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Martian Hydroelectric Concept


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Snow in the mountains imperceptibly melting, added together results in a raging river downstream.

I can imagine something similar taking place in the pipe, where there is little action to be seen near the extreme interfaces but a considerable surge in between.

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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.

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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. 

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7 hours ago, sethoflagos said:

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

 

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.

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.

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

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.

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 ?

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10 hours ago, sethoflagos said:

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

IOW, you are in a steady-state condition.

10 hours ago, sethoflagos said:

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.

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

10 hours ago, sethoflagos said:

It's under operator control.

I seriously doubt that.

10 hours ago, sethoflagos said:

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. 

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

 

10 hours ago, sethoflagos said:

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

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

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

 

 

 

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Posted (edited)
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).

Edited by sethoflagos
small clarification
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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 ?

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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

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3 hours ago, sethoflagos said:

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

No ridicule, but I'm pretty sure I didn't say your second line as you have rephrased it.

 

I'm also pretty sure you fully understand the difference between power and energy.

 

So what I said was that the solar energy incident on that area over a period of one second was eactly the amount you need to thaw a cubic metre of ice, if you could collect all of it and add it all to that ice.

 I was temporarily ignoring the inefficiencies of collection, distribution, storage etc

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.

Given my own knowledge of this sort of thing from deicing of bridge structures I am doubting the practicality of that operation.

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Posted (edited)
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.

Edited by sethoflagos
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40 minutes ago, sethoflagos said:

Thinking about it,

 

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

42 minutes ago, sethoflagos said:

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

Because your ice/water interface must have moved on many metres in that one second, depending upon the interface speed you take.

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.

Also if you have used all the solar energy in your 100m strip for thawing the ice, how are you going to find the power to pump stuff around ?

Finally all this area of solar collector, pipeline, pumps, controls, turbines, output electricl gear and storage must weigh millions of tonnes.
How are you going to get it all there ?

 

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4 hours ago, sethoflagos said:

Argument from incredulity? 

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

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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)

 

 

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1 hour ago, sethoflagos said:

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.   

You have yet to demonstrate there will be a freeze - thaw.

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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.

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1 hour ago, studiot said:

You have yet to demonstrate there will be a freeze - thaw.

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.

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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.   

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33 minutes ago, sethoflagos said:

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.

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.

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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?

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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.

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