Emission Factors

Over the last few years I have performed a lot of emission calculations and have found it frustrating that I had to spend hours searching for emission factors. I recently decided to compile a few short tables and carry them with me on a USB stick for easy access. Here they are. All the values can be found on the IPCC’s EFDB site or through a search engine. Different sources may have different values, but they should not be too far off. I cannot recall and reference all the websites I have used, but if I remember correctly, the values are from the late 90’s and early 2000’s. Since I have to do some editing and don’t have much time, I will break up the factors into a couple of posts.

Using the tables

CO2 = (Amount of fuel used) * (Calorific Value) * (Carbon Content) * (Oxidation) * 3.66

Oxidation values are usually given in percentages and are around 98% to 100%. I usually use 99% to be safe. 3.66 comes from the ratio of Carbon to Oxygen in CO2. If you do not understand why you need 3.66, try doing a line calculation and write out all the units. You are calculating the emission with respect to carbon, but don’t forget that you are emitting CO2 and not just C. Chances are you have come across simple conversion factors which take you from Fuel Used to CO2 emitted. The above calculation takes into account a few more factors and may vary slightly from the simple conversion factors. When doing the calculation remember to keep in mind the units.

Calorific Values, Natural gas-based fuels

Fuel	(GJ/kg)	(MMBTU/lb)
Natural Gas	0.051	0.022
Methane	0.0556	0.0239
Ethane	0.0519	0.0223
Propane	0.0504	0.0217
Butane	0.0495	0.0213
Isobutane	0.0495	0.0213
n-Butane	0.0494	0.0213
LNG	0.052	0.022

Carbon Content, Natural gas-based fuels

Fuel	(kg C/GJ)	(Thousand BTU/lb)
Natural Gas	14	33
Methane	13.5	31.3
Ethane	15.4	35.8
Propane	16.2	37.7
Butane	16.7	38.9
Isobutane	16.7	38.9
n-Butane	16.8	38.9
LNG	16	38

Hall-Heroult Process

Hall-Heroult Process 

The Hall-Heroult process is a continuous industrial process for the production of aluminum. The process is carried out in a molten cryolite salt bath. For more information on how the process works take a look at the wikipedia entry here. To calculate Al production, there are several reactions that one must look at. The first set of reactions is the dissolution of Al2O3 in the bath: 

Na₃AlF₆ -> 3Na+ + AlF₆^3- 

and 

4AlF₆^3- + Al₂O₃ -> 3Al₂OF₆^2- + 6F^- 

The production of Al is at the cathode through: 

Al₂OF₆^2- + 6e^- -> 2Al^o + 6F^- + O^2- 

And at the anode: 

C + 2O^2- -> CO₂ + 4e^- 

For the ease of calculating Al production, the reactions at the cathode and anode can be simplified to: 

Al3- + 3e- -> Al^o at the cathode 

and 

C + 2O^2- -> CO₂ + 4e^- at the anode 

Aluminum production: 

Textbooks on Electrochemistry may use different notation for the stoichiometry so be careful. The notation used for this case is:

a_j = species on opposite side of e^-

b_j = species on the same side as e^-

n_e = electrons 

Keeping this in mind you will have a_Al = 1, b_Al3- = 1 and n_e = 3. The Faraday`s law can be written in several different ways, but the one used here for the production of Al is:

I = -(n_e/a_Al)*F*(dn_Al/dt) 

where:  I = current at the electrode (the cathode is this case)

            F = Faraday`s constant, 96,485.3 C mol-1

            dn_Al/dt = the production of Al for a given time period 

Note: there is a negative sign, because electrons are used up. a_Al in the denominator has to be substituted with b_Al3- if calculating the required Al^3- ions for the production of Al^o. 

The equation can be changed to give the production of Al to:

dn_Al/dt = (|- I_cathode|*a_Al)/(n_e*F) in mol/s 

Why is it in mol/s? Take a look at the I/F term. The units for current are Amperes, which are C sec-1 and C mol-1 for the Faraday`s constant. So (C sec-1)/(C mol-1) = mol sec-1. Also, why is the current |-I_cathode|? It is that way, because we know that I = I _anode = |-I_cathode|. 

At this point you can put all of the knowns in the above equation with I being the only unknown: 

dn_Al/dt = (I*1)/(3*96,485.3) = I/289,455.9 mol/s 

Using the molecular weight of Al, we can convert the units to g/s: 

dn_Al/dt = (I/289,455.9)*26.9815 g/mol = I/10,727.94 g/s 

For example, if we have current of 150,000 amps, the production rate will be: 

dn_Al/dt = 150,000/10,727.94 = 13.98 g/s 

Don`t forget that there might be some side reactions taking place at the cathode, which will reduce the current at the cathode used in the production of Al. If the current efficiency is given, multiply the total current by it to obtain the cathodic one. 

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