Maximum amount of energy from a gallon of gasoline

Whenever the discussion of the mythical “300mpg carburetor” comes up, I like to inject the comment that there is only so much energy that be derived from a gallon of gasoline. Of course, this kills the conversation.

Well, just how much energy IS available from a gallon of gasoline? Assume 87 octane.

I know there are a lot of factors to consider as far as cars are concerned. A Geo Metro could get 45mpg while a Ferrari would only get 9mpg. Then again, a Metro may weigh only 2000lbs and have an engine that makes less than 90hp while the Ferrari could weigh over 3000lbs and have an engine capable of 300+ hp.

How would you figure such a thing?

From http://www.spinglass.net/scooters/thumb.html

Gasoline: 125000 BTU/gallon (HHV)

There have been cars built that can do 200MPG and more. Problem is, they carry one person, no cargo, and go 20MPH.

What you should point out is that the energy losses in gasoline aren’t in the carburater - the fuel which enters the cylinder is already being pretty much completely burned. The energy losses which prevent the car from getting better mileage are in drag and rolling-friction losses. Mileage can be improved - at the cost of cargo room, carrying capacity, handling, safety, and increased purchase cost.

According to my Bosch Automotive Handbook, gasoline has a calorific value of about 42 megajoules per kilogram. Its density is about 0.75 kg per liter. For value in calories, 1 kilocalorie = about 4.2kJ.

I apologize, I’m tired and late for dinner, so I’ll let someone else figure out how many joules or calories per gallon (= about 3.78 liters).

I think the thing that limits how much you can get out of fuel is mainly the thermodynamic efficiency. This is the temperature of the hot gas - temperature of exhaust gas divided by temperature of the hot gas.

If the high temperature is 1500[sup]o[/sup]F and the exhaust temperature is 750[sup]o[/sup]F the thermodynamic efficiency is 30% which is about what this number always is.

Assume E=MC^2. Figure the mass of a gallon of gasoline and solve the equation.

Well, yeah, if you could convert every portion of its mass to energy! You could probably supply the energy needs of Cincinnati for a respectable interval if you could do that.

I think this is incorrect even if you could get gasoline to undergo a nuclear reaction.

As AHunter3 implied, e = m*c[sup]2[/sup] applies to the mass that is lost in such a reaction and not to the total mass of the input to the reaction. That is, a nuclear reaction has an input mass and an output mass + energy. The energy equals c[sup]2[/sup] times input mass - output mass. Only if all the gasoline atoms disappeared would RealityChuck’s answer hold.

You can also inject water along with the gas. The result is like a steam engine: the exhaust ends up cooler, and the expanding steam drives the piston harder. Why don’t engine designers use this effect? Maybe it rusts out the engine?

There is an optimal fuel-air ratio for gasoline, called the stoichiometric ratio, it’s somewhere around 17 to 1 or in there somewheres. That’s why I don’t believe that miracle carbs exist.

17 miles to the gallon with a 700 pound block and 1500 pound body is really pretty good, considering.

Of course, if we could come up with a gallon of anti-gasoline, then E=mc[sup]2[/sup] would work nicely.

hmmm… (4kg)(3x10[sup]8[/sup]m/s)[sup]2[/sup] = 3.6x10[sup]17[/sup] joules. Enough for a few rides around the block, I’d imagine.

My (diesel) vehicle took some water into the engine when it was splashed (read inundated) by a large sheet of water thrown up by a cement truck driving at speed through a foot-deep puddle; the result was explosive compression that did all sorts of nasty irreversible damage to the engine (but of course that’s primarily because the engine wasn’t designed with this in mind).

OK guys! This is a great thread. I’m gonna get a bit techo here, but fear not, I write well and I’ll keep it interesting the whole way thru.

The real critical issue with an internal combustion engine is this - what is the peak energy conversion efficiency obtained when liberating the potential energy within fuel and then converting that into mechanical or ‘kinetic energy’.

Some of the threads above touched on the subject - and remember, Newton’s Third Law states that all energy is maintained - if only placed else where in the form of heat.

It turns out that even the most technologically advanced internal combustion engine is lucky to convert greater than 38% of it’s fuel into at-the-crankshaft horsepower. Everything else is lost purely to the cooling system which attempts to prevent the engine block turning into a lump of molte metal. And it would you know - gasoline in particular burns at a hot enough temperature to melt almost any metal. Ethyl Alcohol burns rather at lower temp but is less powerful as a result.

In Formula One, they have a really unique measurement which is of vital interest here in this article. They call it the specific power consumption measurement and it works thus - the number of grams of gasoline burnt per horsepower per hour during engine operation. Obviously, the higher the engine rpm, the greater the horsepower, but the specific power consumption figure does not.

