I want to figure out how to calculate the intake charge temperature reduction for propane, water and nitrous oxide. Knock / detonation characteristics aside I still want to know how much each of these are reducing an intake charge temperature.
If my research is correct (please correct me if Im wrong) then:
Propane:
Vaporization temp= -44*F
Specific heat= .3885 cal/gram
Water:
Specific heat= 1.0 cal/gram
Nitrous:
Vaporization temp= -129*F
specific heat= .2098 cal/gram
If water has a caloric content 2.5 times greater than propane, will it require 2.5 times as much propane to have the same cooling effect as 1 unit of water? On average what is the volume of water or propane injected? (grams/min?) What about the latent heat of vaporization? Since nitrous and propane are already a vapor, will water absorb even more energy (temp) from vaporization in the combustion chamber? How do I calculate in the vaporization temperature for each?
Have any of you ever calculated this?
Or am I thinking about it way too hard?
If my research is correct (please correct me if Im wrong) then:
Propane:
Vaporization temp= -44*F
Specific heat= .3885 cal/gram
Water:
Specific heat= 1.0 cal/gram
Nitrous:
Vaporization temp= -129*F
specific heat= .2098 cal/gram
If water has a caloric content 2.5 times greater than propane, will it require 2.5 times as much propane to have the same cooling effect as 1 unit of water? On average what is the volume of water or propane injected? (grams/min?) What about the latent heat of vaporization? Since nitrous and propane are already a vapor, will water absorb even more energy (temp) from vaporization in the combustion chamber? How do I calculate in the vaporization temperature for each?
Have any of you ever calculated this?
Or am I thinking about it way too hard?