3D Printed Piston Lower Mass = More Power? 8

[novateague]

This is a piston I had DMLS printed from AlSi10mg - the geometry is derived from generative design using some basic inputs and a starting shape. It's really just an experiment whether a consumer grade design and print can be bolted in and work for a time.

After machining, it is predicted to weigh ~50 grams vs the OEM 79 grams (1970's Honda XR75). The question is, will this lighter piston result in a small power increase as less work done by reciprocating the piston? Theoretically, the rate at which work is done will be increased?

With the reduced mass, the redline could be increased slightly due to less inertia load, but besides that? Of course, the engine should be more responsive (quicker to spin up) and with a much smaller skirt area, less friction too.

By my calculations, at maximum piston acceleration (~TDC) @ 11,000 RPM, the inertia forces on the piston are about 33% less.

 

[3DDave]

Look really close and see that is a non-zero based graph.

Those peaks are transient. Integrate the area under the curves to see what the energy delivered is. The instantaneous difference is nearly 4hp, but because it is such a narrow peak the realized amount reported is only 1.5 average, dropping to 0.4 hp and that appears to be from the higher RPM as shown on the left hand graph of figure 24, about 200 RPM higher on the lighter rod.

200RPM/9000RPM is a 2% increase in fuel burn rate - coincidentally it should mean a 0.6 hp increase on a base of 30 hp, where they only got 0.4 hp increase.

The drop off in RPM is slightly higher through the cycle with the lighter rod because it hasn't got as much inertia to drive the non-combustion cycles, which is what I said. The dips in power with the lighter rods is also greater as there isn't as much inertia to pull it through.

 

[gruntguru]

gruntguru, SwinnyGG is right that there is work done reciprocating the piston.

No - the question is whether work is done reciprocating the piston "AVERAGED OVER THE CYCLE" (or any significant period of time.
The answer is NO.

je suis charlie

 

[gruntguru]

The net work done reciprocating the piston per cycle is ZERO. Every Joule used to accelerate the piston is recovered when the piston decelerates. Absolutely incontrovertible.

For the one millionth time..
This entire thread is about an effect so small that in order to even attempt to observe it, you cannot disregard friction.
What you're saying is only true if you disregard friction. Period.

By definition, what I am saying must disregard friction. If you think the "work done accelerating the piston" should include friction, you are mistaken. There is work done accelerating the piston and there is work done overcoming friction. No overlap.

je suis charlie

 

[gruntguru]

3DDave. The error made in interpreting the "dyno" graph is not one of integration. If you integrate the two curves, the blue curve clearly indicates higher "power".

The error lies in the use of the term "power" when displaying results from an inertia dyno. "Brake Power" is power absorbed by the dynamometer brake - something an inertia dyno cannot do. The blue curve is not showing higher brake power. It is a result of less energy being absorbed by the inertia of the engine itself, so more energy can be absorbed by the inertia-dyno-flywheel. As Greg Locock pointed out above, the lighter piston is producing a similar result to a lighter engine flywheel.

je suis charlie

 

[3DDave]

The specific reference was to "higher peak power AND lower dips," which is what I countered.

 

[gruntguru]

Cheers Dave. (I was making the point mostly for the benefit of the OP anyway.)

je suis charlie

 

[novateague]

The instantaneous difference is nearly 4hp, but because it is such a narrow peak the realized amount reported is only 1.5 average, dropping to 0.4 hp and that appears to be from the higher RPM as shown on the left hand graph of figure 24, about 200 RPM higher on the lighter rod.

If you tortured that data any more, it would confess to anything! That 200 rpm higher likely did have some effect on the power gains - it also confirms my suspicions that lower reciprocating mass would allow the engine to make power faster:

having a higher rate of angular acceleration of the crank should allow the engine to reach the powerband faster, even if it is an infinitesimally small amount?

The error lies in the use of the term "power" when displaying results from an inertia dyno. "Brake Power" is power absorbed by the dynamometer brake - something an inertia dyno cannot do. The blue curve is not showing higher brake power. It is a result of less energy being absorbed by the inertia of the engine itself, so more energy can be absorbed by the inertia-dyno-flywheel. As Greg Locock pointed out above, the lighter piston is producing a similar result to a lighter engine flywheel.

So, the blue line (light weight con rod and wrist pin) which shows increased power, is not power because it was measured using an inertia dynamometer?

 

[TugboatEng]

Equal time spent above and below zero (figuratively). It does not matter what the peak is, only the average. Take the integral to get meaningful information.

I always here people who don't take higher level math complain about how it relates to the real world. Well it's calculus that does that.

 

[novateague]

Equal time spent above and below (figuratively). It does not matter what the peak is, only the average. Take the integral to get meaningful information.

