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R.I.D.E. 05-14-2008 06:42 PM

Pumping losses
 
I was wondering what everyone here is trying to describe when they use the term "pumping losses". I understand that an engine basically moves air in much the same way as a compressor although less efficiently of course.

I am also aware that if you merely ground two opposing lobes on a camshaft a basic long block becomes a much more efficient compressor.

With the throttle closed "pumping losses" are greatly reduced since the force on the piston is 10X atmospheric (approximately depending on compression) with no throttle restriction, the same reason compression tests require wot to be accurate.

Pumping losses would probably be best listed in two cateogries.

First;
The actual losses due to moving the air into the engine, compressing the charge, and pushing out the exhaust.

Second;
losses due to continuous operation of the engine necessary in normal vehicle operation.

I will list a few, add anything you can think of.

All belt driven accesories
Reciprocation of pistons, pins and part of the connecting rod mass (include rings).
Oil pump
valve train
Flywheel
Increased load on water pump due to thermostat restriction to maintain operating temp
Exhaust restriction

regards
badger

theholycow 05-15-2008 05:42 AM

Quote:

Originally Posted by R.I.D.E. (Post 99874)
With the throttle closed "pumping losses" are greatly reduced since the force on the piston is 10X atmospheric (approximately depending on compression) with no throttle restriction, the same reason compression tests require wot to be accurate.
[...]
The actual losses due to moving the air into the engine, compressing the charge, and pushing out the exhaust.

With the throttle closed, the engine has to struggle to pull in air. It uses lots of power trying to pull air through a closed throttle. Try it yourself with a jumbo drinking straw, a regular drinking straw, and a coffee stirrer; you'll move a lot more air with a lot less effort through the jumbo straw than the coffee stirrer.

The bit about the force on the piston and compression tests is a logic failure. The force on the piston is not a loss, it's the actual source of the power. Everything else everywhere else is the loss.

theholycow 05-15-2008 05:44 AM

Ok, after posting that it led me to this question that I've never quite understood: Why does a more free-flowing exhaust reduce torque? What I've heard, though not experienced for myself or learned the theory behind, is that backpressure is required to make low-end torque, but I have no idea why.

GasSavers_RoadWarrior 05-15-2008 06:00 AM

It's not backpressure it's velocity, too large an exhaust too soon can kill velocity thus impeding scavenging. A correctly sized exhaust, including header and downpipes will help scavenging. Once you get everything right with the header and downpipes, really you want the best flowing exhaust possible after that, which shouldn't harm anything. However when the pipes are badly sized at the header and downpipe end, the only thing keeping the velocity up maybe is the restriction in the muffler, so that's when a bigger muffler messes up your scavenging.

dkjones96 05-15-2008 06:07 AM

Quote:

Originally Posted by theholycow (Post 99944)
Ok, after posting that it led me to this question that I've never quite understood: Why does a more free-flowing exhaust reduce torque? What I've heard, though not experienced for myself or learned the theory behind, is that backpressure is required to make low-end torque, but I have no idea why.

Back-pressure is ALWAYS going to hurt performance and mileage.

People used to think back-pressure was what caused the increased torque and fuel economy and to get that they put in smaller pipes. The smaller pipes increased EGV (exhaust gas velocity) which is what really made increased torque. Small pipes with straight through mufflers are optimal.

That's actually a perfect example of why the automotive world is stuck with cars that don't all get 40mpg or more. No offense intended.

theholycow 05-15-2008 07:31 AM

That partially clears it up, but how does exhaust gas velocity help?

GasSavers_RoadWarrior 05-15-2008 07:38 AM

Keeping the velocity and thus momentum high will mean that there is a slight negative pressure in the chamber as the piston comes up to TDC and the intake valve opens, this helps transfer momentum to the intake charge. Also, during evacuation of the chamber gas that has slowed down in the exhaust system will need to be pushed from behind by the following gas, this push has to come from the piston, robbing power. Correctly phased pulse tuned headers will offer optimum extraction and scavenging due to the negative pulses from previous cylinders helping scavenge the succeeding cylinders.

theholycow 05-15-2008 08:04 AM

Very well said, I think I understand it pretty well now.

