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 DRHopton Nov8-12 05:21 AM

Power of a thrust to maintain an extended spring

Hi,

I have a question about working out the power to maintain equilibrium with a spring:

I attach a rocket to a horizontal spring and fire the rocket horizontally. This is done on a track or something similar. The spring will extend until the tension in the spring equals the thrust of the rocket (assuming the thrust is constant) at which point the system will be in equilibrium.

The rocket is clearly doing work to maintain this situation, but how do I calculate this work?

W=Fx doesn't seem right as nothing is moving any distance. While the spring was extending the rocket was putting work in to it equal to 1/2 Kx^2, but I'm not interested in that. I want to know the power required to maintain the equilibrium.

Thanks

Dan

 jbriggs444 Nov8-12 05:42 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 Quote by DRHopton (Post 4150398) W=Fx doesn't seem right as nothing is moving any distance. While the spring was extending the rocket was putting work in to it equal to 1/2 Kx^2, but I'm not interested in that. I want to know the power required to maintain the equilibrium.
The exhaust stream is moving. What is the exhaust velocity?

 DRHopton Nov8-12 06:35 AM

Re: Power of a thrust to maintain an extended spring

Exhaust stream velocity is not known, or at least beyond my ability to measure accurately.

For a bit more background, I want to measure the efficiency of an Aeolipile. Mine is very similar to this one, but it hangs from a bearing over a bunsen instead of being mounted on a turntable. I have done this in the past by trying to measure the angular acceleration as the thing spins up to speed. Knowing the moment of inertia, which I have measured, I can therefore work out the resultant torque (Thrust - friction). Measuring the angular deceleration as it comes to rest allowed me to estimate the friction. Then I just applied W=Fx while it was spinning up to speed.

I made the measurements with a light gate which registered every half rotation, which wasn't a very precise method as the thing is at full speed in a few rotations (<5). Plus the fact that letting the thing spin down to estimate the friction includes all sorts of assumptions about that friction that are probably dubious (e.g. that the friction is is constant and independent of rotational speed). Was there a better way of measuring the work done? I wondered whether it could be done by hanging the Aeolipile from a torsion spring, hence my original question.

However, I'm starting to think that I just don't have enough info. to solve this for the work done. The set up would allow me to measure the thrust of the Aeolipile, but not the power, right?

Could I then let the Aeolipile spin up to a constant angular velocity, measure that velocity, and then apply P=Fv using that measured thrust?

Thanks

p.s. - In case you are interested, the power input is measured by timing how long it takes to boil away a known quantity of water and working out the energy needed to boil that water using water's latent heat of vaporization.

 mfb Nov8-12 08:26 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 Quote by DRHopton (Post 4150446) Exhaust stream velocity is not known, or at least beyond my ability to measure accurately.
You need that, or similar values, to calculate the required power. Mass flow would be fine, too, if you know the force from the spring and the mass flow rate you can calculate the velocity based on momentum conservation.

Quote:
 Could I then let the Aeolipile spin up to a constant angular velocity, measure that velocity, and then apply P=Fv using that measured thrust?
Probably.

 jbriggs444 Nov8-12 11:18 AM

Re: Power of a thrust to maintain an extended spring

So you want to determine the "efficiency" of your Aeolipile. Specifically, you probably want to know how much usable mechanical energy you get out as a fraction of the heat energy that you are putting in.

Right?

At maximum speed you get no usable energy out. There is zero torque available at the PTO (power take off). At zero speed you also get no usable energy out. There is zero rotation rate at the PTO. Your maximum efficiency is clearly going to occur somewhere between those extremes.

One crude estimate could be derived by guessing that available torque decreases linearly with increasing rotation rate. By inspection you will get max power at 1/2 of max rotation rate with 1/2 of max torque. So measure max rotation rate and max torque, take the product and divide by four. That's your estimated max available power at the PTO. Divide by heat input rate and you have your efficiency measure.

[This is conditioned on an assumption that heat input rate is independent of rotation speed].

 DRHopton Nov12-12 04:20 AM

Re: Power of a thrust to maintain an extended spring

Thanks for the replies.

For a bit more background, I am a technician at a sixth form college in the UK (16-18 year olds). This is purely an academic exercise for use in teaching students about heat engines and the theory of working out efficiency. The conclusion of the exercise is intended to be: This is a simple heat engine, but people do not use it because it is very inefficient. They can then discuss why it is inefficient compared to say a Carnot engine.

Quote:
 Quote by jbriggs444 (Post 4150701) Specifically, you probably want to know how much usable mechanical energy you get out as a fraction of the heat energy that you are putting in. Right?
Interesting. I actually hadn't thought about the usable energy, but in fact was thinking about the energy required just to overcome friction i.e. the efficiency of turning heat energy in the water in to kinetic energy of the rotating Aeolipile at maximum rotational velocity. There is no PTO on the device as it is purely for demonstration, but it would certainly be a good teaching point to discuss how you can't extract any useful energy at maximum rotational velocity. And applying your linear model will obviously reduce the efficiency compered to the max. velocity situation.

