“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)

Sunday, September 27, 2009

Acceleration - the final word?

In my efforts to drive in the most fuel efficient way possible, I've examined many factors. For most of these, it's easy to determine how the parameters involved affect fuel economy. The exception is rate of acceleration. I've posted on this before, here and here. And it's a topic of discussion at one of my most frequented sites, ecomodder. The developer of that site has another site, at which he posted the results of an experiment to determine this, at least for his vehicle. He concluded that quick acceleration was most efficient, but also believes that this gets him to "pulse and glide" mode more quickly and is most efficient for that reason. Pulse and glide techniques are not applicable to my LR3 with its automatic transmission.

The discussion at ecomodder revolves around engine maps, gearing, etc. and of course, these are a crucial factor in the analysis. One thing I don't see, though, is the discussion of kinetic energy addition I mentioned in my first post on the subject. The fact is, if I bring my Land Rover LR3 HSE from 0 joules of kinetic energy (i.e., standing still) to about 800,000 joules (the kinetic energy at 55 m.p.h.) at a given rate of acceleration, and at half that rate, I will travel twice as far in the latter case. Since addition of kinetic energy is a matter of converting the chemical potential energy of gasoline, I've used a given amount of fuel over a longer distance. Further, I've gone a greater distance at a lower level of aerodynamic drag.

On the other hand, I've spent a greater amount of time at a speed that is less efficient. It's less efficient for two reasons: low r.p.m. and low torque is a very inefficient part of the engine map, i.e., has a high brake specific fuel consumption; and I've spent a longer time down near 0 m.p.h., where all my fuel is going to turn the engine and not overcome external forces.

Another complicating factor is "torque converter lock up." For those who don't know, a car with an automatic transmission has a torque converter between the flywheel and the transmission. This fluid coupling enables the car to stop in gear without killing the engine and serves a purpose similar to the clutch in a vehicle with manual transmission. It consists (very basically) of a case containing a fluid, a pump attached to the casing that turns with the flywheel, and a turbine that is turned by the fluid forced onto it by the pump. There are losses in the fluid coupling for various reasons, so most vehicles have a solenoid that locks the pump and turbine together when they are turning at similar speeds. This prevents the losses in the fluid coupling but is only active at speed, thus increasing the likelihood that getting to speed faster (i.e., accelerating more quickly) will save fuel.

That must be all then, right? No, of course not. There's also the ECU, or engine control unit. If the throttle is pushed to the floor, most ECU's will operate in so-called "open loop mode." Here, the computer doesn't take feedback from the oxygen sensor and estimates how much fuel to inject based on outside conditions (temperature and pressure) and throttle position. It tends to supply a very rich mixture, thus hurting fuel economy and suggesting that slow acceleration is best.

So, does one method prevail, i.e., as slowly as possible or floor it? Or is there an optimal "Goldilocks" rate (not too quick, not too slow) of acceleration? I determined to find an analytical solution. Many estimates and assumptions are necessary since I don't have an engine map for my vehicle and I don't know the parameters of the ECU. Such considerations have never stopped me before though.

I'll start with a terrific document prepared by Sierra Research for Environment Canada entitled "Alternative and Future Technologies for Reducing Greenhouse Gas Emissions from Road Vehicles." It's linked and can be downloaded as a pdf file here. This document discusses (among a huge variety of other topics) "brake specific fuel consumption" and provides a generic engine map. This map shows engine load in brake mean effective pressure as a function of engine r.p.m. and the isopleth contours show lines of equal brake specific fuel consumption. It's a composite of 1995 normally aspirated, fuel injected, two valve per cylinder engines and thus is similar but not the same as my engine, but it's the closest I can find.

Now, it's well known that losses related to heating the engine block and throttling losses as the engine turns slowly at low r.p.m. and low demand, and heating oil due to friction at high r.p.m. lead to a an island of lowest specific fuel consumption at high engine demand and mid-range r.p.m. I don't have an engine map for my Land Rover LR3 HSE so I'll use the features of the map in the Sierra Research document adjusted for the few calculated points I have for my vehicle. This, together with the calculated force required to add kinetic energy and overcome external forces, will enable me to numerically calculate (estimate) the fuel used in going from 0 to, say, 55 m.p.h. at various acceleration rates.

I'll start with what I do now, which takes me from 0 to 55 m.p.h. in about 45 seconds and compare with a much brisker rate taking me to 55 m.p.h. in 10 seconds. The vehicle is rated to get to 60 m.p.h. in 8 seconds at wide open throttle. I've made a spreadsheet to analyze the energy required to add kinetic energy and to overcome external forces for each tenth of a second.

