DIY water injection
What happens if you combine a water injection system with an intercooler. Will it have any adverse effects. Would it cool things down too much or could you use it in the higher boost levels when the intercooler starts loosing some of it's cooling effects. Or would it be much better to have a larger mist system set to spray on all the oilcoolers, IC and radiator and cool them down instead.
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Cookiemonster
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AB
- Mine is bigger than yours
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Alright, explaining this may take several posts but I will make a start at it now.
First off, contrary to what has already been said, just because water is at 0°C dosen't mean it is instantly turned into ice, in the same way your pot of water on the stove dosen't all instantly turn to steam as soon as it boils. This is because it takes a very very large amount of heat to change states, ie from solid to liquid to gas or vice versa.
There is another property we need to look at called the triple point of water. This tells us that water at 0°C in a sealed vessel can exist in all three states at once, as a solid, a liquid and a gas. By varying the amount of heat you put into or take away from the vessel you change the proportions of solid, liquid and gas.
The next important property is called the specific heat or heat capacity. This tells us how much heat is needed to raise a certain mass of substance by 1°C. Water has a very high specific heat, 4.186kJ/kg.°C and air has a very low specific heat, 1.0 kJ/kg.°C. This tells us is will take 4.186 times as much heat it raise one kg of water (one litre at 20°C) by 1°C as it will to raise one kg of air (833 litres at 20°C) by 1°C. Put another way, it takes the same amount of heat to raise one litre of water by 1°C as it does to raise 3500 litres (3.5 m³) of air by 1°C.
Right, that's enough of a physics lesson for now. Next post I'll try to explain how this applies to cars.
First off, contrary to what has already been said, just because water is at 0°C dosen't mean it is instantly turned into ice, in the same way your pot of water on the stove dosen't all instantly turn to steam as soon as it boils. This is because it takes a very very large amount of heat to change states, ie from solid to liquid to gas or vice versa.
There is another property we need to look at called the triple point of water. This tells us that water at 0°C in a sealed vessel can exist in all three states at once, as a solid, a liquid and a gas. By varying the amount of heat you put into or take away from the vessel you change the proportions of solid, liquid and gas.
The next important property is called the specific heat or heat capacity. This tells us how much heat is needed to raise a certain mass of substance by 1°C. Water has a very high specific heat, 4.186kJ/kg.°C and air has a very low specific heat, 1.0 kJ/kg.°C. This tells us is will take 4.186 times as much heat it raise one kg of water (one litre at 20°C) by 1°C as it will to raise one kg of air (833 litres at 20°C) by 1°C. Put another way, it takes the same amount of heat to raise one litre of water by 1°C as it does to raise 3500 litres (3.5 m³) of air by 1°C.
Right, that's enough of a physics lesson for now. Next post I'll try to explain how this applies to cars.
ahem, I could outdo you and AB both w.r.t Physics but I won't bother :D .......overall the whole system is soo technical and affected by so many internal/external conditions that it's just best to stick to the basics.
To get the most from Water injection, you have to be able to control the amount of water injected (i.e. like a Fuel Injector) w.r.t Engine load across rpm and adjusted to varying intake temps
This will give you 100% max torque/power all the way through rather than just setting it up to prevent detonation at 20psi etc at the loss of only gaining XX% of the 100% available to be gained.
Once you have the setup right, then and only then should you start playing with Equations and seeing how the results compare to the dyno charts :beer
To get the most from Water injection, you have to be able to control the amount of water injected (i.e. like a Fuel Injector) w.r.t Engine load across rpm and adjusted to varying intake temps
This will give you 100% max torque/power all the way through rather than just setting it up to prevent detonation at 20psi etc at the loss of only gaining XX% of the 100% available to be gained.
Once you have the setup right, then and only then should you start playing with Equations and seeing how the results compare to the dyno charts :beer
Well hows this.
I never did physics at school, left year 11 after doing biology as a science.
I work in a factory
I DO still know what im on about (learned through listening to people and working stuff out, i actually set myself little assignments to learn stuff, or if people tell me something i dont understand i research it so i do understand. learning because you want to, rather then learning because you have to is a very powerfiul thing and you absorb alot more knowledge.
