Turbo Efficiency Discussion

[TGP37]

disassemble the BIN as a motorola 68332 BIN, take about ~1 year to completely understand the program if starting from nothing, then you'll be close to "exact".

it's not an easy process.

It will be close enough, thanks Robertisaar. "motorola 68332", that I didn't know yet I know exactly what to do with that info.

edited: Had to clip out false info, the VIN is in the binary dump file, so is the PCM file but it is chopped up into pieces.

[robertisaar]

One item that has evaded me for some time is the exact formula my 97 PCM uses to determine IPW. If I could dissect the formula, I can understand how AE is determined and make adjustments from real data versus trial and error.

disassemble the BIN as a motorola 68332 BIN, take about ~1 year to completely understand the program if starting from nothing, then you'll be close to "exact".

it's not an easy process.

[TGP37]

I don't do this exact stuff for a living. I'm an engineer, but really all this stuff follows a very similar set of rules and math. To me I guess this is an extreme hobby. I'm building a 3.9L that I want to run to 8500 RPM. So getting all this stuff right is important to me. I suppose to an extent I enjoy this type of thing as much as the actual build.

As for the math. If you are really interested in it pickup a college level engineering thermodynamics book. Most of these equations are in there as well on general examples on how to use it. The tough part is applying them right. Which really just takes some algebra a bit of occasional calculus and a good understanding of what is actually happening.

In reality you'll probably never get an exact solution as there are simply to many variables to throw in the mix, but you can get a good approximation that you can correlate to test data and use to predict trends.

Most of the desktop dyno like programs use similar math (though without water injection) to generate a somewhat accurate estimation of performance. These codes are called 1D or 2D codes by the professional engine designers. The professionals use some super high end codes to look at things we probably have never heard of. Many of these codes are full 3D (3 dimensional) that can predict important engine performance parameters. Unfortunately the 3D codes are typically proprietary or very expensive and require some highly trained people to run them.

8500 rpm is nice!

I have some college physics books I read on occasion. Many times I find a formula and I don't know how to work it, the INet is a great resource these days.....can learn anything within reason.

One item that has evaded me for some time is the exact formula my 97 PCM uses to determine IPW. If I could dissect the formula, I can understand how AE is determined and make adjustments from real data versus trial and error.

[NateD4]

yes, I understand what you posted. Do you do this stuff for a living? Or is it an extreme hobby/interest?

Thanks for the math, I am moving this info to my excel file of formulas.

It's like this, a lot of people tune their AE by trial and error over time. It works well enough but I always attempt to grasp the solid mechanics/math behind a system before I mess with it. Knowing the finer details of water vapor presence is gonna help when I start injecting water/meth. (AE being an example)

ehh, I'm a geek! and proud of it, lol. Thanks again!

I don't do this exact stuff for a living. I'm an engineer, but really all this stuff follows a very similar set of rules and math. To me I guess this is an extreme hobby. I'm building a 3.9L that I want to run to 8500 RPM. So getting all this stuff right is important to me. I suppose to an extent I enjoy this type of thing as much as the actual build.

As for the math. If you are really interested in it pickup a college level engineering thermodynamics book. Most of these equations are in there as well on general examples on how to use it. The tough part is applying them right. Which really just takes some algebra a bit of occasional calculus and a good understanding of what is actually happening.

In reality you'll probably never get an exact solution as there are simply to many variables to throw in the mix, but you can get a good approximation that you can correlate to test data and use to predict trends.

Most of the desktop dyno like programs use similar math (though without water injection) to generate a somewhat accurate estimation of performance. These codes are called 1D or 2D codes by the professional engine designers. The professionals use some super high end codes to look at things we probably have never heard of. Many of these codes are full 3D (3 dimensional) that can predict important engine performance parameters. Unfortunately the 3D codes are typically proprietary or very expensive and require some highly trained people to run them.

[TGP37]

yes, I understand what you posted. Do you do this stuff for a living? Or is it an extreme hobby/interest?

Thanks for the math, I am moving this info to my excel file of formulas.

It's like this, a lot of people tune their AE by trial and error over time. It works well enough but I always attempt to grasp the solid mechanics/math behind a system before I mess with it. Knowing the finer details of water vapor presence is gonna help when I start injecting water/meth. (AE being an example)

ehh, I'm a geek! and proud of it, lol. Thanks again!

