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The "Ice Bomb" thermal engine


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1 hour ago, Tom Booth said:

Here, I would have loved to answer Swansont's last question on the thread but got locked out before I could answer.

?

You’re answering now. You haven’t been locked out.

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Apparently because I posted a video that supported my statement that a Stirling engine is functionally, exactly the same as  a Stirling cryo-cooler:

There was no mention of Stirling engines or coolers in this thread before this, so I don’t know how this pertains to the thread. If you think your proposal isn’t slow and inefficient, you should be able to make your case without bringing up your closed thread.

(And no, that’s not an accurate summary of why the thread was closed. The subject matter had nothing to do with it)

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What? It's friggin' common knowledge except maybe for (some) in the industry who try to guard it like a trade secret. Which to some degree it (sort of) is. But do you read: "standard in the natur

The expansion on freezing is only about 9% volume, so the work done in expansion when the pressure is released is not that much - enough to bust the container but not much more. There is very little s

I think we should leave urinary infections out of this. 😆

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3 minutes ago, swansont said:

?

You’re answering now. You haven’t been locked out.

There was no mention of Stirling engines or coolers in this thread before this, so I don’t know how this pertains to the thread. If you think your proposal isn’t slow and inefficient, you should be able to make your case without bringing up your closed thread.

(And no, that’s not an accurate summary of why the thread was closed. The subject matter had nothing to do with it)

The thread was locked, and is still locked.

No specific reason was given so I can only speculate, but posting that video seemed to be the immediate issue, or my argument that it was certainly Germain to the conversation.

It supported my argument from what should, I think, have been considered a reliable and authoritative source on the subject.

You  may not know how these threads relate to one another, but they most certainly do.

 

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10 minutes ago, Tom Booth said:

The thread was locked, and is still locked.

We’re in this thread, not that one. I made my comment earlier today, so a thread locked 9 months ago cannot be an issue.

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No specific reason was given so I can only speculate

You need to reread the modnotes and perhaps hone your comprehension skills.

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You  may not know how these threads relate to one another, but they most certainly do.

So you have re-opened a closed thread, and also wish to use speculation to support your idea? 

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Posted (edited)

Sorry, my mistake. The video about Stirling cryo-coolers was another thread that was locked and moved to "speculations".

 

1 hour ago, swansont said:

We’re in this thread, not that one. I made my comment earlier today, so a thread locked 9 months ago cannot be an issue.

You need to reread the modnotes and perhaps hone your comprehension skills.

So you have re-opened a closed thread, and also wish to use speculation to support your idea? 

How is the video "speculation"? It was produced by Phillips Cryogenics, a company that builds the things. I should think they would know.

 

1 hour ago, swansont said:

We’re in this thread, not that one. I made my comment earlier today,...

I was simply saying the thread was locked. The final post was yours and asked me a specific question.

It would have been nice to have had the opportunity to answer you. That is all.

 

This was the context of the question. I'm not trying to reopen the thread, just FYI. I don't want to be rude by not responding, or your thinking I didn't answer.

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But, heat cannot be "rejected" at a temperature equivalent to the sink.

Why?

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Setting that fact aside for the moment we have the fact that my engine only needs to have a Carnot efficiency of 18.9% (not 100%) to no longer have a condition where heat is being "rejected".

No...

If no heat were rejected, your efficiency would be 100%

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At 18.9% efficiency heat is no longer being rejected. Temperatures are equivalent, so no "flow" can result.

You need to explain why you think this is so. IOW, explain why your ice melts, if no heat is flowing

 

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An efficiency of a mere 19% would have heat flowing backward from the ambient sink back into the engine.

Therefore, to say the engine must reject heat because no engine is 100% efficient is apples to oranges.

It is not necessary to have 100% efficiency on an absolute scale, or bring the temperature down to absolute zero, rejection to the sink ceases at any efficiency >= 18.9%

Expand  

Again, you need to explain why you think this is so

Actual thermodynamics doesn’t work this way.

 

Oh, in that other thread I was told

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"Do not start another thread on this subject"

But, a lot of ground was covered, and exactly what "this subject" was specifically, who knows.

My experiments? Thermodynamics generally?

Anyway, I stayed in the corner for nearly a year.

It seems obvious however, the issue was with the topic itself, apparently.

What specifically? Who knows.

Anyway,

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But, heat cannot be "rejected" at a temperature equivalent to the sink.

 

Why ?

 

 

Because in a heat engine, heat is "rejected" to the sink through a metal plate, a heat exchanger.

