[Mgs] Engine maths...and spare time
Barrie Robinson
barrob at bell.net
Sat Mar 28 08:02:33 MDT 2020
Hello Richard,
One must take into account the retrograde wave peak when reaching a
maximum pressure in the cylinder. Another consideration is the
displacement of the primary components when the pressure wave is at its
maximum value. Using the Laws of Charles Breindigger then one has to
assume that the MEP has peaked at top dead centre and thus decline
occurs in burn speed but reduced in proportion to the ambient
temperature of the incoming air. Failing considerable testing it must
be assumed that the burn will reduce to zero when mechanical components
phase into total inactivity. But then this is just thinking off the top
of my head.
Cheers - just a thought:-)
Barrie
On 3/28/2020 9:48 AM, Richard Lindsay wrote:
> Too fun Barrie! Thanks. The biggest unknown is dynamics of pressure
> rise versus piston position.
>
> On Sat, Mar 28, 2020, 8:38 AM Barrie Robinson <barrob at bell.net
> <mailto:barrob at bell.net>> wrote:
>
> Hello Richard,
>
> My oath, I have just got out of bed - so pre-breakfast & coffee.
> I read your dissertation with bleary eyes and a befuddled brain
> but I followed your augment but just down to the last
> paragraphs. So decided that breakfast was more important than
> erudite discussions - thus am going to put down my teddy bear and
> discard my night cap and make breakfast with the help of this
> chick who slept with me last night .
>
> So cheers everyone and I look forward to academic messaging.
>
> Barrie
>
> On 3/28/2020 8:54 AM, Richard Lindsay via Mgs wrote:
>> Hello friends,
>>
>> When one is a geek, one thinks of geeky things. I am a geek
>> and this house-bound morning I woke up thinking about ignition
>> timing. Here are the details.
>>
>> We know that the charge (fuel plus air) in a cylinder doesn't
>> burn instantly, despite our perception to the contrary. Rather,
>> it takes a finite length of time from the occurance of the
>> 'spark', the flame front to cross the combustion chamber, and to
>> raise the MEP (Mean Effective Pressure) to a maximum - the point
>> where it does the most work. But how much time?
>> Physics problems always start by listing the 'known' and the
>> property to 'find'. So in this case,
>>
>> KNOWN:
>> Idle speed: 900rpm
>> Idle timing advance: 4° BTDC
>> Speed at maximum advance: 3500rpm
>> Maximum timing advance: 32° BTDC
>>
>> FIND:
>> Time from spark to MEP
>>
>> The first thing one might know is that the goal at idle is not
>> to produce maximum power. In fact, at idle 100% of the available
>> power is used to overcome the friction and other forces that
>> exist at idle speed. Stated another way: Idle speed is the
>> fastest the engine can achieve given the available charge. That
>> fact is evident (with carbureted engines) when one notices that
>> engine speed gradually increases, even for a fixed throttle
>> setting, as the engine warms and friction forces decrease. But
>> back to the problem.
>>
>> Because the goal at idle is smooth running and progression off
>> of idle (e.g. speeding up), not maximum power, the calculated
>> wavefront speed may not be correct at idle. But let's see.
>>
>> At idle speed, 900rpm in this MG TD example, the XPAG engine
>> is turning 900rpm or 900rpm / 60mps = 15rps (revolutions per second).
>>
>> Distributor speed is 1/2 engine speed so at idle the
>> distributor is turning only 7.5 revolutions per second. But
>> timing numbers are specified in degrees of crank rotations so we
>> will stick with 15rps.
>>
>> We don't know how fast the flame front travels across the
>> combustion chamber but we do know that maximum work occurs when
>> the piston is half way down the cylinder. And we also know that
>> work isn't an instantaneous parameter so it must begin before the
>> half way point and last past that point. Lots of unknowns and
>> theory doesn't always work in practice. But if we use the average
>> piston position at half-way down the bore, where most work is
>> most effective, and the MEP (Mean Effective Pressure), since Mean
>> is average, calculations begin.
>>
>> A single revolution is 360° so half-way down the power stroke
>> is 90°. Add the idle spark timing of 4° BTDC (Before Top Dead
>> Center) and we get 94° of crank rotation from spark to MEP at
>> half-way down. That's 94/360 or about 0.26 of an engine
>> revolution. And the engine is turning 15 revolutions per second
>> or 67ms (milliseconds) per revolution. So 67 x 0.26 = 17ms from
>> spark to MEP at half-way down the power stroke, at idle.
>>
>> If we repeat the calculations for operating engine speed and
>> at maximum advance, we get 3500rpm / 60mps = 58rps (revolutions
>> per second). Maximum advance is 32° BTDC so 90° + 32° = 122°,
>> spark to MEP or 122°/360° = 0.34 of a revolution
>>
>> 58rps is 17ms/r so 17ms/r x 0.34r = 5.78ms from spark to MEP
>> at half-way down the power stroke. This is a more representative
>> number than the 17ms at idle. One might even divide the idle
>> elapsed time minus the optimal time across the strike's midpoint.
>> Doing so would mean at idle, the pressure at idle becomes most
>> effective 5.6ms before half-way and for another 5.6ms after
>> midpoint. Interesting that the idle pressure application time is
>> about the same as the maximum pressure application time, or is
>> that circular logic?
>>
>> Yes everything above is ripe with assumptions and perhaps even
>> apocryphal and resplendent with errors, but it is only 7am after all.
>>
>> Anyone with extra house-bound time on their hands, please
>> check my maths and share your corrections, including the logic of
>> the whole experiment...or perhaps even why geeks think these ways!
>>
>> Rick
>>
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