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| pomocnik89 |
Posted: Apr 17 2012, 08:25 AM
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Newbie ![]() Group: Members Posts: 2 Joined: 17-April 12 Positive Feedback: 0% Feedback 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 !!! |
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| Robittybob1 |
Posted: Apr 17 2012, 06:29 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Power Member Posts: 6026 Joined: 15-October 11 Positive Feedback: 0% Feedback Score: 0 |
are you planning to burn the methane in the air? Or what? |
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| pomocnik89 |
Posted: Apr 17 2012, 07:13 PM
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Newbie ![]() Group: Members Posts: 2 Joined: 17-April 12 Positive Feedback: 0% Feedback Score: 0 |
... nope, I don't know why I wrote 'air', really stupid example ... I meant if I put 2 'random' gases with known caloric values and how would the total caloric value of the mixture change. E.g. I need to have a constant caloric value of the mixture, lets say 30 MJ/kg and i can control the amount of the two gases, so in what proportions I have to mix them to keep the total value of 30MJ/kg ... |
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| Robittybob1 |
Posted: Apr 17 2012, 07:29 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Power Member Posts: 6026 Joined: 15-October 11 Positive Feedback: 0% Feedback Score: 0 |
So a kg of this gas mixture is only the two gases and nothing else? It sounds like your are going to have to alter the pressure as well. For if one gas is Low and the other high, only one combination of low + high will give you that one result from 1 kg of the mix. Any variation will give you a different result, unless you can change the total amount e.g. by going over the 1 kg if the other mixture has less energy. This could be done if there was a third filler gas that didn't get involved with the reaction, that 3rd gas would allow you to adjust the total weight. |
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