Naked Science Forum
General Science => General Science => Topic started by: Nikita Strauss on 26/09/2011 19:30:06
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Nikita Strauss asked the Naked Scientists:
Hi Chris
I'm curious, is there a difference in fuel consumption on a full or empty tank of gas?
Would appreciate your answer!
Thanks
Nikita Strauss
What do you think?
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The amount of fuel in the tank has no effect on the efficiency of the engine. The mass of the fuel in the tank does have an effect on the fuel consumption of the vehicle, but not very much.
Also, it's really the average amount of fuel in the tank that matters. If you always completely fill it then only refill when it's almost empty, the effect is about the same as trying to keep it half full with frequent fuel stops, except that you'll waste more fuel with all the stopping and and starting.
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In the motor racing game, the weight of the fuel on board makes a tiny, but significant difference to lap times -- maybe tenths of a second.
Same power driving a lighter weight means faster speeds possible.
Only other consideration is that during the lifetime of a vehicle heavier impurities, condensation (=water), and sludge in the fuel can accumulate at the bottom of the tank. So with older cars that have done a lot of running, it is probably a wise idea to avoid running on the last 2 litres in the tank.
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Same power driving a lighter weight means faster speeds possible.
Not necessarily faster speeds, but higher average speeds because of greater acceleration.
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A full tank would have slightly lower fuel efficiency due to the added mass of the liquid itself. Although you would not notice a difference.
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As has been said, there is a difference, but it is not significant, unless you have a particularly large fuel tank. For most cars a full tank will weigh about the same as a slim woman. So you can run on a tank virtually empty and be able to pick up a hitch hiking piece of totty without having an adverse effect on your fuel consumption. Trouble is, the piece of totty might not be too keen on pushing the car to the next filling station when you run out of petrol (gas). Also, if the Mrs finds out you've been picking up totty in the car, you may discover how far and fast she can run and kick you up the butt at the same time.
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I'm curious, is there a difference in fuel consumption on a full or empty tank of gas?
The only difference will be the exrra "deadweight" of a full tank. For longer journeys at near constant-speed this will make very little difference, but if you're driving about town with lots of stopping and starting at lights/junctions, pulling inbetween parked cars to let other pass etc then the added mass that you keep accelerating will detriment the full economy to some extent. Similarly if you're doing a lot of driving in hilly areas so you're lifing that deadweight up the hills.
A typical UK car has a tank of 10 gallons (45 litres). If petrol has a density of 0.7kg/litre, thats just over 30kg of fuel.
If a small car + one occupant (driver) weighs say 1200kg, thats an additional 2.5% on the weight of the car. If we do a simplification and a worst case scenario that you spend all your time accelerating and braking (and never travelling at constant speed where energy is required merely to overcome friction, rolling resistance, wind resistance etc), then (since energy required to accelerate = (1/2)*m*v^2 ) this might worsen your fuel consumption by roughly 2.5%. Similarly for potential energy (hills) potential energy (m*g*h) scales with mass, so the deadweight makes a difference.
This is a simplified worst-case analysis. For normal driving the difference is probably quite a bit less, although if you drive particularly hard on the accelerator (where the engine efficiency probably drops anyway) the economy-impact of carrying extra deadweight could be disproportionate.
Much the same reasoning explains why the motoring organisations will exhort drivers not to habitually carry all the accumulated junk everywhere with them in the boot!
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A typical UK car has a tank of 10 gallons (45 litres). If petrol has a density of 0.7kg/litre, thats just over 30kg of fuel.
If a small car + one occupant (driver) weighs say 1200kg, thats an additional 2.5% on the weight of the car. If we do a simplification and a worst case scenario that you spend all your time accelerating and braking (and never travelling at constant speed where energy is required merely to overcome friction, rolling resistance, wind resistance etc), then (since energy required to accelerate = (1/2)*m*v^2 ) this might worsen your fuel consumption by roughly 2.5%. Similarly for potential energy (hills) potential energy (m*g*h) scales with mass, so the deadweight makes a difference.
Yes, but you are overlooking an important detail. The amount of fuel is not constant. The calculation should be based on the average amount of fuel in the tank, which is probably more like 55% of the tank capacity in the case of vehicles that are fueled to capacity.
The other problem with frequent refueling is that you waste a lot of fuel (and time) at every refueling stop.
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Rolling friction of a car dominates up to about 50 miles per hour or so; and is the main place that the power of the engine ends up (up to that speed).