Apparently, in 1982 when turbos were really kicking in, Ferarri were consuming 310grams per horsepower per hour. They did this because they had to run their engines incredibly rich in fuel air mixture to keep the engine temps low enough to stop them from self immolating. The heat sink of the turbos was murder I’m told. Now extrapolate that figure to a circuit like Monza. (I’m getting to a real good point here by the way). At Monza, the race ran for 1hr20mins in 1982 - that is, 1.33 hours. (roughly)

The Ferrari engine in 1982 produced 650bhp from a 1.5 litre turbocharged v6 - pretty god I think you’d agree. At 310 grams per hp per hour, that was 650 x .310kgs of fuel burnt per hour. That is, 210kgs of fuel! And the race ran for 1.33 hours. Those poor Ferraris which weighed only 600kgs at the time without fuel had to carry at least 250kgs of fuel to finish that race! Almost half their weight again in fuel… no wonder their lap times got lower as the race proceeded. Remember, back then, no fuel stops…

So… then, as the 1980’s wore on, fuel consumption rules were introduced into F1 - something which left everyone aghast at the time - but it was done to stop some engine manufacturers just running their engines at obscenely rich levels and thereby driving around in fuel bombs. So, the engine manufacturers introduced amazingly intricate engine management systems along with telemetry where they could ‘tune’ the engine in mid flight to meet the fuel restrictions. By the way, the weight limit was 220kgs for the race, so it was still PRETTY obscene how fuel consumptive they were.

Nonetheless, by the end of 1988 - the last year of the turbos - Ferrari had dropped their specific fuel consumption from 310 grams per hp per hour to just 205 grams - a truly impressive reduction in consumption - and yet, their engines were now producing close on 800 hp apparently.

Conversely, in passenger cars, rare is the road car capable of getting under 300 grams per hp per hour. They are very innefficient compared to an F1 engine - at least in terms of liberating potential energy and turning it into mechanical movement. Also, rare is the road car which is capabale of rising above 32% liberation per litre of fuel burnt. This compares to race engines which nudge 38%. The rest is lost purely to heat.

So how do F1 engines do it? Simple - they run very, very hot with oodles of research into lubricants which function at such high temperatures. The hotter an engine runs, the less energy is lost to the cooling system - thereby leaving more behind for ‘mechanical energy’.

We naturally worry immensely about a road car running hot - it usually signals a big maintenance bill just around the corner, but in racing, hot is good. Indeed, most F1 engines have pre prgrammed computer warm up routine where the engine just does weird rev patterns for 15 minutes after starting even before turning a wheel.

In summary - the measurement that REALLY counts - the one which truly demonstrates how advanced your engine is - is the ‘specfic fuel consumption’ measurement - that is, grams per horsepower (kw if you’re metric) per hour. This is the one that counts. Then, the only other issue is how much horsepower do you need? The less horsepower required, the lower the overall consumption over a given distance.

Obviously, a Mack Truck needs greater horspower over a 100 mile journey say than a moped because it has massive amounts greater friction and wind resistiance.

Conversely, a car which is doing 220kph also needs massive horsepower to because it too has massive friction as well - this time in the form of wind resistance.

Apparently, in F1, if you lift off the the throttle at 300kph and don’t even touch the brakes, the negative retardation in speed is over -1G just due to the wind resistance of the rear wing alone - such is the downforce generated at high speed.

Thanks for reading this far guys. I really love this subject - even though I’m actually a database designer by trade!

Well written Boo Boo Foo!
I’m glad to see tht the board is attracting competent people, and not only teenagers with poor spelling and a wish to be heard.
It’s a pleasure to read such a well-written and obviously well-researched post. Please hang around!

Now back to the subject:
Your unit of mg/(bhp*h) ought to translate into the efficiency of the fuel-to-motion conversion.
Assuming energu content according to Gary T: 42MJ/kg, and using 1.34Hp/kW, 3600s/h. I get that a 100% efficient 1hp engine would only consume 2.7Mj per hour, which corresponds to 62g of petrol.

The Ferrari you mention with a figure of 205g/(hp*h) seems to be just at 30%. (assuming of course that they use standard fuel - which they don’t, but I would guess that the energy content of what they burn is about the same.)

I have heard rumours that the new lean diesels are past the 30% efficiency, but have so far not seen any hard data to back it up.

“Gasoline is pretty amazing stuff. First, it is a true fuel, clearly delivering net BTU’s of energy to the on-the-books economy. Second, it offers an incredible energy storage of 9000 watt hours per liter and 13500 watt hours per kilogram. Compare this to the best of today’s lithium batteries at 300 watt hours per liter and 150 watt hours per kilogram. Hydrogen? At its normal STP pressures (gaseous state) hydrogen offers an outstanding 39,000 watt hours per kilogram, but only a pitiful 3.5 watt hours per liter. And even if you compress the hydrogen (which adds significant inefficiencies), there still is four times less hydrogen in a gallon of liquid hydrogen than there will be in a gallon of gasoline.” - Don Lacaster http://www.tinaja.com/

Close–14.5 to 1, by weight (i.e., approx. 14.5 kg of air are required for complete combustion of 1 kg of gasoline).

