Ballpark Intake Temperatures?

[Begle1]

Or I could just stop injecting the water as soon as it stops lowering the temperature and boost pressure...

[wannadiesel]

That would be the old-school cut and try method.

[mhuppertz]

There are several principles that govern the vaporization point of a liquid, in this case water.
Boyle's Law:
The volume of a gas is inversely proportional to pressure when the amount of gas and the temperature are held constant.
V = k / P
Therefore, P1 * V1 = P2 * V2

In general we can say that as boost pressure goes up, water wants to stay a liquid at a higher temp.

Also water has a "latent heat of vaporization" of 400 degrees F, which means that when water reaches vaporization temperature at sea level, it takes an additional 400 deg F to push it over the edge because state change from liquid to vapor is endothermic.

If you do the math, it looks like at high boost pressures will keep a lot of the water injection spray in a atomized, liquid state until the diesel explodes.

As the flame front propogates, the water evaporates slowing the pressure rise rate creating a condition where more of the fuels energy is converted to mechanical force and less wasted as heat. That's why exhaust temps are lower with wter injection.

[Begle1]

Times like this I wish I got something higher than an "F" in my Chemistry class...

One step up from Boyle's law is the ideal gas law, PV=nRT.

Assuming that I know the intake pressure and temperature, and I know the volume of the cylinders, I can calculate the number of mols of air in each cylinder.

(Pressure in PSI)x(Volume in Liters)=(number of moles)(Temperature in Kelvin)(1.20)

Number of Moles=(Boost Pressure (in PSI) x Cylinder Volume (in L))/(Intake Temperature (in K) x 1.20)

If I know the number of moles of gas in the cylinder and the temperature of that gas, then I can use the Kinetic-Molecular theory to determine the amount of energy in each of those moles. (This is one of the parts that I don't have much of a clue on.)

Average Energy per Molecule (in Joules) = (a Constant of proportionality) x (Temperature in Kelvin)

The constant of proportionality is (judging from Wikipedia):
(1/2) x (the degrees of freedom of light diatomic gas) x (Boltzman's Constant (in Joules / Kelvin))

There are supposedly 5 degrees of freedom in atmospheric air. So the constant of proportionality is (judging from Wikipedia):
(1/2) x 5 x 1.380E-23, or 3.45E-23

Substituting that value in the original equation:
Average Energy per Molecule = 3.45E-23 x Temperature (in K)
There are 6.02E23 molecules in one mole, so:
Energy per Mole = 3.45E-23 x Temperature x 6.02E23
Energy per Mole = 20.8 x Temperature (in K)

So that means I can calculate the energy in the air going into the cylinder. Based on the specific heat and latent heat of vaporization of water, I think I can determine exactly how much water that the intake air is capable of vaporizing into a gas.

BUT, water injection systems continue to work even after they're fumigating water that isn't vaporized. As said above, a lot of times the water remains a liquid until the fuel is injected; this means that there is liquid water present in the intake horn and throughout the compression stroke.

So all the math above, assuming that I didn't screw it up anyways, is nothing more than a starting point- you still are going to get performance increases by flirting with hydrolock and putting liquid water into the cylinders.

The $6000 question still is, I think, how much liquid water is the engine capable of handling before you get hydrolock. And I have no clue how to figure that out beyond experimental evidence. The good news about it, is once you know how much water locks the engine, that number shouldn't change much.

So who wants to sacrifice a few connecting rods in the name of science?

[seeker1056]

there is more to this than just latent energy/heat conversion

The answer yuor lookin for likely is when water turns to steam in the combustion chamber yuo are also breaking up the water molecule to some degree and you then have free hydrogen and oxgen if memory serves, one being an accelerator or enhancer of combustion and the other being a fuel.

Not so sure you can calculate for that one ;o)

[mhuppertz]

there is more to this than just latent energy/heat conversion

The answer yuor lookin for likely is when water turns to steam in the combustion chamber yuo are also breaking up the water molecule to some degree and you then have free hydrogen and oxgen if memory serves, one being an accelerator or enhancer of combustion and the other being a fuel.

Not so sure you can calculate for that one ;o)

Any hydrogen and oxygen created by cracking the H2O molecule is a net loss because it takes more energy to crack the bonds than can be obtained from oxidizing the hydrogen molecule. That why we don't all drive hydrogen cars right now...

[seeker1056]

who said anything about cracking the molecules??

and seperating water to its components is very easy all things considered - hell we used to do it in science class as a demonstration. a little electricity and a couple anodes - something your motor has by its very nature - dont beleive me - that s why all rad n heatercore manufacturers insist on makin sure the engine is properly grounded so natural electrolysis is prevented.

However - what I said was that in the conversion of water to steam, "some", (not all), free hydrogen and oxygen is created as part of the process.

So you have latent heat reduction, you have steam expansion (see steam engine principles), some free hydrogen and oxygen, and fuel ignition, all in varyin amounts which together make for more power.

