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The "Ice Bomb" thermal engine


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An "Ice bomb" is made by filling a very strong iron container with water and freezing it.

As water freezes it expands in volume exerting a force of some 100,000 pounds per square inch.

The "ice bomb", usually a grenade shaped cast iron container, when frozen, will explode with tremendous force. If contained within a wooden box when it explodes, it will destroy the box.

A conventional hot air or heat engine similarly utilizes the property of a gas that expands and exerts a force with a change in temperature. A gas is heated in a cylinder and the expanding gas expands and drives a piston.

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?

We know that the efficiency of a heat engine is represented by the formula: efficiency = W/Q hot or the work output of the engine divided by the heat input.

In the case of the theoretical ice bomb engine, however, Q hot is negative as heat is taken away to cause the ice to expand and output useful work. Therefore, is the efficiency of the "Ice bomb" engine also negative?

Can the efficiency of a thermal engine be less than zero, yet still operate?

It seems clear to me that by containing water as it freezes and expands, useful work could be extracted.

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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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11 minutes ago, iNow said:

What is the energy source for cooling the water below freezing point?

A cold reservoir. Not an energy source. An energy sink.

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3 hours ago, Tom Booth said:

Can the efficiency of a thermal engine be less than zero, yet still operate?

You have not analysed your heat engine correctly since you have missing elements.

3 hours ago, Tom Booth said:

water and freezing it.

The apparatus to cause this should be included

So yes

3 hours ago, Tom Booth said:

useful work could be extracted.

If you allowed the ice to expand in a non destructive way by compressing something, this compression could be partly extracted by causing it to do useful work.

But you analysis should include the work input to the freezing apparatus which will be greater than the work recoverable by the compression.

 

Remember also that the classical second law applies to a cyclic process, and may be 'violated' in part of a cycle.

A bomb is not a cyclic process.

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3 hours ago, Tom Booth said:

An "Ice bomb" is made by filling a very strong iron container with water and freezing it.

As water freezes it expands in volume exerting a force of some 100,000 pounds per square inch.

The "ice bomb", usually a grenade shaped cast iron container, when frozen, will explode with tremendous force. If contained within a wooden box when it explodes, it will destroy the box.

A conventional hot air or heat engine similarly utilizes the property of a gas that expands and exerts a force with a change in temperature. A gas is heated in a cylinder and the expanding gas expands and drives a piston.

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?

We know that the efficiency of a heat engine is represented by the formula: efficiency = W/Q hot or the work output of the engine divided by the heat input.

In the case of the theoretical ice bomb engine, however, Q hot is negative as heat is taken away to cause the ice to expand and output useful work. Therefore, is the efficiency of the "Ice bomb" engine also negative?

Can the efficiency of a thermal engine be less than zero, yet still operate?

It seems clear to me that by containing water as it freezes and expands, useful work could be extracted.

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 stored energy. You get a lot more stored energy in compressed gases than you do in compressed liquids and solids, because gases expand to many times the confining pressure, doing a lot more work (= energy). That's why heat engines rely on gases. 

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9 minutes ago, studiot said:

 

You have not analysed your heat engine correctly since you have missing elements....

But you analysis should include the work input to the freezing apparatus which will be greater than the work recoverable by the compression.

That assumes some cooling apparatus. As already stated, that is not the case.

The process could, for example be taking place in winter, or at the north pole where below freezing temperature is the prevailing condition.

To have a complete cycle, it can be assumed the engine must contact a  higher temperature reservoir to re-melt the ice, but no energy is expended in freezing or thawing as far as calculating efficiency is concerned.

The Carnot model of an ideal heat engine assumes the pre-existence of two "reservoirs" of different temperature, neither of which needs to be created.

"Efficiency" depends "only" on the difference in temperature and does not address in what way that difference came into existence.

55 minutes ago, exchemist said:

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 stored energy. You get a lot more stored energy in compressed gases than you do in compressed liquids and solids, because gases expand to many times the confining pressure, doing a lot more work (= energy). That's why heat engines rely on gases. 

I don't think expansion volume is a determinant of the available energy.

For example, if I lift a ping pong ball ten feet and a ten ton block of concrete 0.5 inches, which one of the two has the greater potential energy?

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15 minutes ago, Tom Booth said:

That assumes some cooling apparatus. As already stated, that is not the case.

The process could, for example be taking place in winter, or at the north pole where below freezing temperature is the prevailing condition.

