CO2 Emission

http://en.wikipedia.org/wiki/Stoichiometry

To all, I apologize for some errors in my previous post. Instead of using atomic weights, I used atomic numbers in part of the explanation. (That'll teach me to write too late in the evening!).

The specifics will need to be corrected, which I will do, but the main idea will be about the same. Now it is even later (and I sleepier); see ya tomorrow.

Thanks to Kyzine for catching this.

Correcting My earlier post #1 by using atomic/molecular weights:

H = 1, C = 12, O = 16 (ignoring the fractions).

Plugging in molecular weights, we have:

2 (58) + 13 (32) ---> 10 (18) + 8 (44); hence

116 propane + 416 oxygen ---> 180 water + 352 carbon dioxide.

Thus, burning 116 mass units of propane gives 352 mass units of carbon dioxide.

Not bad! only thing now is C₄H₁₀ is butane, not propane. Propane is C₃H₈.

Cheers.......Codey

Dear friend

We can simplified the combustion reaction by assuming

1. The hydrocarbon flared gas is paraffinic

2. hydrogen gas is negligible

With above assumption, the kg/h of CO2 released can be approximated by equation below:

W = [(MW - 2)/14] x [44/22.414] x Q

where

MW is the gas mixture molecular weight

Q is the flared gas flowrate in NM3/h

W is the flowrate of CO2 in kg/h

I purposely left the original numbers intact so that you can figure out what the numbers stand for.

Hope the above is what you are looking for.