Let us take a quick look at ignition. Those who have a Heywood can look it up
- mines on loan so going by memory. The first thing that happens is a plasma
cloud is formed by the arc consisting of super heated electron stripped atoms.
When this cloud "explodes" a ball of high energy particles is shot outward.

The highest energy particles are the hydrogen atoms - and they penetrate the
charge about 5 times as far as the rest of the particles. As they lose energy
and return to normal temps - about 5000 k - they begin to react chemically
with any surrounding fuel and oxygen particles. The effectiveness of spark
ignition is directly related to the availability of free hydrogen. Molecules
containing tightly bound hydrogen such as methanol, nitromethane, and methane
are far more difficult to ignite than those with less bonds.

During combustion - water - H2O ( present and formed ) is extremely active in
the oxidation of the hydrocarbon. The predominate reaction is the following:

OH + H ==> H2O
H2O + O ==> H2O2
H2O2 ==> OH + OH
Loop to top and repeat.

The OH radical is the most effective at stripping hydrogen from the HC
molecule in most ranges of combustion temperature.

Another predominate process is the HOO radical. It is more active at lower
temperatures and is competitive with the H2O2 at higher temps.

OO + H ==> HOO
HOO + H ==> H2O2
H2O2 ==> OH + OH

This mechanism is very active at both stripping hydrogen from the HC and for
getting O2 into usable combustion reactions.

Next consider the combustion of CO. Virtually no C ==> CO2. Its a two step
process. C+O ==> CO. CO virtually drops out of early mid combustion as the O
H reactions are significantly faster and effectively compete for the available
oxygen.

Then consider that pure CO and pure O2 burns very slowly if at all. Virtually
the only mechanism to complete the oxidization ( Glassman - Combustion Third
Edition ) of CO ==> CO2 is the "water method".

CO + OH ==> CO2 + H
H + OH ==> H20
H2O + O ==> H2O2
H2O2 ==> OH + OH
goto to top and repeat.

This simple reaction accounts for 99% + of the conversion of CO to CO2. It is
important in that fully two thirds of the energy of carbon combustion is
released in the CO ==> CO2 process and that this process occurs slow and late
in the combustion of the fuel. Excess water can and does speed this
conversion - by actively entering into the conversion process thru the above
mechanism.

The peak flame temperature is determined by three factors alone - the energy
present and released, the total atomic mass, and the atomic ratio - commonly
called CHON for Carbon, Hydrogen, Oxygen, and Nitrogen. The chemical
reactions in combustion leading to peak temperature are supremely indifferent
to pressure. The temperatures and rates of normal IC combustion are
sufficient to cause most of the fuel and water present to be dissociated and
enter into the flame.

As can be seen above, water is most definitily not only not inert but is a
very active and important player in the combustion of hydrocarbon fuel.
Ricardo and others have documented that under certain conditions ( normally
supercharged ) water can replace fuel up to about 50% and develop the same
power output, or that the power output can be increased by up to 50% addition
of water. This conditions were investigated by NACA and others for piston
aircraft engines. It is important to note that these improvements came at the
upper end of the power range where sufficient fuel and air was available to
have an excess of energy that could not be converted to usable pressure in a
timely manner.

As a side note - Volvo recently released some SAE papers documenting the use
of cooled EGR to both reduce detonation and return to a stoic mixture under
boost in the 15 psi range - while maintaining approximately the same power
output. Notice - they reduced fuel and still get the same power output.

When you consider that EGR consists primarily of nitrogen, CO2, and water ( to
the tune of about two gallons formed from each gallon of water burned ), you
might draw the conclusion that it also was not "inert". They peaked their
tests at about 18% cooled EGR - which would work out to about 36% water
injection and got about the same results under similar conditions that the
early NACA research got.


