Slart

Hi,

I'm building the fuel tank for the methanol of my nitrous/methanol system, and I'm trying to figure out how large the methanol fuel tank needs to be. What I want is to fill up the methanol tank, and have it not run out before the nitrous tank runs out.

I am trying to calculate how large the fuel tank needs to be, but I dont know enough about chemistry to complete the calculations. I've made a guess, but I may be way off base. Here's what I have so far:

Methanol is CH3OH, with a weight of 32 amu (gm/mol).
Nitrous is N2O (duh ), and has a weight of 44 amu.

Combustion of methanol with oxygen is 2 CH3OH + 3 O2 -> 2 CO2 + 4 H20.

In the above example, the 2 CH3OH has a weight 64, and the 3 O2 has a weight of 48

So, does this mean that for stoich I need to have a 64:48 (4:3) ratio of methanol to oxygen?

Since nitrous weighs 44 gm/mol, and the oxygen component of that is 16 gm/mol, using nitrous to get the oxygen for the above reaction means it will take 2.75 times as much as otherwise, so now I should be using 64:132 (64 grams of methanol to 132 grams or n2o).

So my, if I have 15 lbs of nitrous, I need 7 lbs of methanol (15 * 64 / 132). So, it would take roughly 1 gallon of methanol to fuel 15 lbs of nitrous at stoich.


I have only one semester of high school chemistry so I above my head here and I would be suprirsed if my calculations are correct. Does anybody here have enough chemistry knowledge to tell me?

Thanks,

Steve

ws6nick

iTrader: (1)

"3 O2" has a Fm of 96 not 48, the 4:3 ratio in this instance is for volume not molecular weight. the molecular ratio for the Rxn of CH3OH an N2O is 2:3:2:4, you are taking into account that N will also react with CH3OH not the O2 alone. So, basically you are using a N20 to get extra O2 into you engine. I dont know your set-up but wont that fry your O2 sensors (using meth)? I am a Chem./Psych major but know nothing about mixing nitro/meth ......not yet atleast. Just trying to help with what I know.

NXRICKY

iTrader: (1)

I need to know how much horse power you are trying to get to. You can never use a 15lb bottle completly. once that bottle uses 7.5 lbs. the flow changes and you do not get the same amount of poundage flow as you did at 15lbs.
Ricky

Slart

Ricky,

I'm going to jet the system for 300hp and PWM it with a nx maximizer so the HP through the system will vary with RPM from 100 to 300hp.

I will be attempting to set up the maximizer so that nitrous flow is proportional to RPM, so the system will add a constant 260 ft/lbs from 2000-6000 (100hp@2000, ramping up to 300hp@6000, using the maxmizer). I know the maxmizer is not designed for this (it varies duty cycle with time, not RPM), but I will be setting up the ramps to get this net effect.

Can you elaborate a bit on how (or why) the flow changes when you use half the bottle? Is this something to do with the empty bottle still containing gaseous nitrous, or something to do with pressure drop due to the evaporation, or something else entirely?


Thanks,

Steve

jimmyblue

iTrader: (11)

-If- you can keep the bottle contents at constant temp
then your pressure will remain constant and flow too.
But as you draw down this may not be so ideal a case.

Anyway maybe you should work your reaction based on N2O
which would be like

CH3OH + 3N2O => CO2+2H2O+3N2

So one mole of methanol for three of N2O. N2O being about 44g/mol
that puts a full 15-lb bottle at about 155 mol N2O so you would want
maybe 60 mol of methanol just for a little slack. Methanol is about
32 g/mol, call it 2kg of methanol. At roughly 0.8 g/cm3 you want
maybe 2.5L; a half-gallon tank would be a bit short, a 1-gallon well
over what you need. I would go with the 1-gallon.

Slart

Jimmy,

Thanks much for your thoughtful answer! You are a very impressive individual.

It's amazing how simple the math is when you put it that way I went through your calculations for my own education and it was very enlightening.

I have another question if you dont mind. When I first saw the reaction equasion you balanced above, my first (non chemistry savvy) reaction was "CO2? What's that O2 doing there? That means I'm not burning the oxygen!".

Of course after I thought about it for a few moments I realized that since I'm not driving a nuclear powered f-body, the oxygen will remain oxygen and what I really care about is that it's an exothermic reaction.

Then I got to wondering, how do I know it's an exothermic reaction? It's common sense that a chemical reaction between nitrous and methanol will make heat, but I'm just taking it for a given.

Is there a way to tell that's not above my head?

Thanks again,

Steve

jimmyblue

iTrader: (11)

I suppose it would be possible to end at other products
(like if you went over-rich you might get CO instead)
but just figuring that the carbon and hydrogen really
want to burn, and nitrogen couldn't care less, those
looked like the probable outcomes.

It's been a long time since chemistry class but there
is a chemical value that I think is called enthalpy,
which is sort of the "bound energy" in any given
molecule. If you got the enthalpies for the left side
ingredients and the right side products (scaled by
quantity), the right side products should add up to
a lower total energy if it is an exothermic reaction
and the difference is what's left over to burn rubber


I used my old CRC Handbook for the first time in
years, on this one. Handbook, that's pretty funny;
10lb of paper and 4" thick.

Z284U2TRY

iTrader: (5)

Just use your winshield washer resevoir for the tank.

NXRICKY

iTrader: (1)

The bottle volume will change how the boiling nitrous will effect the flow. When a bottle is full there is less area for the expanding nitrous or boiling nitrous to go, increasing the flow. A bottle that is only 1/2 full will have a much high volume of boil there fore moving less nitrous down the line. We have seen several bench tests very between a full bottle and a used bottle. This variance will depend on how much nitrous (shot) you are using.

The amount of fuel you will need for 300hp on alcohol is 251 pounds an hour,so divide by 60 by 60 you come up with .069 lbs. per second this only works at the 300hp setting. to Equal that fuel you will need 756 lbs. of nitrous an hour.
I think you diffently will have your work cut out for you. Trying to make a time based system match RPM.

Ricky
.

92CamaroReallySlow

CH3OH change in heat is -201.2 kJ/mol
3N2O change in heat is +244.8 kJ/mol
CO2 change in heat is -393.5 kJ/mol
N2 and H20 i think are both 0

So Heat of Products - Heat of reactants = Change in heat

-393.5 kJ/mol - 43.6 kj/mol = -437.1 kJ/mol meaning its exothermic

Then again I hate chemistry and its probably wrong

Slart

Thanks jimmy, ricky, and 92crs for your help.

Ricky,

can you verify that my understanding is right? At high rates of flow, as the bottle get emptier (has more empty space), it cant boil fast enough to keep pressure in that empty space?

Thanks,

Steve

jimmyblue

iTrader: (11)

Figure that all of the energy to boil has to come in
through the tank wall. Your volume draw-down rate
implies a cooling energy (power). The compensating
heating energy coming in through the wall, is going
be coupled less well, or well-coupled to less of the
N2O as the level drops. If you are applying no heat
input then volume discharge just drops the liquid
temp reducing vapor pressure. It's not the empty
space, I believe, but the lack of thermal make-up
for the expansion cooling in the short term.

I guess this is why you want a pressure-controlled
bottle warmer and, probably, to position it toward
the bottom of the bottle so it remains well coupled
to the juice. A thermostatic warmer does not control
to the real value of interest, a pressure-switch bottle
warmer is "closed loop" pressure control.

NXRICKY

iTrader: (1)

Correct.
Ricky

Slart

Thanks again for the help guys!

Steve