Rolling friction is proportional to the mass of the car, including the fuel, so adding a gallon of fuel is the same as adding roughly 4 kg to a tonne vehicle (less in America, due to their smaller gallons); so ten gallons is about 4% of the weight of the vehicle and adds 4% to your fuel bill (if you spend most of the time below 50, less if you spend more time cruising at high speed, since the extra weight doesn't affect the aerodynamic drag).
There's also losses during acceleration; if you accelerate to a given speed, you need to use proportionately more fuel to do that, because the weight of the fuel slows you down, and you have to accelerate for longer, with the same, higher, throttle setting.
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Rolling friction of a car dominates up to about 50 miles per hour or so; and is the main place that the power of the engine ends up (up to that speed).
Not really. See FIGURE 3-1. http://onlinepubs.trb.org/onlinepubs/sr/sr286.pdf
At lower speeds, rolling resistance only accounts for about 4% of the fuel consumed.
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Careful here, that's 4% of the 13% of energy from the original fuel that makes it through to drivetrain.
The other bigger part is the 6% you lose in braking (which will depend on how you drive), so that's 10% of 13% or 77% of the propulsive energy that is affected by greater vehicle weight.
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Careful here, that's 4% of the 13% of energy from the original fuel that makes it through to drivetrain.
The other bigger part is the 6% you lose in braking (which will depend on how you drive), so that's 10% of 13% or 77% of the propulsive energy that is affected by greater vehicle weight.
Er, no it ain't [;D]
It's 4% of the total energy from the fuel, and it's not 4% of 13%.
Even if you completely eliminated rolling resistance, you would only reduce the fuel consumption by 4.1%, so it's a bit unlikely that adding a few kilograms in weight will increase your fuel bill by 4%.
The coefficient of resistance for a rolling tire is only around 0.01
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No, the diagram is showing that of the original energy in the original fuel only 13% makes it to the back wheels.
Of that:
4 parts out of 13 is lost in rolling energy
6 parts out of 13 go into the brakes
3 pars out of 13 go into airdrag
Making 13%.
If you were to get rid of all rolling friction you would save 4/13 of your fuel bill, because that 13% is the reason you are burning the fuel in the first place (neglecting the energy to run subsidiary components, lights etc.) So if that energy goes up by a percentage, you need that much more enegy in general.
So if you had twice as much of rolling losses, braking and airdrag, you would (nearly) double your fuel bill.
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If you were to get rid of all rolling friction you would save 4/13 of your fuel bill, because that 13% is the reason you are burning the fuel in the first place (neglecting the energy to run subsidiary components, lights etc.) So if that energy goes up by a percentage, you need that much more enegy in general.
So if you had twice as much of rolling losses, braking and airdrag, you would (nearly) double your fuel bill.
Erm, he he, I'm sorry, but have you considered a career in politics? ([;D] [;D])
You are calculating it as if you didn't have to pay for the gigantic amount of energy that simply goes into heating the atmosphere. Relative to that, the loss to rolling friction is peanuts.
To affect the fuel consumption by 4% by changing the weight as you suggest, you would have to double the weight of the vehicle. By eliminating rolling friction entirely, the best you could hope for would be a 4.2% improvement (100/96*100 = 104.2%)
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No, sorry, you've got it wrong.
The 13% is the useful bit. In other words for every kilojoule you get at the back wheels, you need 7.7 kjoules of petrol.
For example if the rolling friction is 0.02, and the car mass is about a tonne and weighs 10000 N, then for every meter you roll you need 200 joules. For every kilometer that's 200 kJ, and hence that's 1.54MJ of energy. The energy density of petrol is about 35 MJ/L, so that's 22km/l; over 100km to the gallon. You only normally get under a third of that, because of air drag and brakes, they're all very, very roughly equal.
So it's not 4% of the energy, it's 4 out of 13 parts of the USEFUL energy; it's completely not negligible.
So if you increase the weight of the car by X%, you increase the fuel consumption by about 10/13 X%, because the air drag is unaffected, but both the braking energy and rolling energy are increased proportionately.
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The 13% is the useful bit. In other words for every kilojoule you get at the back wheels, you need 7.7 kjoules of petrol.
Wolfekeeper, you are probably going to hate this, but it doesn't matter which bit is "useful". The fuel is consumed whether it does useful work or not. If we measured vehicle fuel consumption in miles per useful gallon, your approach might work, but some poor schmuck would still have to cough up the money to cover the cost of the non-useful part.
