“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 19, 2006

Mixed messages

After my fill-up on August 10 (calculated miles per gallon of 23.90 so climbing back up where I had grown to expect) I installed a K&N high performance air filter. I wanted to see if such a product could increase my fuel efficiency. I was amazed to see that the instant mileage indicator, which I consult frequently when driving, seemed to show a distinct and significant increase. For example, on normal, relatively level stretches of freeway at 55 miles per hour I typically see the indicator jump between 31 and 32 miles per gallon. After installing the K&N filter, the same stretches yielded indications of 33 and 34 miles per gallon. Amazing and quite significant. This should be my best tank full yet, huh?

Think again. Though the driving I did for this tank full was quite typical of my average driving regime, the fill-up from yesterday (August 18) yielded 21.16 miles per gallon. This surprised me as the instant mileage indications led me to expect something like 24 miles per gallon early in the tank. As the fuel quantity indicator went down though, it was clear that I wasn't going to see a good number.

I surmise that one of the major contributors to the variance I see in my results (current standard deviation, the square root of variance, in calculated miles per gallon evaluated at fill-up is 1.71) is inablility to fill the tank to a precise level. That is, if my previous fill-up was short of normal and the current one is above, I'd calculate a lower mileage. The opposite is also true. Such effects clearly average out over a period of time and thus I follow my five tank and ten tank moving average to account for this.

But it's hard for me to see how it could be the entire explanation of this particular fill-up. I put 19.104 gallons in the tank after driving 404.2 miles, thus getting 21.16 m.p.g. For me to have gotten 23 m.p.g. I would have to have only needed 17.565 gallons. That's a difference of 1.539 gallons. It could have happened if the earlier fill-up was, say, about 0.77 below the "average" spot and the most recent was 0.77 above that spot. But it seems like that's an awful lot of variation.

Because I'm aware that the fill-up point is so important to the calculated mileage, I fill it absolutely as full as possible because that seems to be the most accurately repeatable amount. Though I know one is not supposed to "top off," I invariably do. I top off until it's only possible to add a couple of cents worth of fuel (say 0.006 gallons) before the automatic shut-off turns it off. I know that if I didn't top off, the variations would average out to an accurate number over time but I'm too impatient to wait for the averaging out process.

But I'm having a hard time reconciling the low mileage calculated for this tank full with the noticeably higher freeway instant miles per gallon readings. I guess I'll give it another tank full or two before I delve into it more deeply, though I don't know what that delving would invlove.

Monday, August 07, 2006

Cruise horsepower

This will be a short one. I cited a web site a couple of posts back that has some very interesting articles on automobile physics. The author uses coefficient of drag, estimates of tire rolling friction, etc. to determine that his Corvette needs about 26 horsepower to cruise the highway at 60 miles per hour.

I can get at this problem from a different point of view. I get about 31.5 miles per gallon at 55 miles per hour. This is using fuel at the rate of 55/31.5 or 1.75 gallons per hour. Those 1.75 gallons contain about 1.75*125,000,000=218,000,000 joules of heat energy in the chemical bonds I release by burning them in my cylinders. About 25% or 54,600,000 joules goes to maintaining my 55 miles per hour, the rest is wasted as heat expelled to the environment (an inevitable consequence of the second law of thermodynamics, i.e., you can't break even).

So I'm using 54,600,000 joules per hour to maintain motion, or 54,600,000/3600=15,200 joules per second, otherwise known as watts. Now a horsepower is 746 watts, so I'm using 15,200/746=20.4 horsepower. Pretty darn close to the web site author's number, especially considering I'm looking at a speed 5 miles per hour lower. I really love it when different approaches to a problem confirm each others' results.

To carry the analysis a little further (at the risk of causing any remaining reader to throw up his or her hands in despair), force X speed = power, thus power/speed=force. So: 55 miles per hour is 24.6 meters per second; 15,200 watts/24.6 meters/second = 618 Newtons (the metric unit of force). 618 Newtons is 139 pounds of force. That's all it takes, applied continuously, to move me down the highway at 55 miles per hour in my Jeep Grand Cherokee Limited. I'm going to take an informal survey. I bet most people think it's a whole lot more.

Sunday, August 06, 2006

More on fuel used to run the engine

I've spent a couple of posts attempting to determine where the heat energy in the cylinders from burning fuel is used. I talked about the air conditioner and about the fuel used to keep the engine turning. I was rather surprised at the amount used for the latter and decided to look into it a bit more.

