Posted: Sat Mar 22, 2008 10:57 am
So, for a 20X mix using stoich. propane + air and assuming "20X" means the final pressure in the chamber is 20ATM.
Originally, the air in the chamber was 79.05% N<sub>2</sub> and 20.95% O<sub>2</sub>. (Ignoring the trace gases like water, etc.) But, we always add propane until the final percentage is 4.02%, so the N<sub>2</sub> and O<sub>2</sub> have to be decreased by 4.02%. This is the same as mutliplying by (1-0.0402)=0.9598.
So, we calculate the partial pressures of the three main gases;
P<sub>N<sub>2</sub></sub> = (0.7905)(20 ATM)(0.9598) = 15.174 ATM
P<sub>O<sub>2</sub></sub> = (0.2095)(20 ATM)(0.9598) = 4.022 ATM
The amount of propane doesn't get decreased by the 0.9598 factor since 4.02% is the final precentage we are aiming for.
P<sub>propane</sub> = (0.0402)(20 ATM) = 0.8040 ATM
Double check that the total pressure comes out to 20 ATM;
P<sub>total</sub> = P<sub>N<sub>2</sub></sub> + P<sub>O<sub>2</sub></sub> + P<sub>propane</sub> = 15.174 ATM + 4.022 ATM + 0.8040 ATM = 20.00 ATM
Double check that the final propane concentration is indeed the 4.02% we were aiming for;
P<sub>propane</sub>/(P<sub>N<sub>2</sub></sub> + P<sub>O<sub>2</sub></sub> + P<sub>propane</sub>) = 0.8040 ATM / (15.174 ATM + 4.022 ATM + 0.8040 ATM) = 4.02% propane
Now, the pressure that controls whether or not the porpane liquifies is the propane partial pressure, P<sub>propane</sub>, which is only 0.804 ATM. We can look at the graph Spudfarm posted and see that the boiling point at that propane pressure is well below 0°C. So the propane will not liquify in this 20X mixture at normal spudgun temperatures.
Originally, the air in the chamber was 79.05% N<sub>2</sub> and 20.95% O<sub>2</sub>. (Ignoring the trace gases like water, etc.) But, we always add propane until the final percentage is 4.02%, so the N<sub>2</sub> and O<sub>2</sub> have to be decreased by 4.02%. This is the same as mutliplying by (1-0.0402)=0.9598.
So, we calculate the partial pressures of the three main gases;
P<sub>N<sub>2</sub></sub> = (0.7905)(20 ATM)(0.9598) = 15.174 ATM
P<sub>O<sub>2</sub></sub> = (0.2095)(20 ATM)(0.9598) = 4.022 ATM
The amount of propane doesn't get decreased by the 0.9598 factor since 4.02% is the final precentage we are aiming for.
P<sub>propane</sub> = (0.0402)(20 ATM) = 0.8040 ATM
Double check that the total pressure comes out to 20 ATM;
P<sub>total</sub> = P<sub>N<sub>2</sub></sub> + P<sub>O<sub>2</sub></sub> + P<sub>propane</sub> = 15.174 ATM + 4.022 ATM + 0.8040 ATM = 20.00 ATM
Double check that the final propane concentration is indeed the 4.02% we were aiming for;
P<sub>propane</sub>/(P<sub>N<sub>2</sub></sub> + P<sub>O<sub>2</sub></sub> + P<sub>propane</sub>) = 0.8040 ATM / (15.174 ATM + 4.022 ATM + 0.8040 ATM) = 4.02% propane
Now, the pressure that controls whether or not the porpane liquifies is the propane partial pressure, P<sub>propane</sub>, which is only 0.804 ATM. We can look at the graph Spudfarm posted and see that the boiling point at that propane pressure is well below 0°C. So the propane will not liquify in this 20X mixture at normal spudgun temperatures.
