Maximillian Posted September 10, 2016 Posted September 10, 2016 CREATION OF ENERGY Experiment Consider a ball of density 1100kgm-3 and mass 1kg. Consider a tank with water of height 10m. Place the ball in the tank so that it sinks to the bottom of the water. While at the bottom of the water the ball is made to displace an additional volume bringing down its density to 900kgm-3 thus it starts rising to the water surface. Its previous volume was (1/1100)m3 and its new volume is (1/900)m3 The volume added is the new volume minus the old volume (1/900) − (1/1100) = 0.000202m3 The energy needed to work against the water pressure and create this new volume is equal to pressure multiplied by volume 𝐸= 𝑃𝑉 𝐸=ℎρ𝑔𝑉 𝐸=10×1000×10×0.000202 𝐸= 20J When the ball gets to the top of the water it will gain a gravitational potential energy of 𝑃𝐸=ℎ𝑚𝑔 𝑃𝐸=10×1×10 𝑃𝐸=100J Therefore we realize that we input only 20J of pressure potential energy but we have gained a gravitational potential energy of 100J. It works when small changes in the volume of the body make it less dense than the fluid it’s immersed in. For more information please follow this link to the filed patent https://www.researchgate.net/publication/307865505_CREATION_OF_ENERGY.
swansont Posted September 10, 2016 Posted September 10, 2016 How much gravitational potential energy does the water lose?
John Cuthber Posted September 10, 2016 Posted September 10, 2016 "While at the bottom of the water the ball is made to displace an additional volume bringing down its density to 900kgm-3"How? 1
Strange Posted September 10, 2016 Posted September 10, 2016 Why is this Experiment violating the conservation of Energy? Because you haven't analysed it correctly. 1
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 "While at the bottom of the water the ball is made to displace an additional volume bringing down its density to 900kgm-3" How? An electric motor mounted inside the ball can stretch the walls out.
John Cuthber Posted September 11, 2016 Posted September 11, 2016 An electric motor mounted inside the ball can stretch the walls out. And that motor requires energy. That's why the conservation laws are intact. 1
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 (edited) And that motor requires energy. That's why the conservation laws are intact. Yes it requires 20J of energy and that will enable the ball become less dense than water hence float to the water surface. How much gravitational potential energy does the water lose? If you think it looses any can you give the value please. Edited September 11, 2016 by Maximillian
John Cuthber Posted September 11, 2016 Posted September 11, 2016 This expression =ℎ is incomplete- as Swansont said. Imagine that you want to recover the energy from the ball as it floats to the surface by tying a string to it and passing that through a pulley at the bottom of the tank As the ball rises it pulls the string. You can use that to do work- for example- to lift a weight outside teh tank (you'd need a couple more pulleys). How much weight can it lift? Through what distance does it lift the weight? How much work is actually done?
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 Because you haven't analysed it correctly. Please this is basically a thought experiment. I have placed all the parameters you need. You can tell me where I have gone wrong.
swansont Posted September 11, 2016 Posted September 11, 2016 And that motor requires energy. That's why the conservation laws are intact. That would be the work done to expand the walls, which was calculated. But the ball on the bottom lifts water up above it, and when it rises to the top, the water's mass shifts down. Each diameter the ball rises, the volume it displaces drops in below it If you think it looses any can you give the value please. It's your setup. But there is no doubt what when you drop a ball in water, the water level rises, and you have not accounted for this. When the ball expands the water will rise, too. Probably by the amount of energy you used in the expansion.
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 This expression =ℎ is incomplete- as Swansont said. Imagine that you want to recover the energy from the ball as it floats to the surface by tying a string to it and passing that through a pulley at the bottom of the tank As the ball rises it pulls the string. You can use that to do work- for example- to lift a weight outside teh tank (you'd need a couple more pulleys). How much weight can it lift? Through what distance does it lift the weight? How much work is actually done? As far as I have researched nothing is missing. All the parameters are available someone who thinks its incomplete should help. The problem here is the equations are showing that the ball will float when given that small energy.
swansont Posted September 11, 2016 Posted September 11, 2016 As far as I have researched nothing is missing. All the parameters are available someone who thinks its incomplete should help. The problem here is the equations are showing that the ball will float when given that small energy. To claim that you have included everything is to claim that if you climb into the bathtub, the water level does not rise. Archimedes disagrees with you, and that's the principle you are applying when you analyze the ball sinking and rising. Clearly, you have not analyzed this effect in your work. "As far as you have researched" needs to include the feedback you have gotten here. And you are being told that you have missed some things. Besides, your conclusion is that energy conservation is violated is a big, fat indicator that you have missed something. The equations, wrongly applied, do not lead you to a valid conclusion. 1
Endy0816 Posted September 11, 2016 Posted September 11, 2016 (edited) I suspect the work needed to create and maintain a vacuum is going to be higher as well. Edited September 11, 2016 by Endy0816
swansont Posted September 11, 2016 Posted September 11, 2016 I suspect the work needed to create and maintain a vacuum is going to be higher as well. A vacuum? What?
