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Old 2008-07-04, 13:30   #1
Mini-Geek
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Default Maximum theoretical MPG

What is the maximum theoretical MPG that can be obtained by burning a gallon of diesel or gasoline? According to http://auto.howstuffworks.com/diesel3.htm,
Quote:
On average, 1 gallon (3.8 L) of diesel fuel contains approximately 155x106 joules (147,000 BTU), while 1 gallon of gasoline contains 132x106 joules (125,000 BTU).
How would you convert that to how many MPG could possibly be obtained from 1 gallon? If you had a 0.01 drag coefficient car with a 99.99% efficient engine and near-frictionless tires (I'd say 0, 100%, and frictionless, but then you'd want to complain about breaking laws of physics), and some weight (I don't know what weight...maybe try it with a few different weights ranging from a light person in one of those tiny smart car things up to a train or something), what MPG could it get?
It would need to accelerate at a reasonable speed (open to interpretation, but I don't want 0-60 in an hour), and we're talking about burning a gallon of diesel or gasoline and the joules mentioned in the article quoted (or a more accurate article, if that one's wrong), not about E=mc² and the huge amount of power you could get from even, say, 1 gram of perfectly converted matter.

While writing this, I'm noticing that with almost 0 friction, you could apply some power then coast the rest of the way, so maybe a more realistic drag coefficient like 0.16-0.30 and a standard tire friction would be better to ask the question with (but keeping the very high efficiency engine).
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Old 2008-07-04, 14:00   #2
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Quote:
Originally Posted by Mini-Geek View Post
What is the maximum theoretical MPG that can be obtained by burning a gallon of diesel or gasoline? According to http://auto.howstuffworks.com/diesel3.htm, How would you convert that to how many MPG could possibly be obtained from 1 gallon? If you had a 0.01 drag coefficient car with a 99.99% efficient engine and near-frictionless tires (I'd say 0, 100%, and frictionless, but then you'd want to complain about breaking laws of physics), and some weight (I don't know what weight...maybe try it with a few different weights ranging from a light person in one of those tiny smart car things up to a train or something), what MPG could it get?
It would need to accelerate at a reasonable speed (open to interpretation, but I don't want 0-60 in an hour), and we're talking about burning a gallon of diesel or gasoline and the joules mentioned in the article quoted (or a more accurate article, if that one's wrong), not about E=mc² and the huge amount of power you could get from even, say, 1 gram of perfectly converted matter.

While writing this, I'm noticing that with almost 0 friction, you could apply some power then coast the rest of the way, so maybe a more realistic drag coefficient like 0.16-0.30 and a standard tire friction would be better to ask the question with (but keeping the very high efficiency engine).
I think drag/friction will be the dominant factor. Once your kinetic energy is obtained you only need a small amount of fuel to maintain speed. If you start at the top of Mt. Everest then you should be able to get a very good MPG, and even infinite if you start with an empty tank from way up there :)

As for more practical values, ballpark figures would suggest approximately 4-5 times what most cars are currently achieving. I base that guesstimate simply on the (perhaps flawed) estimate that current vehicles are about 20-25% efficient in the engine.
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Old 2008-07-04, 17:26   #3
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The most current maximum achieved value would be 12,665 MPG !
(See here)

Last fiddled with by michaf on 2008-07-04 at 17:26
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Old 2008-07-04, 17:27   #4
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About 15 years ago, some grad students in Germany built a vehicle that got 1600 mpg. It had a single cylinder diesel engine with a very long stroke and the 'vehicle' was an ultralight chassis riding on 4 bicycle wheels.

Friction is the major problem with any effort at improving mileage. There are numerous sources of friction in any typical Internal Combustion Engine.

1. Engine friction (bearings, pistons, etc)
2. Drive train friction (transmission, univeral joints, bearings, axle, etc)
3. Tire friction from contact with the road.
4. Air friction, expecially at speeds over 50 mph.
5. Gravity effects if you are going up and down hills.

The most important of these is air friction. Gravity in and of itself is not the problem, rather, climbing a hill takes more energy to overcome gravity which means the engine has to work harder which increases friction all through the drive train.

Another effect to consider is that all gasolines are not created equal. Different formulations have more or less extractable energy. Also, the condition of the engine has a major impact with older engines typically getting significantly less mileage than identical new engines.

You might look at current efforts to build a solar powered car. There are some fairly good ones that can drive at speeds of 40 mph as long as the sun shines. They are limited by the current technology of electric motors which max out at about 92% efficiency.

