Could otters be trained to do this bit ?
Because to reinsert the float into the column, you have to push the water out of the way. That’s really all there is to it.
No, I’m not. A PM machine does not extract energy from the surroundings.
Okay. This machine is designed to extract energy from the surroundings via gravitation and buoyancy, in some manner that’s not described more deeply than that. But assume it’s true.
If we decide it isn’t a PM, and we’ve eliminated the terminology confusion, are you suggesting this machine works as described? Clearly it’s got bigger problems than simply what we call it.
This seems to be the right answer. You have to lift up water to put the float in – that’s adding the potential energy to the system. When the float bobs up, the water falls back down, and that’s where you recapture the potential energy.
As well as removing it from the column of water at the top. You can’t just move the object horizontally out of the water column. You’ll need to lift it up, out, and over with some method requiring energy.
I must fail to understand your problem. In theory you know you will have to invest as much energy to move the float as it can possibly produce.
The inventor believes he can outsmart the system. And maybe he can (google DWFTW). His problems will be practical not theoretical.
I don’t know that. I believe that. I was hoping for some way to state the drawbacks clearly and intuitively. But I am starting to think I’ll have to draft up a force balance diagram, and I haven’t sat in a physics class in 30 years. And I don’t want to waste time on this, but I probably will, because now it’s annoying me.
I’m still not clear on this, but AIUI either a) there’s a single tube of liquid that the object floats up and then magically sinks to float again - a neat trick, or b) there’s a tube filled with liquid for the object to float up, and then an empty tube for it to fall back down. I’m guessing it’s supposed to be B.
While I’m neither an engineer nor a physicist, I know from observation that if you connect a full tube and an empty tube at the bottom, you’ll wind up with two half-full tubes. And whatever method is used to keep one tube empty and one tube full while you transfer the object, it will require energy.
That, too. Energy needed at the top and bottom of the tubes.
Generally the inventor has a salient feature, chamber, valve, linkage, that circumvents the problem. That’s the part to discuss. Is that the case here?
Because the thing (whatever it is) floats to the top of the water, it is less dense than water. When it is inserted into the tub of water an equal volume of water must be raised, but as the water is denser than the thing, the weight of water raised is more than the weight of the thing, So it requires more work to insert the thing into the water than you recover by its floating to the top.
OK, I haven’t done this in forever, but I muddled through it some numbers. Spoiler - there is no free lunch. It costs110kJ of energy to prime the system for 1 cycle, but operating it only yields 103kJ of energy.
Can someone check that I’ve decomposed the problem correctly (or at least sufficiently) and I’m thinking in the right units?
Given normal fresh water at STP, using a 10 meter column of water, and a 1 m³ ball with density of 50 kg/m³, we will extract energy from the force of the rising buoyant object, shift so that it will fall in air and pull a chain to recapture that energy. Then we will insert the ball at the bottom and start over.
Buoyant force = density of liquid in kg/m³ * volume of object * gravitational acceleration.
Our 1m³ ball has a buoyant force of 9807N. Traveling over a distance of 10 meters, we have done 98,070J of work. Assume we store it all.
Falling force = F*ma. Our 1m³ ball of polystyrene weighs 50kg. That’s 490.35N of force, over 10 meters again, we end up with 4904J. Assume we store it all.
Our combined stored energy is now at 102974J. We cannot exceed this budget as we are resetting the system. How much energy does that require?
The diameter of our 1m³ ball is 1.24m. Again, the buoyant force from the earlier calculation is 9807N, so to submerge it, multiply those two. 12160J merely to put that ball underwater, disregarding pressure.
To reset the system, we must lift the displaced 1 cubic meter of water back to the top.
F = 1000kg * 9.807 = 9807N
work = 98,070J
Adding the cost of submersion plus cost of refill:
98070J + 12160J = 110230J.
Total energy captured via rising buoyancy and falling object = 102974J.
The stored energy has a deficit of 7256J without even factoring friction. This system does not generate enough energy to reset itself for another cycle.
Yep - there’s the catch. In order to reinsert the ball into the bottom of the water column, you have to displace a volume of water equal to the volume of the ball, and you can only do that by pushing the water in the column up*, out of the way, which costs at least the same amount of work that you get from the ball’s buoyancy.
*If you don’t do this, you’re draining water from the column, so the level falls on each cycle, and anything you’re getting out of the system is just like using a head of water to drive a hydroelectric turbine.
The bouyant force minus the submersion force is zero. That should be sufficient.
This is exactly it.
We have a beach ball 1m^3 volume. To insert it through an airlock at the bottom of the column, you open the empty airlock, put in the beach ball, then close the air door, fill the airlock with water - from the column, and open the water side door… push the beachball in and attach the chain.
So the surface of the water column has gone down to accommodate the 1m^3 of water that replaces the beachball in the airlock. The beachball goes to the surface, floats, is transferred to the air column, and attached to another chain, drops to the bottom doing work.
But now, before it can go into the airlock, the airlock water has to be pumped out - which raises the water column by 1m^3. So essentially, the system has to pump that volume of water the equivalent distance it dropped when the airlock filled.
If you pump water in the bottom of the column, it effectively pushes the whole column upward by that distance. that’s work done to the water in the column. If you use a perpetual source of water at the level of the surface, and drain the bottom airlock to the lower level, you have essentially created an highly inefficient hydroelectric dam, using whatever means produces that water source (solar evaporation and the precipitation cycle?)
If I had made a diagram, you would see the buoyant ball rises 10m, but when I submerge it, I’m only pushing it down 1.24m (the diameter of the ball) because the lock is at the bottom of the column. So although the force is same, the magnitude of work is different. Otherwise, point taken.
When you place the object back at the bottom of the column, the object displaces water. You must push that equal volume of water up.
Right–there are actually two methods here. One is that you push the object into the water column, which requires lifting up the entire column. The second is that you swap the object with an equivalent volume of water. The second method doesn’t require any energy, but either the water column drops over time, or you have to pump the displaced water back up to the top.
I’m not seeing what the OP is not seeing. The correct answer has been mentioned repeatedly already in this thread.
Granted, let’s assume a hypothetical system where all the extraneous work is eliminated – the need to move the float laterally out of the top of the column; the work involved in operating the water-lock gate and the loss to friction, etc.
In the theoretically-perfect system, you have to lift some water up out of the way to stuff the float back in at the bottom, and it is exactly the potential energy of that raised water that is recovered when the float rises. Then you can recover the potential energy from the float as it drops in the air, but you’ll just use up that energy to put the float back into the bottom of the water. That alone, without any other extraneous factors, makes it not work.
If it were a perfectly friction-free, entropy-free system, the that could perhaps continue indefinitely in true “perpetual motion” fashion. But no wanna-be PM machine is fully friction-free and entropy-free, so they always grind to a halt. And in any case, you could never extract any useful work from the system, as that would grind it to a halt just the same.
Never mind that post with all the arithmetic. OP wants an “intuitive” debunking.
Something loosely apropos: Time Crystals
Basically, particles that arrange themselves in a crystal pattern in time rather than in space. This appears to be some kind of perpetual motion system with the particles oscillating in some kind of repetitive pattern over time, without loss of energy.
But you can’t extract any useful work from it!
And yet . . . there is this paragraph:
Waitaminnit. If it glows, that means there is energy emanating from the system? Something ain’t right there. I shall have to re-read a little more carefully.