I should think I now have enough info. to launch in to some measurements. This is my first post on this forum and I must say it has been very useful. Thanks!

 sophiecentaur Nov12-12 10:29 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 Quote by DRHopton (Post 4150398) Hi, I have a question about working out the power to maintain equilibrium with a spring: I attach a rocket to a horizontal spring and fire the rocket horizontally. This is done on a track or something similar. The spring will extend until the tension in the spring equals the thrust of the rocket (assuming the thrust is constant) at which point the system will be in equilibrium. The rocket is clearly doing work to maintain this situation, but how do I calculate this work? W=Fx doesn't seem right as nothing is moving any distance. While the spring was extending the rocket was putting work in to it equal to 1/2 Kx^2, but I'm not interested in that. I want to know the power required to maintain the equilibrium. Thanks Dan
You have a conflict of Units here, I think. The spring requires a certain Force that is not Power. If you are not moving the spring past its required displacement length then no more work is being done on it. Using jets to provide a force with no movement is a very inefficient exercise (zero efficiency, aamof).
This thread has usefully moved you away from the original question but I thought it worth while pointing out 'why'.
There is one measurement that you could do to give an idea of the power involved and that would be to measure the Heat Power supplied by the bunsen burner flame. That would give you a maximum possible power of the engine (way more that the real value). You could find the power of the flame by measuring the temperature rise of water in the sphere (or a similar sized container. Then, Power( dW/dt) = mσ dT/dt

Unfortunately, those turbines deliver so little mechanical power that it's pretty difficult to measure. A 'brake' system is common for measuring power (as in brake horsepower) but I guess you couldn't couple the thing to any shaft, easily, without actually stopping it.

 DRHopton Nov13-12 08:25 AM

Re: Power of a thrust to maintain an extended spring

sophiecentaur thanks for your input. I realize that power and force are different units, but in the rocket on a spring situation the rocket is clearly doing work to maintain the situation and I wondered how this could be measured. What I overlooked is the fact that the rocket is not doing work on the spring, but on the exhaust gases. As jbriggs444 said I need to know the exhaust velocity, which I don't. This setup would however allow you to measure the thrust of the rocket.

Thinking now of my Aeolipile, I intend to attach it to a torsion spring and heat it at a certain rate. When the torque produced by the Aeolipile is equal and opposite to the torque of the spring I should be able to measure the angular displacement and therefore work out the torque produced by the Aeolipile at that rate of heating.

Next I will let the Aeolipile spin freely at the same rate of heating. When the torque is equal to the friction it will spin at constant angular velocity (ω). Measuring that velocity, I will apply P=Tω using the previously measured torque (assuming that the torque produced while spinning is the same as that produced when stationary and pushing against the spring). This should tell me the power required to maintain that ω against friction. jbriggs444 noted that no useful work can be extracted at this maximum ω, but I can still use it to work out the efficiency of converting heat energy to rotational kinetic energy of the Aeolipile. As I have said this is purely a demonstration to students so this will still be a useful demonstration of the inefficiency of an Aeolipile.

if you see and obvious problems with this please let me know.

 mfb Nov13-12 08:56 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 (assuming that the torque produced while spinning is the same as that produced when stationary and pushing against the spring)
I don't think this is true. Torque depends on the angular velocity, with maximal torque at rest (highest exhaust velocity relative to the environment).

 DRHopton Nov13-12 09:00 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 Quote by mfb (Post 4157473) I don't think this is true. Torque depends on the angular velocity, with maximal torque at rest (highest exhaust velocity relative to the environment).
OK, so I guess my assumption is that the exhaust steam velocity >> the instantaneous linear velocity of the nozzle as it rotates. Maybe not the best assumption, but without it I'm stuck!

 mfb Nov13-12 09:39 AM

Re: Power of a thrust to maintain an extended spring

What about the mass flow rate? Heat it (with attached spring) for a while, measure the mass before and after some time passed.

Measuring the deceleration at the maximal rotation frequency would allow to estimate the torque there (together with the inertial moment, which can be estimated with the acceleration from rest at known torque for example).

 DRHopton Nov14-12 04:10 AM

Re: Power of a thrust to maintain an extended spring

I've uploaded a video of the Aoelipile so that you have a better idea what I'm talking about: http://www.youtube.com/watch?v=9Muh55fX9gQ

Quote:
 Quote by mfb (Post 4157517) Measuring the deceleration at the maximal rotation frequency would allow to estimate the torque there
You mean the deceleration of the steam after it leaves the nozzle? I'm afraid that is beyond me with the equipment I have.

Or do you mean the deceleration of the Aeolipile when the steam runs out? Which would also be difficult as that does not happen instantly, but rather the steam fades out as the last of the water contacts the hot part of the boiling tube.

Quote:
 Quote by mfb (Post 4157517) (together with the inertial moment, which can be estimated with the acceleration from rest at known torque for example).
That's a better method of finding the moment of inertia than I used before we thought of using the spring to measure the torque. Before I compared angular oscillations of the Aeolipile attached to a spring to those of a plastic bar and used
http://upload.wikimedia.org/math/6/9...d750276d06.png
to calculate the moment of inertia of the bar (Ib). I then had two shm equations with the same torsional spring constant and one unknown moment of inertia. Of course the moment of inertia of the Aeolipile (Ia) changes as the water evaporates, but I am ignoring that on the grounds that the water is close to the rotation axis and accounting for it would add another layer of complication. I'll just have to measure Ia with a known amount of water and then try to measure ω when there is that much water left.

 sophiecentaur Nov14-12 05:52 AM

Re: Power of a thrust to maintain an extended spring

Oscillation method: great idea! Wish I'd thought of it.

 mfb Nov14-12 07:45 AM

Re: Power of a thrust to maintain an extended spring

Quote:
 Quote by DRHopton (Post 4158667) You mean the deceleration of the steam after it leaves the nozzle? I'm afraid that is beyond me with the equipment I have. Or do you mean the deceleration of the Aeolipile when the steam runs out? Which would also be difficult as that does not happen instantly, but rather the steam fades out as the last of the water contacts the hot part of the boiling tube.
I would force rotation without water, and measure deceleration.

Oscillation is a great idea.

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