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Author Topic: Engine Displacement  (Read 17203 times)
Rabbid
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« on: November 24, 2012, 12:41:04 PM »

Hi All

I'm looking to try and work out what is used for the reference cylinder displacement when calculating load etc. Main reason is for asking is a 2.0 engine rebored or stroked out to a larger size worth writing home about.

I've searched so far but have been unsucessful.

Any pointers for dealing with changed displacement?

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Rick
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« Reply #1 on: November 24, 2012, 12:59:27 PM »

MLMAX - do a search Smiley
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Rabbid
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« Reply #2 on: November 24, 2012, 01:21:48 PM »

Thanks but that is maximum possible air flow for load calculation, which would be adjusted on boosted cars for example.

I'm on about the actual calculation of load

i.e

Load = Mass ingested over Mass Standard. From what I understand mass standard is displacement in cc multiplied by 0.00122521 g[cc]

http://s4wiki.com/wiki/Load

I am trying to workout where the ECU would get the displacement cc from

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Rabbid
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« Reply #3 on: November 24, 2012, 01:23:34 PM »

So far I've found posts relating to KUMSRL and KISRM but I'm not 100% sure on how these would relate to engine load.
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Bische
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« Reply #4 on: November 25, 2012, 11:31:42 PM »

KISRM converts rlroh_w to ps_w and KFURL converts ps_w into rl_w.

I believe you should scale KISRM to get ps_w in line with boost pressure(I dont remember the variable name off my head) and thus getting rl_w back in line. I could be wrong but that is how I would approach it.
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nyet
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« Reply #5 on: November 25, 2012, 11:52:51 PM »

KUMSRL converts kg/hr of air into load per rpm

It is inversely proportional to displacement.
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Rabbid
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« Reply #6 on: November 28, 2012, 02:16:28 AM »

Seems KRKTE uses displacement in its calculationt too.
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ABCD
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« Reply #7 on: November 28, 2012, 02:51:21 AM »

KUMSIRL=Vh/2578, where Vh is cylinder volume in Litre

KRKTE = Vh*50.2624/Qstat, where Qstat is Injector static flow in gm/min.
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ABCD
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« Reply #8 on: November 28, 2012, 02:52:25 AM »

Quote
It is inversely proportional to displacement.

Wrong

KUMSIRL in directly proportional to displacement.
« Last Edit: November 28, 2012, 11:23:41 AM by nyet » Logged
Rabbid
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« Reply #9 on: November 28, 2012, 07:09:04 AM »

Certainly seeming alot more difficult than a standalone ecu which just asks for engine displacement and number of cylinders  Cheesy
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nyet
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« Reply #10 on: November 28, 2012, 11:33:20 AM »

Wrong

KUMSIRL in directly proportional to displacement.


HMM. I rechecked this. As far as I can tell, KUMSRL (not KUMISRL) is, in fact, inversely proportional to displacement. I don't actually know what KUMISRL is.
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nyet
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« Reply #11 on: November 28, 2012, 11:40:56 AM »

Here you go. You were right, KUMSRL is proportional to displacement:

(((.001072 (kg / hr)) / (1.27500 (kg / (m^3)))) / (0.5 * (1 / min))) * 100 = 2.80261438 liters

Not sure why 2.8 though, and not 2.7

Am i using the wrong density for air?
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nyet
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« Reply #12 on: November 28, 2012, 12:05:47 PM »

Certainly seeming alot more difficult than a standalone ecu which just asks for engine displacement and number of cylinders  Cheesy

In any case, I think ABCD definitely has it right.

displacement/2578 will get you in the ballpark.
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ABCD
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« Reply #13 on: November 28, 2012, 11:55:26 PM »

Here you go. You were right, KUMSRL is proportional to displacement:

(((.001072 (kg / hr)) / (1.27500 (kg / (m^3)))) / (0.5 * (1 / min))) * 100 = 2.80261438 liters

Not sure why 2.8 though, and not 2.7

Am i using the wrong density for air?


Hi nyet,

I cud not get what have you calculated here. Pls elaborate.

density of air = 1.293 gm/L at 273K
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nyet
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« Reply #14 on: November 29, 2012, 12:07:45 AM »

(((.001072 (kg / hr)) / (1.29300 (kg / (m^3)))) / (0.5 * (1 / min))) * 100

Let google do the conversions for you Smiley

the .5rpm (1/min) is for 4 stroke (1 cyl filling = 2 cycles)

100 is for conversion to %
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