Why does it take more fuel to accelerate a body to a higher speed?

[Elliot Webb]

I apologise for what is probably merely stupidity on my part, but has been vexing me for rather a while, and no one I ask can provide any insight.

A body undergoes two accelerations. The first is from 0 to 10 m/s, the second from 10 to 20 m/s. For simplicity's sake I will say F= 1N, a= 1 m/s² and m = 1kg, for both of the accelerations.

So in both cases, it takes 10 seconds for the particle to accelerate, and so if it were powered by some kind of engine, the time it is running would also be 10 seconds. As the force it produces is constant, the amount of fuel burnt in each case would logically be the same. But apparently it's not.

Here is the problem:
The distance traveled in the first instance is 50m, with u=0, v=10, a=1. As work done= Fs, the energy transferred is therefore 50J. So 50 joules worth of fuel is burnt.

For the second acceleration, u=10 and v=20, so the distance traveled is 150m. Hence the energy transferred the second time is 150J.

So despite the engine providing the same amount of force for the same time, it has used three times as much fuel. Unless it knows how fast it is going (with relation to what, anyway?), how could it take more energy to accelerate the second time?

Thank you for you patience and (hopefully) your help.

 

[CWatters]

As the force it produces is constant, the amount of fuel burnt in each case would logically be the same.

Nope that's not logical. A fridge magnet applies a constant force but consumes no energy.

Perhaps start with..

Work = force * distance

Divide both sides by time and you get..

work/time = force * distance/time

or

Power = force * velocity

It's logical that it takes more power to go faster even with a constant force.

 

[Elliot Webb]

OK, thanks for replying.
If it takes more power, how does the engine 'know' how fast it is going in order to be less efficient? If you were to just leave it running at the same power, assuming no external resistance, the force produced would gradually decrease as the body accelerates. I don't see how this could be possible. If it is, then what is the velocity in relation to? If something is in orbit, the thrusters etc. will have exactly the same effect as if the object was stationary on earth, and yet the changes in kinetic energy would be massively different - with the same amount of chemical energy being used up (fuel).

 

[russ_watters]

If it takes more power, how does the engine 'know' how fast it is going in order to be less efficient?

None of this has anything to do with efficiency.

For an internal combustion engine, you can consider the force applied and fuel used to be roughly constant for every piston stroke with the gas pedal in the same position. But if you are moving faster, the engine is spinning faster, so you have more piston strokes in the same amount of time. So more power and more fuel used.

 

[Elliot Webb]

Until you change gear, and then the engine slows down again.

 

[russ_watters]

Until you change gear, and then the engine slows down again.

True. And what else changes when you change gears...?

 

[A.T.]

...would logically be the same...

...that's not logical... ...It's logical that...

I think it's less a matter of logic, than just of how those quantities are defined in physics vs. the intuitive ideas based on common language usage of the same words.

 

[Elliot Webb]

I do understand how these are defined, but that is what has caused the problem. I don't see how there can be a massive increase in energy for something moving in orbit (for instance), compared to something that is stationary. The KE increase would be thousands if times greater, even though the same amount of fuel would be used up. The energy stored in the fuel can't change, and so where does this come from?

Also, can anyone direct me to a proof of W=Fs that doesn't start with the kinetic energy formula - thanks

 

[A.T.]

I do understand how these are defined, but that is what has caused the problem. I don't see how there can be a massive increase in energy for something moving in orbit (for instance), compared to something that is stationary. The KE increase would be thousands if times greater, even though the same amount of fuel would be used up. The energy stored in the fuel can't change, and so where does this come from?

Did you compare the change in KE of the fuel when it's expelled by the rocket in both cases?

 

[Lsos]

I do understand how these are defined, but that is what has caused the problem. I don't see how there can be a massive increase in energy for something moving in orbit (for instance), compared to something that is stationary. The KE increase would be thousands if times greater, even though the same amount of fuel would be used up. The energy stored in the fuel can't change, and so where does this come from?

It might not be "logical" at first glance, but when you start to study the everyday things around you you will see that this is how it works. For example, if you jump out of of a 2nd story window, you will NOT hit the ground 2x faster than from a 1st story window (0 story being the ground floor).
It's also why it takes 8x more power/ and hence 8x more fuel to drive 200 mph as opposed to 100 mph. Fuel efficiency drops off really fast above highway speeds, much faster than air resistance increases.
It's also why military jets produce many tens of thousands, if not hundreds of thousands of horsepower, despite only being able to go like 6-10x faster than a car (at sea level).
It's also why crashing into a wall at 20 mph is 4x more destructive than at 10 mph.
The examples are all around you.

How does an object KNOW how fast it is moving? It knows because it has to push against something in order to provide a force. A car has to push against the ground. The faster the car moves, the faster the ground recedes, and you need to change gears just to keep up, trading force for distance.

