Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Tomcat on 11/05/2010 14:30:02
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Tomcat asked the Naked Scientists:
Question to the rocket scientists:
If you take a rocket in deep space far from any planet, if you fire the rocket engine, would a rocket continuously accelerate tille burn out or would it end accelerating when its forward speed equals the speed the hot gases leave the nozzle of your rocket engine? thanks
What do you think?
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It would keep accelerating. Assuming the rocket exhaust velocity is the same relative to the rocket this is independent of the rocket's speed relative to a stationary observer. Even when the rocket has accelerated to the point where the rocket gases are also moving in the same direction as the rocket relative to the observer, the rocket will still be accelerating.
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When a rocket accelerates it also accelerates its remaining fuel, this means that the speed the rocket can push the exhaust gasses out is relative to the rocket, not any notional stationary reference frame, so it will keep accelerating. It is also why you need such a big rocket to go any distance - the rocket has to lift the fuel to lift the fuel to lift the fuel to lift the payload.
A jet engine on the other hand takes in air which is 'stationary' and pushes it out the back, so a jet plane can't go faster than it can push the gasses out of the back.
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A jet engine on the other hand takes in air which is 'stationary' and pushes it out the back, so a jet plane can't go faster than it can push the gasses out of the back.
I'm not sure there is much difference between a rocket and a jet Dave. Don't they both develop thrust my accelerating the exhaust gas? After that, the thrust is simply a function of the mass of the exhaust times the acceleration.
With bypass jets, some of the thrust is produced by air that was accelerated around the engine.
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I'm not sure there is much difference between a rocket and a jet Dave. Don't they both develop thrust my accelerating the exhaust gas? After that, the thrust is simply a function of the mass of the exhaust times the acceleration.
Isnt the case that in practical term jets reach a point where they can no longer make an appreciable difference between the speed of the air going in and the speed coming out. I always thought that this was the idea of the afterburner; which is a bit like a rocket.
matthew
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Yes the difference is that the reaction mass for the jet remains 'stationary' where as in a rocket you keep accelerating it. So effectively if there is a maximum speed the jet can take air through its tube, and the jet certainly can't go any faster than this.
So if a jet plane is going near to the speed the air is coming out of its jet, the amount it accelerates the air is less, hence less force. This is why jet engines are different shapes for different speeds of aircraft. Airliners have big short fat engines which push a lot of air backwards realtively slowly (an efficient process but with a low maximum speed, where as fighter jets push less air out much faster which is less efficient but has a higher maximum speed.
The afterburner is effectively dumping fuel into the back end of the jet, and burning with air that has bypassed the burners and has oxygen left in it. This gets hot and expands, increasing the exit speed of the air. This is particularly useful as noone has got a practical jet engine where the air flows through supersonically.
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This gets hot and expands, increasing the exit speed of the air. This is particularly useful as noone has got a practical jet engine where the air flows through supersonically.
http://en.wikipedia.org/wiki/Scramjet
- may be practical one day, huh?...
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So if a jet plane is going near to the speed the air is coming out of its jet, the amount it accelerates the air is less, hence less force.
Well, yes. If it's not accelerating the exhaust much, it's not going to develop much thrust. But the limitation of a jet does not lie in its inability to accelerate mass, I think it lies more in its inability to inhale sufficient oxygen to oxidize the fuel. Hence, as Peppercorn points out, the Scram jet.
Rockets solve this problem by schlepping the oxidizer with them.
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We discussed this question on our show
Andrew - In fact, it would just carry on accelerating forever until it ran out of fuel because the reason that it’s accelerating is not so much to do with measuring particular speeds. It’s more just to do with the fact that it’s throwing out lots of mass out of its rear end, to put it nicely, and when it does that, it feels a little push. This is one of Newton’s laws – every force must have an equal and opposite force. It receives that equal and opposite force, and as a result, it accelerates.
Chris - This reminds me of a question we had in the Naked Scientists awhile back which was, how hard would you have to pee to push yourself over. I think the answer we’ve worked out was, you would have to be able to pee and produce a fountain, more than 20 metres high in order to achieve sufficient force that would have any kind of backward propulsion, assuming a modestly weighted man. Similar physics, Different situation.
