Electriflying

On 2011-06-25, Orval Fairbairn wrote:
Well, Vaughn, all you have to do is a simple back-of-the-envelope
calculation to see it.

First, calculate the fuel and air consumed in an internal combstion
engine:
1. fuel, at 6 lb/gal
2. air, at f/a ratio of 1:15.

This means that, for every pound of fuel, you consume 15 pounds of air.
You may use kilograms, if you please.

Not correct, you don't consume the nitrogen which is around 80% of air.

If you are using batteries, you now have to carry the equivalent of both
the fuel and the consumable air to do the same job, or almost a ton of
consumables.

But you're comparing apples and oranges! Batteries do NOT work on oxidation.
The thing that charges the battery may indeed work by oxidising the
fuel (but equally it may not, it may be a nuclear power plant, a wind
turbine, a solar panel or whatever). You cannot compare how a battery
works to how fuel is burned. They just aren't the same thing at all.

Additionally your calculation on the air consumed is quite significantly
wrong. Not just because you forgot about the nitrogen, but when calculating
how something burns you must consider how many moles of a substance
is reacting, i.e. its molecular mass.

Gasoline is a mix of quite a lot of chemicals, but mostly things like
C6H12 and these numbers will come out the same for any alkane.

To burn one mole of C6H12, we need 12 moles of O2
1 C6H12 x 12 O2 = 6 CO2 + 6 H2O

6 of the O2 molecules coming in make 6 carbon dioxide molecules,
the remaining 6 O2 molecules are used to make the 6 water molecules.

So we use 12 moles of O2 per one mole C6H12.

One mole of O2 is 32 grams.
One mole of C6H12 is 84.2 grams.

So for every 84.2g of fuel we need 320g of oxygen. This means the ratio
by mass of oxidiser to fuel is around 3.8 parts oxidiser to each part
of fuel -- NOT 15 parts oxidiser to each part fuel.