Heat only flows spontaneously from hot to cold.

If the gas in the engine has reached the same temperature as the sink, (the ambient air outside the engine) how is any heat transfered to the sink?

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At 18.9% efficiency heat is no longer being rejected. Temperatures are equivalent, so no "flow" can result.

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You need to explain why you think this is so. IOW, explain why your ice melts, if no heat is flowing

There is a little confusion between different experiments, 1) heat engine running on hot water 2) heat engine running on ice (actually ambient heat)

Regardless, If heat flows into the engine to be  converted into mechanical energy bringing the temperature of the gas within the engine down to the same temperature as the ice (the sink), no matter how well insulated, given the limitations of my little home laboratory in my kitchen, the ice, OUTSIDE OF THE ENGINE, is still subject to some heat infiltration from the environment, and so will eventually melt, even if no heat is reaching the ice through the engine itself.

As insulated as I could get it, the engine (running on ice) ran for 33 hours continuously on a single coffee cup size mug full of ice. ("Running on ice" being a misnomer of course. The engine, properly speaking, was running on ambient heat)

Obviously the ice, serving as the "sink" could not be isolated from the heat of the environment 100% even if the engine was actually cooling it slightly. Granted, this may not be so obvious to anyone not as familiar with Sterling engines.

But 33 hours is a very long time. Typically the engine would run no more than a few hours on a cup of boiling water. even a similarly very well insulated cup of boiling water.

The heat from the boiling water enters the engine and is converted into work and eventually all the heat is consumed, or used up.

Running on AMBIENT HEAT, the heat enters the engine and is converted into work. But is it possible for the ambient heat to get used up or fully consumed by the engine like the finite source of heat represented by the cup of hot water?

Not as long as the sun is shining.

A control using the same set up with the engine just sitting on a cup of ice, not running, resulted in the ice melting in 27 hours.

The engine apparently kept the ice cool and prevented it from melting for about 5 hours by converting the incoming  heat into work.

This is similar to the "Ice Bomb" engine converting heat into work in the freezer, so the freezer has to run less often to keep the ice cool. Though the Ice Bomb example is only a thought experiment, it is intended to illustrate the same principle as the actual Stirling Engine experiments.

I'm sorry if this may be perceived as breaking the rules here by resurrecting a closed topic, but I think this is important and I believe it worthwhile to put this out to whomever might listen, especially the scientific community.

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Posted (edited)
10 hours ago, Tom Booth said:

                ...[snip]....

That's open for debate I guess.

I don't see it as quite normal because positive work is carried out during the portion of the cycle when heat leaves the system.

That is the opposite of a conventional heat engine that produces positive work while heat is being added.

I think, though I'm not entirely certain, that this reversal results in some kind of corruption of the accepted formula for calculating heat engine efficiency, though I haven't actually worked that out mathematically.

Earlier Swansont made a statement which I latched onto as confirmation that the usual formula breaks down or doesn't apply, because the way water expands when frozen is a deviation from the expected norm.

What I mean by normal is that the principle of the Carnot cycle still applies. You have heat flowing from higher temperature T1 to a lower temperature T2, and some of it gets converted to mechanical work.

The Carnot cycle says nothing about the means you choose to use to make the conversion. It just lays down the maximum efficiency achievable in theory, over any heat engine cycle. There is no "corruption" of anything that I can see, nor any reason to suppose that thermodynamics ceases to apply, just because we are dealing with a substance that expands on cooling.  

I can see, though, that it could be interesting, in an academic sort of way, to work out exactly how the efficiency limitations manifest themselves, in this unusual form of heat engine. 

Edited by exchemist
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11 hours ago, Tom Booth said:

How is the video "speculation"?

Nobody said it was.

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What specifically? Who knows.

Strange was pretty specific. 

But this is all off-topic.

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8 hours ago, exchemist said:

What I mean by normal is that the principle of the Carnot cycle still applies. You have heat flowing from higher temperature T1 to a lower temperature T2, and some of it gets converted to mechanical work.

I'm trying to work this out on paper, to figure out what might actually happen through the cycle where the "ice engine" in (and out of) the freezer has a ratchet, so does work in both directions.

First heat is removed, the ice swells and lifts or pushes whatever mechanism doing work. Pretty straightforward.

Where I'm getting hung up is when the machine is removed from the freezer and the ice melts and returns to a liquid. Atmospheric pressure contributes to the reduction in volume, which also winds the ratchet doing additional work.