Click to visit the show page for the podcast in which this question is answered. (http://www.thenakedscientists.com/HTML/podcasts/show/2010.06.27/) Alternatively, [chapter podcast=2752 track=10.06.27/Naked_Scientists_Show_10.06.27_6612.mp3](https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2FHTML%2Ftypo3conf%2Fext%2Fnaksci_podcast%2Fgnome-settings-sound.gif&hash=f2b0d108dc173aeaa367f8db2e2171bd) listen to the answer now[/chapter] or [download as MP3] (http://nakeddiscovery.com/downloads/split_individual/10.06.27/Naked_Scientists_Show_10.06.27_6612.mp3)
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That's how I see it too, the fuel is stationary versus its chamber and when it expands it will 'push on that said chamber until it finds it's way out where the resistance is least.As for the difference between a jet and and a rocket I would expect it to be gravity, fuel efficiency, engine construction and air resistance limiting a jet, as compared to a rocket? The principle is otherwise much the same, isn't it?
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The jet engine is similar to the rocket, in that the hot gas is expanded out the back to provide the thrust to move the unit forward.
You very rarely find a pure jet engine nowadays, most commercial airlines use what is called an ultra high bypass turbine, which is a form of ducted fan. This has the advantage of being a lot quieter and using a lot less fuel, as the fan is driven by an engine ( a jet engine inside that provides power to turn a shaft more than thrust for the plane) providing a massive amount of air moving relatively slowly.
The rocket engine has a high pressure gas ( either hot from combustion or cold for low power thrusters) expanding in an open ended chamber. This provides reaction via Newtons law, and is somewhat different from a plane.
The major difference is thae jet takes in air, slows it down and compresses it then adds energy via the fuel to get a higher volume of gas than what entered. This is limited by having to slow the inlet air to below supersonic speed, even though the aircraft is supersonis. Scramjets burn supersonically, but have a small issue in that they cannot start from stationary, and need to be running at well over the speed of sound before they do anything other than burn fuel. Rockets have none of these issues, just needing an almighty amount of fuel per kilogram of payload to perform.
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Jets and rockets both produce thrust by accelerating mass. You can either accelerate a small mass by a large amount, or a larger mass by a smaller amount of acceleration to produce the desired amount of thrust.
As Sean says, modern commercial jets opt for the latter approach for several reasons. I suspect this method is also more efficient thermodynamically.
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The equation for net airbreathing jet thrust is: F = m_dot * (c-v)
where v is the speed of the aircraft, c is the exhaust speed, and m_dot is the mass of air going through the engine per second. As you can see if you put c=v, you've get no net thrust. If v > c then you're slowing down (this is actually used that's what a lot of jet 'reverser' cups do- they throw the air sideways, which has the same effect as setting c to 0).
For rockets the equation is just F = m_dot * c, and it's independent of the rockets mass and speed.
One weird thing about rockets; if you calculate the power the rocket generates- it *increases* with time/speed, even though the thrust is the same (energy is force x distance, and distance per second is increasing), and the propellant flow rate is constant. [???] [;D]
Puzzle: explain why this is!
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Yeah I forgot that one, the rocket brings it's own oxygen with it bound in it's fuel while ordinary jets take it from the air around it :)
Nice explanation Sean, didn't know about ultra high bypass turbines.
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One weird thing about rockets; if you calculate the power the rocket generates- it *increases* with time/speed, even though the thrust is the same (energy is force x distance, and distance per second is increasing), and the propellant flow rate is constant. [???] [;D]
Puzzle: explain why this is!
Power is work done in time. Work is force times distance. The greater the distance in time, the greater the power.
A stationary aircraft might be generating a lot of force (thrust), but if it's not moving, it's not producing power (although it might be consuming a lot of energy).
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The equation for net airbreathing jet thrust is: F = m_dot * (c-v)
where v is the speed of the aircraft, c is the exhaust speed, and m_dot is the mass of air going through the engine per second. As you can see if you put c=v, you've get no net thrust. If v > c then you're slowing down (this is actually used that's what a lot of jet 'reverser' cups do- they throw the air sideways, which has the same effect as setting c to 0).