[NateD4]

Thank you Nate, I read it twice and saved the formulas for later use.

I like how water can replace the excess fuel to cool the pistons. In any engine finding that perfect AFR for power is ideal, having to dump extra fuel to keep the engine safe is counter productive power wise imho.

Is it reasonable to assume the water vapor (which just absorbed small amounts of combustion heat energy) in the exhaust condense, releasing heat promoting more exhaust velocity? As we know, heat is released when water condenses. As if the water is like a heat battery which puts the heat energy to better use in the exhaust.

Knowing the mass of water injected over time I believe I can mathematically figure out exactly how much heat is taken from the valves/chamber and calc how much extra heat is adding to exhaust velocity.

TGP37, water vapor has an equilibrium point at a specific pressure and temperature (called steam quality, which is expressed as a ratio of liquid to vapor; ie. a percentage).

Expansion (power extraction) results in lower pressure as the piston moves down the stroke. As the pressure drops so does temperature does too. So too does steam quality. Meaning more liquid and less vapor at lower pressure and temperatures. You call this condensation, however remember exhaust still can be 700+ degrees F leaving the combustion chamber.

In a perfect engine you could over-expand the gas with a super long stroke. This means the exhaust pressure would be lower than ambient, temperature could therefore also could be below ambient. Of course practically this would require a huge, heavy engine and a vacuum pump of some sort to create the lower pressure. The result would be higher thermal efficiency.

In theory headers can create lower than ambient pressure by kinetic energy effects. However not under boost as the exhaust pressure needs to be positive (relative to air) to drive the flow for the turbine wheel.



You can figure out energy over time using power=enthalpy times mass flow/unit time. The resulting units are energy/unit mass * mass/unit time = energy/unit time (the mass cancels out as mass/mass =1 in the same way 8/8=1 for those unfamiliar with algebra).

The exhaust should be cooler with slightly less pressure (compared to no water injection) as the water injection helps to extract more energy from the pistons downward expansion.

Does that make anything clearer?

[NateD4]

The water injection is typically injected either at the inlet to the turbo to atomize and cool the mixture or after the turbo. It can be both.

It does two things. Cools the intake air and helps to reduce detonation by absorbing fast compression heat (which typically aids detonation) and converting it to a slower release steam pressure which adds to the cylinder pressure while cooling a bit.

There is a point where water injection impedes combustion. So too much can hurt both performance and efficiency.

Water injection can help spool a turbo, but, that's not it's main intention.

The main idea is to cool the inlet air and help prevent detonation.

[TGP37]

Thank you Nate, I read it twice and saved the formulas for later use.

I like how water can replace the excess fuel to cool the pistons. In any engine finding that perfect AFR for power is ideal, having to dump extra fuel to keep the engine safe is counter productive power wise imho.

Is it reasonable to assume the water vapor (which just absorbed small amounts of combustion heat energy) in the exhaust condense, releasing heat promoting more exhaust velocity? As we know, heat is released when water condenses. As if the water is like a heat battery which puts the heat energy to better use in the exhaust.

Knowing the mass of water injected over time I believe I can mathematically figure out exactly how much heat is taken from the valves/chamber and calc how much extra heat is adding to exhaust velocity.

[robertisaar]

I know this isn't a full analysis of how to do every step but hopefully the concepts make more sense.

please don't post that.

can you make a "i am dumb" summary of all of that? i understood the last 4 paragraphs perfectly. and what i got out of it was that water injection in an engine that is already dealing with a lot of pressure in the cylinders can be quite useful. personally, if i was dealing with a turbo motor, it looks like a quick burst of water when boost is desired would be a quick way to spool up the turbine?

[NateD4]

I understand where I went wrong, lol. I was visualizing on a molecule basis and overlooked the density of molecules per cubic inch.

But still, researching more, I found objects in motion will travel further in humid air versus dry air, assuming other constants are the same. But I think that is because energy is required to change the direction of travel of a specific molecule and heavier molecules provide higher resistance. Like plowing through balloons versus lead balls. Instead of my earlier perception of density.

After reading more and started to learn Vanderwaals ideas. I wonder if the deviation from the Ideal Gas Law, roughly 1% for water and 0.3% for air, is expressing measurable results in the combustion process. As small as the deviation is, across billions of molecules, should express an even constant just as fission has measurable results from a seemingly random, chaotic process.