You realize we're referencing this graph, right? There's no time spent below zero, and if you did find the area under the blue curve, it would be higher than the red curve. Meaning, the lighter weight reciprocating components make more power, faster, than the heavier OEM Honda parts on average.

zero

 

[TugboatEng]

I said figuratively. You can fit a trend line through the data and you'll find that it averages it out quite completely. As others have said you need to integrate the data. Thats going to give you a number instead of a visual representation.

 

[SwinnyGG]

By definition, what I am saying must disregard friction. If you think the "work done accelerating the piston" should include friction, you are mistaken. There is work done accelerating the piston and there is work done overcoming friction. No overlap.

I've read enough of your threads and comments to know that you're smarter than this.

Work done accelerating the piston results in a certain amount of kinetic energy in said piston. From the point of zero piston kinetic energy (at TDC) to the point of highest piston kinetic energy, piston acceleration is positive, and kinetic energy is being added to the piston from the combustion above. From max ke to BDC, the piston is decelerating, as its kinetic energy is transferred to other places.

Some of that kinetic energy is consumed by friction, acting directly on the piston itself through the wrist pin, ring pack, and direct contact with the cylinder wall. The consumed energy is not available to help return the piston back to TDC on the following half rotation of the crankshaft.

There's also the interaction with the crankshaft. In order for the piston's bank of kinetic energy to be available to re-accelerate the piston after BDC, it must first be transferred to the crankshaft, and then from the crankshaft back to the piston again. Between the crankshaft and piston, there are two bearings - meaning there two sources of drag - which in turn means that the transfer of energy from the piston to the crank/flywheel cannot be 100% efficient.

So, for the million-and-oneth time, you can't disregard friction when you're talking about this effect.

 

[TugboatEng]

Here is a diagram of friction vs crank angle of the piston and cylinder wall.

I think most of that friction is from the ring pack so the piston weight is mostly irrelevant.

 

[novateague]

I said figuratively. You can fit a trend line through the data and you'll find that it averages it out quite completely. As others have said you need to integrate the date. Thats going to give you a number instead of a visual representation.

You don't need a trendline or integration to see that the lighter reciprocating components have higher average power on that chart. The thread is not about how much power is freed up with lower piston mass, but is there more power with a lighter piston - the answer is YES, lighter pistons do give higher output power at the crank.

 

[TugboatEng]

If you look at your chart, the red data has a slightly increasing delay for each peak indicating the test was run at a lower rpm. This would explain the difference in power between the two data sets, more so than piston mass.

 

[SwinnyGG]

I think most of that friction is from the ring pack so the piston weight is mostly irrelevant.

We're getting closer...

I don't know the exact split, but side forces can be pretty high, especially in oversquare/short stroke configurations.. so I wouldn't disregard skirt-to-wall friction completely. But, the effect of piston mass on skirt-to-wall friction is going to be very small. So it's all basically the same thing, ie:

'mostly irrelevant' does not equal zero. I've been saying all along: the effect due purely to mass is very small but not zero

 

[GregLocock]

I don't see much maths going on here, integrating by eye is not reliable. You need one full cycle, not selected snips, obviously.

https://en.wikipedia.org/wiki/Slider-crank_linkage

has the main equation

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[TugboatEng]

How does friction and force relate under fully hydrodynamic lubrication conditions? This is the case at all times except the few degrees around TDC and BDC

 

[gruntguru]

SwinnyGG if you motor a frictionless crank-slider mechanism (eg crank + rod + piston) at constant speed, the power (and work) required from the motor is positive while the piston is accelerating and negative while the piston is decelerating. This represents the work required to accelerate the piston. Over a complete cycle (one revolution) the the motoring power (and work) averages to zero. (I know we agree so far).

Now introduce friction. Superimposed on the plot of reciprocation work is a plot of friction. The friction plot is independent of the reciprocation work plot. Find me a single publication that refers to any portion of the friction loss as "work done RECIPROCATING the piston".

Sure its semantics but the distinction is necessary to dispel any inkling in this thread that piston mass might affect brake power directly due to effects other than friction.

je suis charlie

 

[gruntguru]

You don't need a trendline or integration to see that the lighter reciprocating components have higher average power on that chart. The thread is not about how much power is freed up with lower piston mass, but is there more power with a lighter piston - the answer is YES, lighter pistons do give higher output power at the crank.

The power shown in that chart is not BRAKE power. You will get a similar result by lightening the flywheel but zero increase in BRAKE power. See my post above.

je suis charlie

 

[TugboatEng]

This thread is definitely a strange read. I think most of us agree on concept but disagree on terms so we're arguing for nothing. Regardless, it is educational. I appreciate a good argument.