To have a complete cycle, it can be assumed the engine must contact a  higher temperature reservoir to re-melt the ice, but no energy is expended in freezing or thawing as far as calculating efficiency is concerned.

The Carnot model of an ideal heat engine assumes the pre-existence of two "reservoirs" of different temperature, neither of which needs to be created.

"Efficiency" depends "only" on the difference in temperature and does not address in what way that difference came into existence.

 

Yes perfomance of the cycle a requirement of the second law.

So another way to put it is that the work obtained from the expansion of the ice will be less than the heat required to complete the cycle and thaw the ice, ready for the beginning of the next cycle.

Such situations were exactly why Maxwell, Clausius and others stated the second law in cyclic format, as I have already noted.

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Posted (edited)
1 hour ago, Tom Booth said:

That assumes some cooling apparatus. As already stated, that is not the case.

The process could, for example be taking place in winter, or at the north pole where below freezing temperature is the prevailing condition.

To have a complete cycle, it can be assumed the engine must contact a  higher temperature reservoir to re-melt the ice, but no energy is expended in freezing or thawing as far as calculating efficiency is concerned.

The Carnot model of an ideal heat engine assumes the pre-existence of two "reservoirs" of different temperature, neither of which needs to be created.

"Efficiency" depends "only" on the difference in temperature and does not address in what way that difference came into existence.

I don't think expansion volume is a determinant of the available energy.

For example, if I lift a ping pong ball ten feet and a ten ton block of concrete 0.5 inches, which one of the two has the greater potential energy?

Eh? Lifting weights expands nothing. 

Work done in expansion is PdV.

Think of it this way: if you have an expanding fluid pushing a piston, the work done is the force, F, on the piston (pressure x surface area) multiplied by the distance, d, the piston travels - which is volume change/surface area.  So Fd = PA x ΔV/A = PΔV. (Since P is likely to change as V increases you need to do it as an integral, hence ∫PdV. ) 

So expansion volume certainly is the determinant, along with pressure, of the available energy. 

 

Edited by exchemist
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2 hours ago, exchemist said:

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 stored energy. You get a lot more stored energy in compressed gases than you do in compressed liquids and solids, because gases expand to many times the confining pressure, doing a lot more work (= energy). That's why heat engines rely on gases. 

To expand on this, as it were, the scenario depends on the vessel being weak enough to rupture. If that's not the case, you just have cold water at high pressure.

 

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3 hours ago, exchemist said:

... You get a lot more stored energy in compressed gases than you do in compressed liquids and solids, because gases expand to many times the confining pressure, doing a lot more work (= energy). That's why heat engines rely on gases. 

It is also why pressure vessel testing is done with liquids (usually water) rather than gas.



Fundamentally, the idea  depends on having a large "cold body" that you can use for cooling.
And if you have that, you can use it to run a "stream engine" with, for example, butane, as the working fluid.
 

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

And if you have that, you can use it to run a "stream engine"

Don't you need a 'wee burn' to run a stream engine  ?

🙄

Edited by studiot
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41 minutes ago, studiot said:

Don't you need a 'wee burn' to run a stream engine  ?

🙄

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

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4 hours ago, studiot said:

 

Yes perfomance of the cycle a requirement of the second law.

So another way to put it is that the work obtained from the expansion of the ice will be less than the heat required to complete the cycle and thaw the ice, ready for the beginning of the next cycle.

Such situations were exactly why Maxwell, Clausius and others stated the second law in cyclic format, as I have already noted.

OK, that makes sense, I think.

So then, as far as using the formula, it doesn't matter that adding heat does not produce useful work in this situation. Rather, removing heat stores potential energy in the water, which if then allowed to freeze, can crystalize and lift a weight as heat is lost? Can the sign just be ignored and change the negative (heat rejected) to positive? Or is it necessary to measure the heat absorbed by the ice in melting and use that number. Or doesn't it matter?

I feel like my brain is getting a whiplash trying to follow the logic of all this.

Then would W/Qh be identical to W/Qc, one can't just be substituted for the other can it?

3 hours ago, exchemist said:

Eh? Lifting weights expands nothing. 

Work done in expansion is PdV.

Think of it this way: if you have an expanding fluid pushing a piston, the work done is the force, F, on the piston (pressure x surface area) multiplied by the distance, d, the piston travels - which is volume change/surface area.  So Fd = PA x ΔV/A = PΔV. (Since P is likely to change as V increases you need to do it as an integral, hence ∫PdV. ) 

So expansion volume certainly is the determinant, along with pressure, of the available energy. 

