Constant calorific value problem

  • Thread starter Thread starter pomocnik89
  • Start date Start date
  • Tags Tags
    Constant Value
AI Thread Summary
To maintain a constant calorific value when mixing air and methane, it is essential to adjust the mole fractions of each gas accordingly. As the volume of air increases, the amount of methane must be calculated based on the desired calorific value in KJ/kg. A simplified formula can be derived from the calorific values of both gases, allowing for adjustments based on the specific volumes used. The key is to ensure that the ratio of methane to air remains consistent to achieve the target calorific value. This approach can be implemented in the Delphi visualization program as needed.
pomocnik89
Messages
1
Reaction score
0
Hello, I've got this problem:

Imagine I mix 2 gases - air + methane (e.g.) and I can change/control quantity/volume of each gas. So if I change the volume of air how much methane should I add to keep a constant calorific value KJ/kg? So for example if I add 200 litres of air I need to know how much methane to add to have a constant calorific value, and then if I add only 150 then how much gas i need to add etc.

Hope I explained my problem ... Just some simplified formula/equation would be ok. I need it for my Delphi vizualization program, so it hasn't to be "physicaly 100% correct" :-) Thank you very very much for your help !
 
Physics news on Phys.org
pomocnik89 said:
Hello, I've got this problem:

Imagine I mix 2 gases - air + methane (e.g.) and I can change/control quantity/volume of each gas. So if I change the volume of air how much methane should I add to keep a constant calorific value KJ/kg? So for example if I add 200 litres of air I need to know how much methane to add to have a constant calorific value, and then if I add only 150 then how much gas i need to add etc.

Hope I explained my problem ... Just some simplified formula/equation would be ok. I need it for my Delphi vizualization program, so it hasn't to be "physicaly 100% correct" :-) Thank you very very much for your help !
You need to maintain the same mole fractions.
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
comparing a flat solar panel of area 2π r² and a hemisphere of the same area, the hemispherical solar panel would only occupy the area π r² of while the flat panel would occupy an entire 2π r² of land. wouldn't the hemispherical version have the same area of panel exposed to the sun, occupy less land space and can therefore increase the number of panels one land can have fitted? this would increase the power output proportionally as well. when I searched it up I wasn't satisfied with...

Similar threads

Replies
19
Views
4K
Replies
23
Views
7K
Replies
12
Views
2K
Replies
9
Views
2K
Replies
32
Views
3K
Replies
7
Views
2K
Replies
14
Views
3K
Back
Top