R.I.D.E. 05-15-2008 11:26 AM

A 3 inch diameter piston in a perfect vacuum (combustion chamber) would have just over 7 sqaure inches of surface area. Multiply that by 14.7 and you get 102.9 pounds of force.

The same piston at 185 pounds of compression has 1295 pounds of pressure on the top of the piston.

When you close the throttle (restrict airflow) in essence you are taking the engine to a high altitude instantly. Lets say you are restricting 90% of the potential airflow through the engine.

Now you have almost no compression and your vacuum is also reducing the suction forces as well.

When you open the throttle to test compression the engine cranking speed drops, due to increased compression even though your suction has almost disappeared.

It's similar to an engine that has jumped timing, but that affects vacuum and compression.

Without vacuum or compression there is a third force that causes losses (not friction) and thats the energy lost in the reciprocation of the mass of the pistons, pins, rings, and a portion of the connecting rod.

Each piston starts at TDC accelerates to 90 degrees, then decelerates to 180, reaccelerates to 270 then decelerates to 360. two complete cycles per psiton per combustion stroke which represents 90 degrees of the complete 720 degree otto cycle.

You are basically condemned to accepting those 8 violations of newtons law of inertia in order to achieve the benefit of a power producing combustion stroke, in every reciprocating engine.

Suction forces are significant, but compression forces are greater. Reciprocation forces are generally not considered because the assumption is they are inevitable, but that is not the case in the WW1 rotary aircraft engine.

If you take a radial engine, bolt the crankshaft to a fixed structure and allow the engines block pistons, cylinders and cylinder heads to freely rotate you have a rotary engine (not exactly but fairly close). The pistons are not reciprocating, instead they are rotating around the fixed big end crank journal along with the connecting rods, unlike a recip where the rods never rotate around the crank journal.

Google animated engines Gnome and you will see a moving illustration of a rotary engine. The motion of the pistons in the cylinders is a function of the different rotational axes of the block and connecting rod big ends. Combustion forces push the head away from the piston exactly the opposite of any reciprocating engine
Without compression you do not have a pump, even though with compression it's not a very efficient pump, all you need is intake and exhaust for a pump.

Even in a perfect Vacuum with perfect lubrication (no loss) you would still have to use energy to change the inertial state of the pistons 4 times per revolution. The only way to eliminate that loss would be to have no mass in your reciprocating components.

At idle speed (600 rpm) you have 10 revolutions per second. say there are 4 cylinders. Thats 160 inertial reversals per second of all the reciprocating mass of the components involved.

Each inertial reverses energy cost is increased by the fact that it goes from acceleration to deceleration. that force is greeater than acceleration to a stop or acceleration from a stop.

regards
gary

dosco 05-15-2008 12:15 PM

Quote:

Originally Posted by R.I.D.E. (Post 100013)
When you close the throttle (restrict airflow) in essence you are taking the engine to a high altitude instantly. Lets say you are restricting 90% of the potential airflow through the engine.

Gary:
I suggest you spend some time looking at a college text with regards to piston engine design - pumping losses are well-documented.

"Along with friction forces, an operating engine has pumping losses, which is the work required to move air into and out of the cylinders. This pumping loss is minimal at low speed, but increases approximately as the square of the speed, until at rated power an engine is using about 20% of total power production to overcome friction and pumping losses."

Diesels do not have a valve in the intake, which is one component of it's superior efficiency over spark-ignition engines.

And if you think about it, take a look at the Otto cycle thermodynamic graph. Do you see anything in there about changing pressure due to a throttle? If you add the effects of the throttle to the Otto chart, and then take into account the losses associated with it, I think you'll clearly see what is going on.

And for what it's worth, the Gnome engine had less in the way of pumping losses due to the fact that it had no throttle - the engine was on or off. As I recall, Bentley was the first rotary manufacturer to unstall a progressive spark cutoff that killed a certain amount of cylinders, a "makeshift throttle" as it were.


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