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ContinuousWave: Whaler Performance
Fuel management 101
|Author||Topic: Fuel management 101|
posted 01-06-2004 12:30 PM ET (US)
This was snt to me from another forum:
One pound of fuel will produce 2 horse power per hour. Gasoline weight is 6 lb/gal. I have 150 Yamaha. Running at 75% is 112.5 Hp. Formula: (112.5 x .5)/6 = 9.38 GPH. Formula works for anything. No mater if it's a lawn mower or airplane.
Find the flaws in this theory!
posted 01-06-2004 01:27 PM ET (US)
Your outboard motor (or lawn mower or airplane) converts energy stored in petroleum fuel into rotating power to move your boat (or mow your lawn or fly your airplane). The efficiency of your outboard (or lawn mower or airplane) needs to come into the equation somewhere. In all internal combustion engines, there is energy lost as heat that does not contribute to converting the energy stored in fuel to rotating power. Also, the fuel must burn completeley to convert all the stored energy into power. The 2 h.p./hour/lb. of fuel assumes an efficiency, but it can't be the same for every motor.
Take your 70 h.p. 4-stroke; at max RPMs it develops 70 hp, the same as your old 70 h.p. 2-stroke did at max RPMs. If you ran them both for one hour, WOT, on the same boats in the same conditions, I think most of us would expect (based on your testimony) that the 70 h.p. 4-stroke would burn less fuel. Your formula suggests that they would burn exactly the same amount of fuel. How can this be? The 70 h.p. 4-stroke is more efficient at converting energy stored in fuel to rotating power than the old 2-stroke was. We know that a lot of the fuel in conventional 2-strokes is not burned completely, but instead it goes out the exhaust, making them less efficient.
posted 01-06-2004 02:02 PM ET (US)
Actually they burn about the same....along with DFI at WOT. The DFI and 4 strokes are more efficient at lower rpm's.
This is not my theory, something somebody sent me, I will post my thoughts later, just want to see if anyone has the same thoughts as me.
posted 01-06-2004 02:29 PM ET (US)
I agree with Andy that the efficiencies of different motors greatly affect fuel consumption. But we can still use this equation as a basic guide for fuel consumption.
To open a can of worms, think about how a 2-stroke and 4-stroke engine operates. A 2-stroke has one combustion stroke on each engine rotation. A 4-stroke has one combustion stroke for every two rotations. If both have the same displacement and are turning at the same rpm, either the 4-stroke is using 1/2 of the fuel of the 2-stroke to produce the given power, or the 2-stroke must be twice as efficient as the 4-stroke to make the same power. Since the efficiency of different engines do not change much and will only fluctuate a few percent, the first theory is probably more correct than the second.
Now, if we assume this to be true, based on Nick's formulas, a 150 hp 2-stroke pushing a boat at 20 knots at 3000 rpm (assume 55% of WOT) will use about 6.8 gph for 3.4 mpg. So, a 140hp 4-stroke turning at 3500 rpm (assume 58% of WOT) and pushing the boat at 20 knots will use about 3.4 gph for 6.7 mpg. You know what? This is fairly close to what I saw and to what I am seeing in respect to mileage for my boat, although both numbers are a little higher (less than 10%) than what I actually get. Of course, I have no idea how a modern DFI or HPDI will fit into this equation...and I didn't calculate prop pitches or gear ratios...
Please rip this assumption up!
posted 01-06-2004 02:51 PM ET (US)
The angle of the dangle....
posted 01-06-2004 04:17 PM ET (US)
Bigshot - whoever gave you that formula also assumed a flow rate. Somebody had to get the unit of time in the equation. That is, 1 pound of gasoline gives about 20,500 Btu and 1 Btu gives about 0.02358 HP. Note that there is no time units. In fact, that flow rate of gasoline was 0.00414 lb/hour - pretty skinny flow. In a nutshell - that formula is not of value!
Further, and as Andy points out, the effeciency has to come into the picture as all engines have different effeciencies.
Forget it ------ Jerry/Idaho
posted 01-06-2004 04:21 PM ET (US)
Let's apply your forumula to an aircraft.
A Citabria I fly has a 115 HP 0235 Lycoming engine.
Considering cruise at 75%, your formula [(86.25 x .5)/6 = 7.1875 GPH] comes right on with the 7 GPH that I've always used for flight planning. And here we're talking a four-stroke.
