Question

The daily production of carbon dioxide from an 880 MW coal-fired power plant is estimated to be 31,000 tons. A proposal has been made to capture and sequester the CO2 at approximately 300 K and 140 atm.

a) Calculate the specific volume of CO2 at these conditions (in m3/kg).

b) Calculate the volume (m3) of CO2 that would be collected during a one-year period.

Answer 1

Hi
We can take the assumption that carbon dioxide generated behaves ideally and thus we can apply the ideal gas equation for it.

PV =nRT
Here P is the pressure.
V is the volume of the gas.
n is number of moles of the gas.
R is the universal gas constant.
T is the temperature in K.
Volume of 1 mole of the gas is known as the specific volume.
Here we can write.
V/n ( Specific volume ) = RT/P
P=140 atm
1 atm =101325 Pa.
T=300K.
Keeping these values we get
Specific volume =8.314*300/(140*101325)
=1.757*10^-4 m^3 /mole

Please note that the specific volume can be in terms of the volume per kg of the gas as well.
For conversion we know one mole of CO2 has mass of 44 g.
So in one kg we will have 1000/44= 22.72 moles
So the volume of one kg =1.757*10^-4*22.72
= 3.95*10^-3 m^3/kg

Part b
Daily production of CO2 = 31000 tons
1 tonn = 1000 kg
So, daily production =31000*1000 kg
=3.1*10^7 kg / day
Annual production =365*3.1*10^7 kg/year
1.13*10^10 kg /year

Volume of the CO2 generated = volume of 1 kg CO2*Number of kgs of CO2 produced.
=3.95*10^-3 m^3/kg*1.13*10^10 kg /year
=4.46*10^7 kg /year

There may be little deviation in the answer due to the taken approximately in the steps.
Hope this helps
Thank you