So what it sounds like is that the internal combustion engine is a relative dead end as far as increased efficiency is concerned. Obviously, future cars will still have friction and air-resistance issues (although these may be mitigated somewhat with new technology, zero air resistance is not possible or practical).

Anyone know the maximum theortical efficiency of a fuel cell?

I remembered a discussion from some time ago about ceramic engines, so a little Google-work came up with:

(link)

Since it seems that a limiting factor on the efficiency of current engines is the heat at which they can operate, this would seem to be a good way to get more out of internal combustion engines. Comments?

Yes, and it appears auto makers are onto this, Cerowyn.

I saw an article a couple of years ago in a car magazine mentioning that one of the automakers is researching this application.

When it comes to getting work out of fuels, calorific value, is not too good a parameter. Let me explain why.

The ideal chemistry of Combustion:

CxHy + (x + y/4)O2 —> xC02 + (y/2) H2O + HEAT

Note that there is exactly the required amount of oxygen here (no heat is wasted in heating excess oxygen).

Consider any hydrocarbon burning as shown above. There are various ways to measure the heat released in this reaction namely : Higher HeatingValue (HHV) Gross Heating Value (GHV) Gross Calorific Value (GCV) Total Calorific Value (TCV) Lower Heating Value (LHV) Net Heating Value (NHV) Net Calorific Value (NCV). These values are different because the products of the combustion carry some heat with them (H2O carries a lot) and different values arise if you take into account the heat in the products can be recovered by cooling them down. Also, if u use excess oxygen you’ll get some heat taken off in that too (Etc. etc.).

The real chemistry of Combustion:

CxHy + zO2 + 3.7N2—> aC02 + bCO + cC + dCnHm + yH2O + eNOx + HEAT

The above shows that the Carbon in fuel may not be completely burnt (seen soot ? ). The Nitrogen in the air will react with the oxygen and thats assuming there is no other contaminants like sulfur etc. in the fuel.

To get the highest temperature, ideally there should be exactly the amount of Oxygen (air) required to convert the CxHy to CO2 and H2O (this is called the stoichiometric amount). However, this never happens in real life because this gives rise to unburnt carbon (soot particles etc. ) SO some excess air is always use (Usually 10% ). Now you cannot use more excess air because after combustion if more free oxygen is available , you’ll have more NOx ( not desirable). Hence, the amount of excess air is an optimization, (less is bad and more is bad too)

The QUALITY of heat:

The quality aspect of heat is something many don’t understand. Putting in layman’s terms - A pound of water at 100° can do more work that 100 pounds of water at 1° - although both have the same amount of energy (well approx). The higher the temperature the better is the available energy or Anergy.

You can think about it in terms of Carnot Cycle too. That is higher the supply temperature the more work can the engine do.

How to improve Enginer performance or why does my car give such poor mileage :

1> Increase the temperature of combustion will be an immediate attractive answer. Well, turns out engine materials cannot withstand these high temperatures hence there’s a limit to that.

2> Use a gas turbine and a high temperature. Yes gas turbines can withstand high temperatures but the problem is that at higher temperatures NOx is formed more and more. So that limits the temperature you can go upto unless you use pure oxygen.

3> Decrease the temperature of the exhaust thus absorbing more of the energy. Again, the problem here is the NOx (also SOx if present). NOx changes to Nitric/Nitrous Acid and SOx to Sulfuric/Sulfurous acid once you cool down the products below the dew point of NOx and SOx. This acid destroys metals. If you live in a colder place you can see your tail pipe being eaten away by this acid. Also, sometimes air leaks cause moisture in engine exhaust and u can see it get eaten off.

4> Recycle the exhaust gases to preheat the air / fuel coming into the engine. This is done in big turbines and furnaces. But heating air is not too economic in small engines.

Thats my big 2 cents

The best solution so far is fuel cells ( if they become commercially feasible ).

:smiley: :smiley: :smiley:

Your engine must have ingested some solid water, although that is hard to undrstand, what with intake filters and the like.

However, aircraft engines in WWII and after were equiped with water injections systems. There is a maximum time limit that an engine can be operated at maximum output. The use of water injection prevented overly high temperatures from developing and allowed running the engine at full output for a longer time. To a fighter pilot in desperate straits, this was a blessing.

I do believe, however, that whenever water injection was used the pilot recorded it on the aircraft mechanical condition report (Air Force Form 1) so that the engine could be thoroughly checked.