 

I should have said, or my intended meaning was: " I don't think expansion volume is a determinant of the available energy" - all by itself.

2 hours ago, swansont said:

To expand on this, as it were, the scenario depends on the vessel being weak enough to rupture. If that's not the case, you just have cold water at high pressure.

 

Of course. I guess.

An engine is a vessel designed to easily "rupture" in such a way as to perform work by means of a moveable wall; the piston, attached to a crankshaft or whatever.

The "ice bomb" engine would have to be similarly constructed. Something like a very strong hydraulic jack.

The impracticality of the small distance traveled might be overcome by attaching the piston to some kind of long extended lever.

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On 5/7/2021 at 10:00 AM, Tom Booth said:

Of course. I guess.

An engine is a vessel designed to easily "rupture" in such a way as to perform work by means of a moveable wall; the piston, attached to a crankshaft or whatever.

The "ice bomb" engine would have to be similarly constructed. Something like a very strong hydraulic jack.

The impracticality of the small distance traveled might be overcome by attaching the piston to some kind of long extended lever.

But the piston is pretty much free to move, so this is less like a bomb and more like a slow expansion as the water freezes. (It could become a tiny bit more like a bomb if the ice itself was strong and you still had liquid in the center of the systems)

But as has already been pointed out, you're expending a whole lot of energy for  <10% expansion. OTOH, a phase change, or chemical reaction that converts a solid or liquid to a gas, gets you ~22.4 L per mole of gas (assuming an ideal gas) at STP. 

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One more thought comes to mind.

If water freezes and expands, and is made to perform some useful work in the process, it would seem to follow that having lost heat energy as work output, the resulting ice should end up extra cold. Heat and work being equivalent.

So an additional amount of heat will be needed to bring the ice formed under high compression back to a liquid state in order to make up for the heat lost to work output during the freezing process.

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Posted (edited)
1 hour ago, exchemist said:

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

John is that rare breed of Scotsman who possesses a grand sense of humour.

And of course, burn is the Scottish for a stream.

The additional double entendre was unintentional and unnoticed by me, until you pointed it out.

 

49 minutes ago, Tom Booth said:

Of course. I guess.

An engine is a vessel designed to easily "rupture" in such a way as to perform work by means of a moveable wall; the piston, attached to a crankshaft or whatever.

The "ice bomb" engine would have to be similarly constructed. Something like a very strong hydraulic jack.

The impracticality of the small distance traveled might be overcome by attaching the piston to some kind of long extended lever.

 

Since no large deflection is required I was thinking in terms of a diaphragm deflecting and maybe pushing a rod.

 

49 minutes ago, Tom Booth said:

OK, that makes sense, I think.

So then, as far as using the formula, it doesn't matter that adding heat does not produce useful work in this situation. Rather, removing heat stores potential energy in the water, which if then allowed to freeze, can crystalize and lift a weight as heat is lost? Can the sign just be ignored and change the negative (heat rejected) to positive? Or is it necessary to measure the heat absorbed by the ice in melting and use that number. Or doesn't it matter?

I feel like my brain is getting a whiplash trying to follow the logic of all this.

Then would W/Qh be identical to W/Qc, one can't just be substituted for the other can it?

The way to calculate this is not from the equations because water is an exception to the normal pattern.
You need to get out a set of 'International Steam Tables'  - a quick look shows mine are in the garage - and read off the entropy change on a T - S diagram to obtain the energy flows easily.

 

I don't have a convenient glacier to effect the freezing part of the cycle or a convenient hot spring/geyser to provide the thawing, and even if I did they would still be part of the system and their entropy changes need to be included.

Edited by studiot
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Posted (edited)

Trying to understand entropy fogs up my brain like a car windshield on a cold morning with no defroster.

I think the Ideal "ice bomb engine" would contain the freezing water long enough, and the weight or load on the piston would be just heavy enough, (like in the compression stroke of an engine) that the piston would fly off from the sudden explosive force leaving a vacuum behind in the cylinder that would cause the water to boil under such low pressure and vaporize, causing further expansion.

Then the gigantic piston would come crashing back down as the super-cooled water vapor (cooled due to the sudden violent expansion) condensed back into water, then as it continued to cool and freeze back into ice, the process would repeat itself.

