Re: [Vo]:zero point query
Everytime the subject of zero point energy comes up, I wanna ask, where's the 'drain', the 'cold side', the 'pressure release'? At the bottom of the ocean, you have tons of pressure per square inch but what good is that? You need an area of reduced pressure to get a flow going. Likewise, hot to cold and uphill to downhill. If zero point energy is squishing everything everywhere, what good is it and furthermore, how can it be measured? Relative to what hole in the cosmic vacuum? I suppose the Casimir force is suggestive (like a bubble underwater relative to surrounding pressure) but how can a continuous action be created?
Re: [Vo]:zero point query
Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Terry On Mon, Dec 14, 2009 at 9:18 AM, Chris Zell chrisrz...@yahoo.com wrote: Everytime the subject of zero point energy comes up, I wanna ask, where's the 'drain', the 'cold side', the 'pressure release'? At the bottom of the ocean, you have tons of pressure per square inch but what good is that? You need an area of reduced pressure to get a flow going. Likewise, hot to cold and uphill to downhill. If zero point energy is squishing everything everywhere, what good is it and furthermore, how can it be measured? Relative to what hole in the cosmic vacuum? I suppose the Casimir force is suggestive (like a bubble underwater relative to surrounding pressure) but how can a continuous action be created?
RE: [Vo]:zero point query
Terry: As most every vortician appreciates, it is not OU if you can only do it one time. Even a few hundred times is not enough if it is say: a magmo that spins down more slowly than expected. Continuing to add ballast at the top eats up all the gain - unless the ballast is itself a phase of water and free . I think the point about the Casimir not having a proper heat-sink for iterative operation is a good one, but it may overlook non-obvious sink arrangements. In fact the proper operation of ZPE may involve a ladder of emission levels and depletion levels, resulting in what is, in effect, step-wise heat sinks; up to where blackbody radiation (and ambient heat or cold), can be employed at the top layer. I posted on a hypothetical ZPE heat sink years ago but cannot find it in the archives. My recollection is that the important dynamic is semi-coherence (superradiance). Instead I found this, which is pretty cool in its own way - i.e. the implications for finding hidden heat-sinks: http://www.mail-archive.com/vortex-l@eskimo.com/msg07965.html From: Terry Blanton Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Terry Chris Zell wrote: Every time the subject of zero point energy comes up, I wanna ask, where's the 'drain', the 'cold side', the 'pressure release'? At the bottom of the ocean, you have tons of pressure per square inch but what good is that? You need an area of reduced pressure to get a flow going. Likewise, hot to cold and uphill to downhill. If zero point energy is squishing everything everywhere, what good is it and furthermore, how can it be measured? Relative to what hole in the cosmic vacuum? I suppose the Casimir force is suggestive (like a bubble underwater relative to surrounding pressure) but how can a continuous action be created?
Re: [Vo]:zero point query
On 12/14/2009 11:11 AM, Terry Blanton wrote: Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? To start the cycle over again, you need to get the rock back to the surface. If you assume an infinite supply of rocks, so you don't need to do that, then you've already got an unlimited energy source available (at least until the ocean fills up with rocks). Terry On Mon, Dec 14, 2009 at 9:18 AM, Chris Zell chrisrz...@yahoo.com mailto:chrisrz...@yahoo.com wrote: Everytime the subject of zero point energy comes up, I wanna ask, where's the 'drain', the 'cold side', the 'pressure release'? At the bottom of the ocean, you have tons of pressure per square inch but what good is that? You need an area of reduced pressure to get a flow going. Likewise, hot to cold and uphill to downhill. If zero point energy is squishing everything everywhere, what good is it and furthermore, how can it be measured? Relative to what hole in the cosmic vacuum? I suppose the Casimir force is suggestive (like a bubble underwater relative to surrounding pressure) but how can a continuous action be created?
Re: [Vo]:zero point query
On 12/14/2009 11:11 AM, Terry Blanton wrote: Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Well, hrph ... that seems to be harder to answer than I expected, after a page of scribbles failed to get me anywhere. Work done on the accordion by the water, in squishing the air in it, is proportional to the log of the ratio of the pressure at the surface to the pressure at the bottom. (But that's just half the problem, of course.) Work done on the (rock+accordion) by (gravity-buoyancy) is related to depth of the ocean and not a lot else; density of the water enters in there somewhere but seems to get messy. Simplest approach is probably to assume neutral buoyancy for the system at the surface; buoyancy decreases as it sinks, of course, as the accordion is squished. If it's not OU then those two quantities of work had better be provably equal and right now it looks like trying to prove apples equal oranges. To make it more interesting, you can replace the rock with a bag of sand, and let it slide down a slope underwater, and then the sand can be left behind little by little as it sinks, so that the accordion+sandbag are neutrally buoyant all the way down. Then the net work done by (gravity-buoyancy) on the system must be zero, but the accordion has still gained energy. Erk? Where'd it come from? Anyhow gotta go do some day-job work, sad to say, this puzzle is more interesting...