I think you are getting mixed up between mechanical efficiency and thermal efficiency. Miles per gallon is a measure of mechanical and thermal efficiency combined. You can't simply toss away the thermal bit.
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No, I'm not tossing anything away, you're assuming that the thermal bit gets spent whether or not the useful bit gets spent.
That makes no sense at all.
The thermal and other losses are percentages of a gallon of fuel that is used for some purpose. If you don't use part of a gallon for any purpose then you don't get the associated thermal losses for that fraction, because it simply wasn't burnt; you don't press the throttle down so hard, and the fuel doesn't get injected, or if those bits go up you press the throttle harder and burn more fuel.
You press the throttle harder or less hard depending on what you want the car to do.
Actually, what you're describing has been a problem with one type of engine; jet engines and gas turbines use more or less the same amount of energy irrespective of how much useful power is being made (not completely but there's fairly heavy consumption the whole time they're spinning). That's why jet engines and gas turbines aren't used much on cars!!! They're much better for aircraft, where you need much more constant power from takeoff to touchdown.
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No, I'm not tossing anything away, you're assuming that the thermal bit gets spent whether or not the useful bit gets spent.
That makes no sense at all.
Let's start at the beginning.
You burn a gallon of gasoline to make your car go. 4% of the thermal energy in the gasoline is used in overcoming rolling friction. (True/False)
Let's say we increase the total weight of the vehicle by 10%. We have increased the rolling friction by 10%. (True/False)
How much of the thermal energy in the gallon of gasoline is now used in overcoming rolling friction?
a) More than 15%
b) About 4.5%
c) At least 8%
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No, I'm not tossing anything away, you're assuming that the thermal bit gets spent whether or not the useful bit gets spent.
That makes no sense at all.
Let's start at the beginning.
You burn a gallon of gasoline to make your car go. 4% of the thermal energy in the gasoline is used in overcoming rolling friction. (True/False)
FALSE!!!!
Just about 4/13 of the energy in the gasoline is spent overcoming rolling friction!
There's big losses that are specific and necessary parts of generating the force to overcome rolling friction, there's 6% losses in the drive train (of which rolling friction is 4/13ths), and the 60% losses in the unrecovered heat in the exhaust, (of which rolling friction is responsible for 4/13ths).
If the rolling friction magically went away; these energy fractions would drop to zero! If the rolling friction doubles, they double too! (The other parts of the thermal fraction and drive trains losses could in theory be completely unaffected).
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FALSE!!!!
Just about 4/13 of the energy in the gasoline is spent overcoming rolling friction!
I'm afraid not. That's almost 31% of the thermal energy in the gas, which is impossible because the engine only delivers 19% of the energy in the gas to the drive-line. It's quite clear that only 4% of the energy in the gas is overcoming rolling friction.
You might want to study that diagram an awful lot a bit more carefully.
Tell you what, why don't you start with the thermal energy in a gallon of gasoline and show us your budget of where all the energy ends up. I think you'll be quite surprised (and at least we'll get a good giggle).
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The amount of fuel in the tank has no effect on the efficiency of the engine. The mass of the fuel in the tank does have an effect on the fuel consumption of the vehicle, but not very much.
Ok, but I think you forgot to say that if go along a (long) road which on average is a slight descent, without having to accelerate, having more mass *decreases* fuel consumption [:)]
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Here you go Wolfekeeper. I did it for you -
Engine Loss 62%
Standby 17%
Accessories 2%
Driveline Losses 6%
Aero 3%
Rolling 4%
Braking 6%
By an amazing coincidence, when you add those numbers together, the result is 100.
On the urban driving diagram, 19% is the energy output from the engine. It's what is left after you subtract 62, 17 and 2 from 100.
The 13% on the diagram is what's left of the engine's energy output after you take off another 6% to account for friction losses in the transmssion and driveshafts. Remarkably, 3+4+6 = 13
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The amount of fuel in the tank has no effect on the efficiency of the engine. The mass of the fuel in the tank does have an effect on the fuel consumption of the vehicle, but not very much.
Ok, but I think you forgot to say that if go along a (long) road which on average is a slight descent, without having to accelerate, having more mass *decreases* fuel consumption [:)]
Well, it might, but, as Wolfekeeper points out, the extra weight would also increase the rolling friction (just not as much as he seems to think [;D]) so it would sort of depend on how steep the hill was.