For rockets the equation is just F = m_dot * c, and it's independent of the rockets mass and speed.
Is this really any different from Force = mass x acceleration?
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UHB turbines were retrofitted to early jets as soon as they became viable and had enough life history to prove reliability. They are quieter than straight jet engines, and consume less fuel per pound of thrust than the equivalent jet engine. As the drive engine is smaller and lighter it has lower maintenance costs, as well as being cheaper to overhaul. The only drawback is that the engines are fatter and have less ground clearance on underwing pods. The main driver was lower fuel consumption and noise during takeoff, a good thing to all living near airports.
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UHB turbines were retrofitted to early jets as soon as they became viable and had enough life history to prove reliability. They are quieter than straight jet engines, and consume less fuel per pound of thrust than the equivalent jet engine. As the drive engine is smaller and lighter it has lower maintenance costs, as well as being cheaper to overhaul. The only drawback is that the engines are fatter and have less ground clearance on underwing pods. The main driver was lower fuel consumption and noise during takeoff, a good thing to all living near airports.
I thought the UHB turbine was the propfan which is actually unducted.
http://en.wikipedia.org/wiki/Propfan
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One weird thing about rockets; if you calculate the power the rocket generates- it *increases* with time/speed, even though the thrust is the same (energy is force x distance, and distance per second is increasing), and the propellant flow rate is constant. [???] [;D]
Puzzle: explain why this is!
Power is work done in time. Work is force times distance. The greater the distance in time, the greater the power.
Yes, but the faster the rocket goes, the more power is being produced, but the chemical energy in the rocket fuel-it is burning it at a constant rate!!!! Where does the extra power come from?
That's the puzzle. [;D]
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For rockets the equation is just F = m_dot * c, and it's independent of the rockets mass and speed.
Is this really any different from Force = mass x acceleration?
Well, technically it's
F = d/dt (mv)
Where mv is the momentum of the exhaust.
Which is slightly different. The f = ma only applies when m is constant.
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For rockets the equation is just F = m_dot * c, and it's independent of the rockets mass and speed.
Is this really any different from Force = mass x acceleration?
Well, technically it's
F = d/dt (mv)
Where mv is the momentum of the exhaust.
Which is slightly different. The f = ma only applies when m is constant.
At constant aircraft velocity I suppose the mass flow rate would be constant. However, F = ma is not so helpful when the aircraft is accelerating because the accelerated mass is a function of the aircraft's speed.
Hmmm...Rockets are a bit simpler. I wonder why they don't say "it's not jet engine science"?
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At constant aircraft velocity I suppose the mass flow rate would be constant.
Actually it would be going down, because the weight of the fuel is going down, which requires less lift, which in turn means that the aircraft generates less drag, which in turn requires less mass flow rate.
Or I suppose you could gain altitude and keep the mass flow rate constant.
Hmmm...Rockets are a bit simpler. I wonder why they don't say "it's not jet engine science"?
Say that after you've solved the puzzle!
Where does the extra power come from?
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Where does the extra power come from?
Oh, well that's easy. If the rate of consuption has not changed, but the power has increased, it must be operating more efficiently which is likely because it does not have to do so much work to compress the air on account of it getting stuffed into the front end, or something like that maybe.
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You know that bit in qi where it goes whoop-whoop, well... you fell into the trap- rockets have no top speed, and their generated power is simply proportional to speed!
However much energy is being generated from combusting the fuel, or even in the fuel, at some speed, the rocket is generating more power than that!
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You know that bit in qi where it goes whoop-whoop, well... you fell into the trap- rockets have no top speed, and their generated power is simply proportional to speed!
However much energy is being generated from combusting the fuel, or even in the fuel, at some speed, the rocket is generating more power than that!
I thought we were talking about jets!
Rockets simply operate on F = ma.
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Well, yeah... but...
Ok, let's take an example.
Exhaust speed 3500 m/s.
Rocket is travelling at 7800 m/s (it's an upper stage, and it's just about reaching orbit).
For the sake of argument let's say it's generating 1000 N of thrust.
