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Calculating combustion temperature of mixture

  1. Jul 16, 2016 #1
    Hey guys, I'm trying to write a small article that very simply explains some of the math involved with a rocket engine. I'm not looking for a very accurate answer so I tried the following method

    1) Stoichiometry on the fuel which was sugar and saltpeter and their respective enthalpies of formation which I have as follows:

    Sugar (sucrose) - 2221.5 kJ/mol (5 mol)
    Saltpeter - 494 kJ/mol (48 mol)
    Water - 285 kJ/mol (36 mol)
    Nitrogen - 0 kJ/mol (24 mol)
    CO2 - 393 kJ/mol (55 mol)
    Potassium Carbonate 1150.8 kJ/mol (24 mol)

    Giving me 24720 joules for 5 moles of sugar combined with 48 moles saltpeter.

    5 moles of sugar and 48 moles of saltpeter weigh about 6.722 kilograms. So I divided 24720 kJ/Mol by that to get:

    3677 Kilo-joules released for every kilogram of sugar/saltpeter mixture.

    2) Now we know the amount of energy that will be released, we can work out how hot it will get. For example, if you switch on the kettle, it takes a certain amount of energy from the wall socket transferred to the water in the kettle to get it up to a certain temperature. The ratio of energy to rise in temperature is called a the heat capacity. And for water it's like 2.2 or something. That means for one kilogram of water to be raised by one degree Celsius, you need 2.2 Kilo-joules of energy.

    Every substance has its own heat capacity, and I'll find the average heat capacity of the burnt hot exit gasses below so we can predict the temperature.

    Heat capacity of

    Carbon Dioxide: 1.247 kilojoules to raise 1kg by a degree

    Nitrogen: 1.04 kilojoules to raise 1kg by a degree

    Water: 2.28 kilojoules to raise 1kg by a degree

    Potassium Carbonate (salt): 1.3768 kilojoules to raise 1kg by a degree

    I'm going to get the overall heat capacity of the exit gass mixture by multiplying the heat capacities of those 4 things by the fraction that they contribute to the total mass output. If we had 1 kg of mixture that was burned, then we will have 0.493 kg salt, 0.36kg Carbon Dioxide, 0.0964kg Water, 0.05kg Nitrogen. (This is the heat capacity at an expected 1800 degree temperature)

    Here is where I have the problem with the result I get:
    So, we have 3677 Killojoules, and we start off with 293 degrees Kelvin (room temperature). We raise it one degree for every 1.4 killojoules so we end up with a temperature of 2899 + 293 K which is almost twice the expected temperature.

    I haven't done these calculations much and I think there may be several things wrong with what I'm doing. Firstly, for simplicity I assumed constant heat capacity but didn't expect the error to be 80%. I may also be doing this completely wrong and have no idea what garbage I'm punching into Mathcad.

    If you guys have any suggestions on where my fault lies it would be a big help.

    Thanks in advance
  2. jcsd
  3. Jul 21, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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