I assumed an adiabatic reaction, so I just need to balance the energy in the system before and after vaporization:

Hlw(Tl0) + Hlm(Tl0) + Ha(Ta0) = Hgw(Te) + Hgm(Te) + Ha(Te)
where

Tl0 = initial temperature of the water/methanol solution
Ta0 = initial temperature of the air
Te = equilibrium temperature of the system
Hlw = enthalpy of the liquid water
Hlm = enthalpy of the liquid methanol
Ha = enthalpy of air
Hgw = enthalpy of gaseous water
Hgm = enthalpy of gaseous methanol

Enthalpy of Air
My first step was to determine the enthalpy of air at a given temperature. See CRC p.6-1 for the thermodynamic properties of air. I assume that the enthalpy curve is linear, that Cp for the pressure and temperature ranges of interest (1-3 bar and 0-200°C, respectively) is constant. I interpolated the table to 2 bar and 100°C, deriving Cp = 1.0165 J/g°K.

Ha(t) = 1.0165 t J/g
Enthalpy of Methanol
Next I did the same for methanol in both liquid and gaseous form. The CRC p.6-118 gives dHvap(25°C) = 37.43 kJ/mol and dHvap(64.6°C) = 35.21 kJ/mol, so I can get a linear vaporization curve as a function of temperature. Combined with Cp@25°C = 2.5312 J/g°C (CRC p.5-27), I use:

Hlm(t) = 2.5312 t J/g
dHvap(t) = -1.7497 t + 1211.97 J/g
Hgm(t) = 0.7815 t + 1211.97 J/g
Enthalpy of Water
Easy so far, eh? Now I looked at the steam tables, which derive enthalpy as a function of temperature and pressure, just like those for air, but appear to be less linear than those for air and methanol. I decided to do some curve fitting to get better H values for water. The steam tables in CRC (p.6-16) are too sparse for my tastes, so I found a more complete set on the internet (no reference). Using Matlab (a mathematics program from The Mathworks), I computed polynomial coefficients for curves approximating the given table values for temps from 0-200°C at 25° intervals. Here are the resulting equations:

Hlw(t) = 8.7193E-04 t2 + 4.0596 t + 2.2801 J/g
Hgw(t) = -1.3143E-03 t2 + 1.8799 t + 2501.44 J/g
A cubic fit actually produced better statistics, but I opted to use the quadratic representations so that I could solve the equilibrium equation more easily (besides, these are within about ą0.1% over the domain of interest so we aren't losing much).

Putting It All Together
The following quantities on the left side of the energy balance equation are constant from initial conditions (I'm adding the system masses at this point). Sum them up to get the total system energy, call it H0.

Hw0 = Mw Hlw(Tl0)
Hm0 = Mm Hlm(Tl0)
Ha0 = Ma Ha(Ta0)
H0 = Hw0 + Hm0 + Ha0
So, now we have the energy balance as a function of the equilibrium temperature:

Mw Hgw(Te) + Mm Hgm(Te) + Ma Ha(Te) - H0 = 0
= Mw (-1.3143E-03 Te2 + 1.8799 Te + 2501.44) + Mm (0.7815 Te + 1211.97) + 1.0165 Ma Te - H0

Calculate the quantities appropriate for finding the roots.

a = -1.3143E-03 Mw
b = 1.8799 Mw + 0.7815 Mm + 1.0165 Ma
c = 2501.44 Mw + 1211.97 Mm - H0

Pump it through the quadratic formula and, voila, equlibrium temperature.

Te = (-b ą sqrt(b2 - 4ac)) / 2a
Relative Humidity
The above assumes dry air, which is almost impossible to achieve. A more realistic model of the intake air would allow you to specify relative humidity, RH, as a parameter to the initial conditions. This can be done by estimating the vapor pressure of H2O at the ambient temperature, taking RH to indicate the proportion of that vapor pressure as the partial pressure of the H2O in the atmosphere and using those proportions for the intake air.

Antoine's Equation, ln p = A - B / (T+C), gives the characteristic curve for vapor pressures of liquids and provides a good approximation for water in mm-Hg with coefficients of A = 18.3036, B = 3816.44, C = -46.13 and T = temperature in Kelvins.

Given ambient temperature and relative humidity, we can compute the partial pressure of water vapor in the atmosphere.

