Free Fall Scenario: 100kg Piston Kinetic Energy After 20m Drop

[beckerman]

Lets say you have a piston falling in a long cylinder. The piston weighs 100kg but only weighs 1kg under water. The cylinder is filled with water, but there is plumbing going from the bottom of the cylinder to the top. As the piston falls the water can move in a continuous loop, from the bottom of the piston to the top. Taking the waters momentum into account, and assuming 100% efficiency what would the kinetic energy of the piston be after a 20m freefall?

 

[Hootenanny]

What are your thoughts on the matter?

 

[beckerman]

I must warn you I am not a scientist, and am pretty ignorant when it comes to physics. My thoughts are the 1kg of mass we have would set the water into motion. Gravity would cause the piston to continue to accelerate. The initial resistance of the water would keep the piston from accelerating at 9.8mpss initially, but once the water was flowing it would accelerate at or near this speed. I think once everything was moving, you would have the kinetic energy of a 100kg object even though it is submerged and weighs only 1kg underwater. I also think that since joules are a measurement that pertains to energy required to stop a moving object, the fact that the water is moving with the piston, and also would need to stop, would have to be figured into the equation.

 

[Chronos]

The weight of the piston is irrelevant, only the mass is of any real consequence.

 

[beckerman]

Could you elaborate please.

 

[DaveC426913]

I'm thinking it would get nowhere near anything like terminal velocity. The biggest factor would be the diameter of the plumbing and resultant flow rate.

(Or, more accurately, the flow rate through the plumbing would define the terminal velocity.)

 

[beckerman]

Lets say the plumbing is very large, larger in area than the cylinder. If the water can flow very freely than what would prevent the piston from accelerating at 9.8mpss?

 

[rcgldr]

Lets say the plumbing is very large, larger in area than the cylinder. If the water can flow very freely than what would prevent the piston from accelerating at 9.8mpss?

The problem is water doesn't flow very freely. The piston is doing work on the water, accelerating the water downwards, and the fluidic drag (did I just invent a term?) is very large.

If it were a tight cylinder, then you'd have the force of gravity accelerating the piston and all the water being circulated by the piston. The density of the piston, the path the water flows, and the viscosity of the water would determine the terminal velocity.

 

[Danger]

Lets say the plumbing is very large, larger in area than the cylinder. If the water can flow very freely than what would prevent the piston from accelerating at 9.8mpss?

You have another problem with that. If the plumbing is of higher capacity than the cylinder, you wouldn't be able to force it all the way back to the top.

 

[beckerman]

You would never have to "force anything back to the top" because the cylinder and the plumbing would be completely filled with water. The inlet for the plumbing would be above where the top of the piston is at the beginning of the descent. This way the piston never has to work against gravity to raise a water level. I think Jeff Ried is on the right track. We can assume a density since we know mass in and out of water. Our piston has a mass of 100kg but displaces 99kg of water when submerged. So we could say the piston has a volume of 9900cc. How to calculate the effect of the water viscocity is beyond me. It would seem to me that there would be a figure for this that could be plugged into a formula as a value for resistance.

 

[Danger]

Okay. I was thinking of the plumbing being empty to start with. That was a bit dumb on my part. Something about the wording of your original post gave me that impression.

 

[Gelsamel Epsilon]

Assuming 100% efficiency then I think it will accelerate at normal g (not factoring in friction, but does 100% efficiency include nil-friction?).

 

[DaveC426913]

1] A rate-limiting factor will be the bottleneck between the piston and the plumbing. Unless the plumbing is the same X-sectional area as the piston - which means the device is simply a hollow, fluid-filled doughnut.

2] Friction is a huge factor.

3] I think the terminal velocity of the falling piston will be limited to however fast the piston would sink in a swimming pool. (In fact, much slower, but that's an upper limit.) Especially considering the initial condition that the piston is very nearly neutrally buoyant. This will dramatically slow the process.

 

[beckerman]

As far as friction is concerned, I was thinking of a piston with a little bit of clearance so there would be a thin layer of water that would travel with the piston, between its outer wall and the inner wall of the cylinder. In comparison to an object sinking in a swimming pool I think in this situation the object would have to descend much more quickly. The mass is near neutral at the very beginning of the descent, but as the water moves from beneath the piston it would change. In a pool the effect the water would have on the mass would remain constant. I see the effect of the piston has on the water as being similar to hooking the suction and pressure lines together on a water pump. Initially there would be energy used to set the water into motion. Once it was moving the amount of energyy required would decrease. Once the piston establishes momentum with the water the amount of work required to keep it moving would be minimal. As the work requirement decreases the piston would then be able to accelerate. As this occurs and the water and piston are moving, we really cannot say we have an object that is near neutral bouyancy either since an object will not float in a water column that is moving straight down.