The timing would have to be very precise so that the piston reached full compression just as the water re-freezes.

Or something like that.

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

Trying to understand entropy fogs up my brain like a car windshield on a cold morning with no defroster.

I'm not suprised if you tried to study it via Carnot cycles. That really is the hard way.

 

Thermodynamics divides things into a system and the rest of the universe (ie the surroundings).
To do this successfully it must always specify the system boundary.
Having done this state variables (preferably directly measurable ones) are introduced for both the system and its surroundings.
Pressure, volume and temperature being the most recognisable.

Now it was found that the area under a P - V plot was equal to the mechanical work that is exchanged across the system boundary between the system and its surroundings.

This leaves temperature and a similar variable was sought to pair with temperature to calculate the heat energy exchanged across the boundary. Entropy was the name given to this variable. So the area under a T - S plot is the heat energy that passes across the system boundary.

Unfortunately entropy, unlike P, V and T is not directly observable so we use measured tables instead.

That's all there is to it.

 

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22 minutes ago, Tom Booth said:

Trying to understand entropy fogs up my brain like a car windshield on a cold morning with no defroster.

I think the Ideal "ice bomb engine" would contain the freezing water long enough, and the weight or load on the piston would be just heavy enough, (like in the compression stroke of an engine) that the piston would fly off from the sudden explosive force leaving a vacuum behind in the cylinder that would cause the water to boil under such low pressure and vaporize, causing further expansion.

Then the gigantic piston would come crashing back down as the super-cooled water vapor (cooled due to the sudden violent expansion) condensed back into water, then as it continued to cool and freeze back into ice, the process would repeat itself.

The timing would have to be very precise so that the piston reached full compression just as the water re-freezes.

Or something like that.

I'm not sure I follow this. It looks as if you envisage a closed container that stretches and then suddenly bursts from the pressure of the ice and that this moves a piston. Depending on what load the piston is connected to, I suppose with a light load it might fly away from contact with the ice, leaving a near-vacuum behind. It won't be quite a vacuum as there will be the vapour pressure of water or ice at 0C present - about 4.5mmHg, so not much, admittedly. A vacuum won't cause the ice to melt. Many comets are made of ice. So there won't be any boiling.

Why is this piston "gigantic", suddenly?  Are you smoking exotic cheroots? 😀

   

Edited by exchemist
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58 minutes ago, Tom Booth said:

If water freezes and expands, and is made to perform some useful work in the process, it would seem to follow that having lost heat energy as work output, the resulting ice should end up extra cold. Heat and work being equivalent.

This is all happening at 0º or thereabouts; it's a phase change, which happens at constant temperature.  

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35 minutes ago, studiot said:

I'm not suprised if you tried to study it via Carnot cycles. That really is the hard way.

 

Thermodynamics divides things into a system and the rest of the universe (ie the surroundings).
To do this successfully it must always specify the system boundary.
Having done this state variables (preferably directly measurable ones) are introduced for both the system and its surroundings.
Pressure, volume and temperature being the most recognisable.

Now it was found that the area under a P - V plot was equal to the mechanical work that is exchanged across the system boundary between the system and its surroundings.

This leaves temperature and a similar variable was sought to pair with temperature to calculate the heat energy exchanged across the boundary. Entropy was the name given to this variable. So the area under a T - S plot is the heat energy that passes across the system boundary.

Unfortunately entropy, unlike P, V and T is not directly observable so we use measured tables instead.

That's all there is to it.

 

I'll have to reread that and study it again and again, but I have before from other texts and, well, it ends up in my mind seeming to be a new name for the abandoned "caloric". The caloric, having been abandoned in theory, but necessary for the math to work out as previously, it was resurrected as "entropy". That's about as much sense as I can make of it, anyway.

20 minutes ago, swansont said:

This is all happening at 0º or thereabouts; it's a phase change, which happens at constant temperature.  

Boiling happens at constant temperature as well, but a constant heat input is still going on to make it happen. Is it not also true that freezing requires continuous heat removal? The freezing process doesn't happen without energy exchange with the environment 

Work output is also energy exchange with the environment that results in a drop in temperature.