Works for me!
posted 01-06-2004 05:56 PM ET (US)
2545 btus per hour = 1 horsepower-hour
so burning a pound of gas (20,500 btus) for an hour produces 8 horsepower-hours, if the engine is 100% efficient.
It appears the formula you got assumes 25% efficiency for the engine. That's probably not too far off, but as many here have already pointed out, it varies.
posted 01-06-2004 06:22 PM ET (US)
A 4-stroke burns approximately twice the amount of fuel as a 2-stroke does for each power stroke. The 2-stroke's inefficiency mostly comes from having an open exhaust port during charging of the combustion chamber with the next fuel/air charge. The direct injection system solves that problem by waiting to inject fuel (again about half as much as a 4-stroke per power stroke) into the combustion chamber until after the exhaust port is closed. Some DI 2-strokes are actually more fuel efficient than comparable 4-strokes at certain engine speeds, most notibly at idle where the 4-stroke valve train appears to be a big drag on operations. Recent performance reports for an E-Tec 75 HP DI 2-stroke on an 16 foot alumimum boat showed a whopping 20+ MPG at idle speed. The best I've seen comparable 4-strokes do is about 12-14 MPG.
posted 01-06-2004 11:04 PM ET (US)
On this forum we've been talking about how each horsepower consumes 0.5 pounds of fuel per hour for the past several years. Same thing as you're saying.
posted 01-07-2004 12:50 AM ET (US)
As others have said, the equation is wrong in that it does not account for the differences in efficiency between different types of internal compbustion engines. I know that I forgot so much of my chemical engineering background that I can hardly balance my checkbook, but this equation has obvious flaws, or at least, limitations.
Based on my experience, here is a better rule of thumb. 2 stroke engines can produce about 10hp-hr per gallon on gas. Smaller 2 strokes are less efficient, as are any 2-strokes when used in low load applications. Four stroke engines get about 12-14 hp-hr per gallon, and don't vary much across the load range. Diesel engines typically get 16-20 hp-hr per gallon, with Detroit diesels being on the lower end of the efficiency range, and four stroke, turbocharged, seawater aftercooled engines being on the upper end of the enfficiency range. Diesel engines get better fuel economy becuase there is 1) more energy in a gallon of diesel than a gallon of gas, and 2) the the compression ignition engine, with it's higher compression ratios, is more efficient at converting the fuel combustion into useful motion, rather than wasted as heat. This is based on a lot of personal research on different engines types.
Try it yourself: if you have a flow scan and a two stroke engine, go to WOT. If you have a 200 hp engine, your flow scan is reading right around 18 - 20 gallons an hour. The proof is in the pudding.
PS: Why did I bother to post this. This was a silly subject.
posted 01-07-2004 11:20 AM ET (US)
Peter....my 70hp Evinrude/Suzuki 4 stroke gets over 20mpg at 1000rpms. At 1000rpms I am doing between 5 and 6mph and burn 0.2gph according to magazines.
posted 01-07-2004 12:23 PM ET (US)
Nick, that's 25-30 mpg. Peter, I wonder if the 4-strokes you were looking at are carbed (i.e. Yamahas)? 1000 rpm is about the worst possible place you can run a carb... it's just beginning to transition off the idle circuit, which is necessarily rich due to low and variable velocity through the venturi and intake manifold runners. A sequential port EFI four-stroke should do as well as the e-tec here.
posted 01-07-2004 12:30 PM ET (US)
Here's the Suzuki 70 test:
posted 01-07-2004 01:16 PM ET (US)
Hadn't seen that Suzuki report before. Looks like similar consumption rates at idle. However, wouldn't the valve train be a greater drag reducing efficiency, everything else being equal?
posted 01-07-2004 01:24 PM ET (US)
In the overall scheme of things, I don't think the valve train robs that much power. Keep in mind the compressed valve springs return most of the energy that was used to compress them, to the camshaft as the lifters ride down the trailing edge of the cam lobe. And a toothed belt drive or chain drive is relatively efficient. Certainly there's some loss in a drivetrain, but I don't think it's all that significant.
posted 01-07-2004 01:33 PM ET (US)
Moe, I would think that frictional losses must be higher unless there is similar frictional losses in the DI mechanisms.
