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   Perpetual Motion (Basic Chemistry)
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James Fingas
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Perpetual Motion (Basic Chemistry)  
« on: Mar 14th, 2003, 1:42pm »
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Perpetual Motion Machine I
 
Construct two pipes, each 12.4 km high, but with one 1m taller than the other. Join these pipes at the bottom through a semi-permeable membrane. Fill the shorter side with salt water (salinity 3.5%), and the taller side with fresh water.
 
The osmotic head between salt water at 3.5% salinity and fresh water is 2.8 MPA. Salt water at that salinity is also 1.023 times as dense as fresh water. From this difference in density, and knowing that the weight of water is 9.807 kPa/m, the pressure difference at the bottom of the apparatus exceeds 2.8 MPA, and therefore water will migrate from the salt water side to the fresh water side. This will cause the taller pipe to overflow. Capture this overflow and direct it so that it turns a water wheel, then flows into the shorter pipe. Now we have a perpetual motion machine, and an endless supply of energy!
 
Perpetual Motion Machine II
 
Construct a 300m-high closed container. Add a 290m-high internal wall, dividing it in two. In the bottom of the wall, install a port connecting the two sections, fitted with a semipermeable membrane. Fill one side with salt water, salinity 3.5%.
 
Because the static water pressure is greater than 2.8 MPa, water will migrate through the membrane and start to fill the other side. During this process, evaporation from the surface of the salt water will drive the equilibrium humidity towards 98%. However, evaporation from the collecting pool of fresh water will drive the humidity towards 100%. This difference in vapour pressure will cause water to migrate through the air from the fresh water side to the salt water side, creating a perpetual motion machine. We can get energy out of this machine by installing a thermocouple pair connecting the surface of the salt water (net condensation causes a temperature rise) with the surface of the fresh water (net evaporation causes a temperature drop).
 
Would these perpetual motion machines work as described? If so, what about the second law of thermodynamics? If not, why not, and what would actually happen?
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Re: Perpetual Motion (Basic Chemistry)  
« Reply #1 on: Mar 16th, 2003, 7:37am »
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In the first case it would at the start appear to work, but the energy will have to come from somewhere, so either the water cools down or the air around it cools down. And once the water freezes the 'perpetual' motion machine wouldn't seem so perpetually moving anymore..
Pretty much the same in the second caes I'm sure..
 
Of course if it's feasable to build such a device they could be quite usefull, since cooling down the environment wouldn't be a problem at the moment (you might even stop global warming Wink
« Last Edit: Mar 16th, 2003, 7:38am by towr » IP Logged

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Re: Perpetual Motion (Basic Chemistry)  
« Reply #2 on: Mar 16th, 2003, 2:55pm »
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The Kelvin-Plank statment of the 2nd law of thermodyamics says it is impossible to have a device operate in a cycle which has no effect other than the output of work and exchange of heat with a single reservoir. So I disagree with towr.
 
I think Machine 1 would produce work at first and then slow down and stop as the column of salt solution reaches equilibrium. A 12 km high column of uniformly mixed salt solution is not in equilibrium. The free energy of the solution would be minimized with the salt concentration lowest at the top. For small, dilute containers of salt solution, the gravitational potential energy doesn't normally overcome the entropy increase you get from having the solute well distributed. For a column 12 km high I think it would not, and a standing solution would settle to have a concentration gradient. The operation of this device (with fresh water coming in at the top, and being pulled out at the bottom) allows the concentration gradient to be reached even more quickly. Eventually the concentration at the bottom would be high enough that osmotic pressure would match the additional pressure on the salt side.
 
Machine 2 will eventually reach equilibrium. There will be no more net flow through the membrane because of a balance between osmotic pressure and pressure difference due to height and density difference of the two sides. Same for a balance between evaporation/condensation from the surfaces. In the salt solution there will be a salinity gradient like in Machine 1. In the air, equilibrium will be reached because of a humidity gradient. Humidity will be greater near the pure water because it enters the air here, and also this surface because it is lower than the surface of the salt solution.
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Re: Perpetual Motion (Basic Chemistry)  
« Reply #3 on: Mar 20th, 2003, 10:00am »
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SWF,
 
I think you've figured out device #1, but I'm not sure you've nailed down device #2 specifically enough. Think about this:
 
1) The column of water isn't high enough to have a significant salinity gradient.
 
2) It doesn't matter where the water vapour enters the air, because it will disperse in the steady state.
 
3) Water vapour is lighter than air, so there's a higher percentage of water vapour positive as you go up (in steady state).
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Re: Perpetual Motion (Basic Chemistry)  
« Reply #4 on: Oct 8th, 2003, 12:51pm »
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This machine that works by osmosis, looks like it may work.
 
However, correct me if I am wrong, during osmosis areas of high concentration of salt move to areas of low concentration.
 
Salt --> Fresh
 
however after a while the cell or in this case the containers reach = librium.  My comment is wouldn't the salt levels begin to equalize and reduce the propetual motion in this machine? Undecided  Thanks for the fun problem!
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Re: Perpetual Motion (Basic Chemistry)  
« Reply #5 on: Oct 8th, 2003, 1:17pm »
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on Oct 8th, 2003, 12:51pm, Todd, j wrote:
This machine that works by osmosis, looks like it may work.
 
However, correct me if I am wrong, during osmosis areas of high concentration of salt move to areas of low concentration.
 
Salt --> Fresh
 
however after a while the cell or in this case the containers reach = librium.  My comment is wouldn't the salt levels begin to equalize and reduce the propetual motion in this machine? Undecided  Thanks for the fun problem!

 
Both machines work by the process of "reverse osmosis". Exactly like you describe, osmosis is the process by which a chemical spreads out from areas of high concentration to areas of lower concentration. Reverse osmosis is the opposite: you force chemicals to move from an area of lower concetration to an area of higher concentration. To do this, you need to apply fairly high pressure on the lower concentration side.
 
You can calculate all of this by considering what's called "osmotic pressure". Given any two concentrations of a chemical, you can calculate the "osmotic pressure" that you would have to apply to keep osmosis from happening at all. Apply less pressure, and osmosis happens. Apply more pressure, and "reverse osmosis" will occur.
 
One way of explaining how reverse osmosis happens is to consider the "energy of mixing". Whenever you mix two chemicals A and B you are releasing energy (usually a very small amount--but not if you're mixing water in with concentrated sulphuric acid!). In order to get chemicals to unmix, you have to put that energy back in. Normally, there's no way to do this, but if you happen to have a semipermeable membrane that lets chemical A through but not chemical B, then you just apply pressure on the mixed side, so that the volume of chemical A that goes through multiplied by the pressure across the membrane is equal to the mixing energy of that volume of chemical A with chemical B.
 
So to answer your question, the semi-permeable membrane lets water through but not salt, so the concentrations on the two sides don't go towards equilibrium.
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