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Well, it might, but, as Wolfekeeper points out, the extra weight would also increase the rolling friction (just not as much as he seems to think [;D]) so it would sort of depend on how steep the hill was.
Certainly [;)]
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Geezer, you've conclusively proved to everyone, apparently except you, that you don't have the slightest clue what those numbers mean; they don't mean that only 4% of the fuel is used to propel the vehicle against rolling friction; they mean that only 4% of the fuel is usefully used to propel the vehicle; there's a huge tail of wasted fuel energy that is used to produce any useful energy; by a factor of 7.7, the extra 6.7 ends up as wasted heat, as hot air.
So the energy you need in the gallon for that 4% is the 4 x 7.7 = 31% percent of each gallon, because you can't generate propulsive energy without wasting 87% of the initial energy. Cars really are that inefficient, overall, and the losses are listed in that table.
It's not like you have a choice, you can't decide to not burn the energy to make drive train losses today to save energy, they're present whatever or however you drive your car, for any propulsive force your car generates, and for whatever it's being used for. You can't even cut the idle losses, your car is stuck in traffic for a certain proportion of the journey, these are supposed to be a typical journey.
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With an emptier tank, the gas sloshes,stirs, & aerates itself?
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See FIGURE 3-1. http://onlinepubs.trb.org/onlinepubs/sr/sr286.pdf
Following the URL on the page:
http://www.fueleconomy.gov/feg/atv.shtml
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.fueleconomy.gov%2Ffeg%2Fimages%2Fenergy_requirements_combined.jpg&hash=8a005d843d3669173c1ba88b5daa1873)
With slightly different numbers for city, hwy, or combined.
Oddly they don't seem to separate acceleration and deceleration, perhaps the mode of acceleration doesn't play as big of a factor as one might think as it would be a difference between a short, intense burst of energy vs a long, gradual energy use.
Anyway, if you increase energy requirements at any one point of the system, then that will cause increases throughout the system.
So, consider it this way.
Say you have 100KW energy input.
70% (70KW) is "lost" by the engine.
5% (5KW) is consumed by the water pump and alternator.
5% (5KW) is lost in friction in the transmission and drive train.
20% (20KW) gets to the wheels and is divided by 10KW wind, 5KW rolling, 5KW braking.
Now, if you increase the "rolling resistance" by a modest amount from 5KW to 6KW (about a 20% change).
This would propagate through the entire system.
Rolling, up by 20%
Wind - Same
Braking - Also up by 20% if weight was the cause of the increased power needs.
Engine - Also up by 20%, as it is working harder to supply energy to the system.
Water Pump, Alternator, etc. May increase slightly. More cooling requirements? Is it electrically or mechanically cooled??
Idling, - About the same. Traffic and congestion doesn't care about how full or empty your tank is.
Anyway, once you sum it all up.
You started at 100KW giving the equivalent of 5KW "rolling resistance".
But, you are up to between 115 and 120KW total energy input to give you that additional 1KW at the wheels (6KW rolling resistance).
Fuel tanks are generally sized to give a typical vehicle 300 to 400 miles range.
Subcompact, perhaps 10 gallon tank.
Mini-Pickup, 15 gallon tank.
Full sized pickup, 20-30 gallon tank.
Semi Truck, perhaps twin 150 gallon tanks.
At 8 lbs to the gallon (density of fuel is actually slightly less than for water), that gives about 80 lbs for the subcompact, 240 lbs for the big pickup, and 2,400 lbs for the semi.
As mentioned by techmind, the weight of the full tank of fuel is probably somewhere around 2% to 5% of the total vehicle weight.
If 100% of that additional weight goes towards increased rolling resistance, acceleration, and braking energy, then it will also mean increased engine requirements and a 2% to 5% increase in fuel consumption.
80 lbs of fuel in a subcompact may not seem like a lot. However, certainly the loaded vs unloaded fuel consumption of a pickup is noticeable. And 240 lbs would be about ¼ the rated capacity of a half-ton pickup.