34 minutes ago, exchemist said:

I'm not sure I follow this. It looks as if you envisage a closed container that stretches and then suddenly bursts from the pressure of the ice and that this moves a piston. Depending on what load the piston is connected to, I suppose with a light load it might fly away from contact with the ice, leaving a near-vacuum behind. It won't be quite a vacuum as there will be the vapour pressure of water or ice at 0C present - about 4.5mmHg, so not much, admittedly. A vacuum won't cause the ice to melt. Many comets are made of ice. So there won't be any boiling.

Why is this piston "gigantic", suddenly?  Are you smoking exotic cheroots? 😀

   

It doesn't "stretch", any more than a regular engine stretches when a compressed fuel/air mixture is ignited.

For my fanciful invention to work, I imagine that the piston would have to be very heavy in order to contain the expanding ice long enough for an explosive force to develop, at that moment when the expansion pressure is at a maximum, the piston passes TDC, (or I think BDC in this case) and begins to move away (due to energy stored in a flywheel) just as the ice explodes!

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

I'll have to reread that and study it again and again, but I have before from other texts and, well, it ends up in my mind seeming to be a new name for the abandoned "caloric". The caloric, having been abandoned in theory, but necessary for the math to work out as previously, it was resurrected as "entropy". That's about as much sense as I can make of it, anyway.

Boiling happens at constant temperature as well, but a constant heat input is still going on to make it happen. Is it not also true that freezing requires continuous heat removal? The freezing process doesn't happen without energy exchange with the environment 

Work output is also energy exchange with the environment that results in a drop in temperature.

It doesn't "stretch", any more than a regular engine stretches when a compressed fuel/air mixture is ignited.

OK, but it stretches a bit. What happens is you stretch it beyond the elastic limit, reach the yield point and then it fails.  

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24 minutes ago, Tom Booth said:

Boiling happens at constant temperature as well, but a constant heat input is still going on to make it happen. Is it not also true that freezing requires continuous heat removal? The freezing process doesn't happen without energy exchange with the environment 

Work output is also energy exchange with the environment that results in a drop in temperature.

Yes, it can. But the work is part of the process of the energy removal during the phase change. It doesn't require an additional temperature drop. Before you have water at 0ºC. After you have ice, which has expanded, at 0ºC. The enthalpy of that phase change tells you the energy removal that's required to freeze the water. If you're doing extra work while it freezes, you will have to remove less energy, because the work being done removes energy from the system (as studiot points out, you have to define your system. Any object you lift is outside. Its energy goes up, so the system's energy goes down). I don't think you can conclude the temperature must drop, which would require even more energy to be extracted. OTOH, this is thermodynamics, and there's almost always more than one way to get from state A to state B; "the temperature must drop" is not the same as "the temperature can drop"

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23 minutes ago, exchemist said:

OK, but it stretches a bit. What happens is you stretch it beyond the elastic limit, reach the yield point and then it fails.  

There is no container within the cylinder. The engine's cylinder is the container.

Sure, there is some "stretching" on some infinitesimal scale, but the piston is designed to move or yield at the critical moment when the expansion force is at a maximum, but not before.

22 minutes ago, swansont said:

Yes, it can. But the work is part of the process of the energy removal during the phase change. It doesn't require an additional temperature drop. Before you have water at 0ºC. After you have ice, which has expanded, at 0ºC. The enthalpy of that phase change tells you the energy removal that's required to freeze the water. If you're doing extra work while it freezes, you will have to remove less energy, because the work being done removes energy from the system (as studiot points out, you have to define your system. Any object you lift is outside. Its energy goes up, so the system's energy goes down). I don't think you can conclude the temperature must drop, which would require even more energy to be extracted. OTOH, this is thermodynamics, and there's almost always more than one way to get from state A to state B; "the temperature must drop" is not the same as "the temperature can drop"

Well, that's what happens with the liquefaction of gas using an expansion turbine.

The gas is compressed and cooled, then allowed to expand through a turbine. There is a load on the turbine, so the gas also has to perform work as it expands.

The resulting energy loss from the gas due to the combined cooling from expansion and cooling from work output causes the gas to cool instantaneously and suddenly condense before it leaves the turbine.

I would say "must" is appropriate. Otherwise the process would not be reliable enough for industrial applications.

Also, I think ice can be much colder than 0°c

And even liquid water can be super-cooled.

Edited by Tom Booth
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3 hours ago, exchemist said:

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

Are you taking the p***?

 

If you supercool water (for example in the cylinder of some "engine") and then induce it to freeze- say by shaking, it warms up to 0C as it freezes.

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