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Why is 3:2 (KNO3:C12H22O11) the best weight ratio for rockets and what is the equation?


bobes

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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.

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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.

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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 by John Cuthber
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---------- This is what CPropepShel tells for 150:100 ratio: ----------

 

CHAMBER THROAT EXIT
Pressure (atm) : 10.000 5.770 1.000
Temperature (K) : 1457.033 1368.268 1154.265

Isp (m/s) : 590.32222 1150.08875
Isp/g (s) : 60.19611 117.27641

 

CH4 7.3695e-006 9.3720e-006 1.6197e-005
CO 2.4772e-001 2.3932e-001 2.1335e-001
CO2 1.2421e-001 1.3140e-001 1.5630e-001
H 1.3607e-006 5.4951e-007 3.6607e-008
HCN 5.8517e-007 3.6625e-007 8.0481e-008
HNCO 2.4423e-007 1.4515e-007 2.7267e-008
H2 1.9358e-001 2.0207e-001 2.2809e-001
HCHO,formaldehy 2.4321e-007 1.4586e-007 2.7891e-008
HCOOH 3.4249e-007 2.0335e-007 3.8719e-008
H2O 2.3380e-001 2.2644e-001 2.0140e-001
K 3.2801e-004 1.5684e-004 1.6416e-005
KCN 1.9045e-006 9.0327e-007 8.1268e-008
KH 9.2841e-007 2.8737e-007 0.0000e+000
KOH 4.2806e-003 2.1437e-003 2.4746e-004
K2 1.4977e-008 0.0000e+000 0.0000e+000
K2CO3 4.8300e-006 2.2049e-006 1.7207e-007
K2O2H2 1.0039e-004 3.5809e-005 1.3074e-006
NH3 2.1401e-005 1.7850e-005 9.2672e-006
N2 9.9170e-002 9.9173e-002 9.9178e-002
OH 4.0358e-008 1.0771e-008 0.0000e+000
Condensed species
K2CO3(L) 9.6771e-002 9.7993e-002 0.0000e+000
K2CO3(b) 0.0000e+000 0.0000e+000 9.9049e-002

 

---------------------- and for the optimum 197:100: ----------------------

 

CHAMBER THROAT EXIT
Pressure (atm) : 10.000 5.821 1.000
Temperature (K) : 1673.300 1598.166 1380.030

Isp (m/s) : 591.87466 1169.40976
Isp/g (s) : 60.35442 119.24661

 

CH4 4.2282e-008 3.6128e-008 2.5866e-008
CO 1.7677e-001 1.7145e-001 1.5293e-001
CO2 1.6845e-001 1.6933e-001 1.7586e-001
COOH 9.4129e-009 0.0000e+000 0.0000e+000
H 1.0385e-005 6.5103e-006 1.1699e-006
HCN 1.2019e-007 7.5784e-008 1.6679e-008
HCO 1.7593e-008 0.0000e+000 0.0000e+000
HNCO 1.2569e-007 7.5346e-008 1.4125e-008
H2 9.2749e-002 9.8295e-002 1.1769e-001
HCHO,formaldehy 7.9391e-008 4.8660e-008 9.7228e-009
HCOOH 2.6167e-007 1.5544e-007 2.8789e-008
H2O 2.9234e-001 2.9100e-001 2.8275e-001
K 3.6727e-003 3.2278e-003 1.5141e-003
KCN 1.4064e-006 9.3495e-007 1.9852e-007
KH 1.0393e-005 6.4167e-006 9.2099e-007
KO 3.3645e-008 1.2641e-008 0.0000e+000
KOH 4.1552e-002 3.4130e-002 1.3626e-002
K2 9.7556e-007 5.4702e-007 4.3780e-008
K2CO3 6.0504e-005 4.7093e-005 1.4805e-005
K2O 3.4613e-008 1.4916e-008 0.0000e+000
K2O2H2 1.6017e-003 1.1131e-003 2.2933e-004
NH3 4.3591e-006 3.3794e-006 1.5265e-006
NO 3.2116e-008 1.2385e-008 0.0000e+000
N2 1.2353e-001 1.2353e-001 1.2353e-001
OH 1.5730e-006 7.4935e-007 5.1560e-008
Condensed species
K2CO3(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 by Enthalpy
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