sethoflagos

58 minutes ago, Tom Booth said:

Yes, I thought so, as I mentioned, something didn't look right. The result of a hasty cut n paste on the Stirling engine forum, which I then repeated here.

Sorry, but I don't buy this explanation. 

Take this example from a previous post:

7 hours ago, Tom Booth said:

If my heat engine utilizes ALL the heat fed into it, and converts every bit of it into useful work so no heat whatsoever "flows out" into the sink, carnot efficiency might be calculated at almost any arbitrary number represented by the temperature difference.

 If there was "no heat whatsoever" flowing into your cold sink, that would imply that the thermal efficiency of your machine was 100% (it can't be, but we'll let that pass for now). In other words, you were extracting every last milliJoule of work allowed by the 2nd Law of Thermodynamics.

However, if your heat engine did indeed "utilize ALL the heat fed into it" from say 1 kg of hot cocoa, then you would not only have your 100% thermal efficiency, but you would also have attained 100% Carnot efficiency. 

Congratulations!

Now explain to me what you did with that 1 kg lump of cocoa at absolute zero. 

Because that is what "ALL the heat" means.

If on the other hand, you actually ended up with luke warm cocoa, the Carnot efficiency simply tells you what percentage of "ALL the heat" was actually available to you. It says nothing whatsoever about the virtues or quirks of your machine - it is an absolute limit for any machine that was fueled by 1 kg of hot cocoa and exhausted 1 kg of luke warm cocoa. And it's a simple function of those two temperatures.

Please pause and reread the last few paragraphs. You've wasted a fair chunk of the last few years through not understanding the difference between the highlighted terms. How much more time can you afford to waste?

1 hour ago, Tom Booth said:

Please site an example.

I've heart this affirmation repeated over and over and over, but that seems to be all it is. I've scoured through the available literature for ten years and find no accounts whatsoever of any detailed experiment demonstrating, for example, the actual heat flow in and out of a Stirling engine, running a Stirling heat engine on ice, insulating the sink, nothing whatsoever that would either add weight to or call into question Carnot's conclusions.

Carnot's equation is a simple algebraic manipulation of the 2nd Law limit - delta S = 0

Experimental verification of the 2nd Law automatically verifies the Carnot limit.

The first item to pop up on Google was: 

"Experiment to verify the second law of thermodynamics using a thermoelectric device", Gupta, V. K.; Shanker, Gauri; Saraf, B.; Sharma, N. K.  American Journal of Physics, Volume 52, Issue 7, pp. 625-628 (1984).

I've not read it, but I'm sure it's fine, and typical of many thousands of similar published papers. It's actually based on a Seeback-effect heat engine, but the underlying principles are just the same.

If you want to see more, just Google "experimental verification of 2nd law of thermodynamics" as I did. There's over 6 million results for you to sift through.

Good hunting.

You're confusing Carnot efficiency, 1 - T_C/T_H

... with thermal efficiency, 1 - Q_C/Q_H

It is entirely consistent with both classical thermodynamic theory, and centuries of detailed empirical observation by the most gifted of experimentalists, for a machine to have a low Carnot efficiency (the fraction of energy in an input stream that is available for conversion to work) and a high thermal efficiency (the fraction of that available work you manage to convert to actual work)

The following demonstrates most clearly that you are totally unaware of this distinction:

1 hour ago, Tom Booth said:

If my heat engine is 500 degrees on one side and 250 degrees on the other than the "caloric" can only fall 250 degrees from 500 down to 250 which is 50% of the "fall" on the way down to absolute zero. (Arbitrary numbers on the Kelvin scale but it works out the same way on any other) That makes my engine 50% efficient, at best, because that is as far as the "caloric" can possibly fall.

With an inescapable effect on the authority of your output:

1 hour ago, Tom Booth said:

It's complete hogwash.

7 hours ago, studiot said:

If this does not happen then you will not have the required water for the hydroturbine to operate. It cannot do so in a slush.

