Inertia parasitic losses 1

[al1]

Can anyone help me better understand just how much parasitic losses there are due to inertia of the reciprocation piston/rod mass assembly? Is there a way calculate these parasitic losses, as I already know how to calculated the inertia loads? Since inertia forces square themselves in proportion to RPM, does the rotating mass more than overcoming these reciprocating higher inertia mass loads or is there still a losses at any RPM?
Al1

 

[strokersix]

Reciprocating parasitic losses are not directly related to component inertia. To illustrate, accelerating the piston absorbs energy from the crankshaft but then decelerating it releases energy to the crankshaft for a net zero effect. Parasitic losses are indirectly related to inertia due to increased friction. For example, increasing inertial load required to accelerate the piston will likely result in increased bearing friction. I suppose you could argue that component deflection under inertial loads causes heating due to hysteresis but that's a stretch (pardon the pun).

 

[ivymike]

ss's answer is basically correct. I'd add that total friction can increase or decrease as a result of increased inertia, depending on how the force history at each interface is changed (inertia can counteract some firing load).

Of course, if you're living in "tunerland" where parasitic loss is misconstrued to mean "a reduction in WOT vehicle acceleration," rather than a net loss in steady-state engine power output, then recip assy inertia does have an effect. Rotating assy inertia is more significant, of course.

For a reasonable estimate, add the entire "large end" mass of the conrod (if you split the rod at the cg) and 1/2 of the reciprocating assy mass (SE of rod + pin + piston + rings + bushes +etc) to the mass of each piston pin, and calculate the change in crankshaft "effective" inertia.

 

[al1]

I understand the concept of in and out inertia energy, i.e. the energy it take to accelerate the piston/rod mass on the power stroke at TDC and the returning of this kinetic energy while stopping the piston/rod mass at BDC. However, There are four 180-degree events on a 4 cycle. There is also the geometric problem of the piston going at different speed from TDC to BDC, so the rate of acceleration and deceleration is different and/or not the same through out the 4 cycles. I know that this area is not commonly looked at as a big area of loses, but there has to be something there.
Al1

 

[ivymike]

I understand the concept of in and out inertia energy, i.e. the energy it take to accelerate the piston/rod mass on the power stroke at TDC and the returning of this kinetic energy while stopping the piston/rod mass at BDC. I know that this area is not commonly looked at as a big area of loses, but there has to be something there.

Your question suggests that perhaps you don't understand the "inertia energy" as well as you think you do. If you accelerate the piston up and down continuously in a frictionless engine, regardless of the acceleration rates up vs down, etc., whatever you put into the piston will have come out completely when it comes to a stop (top or bottom). The fact that there is a varying rate at which you put the energy in or take it out doesn't make a difference- the stopped piston hasn't accumulated any energy (where would it store it?).

If you consider friction, hysteresis, etc., then you can have energy losses, and the piston can get a little hotter.