I recommend you don't think about rockets until you understand the above, because a rocket is a special case and can get confusing. But the jist of it is this: the difference in a rocket is that it pushes against its own exhaust. It essentially brings the ground with it. So yes, if a rocket travels fast enough it can indeed gain more energy than what is stored chemically in its fuel, but that energy didn't come from nowhere. It came from the kinetic energy that was put into the fuel from the beginning of the flight. After all, an orbiting rocket's fuel has ALOT of kinetic energy in it, much more in fact than its chemical energy. That energy was "invested" into the fuel from liftoff, and that energy gets returned at high speeds. This is called the Oberth Effect btw.

 

[Elliot Webb]

Yes, and it makes the problem even worse. With the first case, say for each of the accelerations 10 kg of fuel is expelled at 1 m/s for 10 seconds. The KE if the fuel is 80 J, so the total KE of the rocket and fuel is 130 J. After the second acceleration, the total KE is much more - 1180 J. If I were to calculate this for real rockets traveling at thousands of km/h then tiny accelerations would consume immense amounts of energy- and the exhaust doesn't help.

 

[jbriggs444]

Go ahead and calculate it. You will find that the apparent discrepancy is covered by a careful accounting for the energy in the exhaust stream.

 

[Lsos]

Again, this is called the Oberth Effect and is a special case for rockets.

But not really, because you must keep in mind that kinetic energy depends relative to what you calculate it. We usually calculate kinetic energy relative to some stationary object (the earth), but a walk in the park could literally be millions of horsepower when compared to some receding globular cluster on the other side of the universe. That energy is not from nowhere though, it was put into you and the Earth and the globular cluster at some distant point in time, by whatever caused them to fly apart from each other in the first place.

 

[A.T.]

There IS a massive increase in energy for something in orbit, but there ISN'T the same amount of fuel used up. This is why a rocket is mostly fuel.

If the same rocket burns for the same time, then the fuel used will be obviously the same.

 

[A.T.]

Yes, and it makes the problem even worse. With the first case, say for each of the accelerations 10 kg of fuel is expelled at 1 m/s for 10 seconds. The KE if the fuel is 80 J, so the total KE of the rocket and fuel is 130 J.

How did you compute that?

 

[Lsos]

If the same rocket burns for the same time, then the fuel used will be obviously the same.

You're right, I removed that entire sentence because I thought OP meant something completely different.

 

[russ_watters]

It's also why it takes 8x more power/ and hence 8x more fuel to drive 200 mph as opposed to 100 mph. Fuel efficiency drops off really fast above highway speeds, much faster than air resistance increases.

Much of that is wrong. Air resistance is a square function of speed and so if you double the speed you get 4x the air resistance. Then combining the air resistance and higher speed, 2x4=8 times the power. And none of that has anything to do with efficiency. If you meant fuel economy, which is sort of a measure of efficiency, it is still wrong since that is in terms of distance, not time. So the 2x speed factor goes away and the drop in fuel economy is due entirely to drag and other frictions... all still assuming no change in thermodynamic efficiency of the engine. In other words, it takes 8x as much power but only 4x as much fuel per mile.

Again, let's keep efficiency out of this. The ideal case of no drag and ignoring engine efficiency issues and focuses on the acceleration from 0-100 and 100-200. Adding efficiency is a separate, confounding factor that won't help answer the original question.

Bottom line is that if you want acceleration to be the same at 200 mph as at 100 you need twice the power because you are covering twice the distance in the same time. And that manifests as the engine spinning twice as fast. Simple.

Or alternately if you want to keep the rpm constant by changing gears... what else must change? (Question I previously asked of the OP that went unanswered). Again, the relationship here between force, rpm, speed, power and acceleration is simple proportions. If one changes another must change by the same amount. It isn't a trick question and no detailed analysis is required.

 

[CWatters]

OK, thanks for replying.
If it takes more power, how does the engine 'know' how fast it is going in order to be less efficient?...

This has nothing to do with efficiency.

The engine doesn't know anything. In order to accelerate from 10 to 20M/S you (the driver of the car) will find they have to increase the amount of fuel fed to the engine compared to 0 to 10M/S. eg If you don't step on the gas it won't accelerate from 10 to 20M/S as fast as it did from 0 to 10 M/S.

 

[Elliot Webb]

Thanks for pointing out the Oberth Effect - I'd never heard of it and didn't consider that the fuel itself has kinetic energy. At some point I'll do some Maths and see if it satisfies the problem. I apologise for mentioning efficiency - I didn't mean it in the traditional sense, I just meant that for the same amount of fuel being used, less force is produced, which to me seems less efficient, but maybe I used the wrong word. However, you keep asking that 'what else changes when you change gear?' Although the wheels may be at a different speed, as long as the engine RPM is the same, surely the same amount of fuel is being consumed in a certain time? It can be seen quite easily that as long as the engine RPM remains the same, the rate of fuel consumption stays constant, irrespective of the speed of the car - otherwise maintaining a high speed would be next to impossible.

I thank you all for helping, but I'm afraid I'm still not satisfied. Also, has anyone got a proof of the Work-Energy principle that doesn't begin with 'assume KE=1/2mv^2'?