Ppw = RH * Pvw
RH = proportional relative humidity, 0.0 to 1.0
Pvw = exp(A - B / (Ta+273.15+C)) * 101.325/760
Ta = ambient air temperature, from previous section
Some trivial algebra on Dalton's Law shows we can compute the mole fraction of the water vapor as the ratio of its partial pressure wrt the ambient pressure. Note that for air temperatures over the boiling point, the vapor pressure of water will exceed the ambient pressure (that's what defines boiling, after all), so you'll have to look out for Ppw > Pa.

moleFraction = Ppw / Pa
Pa = ambient pressure in kPa

This model answers many questions, such as the following.

What is the maximum amount of water that can be vaporized given a particular inlet air temperature?

How much cooling effect will result from maximum vaporization?

How much dilution of the intake charge results from the vaporized fluid? (In other words, how much oxygen is displaced by water vapor?)

What is the net effect of cooling and dilution? Is there a charge density benefit from water injection?

Limitations
Before we try to answer these questions, we should make sure we know the major shortcomings and assumptions of the model. First, it assumes complete vaporization is possible when injecting the specified mass of liquid. In reality, less than the saturation amount will be vaporized. How much? If I knew, I'd put it in the model. (Maybe I'll parameterize that, too, so it can be an input). This effect is probably benign, in that it reduces all effects proportionately.

Second, it assumes that the energy for vaporizing the fluid all comes from the intake air. This is almost always incorrect, as some of the liquid will almost certainly wet a hot surface in the intake tract, say the manifold, and pick up energy there. This will cause an increase in charge volume without its beneficial cooling effect, so is a bad thing and the model is overly optmistic in this regard. How much error will be caused by this effect? This is completely installation dependent, so there is no theoretical way to estimate how large the error will be; it must be determined empirically for each case.

If you have good atomization, this will help with vaporization considerable and avoid these first two problems.

Third, it assumes that the vapor pressure for methanol is infinitely higher than that for water, and thus assumes complete vaporization of the injected liquid, even for huge quantities of methanol. This is a shortcoming in the model that may be addressed in the future, but for now you must avoid high proportions of methanol in the intake liquid mixture.

Analysis
What is the maximum amount of water that can be vaporized given a particular inlet air temperature? See the column labelled Mass H2O. This indicates in grams how much water can be vaporized into 100 grams of air at the given inlet temperature. Way over on the right is the air-liquid ratio column, indicating the relative mass ratios of the injected liquid and air. Notice that for cool air (Ti < 25°C), you are already near saturation in terms of ALR, even in dry air; but for hot non-intercooled air, you can squirt in quite a bit.

How much cooling effect will result from maximum vaporization? The dT (delta temperature) column indicates how much the temperature drops due to vaporization of the injected fluid. Look at the O2 Cool column. This is the fractional improvement in charge density due to this cooling effect. All other things held constant, this is the increase in HP you would feel.

How much dilution of the intake charge results from the vaporized fluid? (In other words, how much oxygen is displaced by water vapor?) Look at the O2 Dil column and you see the fractional decrease in charge density due to air being displaced by water vapor. Note that this is referenced to ambient air, not corrected back to standard conditions. Dry air provides the worst case (we are vaporizing the most water per volume of air), so look at the top table. We see that HP would be reduced to 86% for an inlet temperature of 300°C with 10:1 water injection.

What is the net effect of cooling and dilution? Take the previous two answers and multiply them out to get the answer, it's in the column labelled O2 Final. For hot intake charges, the effect is quite considerable, so you can readily see why water injection works so well on non-intercooled engines or those with marginal coolers.

As an example, using the 50% RH table, if you had a non-intercooled engine with post-turbo charge temperatures of about 200°C producing 200 WHP, you could inject a 16:1 air to water ratio mix and get 200*1.221 = 244 WHP.