The temperature of the ice, doesn't actually change during this phase, right?

The volume is being reduced as the ice melts, but work is also being performed, so...

The ice is being worked on by the atmosphere? Or the ice itself is doing work to wind the ratchet?

The heat taken in is converted to work? Or not?

Is it possible to draw up some sort of PV or Entropy diagram of this?

At this point the heat/work/pressure starts getting muddled up and I'm not quite able to make sense of it.

I question wether or not the heat that melts the ice actually ever does any external work.

True the ice takes in heat which is used to rearrange it's molecular structure, but that is internal energy storage not work.

When heat is taken away and the ice is formed, what does the work exactly?

Maybe it could be thought of as, heat entering the liquid H2O is like weakening a spring, allowing it to be compressed. Then when the heat is taken away, the spring gets back it's strength and expands.

Take heat away and that releases the spring. But where exactly is the energy for the spring to do work come from when it gets back it's strength due to the oppressive heat having been removed.

In other words, where does the spring strength actually come from.

Not the heat, because the heat is what took away it's strength.

Now I'm thinking about Nitinol. Does H2O have a kind of "shape memory".

Some types of memory metal can be trained to work both ways. That is, it takes on a shape, both when heated and when cooled, so can be made to do work to run a heat engine either way or both. 

After all that rambling, I'd say, tentatively, I guess, maybe. As the ice is melting. Sort of, but then again, would the volume of the H2O be reduced as it melts, without atmospheric pressure?

Can we set this up in a vacuum?

And if the temperature is constant, as the H2O changes state from ice to water, is the heat really doing any work, or is it all used to just "weaken" the structure of the ice so it collapses under the weight of atmospheric pressure, so atmospheric pressure is actually doing all the external "useful work" not the heat.

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40 minutes ago, Tom Booth said:

I'm trying to work this out on paper, to figure out what might actually happen through the cycle where the "ice engine" in (and out of) the freezer has a ratchet, so does work in both directions.

First heat is removed, the ice swells and lifts or pushes whatever mechanism doing work. Pretty straightforward.

Where I'm getting hung up is when the machine is removed from the freezer and the ice melts and returns to a liquid. Atmospheric pressure contributes to the reduction in volume, which also winds the ratchet doing additional work.

The temperature of the ice, doesn't actually change during this phase, right?

The volume is being reduced as the ice melts, but work is also being performed, so...

The ice is being worked on by the atmosphere? Or the ice itself is doing work to wind the ratchet?

The heat taken in is converted to work? Or not?

Is it possible to draw up some sort of PV or Entropy diagram of this?

At this point the heat/work/pressure starts getting muddled up and I'm not quite able to make sense of it.

I question wether or not the heat that melts the ice actually ever does any external work.

True the ice takes in heat which is used to rearrange it's molecular structure, but that is internal energy storage not work.

When heat is taken away and the ice is formed, what does the work exactly?

Maybe it could be thought of as, heat entering the liquid H2O is like weakening a spring, allowing it to be compressed. Then when the heat is taken away, the spring gets back it's strength and expands.

Take heat away and that releases the spring. But where exactly is the energy for the spring to do work come from when it gets back it's strength due to the oppressive heat having been removed.

In other words, where does the spring strength actually come from.

Not the heat, because the heat is what took away it's strength.

Now I'm thinking about Nitinol. Does H2O have a kind of "shape memory".

Some types of memory metal can be trained to work both ways. That is, it takes on a shape, both when heated and when cooled, so can be made to do work to run a heat engine either way or both. 

After all that rambling, I'd say, tentatively, I guess, maybe. As the ice is melting. Sort of, but then again, would the volume of the H2O be reduced as it melts, without atmospheric pressure?

Can we set this up in a vacuum?

And if the temperature is constant, as the H2O changes state from ice to water, is the heat really doing any work, or is it all used to just "weaken" the structure of the ice so it collapses under the weight of atmospheric pressure, so atmospheric pressure is actually doing all the external "useful work" not the heat.

No I think you re complicating this unnecessarily. The way I envisage this, for simplicity, is that the ratchet acts to stop the load pressing on the ice the moment it starts to melt, i.e. it engages immediately to support the load. So NO work is done on it by the load during what you might call the "return stroke" of the engine.  So ignore the ratchet: its function in this scenario is to make things simple.  

Regarding the atmosphere, yes, a tiny bit of work is done by the atmosphere when the ice melts, sure. But we can forget that too, since it is (i) very small and (ii) an equal and opposite amount of work is done against it during the expansion "power stroke". The ice has not only to push up the load, but also push back the atmosphere too, right?  So it makes no net difference to anything. So again, forget it, to make things simple.

