A Carnot engine between the Earth's poles and equator

[artis]

I was imagining a significant delay between when(and where) the heat was stored/released, which is why I had originally considered a large heat capacity to be important. I calculated my estimates based on the ability to store a day's power output.

Assuming Conductors can flow heat, keeping hot and cold reservoirs at a fixed temperature. The reservoirs should themselves need high thermal capacity right ? (The reservoirs are heated at the equator and cooled at the poles.)

I will rethink this also however.

Well @Prathyush here is the main difference I want you to understand. Think of a central heating system within a city. You have a power/heating plant , some pipes , water in them and then the radiators in the apartments and houses. You heat the water at the plant then use pumps to circulate it and then it gives off it's heat along the pipes and in the radiators. The heat lost in the pipes is considered loss.
This is considered active heat transfer using a physical medium (water in this case) to store and transport heat/energy.

A passive heat transfer system in its purest form is basically a metal rod. The problem is this. Even if the rod is of the best heat conducting metal , if made long enough it will not be efficient in fact it will not be able to make a heat engine work at all. Imagine having a 100 metre long rod, even if you heat one end up to red hot glowing temperature the other end will be room temperature, why? Because passive heat transfer means the heat is directed within the material evenly due to random motion of atoms/molecules , this process is not lossless or ideal. As you go along the length of the rod the temperature starts to decrease gradually and if the rod is long enough at some point you won't even be able to detect the heat supplied to the other side of the rod, why? Because the heat will have managed to dissipate away along the rod.
No matter how great your insulation is , it will still dissipate.

 

[Keith_McClary]

wow a 20m circumference for a pipe , that is more like a tunnel and less a pipe.

Saudi Arabia’s Saline Water Conversion Corporation (SWCC) has completed the longest water pipeline tunnel project in the world, with a diameter of 8.4 meters and a distance of 12.5 km.
https://idadesal.org/swcc-completes-worlds-longest-water-pipeline-tunnel-project/

 

[berkeman]

https://idadesal.org/swcc-completes-worlds-longest-water-pipeline-tunnel-project/

Wow. I wonder if they coat/line the interior of the tunnel with anything to lower loss of the water into the walls of the tunnel, or if they rely on it having been bored out of mostly solid rock...

 

[Prathyush]

As you go along the length of the rod the temperature starts to decrease gradually and if the rod is long enough at some point you won't even be able to detect the heat supplied to the other side of the rod, why? Because the heat will have managed to dissipate away along the rod.
No matter how great your insulation is , it will still dissipate.

If you want to transport 1 GW of heat across a temperature difference of 10C using a copper conductor assuming perfect insulation, over a distance of 1KM. You will need a surface area of 500m*500m.

If you lower the surface area to something modest like 1m*1m over a 1 Km same 10C, you will transport about 4 watts of heat.

and area of 1m*1m over 1 Km and 1000C temperature difference, you will transport 400 watts of heat.

This is assuming perfect insulation. Conduction is really bad over long distances, even with perfect insulation.

 

[sophiecentaur]

I think turbines in the Gulf Stream might make more sense.

I used to be in favour of water turbines, driven by tides (or anything) but stuff grows on stuff under the sea and soon kills the efficiency. Annual (or more often) cleaning of underwater moving parts would be a major item, as opposed to wind turbines which are subject to much less surface coverings.

Someone already mentioned the existing thermal to mechanical Energy transfer that the weather achieves for us and efficiency is much less relevant because we don't have to build that system.

 

[sophiecentaur]

I used to be in favour of water turbines,

As it happens, I heard a BBC News item that they have just installed a tidal turbine in the North of Scotland where the tides are fierce so somebody must think it's worth while. It's not very big though. I must look for a hard copy of the info.

 

[Tom.G]

I must look for a hard copy of the info.

(with video)
https://www.bbc.com/news/uk-scotland-north-east-orkney-shetland-57991351

And from April 2021:
https://www.energylivenews.com/2021...idal-turbine-ready-to-make-waves-in-scotland/

 

[NTL2009]

I was thinking there must be a much easier way to find a cool heat sink for the solar collectors to drive a Stirling engine. How about the colder ocean water? Instead of getting your 'cool' from 10,000km away, you could go just a single km down, and find a massive heat sink at just ~ 4~5 C.

https://www.amnh.org/learn-teach/curriculum-collections/deep-sea-vents/global-ocean-circulation-and-deep-sea-temperatures

In fact, no matter how warm it is up top, by the time the sub has sunk to a depth of just 1,000 meters (3,280 feet), the water temperature is about 40ºF. From there it continues to drop until, 7,200 feet down, where the smokers are, the temperature is only about 2º to 3ºC (35ºF)—just above freezing!