John Cuthber Posted September 11, 2016 Posted September 11, 2016 (edited) As far as I have researched nothing is missing. All the parameters are available someone who thinks its incomplete should help. The problem here is the equations are showing that the ball will float when given that small energy. I tried to help. I tried to explain that you need to use the right equation rather than plugging numbers into the wrong one. I tried to get you to calculate how much energy you could get out of the system and I got drivel. So, why not do what i suggested? Calculate the buoyancy force on the ball, multiply it by the distance it can rise and find out just how wrong this PE = mgh is. Edited September 11, 2016 by John Cuthber
Endy0816 Posted September 11, 2016 Posted September 11, 2016 (edited) Volume increase of the ball. Same amount of air inside the whole time. Edited September 11, 2016 by Endy0816
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 (edited) To claim that you have included everything is to claim that if you climb into the bathtub, the water level does not rise. Archimedes disagrees with you, and that's the principle you are applying when you analyze the ball sinking and rising. Clearly, you have not analyzed this effect in your work. "As far as you have researched" needs to include the feedback you have gotten here. And you are being told that you have missed some things. Besides, your conclusion is that energy conservation is violated is a big, fat indicator that you have missed something. The equations, wrongly applied, do not lead you to a valid conclusion. Look at the equation again E=hρgV This is actually equal to the gain in the water gravitational potential energy PE=ℎmg since m=ρV It already includes the water gravitational potential your want to add. I tried to help. I tried to explain that you need to use the right equation rather than plugging numbers into the wrong one. I tried to get you to calculate how much energy you could get out of the system and I got drivel. So, why not do what i suggested? Calculate the buoyancy force on the ball, multiply it by the distance it can rise and find out just how wrong this PE = mgh is. The problem here is the density the ball has gained will force it to attain that gravitational potential energy, so PE = mgh is correct. Edited September 11, 2016 by Maximillian
swansont Posted September 11, 2016 Posted September 11, 2016 Look at the equation again E=hρgV This is actually equal to the gain in the water gravitational potential energy PE=ℎmg since m=ρV It already includes the water gravitational potential your want to add. So they cancel. IOW, energy is conserved.
imatfaal Posted September 11, 2016 Posted September 11, 2016 Surely the problem is that you are doing a conservation of energy audit and not returning to the same situation? You start with the ball in the water and then claim you have more potential energy gained than you expended in raising the water column / decreasing the density of the ball. But you should start and end with the ball in the same position. Let us take a start point as the ball on the floor next to the column - that either involves 1. lifting the ball to the top of the column in order to drop it in the top 2. Pushing it in the bottom which means raising the water column by a lot more than the decrease in density causes.
swansont Posted September 11, 2016 Posted September 11, 2016 Volume increase of the ball. Same amount of air inside the whole time. Who said there was air inside? Maybe it's solid. Maybe the expansion is from a chemical reaction, either a solid or foam, or one that creates gas from a solid or liquid.
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 (edited) That would be the work done to expand the walls, which was calculated. But the ball on the bottom lifts water up above it, and when it rises to the top, the water's mass shifts down. Each diameter the ball rises, the volume it displaces drops in below it It's your setup. But there is no doubt what when you drop a ball in water, the water level rises, and you have not accounted for this. When the ball expands the water will rise, too. Probably by the amount of energy you used in the expansion. If I asked you the height an ocean will increase by if I jumped in it. Its negligible. That is my assumption. Edited September 11, 2016 by Maximillian
swansont Posted September 11, 2016 Posted September 11, 2016 If I asked you the height an ocean will increase by if I jumped in it. Its negligible.It's not zero, though. The energy change would be the same. You decrease the height change by increasing the area and raising more mass. Or you can increase the height change by decreasing the area. What you can't do is ignore it in your analysis.
Strange Posted September 11, 2016 Posted September 11, 2016 As far as I have researched nothing is missing. Well, as you get the wrong result, you have obviously missed something. Dismissing the suggestions made by people here won't help you identify your error. (It may not be a single error.)
Maximillian Posted September 11, 2016 Author Posted September 11, 2016 (edited) Surely the problem is that you are doing a conservation of energy audit and not returning to the same situation? You start with the ball in the water and then claim you have more potential energy gained than you expended in raising the water column / decreasing the density of the ball. But you should start and end with the ball in the same position. Let us take a start point as the ball on the floor next to the column - that either involves 1. lifting the ball to the top of the column in order to drop it in the top 2. Pushing it in the bottom which means raising the water column by a lot more than the decrease in density causes. The problem is I had to use 100J of energy to carry the ball and drop it into the tank of water. The ball then sinks to the bottom of the water. Then this time I only use 20J to achieve the 100J I used previously. So they cancel. IOW, energy is conserved. They don't cancel its clear that they have different magnitudes as shown. Edited September 11, 2016 by Maximillian
John Cuthber Posted September 11, 2016 Posted September 11, 2016 The problem here is the density the ball has gained will force it to attain that gravitational potential energy, so PE = mgh is correct. No it is not. Stop ignoring reality. This PE = mgh is wrong. If I asked you the height an ocean will increase by if I jumped in it. Its negligible. That is my assumption. It's small- but you need to multiply it by the surface area of the ocean- which is huge- to get the change you are looking at.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now