At those low flow rates, the measurement error is probably significant. The accuracy of most flow gauges is rated on a full scale basis meaning that the error is more significant at the low end.
In any event, the DI 2-strokes have caught up to the 4-strokes in terms of fuel efficiency because they are not pouring fuel out the exhaust port anymore. If everything else is equal, they should be considered a little more energy efficient because they need not push the extra weight of the valve train around.
posted 01-07-2004 03:30 PM ET (US)
My hull is a 200lbs lighter but numbers are close. I run about 37-3800 at cruise giving me about 25-27mph depending on conditions. My top end is also 39+. The jackplate might improve the efficiency being less drag....who knows. I also burn less than the 3+/hour that would be in that rpm range. I ran 4.2 hours on a 6 gallon tank on Sunday and did not run it dry. I would say that roughly 1 hour or so was off plane, rest was running at cruise. That engine absolutely friggin amazes me.
posted 01-07-2004 04:44 PM ET (US)
Peter, IMHO, there are too many other differences to say other things being equal. The proof has to come in real-world tests, and those are generally showing the DFI two-strokes have caught up to the EFI four-strokes in mileage and emissions, and in some cases, they're now pretty close in weight.
The important thing is that there's still a choice for proponents of each.
posted 01-07-2004 10:18 PM ET (US)
As a rule of thumb, Bigshot's formula is fine. And as jimh stated, it's the same as .5 lbs/hr/hp. It is however only a rule of thumb. I wouldn't base my fuel calculations on it but as Moe pointed out, 25% efficient is a reasonable estimate for gasoline powered engines. BTW, losses due to internal friction might be higher than you realize. I don't recall exactly but I think five percent mechanical loss would be a reasonable number. I used to know alot of these numbers when I designed cogeneration plants - almost twenty years ago.
According to the DOE, gas engines are about 30% efficient, diesel's are around 45%. Check out the following, nothing spectatular but interesting info about direct injection diesels. Kind of funny "... tiny computers..." like that's something new. No mainframes on that puppy!
I also fly a citabria, 1968 7ECA (I think that makes three citabria pilots on the whaler forum). That makes more citabria pilots on the whaler forum than on the usenet group, rec.aviation.owners:) Strange but probably true. If any of you citabria pilots need citabria information as it relates to boston whalers, check out the yahoo group for citabria pilots. You'll never figure out which one I am.
posted 01-08-2004 09:51 AM ET (US)
The main problem with this formula that I see is how do you know where 75% of your HP is? I think most people think that 75% is 3/4 throttle but we all know better. 4 strokes make less hp at 3/4 throttle than a 2 stroke BUT make more torque. What makes a vehicle go....torque of course, hp dicatates how fast it will get there(rough definition). The only way this formula will work is if you had an onboard dynometer, I think a flo-scan might be easier to hook up. Also what type of fuel are we talking about here? Gas, Diesel, Methanol, gasohol(corn), Propane, etc. I think the fuel media has a big factor on it as well but I could be wrong.
posted 01-08-2004 10:44 AM ET (US)
Did you mean where does a certain percentage of your horsepower occur?
Horsepower = (torque X rpm)/5252
If you have one of those relatively flat torque curves many engine tuners would die for (like the one on your Suzuki), horsepower is going to be basically linear with rpm.
But if you have a peaky torque curve, the slope of the horsepower will be steeper before the torque peak and shallower after it.
Remember, this fuel thing is a ROT so a SWAG at horsepower for the input oughta do. :-) I just wish all motor manufacturers would publish torque and horsepower curves.
posted 01-08-2004 10:53 AM ET (US)
You take a young horse. Foist thing in the mornin', you can get that colt to run hard for a hour on nothin' but a good piss and a sugar cube to start his day. But your old cart horse, now he ain't goin' nowhere if he hasn't had a three pound bucket of oats, a good waterin' and a complete relief before he's been hitched into the leads. You'se gots t'figure the age o'the beast into the equation somewheres. Now later in the day, that young colt's goin' t' take HIS two-pound bucket and a nap before he's any good. And that old horse, he'll pull all day on just what he started with, just catchin' a wink between stops. So you's gots to consider the habits and personality of your horse as well. Ah don't see ANY way to find one cypher that'll do for all kinds o' horsepowers, so why bother?
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