Water Injection Thermodynamics
Copyright © 2003 Eric Fahlgren
Last updated 2007-09-18 11:16 PDT

Water Injection Thermodynamics

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.

Energy Balance

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)


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 ±0.1% 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 ± 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

To be continued...

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