Is there a charge density benefit from water injection? The answer is almost always yes. To answer this question definitively, you must examine the assumptions above and have reasonable input from them. For example, if your injector sprays the water stream directly on a hot part of the manifold, you'll see very little direct charge cooling effect, but you'll see a considerable increase in the H2O portion of charge volume. If this is bad enough, it can overwhelm the benefits, in very much the same way as running a turbocharger at a very high pressure ratio.

T ambient = 21.0°C
P ambient = 101.325 kPa
Methanol = 0.0% by mass
Temp water = 21.0°C
RH = 0.0%
Mass air = 100.0 g (100.00g air, 0.00g H2O)

Ti Mass Mass Te dT Temp O2 O2 O2 H2O A:L
°C H2O Meth °C ° Frac Dil Cool Final VP Ratio
--- ----- ----- --- --- ----- ----- ----- ----- ----- -------
0 0.24 0.00 -6 -6 0.979 0.996 1.017 1.013 0.39 419.7:1
25 0.68 0.00 9 -16 0.945 0.989 1.046 1.035 1.10 146.4:1
50 1.31 0.00 18 -32 0.902 0.979 1.085 1.063 2.09 76.2:1
75 2.05 0.00 25 -50 0.858 0.968 1.129 1.093 3.23 48.8:1
100 2.85 0.00 31 -69 0.815 0.956 1.174 1.122 4.43 35.1:1
125 3.68 0.00 35 -90 0.774 0.944 1.219 1.151 5.67 27.1:1
150 4.55 0.00 39 -111 0.737 0.932 1.264 1.178 6.91 22.0:1
175 5.43 0.00 42 -133 0.703 0.920 1.308 1.203 8.14 18.4:1
200 6.33 0.00 45 -155 0.672 0.908 1.352 1.227 9.36 15.8:1
225 7.24 0.00 47 -178 0.643 0.896 1.394 1.249 10.57 13.8:1
250 8.16 0.00 49 -201 0.616 0.884 1.435 1.269 11.75 12.3:1
275 9.09 0.00 51 -224 0.591 0.873 1.476 1.288 12.92 11.0:1
300 10.02 0.00 53 -247 0.568 0.861 1.515 1.305 14.06 10.0:1

Methanol = 0.0% by mass
RH = 50.0%
Mass air = 100.0 g (99.24g air, 0.76g H2O)

Ti Mass Mass Te dT Temp O2 O2 O2 H2O A:L
°C H2O Meth °C ° Frac Dil Cool Final VP Ratio
--- ----- ----- --- --- ----- ----- ----- ----- ----- -------
0 0.00 0.00 0 0 1.000 1.000 1.000 1.000 0.59 0.0:1
25 0.37 0.00 16 -9 0.971 0.994 1.024 1.018 1.82 273.3:1
50 1.08 0.00 24 -26 0.920 0.983 1.069 1.051 2.96 92.5:1
75 1.87 0.00 30 -45 0.870 0.971 1.115 1.083 4.19 53.5:1
100 2.70 0.00 35 -65 0.824 0.958 1.162 1.114 5.45 37.0:1
125 3.56 0.00 38 -87 0.782 0.946 1.209 1.143 6.72 28.1:1
150 4.45 0.00 42 -108 0.744 0.933 1.255 1.171 7.99 22.5:1
175 5.35 0.00 44 -131 0.708 0.921 1.300 1.197 9.25 18.7:1
200 6.26 0.00 47 -153 0.676 0.909 1.344 1.221 10.49 16.0:1
225 7.18 0.00 49 -176 0.647 0.897 1.386 1.243 11.71 13.9:1
250 8.10 0.00 51 -199 0.619 0.885 1.428 1.264 12.91 12.3:1
275 9.04 0.00 53 -222 0.594 0.873 1.469 1.283 14.08 11.1:1
300 9.97 0.00 54 -246 0.571 0.862 1.508 1.300 15.23 10.0:1

Methanol = 50.0% by mass
RH = 50.0%
Mass air = 100.0 g (99.24g air, 0.76g H2O)