I got to think more on that.........

You might want to re-read what you wrote. Objects in humid air will not travel further in heavier air (unless you have some empirical data to back that one up).

I don't think the discussion of water injection requires quantum physics, at least at this level. You are more than welcome to calculate the molecular energy levels, however, a simpler approach is much easier.

The ideal gas law governs the relation between an ideal gas's density, pressure and temperature.

The common expression of course is PV=nRT where n is the mole weight. In aerodynamics we use PV=rhoT where rho is the mole weight of air and R is the proper constant for this case (as opposed to nR's constant R).

There is also a non linear temperature rise equation for compressible flow (which a turbo would use). In that case where you relate the total (dynamic) values to static values. In this case dynamic values are air's pressure, temperature and density while in mostion.

In the simplest form the Pressure measured in front of the car while moving is called Total pressure, which consists of static pressure + dynamic pressure. Each of these has a total, static and dynamic temperature and density associated with it. It might be easiest to think of these as non physical properties (though you can measure them).

( http://www.grc.nasa.gov/WWW/BGH/isentrop.html )

In order to relate these quanities to each other the following equation is used:

p/pt = (rho/rhot)^ gamma = (T/Tt)^(gamma/(gamma-1))

In that case gamma is 1.4 for air which is defined as the ratio of cp/cv where cp is the specific pressure coefficient and cv is specific volume.

This is where things get tricky. Air is easy as it has a gamma of 1.4 . combustion products have a different gamma which has to be calculated via some complicated combustion chemistry (which may actually involve a bit of quantum math). the end result is a function that relates gamma to air/fuel ratio.

Now to FURTHER confuse things water doesn't follow the ideal gas law at all. It has some experimentally determined equations that generate the Mollier diagram (or temperature pressure enthalpy diagram for water).

http://www.steamtablesonline.com/images/steam%20tables%20p-h%20diagram%20(large).png

In this diagram it is often easiest to look up the values. There are a number of ways to use it, but it basically relates pressure, temperature, internal energy and density of water and steam.

With the ideal gas equations a fuel injection computer can calculate mass flow if it knows the velocity, temperature and pressure of the intake air.

If you want to factor water injection into the mix to see how it's cooling effect you need to also look at the internal energy balance. Internal energy is called enthalpy (h on the Mollier diagram).

Interestingly enough you can calculate the internal energy of air using the equation h = cp*T, where h is enthalpy (measured in BTU/Lbsmass or KJ/KG). cp is the constant pressure constant and T is the current temperature.

(http://www.grc.nasa.gov/WWW/k-12/airplane/specheat.html)

So for a given throttle position and turbo boost I have the following data at my disposal: Temperature, Pressure, Velocity.

Now I need mass flow (lbsm/time or kg/time) of air. To calculate this I can use mass flow = rho * V * Area. If I know density (which is calculated by PV=rhoRT (again with a special value of R that contains the moles term). All I need is a known area, density and speed.

Once I have mass flow, pressure and temperature I can use cp of air and temperature to calculate h.

If I then take the enthalphy (h value) for air entering the intake (with or without turbo), I can then use the following:

h inlet *m air + h water injection (liquid) * m water = h intake * m air + h water (liquid) * m water + h water (vapor) * m vapor = h final * m total.

we can then use the pressure of the intake, and temperature of the water (as liquid) to calculate its initial enthalpy then find the steam & water equilibrium point (called steam quality) and calculate the portion of water that is liquid and portion that is steam.

The way this works is that to go through a phase change the water has to absorb energy from somewhere. In this case you have to guess a few times to find the steam quality and energy balance.

Once you find the balance you can figure out how much energy the air gave to the water and use the data to calculate the new air and water density. The result is a change in density, pressure and temperature of the entire mixture of water and air.

Without water injection it is common to cool the mixture with fuel (which of course is expensive and wasteful compared to water).

So with these equations and some trial and error you should be able to approximately calculate the change in air density and pressure based on the amount of water injected and the inlet air temp and pressure.





Water injection will increase mixture density and reduce detonation. It will also displace a small fraction of air.