Please give me a credible mechanism for the creation of a slush (suspension of ice particles in water) in the system I have described. I currently regard the idea as a deus ex machina, but would be delighted to be shown otherwise.

35 minutes ago, MigL said:

I'm not convinced of that either.
A Martian day is similar to an Earth day, which might not be enough time to freeze or liquify completely, leading to a perpetually liquid or ice ring around the planet.
Or possibly a state of 'slush' as a large buffer between the ice and water sections.

I find it interesting as an idealized mental exercise, but the 'realities' which Studiot mentions, make it impractical, if not impossible.

Average low night temperatures on mars are below 200 K throughout the year. There is next to no atmospheric blanketing and the (blackened) pipes are shielded from the ground by mirror finish parabolic troughs. They are in thermal communication with nothing but the empty vacuum of space. 

You may have noticed that I switched from an initial single 48" ND pipe to multiple 8" ND pipes. This was specifically directed at providing the necessary 'A' in sigma.A.T^4 to meet the night side heat shedding load.

You may also have noticed that I've switched to extracting the 18 m^3/s via multiple tapping points (oto 10,000 spread over 1,000 km) so that it is drawn off at practically zero velocity. If there is any ice nucleation in the body of liquid (rather than at the annular interface), then it will float upwards as far as it can. There is insufficient fluid shear to overcome buoyancy forces. And certainly insufficient fluid shear to start ripping consolidated phase Ih ice away from the pipe wall.

And yes, the project is impractical if not impossible. That was never in doubt.

Of course, I'd like to carry everyone along with me on this. But it's as clear as day that there are a few individuals who will never concede as a matter of principle. And bottom line is, I don't see I'm under any real obligation to convince anybody. Except possibly myself.  

32 minutes ago, swansont said:

I don’t recall these being described in the OP. You had a tube and a turbine. I think my disbelief was well-founded, based on the available information.

I'm sure your disbelief was exemplary.

But now that you have updated information, do you agree that the interface pressures can be subject to operator control?

2 minutes ago, swansont said:

That wasn’t one, so...

... So... Well you expressed disbelief that the interface pressures could be controlled by the operator. My second to last post sort of covered that, but maybe I could flesh out some details for you.

It is relatively easy to set the pressure of a ringmain. A very simple example would be to fit it with an open header tank set at the appropriate elevation and sized to accommodate any expansion/contraction or temporary fluctuations in inventory.

Something a little more sophisticated would be called for here. Maybe an underground reservoir with a substantial gas blanket to absorb the fluctuations within a tight pressure band. From a functional point of view, it's identical to the elevated tank, but without the open connection to the martian atmosphere.

So we can create two ringmains with tightly controlled operating pressures.

We now install tapping lines equipped with one-way flow valves (check valves) into the freeze/thaw system. Those connected to the high pressure ringmain would have their check valves so oriented to only allow flow into the ringmain. These will draw flow from the freeze/thaw system only when its pressure at that point exceeds that of the high pressure ringmain - ie in the vicinity of the freezing interface. Conversely, those tapping lines connected to the low pressure ringmain would be oriented in the opposite sense, feeding the freeze/thaw system only at those locations at a pressure below that of the low pressure ringmain - ie in the vicinity of the thawing interface.

Obviously, the ringmains and tapping lines would be fully insulated and traced to prevent them from freezing up.

Once the high pressure ringmain is pressurised, the turbine sluices can be opened, controlling the high pressure ringmain to a setpoint somewhere in the middle of its design operating band, this will automatically feed the low pressure ringmain with precisely the volume required to feed the low pressure injection tappings.

Since it's a closed, very nearly constant volume system, fluctuations should be very small, and self-regulating.

I trust this rather long and detailed post is sufficient to quell your disbelief.   

26 minutes ago, studiot said:

I'm glad you are beginning to do some thinking about it.

16 minutes ago, swansont said:

You made an assertion without backing it up, so thou doth protest too much, methinks.

Not that interested in ad hominems.

Guess we're done.