Keep in mind that extra trips to the gas station may not be efficient. Do you go out of the way to get gas? Do you end up idling the car while waiting in line to fill up, or does your car konk out 1/2 mile from the gas station due to being out of gas. You coast half that distance, and end up getting out and pushing it the last ¼ mile, saving additional gas. [;)]
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they don't mean that only 4% of the fuel is used to propel the vehicle against rolling friction;
When did I ever say that only 4% of the fuel is used to propel the vehicle? Of course that's ridiculous. What I said was that only 4% of the thermal energy released by burning the fuel goes into overcoming rolling friction.
And, if you had even bothered to read the link I posted, you would have found this on page 39;
"For urban driving, only 10 to 15 percent of the fuel energy is ultimately transmitted as power to the wheels."
and
"For both urban and highway driving, the mechanical energy that does make its way through the driveline to turn the wheels is consumed by three (energy) sinks: aerodynamic drag, rolling resistance, and braking."
Your loud assertion that "Just about 4/13 of the energy in the gasoline is spent overcoming rolling friction!" is just plain wrong. In the urban model it is simply 4% of the total thermal energy in the gasoline.
If you do the math you will discover that 4% of the thermal energy in a quantity of the fuel put into the engine will be very similar to the energy consumed in overcoming rolling friction (which, BTW, is mainly dissipated as heat too.) You'll have to assume a value for the coefficient of friction of course, but 0.01 might be about right.
The other mistake you are making is that you are getting all hung up on the term "useful". Why is the energy that is dissipated as rolling friction, aerodynamic drag (friction) and braking (friction) any more useful than the heat produced by the engine while it is doing work? The only really "useful" bit is the work done by the engine, but ultimately that is all converted into "useless" heat in propelling the vehicle.
It's really much more satisfactory just to analyse where the energy released from the gasoline ends up, and the diagrams on page 40 do a very good job of explaining that.
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70% (70KW) is "lost" by the engine.
5% (5KW) is consumed by the water pump and alternator.
5% (5KW) is lost in friction in the transmission and drive train.
20% (20KW) gets to the wheels and is divided by 10KW wind, 5KW rolling, 5KW braking.
Now, if you increase the "rolling resistance" by a modest amount from 5KW to 6KW (about a 20% change).
This would propagate through the entire system.
Rolling, up by 20%
Wind - Same
Braking - Also up by 20% if weight was the cause of the increased power needs.
Engine - Also up by 20%, as it is working harder to supply energy to the system.
Water Pump, Alternator, etc. May increase slightly. More cooling requirements? Is it electrically or mechanically cooled??
Nice link Clifford, however, I see a slight problem.
Rolling resistance is simply the resistance to rolling! It's only a function of the tires and the weight they are carrying, and it is also influenced by the nature of the road surface. Wind resistance (drag) has nothing to do with rolling resistance, so it does not follow that if one changes the other will also change.
For example, you can increase the rolling resistance by letting some air out of the tires [:D]
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Clifford,
Perhaps you are saying that if one quantity changes, the other will also have to change too. I'd agree with that.
If you upped the rolling resistance from 5 to 6kW, it's no longer 20% of the power that is getting to the wheels. 1 kW is 1% of the power, so, it's now 21% that is going to the wheels, but if we simply do that, we will end up with a total power of 101%, which is obviously wrong.
To fix this, we will need to adjust all the values so that the total power still sums to 100%, if you see what I mean.
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Wind - Same
Nice link Clifford, however, I see a slight problem.
Wind resistance (drag) has nothing to do with rolling resistance, so it does not follow that if one changes the other will also change.
Precisely what I said...
Wind resistance would be essentially unchanged by an increase in internal mass.
So, in fact, by changing the internal mass, you will change the actual energy consumption of certain aspects of the car (rolling resistance, acceleration, braking, engine, etc). Other aspects will remain largely unchanged for energy consumption (wind resistance, car stereo, AC, idling, etc).
If you increase your rolling resistance and braking by about 20%, you will increase your overall energy consumption by about 20%.
I.E. in my example, changing your energy consumption from about 100KW to just under 120KW (perhaps 118KW).
It will also shift the relative percentages slightly. The power lost to wind resistance will stay the same, so the relative percentage would actually drop slightly.
However,
We've concluded that a full fuel tank is only between 2% and 5% of the total vehicle weight. So, the savings of an empty tank vs full tank would likely be somewhat less than 5%.
If you regularly top off your tank, then you might realize the cost of a continuously full tank. However, if you drive from full to empty, then refill, your average mass would be just over half a tank (1 to 2.5% of the cars weight).
