# Commission Delegated Regulation (EU) 2017/654 of 19 December 2016 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, Appendix 2

Extent: E+W+S+N.I.

Calculation of λ-Shift factor (Sλ)
1.
Calculation
The λ-shift factor (Sλ) shall be calculated by means of equation (9-1):
S λ = 2 1 − inert% 100 n + m 4 − O * 2 100
(9-1)
Where:
Sλ 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 CnHm representing the fuel hydrocarbons, i.e.:
n = 1 × CH 4 % 100 + 2 × C 2 % 100 + 3 × C 3 % 100 + 4 × C 4 % 100 + 5 × C 5 % 100 + .. 1 − diluent% 100
(9-2)
m = 4 × CH 4 % 100 + 4 × C 2 H 4 % 100 + 6 × C 2 H 6 % 100 + …8 × C 3 H 8 % 100 + .. 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.).
2.
Examples for the calculation of the λ-shift factor Sλ:
Example 1: G25: CH4 = 86 %, N2 = 14 % (by volume)
n = 1 × CH 4 % 100 + 2 × C 2 % 100 + .. 1 − diluent % 100 = 1 × 0,86 1 − 14 100 = 0,86 0,86 = 1
m = 4 × CH 4 % 100 + 4 × C 2 H 4 % 100 + .. 1 − diluent % 100 = 4 × 0,86 0,86 = 4
S λ = 2 1 − inert% 100 n + m 4 − O * 2 100 = 2 1 − 14 100 × 1 + 4 4 = 1,16
Example 2: GR: CH4 = 87 %, C2H6 = 13 % (by vol)
n = 1 × CH 4 % 100 + 2 × C 2 % 100 + .. 1 − diluent % 100 = 1 × 0,87 + 2 × 0,13 1 − 0 100 = 1,13 = 1,13
m = 4 × CH 4 % 100 + 2 × C 2 H 4 % 100 + .. 1 − diluent % 100 = 4 × 0,87 + 6 × 0,13 1 = 4,26
S λ = 2 1 − inert% 100 n + m 4 − O * 2 100 = 2 1 − 0 100 × 1,13 + 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 = 1 × CH 4 % 100 + 2 × C 2 % 100 + .. 1 − diluent % 100 = 1 × 0,89 + 2 × 0,045 + 3 × 0,023 + 4 × 0,002 1 − 0,64 + 4 100 = 1,11
m = 4 × CH 4 % 100 + 2 × C 2 H 4 % 100 + 6 × C 2 H 6 % 100 + .. + 8 × C 3 H 8 % 100 1 − diluent % 100 = 4 × 0,89 + 4 × 0,045 + 8 × 0,023 + 14 × 0,002 1 − 0,6 + 4 100 = 4,24
S λ = 2 1 − inert% 100 n + m 4 − O * 2 100 = 2 1 − 4 100 × 1,11 + 4,24 4 − 0,6 100 = 0,96
As an alternative to the above equation, Sλ 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 (Sλ) 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:
n O 2 n CH 4 = 2
Where:
nO 2 number of molecules of oxygen
nCH 4 number of molecules of methane
The oxygen demand for pure methane is therefore:
nO 2 2 ·nCH 4 with a reference value of [nCH4 ]= 1 kmol
The 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 = n O2 n CH 4 n O2 n blend = 2 n O2 blend
(9-5)
Where:
nblend number of molecules of the fuel blend
(nO 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 Lst of any fuel shall be calculated by means of equation (9-6):
L st,fuel = n O2, fuel 0,21
(9-6)
Where:
Lst,fuel the stoichiometric air demand for the fuel
nO 2 fuel the stoichiometric oxygen demand for the fuel
Consequently the value of Sλ 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 = n O 2 n CH 4 ∕ 0,21 n O 2 n blend ∕ 0,21 = n O 2 0,21 CH 4 n O 2 0,21 blend = 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.

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Source: https://www.legislation.gov.uk/eur/2017/654/annex/IX/appendix/2 (the official version)
Page: https://openlegislation.uk/eu-regulation-2017-654/annex-IX/appendix-2
Contains public sector information licensed under the Open Government Licence v3.0. Source: legislation.gov.uk, The National Archives.