20 hours ago, swansont said:

How big of a gradient are you expecting?

I'm comfortable with a waterside delta P of ~100 bar (10^7 Pa).

Over 10,000 km, this equates to a pressure gradient of 1 kPa/km.

This gradient is compatible with a line velocity of 0.15 m/s in 8" double extra strong (XXS) linepipe.

This is less than 1% of the flow generated by the freeze thaw cycles making ~ 18 m^3/s available to be tapped off into a high pressure ringmain operating at say 10 bar less than the high pressure (freezing) interface.

Similarly, a low pressure ringmain operating at 10 bar above the pressure of the low pressure (thawing) interface will return the 'borrowed' 18 m^3/s back into the thawing zone.

The residual 0.15 m/s water velocity in the freeze/thaw system will be naturally maintained as a consequence of the imposed delta P.

Hydroelectric generators linking high and low pressure ringmains will utilise the 18 m^3/s flowing between them with a delta P of 80 bar (8 x 10^6 Pa) to yield:

Est. Power Output = 0.9 x 18 x 8 x 10^6 = 129.6 MW (continuous)

42 minutes ago, studiot said:

and simply asking how you would transfer that amount of energy into a block of ice in one second, since you would need to do the same again with the next block of ice in the next second and so on.

Why would any particular m^3 of ice need to be thawed in 1 second?

So long as it's fully thawed by mid afternoon, say, before the heat input has reduced to the point where it starts to refreeze, then it's done its job. 6 hrs to thaw = ~5,000 km of the collection array doing the thawing. Actually, if your figure of 590 W/m^2 is good, a high efficiency collection strip 100 m wide will do the job over ~1,200 km or about an hour and a half. So there's a fair safety margin to play with.   

PS. Thinking about it, since I'm going to be reinjecting the somewhat warmish low pressure discharge from the water turbines back into the thawing zone to meet the contraction demand, that fact in itself should significantly accelerate the thawing process.

29 minutes ago, MigL said:

So, using water which expands on freezing, you expect a higher pressure on th day-to-night freezing interface, and a lower pressure on the night-to-day thawing interface, leading to a pressure gradient, and a net flow of water, from which you hope to generate power.

Obviously if a liquid other than water was used, that doesn't change density on freezing, you would have no pressure gradient, and no flow. The frozen section would just move around following the night-side.

But what if you had any other liquid that contracts on freezing. Your analysis would then indicate a reverse pressure gradient, and a flow in the opposing direction ?

Exactly put. +1

8 hours ago, studiot said:

So let me get this straight.

You are envisaging using the entire output from a 100m wide strip, one quarter of the way round the martian globe to melt the ice in one metre of the pipeline ?

Why the tone of ridicule?

You state that 6 x 10^5 m^2 of collected solar radiation will melt 1 m^3 of ice in one second

So 240 x 6 x 10^5 = 1.44 x 10^8 m^2 will melt 240 m^3 of ice per second, the thermal duty we are looking for.

My order of magnitude guess of a 100 m strip around the planet seems to meet the requirement several times over. 

5 hours ago, swansont said:

IOW, you are in a steady-state condition.

Yes

5 hours ago, swansont said:

The gradient causes flow, it is not preserved by it. 

They come as an indivisible pair. The one leads to the other and vice versa.

5 hours ago, swansont said:

I seriously doubt that.

 

 Argument from incredulity? 

5 hours ago, swansont said:

Which is what I expect will happen if you tried this.

The safety systems could well be a challenge

5 hours ago, swansont said:

So the pressure will equilibrate much faster than any flow you are expecting.

Following a dynamic peturbation (passing dust cloud, for example), steady state will reestablish itself (if this is what you mean by 'equilibrate') not by transmission of pressure waves as such, but rather by their attenuation due to viscous dissipation, which can take a significant length of time. With such a long pipeline, water hammer effects would be a significant concern (because spontaneous disassembly again).

