Evaporating water is well worth doing:
The specific heat of water is 4.2 kJ / kg / K.
The latent heat of vapourisation of water is 2260 kJ/kg.
So even if you were taking liquid water from 0C to +100C (the maximum available temp span at MSL) you are still looking at just 420 kJ/kg versus the 2260 kJ/kg which you get rid of by evaporating it.
It gets slightly better at altitude e.g. at FL180 (500mb pressure) water boils at about +80C.
You can’t start below 0C because it will freeze, unless you depress the freezing point with antifreeze, but then you can’t evaporate it because you would need a constant supply of antifreeze 
The downside of evaporating water is that you need a constant supply of it, too…
Thank you for the numbers Peter.
The issue I an trying to address is really located between FL100 and FL200 – and the fact that we are dealing with overheating means that we can always assume tgat the radiators’ surface is above the boiling point.
I am getting motivated to give it a shot…. 5 kg of water, 5 minutes of effective operation – sounds like a fair deal…
Wasting 1kg of water per minute would dump 37.6kW if I worked it out correctly. That’s quite a lot.
Actually it would be slightly more because you would also have the warming up of the water prior to it boiling off.
Indeed! Engine output during climb is around 200 KW, so it’s about 20%
Boiling (or simply heating up water) and evaporation are two very different things. The radiator is designed specifically for air as the cooling medium. Cooling it with water is not the same as cooling it by evaporating water onto it. The largest effect of evaporation, I would think, would be to evaporate water by the air before entering the radiator. You cool the air before entering the radiator. How well this works is dependent on pressure, humidity and temperature of the air as well as air speed.
At the altitudes we are discussing, OAT will be near or even below freezing, except inside the radiator channels. I am pretty certain that there will not be enough heating power available upstream of the radiator for the change of state to happen. Actually, we might even get some kind of FZFG!
So for this idea to work, we need the vaporised water droplets to change state while passing inside these channels and extract energy from the metal of the fins.
I understand this is highly dependent upon the fineness of the mist, A basic water squirt won’t deliver nearly enough vaporisation for the state change to happen. If would be a similar situation to what Antonio observed : when flying in rain, there is no obvious CHT drop .
This is why I intend to use a high pressure (13 bar- 200psi) pump and precision nozzle(s)…
Will that work? No idea. But this debates makes me want to try.
I am pretty certain that there will not be enough heating power available upstream of the radiator for the change of state to happen.
Evaporation is per def a change of state without applying heat. Applying heat, and we are talking about simple heat transfer.
This is not to say that spraying water won’t help. This has nothing to do with evaporation though. It’s ordinary heat transfer.
What would have worked insanely well is to use some very efficient evaporating liquid up front. Gasoline would do for instance. Then, what you would get through the radiator is cool, dense air. But spraying some highly volatile liquid around the engine is perhaps not such a vise thing. Maybe water also would do the trick, but it is very dependent of the humidity, temp and pressure of the ambient air.
Something similar is done in turbine powered boats, but for different reasons. Water is injected into the exhaust to cool it, massively. This has the effect of increasing the efficiency, but most of all to reduce noise because the exhaust velocity is decreased equally massively.
Just thinking out loud here I would probably think the largest effect of your water would be to cool the exhaust air from the radiator. Thereby reducing the velocity of this air (it gets cold and dense). This enables you to get more air through the radiator, and therefore more cooling. After the radiator the air is hot, and therefore also much more susceptible to both cooling due to heat transfer and evaporation of the water.
It’s hard to tell without doing the math what exactly would be the best solution. Way too many unknown variables here. Better just try it
Evaporation is per def a change of state without applying heat. Applying heat, and we are talking about simple heat transfer.
Really?
Put some water into a kettle and see how much power is needed to boil off all the water
If you can get water to evaporate by using powerful air velocity, the airflow in which the evaporation is taking place will be dramatically cooled.
The problem is evaporator icing. In standard air to water heat pumps this is a huge issue even at say +30C and accounts for why heat pumps don’t achieve their claimed COP in real life. They probably test them at +30C and a dewpoint of -10C In this proposed scenario it will be harder because the ambient will likely be below 0C and it will be really hard to do this. The nozzle will need to be heated to above 0C otherwise the water will freeze right there.
Flyingfish wrote:
Thoughts ?
In addition to what was already mentioned, I think Icing will become a major issue – not so much on the intercooler itself, but water that tases the intercooler (or in an optimal case is even evaporated there) will eventually condense and freeze somewhere downstream if at high altitudes/temperatures.
Don’t think you want to install an automated airplane icing system…
Really?
Yes. Heat energy is not the same as the kinetic energy of moving air.
Boiling is the process of applying external heat to evaporate a liquid at a given pressure. Evaporation is the process of evaporating a liquid without applying external heat, cooling will occur. Similar in principle, but two very different processes.
It requires lots of heat to boil water, but there is no boiling unless the temperature in the liquid reaches the boiling point.
