# Nitrous oxide questions

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1. Sep 13, 2016

### Metals

Now I'm aware of how at around 300 degrees C N2O breaks apart, resulting in more oxygen in the cylinders.

I'm also aware that N2O cools down the air fuel mixture when it evaporates, increasing density and resulting in more volumetric efficiency.

What I don't get, is why not just take in more air? Is it because there is maximum cylinder displacement, so no more can be taken in, and using nitrous oxide is the only other way to add more oxygen when at maximum volume?

Is extra fuel automatically added to match the increased oxygen?

Where is the N2O sprayed? As in, at which point?

Many thanks.

2. Sep 13, 2016

### cosmik debris

The amount of air taken in is usually a little less than the volumetric displacement of the engine, this can be increased a little by designing the intake tract and the exhaust to slightly overfill the engine. To get any more air in you need to use a supercharger or turbocharger which compresses the air to get more in. Nitrous Oxide being 30% oxygen has a similar effect and is often called chemical supercharging. You could use pure oxygen of course but it causes very erratic burning of the fuel and can cause engine damage.

More fuel must be added at he same time as the N2O otherwise the engine will run lean. There are various ways of doing this if the engine is fuel injected as the solenoid valve opens and lets in the N2O you can increase the fuel pressure to suit. On carburettor engines a fuel pump and an extra fuel jet are used and the fuel and N2O fed to a spray plate under the carb or spray nozzles directly into the intake manifold.

3. Sep 13, 2016

### jack action

The N2O-fuel ratio is 64% of the equivalent air-fuel ratio. For example, if the air-fuel ratio is 14.7:1, then the N2O-fuel ratio for that same fuel is 9.4:1. This is a mass ratio.

The density of air is 1.225 kg/m³ and that of gaseous N2O is 1.977 kg/m³. So you can compare the volume of N2O to an equivalent volume of air to burn the same amount of fuel:

$$\frac{V_{N_2O}}{V_{air}} = \frac{\frac{m_{N_2O}}{\rho_{N_2O}}}{\frac{m_{air}}{\rho_{air}}} = \frac{m_{N_2O}}{m_{air}}\frac{\rho_{air}}{\rho_{N_2O}} = \frac{0.64}{1}\frac{1.225}{1.977} = 0.397$$

So N2O takes only 39.7% of the volume taken by air to burn the same amount of fuel. This mean you can have more oxygen in your cylinder. That is in its gaseous state. If you manage to inject the N2O in the cylinder in its liquid state ($\rho_{N_2O}$ = 1230 kg/m³), then the volume taken is about 0.06% of the equivalent volume of air.

Yes. You need fuel with your oxygen to create combustion.

4. Sep 14, 2016

### Kevin McHugh

You do not EVER want a lean condition of fuel and N2O. It will explode. I had an ancient flame AA spectrophotometer that was a little buggy when switching from air to N2O. Every once in a while the flame head would blow up because it switched over too lean.

5. Sep 14, 2016

### Kevin McHugh

PS: Keep your compression ratio at about 9:1, or you run the risk of severely damaging your engine. N2O injection doesn't like high compression ratios.