I noticed that O2 dilution due to the presence of water vapor (or methanol vapor) entering the combustion chamber, is one of the least discussed and understood mechanism on this forum.

While O2 dilution should not be considered in isolation - cool down from evaporation causes density to increase and therefore more O2 in the mixture - calculations show this is a mechanism that should not be dismissed as marginal, especially at higher temperatures.

I am making an attempt here to summarize how the applications of the formulas governing the ideal gas laws can be used to estimate the % amount of O2 dilution as a result of the presence of water vapor in the air intake flow, compared with 0% humidity.

Water vapor approximately acts as one of the gases constituting an air-water vapor mixture. Approximate results can be derived by applying the ideal gas laws.

Note: all the quantities in the formulas are in multiples derived by consistently applying the International System of Units http://en.wikipedia.org/wiki/Interna...ystem_of_Units

Ideal gas law http://en.wikipedia.org/wiki/Ideal_gas_law

This law states that the product of volume and absolute pressure of a given amount of gas is proportional to absolute temperature.

(1) P x V = n x R x T

where

P is the absolute pressure of the gas in Pascal [Pa]; http://en.wikipedia.org/wiki/Pascal_%28unit%29

V is the volume of the gas in cubic meters [m3];

“n” is the amount of substance of the gas, measured in moles [mol]; http://en.wikipedia.org/wiki/Mole_%28unit%29

T is the absolute temperature in Kelvin [K]; http://en.wikipedia.org/wiki/Kelvin
Note: [K] = [C] Centigrade degrees + 273.15

“R” is the gas constant, which is independent from the other variables (P, V, n, and T) and applies to all gasses.

R = 8.314472 [J/(K x mol)] for al gasses

J = Joule http://en.wikipedia.org/wiki/Joule

The number of moles “n” is equal to the gas mass “m” divided by the gas molar mass “M” http://en.wikipedia.org/wiki/Molar_mass

(2) n = m/M

(3) For air, M = 28.97 [kg/kmol]; [kmol] = 1,000 x [mol]

(4) For water vapor, M = 18.02 [kg/kmole]

By replacing “n” in (1) with (2), we get:

(5) P x V = (m/M) x R x T

From where

(6) P = (m/V) x R x T x 1/M = d x R x T x 1/M

Where d = density = m/V

From where we can calculate the density of the gas

(7) d = (P x M) / (R x T)

Note: This variation (7) of the ideal gas law, links pressure, density, and temperature independently from the quantity of the considered gas.

Vapor pressure and vapor pressure of water http://en.wikipedia.org/wiki/Water_vapor_pressure

All liquids and solids have a tendency to evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid form.

Vapor pressure or equilibrium vapor pressure, or saturation pressure, is the pressure of a vapor in thermodynamic equilibrium with its condensed phase, in a closed container.

Examples of vapor/saturation pressure values for various substances are in this table http://en.wikipedia.org/wiki/Vapor_pressure#Examples

The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid. A substance with a high vapor pressure at normal temperatures is referred to as volatile.

Evaporation should not be confused with boiling http://en.wikipedia.org/wiki/Boiling
Evaporation happens at any temperature, even below the liquid boiling temperature.

The vapor pressure of any substance increases non-linearly with temperature according to the Clausius–Clapeyron relation http://en.wikipedia.org/wiki/Clausiu...eyron_relation which is only dependent on temperature and latent heat. Vapor pressure of water increases dramatically with temperature.

There are a number of equations in literature that can be used to calculate the vapor pressure of water http://en.wikipedia.org/wiki/Water_vapor_pressure

The application of the Antoine equation http://en.wikipedia.org/wiki/Antoine_equation , which I used in my spreadsheet, gives these results:



Note: saturation pressure increases exponentially with temperature

Dalton’s law of partial pressures http://en.wikipedia.org/wiki/Partial...tial_pressures

In a mixture of ideal gases, each gas has a partial pressure which is the pressure the gas would have if it occupied the volume alone. This is because ideal gas molecules are so far apart that they don't interfere with each other at all. Actual real-world gases come very close to this ideal. The total pressure of a gas mixture is the sum of the partial pressures of each individual gas in the mixture.

(8) P = P1 + P2 + …

For an air-water vapor mixture, this can be expressed as

(9) P = Pw + Pa, where

P is the absolute pressure of the mixture, e.g. what you measure with the MAP sensor.