So the fuel burn rate is 1000/3500 = 0.29 kg/s.
For the sake of arguments let's say the fuel's energy is 14MJ/kg. (It's about right for a LOX/Kero rocket propellant mix to give an exhaust velocity of 3500m/s).
So the rocket is generating 1000 * 7800 = 7.8MW
But the chemical power generated is 0.29*14e6 = 4MW
So there's a mysterious 3.8MW that's appearing from nowhere?
Where is it coming from?
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Well, yeah... but...
Ok, let's take an example.
Exhaust speed 3500 m/s.
Rocket is travelling at 7800 m/s (it's an upper stage, and it's just about reaching orbit).
For the sake of argument let's say it's generating 1000 N of thrust.
So the fuel burn rate is 1000/3500 = 0.29 kg/s.
For the sake of arguments let's say the fuel's energy is 14MJ/kg. (It's about right for a LOX/Kero rocket propellant mix to give an exhaust velocity of 3500m/s).
So the rocket is generating 1000 * 7800 = 7.8MW (power = force * distance per second)
But the chemical power generated is 0.29*14e6 = 4MW
So there's a mysterious 3.8MW that's appearing from nowhere?
Where is it coming from?
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A rocket in orbit does not need to do any work (well, hardly any) to travel at constant velocity in orbit.
Power is the rate of doing work. No work, no power.
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Whenever a rocket is being accelerated work is being done.
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Not if that acceleration is to keep it in a circular path at a constant speed.
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Ok, you have a point about that at least, but here we're talking about rockets accelerating under their own power.
The rocket power is proportional to speed (from force x distance) but that implies that the rocket's power increases apparently without limit, depending on how fast it happens to be going.
This is an apparent paradox. But of course in physics there are no true paradoxes!
So how can a rocket produce ever increasing power?
(FWIW the early rocket pioneer that discovered this thought that he had broken physics, but later realised what's happening.)
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Hint: there's a hidden source of energy, other than chemical energy of the fuel.
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Hint: there's a hidden source of energy, other than chemical energy of the fuel.
Yes, well considering when all of this was going on, it would not surprise me in the least to learn that many of the "rocket scientists" thought magic mushrooms were the source of the apparent energy.
At any point in time, f = m.a Does anyone care about the apparent power?
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Actually energy is really important in rocketry.
For example, if you want to go to Mars from the surface of the Earth.
You can either:
a) takeoff, and burn to an extremely high orbit, check things out, then do a burn to go to Mars.
b) takeoff, and burn to a low orbit, check things out, then do a burn to go to Mars.
You should always do b), because in a low orbit you're moving faster and the energy you're making from the rocket burn is higher, even though F=ma is the same. Doing this reduces the amount of propellant you need by several times!
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Anyway, I guess you've given up.
The secret is in the propellant. It has the chemical energy, but it also has kinetic energy. That's the hidden source.
Before it's burnt any kilogram of fuel has:
14MJ/kg of chemical energy
0.5 m v^2 of kinetic energy
At high enough speed the kinetic energy is bigger.
When it gets thrown out the back of the rocket, the kinetic energy of each kilogram ends up as 0.5 * 1 * (v-c)^2
So if, m = 1 kg, it changes by:
0.5 * 1 * (v - c)^2 - 0.5 * 1 * v^2
= 0.5 * ( v^2 - 2 v c + c^2 ) - 0.5 * v^2
= 0.5 v^2 - v c + 0.5 c ^ 2 - 0.5 * v^2
= 0.5 c^2 - v c
The 0.5 c^2 is just 0.5 m c^2 where c is the exhaust velocity- it comes from the chemical energy, that's just the fixed chemical energy (14 MJ/kg - losses) used by the rocket to make the jet, and it's constant. the -vc is more interesting; that's extra energy that's lost by the exhaust, an amount directly proportional to the rockets speed v, and from conservation of energy it all ends up in the rocket; that's where the extra energy you can calculate from W = f x d comes from.
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Ah! Right. As a in f=ma is the amount of acceleration applied to the exhaust relative to the rocket, would f=ma not take the kinetic energy of the propellant into account?
