,86}{1 - \frac{14}{100}} = \frac{0,86}{0,86} = 1

m = \frac{4 \times [\frac{{CH}_{4} %}{100}] + 4 \times [\frac{C_{2} H_{4} %}{100}] + ..}{\frac{1 - diluent %}{100}} = \frac{4 \times 0,86}{0,86} = 4

S_{λ} = \frac{2}{(1 - \frac{inert%}{100}) (n + \frac{m}{4}) - \frac{O_{*}^{2}}{100}} = \frac{2}{(1 - \frac{14}{100}) \times (1 + \frac{4}{4})} = 1,16

Example 2: G_R: CH₄ = 87 %, C₂H₆ = 13 % (by vol)

n = \frac{1 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} %}{100}] + ..}{\frac{1 - diluent %}{100}} = \frac{1 \times 0,87 + 2 \times 0,13}{1 - \frac{0}{100}} = \frac{1,13}{} = 1,13

m = \frac{4 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} H_{4} %}{100}] + ..}{\frac{1 - diluent %}{100}} = \frac{4 \times 0,87 + 6 \times 0,13}{1} = 4,26

S_{λ} = \frac{2}{(1 - \frac{inert%}{100}) (n + \frac{m}{4}) - \frac{O_{*}^{2}}{100}} = \frac{2}{(1 - \frac{0}{100}) \times (1,13 + \frac{4,26}{4})} = 0,911

Example 3: USA: CH₄ = 89 %, C₂H₆ = 4,5 %, C₃H₈ = 2,3 %, C₆H₁₄ = 0,2 %, O₂ = 0,6 %, N₂ = 4 %

n = \frac{1 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} %}{100}] + ..}{\frac{1 - diluent %}{100}} = \frac{1 \times 0,89 + 2 \times 0,045 + 3 \times 0,023 + 4 \times 0,002}{1 - \frac{0,64 + 4}{100}} = 1,11

m = \frac{4 \times [\frac{{CH}_{4} %}{100}] + 2 \times [\frac{C_{2} H_{4} %}{100}] + 6 \times [\frac{C_{2} H_{6} %}{100}] + .. + 8 \times [\frac{C_{3} H_{8} %}{100}]}{\frac{1 - diluent %}{100}} = \frac{4 \times 0,89 + 4 \times 0,045 + 8 \times 0,023 + 14 \times 0,002}{1 - \frac{0,6 + 4}{100}} = 4,24

S_{λ} = \frac{2}{(1 - \frac{inert%}{100}) (n + \frac{m}{4}) - \frac{O_{*}^{2}}{100}} = \frac{2}{(1 - \frac{4}{100}) \times (1,11 + \frac{4,24}{4}) - \frac{0,6}{100}} = 0,96

As an alternative to the above equation, S_λ may be calculated from the ratio of the stoichiometric air demand of pure methane to the stoichiometric air demand of the fuel blend supplied to the engine, as specified below.

Lambda-shift factor (S_λ) expresses the oxygen demand of any fuel blend in relation to oxygen demand of pure methane. Oxygen demand means the amount of oxygen to oxidise methane in a stoichiometric composition of reaction partners to products of complete combustion (i.e. carbon-dioxide and water).

For the combustion of pure methane the reaction is as set out in equation (9-4):

1 · CH ₄ + 2 · O ₂ → 1 · CO ₂ + 2 · H ₂ O

(9-4)

In this case the ratio of molecules in stoichiometric composition of reaction partners is exactly 2:

\frac{n_{O_{2}}}{n_{{CH}_{4}}} = 2

Where:

n_O ₂

number of molecules of oxygen

n_CH ₄

number of molecules of methane

The oxygen demand for pure methane is therefore:

n_O ₂

2 ·n_CH ₄ with a reference value of [n_CH4 ]= 1 kmol

F1The value of S _λ may be determined from the ratio of the ratio of the stoichiometric composition of oxygen and methane to the ratio of the stoichiometric composition of oxygen and the fuel blend supplied to the engine, as set out in equation (9-5):

S_{λ} = \frac{(\frac{n_{O2}}{n_{CH 4}})}{(\frac{n_{O2}}{n_{blend}})} = \frac{2}{{(n_{O2})}_{blend}}

(9-5)

Where:

n_blend

number of molecules of the fuel blend

(n_O ₂)_blend

the ratio of the molecules in the stoichiometric composition of oxygen and the fuel blend supplied to the engine

Because air contains 21 % oxygen the stoichiometric air demand L_st of any fuel shall be calculated by means of equation (9-6):

L_{st,fuel} = \frac{n_{O2, fuel}}{0,21}

(9-6)

Where:

L_st,fuel

the stoichiometric air demand for the fuel

n_O ₂ _fuel

the stoichiometric oxygen demand for the fuel

Consequently the value of S_λ may also be determined from the ratio of the stoichiometric composition of air and methane to the ratio of the stoichiometric composition of air and the fuel blend supplied to the engine, i.e. the ratio of the stoichiometric air demand of methane to that of the fuel blend supplied to the engine, as set out in equation (9-7):

S_{λ} = \frac{(\frac{n_{O 2}}{n_{CH 4}}) ∕ 0,21}{(\frac{n_{O 2}}{n_{blend}}) ∕ 0,21} = \frac{{(\frac{n_{O 2}}{0,21})}_{{CH}_{4}}}{{(\frac{n_{O 2}}{0,21})}_{blend}} = \frac{L_{{st,CH}_{4}}}{L_{st,blend}}

(9-7)

Therefore, any calculation that specifies the stoichiometric air demand may be used to express the Lambda-shift factor.

Annotations

Textual amendments

F1: Substituted by Commission Delegated Regulation (EU) 2018/989 of 18 May 2018 amending and correcting Delegated Regulation (EU) 2017/654 supplementing Regulation (EU) 2016/1628 of the European Parliament and of the Council with regard to technical and general requirements relating to emission limits and type-approval for internal combustion engines for non-road mobile machinery (Text with EEA relevance).

Footnotes

f1: Stoichiometric Air/Fuel ratios of automotive fuels — SAE J1829, June 1987. John B. Heywood, Internal combustion engine fundamentals, McGraw-Hill, 1988, Chapter 3.4 ‘Combustion stoichiometry’ (pp. 68 to 72).