addressing uncertainty in oil and natural gas industry greenhouse

E CO
2
6
500×
10 scf gas lbmole gas 44.01 lb CO2
tonne
= × × × ×
yr 379.3 scf gas lbmole CO 2204.62 lb
⎡⎛0.80 lbmole CH4
1 lbmole C 0.042 lbmole C2H6
2 lbmole C
⎢⎜
× + ×
⎢
lbmole gas lbmole CH4 lbmole gas lbmole C2H
⎜
6
⎢⎜
0.013 lbmole C3H8
3 lbmole C 0.004 lbmole C4H10
4 lbmole C
⎢⎜+
× + ×
⎢⎝
lbmole gas lbmole C H lbmole gas lbmole C H
⎢ 0.98 lbmole CO
2
formed 0.12 lbmole CO2
⎢× +
⎣ lbmole C combusted lbmole gas
27,400 tonnes CO /yr
E CO
2
=
2
2
3 8 4 10
⎞⎤
⎟⎥
⎟⎥
⎟⎥
⎟⎥
⎠⎥
⎥
⎥
⎦
For the uncertainty aggregation, start first with the uncertainty for the moles of carbon in the flared gas
stream (the terms in parenthesis) by applying Equation 4-4 and using the absolute uncertainties. The
uncertainty in the gas stream composition is 4%.
∑
U ( abs) = U ( abs)
∑lbmoleC
2
mole%
U ( abs) = (0.04× 0.80× 1) + (0.04× 0.042× 2) + (0.04× 0.013× 3) + (0.04× 0.004×
4)
∑lbmoleC
U ( abs) = 0.0322
∑lbmoleC
0.0322
Urel ( ) = 100% = 3.43%
∑lbmoleC
0.939
2 2 2 2
Then calculate the uncertainty of the product of the composition and the combustion efficiency by applying
Equation 4-4 and using the relative uncertainty.
U rel U rel U rel
2 2 2 2
( )
1
= ( )
composition
+ ( )
CombustionEff
= 3.43 + 20 = 20.3%
Next, account for the uncertainty of the CO 2 present in the gas by applying Equation 4-4 and using the
absolute uncertainty values. This will aggregate uncertainty for all the terms in the brackets.
U ( abs) = U ( abs) + U ( abs) = (0.203× 0.939× 0.98) + (0.04× 0.120) = 0.187
2 2 2 2
2 1 mole%
0.187
Urel ( )
2
= 100% × = 18.0%
0.939× 0.98 + 0.120
( )
Finally the uncertainty of the emissions is calculated by applying Equation 4-6 and using the relative
uncertainty values. For the CO 2 that is in the flared gas, the uncertainty in the activity factor is 15% based
on expert judgment.
Urel ( ) = Urel ( ) + Urel ( ) = 15 + 18.0 = 23.4%
emissions
2 2 2 2
AF
2
Pilot Version, September 2009 F-15