2 minutes ago, Tom Booth said:

I don't know where you're coming from but Turbo-expanders are WIDELY used for gas liquefaction. It has practically replaced every other method there is in every industry involved in gas liquefaction all around the world.

First you try to deny it exists, apparently because it wasn't on Wikipedia from which you were apparently cp'ing, now you are trying to demonize a standard industrial process, used all around the world.

What's your game dude?

Seriously?

Non-cowboy operations condition their gas in a proper gas plant with the full demethaniser, deethaniser, depropaniser and debutaniser set to maximise LPG extraction and ensure their sales gas output is fit for purpose.

Cowboy operations cherry pick a rough LPG cut with a single stage J-T or turboexpansion stage and more often than not screw up the national sales gas supply grid with intermittent slugs of condensate. 

Don't confuse typical US practice for global practice. Most of the world falls into the first of these two categories.

1 minute ago, Tom Booth said:

What?

It's friggin' common knowledge except maybe for people in the industry who try to guard it like some kind of trade secret or other.

It isn't right though is it, Tom.

Right would be investing in the appropriate refrigeration system to take out the condensate cut you want in a conventional condenser. Just like the textbooks say.

Just sayin'

37 minutes ago, Tom Booth said:

How about you stop devolving the conversation into petty attempts at character assassination.

If that's how you read my posts then, I'm sorry, it was not my intent.

37 minutes ago, Tom Booth said:

If you are unfamiliar with such a use for turbo-expanders, I can provide references...

Having spent the last 22 years in the West African oilfields, I am unfortunately more familiar with such malpractices than you can possibly imagine. Unless that is you've done time with Shell Petroleum Development Company of Nigeria which would put us on a par. Using a turboexpander as a souped up J-T valve is simply something you should not be broadcasting to the world in my view. At best, people won't have a clue what you're on about, and those who do understand will assume you've worked for Shell Petroleum Development Company of Nigeria. Lose-lose.

12 minutes ago, Tom Booth said:

Your referring to a different use case.

That's like saying the evaporator in a refrigerator is undesirable because it doesn't produce heat or pressure.

A turbo-expander when used to liquefy gases is not meant to produce work, it's meant to liquefy gases.

That it succeeds in liquefying gas does not "reduce it's performance" because it does less work. Doing work is not it's purpose.

This is about as a valid a use case as calling your car a tractor to explain why its upside down in a potato field.

Turboexpanders, if they were in the slightest way relevant to your OP which they are not, are NEVER designed for the purpose you describe and to infer that they are serves no purpose other than to mislead the membership of this site. 

49 minutes ago, Tom Booth said:

It goes back to my statement that the "ice bomb engine" does NOT use ambient heat to do work at all.

You said: "yes it does".

I had to check back through my activity record, but I can say with confidence that I've passed no comment on the thermodynamics of your 'ice bomb' whatsoever.

I have passed comment on the thermodynamics of your Stirling engine (which extracts a percentage of the heat flow between a hot source and cold sink to create shaft work)

And also your refrigerator machine which essentially employs a compressor to lift a weight a few millimetres.

1 hour ago, Tom Booth said:

You then go on:

Quote

This potential energy then flows out when the temperature is reduced

 

What? 

This is most definitely someone else's words you're quoting.

Please try and keep track of who you are addressing, who you are quoting, and the true context of each quote. 

For info. from a hydraulics point of view, 18 m/s is a really high velocity for a liquid pipeline. A major issue from my perspective would be that the pressure gradient necessary to maintain that kind of flowrate would be untenable beyond a few hundred metres at most.

One possible solution would be to run a much larger pipeline system in parallel, fully insulated and traced to prevent freezing, to carry the major part of the water flow at a much lower velocity (<1 m/s). 

The original 'ice pump' circuit could then be crosslinked to the larger pipeline every 100 m say so we transfer all the energy generated by the ice pump circuit into a much lower velocity system that can transport the water up to say 100 km to a generating station with minimal hydraulic losses. Most of the energy would actually be transmitted as regular pressure surges which are practically lossless (essentially a controlled 'tsunami'). We would have the opportunity to run a much higher pressure drop across the turbines, maybe 64 MW rather than 16 MW to compensate for the additional capital expenditure. 