Pw is the partial pressure from water vapor.

Pa is the partial pressure from air.

Pw is dependent upon the amount of water vapor present, according to the general formula

(6) P = (m/V) x R x T x 1/M

However, Pw can in no case exceed the water vapor/ saturation pressure, which only depends on temperature. In the table above, I used the heading PwMAX for vapor/ saturation pressure to indicate this.

Note: the term “relative humidity” in an air-water mixture is formally defined as the ratio of the partial pressure of water vapor in the mixture to the saturated vapor pressure of water. http://en.wikipedia.org/wiki/Relative_humidity

At 100% relative humidity, a special case of the formula (9) becomes

(10) P = PwMAX +Pa

Note: any water in excess of the amount causing the partial pressure from water vapor to equal the water saturation pressure will be present as a liquid, not as a vapor.

We can now evaluate the maximum dilution of dry air (and, therefore O2) in an air-water vapor mixture.

If you are running at – say – 200 kPa (29.01 psi abs @ 14.3 psi gauge), and 50 deg C at the intake, the saturated water vapor pressure is 12.31 kPa (see table above). This will cause a 6.2% reduction in O2… as shown below:

From formula (7) d = (P x M) / (R x T)

Air density at 200 kPa (abs) and 50 C (@ 323.15 K) and 0% humidity

(11) d (0% humidity) = (200 x 28.97) / (8.31447 x 323.15) = 2.156 [kg/m3]

With 100% humidity at the same intake pressure of 200 kPa, partial pressure from air can be calculated from formula (10) as

Pa = P - PwMAX = 200 – 12.31 = 187.69 [kPa], where 12.31 kPa is the partial pressure from saturated water vapor.

From where:

(12) d (100% humidity) = (187.69 x 28.97) / (8.31447 x 323.15) = 2.024

Therefore, 100% humidity at 50 C and 14.3 psi causes a reduction in mass of air from 0% humidity of

(2.156 – 2.024)/2.156 = Pw/P = 6.1%

Obviously, this reduction applies proportionally to all constituents of dry air, including O2

Note: % air/O2 dilution can also be calculated directly from the ratio of vapor partial pressure “Pw” to total pressure “P”

(13) % O2 dilution = Pw/P, with the special case

(14) % O2 dilution = PwMAX/P at 100% humidity

Below, are a couple of tables showing the results of similar calculations at 200 kPa and 300 kPa absolute pressure, or 14.3 psig and 28.8 psig, respectively





Note how dilution increases with higher temperatures and decreases with higher pressures.

While the considerations above are for water, they apply to methanol as well. Methanol is a volatile substance. E.g. at 50 C, vapor pressure is 55 kPa v. 12 of water. http://www.engineeringtoolbox.com/me...es-d_1209.html
Everything else equal, methanol can displace more O2 than water.

- Sandro

 

That sounds like a semester research project. Your next project is integrating a hygrometer into a standalone

 

Very good info.

So for the non-number crunchers out there, in short, depending on the pressure and temp (on the chart), our A/F gauges will actually read leaner A/Fs then what the actual A/F mixture is?

 

I guess it depends. A/F can read leaner, richer or no change depending on the interaction of various effects.

Without AI, humidity on its own will cause the mixture entering the air intake (I am talking about the intake of ambient air) to contain less % of O2. Car will produce less power and the A/F meter should read richer (excess of unburned fuel because of less O2).

There are calculators around to estimate this. E.g. http://www.csgnetwork.com/relhumhpcalc.html
These are derived from SAE J1349 Revision, June of 1990, and include the effect of relative humidity.

With AI, on the other hand, if evaporation takes place before the mixture entering the combustion chamber, phase change will absorb heat and cause increase of density and therefore more O2.

This simplified model http://not2fast.com/thermo/water_inj...opt_mass.shtml predicts that if the latent heat of evaporation is entirely absorbed by the air mixture, the resulting cooldown in the air mixture will cause the density to increase "more" than the dilution caused by water vapor. In this case, the O2 concentration would increase and the A/F monitor should read leaner. But it is not that simple. E.g.some of the evaporation may be caused by absorbing heat from wet surfaces (piping, manifold). That would cause generating vapor (therefore dilution) without the benefit of cooling down the mixture (only the walls).

