My Turbo Calculator allows you to include a water injection model into the overall engine model. There are several user-defined parameters driving the system, including the water injection rate, the water temperature and the quantity of methanol in the "water" solution. The principal effect of the water injection model is reduction in charge temperature and its consequent increase in charge density. (Although this note only deals with the thermo effects of water injection, you can find out about the combustion chemistry by reading Robert Harris's post to DIY-EFI list.)

Here's how I went about computing the equilibrium temperature
of a solution of gaseous water, methanol and air, given a liquid
solution of water and methanol, and a quantity of air at different
temperatures. All CRC references are to the *CRC Handbook of
Chemistry and Physics, 77th Edition*, CRC Press, 1997, ISBN
0-8493-0477-6. All calculations are centered about the triple
point of water, roughly 0°C, since the steam tables discussed
later are defined at that reference temperature.

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

Hlw(Tl_{0}) + Hlm(Tl_{0}) + Ha(Ta_{0}) = Hgw(Te) + Hgm(Te) + Ha(Te)

where

Tl_{0}= initial temperature of the water/methanol solution Ta_{0}= 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

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

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

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 t^{2}+ 4.0596 t + 2.2801 J/g

Hgw(t) = -1.3143E-03 t^{2}+ 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 ±0.1% over the domain of interest so we aren't losing much).

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 `H _{0}`.

Hw_{0}= Mw Hlw(Tl_{0})

Hm_{0}= Mm Hlm(Tl_{0})

Ha_{0}= Ma Ha(Ta_{0})

H_{0}= Hw_{0}+ Hm_{0}+ Ha_{0}

So, now we have the energy balance as a function of the equilibrium temperature:

Mw Hgw(Te) + Mm Hgm(Te) + Ma Ha(Te) - H_{0}= 0

= Mw (-1.3143E-03 Te^{2}+ 1.8799 Te + 2501.44) + Mm (0.7815 Te + 1211.97) + 1.0165 Ma Te - H_{0}

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 - H_{0}

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

Te = (-b ± sqrt(b^{2}- 4ac)) / 2a

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 `H _{2}O` at the ambient temperature, taking RH
to indicate the proportion of that vapor pressure as the partial
pressure of the

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

** To be continued... **