# Calculating the mass of CO2 from stack emissions

I am working on calculations for an LCA of an operation. The plant operator has provided me a report from an engineering firm who has measured the emissions from their stack. I am trying to estimate the amount of CO2 produced per year.

The total volume of gas emitted was 2.08Rm3 (25 degrees C, and 101.3 kPa) every two hours. The gas contains 1.30% CO2. and the plant runs 1450 hours/year.

I tried to find the density of CO2, which is 1.78g/L and multiplied volume (1,508,000 L/year) by (0.013) but I am unsure if this is correct.

The plant is using wood chips in a suspension burner and undergoing complete combustion. But in their report there is 10.40000 kg of carbon monoxide being produced.

Well the answer is different depending on what 1.30% $$CO_2$$ is. Is it by mass or by volume? I would think the most likely is volume, but with out knowing I'll just do both.

## Volume

Our volume of $$CO_2$$ would be: $$\frac{101300\ Pa*2.08\ m^3*0.013}{8.314\ m^3\cdot Pa\cdot K^{-1}\cdot mol^{-1}*298\ K}=1.106\ mol\ CO_2$$ The molar mass of $$CO_2$$ is $$44.009\ g/mol$$, so that would give us: $$1.106*44.009=48.66\ g\ CO_2$$ for every 2 hours.

## Mass

Assuming an atmospheric density of $$1183.9\ g/m^3$$ our mass of $$CO_2$$ would be: $$2.08\ m^3*0.013*1183.9\ g/m^3=32.01\ g \ CO_2$$ for every 2 hours.

Then we just multiply the values by $$1450\ hr/2*1\ kg/1000\ g$$ to get the kg of $$CO_2$$ per year. $$\frac{48.66\ g}{2\ hr}*\frac{1450\ hr}{year}*\frac{1\ kg}{1000\ g}=35.26\ kg/year\ CO_2$$ $$\frac{32.01\ g}{2\ hr}*\frac{1450\ hr}{year}*\frac{1\ kg}{1000\ g}=23.21\ kg/year\ CO_2$$

As a proofing method you can use the input of fuel to estimate the mass of CO2 produced.

First get the percentage mass of carbon in woodchips (around 50% according to a quick bit of searching)

Then multiply the mass of woodchips burned by that percentage and by

$$\frac{molar\ mass\ CO2}{molar\ mass\ C} = \frac{12+16+16}{12} = 3.66$$

If there is full combustion then this should come to the same number.