 

[rcgldr]

Initially there would be energy used to set the water into motion. Once it was moving the amount of energy required would decrease.

The water would be accelerating just as fast as the piston, and due to visocity, the amount of engery required to maintain a constant acceleration would increase. Since the force is fixed, the piston and water would reach a terminal velocity if the cylinder is tall enough.

 

[beckerman]

But would the velocity over a 20m freefall be the same as it would be if falling through the air? Or would the object reach terminal velocity at a low speed and in a short distance?

 

[rcgldr]

But would the velocity over a 20m freefall be the same as it would be if falling through the air? Or would the object reach terminal velocity at a low speed and in a short distance?

No, the only way to get true free fall is in a vacuum with no air. The air provides resistance, and water provides much more resistance than air.

 

[Danger]

No, the only wall to get true free fall is in a vacuum with no air.

Might I assume that you meant 'way'? That would maintain my respect for you.

 

[Chronos]

Turbulence is the keyword. Think about it.

 

[rcgldr]

Might I assume that you meant 'way'?

Oops, editted the post. Was listening to something while typing, apparenly my multiplexing paths got crossed.

 

[beckerman]

No, the only way to get true free fall is in a vacuum with no air. The air provides resistance, and water provides much more resistance than air.

I understand water provides more resistance than air, but we are not falling through water. Not in the same sense that we would be falling through air if something fell out of a tree for instance. When an object falls through a stationary fluid it has to push the stationary molecules of that fluid out of the way to move. As the object moves past these molecules frictional forces present themselves to all sides of the object creating drag. In this case rather than pushing past any molecules the piston is pushing the molecules through the plumbing. The water moves with the piston as opposed to around the piston. I guess my question is if the water presents so much resistance to allowing the piston to descend in this scenario then why? The only possibility I see is friction between the water molecules and the walls of the piston.

 

[Danger]

Was listening to something while typing, apparenly my multiplexing paths got crossed.

Pink Floyd is a valid excuse. Carry on.

 

[DaveC426913]

The only possibility I see is friction between the water molecules and the walls of the piston.

Friction between the water and the walls of the entire construct. Realize that ALL the water is moving. Also turbulence, since the construct has twists and bends.


Really, you seem to be imagining that it will fall as if in air. I think it will fall as if in water. Slowly.

 

[Danger]

Yeah... it's not going to hit anything like the terminal speed that it would have in air. I've tried looking at it in reverse, as well. If there's a rope attached to the top of the piston, you're not going to be able to pull it back up as quickly as you could in air.

 

[beckerman]

I agree that it would not move as fast as in air, I also believe the terminal velocity would be slower than it would be in air. I do not agree the descent would be "slow" as it would be in water. Lifting the piston is an interesting way to look at this. The thing that I think is important to remember is that the weight of an object depends on where it is. An object can weigh one amount on Earth and a different amount on the moon for instance. This object weighs 1kg under water. If you were lifting the piston and let go of the rope, would you say it weighed 1kg during the period where the waters momentum was still carrying it upwards? I think the opposite is true also. The piston weighs 1kg in still water, not in water that is moving in the same direction that gravity is pulling it anyway. Hence I think the force we have to overcome effects like turbulence and friction (which I think could be minimized quite a bit in the design of the device anyway) would be similar to a 100kg object not a 1kg object. Either object is going to accelerate to Earth at the same velocity, however the heavy object will be affected less by the drag of the water.

 

[Danger]

By that same reasoning, the oxygen and nitrogen bubbles in the water are more affected by the 'drag' of the water; they do not, however, travel downward at the same acceleration as the other things. In fact, they don't travel down at all.

 

[beckerman]

Oxygen and nitrogen bubbles float. An object that weighs 1kg under water sinks. I do not really understand your comment. Maybe I am missing something.

 

[Danger]

Merely pointing out that the following is a self-contradicting statement.

Either object is going to accelerate to Earth at the same velocity, however the heavy object will be affected less by the drag of the water.

They won't accelerate at the same velocity because the drag gives each a different interaction. Also, things don't accelerate at a velocity; they accelerate at a rate of velocity change.

 

[NoTime]

I see the effect of the piston has on the water as being similar to hooking the suction and pressure lines together on a water pump. Initially there would be energy used to set the water into motion. Once it was moving the amount of energyy required would decrease.

Seems to me, if you totally neglected friction, that you could treat this as an unbalanced wheel.
However, friction/turbulance between the walls of the tube and the water are going to have a major impact.
OTOH I suppose you could freeze the water, get rid of the tube, and mount it on a low friction bearing

 

[DaveC426913]

Seems to me, if you totally neglected friction, that you could treat this as an unbalanced wheel.

