Calculating the mass of CO2 from stack emissions

Question

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.

Answer 1

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.

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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.

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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.

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

Answer 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.