If, on the other hand, your fuel miser drives the car in the range ½ to empty, then the average would be about ¼ of a tank, or about ½% to 1.25% of the total vehicle weight. And, thus the actual average weight & fuel savings of the fuel miser would be on the order of ½% to 1.25%.
So, from my notes above, the subcompact with a 10 gallon tank would have a maximum capacity of about 80 lbs fuel. But, the person driving from full to empty would average 40 lbs of fuel, and the person driving half to empty would average 20 lbs fuel. The savings is thus, only about an average of about 20 lbs.
For many of us, we might be better off just dieting [xx(], and filling the tank to the max [^]
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Oh! When you said "same", I thought you meant it would also increase by 20%, but what you meant was it is unchanged! I see what you mean.
Yes, a 20% increase in rolling resistance and braking friction will have a subsantial impact on fuel consumption, although precisely how much will depend on how fast the vehicle is travelling because the energy that goes into overcoming drag is unchanged (assuming the same set of conditions), so it might be a bit less than 20%.
Rather than adjusting percentages, I suspect it's safer to use a model and apply real values, then derive the percentage values, otherwise things tend to go pearshaped very quickly.
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Time for a bit of science: the original question was Does the amount of fuel in the tank affect petrol consumption?
The discussion has moved on to How much does the amount of fuel in the tank affect petrol consumption?
There is a consensus in the discussion, and good reason to believe that (apart from dirty petrol and condensation as I earlier pointed out) the only significant factor is the mass of fuel involved. If that is the case then the mass does not have to be petrol. There is a simple experiment that can be done.
Run your family car for a month, filling the tank quite full at the beginning and the end. Get an accurate check on fuel consumption.
Then, put a 20 kg sandbag in the boot, and leave it there for the next month. Take a little trouble to see that you mostly do a similar amount and style of running around in the car, including issues like how much you use the air conditioner, and how much driving is done on wet or dry roads. Compare the fuel consumption for the second month with the first, and then think carefully about the amount of experimental error involved.
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Time for a bit of science:
Oi! We must have missed the bit where you came riding in on your gleaming white stallion. [;D]
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Wolfekeeper, old pal [;D],
I think I understand your problem - it looks as if you are somehow equating power and energy. The diagrams in that paper are all about energy. What you are concerned about is power. It takes power to overcome rolling friction and air friction, and power requires the conversion of energy in time.
As you implied, the mechanical power produced by the engine is only a fraction of the "thermal power" being produced by the fuel, so any increase in the mechanical power consumed by rolling friction has a corresponding multiplication effect on fuel consumption to generate the required amount of mechanical power, and the multiplication factor will depend, to some extent, on the thermal efficiency of the engine.
I say "to some extent" because it's not that simple. In urban driving, the vehicle's speed is anything but constant, so a lot of the rolling resistance occurs while the vehicle is slowing down and the engine is developing hardly any power. During acceleration, the power produced by the engine to accelerate the vehicle "swamps" the power required for rolling friction, and the engine's thermal efficiency during acceleration is also very different. The energy values in the paper average all that out for us.
Ah-ha!, you might say, so why don't we keep the conditions constant. Well, yes, we can. That's when we might use the highway driving model, in which case the rolling resistance is much less than the air resistance, so that throws everything askew as well.
The "bottom line" is that you can't simply equate a percentage increase in vehicle weight with a percentage increase in fuel consumption without specifing the conditions in which the vehicle is operating.
EDIT: BTW, I realized that your 4/13ths estimate might actually be a bit conservative. You shouldn't really count the braking component because it's unlikely to occur while the engine is producing power [:D]. On the other hand, a lot of the rolling resistance happens while the engine is in standby, so that takes things in the other direction. It's very complicated.
The energy analysis in the paper is useful as a means to determine, averaged over a long interval, where all the energy goes, but it's not much use as a real-time model.
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Then, put a 20 kg sandbag in the boot, and leave it there for the next month. Take a little trouble to see that you mostly do a similar amount and style of running around in the car, including issues like how much you use the air conditioner, and how much driving is done on wet or dry roads. Compare the fuel consumption for the second month with the first, and then think carefully about the amount of experimental error involved.
Yes,
A 10 or 20 kg weight would be representative of the different driving styles.
If you put 20 of the in the trunk (400 kg), the differences should be more obvious, but not as representative of the actual fuel.
Experimental Error?
Good point.
And, it is likely low enough that you could only tease the difference out (for the 10 or 20 kg weight) with an average over many tankfuls.
