"HLS" <nospam@[EMAIL PROTECTED]
> wrote in
news:q5XTj.14417$V14.2005@[EMAIL PROTECTED]
>
> "Tegger" <tegger@[EMAIL PROTECTED]
> wrote in message
>>
>> Then MTBE is about twice as efficient as ethanol, so the mixture can
>> be leaner?
>>
>>
>>
>
>
> "Lean", to me, indicates a mixture which does not have an excess of
> fuel, or may even have an excess of the oxidant.
Then maybe I am using "lean" incorrectly.
My intent in the use of that word was to describe a mixture that
contained closer to the traditional emissions-desired stoich of 14.7:1
air/fuel, versus a "rich" mixture that was lower than 14.7:1 (more fuel,
less air).
>
> It requires more oxygen to burn a unit amount of MTBE than it does to
> burn the same unit amount of ethanol, and the energy released is
> consequently greater.
Because MTBE does not contain as much oxygen as ethanol...?
I think you can tell I'm no chemist.
>
> Most of the energy is derived from the oxidation of the hydrogen atoms
> in a hydrocarbon
> to form water. Less energy is derived from oxidation of the carbon to
> give CO2.
My faulty long-term memory coughs up something here... As I recall, the
original purpose of oxygenates was to reduce CO by increasing C02, this
to be achieved by making more oxygen available to the engine by packing
it directly into the fuel.
>
> The oxygen (in oxygenates such as alcohol and MTBE) just occupies
> space and doesnt
> contribute to the energy derived from combustion. So, since the
> ethanol has a higher percentage of oxygen in the molecule, it has a
> lower amount of energy that it can contribute
> upon combustion.
Hmmm. Then it appears to make a difference in combustion whether the
oxygen is contained within the intake air or contained within the fuel.
Why?
--
Tegger
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