6 minutes ago, swansont said:

And why won’t the pressure just equalize?

Because the imbalance is being generated continuously by expansion at one end and contraction at the other.

9 minutes ago, swansont said:

Once it has propagated, the whole tube is at pressure. You’re all done. 

The pressure gradient across any given pipeline section is (in the absence of fluid acceleration) balanced and preserved by the nett hydraulic shear forces due to fluid flow in that section.

22 minutes ago, swansont said:

 How big of a gradient are you expecting?

It's under operator control. If you fully throttle fluid flow by shutting an inline valve, the ice does not have the 8% space it needs to expand freely generating a theoretical pressure spike ~ 8% of its bulk modulus (8.4 GPa) = 672 MPa. It would never actually get that high due to spontaneous disassembly of its containment. 

Desirable operating pressures are set by dialling in the appropriate resistance to flow.

38 minutes ago, swansont said:

How fast does the pressure differential propagate? 

Depends on context. Disturbances to steady state would propagate at sonic velocity (~1,400 m/s at 0 C)

36 minutes ago, studiot said:

As an obviously competent engineer I am disappointed with the obviously politician's brush off when pressed for hard detail, more especially as you invited comment.

Here is the problem I am trying to reconcile.

 

You have mentioned several different pipe sizes, and somewhere a square metre of cross section.

So let us consider 1 m² section of pipe 1 metre long, at the frozen stage.

This has a volume of 1 cubic metre.

Ignoring, for the moment,  the small difference in density between ice and water, this has a mass of 10³ kg

So to calculate the approximate energy reuqirement to raise this from ice at -4 C to water at  +1 C  ie to melt it we require

10³(4*2050 + 334000 + 4200)  = 3.464 10⁵ x 10³ Joules per m³

Now the rate of insolation on the mars is 590 w/m²

or 5.9 x 10² J/m² per second

So it requires ( 3.464 x 10⁸ ) / (5.9 x 10² ) = 6 x 10⁵ square metres of martian surface to receive this energy every second multiplied by the rate of movement of the ice/water interface  as this was calculated on a 1m/s basis and assuming perfect energy conversion.

 

You got ahead of me. Though I have no intention whatsoever of detailing this idea out (other than the highly unlikely event I was paid union rates for it!)

What's the area of a 5,000 km strip 100 m wide? 5 x 10^8 m. A bit generous for a pipeline RoW, but it isn't as if we were displacing indigenous residents.

2 minutes ago, Endy0816 said:

I mean water will also sublimate on Mars at Martian atmospheric pressure, depending on temperature.

It's in a pressurised, fully contained system. There should be no physical contact with the martian atmosphere.

1 minute ago, Endy0816 said:

What is the plan for sublimation?

That would be Revision 2, the CO2 version,

2 minutes ago, swansont said:

No, this does not follow.  Why wouldn’t the pressure just go up?

But the pressure does go up in the zone where ice expansion is displacing the incoming water, just as it drops where melting contraction creates space for water to flow into.

7 minutes ago, swansont said:

Also you seem to assume instantaneous freezing and melting, and that it would happen along the direction of the pipe, and not in the radial direction, from outside in.

No, I don't assume that at all. Quite clearly freezing (and thawing too) progress from the pipewall to the centre over a significant period. I think I said in an earlier post that I envisioned the interface to be a deep taper, (probably thousands of kilometres long).

Note that as the taper narrows, the expanding ice will squeeze that 8% excess volume of water back the way it came just like a tube of toothpaste.

24 minutes ago, swansont said:

The thing is, some distance away, you are melting ice and having a corresponding collapse of 240 m of ice into 222 m of liquid water, which means there could just be a certain pressure increase, which remains static. No motion relative to the ground. I’m not seeing a net impulse exerted to the water.