Several posts back I wrote about the web site of Dr. Steven Dutch, a Professor at the University of Wisconsin at Green Bay. He has an article debunking the fantasy 200 mile per gallon car. To be clear, he doesn't claim that no such vehicle exists or is possible, only that it's not possible to use simple bolt-on parts or additives to achieve this kind of mileage with "off the shelf" cars.

In any case, Dr. Dutch uses several strategies to infer the force required to turn the engine against the friction of the moving parts (probably primarily the pistons in the cylinders I would guess). The one I'm considering is his analysis involving the cranking power of an automotive battery. He concluded that turning the engine over requires 3600 joules per second or 3600 watts.

If that's true, it means that the energy needed to turn the engine for an hour is 3600*3600 or 12,960,000 joules. However, the engine only uses about 25% of the heat energy in gasoline to do useful work, the rest is wasted as heat expelled to the environment. In fact, some of the "useful work" is turning the fan and the water pump to dissipate the heat. In any case, this means I need the energy of 4*12,960,000 or 51,840,000 joules of heat. This is the amount available in 51,840,000/125,000,000 or about 0.41 gallons of gasoline.

Let me repeat that. Dr. Dutch's calculations imply that I consume about 0.41 gallons per hour to turn the engine. My observations on the road lead me to conclude it's about 0.38 gallons per hour. Absolutely amazing that the agreement is so close. And it reinforces my conclusion that a large amount of the fuel burned in a car is used to operate the engine.

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.

Friday, August 04, 2006

Air conditioning?

Though I doubt anyone reads this blog so carefully (hidden assumption: anyone reads this blog at all), it's possible that someone could have wondered, reading earlier posts, how I estimated the idling fuel consumption of my Grand Cherokee. When coasting in neutral down a mild grade, the car will sometimes reach an equilibrium speed. When this happens for long enough for the instant gas mileage reading on the display to stabilize, the speed in miles per hour divided by the mileage in miles per gallon gives the consumption in gallons per hour. I think this is reasonably accurate since the car is only burning fuel to maintain idling r.p.m. (650).

The number is typically pretty close to 0.38 gallons per hour. However, coasting down the typical grades where I check that, I've noticed lately that I'm burning closer to 0.48 gallons per hour. Now, although my most recent fill-up indicated an improved mileage of 22.1, my previous post was pretty much of a rant about my declining gas mileage of late. Is the increase in idling fuel consumption indicative of a systemic problem that could cause my overall decrease in efficiency? Perhaps it's a symptom of dirty injectors or a clogged air filter.

But another possibility is air conditioning. I try to minimize its use, but it's been quite hot in Southern California and I've had it on quite a bit of late. Could it be that this is the difference I'm seeing? If so, it would indicate that the air conditioner uses about 0.1 gallons per hour when it's on. It certainly seems reasonable that air conditioning fuel use would be a constant function of time and independent of vehicle speed, therefore, the faster I'm going, the less deleterious the effect would be on my fuel consumption per hour.

Let's see if this makes any sense at all. A gallon of gasoline contains heat energy of about 1.25*10^8 joules. Thus, 0.1 gallons per hour would be 1.25*10^7 joules per hour, or about 3500 watts. But the internal combustion engine wastes most of that heat energy - something like 75%. So that means only about 25%, or 875 watts (a tad over 1 horsepower) would be available as mechanical energy to cool the air. This squares niceley with my "gut feeling" about air conditioning.

If it is the air conditioner, I can calculate the effect of a.c. on my gas mileage at , say, 55 miles per hour. Let's see what we find: when driving at 55 m.p.h. on stretches of freeway that appear to be level (i.e., 0% grade) I typically see the instant mileage display oscillate between 31 and 32 miles per gallon. Let's say it's 31.5 m.p.g. This indicates I'm burning 1.75 gallons per hour. But if the air conditioner then burns an additional 0.1 gallons per hour for a total of 1.85 gallons per hour, it should reduce my gas mileage to about 29.7 m.p.g.

This number should be quite noticeable and I don't notice it. It could mean that the 0.1 gallons per hour useage noted above comes from some other cause (e.g., the clogged injectors or air filter) or it could mean that the fuel consumption of the air conditioner is not merely a constant function of time. Or finally, it could be an indication of the error in my determination of the relevant numbers while also performing driving tasks (tuning the radio, answering the cell phone, checking my schedule on the pda, etc.).

I guess I need to get a tune up, see about getting the injectors cleaned, and change the air filter. Then it would be reasonable to check the numbers, followed by a day trip out to the desert somewhere to do some experimentation under more controlled conditions. Stay tuned...