Originally Posted by FormerDatsun510Man

Interesting Corky, then based on what you have said I know my calculations aren't far off. Here is the longhand.

Taking for example a Coldside MP62 running a 105/65 pulley at 7000rpm:

cfm = CID(blower) * VE * RPM(blower) / 12^3
cfm = 62 * 75% * (7000*105/65) / 12^3
cfm = 395.6
m^3/min = 395.6 * .3048^3
m^3/min = 11.2
airflow gram/min = 11.2 * airdensity
airflow gram/min = 11.2 * 1200g/m^3
airflow gram/min = 13441.7

For a 12:1 a/f ratio this would mean we would inject a total of 13441.7/12 grams of fuel per min.

fuelflow gram/min = 13441.7/12
fuelflow gram/min = 1120.1

It takes 1 Joule to raise the temp of 1 gram of air 1 deg C. Likewise, removing 1 Joule of energy from 1 gram of air reduces its temperature 1 deg C. So for our massflow of the air above there is a total of 13441.7 Joules of heat energy needed to be removed per minute to lower the temperature 1 deg C.

The latent heat capacity of gasoline is 349 Joules/gram, meaning 1 gram of gasoline absorbs 349 Joules of heat energy in going from a liquid to a gas. So for the total fuel flow of 1120.1 gram/min this comes out to:

Joules absorbed = 1120.1 gram/min * 349 Joules/Gram = 390914.9 Joules/min

Dividing this by the Joules per deg C per minute for the air massflow comes out to:

Temp drop Deg C = 390914.9 Joules/min / 13441.7 Joules/min-deg C
Temp drop Deg C = 29.1

Converting to Deg F:

Temp drop Deg F = 29.1 deg C * 9/5
Temp drop Deg F = 52.4 deg F

So with a 12:1 a/f ratio it looks like the total amount of fuel going into vapor state would cause a drop of air temp of around 52 deg F.

Now the thing I am interested in is what temp drop would E-Cool (aka the extra injector) would cause in the intake manifold. I am figuring at most the extra injector would inject 1/3 of the total. So at most, in the intake manifold, the temp drop because of E-Cool would be around 52.4 deg F / 3 = 17.5 deg F

So, something else must be the cause for an observed 100 deg F temp drop from E-Cool?

In comparison to gasoline with 349 Joules/gram, water has a latent heat capacity of 2260 Joules/gram... about 6.5 times as much! Running water injection for the above calculation with a typical 15% water to fuel ratio yields:

fuelflow gram/min = 1120.1
waterflow gram/min = 1120.1 * 15%
waterflow gram/min = 168.0
Joules absorbed = 168.0 * 2260 Joules/min
Joules absorbed = 379680 Joules/min
Temp drop Deg C = 379680 Joules/min / 13441.7 Joules/min-deg C
Temp drop Deg C = 28.2
Temp drop Deg F = 28.2 deg C * 9/5
Temp drop Deg F = 50.8 deg F

Interesting the total temp drop from injecting water at a 15% water to fuel ratio is slightly less than the total temp drop from injecting the fuel... however you inject both so you get a double temp drop. With the E-Cool it looks like you would get 18 deg F temp drop in the intake manifold and then the remaining temp drop of 34 deg F in the combustion chamber (before ignition) for a total of 52 deg F temp drop. With WI you would see a 51 deg F temp drop in the intake manifold and then a 52 deg F temp drop in the combustion chamber from the fuel injected (before ignition) for a total of 103 deg F temp drop. The thing that confuses me is from this analysis it looks like the temp drop would be the same for a larger injector setup vs. an E-Cool setup if they are running the same a/f ratio. Unless, not all of the fuel is vaporized that is injected in the combustion chamber?

Bill

Originally Posted by TurboTim

Good questions that I also want to know the answers to.

Where did you read this article about the 10%?

Originally Posted by MiataNuTca

Reading an article recently on water injection (I'll keep methanol out of this for simplicity). It says you should try and match 10% flow of water to the injector flow of fuel.

I currently have 305cc injectors. The nozzle I have installed is a 3GPH nozzle.