Also ignore taking machines in and out of freezers and all that jazz. You have a warm reservoir at T1 and a cold one at T2 and you put the engine in contact with each alternately: it does not matter how.  

Forget the engineering. It does not matter a toss and it is preventing you seeing the essence of the scenario clearly. The point is that, stripped to its essentials, you have a standard heat engine, to which the Carnot cycle principle can be applied, as it can to any other heat engine. 

As to what does the work during the expansion "power stroke", it is chemical energy. As ice crystallises, chemical bonds are formed as the molecules move to a lower energy state. This releases energy, which emerges in the form of both heat and in any work that has to be done to push the molecules apart, as they start to take up the more open but bonded, rigid structure they have in solid ice. 

During the return stroke, heat has to be added to give the molecules the thermal kinetic energy to break those bonds again.

 

 

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Posted (edited)
8 hours ago, exchemist said:

(...)

Regarding the atmosphere, yes, a tiny bit of work is done by the atmosphere when the ice melts, sure. But we can forget that too, since it is (i) very small and (ii) an equal and opposite amount of work is done against it during the expansion "power stroke". The ice has not only to push up the load, but also push back the atmosphere too, right?  So it makes no net difference to anything. So again, forget it, to make things simple.

(...)

As to what does the work during the expansion "power stroke", it is chemical energy. As ice crystallises, chemical bonds are formed as the molecules move to a lower energy state. This releases energy, which emerges in the form of both heat and in any work that has to be done to push the molecules apart, as they start to take up the more open but bonded, rigid structure they have in solid ice. 

During the return stroke, heat has to be added to give the molecules the thermal kinetic energy to break those bonds again.

 

I'm hesitant to discard atmospheric pressure as a factor. Equal and opposite or not.

During expansion as ice freezes, there is no temperature change. So heat/energy is involved in the transformation, but solids and liquids have forms of energy to draw on other than heat.

When gas is liquefied through an expansion turbine the heat previously removed by compression cannot return. Nor can ambient heat return, as the turbine housing is kept cold and insulated, but the gas does work powering the load on the turbine so looses energy and liquefies. Where did the energy to do the work to turn the turbine come from? Molecular "spin" or something?

The point is, NOT external heat/energy. The gas is forced to do work in thermal isolation.

And no one has answered the question about, can we thaw out the ice in a vacuum?

We are all aware, I should think, that at typical ambient temperatures, water, subject to a vacuum freezes.

So, after some thought, I'm hazarding a guess that no, we can't.

So, whats left, the weight of whatever was lifted, other than atmosphere, pressure pressing down, if sufficiently strong, might provide energy to produce the transformation, but that is already at a balance, practically speaking, we don't want to add more weight, so we are back to atmospheric pressure.

No matter how long we leave the ice in ambient heat, while in a vacuum, it will not utilize the heat for effecting the transformation from solid to liquid. So that atmospheric pressure, I think, is not insignificant.

Remove the atmospheric pressure and the ice cannot melt. Bring it back, and the ice melts.

A predictable result, if atmospheric pressure is the source of energy effecting the transformation, along with the weight that was lifted.

So rather than the atmospheric pressure being insignificant, on the contrary, it can be demonstrated experimentally that the ambient heat is a relatively insignificant part of the process.

The work done by the atmosphere is not "tiny". Isolated as a variable. Vacuum vs. no vacuum. Will melt, won't melt.

If our freezer box size engine contained about a cup of water, to make calculation easy say cube 3 inches on each side, at 15psi atmospheric pressure on all sides could account for as much as roughly 800+ lbs.

The weight of say four people up to 200 lb each.

That many people could hardly fit in my ice box.

Edited by Tom Booth
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Posted (edited)

This guy is lifting a mere 500 ponds. We need to add another 300 lbs.

I don't think even those relatively compact weights could fit.

The freezer could accommodate maybe at most, a 50 lb block of cast iron or something. Unless I get a bigger freezer. An ice chest model maybe, but my kitchen refrigerator ice box is not big enough to hold anywhere near the equivalent weight of the 1 ATM on a cup of water.

Insignificant?

After watching several "Ice Bomb" videos's I'm begining to question something else.

One would expect, I think, to find a ball of ice amongst the cast iron fragments let behind by the ice bomb.

Like a big hailstone or something, or at least fragments.

