air - fuel ratio

What are the thoughts on the air-fuel ratio values using
todays fuels. How much does the (varied) alcohol content of
todays fuels modify the 'de facto' standard of 14:1 as the
target mixture?

Dyno wrote:
> BobJ wrote:
>> What are the thoughts on the air-fuel ratio values using todays
>> fuels. How much does the (varied) alcohol content of todays fuels
>> modify the 'de facto' standard of 14:1 as the target mixture?
> Virtually all newer vehicles use three-way catalyst exhaust
> after-treatment systems. For maximum catalyst efficiency the A/F target
> average is stoichiometric (or the chemically correct mixture ratio). For
> 100% gasoline this is nominally 14.6:1. Most engine control systems
> purposefully dither on either side of stoich to maximize catalyst
> efficiency.
>
> As alcohol(ethanol)is added, the feedback ECU controls will adjust the
> fuel system to keep the mixture at the mixture's stoichiometric value.
> Here's a simple table to illustrate how stoich varies with %EtOH
>
> %EtOH Stoich A/F
> 0 14.6
> 5 14.16
> 10 13.74
> 15 13.35
> 20 12.98
> 85 9.55
> 100 9.00

Excellent info, thanks...

Dyno > wrote in news:481dcd15$0$20205
:

> BobJ wrote:
>> What are the thoughts on the air-fuel ratio values using todays fuels.
>> How much does the (varied) alcohol content of todays fuels modify the
>> 'de facto' standard of 14:1 as the target mixture?
> Virtually all newer vehicles use three-way catalyst exhaust
> after-treatment systems. For maximum catalyst efficiency the A/F target
> average is stoichiometric (or the chemically correct mixture ratio). For
> 100% gasoline this is nominally 14.6:1. Most engine control systems
> purposefully dither on either side of stoich to maximize catalyst
> efficiency.
>
> As alcohol(ethanol)is added, the feedback ECU controls will adjust the
> fuel system to keep the mixture at the mixture's stoichiometric value.
> Here's a simple table to illustrate how stoich varies with %EtOH
>
> %EtOH Stoich A/F
> 0 14.6
> 5 14.16
> 10 13.74
> 15 13.35
> 20 12.98
> 85 9.55
> 100 9.00



Interesting information. But no road-legal US/Canadian gasoline is 100%
gasoline.

How do the above numbers compare to fuels with other oxygenates, such as
MTBE? And what if the fuel contains MMT or TEL, neither of which are
oxygenates?

--
Tegger

Tegger wrote:
> Dyno > wrote in news:481dcd15$0$20205
> :
>
>> BobJ wrote:
>>> What are the thoughts on the air-fuel ratio values using todays fuels.
>>> How much does the (varied) alcohol content of todays fuels modify the
>>> 'de facto' standard of 14:1 as the target mixture?
>> Virtually all newer vehicles use three-way catalyst exhaust
>> after-treatment systems. For maximum catalyst efficiency the A/F target
>> average is stoichiometric (or the chemically correct mixture ratio). For
>> 100% gasoline this is nominally 14.6:1. Most engine control systems
>> purposefully dither on either side of stoich to maximize catalyst
>> efficiency.
>>
>> As alcohol(ethanol)is added, the feedback ECU controls will adjust the
>> fuel system to keep the mixture at the mixture's stoichiometric value.
>> Here's a simple table to illustrate how stoich varies with %EtOH
>>
>> %EtOH Stoich A/F
>> 0 14.6
>> 5 14.16
>> 10 13.74
>> 15 13.35
>> 20 12.98
>> 85 9.55
>> 100 9.00
>
>
>
> Interesting information. But no road-legal US/Canadian gasoline is 100%
> gasoline.
>
> How do the above numbers compare to fuels with other oxygenates, such as
> MTBE? And what if the fuel contains MMT or TEL, neither of which are
> oxygenates?

Not sure about MTBE, but MMT and TEL are added in such tiny percentages
to do their jobs that they really don't affect the stoichiometric ratio
in any measurable way. They're fractions of a percent by volume, not the
10% of ethanol in E10 fuel.

Steve > wrote in
:

> Tegger wrote:

>>
>> How do the above numbers compare to fuels with other oxygenates, such
>> as MTBE? And what if the fuel contains MMT or TEL, neither of which
>> are oxygenates?
>
> Not sure about MTBE, but MMT and TEL are added in such tiny
> percentages to do their jobs that they really don't affect the
> stoichiometric ratio in any measurable way. They're fractions of a
> percent by volume, not the 10% of ethanol in E10 fuel.
>
>
>


That's the answer I was looking for.

If volume is the determining factor, I'd guess MTBE would have a similar
effect on stoichiometric as ethanol, since it's added in concentrations up
to 15%.


--
Tegger

"Tegger" > wrote in message
> That's the answer I was looking for.
>
> If volume is the determining factor, I'd guess MTBE would have a similar
> effect on stoichiometric as ethanol, since it's added in concentrations up
> to 15%.

I dont think so, Tegger... Methyltertiarybutylether has a different ratio of
hydrogen to
carbon to oxygen.. It should not have the same stoichiometry as ethanol.

Ethanol is almost 35% oxygen by weight. MTBE is only about 18% oxygen.

"HLS" > wrote in
:

>
> "Tegger" > wrote in message
>> That's the answer I was looking for.
>>
>> If volume is the determining factor, I'd guess MTBE would have a
>> similar effect on stoichiometric as ethanol, since it's added in
>> concentrations up to 15%.
>
> I dont think so, Tegger... Methyltertiarybutylether has a different
> ratio of hydrogen to
> carbon to oxygen.. It should not have the same stoichiometry as
> ethanol.
>
> Ethanol is almost 35% oxygen by weight. MTBE is only about 18% oxygen.
>
>



Then MTBE is about twice as efficient as ethanol, so the mixture can be
leaner?


--
Tegger

"Tegger" > wrote in message
>
> Then MTBE is about twice as efficient as ethanol, so the mixture can be
> leaner?
>
>
> --
> Tegger


"Lean", to me, indicates a mixture which does not have an excess of fuel,
or may
even have an excess of the oxidant.

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.

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.

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.

The "octane rating" takes into account different properties of the fuel, not
directly related to
the stoichiometry.

jim wrote:
>
> HLS wrote:
>
>>"Tegger" > wrote in message
>>
>>>Then MTBE is about twice as efficient as ethanol, so the mixture can be
>>>leaner?
>>>
>>>
>>>--
>>>Tegger
>>
>>"Lean", to me, indicates a mixture which does not have an excess of fuel,
>>or may
>>even have an excess of the oxidant.
>>
>>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.
>
>
> Not according to the EPA. They say it contains less energy than ethanol
> per gallon. Density might be different but doesn't matter - no one
> purchases liquid fuel by weight.
>
>
>
>>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.
>>
>>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.
>
>
>
>
>
>>The "octane rating" takes into account different properties of the fuel, not
>>directly related to
>>the stoichiometry.
>
>
> But octane is far more important to fuel economy and performance. That is
> more important to how much energy goes to the wheels.
>
> According to the EPA Ethanol and MTBE are almost the same for oxygen
> content, octane and energy content. According to the EPA it takes 10
> percent ethanol to get the same oxygen as 11 percent MTBE.
>
> -Jim
>
>
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the amount of oxygen required to stoiciometrically burn a unit of fuel
is not directly related to the energy yielded by that process. those
are two different properties of a fuel.

nate

--
replace "roosters" with "cox" to reply.
http://members.cox.net/njnagel

"HLS" > wrote in
:

>
> "Tegger" > 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