bobes Posted July 20, 2013 Posted July 20, 2013 Hi, please can someone help me? I can´t explain why is 3:2 the best ratio for rocket. It is insane excess of sucrose, so HOw does it work? What equation describe this reaction? Thank you.
John Cuthber Posted July 20, 2013 Posted July 20, 2013 Are you familiar with this sort of thing? http://en.wikipedia.org/wiki/Stoichiometry
bobes Posted July 20, 2013 Author Posted July 20, 2013 Yes, but I have seen so many equations of this reactions with different products and I don´t know what products are the right ones.
Enthalpy Posted July 22, 2013 Posted July 22, 2013 This is a typical hot reaction. Combustions in air tend to make CO2 because 80% nitrogen cool the flame a lot. Not so with pure oxygen, nor with oxygen-rich reactants like KNO3. Heat decomposes CO2 partially, so the reaction makes both CO2 and CO. The best proportion, that brings enough oxygen to produce favourable CO2 but not excessive O2, is not really accessible to hand computation. More energetic reactions would also decompose H2O. Software exists because of that. You can try RPA and Propep: http://www.propulsion-analysis.com/downloads.htm http://www.dark.dk/download (seems 404, search alsewhere for CPropepShell) From CPropepShell, taking with no good reason 10 atm in the chamber and 1 atm down the nozzle, I get 197:100 mass nitrate:sucrose, with 0.15*CO for 0.18*CO2 down the nozzle. A higher chamber pressure allows more oxidizer. I suppose the fabrication process limits the proportion of nitrate. That's true for big launchers as well.
John Cuthber Posted July 22, 2013 Posted July 22, 2013 (edited) Very roughly, C12 H22 O11 + KNO3 ---> K2CO3 + CO2 +N2 +H2O Well, that equation is a pig, so I'm going to pretend that the reaction happens in two steps (it's not utterly unrealistic, there will certainly be several steps, and I'm only using it as a handy accounting trick.) First I assume that the sugar decomposes to carbon and water. C12H22O11 ---> 12 C + 11 H2O (I like carbohydrates-that equation is easy to balance) Now I will pretend that the carbon reacts with the nitrate cleanly to give potassium carbonate, co2 and water so here's the unbalanced equation. C + KNO3 --> K2CO3 + CO2 + N2 The potassium and oxygen are not likely to change their oxidation state. So the redox rection involves just the C and the N The N gains 5 electrons per atom and the carbon loses 4 So there must be 4 nitrogens for each 5 carbons (so that 20 electrons are lost and 20 are gained) 5C + 4KNO3 --> K2CO3 + CO2 + N2 The oxygens end up in two places, and so do the carbons so they are a bit hard to tally at the moment. Still, it's clear that the potassium atoms have to balance. 5C + 4KNO3 --> 2 K2CO3 + CO2 + N2 So we start with 12 Oxygens and 6 end up in the carbonate so the other 6 must be in the CO2: 2 oxygens each tells me there must be 3 CO2 5C + 4KNO3 --> 2 K2CO3 + 3 CO2 + N2 Tidy up the nitrogens (I could have done that earlier if I had liked) 4 Nitrogens on the left so there must be 2 nitrogen molecules on the right 5C + 4KNO3 --> 2 K2CO3 + 3 CO2 + 2 N2 Good! it all works out. So, I need to use 4 KNO3 molecules for every 5 carbon atoms And the sucrose provides me with 12 atoms of carbon. Ugh! Ok, it's a bit like finding lowest common denominators. If I have 60 carbon atoms I can share them out in bunches of 5 or 12. So, Multiply the second equation 5C + 4KNO3 --> 2 K2CO3 + 3 CO2 + 2 N2 by 12 to give 60 C + 48 KNO3 ---> 24 K2CO3 + 36 CO2 + 24 N2 And the first equation C12 H22 O11 ---> 12 C + 11 H2O multiplied by 5 gives me 5 C12H22O11 ---> 60C +55 H2O Add the two together, cancel out the carbons, and get 5 C12H22O11 + 48 KNO3 ---> 24 K2CO3 + 36 CO2 + 24 N2 +55 H2O Beurgh! Best check it balances Carbons 12*5 = 24*2+36 check Potassium 48 =24*2 Check Nitrogen 48 =24*2 hydrogen 22*5 = 55*2 Oxygen 11*5 =55 Thank [deity of choice] for that OK so we now have a balanced equation. 