Diff between theory and practice

[Mazda Ebrahimi]

> The engine is most efficient at peak torque
> so the amount of fuel needed is at its lowest point. At the higher RPM
> levels the efficiency drops off and the engine requires more fuel.

The engine IS most efficient at peak torque, but that is not why he needs more fuel at 
higher RPMs.  If he needed more FUEL PER HORSEPOWER produced, then you are right.  But 
what he is indicating is that he is maintaining constant air fuel ratio (constant CO), 
and is using more fuel at 6000 RPM.  This simply means he is flowing more air which 
means that:

	(vol eff @ 6000)* 3000 * disp > (vol eff @4800) * 2400 *disp

therefore, if  (vol eff @ 6000) is 80% or greater than (vol eff @ 4800), he will need 
more fuel at 6000 RPM.  According to the numbers provided, he used 170 units at 6000 RPM 
and 155 at 4800 RPM.  This means (assuming constant A/F), the engine flowed 9.6%  (which 
is (170/155) - 1)more air.  If volumetric efficiency was constant between the two 
speeds, the engine would have flowed 25% (which is (6000/4800) - 1)).

Given the above info., VE at 6000/VE at 4800 = (4800*170)/(6000*155) = 0.877 which means the 
volumetric efficiency dropped 13% at 6000 RPM compared to 4800 RPM.

This is assuming engine temp. and everything else was constant.