And I figured I was not the first to think of this.

https://en.wikipedia.org/wiki/Stirling_engine

Heat sink
The larger the temperature difference between the hot and cold sections of a Stirling engine, the greater the engine's efficiency. The heat sink is typically the environment the engine operates in, at ambient temperature. In the case of medium- to high-power engines, a radiator is required to transfer the heat from the engine to the ambient air. Marine engines have the advantage of using cool ambient sea, lake, or river water, which is typically cooler than ambient air.

 

[FinBurger]

Saudi Arabia’s Saline Water Conversion Corporation (SWCC) has completed the longest water pipeline tunnel project in the world, with a diameter of 8.4 meters and a distance of 12.5 km.
https://idadesal.org/swcc-completes-worlds-longest-water-pipeline-tunnel-project/

What about this? https://www.water-technology.net/projects/new-york-tunnel-3/
60 miles of tunnel to funnel clean water into New York City.

 

[bob012345]

It's just a shame there is no way to capture a little of the Earth's rotational energy other than tidal power. Or is there?

 

[Vanadium 50]

It's just a shame there is no way to capture a little of the Earth's rotational energy other than tidal power.

A. That's not what this thread is about.
B. Be careful what you wish for. The Earth's rotational kinetic energy is only enough to power civilization for about 13 years.

 

[NTL2009]

...
B. Be careful what you wish for. The Earth's rotational kinetic energy is only enough to power civilization for about 13 years.

That's fascinating. I would have guessed much more. And the phrase that popped into my head was "And where she stops, nobody knows!" :) The price of real estate at the horizons would go nuts, everywhere else would be freezing or hot as Hades!

 

[FinBurger]

A. That's not what this thread is about.
B. Be careful what you wish for. The Earth's rotational kinetic energy is only enough to power civilization for about 13 years.

Rotational energy of Earth = 2e29 J
Energy consumption of humanity = 1.2E20 J/yr
Rotational energy would last for >1e9 year. (give or take - that was a really loose calculation)

 

[sophiecentaur]

Rotational energy of Earth = 2e29 J
Energy consumption of humanity = 1.2E20 J/yr
Rotational energy would last for >1e9 year. (give or take - that was a really loose calculation)

I haven’t done any sums but @Vanadium ‘s figure is hard to credit. 13 years’ energy would make loads of daft projects possible and I’ve always felt safe from such nonsensical futures. 1e9 makes me feel much safer.

 

[bob012345]

I haven’t done any sums but @Vanadium ‘s figure is hard to credit. 13 years’ energy would make loads of daft projects possible and I’ve always felt safe from such nonsensical futures. 1e9 makes me feel much safer.

I don't wish to make you feel less safe but at the current rate of energy which is about $6 x 10^{20}J$, I figure we only have 369 million years.

 

[Vanadium 50]

Energy consumption of humanity = 1.2E20 J/yr

I stand corrected. My Googling gave me J/s and not J/year, so I am off by π x 10⁷.

 

[Prathyush]

I was thinking there must be a much easier way to find a cool heat sink for the solar collectors to drive a Stirling engine. How about the colder ocean water? Instead of getting your 'cool' from 10,000km away, you could go just a single km down, and find a massive heat sink at just ~ 4~5 C.

You are right, is that is (possibly) a valid sink. It has different challenges though. 1 KM deep you are talking about 100 atmospheres of pressure.

 

[hutchphd]

And the planet keeps it cold
https://www.ontrack-media.net/gateway/science8/g_s8m3l10s5.html

 

[NTL2009]

You are right, is that is (possibly) a valid sink. It has different challenges though. 1 KM deep you are talking about 100 atmospheres of pressure.

