[Mgs] Engine maths...and spare time
Barrie Robinson
barrob at bell.net
Sun Mar 29 09:21:34 MDT 2020
Hello folks,
I am really surprised that no one has blasted me for sending out a
message that was all (deliberately) gobbledygook. I had at least
expected an enquiry as to who Charles Breindigger was.
Cheers
Barrie
On 3/28/2020 7:30 PM, Richard Lindsay via Mgs wrote:
> Excellent. Thank you.
>
> On Sat, Mar 28, 2020, 6:15 PM Barney Gaylord <barneymg at mgaguru.com
> <mailto:barneymg at mgaguru.com>> wrote:
>
> I like the chart. Pressure peaks at 12d ATDC, and by 90d ATDC it
> is 90% gone.
>
> If you multiply sine of the angle by the pressure all alog the
> curve (after TDC), you get another curve representing progression
> of torque. Not a lot of torque yet at 10d, pretty good torque by
> 20d, peaking around 30d ATDC. Still a fair amount of torque at
> 60d, but by 90d (half stroke) the torque is nearly gone along with
> the pressure.
>
> Work being done would be represented by the area under the torque
> curve (not on the chart). Since both pressure and torque are
> nearly exhausted by 90d ATDC, there is very little energy (work)
> left to be harvested after mid stroke. That's why we like to open
> the exhaust valve early, to let what remains of the mostly useless
> fumes out of the cylinder, encouraging best intake of fresh air
> and fuel mix half a turn later.
>
> Barney
>
>
> At 03:32 PM 3/28/2020, Richard Lindsay wrote:
>> ...from Campbell's book.
>> ....
>> Attachment: 20200328_113435.jpg
>
>
>> On Sat, Mar 28, 2020, 2:07 PM Barney Gaylord
>> <barneymg at mgaguru.com <mailto:barneymg at mgaguru.com>> wrote:
>>
>> Rick, ---- Okay, time to spare, so I'll bite.
>>
>> You do a good job of calculating time from spark
>> event to half stroke (about 6-ms at road speed),
>> but I think you were asking a different
>> question. I thought you were asking how much
>> time to complete combustion to get to maximum
>> pressure. That is, how much time for the flame
>> front to propagate all the way across the combustion chamber?
>>
>> And you also said. "we do know that maximum work
>> occurs when the piston is half way down the
>> cylinder", which is not true. Most of the work
>> has already been done before the piston gets half
>> way down, and maximum torque on the crankshaft
>> happens significantly higher in the stroke before
>> pressure is lost to expansion.
>>
>> For best power and efficiency, combustion should
>> be completed at or slightly after TDC. But since
>> there is very little motion of the piston
>> immediately after TDC, It works just about as
>> well if max pressure comes just a little later,
>> like maybe 10 to 20d ATDC. That little delay can
>> allow use of higher compression ratio for better power and
>> efficiency.
>>
>> I like to use 3600 rpm for road speed, because it
>> divides evenly into 360 degrees rotation for nice
>> round numbers. And 900 rpm idle speed will be
>> exactly 1/4 of road speed. If you make spark at
>> 32d BTDC at road speed, it takes 1.5-ms to reach
>> TDC. 10d ATDC is at 2-ms, and 30d ATDC would be
>> 3-ms (after spark). So the flame front
>> propagation to complete combustion is in the 2 to
>> 3-ms range. I suppose this is the answer to your
>> question, "how much time" for the flame front to cross the
>> combustion chamber.
>>
>> Distance from the spark plug to far side of the
>> combustion chamber is about 2-1/2 inches, which
>> it does in about 2-1/2 ms, so flame front speed
>> is about 1 inch per ms, or 1000 inches per
>> second, which is fairly close to 60-mph. And you
>> night notice that I did not use "MEP" in that entire discussion.
>>
>> Barney
>>
>
>
>> At 08:54 AM 3/28/2020, you 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.
>> >
>> >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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