I was thinking along the same lines when I was moving toward a "hollow donut" idea.

Perhaps this should be addressed to the OP. A device operating on an "unbalanced wheel principle" cannot be used as a power source, because you expend as much energy lifting the weight back to the top.

Even though OP's idea is somewhat far from an unbalanced wheel, the principle still applies. He's still simply "storing" the energy as potential energy. Drop the weight to convert it into useful energy.

But it's still nothing more than a storage device, one that has to be "recharged".

 

[beckerman]

Merely pointing out that the following is a self-contradicting statement.



They won't accelerate at the same velocity because the drag gives each a different interaction. Also, things don't accelerate at a velocity; they accelerate at a rate of velocity change.

Ok poor wording on my part. But all freefalling objects accelerate at a rate of 9.8mpss. I still do not understand your comparison with the bubbles. I was stating that if you have two objects of identical size, and one is heavier than the other, and they were both dropped, the heavy one would reach the bottom of the drop quicker because its weight allows it to overcome drag forces more effectively. So if we can say our piston weighs 1kg at the beginning of the descent but its weight increases as the water is set to motion, then perhaps we would have a 50kg piston after a couple meters of drop. As the water moves it has less and less of an effect on the pistons bouyancy.And as the piston becomes less bouyant and more heavy its acceleration would be hindered less and less by drag.

 

[beckerman]

I was thinking along the same lines when I was moving toward a "hollow donut" idea.

Perhaps this should be addressed to the OP. A device operating on an "unbalanced wheel principle" cannot be used as a power source, because you expend as much energy lifting the weight back to the top.

Even though OP's idea is somewhat far from an unbalanced wheel, the principle still applies. He's still simply "storing" the energy as potential energy. Drop the weight to convert it into useful energy.

But it's still nothing more than a storage device, one that has to be "recharged".

I suppose this is similar to an unbalanced wheel. Also your hollow doughnut applies, if it was oval shaped with long straight sides, so one side could be used as our cylinder. Call it what you will. Does anyone agree with me that the piston will fall "fast" and not "slow?" Can anyone prove either way?

 

[NoTime]

Ok poor wording on my part. But all freefalling objects accelerate at a rate of 9.8mpss. I still do not understand your comparison with the bubbles.

The keyword here is freefall.
The only time you get freefall is in a vacuum.
A feather obviously does not fall as fast as a BB.
And a hot air balloon with people in it, having a mass much greater than a feather, doesn't fall at all.

I was stating that if you have two objects of identical size, and one is heavier than the other, and they were both dropped, the heavy one would reach the bottom of the drop quicker because its weight allows it to overcome drag forces more effectively.

This is simply not true.
Take a roll of aluminum foil, drop it.
Now unroll the foil and drop it.
Same mass, but in the unrolled state, it's going to take a long time to drop.
It's all about density here.

So if we can say our piston weighs 1kg at the beginning of the descent but its weight increases as the water is set to motion, then perhaps we would have a 50kg piston after a couple meters of drop. As the water moves it has less and less of an effect on the pistons bouyancy.And as the piston becomes less bouyant and more heavy its acceleration would be hindered less and less by drag.

Consider my frozen water example.
The water/piston is one system. Not two.
Bouyancy has no meaning here.

 

[NoTime]

I suppose this is similar to an unbalanced wheel. Also your hollow doughnut applies, if it was oval shaped with long straight sides, so one side could be used as our cylinder. Call it what you will. Does anyone agree with me that the piston will fall "fast" and not "slow?" Can anyone prove either way?

No, I don't agree.
Look up plumbing. Read up on how pipe diameter affects water flow.

 

[beckerman]

I said two objects of identical size with different mass. That means the density of the two objects are different. Your foil scenario is the same as my scenario. There both right. Read and think before you make corrections. Also are you picturing a 100kg object in a piece 3/4" pipe? Pipe diameter was never specified because I was hoping you and others would visualize a best case scenario and not what you might be limited to at Home Depot.

 

[NoTime]

I said two objects of identical size with different mass. That means the density of the two objects are different. Your foil scenario is the same as my scenario. There both right. Read and think before you make corrections. Also are you picturing a 100kg object in a piece 3/4" pipe? Pipe diameter was never specified because I was hoping you and others would visualize a best case scenario and not what you might be limited to at Home Depot.

Yep, I mispoke myself.
Substitute the word aerodynamic for density.

Pipe is pipe. Doesn't matter where you buy it.
And Home Depot sells pipe up to 4" in diameter.

You don't seem to be paying attention.
Your question is no longer interesting.

