Ph4tty said:
Its the amount of pressure between the engine and the turbo. From what little I understand, the problem is that when drive pressure is significantly higher than boost pressure, when the exhaust valve opens some of the exhaust is fed back into the cylinder causing a loss of power.
Not sure that the ISB has any valve overlap?
The problem really with a higher drive pressure has to do allot with pumping losses. Keep in mind that any internal combustion engine at its very soul is an air pump. This means that it simply moves air from point A to point B. It takes a significant amount of Hp just to rotate the engine with no fuel added at all. As the rpm increases the neccessary Hp increases at a much faster rate.
Basically the the amount of rotation Hp required is determined by the bore / stroke as well as the compression ratio, and of course the rpm. But in essence, you are compressing the air then decompressing the same air. You are sucking it in at one pressure (inlet) and exhausting at another pressure (outlet).
Now ask "The inlet and outlet pressures are equal right?" because the engine is on planet earth lets say at sea level. So one would think that it requires zero Hp to rotate the engine. This is because the Hp needed to compress the air is returned when you decompress the air, as the two forces are equal. All that being true it still requires
significant Hp just to rotate the engine. The key thing to remember is that it is
an air pump, moving air from point A to point B.
This rotational Hp is often called pumping losses. When the engine burns fuel to make a usable shaft Hp it must also make the rotational Hp required to equal the pumping losses. Basically the shaft Hp is the gross Hp minus the rotational Hp (pumping losses) minus the exhaust heat rejection Hp minus the coolant heat rejection Hp minus the friction Hp losses minus the mechanical Hp losses (valve train) minus the oil pump Hp requirements etc etc etc....
The actual amount of fuel burned is more directly related to gross Hp, not shaft Hp. This is the whole foundation for efficientcy of the engine.
Still speaking from an air oump stand point and thinking about raising or lowering the inlet / outlet pressures. Say if the engine were in a pressurized room that could raise and lower the pressure. The pumping losses would be the same for all pressures. This is true as long as the inlet and exhaust pressures were the same. This would also mean that the pumping losses were at their minimum. But if the outlet pressure was greater than the inlet pressure then the pumping losses would increase. This means the rotational Hp required to overcome the pumping losses also goes up.
This is all true for a turbocharged engine. Where the inlet pressure is boost and the outlet pressure is the turbine drive pressure. If inlet (boost) and outlet (drive) pressures are equal then pumping losses are the same for all pressures. If this is the case, then we are getting boost for free and the engines pumping losses are minimum. (As the "exhaust heat rejection Hp" of the engines gross Hp is powering the turbo)
If the outlet (drive) pressure is higher than the inlet (boost) pressure then the pumping losses will increase.
So lets say that the engine is already running at a maximum fuel rate or the maximum gross Hp. With increased pumping losses then the shaft Hp will decrease, because it now requires more Hp to rotate the engine.
On the flip side for the engine to make the same shaft Hp the gross Hp must increase. The only way to increase the gross Hp is to increase the fuel rate. The increased fuel rate drops the air fuel ratio which causes the EGT to increase.
In just about any case you end up with higher EGT and / or lower shaft Hp and a less efficient engine.
It is a vicous circle. *bdh*
Jim