109 e3b against spitfire II

[Crumpp]

Why do you think it's better?

It is better Viper. An intake manifold cannot precisely meter fuel to the cylinders irregardless of having supercharger on it or not.

Have you ever flow an piston engine aircraft with individual EGT/CHT? The CHT and EGT will be vastly different if the fuel metering system is not direct injection.

Each cylinder is being meter a different amount of fuel. That means power loss just in the thermal differences!

Not to mention that none of them are a stoichiometric mixture. Add to that, it is impossible to optimize the timing advance. All of your cylinders are developing different power levels and none of them are optimal.

There is no why to precisely control how much fuel goes to each cylinder in an intake manifold.

With direct injection, you can not only optimize timing advance to the power curve, you can maintain a stoichiometric mixture. The thermal losses are eliminated because your CHT/EGT's are the same.

In any case, you're always going to gain more by reducing temperature as early in the compression process as possible, because compressors (whether steady-flow or non-flow) produce temperature ratios in exchange for pressure ratios, whilst the absolute work required for the compression process is proportional to deltaH, i.e. Cp*deltaT.

Add to that, the fuel cools the combustion chamber much more efficiently than cold air.

A simple illustration of that basic principle.

1006 J/kgC

460 J/kgC

2100 J/kgC

To change the temperature of a mass of 1 Kg of each by 2 degrees….

Air = 1006 J/kgC * 1kg* 2 C = 2012J
Fuel = 2100J/kgC*1kg*2 C = 4200J
Steel = 460J/kgC * 1kg * 2 C = 920J

Our 4200J of fuel energy goes to cool the 15C air…

4200J * 1kg /1006J/kgC = Change in T = 4.17 C

15C - 4.17C = 10.83C

Now let us dump our fuel on the hot steel of our combustion chamber.

4200J * 1kg / 420J/kgC = Change in T = 10 C

15C - 10C = 5C

5 degrees Celsius is much colder than 10 degrees Celsius.