The whole idea though is to keep the water just below the boiling point temperature until the piston starts to come up and heats it to the boiling point. Then when the combustion event happens some of the combustion heat is turned into steam pressure.


Interestingly enough if you compare two temperature rises between air and water. Say start from 150 F and raise the temperature to 500F. The difference in pressure between air and water & steam is large for equal temperature change..

That is the benefit of water injection. You give up a small volume of air when water is liquid in the chamber and end up with a HUGE volume of steam after it has phase changed, this adds to the total pressure in the engine.


This also allows you to run a higher compression ratio which means you can extract more energy from it (Work = pressure * volume * dVolume). Work * time is power.

I know this isn't a full analysis of how to do every step but hopefully the concepts make more sense.

[TGP37]

That's great Joseph, glad to hear things are panning out nicely.

The brief back pressure recorded from tip in, the data you collected, could be helpful somehow retuning the accel enrichment tables. I'm thinking the Decay Profile might be adjusted according to the transient back pressure changes....and maybe a little tweeking on the low TPS% side since the turbine represents a brief restriction, that restriction should take more time to overcome with less exhaust mass overcoming it.

Either way, my AE is good but I know it can be better. Adding a turbo should have some effect on AE but knowing exactly how could really tighten it up.

I don't understand why your trying to re-write the idea that hot compressed air is better than cool compressed air...

Unless I'm just reading this wrong.

Not so much hot air is better, rather how adding water vapor changes the molecular weight of the over all mass entering the engine. Specifically how molecular weight still needs to overcome normal (i think Newtonian physics its called) as the engine operates. It should, in theory, take less parasitic loss in VE to pull a lighter molecule into the cylinder. Replacing heavier air with lighter h2o should ease the energy required to draw in the intake charge.

Adding to the complexity of it all, polarity of the water seems to slightly disobey the Ideal Gas Laws more so then dry air (more like evade us to our understandings with current science). I'm trying to keep track of my thoughts, lol.

Water is a very mysterious force. I'm attempting to lay out the process and interaction on an atomic/molecular level.

[3400-95-Modified]

I don't understand why your trying to re-write the idea that hot compressed air is better than cool compressed air...

Unless I'm just reading this wrong.

[TGP37]

I understand where I went wrong, lol. I was visualizing on a molecule basis and overlooked the density of molecules per cubic inch.

But still, researching more, I found objects in motion will travel further in humid air versus dry air, assuming other constants are the same. But I think that is because energy is required to change the direction of travel of a specific molecule and heavier molecules provide higher resistance. Like plowing through balloons versus lead balls. Instead of my earlier perception of density.

After reading more and started to learn Vanderwaals ideas. I wonder if the deviation from the Ideal Gas Law, roughly 1% for water and 0.3% for air, is expressing measurable results in the combustion process. As small as the deviation is, across billions of molecules, should express an even constant just as fission has measurable results from a seemingly random, chaotic process.

I got to think more on that.........

[NateD4]

Your understanding of thermodynamics appears to be off. Density and pressure are related but not in the way you portray. The mole mass may be lower, however the effect the energy density has on the pressure is not directly related to the pressure.

I think what you are getting at is combustion reagent displacement. Whereby the volume of water displaces a volume of air which means less fuel to mix with the air. However the fault I think I see is in the phase change density difference during/after combustion.

[TGP37]

Interesting info I found pertaining to water/meth injection. As it is in physics, water has a molecular mass of 18 while air we breath has a molecular mass of 29. The water, h20, has two hydrogen atoms which is far lighter then oxygen and means water vapor is less dense then the air our engines ingest.

So injecting water vapor will actually reduce the density of the intake charge over all. The common misconception deals with waters inability to compress well due to it's state of matter regarding temperature.

So air will become less dense as humidity rises. I checked up on it, sure as anything it's true.

Injecting Water will not increase compression when the water is in vapor form. Maybe this is the mechanics behind the cooling effect it has? Seeming how temperature is proportional to density delta rate of changes (considering ambient losses over time). A drop in density would bring about a drop in temps. The air (higher density) will act on the water vapor (lower density) and results in the air decompressing against the vapor being compressed. So the water inherits temps and the air reduces temp in the balancing of differences. Then the water releases heat when condensing in the exhaust allowing to keep gases hotter for better velocity.

What do you guys think? It seems reasonable.