The result using slow acceleration is that I burn about 0.27 pounds of fuel in accelerating to 55 m.p.h. over a distance of 550 meters. Accelerating quickly, I burn about 0.19 pounds of fuel to get to 55 m.p.h. over 120 meters. To this I need to add the fuel used in going the additional 430 meters to get to where the slow acceleration regime took me. Using the figures from previous calculations on highway cruising fuel use, I'll use about 0.072 pounds to go 430 meters for a total of 0.262 pounds. This indicates that the quick regime is very VERY slightly more efficient. It's doubtful that my various estimates and interpolations are accurate enough to have much confidence in this result with respect to the specific numbers, but I do think it's safe to say that there isn't much difference.

Now, in gathering this data and calculating, it's clear that the poor economy is at the very low r.p.m.'s and speeds, and at r.p.m.'s above about 2600. So I believe the key is to accelerate briskly to second gear and then back off to a moderate rate to get to speed. I'll take the time to plug the numbers in for such a regime in the coming days, but I wanted to get something posted and the data gathering and calculations resulting in the conclusions above took about eight hours. So I guess the question mark in the title of this post is there for a reason, this isn't quite the final word.

Monday, September 07, 2009


I made a post in which I estimated the total energy use in my family. It was disturbing, in that we use a LOT of energy and there aren't a lot of places where cuts are easy (hypermiling my 3 ton SUV notwithstanding). Over 18 months have gone by since I wrote that article, and I've learned a few things. None would change the overall thrust of the article though; the main modification would be in the energy content of purchased items (the so-called "embedded energy"). I estimated one third of the cost of the average purchased item went for the total energy used in producing it, I now think that's probably too high.

But the last couple of weeks have been quite hot (around 100 degrees F for a daytime peak) and I've had the air conditioner on quite a bit. Our air conditioning system was manufactured in 1995 by Goodman Manufacturing and is a model CK60-18. As best I can tell, this means it's rated at a little under 60,000 BTU/hour (about 56,000) cooling capability at a Seasonal Energy Efficiency Ratio ("SEER") of about 10.5. This is typically found by adjusting the Energy Efficiency Ratio ("EER") which is defined as the cooling capacity over a period (e.g., BTU/hour) divided by the power used during that period in kilowatts at a particular outdoor temperature. Thus, EER is a mixed unit, BTU/hour/kilowatt. In dimensional terms, it's unitless but it's expressing how much energy it takes to move any particular amount of heat from inside to outside in a particular amount of time. Residential central air conditioners installed in the United States after 2006 are required to have a SEER of at least 13.

Air conditioner cooling capacity can also be rated in "tons," equivalent to the ability to move 12,000 BTU/hour from inside to outside in an hour. This unit hearkens back to the days when ice was used for cooling and the melting of a (short) ton of ice removes about 288,000 BTU from the environment, so doing so in a day uses 12,000 BTU per hour. My air conditioner thus provides the equivalent cooling capacity of about 5 tons of melting ice and hence is a five ton unit. Furlongs per fortnight anyone?

Now interestingly, working it out, my air conditioner will move 56,000 BTU of thermal energy from the inside my house to the outdoors in an hour. 56,000 BTU/hour is energy divided by time or power and is equivalent to about 16,400 watts. But the EER is about 9.5 so, since EER=(btu/hour)/watts, the electrical energy input is (BTU/hour)/EER or 5,890 watts. That's nice, my air conditioner produces about 2.8 times as much heat output as electrical energy input. Has Goodman Manufacturing succeeded in defying the law of conservation of energy and the first and second laws of thermodynamics? And by specifying a SEER no less than 13 is the U.S. government requiring changing the laws of physics?

No. The electrical input is used to compress a fluid and pump it around a circuit (absorbing heat from room air by changing from a liquid to a gas in the evaporator coil, then releasing it by changing back to a liquid) and to power a fan to blow the air cooled by the coil into the rooms of the house. Thus, the work being done is the movement and compression of fluids. The entropy of the air inside the house is decreased, that of the air outside is increased more in strict accordance to the second law.

The functioning of an air conditioning system is the same as that of a refrigerator and the most wonderful television series of all time, "The Secret Life of Machines," covers the refrigerator (among many other fascinating topics), here. A series of three YouTube videos comprising that episode are embedded below, but all episodes can be downloaded in their entirety. I heartily recommend doing so.

If my air conditioner is operating at a SEER of 9 (it's 14 years old after all), and I replace it with a new one with an SEER of 14, what can be saved? Well, I'll use 9/14 as much electrical energy, and I estimate that I'm using 560 hours per air conditioning season at about 5 kilowatts costing about $0.125/kilowatt hour. This will cost me about $350. If I buy the new unit, I'll spend (9/14)*$350 or $225, saving $125. It will take a very, very long time to pay for the new unit at that rate.