 

[beckerman]

Consider a 20cm diameter cylinder. 1 linear meter = 3.14 cubic m. So for the piston to descend 1 meter, 3.14 m3 of water must exit. Likewise for the piston to descend 1m in 1 sec the water would have to flow at 3.14 cubic meters per second. In a 20 cm wide pipe I really don't think the back pressure generated to push that volume would be that great. The volume or pipe diameter of the plumbing used to rout the water back to the top of the cylinder can be any size. It does not have to match the cylinder size. Therefore I feel the resistance in this part of the system could be kept quite small. If I was convinced 100% I knew what I was doing I would not be here. I appreciate everyones help, and hope you realize if I am arguementative I am just trying to be certain.

 

[Danger]

You're entitled to your opinion, of course, but you're quite mistaken about NoTime. Everyone gets a bit frustrated at finding new ways to say the same thing over and over again. That naturally leads to an occassional slip-up in wording.
I'm going to have to bail out of this thread as well. It's not because I'm no longer interested; rather, I've exhausted my knowledge in the subject. It looks like a job for Clausius.

 

[NoTime]

Consider a 20cm diameter cylinder. 1 linear meter = 3.14 cubic m. So for the piston to descend 1 meter, 3.14 m3 of water must exit. Likewise for the piston to descend 1m in 1 sec the water would have to flow at 3.14 cubic meters per second. In a 20 cm wide pipe I really don't think the back pressure generated to push that volume would be that great. The volume or pipe diameter of the plumbing used to rout the water back to the top of the cylinder can be any size. It does not have to match the cylinder size. Therefore I feel the resistance in this part of the system could be kept quite small. If I was convinced 100% I knew what I was doing I would not be here. I appreciate everyones help, and hope you realize if I am arguementative I am just trying to be certain.

Thanks Danger.

Ok, If you are interested in learning.

1) calculate the correct volume of 20cm diameter by 1m length of water. You are off by a lot.
Show all of the math you use.

2) give the mass (weight) of 3.14 m^3 of water. Your current answer.
If you live in the US give the answer in lbs.
Show all of the math you use.

3) give the length of your 20cm diameter piston.
Show all of the math you use.

 

[beckerman]

Fair enough,
r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

 

[Danger]

r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

Wanna recheck that?

 

[beckerman]

The 314 square centemeters = 3.14 sq meters part? Would I not divide by 100 since there are 100 centemeters in a meter? Help me out here.

 

[Danger]

There are 100 centimetres to a metre, but you forgot to square it. A square metre equals 10,000 square centimetres.

 

[beckerman]

Of course thank you.

 

[beckerman]

Another way to look at this scenario would be to compare the cylinder and plumbing to a balanced see-saw. If you had two tubes standing vertical with their bottoms and tops attatched, the water in one would balance the water in the other. Moving a cubic meter of water down in one would mean moving one up in the other, and vise versa. There would be no force required because opposite the cubic meter of water in the left tube, there is a cubic meter in the right tube that is directly opposite it. Moving any mass in either tube would be like pushing one side of a perfectly balanced see-saw. If you had a perfectly balanced see saw and threw 1kg on 1 side it would push that one side down very quickly.

 

[DaveC426913]

If you had a perfectly balanced see saw and threw 1kg on 1 side it would push that one side down very quickly.

No, it wouldn't. You're forgetting about inertia. How fast do you think you could set 2,000kg of water in motion? Can you see a seesaw with 1,000kg sitting on each end bouncing up and down as if empty?

 

[beckerman]

No, of course not. What I see happening is consistant with what I have been saying all along. The piston would accelerate slowly at first (to break the inertia of the stationary water) but once it set the water into motion the piston would accelerate at a normal 9.8mpss (less losses due to friction between the water and the plumbing)

 

[NoTime]

Fair enough,
r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

1) Danger helped you on this one, but I havn't seen your corrected answer anywhere.

2) 314 cc of water does not weigh 3.14Kg.
Neither does 3.14 cubic meters.

cc usually stands for cubic centimeter, also written as cm^3.
How did you get from 3.14 cubic meters (m^3) to 314 cubic centimeters (cm^3).
Pay attention to what Danger told you here.

How long is a meter in feet.
hint: It's more than a yard.
Just how big is a chunk of water 3.14meter x 3.14 meter x 3.14 meter (or 3.14 cubic meters) in cubic feet?

3) You gave "The piston weighs 100kg but only weighs 1kg under water." in the original post.
The 20m was for the pipe.

I asked how long is the piston.
It's not 20 meters.
With a 20cm diameter it has a specific length.
You might shout Eruka! when you figure it out.

 

[beckerman]

Dear No Time,
When an object descends in a cylinder filled with water, which is directly linked at top and bottom by an identical cyliner which is also filled with water, describe the chain of events that must take place for 1chunk of water measuring 1 cubic foot to move 1 linear hand!

 

[beckerman]

Erukaaaaaaaaa!