Yes, but I'm not sure (emphasis on "not sure") that's a huge problem. 1500 PSI is pretty routine for small diameter tubing (obviously, this would need to be large for any meaningful power), and Nitrogen cylinders are about 2200-2400 PSI, and in common use. But I'm not sure that you would have that pressure difference - I'm thinking the working fluid would also be that deep, and have the same pressure in and out? In the same way that we don't crush under the ~ 14.7 PSI atmosphere.

The challenge with manned vessels is that something close to normal atmosphere pressure must be maintained inside, so there is a pressure delta. And this manned vessel went down to 10,911 meters, way back in 1960.

https://en.wikipedia.org/wiki/Trieste_(bathyscaphe)

That doesn't mean it would be practical for a Stirling Engine, but I'm thinking that pressure issue may not be a barrier.

 

[Prathyush]

obviously, this would need to be large for any meaningful power

This is a rough estimate for quantity of fluid transported want to generate 1 GW of electricity.

Assuming a heat capacity of 4KJ per Kg per degree. We have heat capacity of 200KJ over 50 degrees.
If we generate electricity at 10 pc efficiency, we need to transport 10 GW of heat.
Which amounts to transport 50000 Kg of working fluid per second.
At a density of 1000Kg/m^3 we have to transport 50m^3/s of working fluid.
With a flow velocity of 10 m/s, we get a cross section area of 5m^2.I am not sure what kind of heat exchangers would be required to absorb/dissipate heat. But if the pressure difference is too high, we need thick heat exchangers, decreasing the rate of heat exchange.
But it gives a rough idea with regard to the scale/nature of the problem.

 

[russ_watters]

I am not sure what kind of heat exchangers would be required to absorb/dissipate heat. But if the pressure difference is too high, we need thick heat exchangers, decreasing the rate of heat exchange.

@NTL2009 is correct that there is no pressure issue here. If the working fluid is water, the pressure is the same inside and outside the pipe/hx, so the pipe/hx is not under any stress.

 

[NTL2009]

@russ_watters confirmed my expectation that the pressures would be equal in/out, so this isn't a design restraint.

This is a rough estimate for quantity of fluid transported want to generate 1 GW of electricity.

Assuming a heat capacity of 4KJ per Kg per degree. We have heat capacity of 200KJ over 50 degrees.
If we generate electricity at 10 pc efficiency, we need to transport 10 GW of heat.
Which amounts to transport 50000 Kg of working fluid per second.
At a density of 1000Kg/m^3 we have to transport 50m^3/s of working fluid.
With a flow velocity of 10 m/s, we get a cross section area of 5m^2.I am not sure what kind of heat exchangers would be required to absorb/dissipate heat. But if the pressure difference is too high, we need thick heat exchangers, decreasing the rate of heat exchange.
But it gives a rough idea with regard to the scale/nature of the problem.

I didn't double check your heat capacity/flow numbers, but if correct, what kind of energy would be needed to pump that volume through ~ 2 km of pipe (down and back up)? That alone might make this an energy sink rather than source? But at least something on the order of a 5m^2 area pipe is not anything crazy (< 3 meter or 10 foot diameter). Supporting over 1 km of it in the ocean could be though!

Though the heat exchanger also might not be all that much of a challenge. The fluid would be losing heat to the ocean progressively on the trip down to that ~4~5C water. You would need to insulate the pipe on the way up, so as not to lose it back again. A 1 km pipe will have a lot of surface area, maybe just a relatively short distance of parallel paths (to increase surface area) horizontally at that 1 km depth would do the job? edit/add - OK, so at 10 m/s velocity, the 1 km uninsulated pipe would 'see' the working fluid for ~ 100 seconds. But those calculations would take me some time/research for even a rough estimate, and impossible on my first cup of coffee. I suspect this is a fairly easy calulation for someone with expertise in this area.

 

[Prathyush]

But those calculations would take me some time/research for even a rough estimate, and impossible on my first cup of coffee. I suspect this is a fairly easy calulation for someone with expertise in this area.

I suspect you can look at the density difference between hot and cold water and get an estimate.

I am not entirely clear about how density of water changes with temperature and pressure.(data is available though and it needs careful analysis)

For the purpose of estimation, I will assume density difference is 4 kg/m^3 based on https://www.open.edu/openlearn/science-maths-technology/the-oceans/content-section-3.2.
(thought it's probably a very rough estimate)

we have to transport 50m^3/s of working fluid

Assuming that estimate is not unreasonable, for a density difference of 4 kg/m^3, effectively 200 kg needs to be lifted by 1 KM every second. Which gives an estimate of 2 Mega watts of power.

