Another math question for you techmath gurus

[Guest bastaad525]

okay... someone humor me here. I know that what I'm asking for probably wouldn't be exact in actual practice, but I'm just looking for an idea. I know some of you guys here know all the equations for this stuff and are a hell of a lot better at math than me so can one of you guys take a crack at this just for the heck of it?

 

I'm trying to figure out, if you are moving a known amount of air, and a known amount of fuel, and you know you're AFR and see that it's too lean, how much LESS AIR would you need to bring it to where you want it.

 

 

Okay, so here are the numbers I would like computed

 

I'm running an L28 motor, so 2.8L displacement. On top of that I was running about 10psi boost. First thing I need to know is how much air the motor would be moving at 5000rpm

 

I'm running the stock injectors. I don't know how many cc's they are but I"m sure I'm the only one who doesn't here Lets just say for arguments sake that at 5000rpm, at WOT and 10psi, they were running 80% duty cycle (feel free to correct this number if you know I'm wrong... I actually doubt they are running 80% at this point but really have no idea). How much fuel would they be flowing?

 

Okay, now here's the meat of my question. On the dyno, at 5000rpm, 10psi, my air/fuel ratio was at 13:1. Now, I plan on getting some extra fuel in there at some point, but until then I've lowered the boost to be safe. But what I want to know is, mathematically, EXACTLY how much LESS air (to be more specific, how much would I have to turn down the boost) would I need to be flowing at WOT 5000rpm, given the exact same amount of fuel (again, whatever the stock injectors flow at 80% duty cycle), to bring my air/fuel down to a much more acceptable 12 or even 12.5:1?

 

Many thanks in advance to anyone who takes the time to do the math on this one

[Guest bastaad525]

wow 26 views so far and not one person wanted to bother with this huh? I know it's kind of useless and maybe a 'waste of time' but come on someone here has got to have a few spare minutes for a fellow Z-er!

[zolorin]

Lets do the math together

 

So you have 10lbs of boost on the L28 engine at 5000rpms and runing 13:1 Fuel to air ratio.

 

So lets say you are 80% VE (fair guess for this motor).

Now lets assume no pressure losses (fairly insignificant) due to the manifold/head restrictions (I know, this is not the case, but for the ease of math).

Since it is a four-stroke only 1.4L consumed each revolution of crank

Compressor efficiency 70%

 

I will assume that you are running an intercooler of 70% efficient, if you are not then see the math bellow

 

So lets do math

Volume air = RPMs * (1.4) = 7000 L / min

PR = (14.7+10) / 14.7 = 1.68 (not bad)

equivalent density from my Propulsion book

 

Pr (295K) = 1.31

Pr (compressed + ideal) = 2.20

enthalpy (ideal) = 40.1 Kj/kg

enthalpy (real) = 40.1 / 0.7 = 57.28 Kj/Kg

 

From the tables again

Temperature of inlet air = 360K (yep it is 190F) -- are you running hot

 

Temp drop from IC = 0.70*(360-295) = 45.5C

Temp in the manifold = 315 K (or 108F)

 

Now use ideal gas law for both conditions simplifies to

 

ro_2 = ro_1 * ( T1*p2)/(T2*p1) = 1.23 kg/m^3 (295*1.68)/(315*1) =

ro_2 = 1.57* ro_1 = 1.94 kg/m^3

 

So mass flow of air

m_dot = V*ro_2 * VE= 10.84 kg/min

 

now you are 13:1 with fuel so

m_fuel = 0.834 kg/min

Now you want 12.5:1 ratio

new mass flow of air = 10.43 kg/min

Need to decrease the air flow by about 4 percent

So from 24.7 psia (absolute pressure) need to go to 23.7 psia or 9lbs of boost (plus minus factor of error)

 

 

Oh yeah have fun (this was an abridged version since I did not include the temperature effects, but this is a good ball park figure

 

So dial at 9lbs and retest the your car

[Guest bastaad525]

holy cow....

 

 

see this is why I post these questions here... I'd never have been able to figure that out even knowing what equations to use

 

So 1psi of boost netted me .5 decrease in air to fuel ratio... that kinda sux

 

Well good enough... that's where I set my boost at after the head gasket went, anyways, so that should be safe, I hope, nice safe 12.5:1.

[zolorin]

I know this was long, and I had to check with my books on this one (tables, but not equations) But why not switch those injectors -- they are not flowing a lot. Also have fun with turbo

[Guest bastaad525]

But why not switch those injectors -- they are not flowing a lot.

 

 

heh it's just one of many mods on my list of things that are just waiting for me to win the lottery

 

injectors are a bit out of reach now and I got other things that need to be fixed first... safety issues so I can't put them off. But I will hopefully soon get an adjustable FPR and up the pressure a bit to get some extra fuel flow and take advantage of the better pump.

[TimZ]

Lets do the math together

 

 

Very impressive use of physics.

 

A couple of observations, though...

 

First, this only holds true if your injectors had gone static (i.e. 100% duty cycle and/or the MAF was pegged and no longer having an effect) to begin with at 10psi and remained so at 9psi. If the injectors are not static, then the ECU would probably decrease the fuel with the decreased airflow, and you'd be in the same place AFR-wise.

 

The ECU might ignore the MAF above some boost level or throttle position (not sure on this) - if so, then the assumptions should hold.

 

Second - it was nice to see such a thorough representation of the airflow, etc., but I think you made this a bit too hard. If you assume the injector duty cycle is the same in both cases, then you don't really need to know what the actual mass flow was - you just need to decrease it by ~4%. Since we are assuming that nothing else changed, the flow into the motor will be directly proportional to the manifold absolute pressure. So...

 

(10 + 14.7)* (12.5/13) = 23.75

 

...subtracting out 14.7psi for atmospheric pressure, you get 9.05psi.

 

Now, neither of us take into account the fact that the compressor will be running a bit more efficiently at the lower boost level and the compressor outlet temp will be lower due to the lower boost pressure, so the actual airflow won't decrease by the full 4%. You could probably get a good idea of the extent of this effect with the with the above equations, although you'd need a compressor efficiency map to get that part. The math does get kind of ugly, though, since now you'd be solving for two equations and two unknowns (final boost, given outlet temp and airflow reduction required).

 

I think it's safe to say that this effect won't be that big since we are only talking about a 4% change. Maybe go to 8.5psi just to be safe...

[Guest bastaad525]

okay, another question just out of curiosity, how efficient is the stock T3 at 10psi, and how efficient at 9psi? Just curios, not to relate it to my original question.

[zolorin]

My assumption was purely based on airflow and fuel flow, the rest can be done by the computer trickery.

 

Now back to the efficiency question (from Ray Hall turbo website)

http://www.turbofast.com.au/turbomap.html

for the T3-60 compressor efficiency for both 9 and 10 lbs is around 70% now for the T3-50 if would be around 60% and for those points. Other trim will not flow that much. So I assumed best case scenario for the turbos (biggest trim, which can be wrong), but the overall math process does not change.

The efficiency remained the same so my argument works for the error of 10-15% (which in heat transfer is considered very good).

So the best way is to go to the dyno, also bastard you have heard of stripping the injectors from Mustangs and others at the JY (very cheap )