DarJones
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Old 2008-07-04, 18:02   #5
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Old 2008-07-13, 17:04   #6
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Gasoline engines in newer, compact cars are approximately 20% fuel efficient. http://automobiles.honda.com/fcx-cla...omparison.aspx However, this is only the conversion to kinetic energy in the engine- to find the actual MPG maximum potential- you need to consider more factors such as tire friction, road friction, wind friction (I agree with everyone else that friction is, by far, the dominant factor). Conversely, newer hydrogen fuel-cell vehicles can achieve 50-60% energy efficiency in the engine.

The Honda FCX Clarity, 60% efficient, delivers 280 miles on a single tank of 4.1kg of Hydrogen compressed at 5,000 psi. Correct me if I am wrong-- I may be forgetting a step, but figuring the mileage is simple for this. Hydrogen gas is diatomic, so 2.0 grams per mole. Therefore, 4,100 grams / 2 grams per mole equals 2,050 moles (psi can be ignored as it is irrelevant). Then, 280 miles / 2,050 moles = 0.14 miles per mole or approximately 140 miles per kilomole.

According to their site, 140 miles per kilomole equates to an average of 72 mpg.

Last fiddled with by Primeinator on 2008-07-13 at 17:15
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Old 2008-07-14, 01:39   #7
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Quote:
Originally Posted by Primeinator View Post
(I agree with everyone else that friction is, by far, the dominant factor).
I disagree. Vehicle weight is the issue. One can build a vehicle that has 1/4th the weight. F=mA, no? There is a gorup working on making ultralight vehicles, basically carbon fibre parts. This would also allow more efficient assembly lines. A door that weighs in at ~2 kilos, can be handled by a worker without fancy machines. Why do you think Boeing is going to CF for it's 787, weight, thus fuel economy.
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Old 2008-07-14, 02:29   #8
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Quote:
Originally Posted by Uncwilly View Post
I disagree. Vehicle weight is the issue. One can build a vehicle that has 1/4th the weight. F=mA, no? There is a gorup working on making ultralight vehicles, basically carbon fibre parts. This would also allow more efficient assembly lines. A door that weighs in at ~2 kilos, can be handled by a worker without fancy machines. Why do you think Boeing is going to CF for it's 787, weight, thus fuel economy.
I disagree. Friction is where nearly ALL the energy goes. When accelerating, you're burning fuel, but increasing the kinetic energy of the car and occupants (minus engine friction, driveline friction, rolling resistance of the tires, and wind resistance/friction). If it's on an uphill incline, you're also increasing the potential energy of the vehicle.

Where does that kinetic energy go when the vehicle comes to a stop on level ground? Whatever isn't dissipated as friction above (due to engine braking, for example), is turned to heat by the brake pads/shoes and discs/drums. Obviously, the heavier the car, the greater such friction losses will be (so of course the weight matters), but the energy loss is still due to friction.

Probably the other biggest loss besides friction would be due to combustion inefficiencies released as unburned fuel (or burned by the catalytic converter), exhaust heat, and cooling system heat.

An aircraft is fundamentally different than a land vehicle, in that it requires lift to stay in the air. Lift is generated by airflow as a side effect of its forward momentum (initially provided as trust by the engines). The heavier the aircraft, the more lift is required, so the more wing area is required, and the more drag this imposes (requiring more thrust to overcome). I guess friction would still be a primary loss of energy (in terms of parasitic drag and induced drag...the latter a result of lift production), but weight has a greater impact on an aircraft than on a car, I think (I could be wrong about that).

Weight DOES matter, but a heavier land vehicle will tend to have more energy losses to friction than a lighter land vehicle due to a heavier drivetrain, larger engine, etc, so they are related to each other. The increased momentum of a heavier vehicle requires greater braking power to slow down, meaning more friction is needed, and more energy is lost than in a lighter vehicle (ever seen a semi truck brake fire?).
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Old 2008-07-14, 05:22   #9
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Especially where aircraft are concerned, the main point
of reducing weight is to reduce the drag while cruising.
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Old 2008-07-14, 22:45   #10
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Quote:
Originally Posted by davieddy View Post
Especially where aircraft are concerned, the main point
of reducing weight is to reduce the drag while cruising.
I totally agree...but let me ask you, could this apply to cars as well? Consider two otherwise equal cars, except one has a 1,000 pound higher gross weight evenly distributed...that increases the bearing load on each hub by 250 pounds, as well as increasing the tire loading that much, so (I'd theorize) that it would increase bearing friction slightly under level cruise conditions, as well as increased energy loss through sidewall flex in the tires. But my guess is that it's not as significant as in an aircraft.
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