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No, it doesn't tell you where the energy ends up. You need to look at the speeds to calculate the kinetic energy, F=ma only gives you the rate of change of speed.
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No, it doesn't tell you where the energy ends up. You need to look at the speeds to calculate the kinetic energy, F=ma only gives you the rate of change of speed.
But surely if you knew how fast you were going in the first place, it should be quite straightforward to determine your current speed from the rate of change of speed.
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Pretty much, but only if you know the speeds at some point (and if the accelerations are known to sufficient accuracy).
The thing is that energy and forces are fundamentally different things, albeit related by numerous equations.
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Pretty much, but only if you know the speeds at some point (and if the accelerations are known to sufficient accuracy).
The thing is that energy and forces are fundamentally different things, albeit related by numerous equations.
Quite so. It is not so easy to convert energy into something useful. If it was easy, we might not find ourselves in the pickle that we seem to have created.
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ah, the fuel?
How do you differ between its kinetic energy and the chemical?
Doesn't the kinetic energy include all energy that leaves the exhaust?
This one was a tricky one.
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Thinking of it, what about those exhausts we all leave at times, bringing those vitamins into the air, uplifting the general temperament of our guests and loved ones, especially after a good long dinner?
Chemical energy that too me thinks? As well as kinetic..
So, will the same formula apply there?
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How do you differ between its kinetic energy and the chemical?
It's the kinetic energy before you burn it. You can tap into that, as well as the chemical energy, which is a constant amount independent of speed.
Doesn't the kinetic energy include all energy that leaves the exhaust?
No, there's heat energy as well (although at a microscopic level this is largely kinetic energy, but it's relative to the bulk flow which has overall kinetic energy).
A rocket nozzle extracts a lot of that heat, but about 30% is still left over.
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Okay thanks Wolfe, I'm not sure I get it still, the difference between chemical and kinetic energy I mean. You mean that it have a 'potential' kinetic energy before getting burned, well, as I think about it, although I won't promise I do, think, that is :)
To me kinetic energy is all those atoms, molecules etc, bouncing around in that chamber as the chemical reaction takes place, burning at a furious rate, throwing them out from the only opening they will find, the exhaust.
But I can see that you consider it two processes, one chemical consisting of a certain energy in itself, and the other one kinetic, getting a boost by the speed of the rocket. How exactly do the speed influence the kinetic energy? It's not like a jet where you can get more oxygen the faster you fly, right? So, where does that extra energy comes from?
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In physics, you always pick a frame of reference. It's usually a good idea to pick the centre of mass frame or the 'zero momentum frame' which is usually the same thing.
In rockets the best frame is usually the one that the rocket takes off from. In that frame, the overall momentum of the rocket is zero at all times (if you include the fuel, and less obviously the exhaust, which you must for the energy/momentum conservation to work out properly).
So the rocket starts going and lobs the exhaust one way, and goes off in the other direction.
At any time the energy of each kilogram of anything is 0.5 m v^2, but m=1, so it's simply 0.5 v^2.
So the kinetic energy of the fuel BEFORE it is burnt is 0.5 v^2, and the potential kinetic energy is just a constant on top.
The maths does work out if you use any other frame, but you have to be a lot more careful.
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Thanks, i have a distinct feeling I will need to come back to this one and reread it all :) I should get some sleep first too I think. "In rockets the best frame is usually the one that the rocket takes off from." And that would be the launch pad in this case, right? I really liked the sound of the 'zero momentum frame', I mean it, it made me think :) And that's a hard thing for me.
"energy/momentum conservation" you say, and that the best frame then would be a 'still' one for that to 'balance out' sort of. Would you say that all uniformly moving frames are 'still' 'zero momentum frames'? As that momentum will be impossible to define inside a black box? Anyway, It made a sort of sense to me. But for the rest I will need to think harder, a lot I'm afraid.. I think I need to start with looking into "energy/momentum conservation" and see how that is expressed, and if i can understand it. Can't say my math is that polished :) But I will look at it, as long as it doesn't attack, that is..
Never trust math that bites back.
But first some sleep :)
Thanks again Wolfe.