I did some more research and found this: https://labdemos.physics.sunysb.edu/i.-thermodynamics/i4.-changes-of-state/ice_bomb

The text reads: 

 

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A short cast iron nipple with caps on each end is filled with water and dropped into a container of liquid nitrogen. Within about one minute the water freezes, expanding sufficiently to break the cast iron pipe with a loud crack and a big cloud of vapor.

A big cloud of vapor?

In the video's no ice remnants are ever displayed. So I'm wondering if the dramatic results are due to ice formation at all.

Maybe, instead, due to the build up of pressure the water never freezes but instead reaches the triple point and sublimates directly into vapor, which, presumably, would have an even greater expansion force than ice.

I don't think that would necessarily nullify the freezer box thought experiment, but, it certainly changes things a bit.

The big "ice bomb" explosion may have nothing to do with the formation or great power of ice at all.

Edited by Tom Booth
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Posted (edited)

I'd like to do the ice bomb experiment using a clear high pressure acrylic chamber instead of cast iron, and video tape the "explosion" with an ultra high speed camera. See what's REALLY going on in there.

Maybe instead of an explosion it is actually a kind of implosion, and all the destruction is just due to concussion waves or who knows.

I'm a little puzzled that often in these experiments ALL the cast iron fragments are just there, easy to find, all in one place, not scattered all over. Maybe implosion, initially, followed by explosion.

At any rate extreme cold also makes cast iron very brittle and much easier to break, so how great really is this apparent explosive force of ice, if it is even really ice at all?

Edited by Tom Booth
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4 hours ago, Tom Booth said:

I'm hesitant to discard atmospheric pressure as a factor. Equal and opposite or not.

During expansion as ice freezes, there is no temperature change. So heat/energy is involved in the transformation, but solids and liquids have forms of energy to draw on other than heat.

When gas is liquefied through an expansion turbine the heat previously removed by compression cannot return. Nor can ambient heat return, as the turbine housing is kept cold and insulated, but the gas does work powering the load on the turbine so looses energy and liquefies. Where did the energy to do the work to turn the turbine come from? Molecular "spin" or something?

The point is, NOT external heat/energy. The gas is forced to do work in thermal isolation.

And no one has answered the question about, can we thaw out the ice in a vacuum?

We are all aware, I should think, that at typical ambient temperatures, water, subject to a vacuum freezes.

So, after some thought, I'm hazarding a guess that no, we can't.

So, whats left, the weight of whatever was lifted, other than atmosphere, pressure pressing down, if sufficiently strong, might provide energy to produce the transformation, but that is already at a balance, practically speaking, we don't want to add more weight, so we are back to atmospheric pressure.

No matter how long we leave the ice in ambient heat, while in a vacuum, it will not utilize the heat for effecting the transformation from solid to liquid. So that atmospheric pressure, I think, is not insignificant.

Remove the atmospheric pressure and the ice cannot melt. Bring it back, and the ice melts.

A predictable result, if atmospheric pressure is the source of energy effecting the transformation, along with the weight that was lifted.

So rather than the atmospheric pressure being insignificant, on the contrary, it can be demonstrated experimentally that the ambient heat is a relatively insignificant part of the process.

The work done by the atmosphere is not "tiny". Isolated as a variable. Vacuum vs. no vacuum. Will melt, won't melt.

If our freezer box size engine contained about a cup of water, to make calculation easy say cube 3 inches on each side, at 15psi atmospheric pressure on all sides could account for as much as roughly 800+ lbs.

The weight of say four people up to 200 lb each.

That many people could hardly fit in my ice box.

This seems to have dissolved into rambling incoherence. I don't think there's much point in me continuing this discussion. 

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3 hours ago, exchemist said:

This seems to have dissolved into rambling incoherence. I don't think there's much point in me continuing this discussion. 

The gist is, atmospheric pressure on a 3x3 inch sample of ice at 1 atmosphere is approximately 800 lb.

Take that away and the ice would not melt in a vacuum is my prediction.

So I don't think it has been established that atmospheric pressure is insignificant.

I apologize for admittedly, thinking out loud. A number of posts were merged as additional thoughts came to mind and posts accumulated without response.

You also made many valid points. I just can't see, at this juncture that atmospheric pressure can be or should be ignored.

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9 minutes ago, Tom Booth said:

The gist is, atmospheric pressure on a 3x3 inch sample of ice at 1 atmosphere is approximately 800 lb.

Take that away and the ice would not melt in a vacuum is my prediction.