 

[jbriggs444]

If engine RPM remains the same and the vehicle goes faster, then you must have shifted into a higher gear. Even if air and rolling resistance remain constant (a dubious assumption), the engine will still have to push harder to make up for the higher gear ratio. You will have to press down harder on the gas pedal to open up the throttle. Fuel consumption will increase accordingly.

 

[Elliot Webb]

But surely the same amount of fuel is going into the engine per stroke of the piston - the volume of it doesn't change, and the explosive force it provides must be the same for every stroke? So if it is a 1 litre engine, surely the amount of gas burnt per minute is always 1 litre * the RPM. How does more fuel fit in the cylinder when it is traveling more quickly?

 

[russ_watters]

But surely the same amount of fuel is going into the engine per stroke of the piston - the volume of it doesn't change, and the explosive force it provides must be the same for every stroke? So if it is a 1 litre engine, surely the amount of gas burnt per minute is always 1 litre * the RPM. How does more fuel fit in the cylinder when it is traveling more quickly?

No one claimed it did. You're still not answering my question and the answer you want is in the answer I want you to give me!

If you change gears to cut the rpm in half, the fuel consumption per unit time naturally also drops by half. I think you already know that.

So again: what else changes? Gears are a mechanical advantage device. They keep power constant (for constant input rpm) while varying output rpm and WHAT?

Or, if the output rpm stays the same and the input rpm is cut in half, WHAT else on the output drops by half?

 

[Elliot Webb]

The force on the output would decrease by half. The wheels will not have as much torque. Is that what you're looking for? But this just shows that if the input rpm is cut in half, half as much fuel is being used, and therefore the fuel used is still proportional to the force. I'm sorry I don't understand this but it really doesn't help.

 

[russ_watters]

The force on the output would decrease by half. The wheels will not have as much torque. Is that what you're looking for?

Yes!

But this just shows that if the input rpm is cut in half, half as much fuel is being used, and therefore the fuel used is still proportional to the force.

The fuel use was not proportional to the force (torque) when accelerating in the same gear. You knew that in the beginning of the thread, but seem to have forgotten it now. The torque and therefore acceleration were constant. I'm not sure if you forgot your initial issue here or what, but here it is again:

A body undergoes two accelerations. The first is from 0 to 10 m/s, the second from 10 to 20 m/s. For simplicity's sake I will say F= 1N, a= 1 m/s2 and m = 1kg, for both of the accelerations.

As the force it produces is constant, the amount of fuel burnt in each case would logically be the same. But apparently it's not...

Unless it knows how fast it is going (with relation to what, anyway?), how could it take more energy to accelerate the second time?

As I said in the beginning, the fuel flow rate is a function of both the fuel burned per stroke (which is constant) and the number of strokes (which is not). You tried to argue your way around that by changing gears:

Until you change gear, and then the engine slows down again.

But now you know that you can't do that: when you change gears, you drop the torque/force applied to the wheels, and in your initial scenario you wanted to keep it constant.

So again: there is no conservation of energy problem here: more fuel is burned in the second acceleration period because the motor is spinning faster. Or, to make a workable scenario that includes changing gears, the torque/force and therefore acceleration must be lower after you change gears.

 

[Elliot Webb]

I see what you mean. Thank you for being so patient, I finally understand what is wrong. I'll do some maths with the rockets and exhaust etc. but I'm sure you'll be right in the end. Thanks.

 

[A.T.]

I see what you mean. Thank you for being so patient, I finally understand what is wrong. I'll do some maths with the rockets and exhaust etc. but I'm sure you'll be right in the end. Thanks.

Note that rockets are different from car engines here:
- For rockets the fuel consumption relates to change in momentum (assuming constant expulsion velocity)
- For cars the fuel consumption relates to change in KE relative to the ground (ignoring losses)

 

[Elliot Webb]

Well, I thought I understood it until you said that. How can the fuel consumption relate to momentum, when there is a finite amount of energy stored in the fuel? Seeing as the problem I had in the first place was with a body NOT connected to the ground, the rocket model is more accurate. So as I originally thought, burning the fuel for a constant time produces the same impulse. But if this is true, then more energy must be released from the same amount of fuel when the rocket travels faster?

 

[A.T.]

How can the fuel consumption relate to momentum,

For small fuel mass (compared to the rest of the rocket), you can approximately say: Burning rocket fuel at a constant rate creates a constant force which means a constant momentum change.

But if this is true, then more energy must be released from the same amount of fuel when the rocket travels faster?

Sure, because the fuel travels faster too, and therefore has more kinetic energy to provide to the rocket, when the rocket pushed off from it. In contrast, the car pushes off the ground, which doesn't travel with it.

 

[Elliot Webb]

OK thanks. It doesn't seem like enough but I'll look deeper into the Oberth Effect, and do some maths, and hopefully it'll make sebse (before, I worked out the mass of fuel, but just assumed the rocket's mass was constant).