Estimating boost from load


OK, In this thread https://www.evolutionm.net/forums/sh...d.php?t=297809, I was trying to find the relation of load to a mass airflow. Bez chimed in and verified that load was indeed a mass airflow value per rev. So, with the help of a few logs here and there, I came up with a quick, rough estimate of g/rev based on load. That was load/95.

Well, I remember a few people asking about boost estimates and with me coming from DSMLink, boost estimate was useful little tool for evaluating VE's, true intake temps (IC efficiencies), tuning GM mafs, as well as just a double check of your boost gauge.

This is going to be a bit of math, but for those who care, this should be able to give you a boost estimate, based on your logged 2-byte load (If you don't want to read all this crap, just look at the bolded equations at the bottom ). This is depending heavily on my original calculation of g/rev from load, but this may prove to be useful to fine tune that as well. Hopefully, this can result in an equation that we can put into our loggers.

Anyway, here goes:

First we need to understand a few basic equations. The first is the volumetric airflow through an engine:
Airflow (CFM) = PR[RPM*V.E.*Cid/3456]
PR=Pressure ratio=(boost in psi+14.7)/14.7
RPM = RPM of engine
V.E. = volumetric efficiency at RPM being measured
Cid=cubic inch displacement= 122 for our 2.0L engines

So, for a quick example, let's take 22 psi of boost at 5000 RPM. I know that our engines are at about 100% VE at this RPM, so I am using this for our example:

First, you find the Pressure Ratio for 22 psi of boost:
(22+14.7)/14.7=2.5 PR

Now, we run the airflow (cfm) equation:
Airflow (cfm) = 2.5[5000*1.0*122/3456]=441.26 CFM


Now, this is where our true intake temperatures come into play. To convert air from a volumetric airflow to a mass airflow, we need to know the temperature, or more specifically it's density. This is because the density of air decreases with increasing temperature. Using the ideal gas law, PV=nRT, we can see that density(n/V)=P/RT.

So, to get our CFM value to a lb/min value, we have to multiply CFM by density. So, we get this equation:

Airflow(lb/min)=Airflow(CFM)*P/RT
P=atmostpheric pressure in PSI
R=ideal gas constant=10.7316 ft3·psi· °R-1·lb-mol-1
T=temperature in R (F +460)

Using our 441.26 CFM and a true intake temperature of 40F (let's say for winter), we get:
441.26*14.7/(10.7316*500)=1.21 lb/min*mol * 29(g/mol for air) = 35.09 lb/min

Using this same 441.26 CFM and 140F true intake temps (for a hot summer day or inefficient IC), we get:
441.26*14.7/(10.7316*600)=1.01 lb/min*mol * 29(g/mol for air) = 29.29 lb/min

That's roughly a 6 lb/min difference from 40F to 140 F intake manifold air temps. That's roughly 60 HP, all at the same boost level! This shows how important good intercooling is and the effects of a cool day compared to a hot summer day if your IC isn't up to the task.


OK, now this is where my original equation that relates mass airflow/rev and load come in. Using those same to mass airflow values that we just calculated, 35.09 lb/min and 29.29 lb/min, we have to convert them to g/rev. That's pretty easy. Just multiply them by 454 and divide by RPM. So, we get:

35.09 lb/min *454/5000=3.186 g/rev
29.29 lb/min *454/5000=2.641 g/rev

Remember, my original correclation was load/95=mass airflow per rev in g/rev So, let's just multiply those two numbers by 95 to see what load our loggers should report:

3.186g/rev*95=303 load
2.641g/rev*95=251 load


So, simply from a change in temperature (the actual intake manifold air temperature), I have shown that 22 psi of boost at 5000 RPM can vary with a load of 251-303, while providing a mass airflow of 29-35lb/min. So, if all of this math holds true and Bez's diassembly is correct, we can't simply correlate boost to load. We have to come up with a single equation using all of the above equations to estimate boost. Sort of working backwards from load with the equations that I just have shown and solving for boost. I went from 22 psi to get to a load number. Now, we have to go from a load number to a boost number.


So, without boring you with more math, I think this equation can give us a boost estimate based on our 2-byte load:

BoostEst = [(load *10.73 * T *3456 *14.7)/(95*454*29*Patm in psi*VE*122)]-14.7
That is with most of the terms kept separate. If the 95 constant proves to be OK and you are 2.0L engine, it can be simplified to
BoostEst=[(load*T)/(P*VE*280)]-14.7
T = actual intake air temp (not MAF temp) in degrees Rankin (degrees F +460)
P = atmospheric pressure in psi at the MAF(baro)
VE = VE of engine at the RPM being measured


I just wrote this quickly and didn't have much time to check, but let's start plugging in some numbers and see what we get from a bunch of different people. You will have to guess your VE, but if you pick 5000PM, then the VE should be about 1, so the VE shouldn't come into play. The only other 'guess' is your intake temp, which isn't the temp at your MAF. Also, remember that the baro will decrease quite a bit during a pull, so put in your value that you logged for baro at that RPM (in psi).

I'll be back later to see how this seems to be coming along.



Eric

 

Insightful!

 

Eric loves math haha

 

OK, just using some fake numbers for a few examples, this is what I'm looking for to see how this equation holds up.

Give me your boost at 5000 RPM and intake air temp (MAF is fine - actual will be higher). If you have your baro reading too at that point, that would be great.

So, the simplified equation at this point is:
BoostEst=[(load*T)/(P*VE*280)]-14.7

So, let's say that we have a load of 310, 5000 RPM, baro of 14.5 psi, and an intake temp of 100F (guessing based on MAF intake temp of 80F):

BoostEst=[(310*560)/(14.5*1*280)]-14.7
BoostEst=28 psi


Let's use the same load and RPM of 310 and 5000RPM, but now this car has a more restrictive intake, so a baro of 14.0 psi and a hotter day or worse intercooler, so an intake temp of 140F.

BoostEst=[(310*600)/(14.0*1*280)]-14.7
BoostEst=32.75 psi


Finally, let's use that same load and RPM again, but a nice cool day with a great intercooler, so an intake temp of 60F and a baro of 14.5 psi again.
BoostEst=[(310*520)/(14.5*1*280)]-14.7
BoostEst=25 psi



Eric