Re: [Vo]:zero point query
On Dec 14, 2009, at 7:11 AM, Terry Blanton wrote: Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Terry Take a look at my deep well or ocean bubble lift idea in Fig. 3 at: http://mtaonline.net/~hheffner/SLVN.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:zero point query
On 12/14/2009 12:45 PM, Stephen A. Lawrence wrote: On 12/14/2009 11:11 AM, Terry Blanton wrote: Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Well, hrph ... that seems to be harder to answer than I expected, after a page of scribbles failed to get me anywhere. OK, here we go (and if my company goes bust as a result of the time I wasted on this, oh well): The rock is a red herring, because work done to sink it and lift it balances out. Ditto the accordion. All we need to worry about is the gas. Imagine that we put the gas in a cylinder with a piston, area(piston)=A, volume of cylinder = A*L where L is the distance between the piston and the back wall. The mass of the cylinder etc is neglectable also because it's the same going down and up; as I said, it's just the gas we need to worry about. The pressure in the cylinder -- and the surrounding water, of course -- is 'P'. Now imagine that we *drag* the cylinder down to the bottom, and then lift it back up. Imagine further that the density of the gas EQUALS the density of the water AT THE BOTTOM. Then we do work all the way down, but it's neutrally buoyant all the way up. So, all we need to worry about is the work to get it to the bottom. In fact we don't need to worry about taking integrals or finding total work; all we need to worry about is the work needed to drag the cylinder *a* *little* *bit* farther down, and the amount of additional energy which will be packed into it by that incremental drag due to additional compression of the gas. If that incremental bit of work balances, then the total work will balance also. Call depth in the water 'h'. Then work done by the water in compressing the gas a little more as it's dragged down a little deeper will be: dW(pressure) = P*A*dL that is, the pressure in the cylinder, times the area of the piston, times the amount the piston moves (ignore sign issues, we can always patch them up later if we feel moved to do so). A little fiddling reduces this to: dW(pressure) = -(nRT/P) * rho * dh where 'rho' is density of the water, and PV = nRT is the ideal gas law. I'm not showing the intermediate steps; if anyone wants to see them I can post them later. Now let's look at the work done to drag it dh deeper in the water. That's the buoyancy, times the distance dragged: dW(buoyancy) = (rho * V - k*n) * dh That is to say, the buoyancy force is equal to the density of the water times the displacement (that's rho*V), *minus* the weight of the gas, which will be some constant, 'k', times the number of moles of gas, 'n'. However, as it happens, we're going to drag the gas back to the surface again, so just like the weight of the rock and the weight of the cylinder, the weight of the gas is neglectable -- work done on it going down and back up will cancel. So, the part we're interested in reduces to dW(buoyancy) = rho * V * dh but we also have V = nRT/P so dW(buoyancy) = rho * (nRT/P) * dh and that's equal to the incremental work we found was done in compressing the gas in the cylinder. So, it appears that the energy gained by squeezing the gas is exactly equal to the energy it took to drag it to the bottom of the ocean. * * * Oh -- one other item -- I assumed the temperature was constant. From the POV of thermodynamics, at least, that's legit, as we can treat the ocean as a single heat source/sink which absorbs or provides heat as needed to keep everything at one temperature.
