“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

## Saturday, August 05, 2006

### Dissipative forces

There's a principle in physics called "conservation of energy" which is of awesome utility. There are also what are termed "conservative forces" (gravity is one). Please note carefully that I am talking about PHYSICS and NOT POLITICS. Anyway, gravity being a conservative force, theoretically, I can get all of the energy I use to climb a hill back when I descend. As we all know, this isn't the nature of our real world. The fly in the ointment is the phenomenon of dissipative forces. Such forces in our example include friction of all types and fluid mechanical drag.

I was reading a site that discusses the physics of automobiles (a very interesting site by the way) wherein the author stated that "Mechanical Drag is due to all the moving mechanisms in the vehicle that have frictional losses, most specifically the wheel bearings, but is actually nearly entirely due to the action of the tires on the road surface. In nearly all actual situations, all the other causes of mechanical drag factors can be ignored, and just the Tire Resistance considered, regarding the Mechanical Drag."

I don't know about that. As I discussed in my last post, my Grand Cherokee burns about 0.38 gallons per hour when idling or coasting in neutral. It idles at 650 r.p.m. I think it's reasonable to assume that: a) all of the fuel burned while idling goes to overcome engine friction and pumping losses; b) these losses are directly proportional to engine r.p.m.

At 55 miles per hour, the engine turns at 1750 r.p.m., so assumption b) above would indicate that I'm burning (1750/650)*0.38 or 1.02 gallons per hour to overcome engine friction and pump fluids. As I also showed in the previous post, I burn 1.75 gallons per hour at 55 miles per hour. This would indicate that 58% of my fuel consumption goes to keeping the engine running at 1750 r.p.m. and 42% goes to overcoming aerodynamic drag, driveline friction and tire rolling friction.

This seems very surprising to me, and I'm sure it would be extremely surprising to the author of the article cited above. Could it be true? Let's suppose that the engine friction and pumping losses are proportional to the square root of r.p.m. If so, I'd use 0.62 gallons per hour at 1750 r.p.m. and about 35% of the fuel burn is to overcome engine friction and pumping losses. If fuel used to overcome these losses doesn't increase at all with r.p.m., an extremely unlikely scenario, it would still mean that almost 22% of the fuel burn at 55 miles per hour is used in keeping the engine going. Very surprising indeed.

It's possible that the fuel is used for something other than overcoming these dissipative forces when idling, but I'm not sure what it would be. Unless there is a governor that absorbs the mechanical energy and turns it into heat, if there is an excessive amount of fuel for the state of energy use in the engine, the r.p.m.'s would increase. In other words, since the r.p.m.'s stay at 650, the system is in equilibrium. So, surprising as it may be, I think the conclusion is accurate, i.e., that a whole lot of fuel is burned to keep the fuel burner burning fuel.

#### 1 comment:

Anonymous said...

Good article. Strange no-one commented yet.

I see the same. Using the instant-consumption-display of my car, I can see that while running on a flat road it will take about 50% of the consumption to keep the engine spinning at the proper RPM. I.E: establish neutral situation where you can pull the manual transmission car into neutral without resistance, and see what the display reads.

Then compare against the consumption needed to keep the car running at the same speed without any speed-blowoff.

In my case: 0.54 l/10km vs 0.25 l/10km