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Water Injection on a NA motor. BS?


Guest Andrew240z

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Hadn't looked at your link, but I guess it answers the question with measured data.

 

His dyno charts were a little confusing to me, but if I am reading them right doing nothing but turning on the WI cost him 15 HP.

 

Anyone else read it that way?

 

But if he advances the timing (retunes), he gets most of that back with a drop in fuel consumption.

 

He also talked about the WI cooling the intake, but concluded the change was probably too small to measure with his set up.

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  • 3 months later...
Ok' date=' I asked my uncle, retired AF colonel, who was a WWII fighter pilot, korea (jet age), and Vietnam (jet age), although in Vietnam he flew B52 missions out of alaska.

 

He said irregardless of whether the plane had water injection or nitrous, the engines had to be rebuilt if the throttles had been pushed beyond a certain point, due to overrevving. He said the H2O injection rarely worked since it would freeze in the winters of England and, but rarely used NOS since they needed every plane they had intact and not down for repairs. He said it was a hell of a plane and the H2O injection wasn't that critical to the planes performance as was the NOS and his understanding from briefings was that the H2O was for suppressing detonation. He also said he personally never needed or used the NOS.

 

Stoichiometrically, would somebody show me how water reacts with gasoline in a controlled combustion to increase HP!

 

I thought it was like this;

 

C8H18 + 12.5O2 --> 8CO2 +9H2O

 

Correct me if I'm wrong here but if you add H2O to the left side then to balance you add H20 to the right side as well, however you are adding (L)H2O to the left and the product on the right is (G)H20 resulting in a net loss of energy from the equation, subsequently you have less BTU given off because it was used in converting the liquid H2O to gaseous H2O. So any benefit has to be from denser inlet air temperature and nothing else. Water does not react with gasoline, never has and never will.[/quote']

 

In efforts to revive this thread, I found some new info that I was looking for. Not the same college paper write up I once found, but it does provide a little insight into what I was talking about earlier. More importantly, hopefully I'm proving myself to be not entirely crazy. Here's the link http://www.innovatemotorsports.com/resources/rich.php Basically, from what I can tell, and what I remember reading previously (and learned from my general college chem course) is that reactions don't happen all at once according to chemical kinematics. It is said in an easy to understand manner in the article linked above. Here's the quote I've been looking for, for the past few months:

"First the gasoline molecules are broken up into hydrogen and carbon. The hydrogen combines with oxygen from the air to form H2O (water) and the carbon molecules form CO. This process happens very fast at the front edge of the flame front. The second stage converts CO to CO2. This process is relatively slow and requires water molecules (from the first stage) for completion. If there is no more oxygen available (most of it consumed in the first stage), the second stage can't happen. But about 2/3 of the energy released from the burning of the carbon is released in the second stage. Therefore a richer mixture releases less energy, lowering peak pressures and temperatures, and produces less power. A secondary side effect is of course also a lowering of knock probability."

 

Now, according to this, adding water actually assists in the chemical process by helping to more thoroughly complete the combustion of the gasoline. Maybe I can find (or someone else...hint...hint) the kinematic process for the combustion of gasoline in an engine.

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