( I will recheck my calculations once again though tomorrow. It feels suspiciously small. Perhaps I should worry about friction losses.)

 

[NTL2009]

I suspect you can look at the density difference between hot and cold water and get an estimate.
...
Assuming that estimate is not unreasonable, for a density difference of 4 kg/m^3, effectively 200 kg needs to be lifted by 1 KM every second. Which gives an estimate of 2 Mega watts of power.

( I will recheck my calculations once again though tomorrow. It feels suspiciously small. Perhaps I should worry about friction losses.)

Ahhh, I had not considered that the water being lifted would be cooler and more dense (by a few %, but could be a significant weight to lift at these scales), I'll try checking your numbers later.

What I was thinking of, but neglected to mention, was those friction losses - that's what I can't calculate w/o some research.

 

[gmax137]

With a flow velocity of 10 m/s, we get a cross section area of 5m^2.

Is there a basis for the 10 m/sec? In design for process piping, a value of 7 ft/sec or 2 m/sec is a typical maximum.

But, I don't know if that thumbrule applies to large concrete pipe. And it would be large: using 2 m/sec would make your area 25 m^2 (diameter 5.6 m or 18.5 ft).

 

[jrmichler]

The best way to calculate pressure drop through long pipes is by using the Moody chart (search the term). This Moody chart is from the Wikipedia article:

You can check your calculations by comparing to the following page from Cameron Hydraulic Data, 16th Edition:

Note that head loss is typically expressed as feet of head per 100 feet of pipe. One foot of head is the pressure from a column of liquid one foot high. Multiply feet of head by density in lbs/ft^3 to get pressure in lbs per ft^2.

It's a very important calculation to get right in OTEC (search the term) conceptual calculations.

 

[Prathyush]

It's a very important calculation to get right in OTEC (search the term) conceptual calculations.

Thank you. A recent paper https://www.mdpi.com/2077-1312/9/4/356 does a review of the technology and possible implementations.

One sentence caught my attention "In a closed-loop OTEC system, the highest cost of implementation lies in the heat exchangers. These devices are responsible for transferring heat from one place to another,
and are separated by a contact wall."

The best way to calculate pressure drop through long pipes is by using the Moody chart

Thank you again, I will make basic calculations estimating pumping energy cost for various configurations.(in a bit)

Is there a basis for the 10 m/sec?

It was an arbitrary choice based on vague impressions.

 

[Prathyush]

At a density of 1000Kg/m^3 we have to transport 50m^3/s of working fluid.

This estimate is not reasonable for a pure OTEC system (where the only heat source is surface warm water. if heat can be harvested from another source and $\Delta T$ is increased, efficiencies can be increased)

Extracting heat from small thermal gradients is a technical challenge. The following paper reviews the topic.
(A review of research on the closed thermodynamic cycles of ocean thermal energy conversion).

I don't understand the details, but it seems typical efficiencies start at around 2.4pc for the simpler cycles and reaches upto 5.1 pc for more complicated multi component cycles. These are theoretical results, only recently obtained.

Assuming a heat capacity of 4KJ per Kg per degree. We have heat capacity of 200KJ over 50 degrees.
If we generate electricity at 10 pc efficiency, we need to transport 10 GW of heat.
Which amounts to transport 50000 Kg of working fluid per second.
At a density of 1000Kg/m^3 we have to transport 50m^3/s of working fluid.
With a flow velocity of 10 m/s, we get a cross section area of 5m^2.

To revise this estimate, We assume efficiency of 2.5 pc and $\Delta T = 20^{\circ}$, volume flow rate increases by 10 times to 500 m^3/s. And if we double the efficiency to 5 pc, it will be 250 m^3/s.

If we take the previously suggested 2 m/s flow rate, we get we get a pipe area between 125-250 m^2(6-9m radius). These calculations are for 1 GW of power generated before pumping costs.

 

[Prathyush]

I ran the numbers for a Carnot engine running between poles and equator and estimate diameter of the tunnel after accounting for pumping costs.

