7730: What does cylinder volume table do?
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From: Export, Pa USA
7730: What does cylinder volume table do?
I noticed this table for the $8D code. Since I have a 396 should I increase that to .81 which is one cylinder volume in liters? What does this table control if the VE tables are set by the me?
Jason
------------------
89 Six Speed FX3 coupe
More Performance 396cid specs:
Callies 4340 crank 3.875 stroke
Lunati Pro-Mod 5.85" rods
SRP Pistons 14cc dish
Comp Cams Custom cam 550 lift 220/229 duration
10.8:1 compression
AFR 195 competition ported heads
Miniram and TPIS 1 3/4" headers
Speed Density conversion in process
Jason
------------------
89 Six Speed FX3 coupe
More Performance 396cid specs:
Callies 4340 crank 3.875 stroke
Lunati Pro-Mod 5.85" rods
SRP Pistons 14cc dish
Comp Cams Custom cam 550 lift 220/229 duration
10.8:1 compression
AFR 195 competition ported heads
Miniram and TPIS 1 3/4" headers
Speed Density conversion in process
SD calculates the amount of air entering the cylinder. It needs the cylinder volume as part of that equation. Here is a brief note on it that I found eduactional... (he addresses SD at the bottom)
<font face="Verdana, Arial" size="2">
It sounds easy, because it really is....
The math in the ECM is fairly simple. For instance, take a look
at MAF sensor data. If we are getting say 192 gps at 3600 RPM in a V8
engine, what is the amount of air in each cylinder for each firing?
It goes like this:
3600 RPM is also 60 RPS.
(192 grams/sec) / (60 rev/sec) = 3.2 grams per revolution.
Now, how many cylinders took in air in one revolution? The answer is
four, so with perfect cylinder distribution, each cylinder took in 0.8
grams of air.
Assuming we are in PE, and the AFR is 12.5 to one, how much fuel
do we need? 0.8 grams of air, divided by 12.5 AFR is 0.064 grams of
fuel. So how much time do we need to inject that much fuel? Easy.
Assume a 2.78 gps fuel injector (22 lb per hour).
0.64 grams * (1 Sec/2.78 grams) = 23.0 mSec. As we are probably
batch firing once per rev, that will be divided in half, and the
injector pw will be 11.5 mSec.
Not really that hard, is it? ;-)
> Rho=Manifold Air Density (g/litre)
> How does one calculate that?.
>From the sensor data.
> > Rho - Manifold Air Density (g/litre)
> > where: Rho = 1.202 * MAP / 101.325
> > MAP = Manifold Absolute Pressure (kPa)
What we really need in speed density is the same thing we
needed in the MAF system. What is the amount of air in the
cylinder? As an example, let's use a V8 engine with a 4 inch
bore and 3.48 inch stroke. ;-)
That works out to be 716.6 cc per cylinder, or 0.7166 L per cyl.
At 20 deg C and 101.3 kPa, air is 1.202 g per liter, so the nominal
air mass of the swept volume of the cylinder is 1.202g/L * 0.7166L,
or 0.861 grams of air in the cylinder at 100% VE, 20 deg C, and 101 kPa.
Assume 95% VE, and we get 0.818 grams of air, and say we get
to 100 kPa in the manifold, that reduces it down to 0.807 grams of air.
If the cylinder temp elevates the air temp some, we now may have just
0.8 grams of air in the cylinder. And you know how to get the right
amount of fuel in there now..... ;-)
Pretty simple, isn't it? What method do you like? :-) To me, it
depends on MAF accuracy, and on the other side, the routine used to
calculate air mass in the cylinder. My SD code needed help in the
equation, but I'll save that for another post.
Scot Sealander</font>
It sounds easy, because it really is....
The math in the ECM is fairly simple. For instance, take a look
at MAF sensor data. If we are getting say 192 gps at 3600 RPM in a V8
engine, what is the amount of air in each cylinder for each firing?
It goes like this:
3600 RPM is also 60 RPS.
(192 grams/sec) / (60 rev/sec) = 3.2 grams per revolution.
Now, how many cylinders took in air in one revolution? The answer is
four, so with perfect cylinder distribution, each cylinder took in 0.8
grams of air.
Assuming we are in PE, and the AFR is 12.5 to one, how much fuel
do we need? 0.8 grams of air, divided by 12.5 AFR is 0.064 grams of
fuel. So how much time do we need to inject that much fuel? Easy.
Assume a 2.78 gps fuel injector (22 lb per hour).
0.64 grams * (1 Sec/2.78 grams) = 23.0 mSec. As we are probably
batch firing once per rev, that will be divided in half, and the
injector pw will be 11.5 mSec.
Not really that hard, is it? ;-)
> Rho=Manifold Air Density (g/litre)
> How does one calculate that?.
>From the sensor data.
> > Rho - Manifold Air Density (g/litre)
> > where: Rho = 1.202 * MAP / 101.325
> > MAP = Manifold Absolute Pressure (kPa)
What we really need in speed density is the same thing we
needed in the MAF system. What is the amount of air in the
cylinder? As an example, let's use a V8 engine with a 4 inch
bore and 3.48 inch stroke. ;-)
That works out to be 716.6 cc per cylinder, or 0.7166 L per cyl.
At 20 deg C and 101.3 kPa, air is 1.202 g per liter, so the nominal
air mass of the swept volume of the cylinder is 1.202g/L * 0.7166L,
or 0.861 grams of air in the cylinder at 100% VE, 20 deg C, and 101 kPa.
Assume 95% VE, and we get 0.818 grams of air, and say we get
to 100 kPa in the manifold, that reduces it down to 0.807 grams of air.
If the cylinder temp elevates the air temp some, we now may have just
0.8 grams of air in the cylinder. And you know how to get the right
amount of fuel in there now..... ;-)
Pretty simple, isn't it? What method do you like? :-) To me, it
depends on MAF accuracy, and on the other side, the routine used to
calculate air mass in the cylinder. My SD code needed help in the
equation, but I'll save that for another post.
Scot Sealander</font>
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