Understanding the theory (long post)

[Gelf]

Ive just finished installing my WI, and trying to get to grips with setting up. So far ive come up with...

Notes:

Very little of the water gets vaporised (specific heat absorption) until it reaches the cylinders, unless inlet temps exceed 100 C. Methanol vaporises at a lower temp than water, so more of it will have vaporised before reaching the cylinders. Unless injected prior to the turbo, then vaporisation occurs in the compressor.

Temperature reductions in the inlet tract are from the effects of the liquid water/methanol reaching equilibrium with the temperature of the inlet charge, the latent heat of absorption of the water/methanol.

Quantity of water vaporised depends on the temperature of the air and relative humidity. I?ve used data from Not2Fast.com about the maximum quantity of water that can be vaporised at given temperatures.

Vaporisation:

At 50/100% RH and at

0 C, 0/0 g (0/0 ml) max of water can be vaporised per kg air

50 C, 10.8/8.9 g (10.8/8.9 ml) max of water can be vaporised per kg air

100 C, 27/25.8 g (27/25.8 ml) max of water can be vaporised per kg air

200 C, 62.6/61 g (62.6/61 ml) max of water can be vaporised per kg air

300 C, 99.7/99.3 g (99.7/99.3 ml) max of water can be vaporised per kg air

If injecting 50/50 with methanol then the same quantity of Methanol is also vaporised doubling the injection amount.

RH does not affect the amount of methanol that can be vaporised.

At what temperature do you calculate the amount of water to be injected :?:

Is it the inlet charge temp or in cylinder temps :?


At maximum torque (265 lb ft @ 3500 rpm), I?m consuming about 9 kg of air per minute. And, an estimated 12.5 AFR / 0.56 SFC, 755 g of fuel.

At 100 C / 100 % RH at least 25.8 g per kg can be vaporised.

25.8 x 9 = 232.2 g

The max amount that can be vaporised is 232 ml of water, using a 0.6 or 0.7 jet, depending on IDC.

This is, 31 % of the amount of the fuel and 2.6 % of the amount of the air used.

Or, 464ml of 50/50 water/methanol, using 2 x 0.6 or 0.7 jets depending on IDC.

This is, 61 % of the amount of the fuel and 5.2 % of the amount of the air used. (Similar fuel ratio to that which was used in a NACA paper, using 70/30 methanol/water)


At maximum bhp (230 @ 5400 rpm), I'm consuming about 12.26 kg of air per minute. And an estimated 12.5 AFR / 0.56 SFC, 977 g of fuel.

25.8 x 12.26 = 316.3 g

The max amount that can be vaporised is 316 ml of water, using a 0.9 or two 0.5 jets, depending on IDC.

This is, 32 % of the amount of the fuel and 2.6 % of the amount of the air used.

Or 632 ml of 50/50 water/methanol, using two 0.9 or 1.0 jets, depending on IDC.

This is, 65 % of the amount of the fuel and 5.2 % of the amount of the air used.

These figures are for the maximum amount of water that can be injected and be vaporised, which may not necessarily be the optimum amounts for maximum power.

I'm starting out using a 0.5 and 0.4 jets, pre turbo and at the throttle. Im going to use an accelerometer to determine the best set up for torgue.

Please correct me if I?ve got any of the above incorrect

[hotrod]

Sorry but your numbers complete lost me.

What does this mean :


0 C, 0/0 g (0/0 ml) max of water can be vaporised per kg air

50 C, 10.8/8.9 g (10.8/8.9 ml) max of water can be vaporised per kg air

100 C, 27/25.8 g (27/25.8 ml) max of water can be vaporised per kg air



10.8 divided by 8.9 grams of what ?? 10.8 what divided by 8.9 ml of what ?? --- I am assuming the grams and ml values are the amount of water, but what is the 10.8 refer to ????

Sorry, maybe your using a british notation format that is obvious to you, but to me its not at all clear what your units are, that go with the numbers.




Even air that is below freezing can hold some water vapor, so the zero grams at 0 deg C is not correct as I read this.

At 100 deg C. and atmospheric pressure you can have unlimited amounts of water in the air.

at 50 deg C, ( 122 deg F) air can hold 85.4 grams of water / kg of dry air.

If you start with 50 % RH then you should be able to evaporate up to about 43 grams of water, if you have enough heat available to hold your air temp at 50 deg C.



I believe your trying to over think this. The typical temperature drop with WI and common alcohol mixes is about 30 deg F ( 17 deg C ) depending on outside temp and humidity. With water only the temp drop is frequently around 20 deg F ( 10 deg C) at low RH initial conditions.

The amount of water vapor air can hold varies both with temperature and with pressure, so depending on your injection point, current weather conditions, and the boost level at that injection site -- all your numbers change.

The basic rules of thumb for injection rates are a whole lot easier to work with IMHO, unless you want to solve a bunch of multi variable equations for each possible boost, initial RH and temperature condition your likely to run into.

If I'm missing the whole point here, please take another shot at it !

Larry

[Gelf]

I apologise for my notation, i simply meant for the forward slash to seperate results for the differing amounts vaporised at either 50 or 100 percent humidity.

I obtained the data from the link below

http://not2fast.wryday.com/thermo/water_injection/opt_mass.shtml

Im no expert, I am just trying to get my head around the theory.

[Richard L]

I understand you concised data, great to put it to the test.

The effect of inlet cooling is self-regulated. Initially, hot air will be cooled by evaporation until the 100% RH is reached. No further drop in temperature is expected.

The un-evaporated water will carry into the cylinders and then further evaporated during combustion - at temperature over 2000C, 100% RH will not be reached that easily. So the only way you can monitor the action of water is by reading the EGT.

If you are using an accelerometer, it will help to determine the force the engine produces at all RPM rather just reading the peak HP (meaningless) as peak HP is limited buy mechical set up.

I woudl test you engine's acceleration in the mid range area rather the top end.