 Well if you see a pressure gradient along the water column then we're more than half way there. All that remains is to be able to visualise the contraction of ice to meltwater as continuously creating a space for water to flow into. The impulse exerted on the water is purely and simply the pressure gradient: the continuous creation of upstream space (and corresponding continuous denial of space downstream) then yields all that is necessary to establish bulk water flow towards the thawing zone.

I do appreciate that dynamic systems in peculiar coordinate systems like this can be hard to visualise with clarity. Especially if you're not particularly predisposed to accept a particular person's viewpoint. 

So while I note that no one has actually stepped forward to say that they've bought into this picture, that's really not an issue. Sometimes that's just the way things are.

Thanks to all who contributed for your assistance.

16 minutes ago, Tom Booth said:

I'm wondering, why not build a small model, a loop of pipe, a heat lamp in a cold room, some flow meters in the pipeline, a stepper motor to keep it revolving.

I guess I've got too used to having my designs constructed to a scale of 1:1. 

52 minutes ago, Tom Booth said:

In this respect it is, I think, worthwhile to consider the actual industrial process of gas liquefaction through an expansion turbine. The phase change draws on internal energy converted to "work" not heat transfer to the environment.

Why do you say this?

Partial condensation within a turboexpander reduces its performance. The phase change does not produce work, it renders some of the potential work output unavailable. It is therefore undesirable, though often unavoidable in some typical applications (eg chilling and depressuring the inlet stream to a demethaniser).

You do not strengthen your posts by pretending expertise in fields where you have limited insight.

1 hour ago, Tom Booth said:

I hesitate to make the claim that ALL these compressed air heat recovery industrialists are lying through their teeth.

I've been doing business with them for 40 years, and have no such hesitation. And not behind their backs either.

1 hour ago, studiot said:

Sadly you are still not answering my question(s)

Let me check back and see what I overlooked:

3 hours ago, studiot said:

Well perhaps you are not the only one being unclear since this post of yours describes exactly what is worrying me and I thought I had stated in my last post. Someone seems to have perhaps understood it, although your answer suggested that you did not catch it.

Is this a question? 

If it is then I've frankly no idea what you're alluding to. 

3 hours ago, studiot said:

What sort of heat transfer coefficients are you envisaging for the pipline and what about the energy flows to accomplish this ?

Again, more detailed engineering design issues than challenges to the underlying physics. It would be a complete waste of time to evaluate individual heat transfer coefficients at this stage of the process, but in general the external coefficients would be derived from the Stefan-Boltzmann Law, while the convective heat transfer coefficient for water would be estimated via the Sieder-Tate correlation. The simple conductive heat transfer coefficients would come from direct integration of Fourier's Law.

The overall heat flow in the preliminary line size I picked (48" ND 2" wall) was around 72 GW (oto half the thermal load for all installed electrical generating capacity in the UK) so not unreasonable for a planet-wide system. A back of envelope calculation indicates that this line size would shed only around 12 GW at night-time, so a practical system would have to have considerably more surface area for the same volume. 16 x 8" ND pipes in parallel may do the trick.  

2 hours ago, Tom Booth said:

Pipe rupture due to water expansion pressure rather than displacement?

In a completely rigid pipe, the water sealed between ice dams,... Lots of potential expansion and contraction issues, maybe?

If the pipe is elastic to avoid rupture, that reduces horizontal displacement leaving a long pipe that just kind of throbs around, possibly rupturing and ripping itself from its mores like a loose firehouse.

But that's just one of those practical challenges.

I had a picture flash in my mind of a Nitinol engine. The kind where a loop of Nitinol wire revolves around some wheels.

In this case, the Nitinol pipe circling the planet.

Eliminate any possibility of rupture, and enhance the displacement effect, by "training" the metal to...

Not sure what, but the super-elasticity might be of some advantage.

 

Again, these are detailed engineering design challenges rather than issues with the underlying physics. Most of the concerns you raise are grist to the mill for a competant pipeline design engineer.