They would beat a path to your door if you cracked the math to this one

;o)

[Murdawg]

The amount of water needed to hydraulic your motor would be in some relation proportional to the cup in the top of the piston and piston to deck clearance. I would think it would be less than the CC's of your compression area (for lack of a better word) because you would need some room for the fuel and air. I would think of it as increasing your compression to a point where the compression for would blow a rod through the side of your block. I don't know if you could calculate how much water would decrease Combustion chamber size to where your compression ratio would be too high and blow your motor apart. I think this would be your down fall and it would not be an actually hydro lock in sense but a raise in compression ratio to a point that the weakest link in your motor would fail. I bet this point would be different for a motor that was built to a stock.
I think it would be hard especially at higher RPM's to be able to get that much water into the motor (unless you are using a garden hose). At lower RPM's where the benefit of the water would not be there, a problem could arise.
Have not hear a lot of this stuff since Chem and Physics back in school.

[Begle1]

Liquid water in the cylinder will act like a higher-compression piston; you are displacing compressable air with a non-compressable liquid. At least, it will do that as long as it remains a liquid.

At BDC, each cylinder has a displacement of 59.9 cubic inches or .982 liters.
At TDC, with the stock 17.5:1 compression ratio pistons, each cylinder has 3.52 cubic inches of displacement, or .0577 liters.

So more than 3.52 cubic inches can never fit into a given cylinder. If you put 1 cubic inch of liquid water into a cylinder, that would be an equivalent to having 24.8:1 compression ratio pistons.

Does anybody know if 5.9's have ever come from the factory with higher-comp pistons? If they made some with 18:1 or 20:1 ratio pistons, that'd be something good to know.

What happens to liquid water at the time of fuel injection is either complicated or controversial.
The water absorbs heat when it turns to steam; that lowers the temperature of the cylinder.
Since the steam is a gas, the number of gas moles in the cylinder increases. More gas causes a higher pressure. However, since the liquid water is no longer reducing chamber dimensions, the more gas is throughout a larger volume. I'll get back to ya'll on what that math looks like.
Then there's guys who say that some of the liquid water (or some of the gaseous water) actually cracks apart into hydrogen and oxygen. How that works without an electric current beats me. But if the molecules did crack apart, that would result in a greater decrease in temperature (due to heat being absorbed to break the bonds) and a greater increase in the quantity gaseous moles (instead of 2 moles of water, there would be 2 moles of hydrogen and 1 mole of oxygen). The increase in moles would definitely result in pressure increases, as you aren't reducing volume to make a gas. And the hydrogen would improve combustion just like propane fumigation does; since it would be distributed throughout the cylinder at the time of fuel injection, it would aid flame propagation. But that is assuming the molecules "burn", which I don't understand.

Unanswered questions:
1. I know the temperature, pressure and number of moles in the cylinder. When I compress the cylinder, both temperature and pressure are going to increase. How do I know how much each is going to increase? If I knew the answer to this one, then I could calculate what my peak cylinder pressures and temperatures would be. I also need to know this to ensure that any liquid that is present at BDC isn't going to turn into a gas sometime before TDC.
2. Does anybody know what the maximum peak cylinder pressure is for the engine? What pressures the headgaskets or connecting rods are spec'd to be good for? If I knew that, then I could use liquid water to increase compression ratio under low boost conditions, and I'd always be able to run at the maximum pressure.

[wannadiesel]

Naturally aspirated 6b's have higher compression pistons.

[seeker1056]

how will you calculate for the flamefront dampening effect and the energy loss of the excess fuel that is the black smoke spewin out the tailpipe

You may know or be able to calculate forthe number of molecules on a theoretical scale, but how will you calculate for the varying losses of incomplete combustion

[Begle1]

how will you calculate for the flamefront dampening effect and the energy loss of the excess fuel that is the black smoke spewin out the tailpipe

You may know or be able to calculate forthe number of molecules on a theoretical scale, but how will you calculate for the varying losses of incomplete combustion

I'm trying to calculate how much liquid water can be in the cylinder, so that at no point during compression stroke the pressures in the cylinder exceed whatever the headgasket or connecting rod breaking point is.

Liquid water vapor is spread evenly throughout the cylinder; it should slow down ignition, because the fuel isn't going to have as much air in proximity.
So water injection alone should have a retarding effect; it is going to increase pressures by increasing compression ratio, but the fuel is going to burn slower. (If you wanted to make fuel burn faster, you would need to put in methanol, hydrogen or propane. Nitrous should slow down burn rate as well.)

So water slows down flame front propagation, and that makes it harder to blow a headgasket. And the affect the water has on compression ratio's is easily calculated.

The part that I cannot know accurately is how much fuel is being added and over how long of a period. If I knew that (say, if I had a CR engine and EFI Live), then I could know how much pressure increase would result from the burning fuel.
But as it stands with my engine, I only worry about how much the water is increasing the compression ratio, and I blissfully ignore the water's retardation affect so that I have some built-in error zone.

If you do it smart, water injection can retard timing and increase compression ratio at low boost and RPM- which is where you want retarded timing and can take the increased compression ratio. Then you wean off on the water as your boost increases. The affect would be most noticeable on a truck with low-comp pistons running lots of alcohol, which is were I'm going with this.

I would really like to know what the maximum operating pressures of our headgaskets are.