But there is even more than that. If you consider the amount of air/O2 flowing through the intake manifold, there is no just density and % O2 but also flow rate to consider. If, everything else equal, you can get more flow, that likely shadows the negative side-effect of O2 dilution (especially if you keep the intake temperature low - remember the saturation pressure of water becomes less relevant at lower temperatures).

This is indeed what pre-compressor WI may do. For instance, look at the spectacular results achieved by rx72c. My theory is that by absorbing heat along the compression process, air becomes cooler. The compressor is a volumetric machine. With the same volumetric flow, if you can now make the air cooler/denser and spit higher mass flow. Additionally, because you now run the compressor more efficiently, you can also increase boost (which rx72c does), which further increases density and mass flow (note also higher boost makes vapor partial pressure less important).

I could not find data to prove this though. Taking a temperature reading at the compressor exit would help to measure the effectiveness of evaporative cooling in the compressor.

- Sandro

 

To further exemplify and add more numerical data, look at the results below I calculated from my spreadsheet, based on the gas laws and humidity formulas.





These cases are for:

- 29 psi boost
- 600 cfm
- and different temperatures

I added columns that show explicitly and compare:

- mass flow rate of air at 0% humidity
- mass flow rate of air at 100% humidity
- difference among the two and associated dilution due to the presence of water vapor
- mass flow rate of water vapor at 100% humidity

and

- WI flow in cc/min (1 kg/m3 density assumed for simplicity)

This is the water you need to inject - and have it evaporated completely - in order to get 100% humidity at the intake manifold. The best chances to achieve this is to inject pre-compressor. The hypothesis is that the compressor would make it for a good humidifier...and that the water injected would evaporate and cool the air down in the process, inside the compressor itself.

So, back to the air dilution. If you compare the two cases at 50 and 40 degrees, you can see that if - everything else equal - a drop of 10 deg C in temperature would be sufficient to compensate for the air dilution.

- Sandro

 

re-posting it with the table...

To further exemplify and add more numerical data, look at the results below I calculated from my spreadsheet, based on the gas laws and humidity formulas.



These cases are for:

- 29 psi boost
- 600 cfm
- and different temperatures

I added columns that show explicitly and compare:

- mass flow rate of air at 0% humidity
- mass flow rate of air at 100% humidity
- difference among the two and associated dilution due to the presence of water vapor
- mass flow rate of water vapor at 100% humidity

and

- WI flow in cc/min (1 kg/m3 density assumed for simplicity)

This is the water you need to inject - and have it evaporated completely - in order to get 100% humidity at the intake manifold. The best chances to achieve this is to inject pre-compressor. The hypothesis is that the compressor would make it for a good humidifier...and that the water injected would evaporate and cool the air down in the process, inside the compressor itself.

So, back to the air dilution. If you compare the two cases at 50 and 40 degrees, you can see that if - everything else equal - a drop of 10 deg C in temperature would be sufficient to compensate for the air dilution.

- Sandro

 

If I am understanding this chart correctly, it also shows that your compressor will flow more if intake air temperature is dropped?

So if I had a 200kg/min compressor (at 50c) if I injected water and it dropped IAT down to 40c, I would see a 2kg/min gain in air flow? Thus more horsepower? Or would that not be an accurate statement because the water injected in the second colum was more?

 

Well, the table does not relate to any compressor rating but to the volumetric flow entering the combustion chamber. 200 kg/min flow through the intake manifold at 50 C and 28.8 psi would make 2,184 cfm!!! which does not sound quite realistic...

Back to table, you have to compare the figures highlighted in blue, that would only give you a modest increase of 0.36 kg/min

Keep in mind, the purpose of the table was to show that the negative effect on air dilution could be more than compensated by the increase in density resulting from evapoarative cooling.

If you tell me how many cfm you would expect at the manifold and which pressure, I can run the numbers for you.

- Sandro

 

WaachBack, on a second thought you could just scale up what is in the table already.

The table shows 54.96 kg/min of air and 0% humidity at 50 C and 28.8 psi relative. If you have 200 kg/min (assuming you can have that...) at the same temperature and pressure, that would be 3.64 time. You could use this factor to scale up the other figures in the other columns. So, the resulting increase in air flow at 40 C would be 1.32 kg/min. Note that this is under the assumption of 100% humidity. Note that would require the complete evaporation of very large amount of water injection.

- Sandro