We don't need to rely on your guesswork.
We know that ice sublimes in a vacuum.


We also know how the melting point of ice changes with pressure. 
Raising the pressure reduces the melting point by about 0.01C per atmosphere of pressure.

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Posted (edited)
29 minutes ago, John Cuthber said:

We don't need to rely on your guesswork.

Here, we are carrying out a (thought) experiment where Ice is formed in a sealed engine cylinder under a 50 lb weight. Not typical vacuum conditions.

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Raising the pressure reduces the melting point by about 0.01C per atmosphere of pressure.

The difference made by 50 lb weight (or whatever could fit in the freezer in this experiment) is relatively insignificant (1/1600th of a degree C or 0.000625 C) when compared with the 800lb of atmospheric pressure IMO

____________

Looking back, launching into the bit about industrial liquefaction of gases with expansion turbines was probably not the way to make a point, as it is a little known process, very counter intuitive and sounds like complete BS.

Edited by Tom Booth
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54 minutes ago, Tom Booth said:

The gist is, atmospheric pressure on a 3x3 inch sample of ice at 1 atmosphere is approximately 800 lb.

How so? 14.7 psi x 9 si  = 132.3 lbs

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8 minutes ago, swansont said:

How so? 14.7 psi x 9 si  = 132.3 lbs

Sorry 793.8 lb.

A cube has 6 sides.

It's debatable if the pressure on the sides and bottom of the vessel can be discounted, but I don't think so. It may depend on the rigidity of the container holding the ice.

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On 5/7/2021 at 7:07 AM, Tom Booth said:

Could a thermal engine be constructed that utilizes the property of water to expand when frozen, taking advantage of the same force that causes the "ice bomb" to explode?

If not, why not?

Of course it can.

Mother Nature has been using ice jacking to turn mountains into sand ever since mountains became a thing.

Considering that about 4% of the globe's electricity output is consumed in turning big rocks into little ones; and that those processes are typically 1% efficient (chemical bond energy/total energy consumed); a high capacity, solar powered alternative technology could be of significant value.

  

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53 minutes ago, Tom Booth said:

Sorry 793.8 lb.

A cube has 6 sides.

It's debatable if the pressure on the sides and bottom of the vessel can be discounted, but I don't think so. It may depend on the rigidity of the container holding the ice.

You had been talking about a piston, which would be exposed to atmosphere on one side, so it's ~132 lbs. (and an additional 50 lbs would not be insignificant)

If you think 6 sides are exposed, you'd have to wonder how the water maintains its cubic shape before it freezes, under atmospheric pressure.

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Posted (edited)
13 minutes ago, swansont said:

If you think 6 sides are exposed, you'd have to wonder how the water maintains its cubic shape before it freezes, under atmospheric pressure.

Water could be frozen in a paper cup. But not if atmospheric pressure was removed from all sides but the top.

Edited by Tom Booth
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5 minutes ago, Tom Booth said:

Water could be frozen in a paper cup. But not if atmospheric pressure was removed from all sides but the top.

That's a poor argument, as it rests upon the fact that paper has poor structural integrity; you tend to not make piston systems out of paper, because you want/need the system to be rigid.

If you had a rigid, non-permeable cup, you could evacuate the region around the sides and bottom, and the water isn't going to notice.

 

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1 minute ago, swansont said:

That's a poor argument, as it rests upon the fact that paper has poor structural integrity; you tend to not make piston systems out of paper, because you want/need the system to be rigid.

If you had a rigid, non-permeable cup, you could evacuate the region around the sides and bottom, and the water isn't going to notice.

 

Atmospheric pressure is omnidirectional. Equal all around 

Therefore the containment vesel only needs to withstand the added pressure of the 50 lb weight. Not atmospheric pressure 

Even a rather substantial steel container would not withstand the pressure if the outside atmosphere were removed from all but the top.

 

 

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8 minutes ago, exchemist said:

....and we are where, now, exactly????:wacko:

Welcome back!

Just been trying to establish what goes on in this theoretical ice power engine, stage by stage, throughout the freezing and thawing, lifting and lowering. Energy flow, etc.

And answer the question, can this engine be viewed as just an ordinary heat engine. Can we use standard formulas, equations, PV diagrams etc. to model it's behavior.

I've been thinking about how removing heat can result in positive work output.

Removing heat might bring the water molecules (which are otherwise repelling each other to some degree), close enough together that some other more powerful intermolecular force results in a rearrangement into a crystal like formation.

The energy to do the lifting then, comes from this intermolecular force.

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