Also, consider a "blinded" trial. Have someone else fill (or empty) the trunk. Then seal it so you won't know whether it is full or empty.
In fact, I think the last time I was watching my fuel consumption, I saw a significant difference between summer and winter. So, as we are approaching winter, the fuel efficiency might naturally drop.
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Then, put a 20 kg sandbag in the boot, and leave it there for the next month. Take a little trouble to see that you mostly do a similar amount and style of running around in the car, including issues like how much you use the air conditioner, and how much driving is done on wet or dry roads. Compare the fuel consumption for the second month with the first, and then think carefully about the amount of experimental error involved.
Yes,
A 10 or 20 kg weight would be representative of the different driving styles.
If you put 20 of the in the trunk (400 kg), the differences should be more obvious, but not as representative of the actual fuel.
Experimental Error?
Good point.
And, it is likely low enough that you could only tease the difference out (for the 10 or 20 kg weight) with an average over many tankfuls.
Also, consider a "blinded" trial. Have someone else fill (or empty) the trunk. Then seal it so you won't know whether it is full or empty.
In fact, I think the last time I was watching my fuel consumption, I saw a significant difference between summer and winter. So, as we are approaching winter, the fuel efficiency might naturally drop.
It's a lot more than experimental error. That would only apply if all conditions were carefully controlled, but in this case there are so many uncontrolled variables that the results could be very misleading. I would think a simulation with accurate input data would yield much more meaningful results.
Clifford, did you see more or less MPG with lower temperatures? I think you might expect slightly more MPG with lower temperatures, but there are a lot of other factors that could take it in the opposite direction.
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Clifford, did you see more or less MPG with lower temperatures? I think you might expect slightly more MPG with lower temperatures, but there are a lot of other factors that could take it in the opposite direction.
It has been a few years since I've watched it, but I thought I was getting lower fuel mileage in the winter than in the summer. For my Ranger, it was something like 20-22mpg in the winter, and 22 to 25 mpg in the summer. But, I could have got that reversed.
NO AC, of course. [xx(]
I'm not quite sure what seemed to be causing the difference. Fuel mixes may have been different, with municipalities often adding more ethanol in the winter. Or, perhaps the colder fuel/air mix didn't burn as efficiently. Obviously there is a warm-up period in the winter, but I think I was also noticing the difference with freeway driving. The Ranger would have had a mechanical water pump and fan, so likely not much change with hot vs cold. Heaters are run off of water, so not much difference there either.
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Clifford, did you see more or less MPG with lower temperatures? I think you might expect slightly more MPG with lower temperatures, but there are a lot of other factors that could take it in the opposite direction.
It has been a few years since I've watched it, but I thought I was getting lower fuel mileage in the winter than in the summer. For my Ranger, it was something like 20-22mpg in the winter, and 22 to 25 mpg in the summer. But, I could have got that reversed.
NO AC, of course. [xx(]
I'm not quite sure what seemed to be causing the difference. Fuel mixes may have been different, with municipalities often adding more ethanol in the winter. Or, perhaps the colder fuel/air mix didn't burn as efficiently. Obviously there is a warm-up period in the winter, but I think I was also noticing the difference with freeway driving. The Ranger would have had a mechanical water pump and fan, so likely not much change with hot vs cold. Heaters are run off of water, so not much difference there either.
Well, at least we know it's not the AC!
I would have thought that you might get a slight thermodynamic advantage because of the lower air temps, but, come to think of it, the bastards probably screw that up by adulterating your gas in the "interests" of air-pollution.
(Don't get me started on expletive ethanol.)
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From Geezer:
It's a lot more than experimental error. That would only apply if all conditions were carefully controlled, but in this case there are so many uncontrolled variables that the results could be very misleading. I would think a simulation with accurate input data would yield much more meaningful results.
Geezer, in our local dialect of the English language "experimental error" means "the uncertainties arising from design, execution, and measurement in an experiment". It is hard to see how the effects that you refer to, which are undoubtedly real ones, are "...much more than..." this.
I put together the outline of a very crude experiment that should probably undergo considerable refinement of experimental design in an attempt to snag most of these "uncontrolled variables" before it would show significant and trustworthy results at a low error level.
But the question has always been about real operation of real vehicles. I would submit that testing of some sort on real vehicles should always be preferred to simulations or models, which are always subject to missing or mis-translation of real operational effects.