Ti Mass Mass Te dT Temp O2 O2 O2 H2O A:L
°C H2O Meth °C ° Frac Dil Cool Final VP Ratio
--- ----- ----- --- --- ----- ----- ----- ----- ----- -------
0 0.00 0.00 0 0 1.000 1.000 1.000 1.000 0.59 0.0:1
25 0.28 0.28 15 -10 0.967 0.993 1.027 1.020 1.69 177.5:1
50 0.81 0.81 21 -29 0.911 0.980 1.075 1.054 2.52 62.0:1
75 1.37 1.37 26 -49 0.860 0.967 1.124 1.087 3.39 36.5:1
100 1.95 1.95 30 -70 0.813 0.953 1.173 1.118 4.26 25.6:1
125 2.55 2.55 33 -92 0.770 0.940 1.220 1.147 5.14 19.6:1
150 3.17 3.17 36 -114 0.731 0.926 1.267 1.173 6.01 15.8:1
175 3.79 3.79 39 -136 0.696 0.913 1.312 1.198 6.87 13.2:1
200 4.42 4.42 41 -159 0.664 0.900 1.356 1.221 7.71 11.3:1
225 5.05 5.05 43 -182 0.634 0.887 1.399 1.242 8.53 9.9:1
250 5.68 5.68 45 -205 0.607 0.875 1.441 1.261 9.33 8.8:1
275 6.32 6.32 46 -229 0.582 0.863 1.482 1.279 10.12 7.9:1
300 6.97 6.97 48 -252 0.559 0.851 1.521 1.295 10.89 7.2:1

Methanol = 0.0% by mass
RH = 100.0%
Mass air = 100.0 g (98.47g air, 1.53g H2O)

Ti Mass Mass Te dT Temp O2 O2 O2 H2O A:L
°C H2O Meth °C ° Frac Dil Cool Final VP Ratio
--- ----- ----- --- --- ----- ----- ----- ----- ----- -------
0 0.00 0.00 0 0 1.000 1.000 1.000 1.000 0.59 0.0:1
25 0.12 0.00 22 -3 0.991 0.998 1.008 1.006 2.65 854.1:1
50 0.89 0.00 29 -21 0.934 0.986 1.056 1.041 3.89 112.2:1
75 1.72 0.00 34 -41 0.881 0.973 1.104 1.075 5.19 58.2:1
100 2.58 0.00 38 -62 0.833 0.960 1.153 1.107 6.49 38.8:1
125 3.46 0.00 41 -84 0.789 0.947 1.200 1.137 7.80 28.9:1
150 4.36 0.00 44 -106 0.750 0.935 1.247 1.165 9.10 22.9:1
175 5.27 0.00 47 -128 0.713 0.922 1.292 1.191 10.38 19.0:1
200 6.19 0.00 49 -151 0.681 0.909 1.336 1.215 11.63 16.2:1
225 7.12 0.00 51 -174 0.650 0.897 1.380 1.238 12.87 14.1:1
250 8.05 0.00 53 -197 0.623 0.885 1.422 1.259 14.07 12.4:1
275 8.99 0.00 54 -221 0.597 0.874 1.462 1.278 15.25 11.1:1
300 9.93 0.00 56 -244 0.574 0.862 1.502 1.295 16.41 10.1:1

Ti = Temperature of input air in °C.
Mass H2O = Optimized ideal mass of water which produces
partial pressure == vapor pressure @ Te. This is
the maximum amount of water that can be vaporized;
any more will be liquid.
Mass Meth = Mass of methanol injected in solution with water.
Te = Equilibrium temperature of evaporated mix in °C.
dT = Change in temperature in °.
Temp Frac = Relative change in input air temperature.
O2 Dil = Concentration due to dilution with water vapor.

The next two are relative to Ti, not an absolute temperature!

O2 Cool = Consequent concentration of O2 due to cooling.
O2 Final = Final concentration of O2 from both factors.
This is the important column, as it tells us
the relative intercooling effect of vaporizing
water injected into the intake tract.

H2O VP = Vapor pressure of H2O at Te.
A:L Ratio = Mass ratio of air to liquid.