Re: [Vo]:zero point query
Oh, dear. I intuitively believed what you have proved. Hope the company is okay still. Terry On Mon, Dec 14, 2009 at 2:47 PM, Stephen A. Lawrence sa...@pobox.com wrote: (at length)
Re: [Vo]:zero point query
Electrolyte included! Terry On Mon, Dec 14, 2009 at 12:56 PM, Horace Heffner hheff...@mtaonline.net wrote: On Dec 14, 2009, at 7:11 AM, Terry Blanton wrote: Hmmm, suppose you have an accordion device that you drop into the ocean with a rock. The device pressurizes air as it sinks and locks into place, then you drop the ballast and it floats to the top. You use the pressurized air to do work. Is possible? If so, is it OU? Terry Take a look at my deep well or ocean bubble lift idea in Fig. 3 at: http://mtaonline.net/~hheffner/SLVN.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:zero point query
Whoa, speaking of strange coincidence ... and ... electrolytes ... and ... err ... PEE Power ;-) ya gotta follow some of this thread on YouTube http://www.youtube.com/watch?v=ITerHqq6XPU -Original Message- From: Terry Blanton Electrolyte included! Terry Horace Heffner wrote: Take a look at my deep well or ocean bubble lift idea in Fig. 3 at: http://mtaonline.net/~hheffner/SLVN.pdf
Re: [Vo]:zero point query
On Dec 14, 2009, at 8:56 AM, Horace Heffner wrote: The following is in regards to the bubble lift idea at: http://mtaonline.net/~hheffner/SLVN.pdf The density of water is 1000 kg/m^3. The pressure of 1 atmosphere is about 10,000 kgf/m^2. At a depth of x meters, the weight of water above a square meter is equal to a column of water x meters high, or x * 1000 kgf/m^3. The pressure of water at depth x is given by: P(x) = (1 atm) + (1000 kgf/m^2) *x = (1 atm) * (1000 kgf/m^2)* [(1 atm)/(10,000 kgf/m^2)] * x P(x) = (1 atm) + (0.1 atm)/m * x So, the volume V(x) of a bubble at depth x, having risen from being at volume V0 at depth D is: V(x,D) = V0 * P(x)/P(D) = (V0/P(D)) * P(x) and the work done per cubic meter of water an increment dx is g* ((1000 kg)/m^3)*dx so the work dW(x,D) by V(x,D) gas moving up an increment dx is: dW(x,D) = V(x,D) * [g*(1000 kg/m^3) * dx] dW(x,D) = [(V0/P(x)) * P(D)] * [g*(1000 kg/m^3) * dx] but with D constant: dW(x) = (V0 * P(D) * g * (1000 kg/m^3)) * 1/((1 (atm) + (0.1 atm/ m) x) dx Let k = (V0 P(d) * g * (1000 kg/m^3) / (1 atm)) so: dW(x) = k * ((1 + (0.1/m) x) dx To obtain total work W(x) integrating dW(x) from x=-D to 0 we have: W(x) = [integral x=D to 0] k * 1/(1 + (0.1/m) * x) dx W(x) = [eval x=D to 0 m] k * ln(|(0.1/m) * x + 1|)* (10 m) W(x) = k [ln(|(0.1/m) * D + 1|)* (10 m)] - k [ln(|(0.1/m) * 0 + 1|)* (10 m)] W(x) = k * (10m) * ln(|(0.1/m) * D + 1|) Example: for v0 = 1 m^3 and D = 2000 m we have: P(D) = (1 atm) + 0.1 atm/m * (2000 m) = 201 atm k * (10 m) = (V0 * g * (1000 kg/m^3) * (P(D) / (1 atm)))* (10 m) k = ((1 m^3) * g *(1000 kg/m^3) * ((201 atm) / (1 atm))) * (10 m) k = 1.9711x10^7 J W(2000 m) = (1.9711x10^7 J) * ln(201) = 1,045x10^8 J = 29 kWh A commercial electrolyzer produces hydrogen using 4.8 kWh per cubic meter (Peavy, Fuel from Water, p. 22). However, we can include an equal volume of O2, so that is 2.3 kWh per cubic meter at the surface. However, at depth, we need P(D)/(1 atm) as much gas, and this takes V0*(2.3 kWh/m^3)*P(D)/(1 atm) energy, so at 2000 m depth, that is 201 * 2.3 kWh = 462 kWh, even if the H2 expansion energy is free. However, this can be offset by recombining the H2 and O2 at the surface to drive the electrolysis. If we don't want to recover the chemical energy and use: COP = [V0*(2.3 kWh/m^3)*P(D)/(1 atm)] / [V0 * g * (1000 kg/m^3) * (P(D) / (1 atm)))* (10 m) * ln (|(0.1/m) * D + 1|) Then COP = (2.3 kWh)/[g*(1000 kg)*10m*ln(|(0.1/m) * D + 1|)] and if we want COP of 1 then: (1000 kg/m^3)*10m*ln(|(0.1/m) * D + 1|) = (2.3 kWh) ln(|(0.1/m) * D + 1|) = (2.3 kWh) / [g*(1000 kg)*10m] = (2.3 kWh)/0.0272 kWh) = 84.4 (0.1/m D * + 1 = exp(84.1) = 4.5x10^26 thus D = 4.5 x 10^27 m. Pretty deep ocean! This is not a very practical idea! Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