So you should run the tests on two days with identical weather conditions with a professional driver on a vehicle testing track with a sealed fuel tank and petrol from the same batch ... etc. The real danger is that the in going down this track, you may produce a poorer simulation of operation of an ordinary family vehicle in ordinary conditions. But there is little doubt in my mind that a computer type model or simulation would be even more artificial, and much more fraught with the danger of overlooking important but unforeseen factors.
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From Geezer:
It's a lot more than experimental error. That would only apply if all conditions were carefully controlled, but in this case there are so many uncontrolled variables that the results could be very misleading. I would think a simulation with accurate input data would yield much more meaningful results.
Geezer, in our local dialect of the English language "experimental error" means "the uncertainties arising from design, execution, and measurement in an experiment". It is hard to see how the effects that you refer to, which are undoubtedly real ones, are "...much more than..." this..
Ah yes, but if I'm not mistaken, you proposed to run your experiment for a month.
So, it might be a teensy bit tricky to control some things that could have a significant influence on the outcome. A few that spring to mind are: Temperature, precipitation, wind speed, wind direction, humidity, road surface, tire wear, tire pressure, route, load, driver, driver's level of intoxication, mother-in-law in the back seat,.......
I'm sure this is a less than exhaustive list and that you will be able to add quite a few more. My point is, by the time you are done, the summation of these variables could easily swamp any meaningful result. That's why I think a simulation could be a lot more useful, although a carefully controlled "rolling road" environment with a real vehicle would probably be even better.
I wonder if that explains why the automotive industry does it that way?
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If you could build an accurate fuel consumption gauge, then you could significantly reduce the time involved in the experiment. And, thus control many other variables, and/or have more trial runs.
I remember a Mythbusters episode where they rigged some kind of a graduated cylinder to the fuel system, so they could run experiments with about 1 liter of fuel at a time.
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Being a glutton for punishment, I took a shot at some actual numbers (I probably mucked it up somewhere too.)
Vehicle weight 1,000 kg
Force on tires 9,810 N
Coeff. rolling friction assumed 0.01
Force to overcome rolling resistance 9,810*0.01 = 98.1 N
Power consumed at 5.55 m/s (20 km/h) 98.1*5.55 = 545 W (0.55 kW)
At 80 km/h (50 mph), power consumed = 2.2 kW (almost 3 HP)
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At 80 km/h (50 mph), power consumed = 2.2 kW (almost 3 HP)
Whew, not as much as I would have thought. But... perhaps representative.
Plus, of course, wind resistance, acceleration/Braking, etc.
With my Ranger EV, I can watch power output.
Acceleration, I usually hit about 200A x 144V, or about 28KW (and it is pretty poky at that).
Cruising at about 45 MPH, I usually hit about 50A to 100A x 144V, or about 7 to 15 KW.
And, it is a pretty heavy vehicle with all the lead in the back.
Unfortunately I only have an instantaneous reading, not a trip reading, but it is not too far out from Geezer's calculations.
One thing in the Electric Vehicle discussions is that with Lead batteries, at some point with a range of about 40 or 50 miles, adding more batteries (equivalent of fuel) no longer significantly extends the range due to the increase of weight.
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Apparently any ODB-II equipped car can give you real-time mileage estimates. Perhaps not as accurate as readings from the pump, but they could be used for comparitive purposes.
http://www.smarthome.com/93001/Linear-Logic-SGE-ScanGaugeE-Fuel-Economy-Gauge/p.aspx
Unfortunately, I don't think I have any ODB-II compatible vehicles, [xx(]
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Yes, I was a bit surprised the power is so small. Maybe there is something wrong with the calculation, but it is quite straightforward (I think!) Perhaps the coefficient is too optimistic, but 0.01 seems to be around the middle of the range.
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Mind you, here's the other side of the equation: Lets' say our 1000 kg vehicle manages to do 30 mpg (US) while travelling at a steady 50 MPH. That means it consumes 50/30 = 1.66 gallons per hour
Switching to SI units, that's 3.7854*1.66 = 6.31 litres per hour, or 1.75 ml per second.
1 ml of gasoline has a thermal energy of 34 kJ, so thermal energy is produced in the engine at 1.75*34 = 59.5 kJ/s, or 59.5 kW (imagine how quickly that would heat your house!)
If we are optimistic, 20% of that, about 11.9 kW (16 HP), might make it to the wheels.