5 C12H22O11 + 48 KNO3 ---> 24 K2CO3 + 36 CO2 + 24 N2 +55 H2O Next question- does anyone actually believe that all the dozens of atoms involved all behave nicely and swap partners in accordance with that equation? Well, if they did the smoke would be white- because it would just be K2CO3 So, at best , it's going to be an approximation but never mind- it might not be too bad. 5 moles of sugar weighs 5 x 342.3 grams i.e. 1.71 Kg And 48 moles of potassium nitrate weighs 48 * 101.1 grams i.e. 4.85 Kg The ratio is about 2.8 to 1 The usual recipe (about 3 to 2 or 1.5 to 1) is quite a long way from stoichiometry - but that's because some of the nitrate doesn't get reduced all the way to nitrogen, it only gets converted to nitrite (KNO2). Odd as it may seem, I don't feel like going through all that calculation again. I'm going to cheat a bit. Firstly, I don't care what the products weigh. I just need the relative proportions of the two reactants. Remember the bit about counting the electrons transfered? Now, the change in oxidation state for the nitrogen is 2 (it goes from 5 to 3) And the change in oxidation for the carbon is still 4 (zero to 4) So there must be twice as many nitrogens as carbons So the reaction must be something like C + 2 KNO3 ---> something and so 12 grams of carbon react with 2*101.1 grams of KNO3 so each gram of nitrate reacts with 202.2/12 grams of carbon about 16.8 grams of nitrate per gram of carbon similarly, each sucrose decomposes to give 12 atoms of carbon 342 grams gives 144 grams 28.5 grams of sugar per gram of carbon so you can cancel the carbons and get 28.5 grams of sugar for 16.8 grams of nitrate. That's about 1.7 to 1: fairly close to 1.5 to 1, or 3:2 The moral of this story is that the 1.5 to 1 is a rough approximation to one possible course for the reaction. It's almost certainly been produced by trial and error. Also, it's quite possible that it melts more easily than the "proper" mixtures. Edited July 22, 2013 by John Cuthber 1
Enthalpy Posted July 23, 2013 Posted July 23, 2013 (edited) ---------- This is what CPropepShel tells for 150:100 ratio: ---------- CHAMBER THROAT EXITPressure (atm) : 10.000 5.770 1.000Temperature (K) : 1457.033 1368.268 1154.265 Isp (m/s) : 590.32222 1150.08875Isp/g (s) : 60.19611 117.27641 CH4 7.3695e-006 9.3720e-006 1.6197e-005CO 2.4772e-001 2.3932e-001 2.1335e-001CO2 1.2421e-001 1.3140e-001 1.5630e-001H 1.3607e-006 5.4951e-007 3.6607e-008HCN 5.8517e-007 3.6625e-007 8.0481e-008HNCO 2.4423e-007 1.4515e-007 2.7267e-008H2 1.9358e-001 2.0207e-001 2.2809e-001HCHO,formaldehy 2.4321e-007 1.4586e-007 2.7891e-008HCOOH 3.4249e-007 2.0335e-007 3.8719e-008H2O 2.3380e-001 2.2644e-001 2.0140e-001K 3.2801e-004 1.5684e-004 1.6416e-005KCN 1.9045e-006 9.0327e-007 8.1268e-008KH 9.2841e-007 2.8737e-007 0.0000e+000KOH 4.2806e-003 2.1437e-003 2.4746e-004K2 1.4977e-008 0.0000e+000 0.0000e+000K2CO3 4.8300e-006 2.2049e-006 1.7207e-007K2O2H2 1.0039e-004 3.5809e-005 1.3074e-006NH3 2.1401e-005 1.7850e-005 9.2672e-006N2 9.9170e-002 9.9173e-002 9.9178e-002OH 4.0358e-008 1.0771e-008 0.0000e+000Condensed speciesK2CO3(L) 9.6771e-002 9.7993e-002 0.0000e+000K2CO3(b) 0.0000e+000 0.0000e+000 9.9049e-002 ---------------------- and for the optimum 197:100: ---------------------- CHAMBER THROAT EXITPressure (atm) : 10.000 5.821 1.000Temperature (K) : 1673.300 1598.166 1380.030 Isp (m/s) : 591.87466 1169.40976Isp/g (s) : 60.35442 119.24661 CH4 4.2282e-008 3.6128e-008 2.5866e-008CO 1.7677e-001 1.7145e-001 1.5293e-001CO2 1.6845e-001 1.6933e-001 1.7586e-001COOH 9.4129e-009 0.0000e+000 0.0000e+000H 1.0385e-005 6.5103e-006 1.1699e-006HCN 1.2019e-007 7.5784e-008 1.6679e-008HCO 1.7593e-008 0.0000e+000 0.0000e+000HNCO 1.2569e-007 7.5346e-008 1.4125e-008H2 9.2749e-002 9.8295e-002 1.1769e-001HCHO,formaldehy 7.9391e-008 4.8660e-008 9.7228e-009HCOOH 2.6167e-007 1.5544e-007 2.8789e-008H2O 2.9234e-001 2.9100e-001 2.8275e-001K 3.6727e-003 3.2278e-003 1.5141e-003KCN 1.4064e-006 9.3495e-007 1.9852e-007KH 1.0393e-005 6.4167e-006 9.2099e-007KO 3.3645e-008 1.2641e-008 0.0000e+000KOH 4.1552e-002 3.4130e-002 1.3626e-002K2 9.7556e-007 5.4702e-007 4.3780e-008K2CO3 6.0504e-005 4.7093e-005 1.4805e-005K2O 3.4613e-008 1.4916e-008 0.0000e+000K2O2H2 1.6017e-003 1.1131e-003 2.2933e-004NH3 4.3591e-006 3.3794e-006 1.5265e-006NO 3.2116e-008 1.2385e-008 0.0000e+000N2 1.2353e-001 1.2353e-001 1.2353e-001OH 1.5730e-006 7.4935e-007 5.1560e-008Condensed speciesK2CO3(L) 9.9253e-002 1.0369e-001 1.1572e-001 so: - 197:100 would have been more efficient, but that little sugar would make the solid brittle. - same reason at launchers. Less polybutadiene would have improved performance. - CO is about as abundent as CO2, even at the optimum ratio. - Much unburnt H2 is left as well - typical for hot combustion. Edited July 23, 2013 by Enthalpy
bobes Posted August 3, 2013 Author Posted August 3, 2013 Oh my god thank you a lot this is awesome ... I will be translating this for a while but thank you.
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