Hence how I worked out WHY axel tramp occurs. (some people might say 'durrr' but im happy i worked it out)
I never did physics at school, left year 11 after doing biology as a science.
I work in a factory
I DO still know what im on about (learned through listening to people and working stuff out, i actually set myself little assignments to learn stuff, or if people tell me something i dont understand i research it so i do understand. learning because you want to, rather then learning because you have to is a very powerfiul thing and you absorb alot more knowledge.
Hence how I worked out WHY axel tramp occurs. (some people might say 'durrr' but im happy i worked it out)
1983 JA Staz
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Cookiemonster
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That statement is too vague, is assuming 100% atomization and generally one would work it out backwards to get a % of injected mixture w.r.t duty cycle.Cookiemonster wrote:I still want to know how water/methanol, at 25 degrees, mixed with air, at say 35 degrees after the intercooler, will get you an intake mixture of 0 degrees.
I'll let AB show you if you want, just remember what happens in theory isn't exactly the case in practice but is a guide on which way to expect the results to fall.
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AB
- Mine is bigger than yours
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- Joined: Wed May 25, 2005 12:27 am
- Location: Wellington, New Zealand
Now for part two.
First we need to work out how much heat we need to remove from the air that is at 35°C to lower it to 0°C. I have no idea how much air these engines cosume per second at x rpm so I'm just going to use 1kg (1000L) a second as a hypothetical number. 1kg and 1.0 kJ/kg.°C tell us we will need to remove 35kJ of heat per second per 1000L of air.
Now we need to look at one more important property, the heat of vaporization. This is a very large number, which tells us it takes a huge amount of heat to change water from a liquid to a gas. That is why it takes so long to completly boil down a pot of water, because you have to use a very large amount of energy. The heat of vaporization of water is 2256 kJ/kg.
The final important point is that ideally the water we are injecting has an extremely large surface area to volume ratio, this allows the water to vaporize extremely easily. The actual ratio will depend on the quality of the nozzle you use, the higher you can get it the more effective the system will be.
Now the final calculation is how much water will we need to inject to remove 35°C of heat from 1000L of air. 35kJ divided by 2256 kJ/kg gives us 0.0155kg of water or 15.5ml per second per 1000L of air per second. This is under ideal conditions, where all the water vaporizes before it reaches the intake valve. In the real world not all the water will be able to vaporize in that time so you need to compensate for this by injecting more water.
I may be missing something here, I'm only half awake right now, hopefully someone can check over this and confirm/deny the accuracy of these calculations.
First we need to work out how much heat we need to remove from the air that is at 35°C to lower it to 0°C. I have no idea how much air these engines cosume per second at x rpm so I'm just going to use 1kg (1000L) a second as a hypothetical number. 1kg and 1.0 kJ/kg.°C tell us we will need to remove 35kJ of heat per second per 1000L of air.
Now we need to look at one more important property, the heat of vaporization. This is a very large number, which tells us it takes a huge amount of heat to change water from a liquid to a gas. That is why it takes so long to completly boil down a pot of water, because you have to use a very large amount of energy. The heat of vaporization of water is 2256 kJ/kg.
The final important point is that ideally the water we are injecting has an extremely large surface area to volume ratio, this allows the water to vaporize extremely easily. The actual ratio will depend on the quality of the nozzle you use, the higher you can get it the more effective the system will be.
Now the final calculation is how much water will we need to inject to remove 35°C of heat from 1000L of air. 35kJ divided by 2256 kJ/kg gives us 0.0155kg of water or 15.5ml per second per 1000L of air per second. This is under ideal conditions, where all the water vaporizes before it reaches the intake valve. In the real world not all the water will be able to vaporize in that time so you need to compensate for this by injecting more water.
I may be missing something here, I'm only half awake right now, hopefully someone can check over this and confirm/deny the accuracy of these calculations.
Dude, you are a knob. This thread is completely over my head, but I can see that AB has spent thought and time explaining this, then you try and refute his argument with a couple of sentences? I have no idea who is "right" here, but if you are going to argue with someone, you should back it up with a few more facts than just "Its obvious that 15 ml a second wouldnt turn into gas". Maybe that passes as a valid argument on the sigma-galant forums, it doesn't here.Chryzla wrote:oh give up and stop trying to look all high and mighty.
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