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Old 06-04-2007, 08:34 PM   #1
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Default 10% of fuel should be water?

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.

1 Cubic Centimeter = 0.000264172052358148 Gallons

So my 3GPH nozzle would be = 11356.235352000017856362404319665/ 60min= 189CC/min

Now my question is. Are my 305cc injectors rated at 305cc's at 100% duty cycle? or the industry standard of 80%? To properly calculate 10% of my fuel flow I'll need to know my fuel pressure, cc's at that duty cycle, and my CC's of my nozzle. This way I can get a good idea on a proper progressive setup.
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Old 06-05-2007, 10:17 AM   #2
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Good questions that I also want to know the answers to.

Where did you read this article about the 10%?
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Old 06-05-2007, 11:27 AM   #3
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probably 80%/43psi. and dont forget to multiply the injectors by 4.
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Old 06-05-2007, 11:55 AM   #4
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they are rated at 100% most of the time. so 305*your max DC say 90% being agressive your FP would be the normal 50 at 0 vac + the number of boost in psi. so say 13 psi. making 63 psi FP. all togethor would give 10% mass a 110 cc/min requirement. it isnt that hard. there was a person on .net who worked it through calories as well that was real interesting.
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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
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Old 06-05-2007, 06:48 PM   #5
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Quote:
Originally Posted by TurboTim View Post
Good questions that I also want to know the answers to.

Where did you read this article about the 10%?
http://web.archive.org/web/200601030...rinjection.htm
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Old 06-05-2007, 07:10 PM   #6
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There are so many variables....the boost pressure in the manifold would cause a pressure differential to the injector pressure. The only way to accurately test this would be to do a long and short shot in a beaker in a simulated environment.....(pressure tank), then weigh the fuel, calculate using the proper fuel density (all fuels have different densities) and figure out 10% of the final volume number of fuel. I guess the easiest way to get a ballpark figure would be to go on the RC engineering website and figure out how much injector you would need for your fuel pressure and rwhp needs, then figure 10% of the CC's of required fuel (all 4 injectors)- should be what water amount you should be at. If you have a dyno sheet, you should be able to get in the ballpark. To get a progressive setup, get 2 points....maybe one a 0VAC and one at your peak boost.

Last edited by MiataNuTca; 06-05-2007 at 07:29 PM.
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Old 06-05-2007, 10:39 PM   #7
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Quote:
Originally Posted by MiataNuTca View Post
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.

If you don't know the pressure at the nozzle, the 3 GPH is nearly worthless. Small changes in delivery pressure will have huge influence in the actual amount of water injected.
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Old 06-06-2007, 09:39 AM   #8
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BS i can use abit more than 1.7 qaurts in 20 minutes of boosting the use is directly dependant on the nozzle pressure, and there is little variance. My system tracks as it should for consumption given time past 4psi and amount used in a very linear way. Remember guys these things are 100% DC all the time.
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Old 06-06-2007, 08:58 PM   #9
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interesting...
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