Calculation is done for an engine generating 1 Terawatt power, running at a temperature difference of 90 degrees, operating between 5 and 95 degrees. We assume 20 pc of the energy generated is used for pumping.

I will write down all the formulas so that it can be checked.

For the calculation I will assume a friction factor of .01. (The actual friction factor is slightly lower, which we will calculate at the end. We are overestimating friction by ~30pc, to leave some headroom)

$$ f = .01$$

The head loss per unit length is given by the formula

$$\frac{\Delta p}{L} = \frac{f \rho v^2}{2 d} $$

We assume the engine works at efficiency at peak power.( $T_h = 95$ and $T_c = 5$)

$$ \eta = 1- \sqrt{\frac{T_c}{T_h}} = .131$$

The potential to do work per Kg of water is ( c = specific heat of water )

$$ w = c \eta \Delta T = 49300 J/Kg$$

A fraction of the energy $p_f = .2$ is used for pumping. We assume pumping efficiency is 75 pc. In one cycle work can be extracted twice from the same mass, one at the pole when cooling, and another at the equator when heating, we will assume it is identical. ($h_{head} = \frac{p_f w}{g}$, L = 10000 Km} )

$$ \frac{\Delta p}{L} = .75 \frac{p g h_{head} }{L} = .739 pascal/m $$

mass flow rate M to generate 1 Terawatt of power is $M = \frac{10^{12}}{w (1-p_f)} = 2.5 \times 10^7 kg/s$ and Volume flow rate $Q = \frac{M}{\rho} = 25\times 10^3 m^3/s$

The formula $$\frac{\Delta p}{L} = \frac{f \rho v^2}{2 d} = \frac{ 8 f \rho Q^2 }{\pi^2 d^5}$$ can be inverted to get the diameter

$$d = \sqrt[5]{\frac{ 8 f Q^2 \rho }{ \pi^2 (\Delta P/L) } } = 93 m$$

velocity of flow is $v= \frac{Q}{\pi d^2} = .92 m/s$

We use Colebrook-White equation iteratively to estimate the friction factor for these flow parameters, with surface roughness $e = .25 mm$ and reynolds number $Re = 3.9 \times 10^8$. We get $f = 0.0067$ In principle the calculation can iterated over, and if that is done we get $d = 86m$. But some head room may be useful.

I will repeat the same for other cases of interest for different kinds of cases and make general comments on a separate post. But is the general method OK for the problem as stated?

 

[jrmichler]

Looking at some of your assumptions, but not your equations...

Per NOAA, the average deep (below 200m) ocean temperature at the North pole is about 4 deg C, and the temperature near the equator is 30 deg C. There will be a temperature difference across the heat exchanger walls. The ocean temperature will vary as it travels through the heat exchanger, while the temperature of the condensing/boiling fluid will be roughly constant. The temperature for the Carnot efficiency is the hot and cold temperatures of the working fluid, not the incoming temperatures of the ocean water. The ocean water temperature leaving the condenser will be colder than the working fluid temperature, and the ocean water temperature leaving the evaporator will be warmer than the vapor temperature.

The average temperature difference across the heat exchanger is the average of the entering and leaving temperature differences. Increasing water flow reduces water temperature rise/fall, while increasing pumping power. Decreasing average temperature difference across the heat exchanger requires a combination of increased water flow and increased heat exchanger area. Pumping power must be calculated for the ocean water lines plus the heat exchanger pressure drop.

OTEC system design is an iterative process. Without fully thinking it through, here's how I would approach this problem:

1) Assume initial values for cold side ocean temperature and depth, warm side ocean temperature and depth, distance between the two, and cold and warm side working fluid temperatures. Assume a design for the heat exchanger, but not the heat exchange area.

2) Using the numbers from above, calculate the required heat exchanger size for different water flow rates. You will need to calculate the heat transfer coefficients.

3) Calculate water pressure loss in the pipelines and heat exchanger on both the hot and cold sides. This is a good place to assume so pipe diameters, and do the calculation for each. There will also be pressure losses in the working fluid.

4) You now have flow rates and pressure drops on the water sides. Calculate pumping power.

5) You now have hot and cold side temperatures of the working fluid at the engine after allowing for the pressure drops between the heat exchanger and engine. Calculate the Carnot efficiency, then search for actual efficiencies as a percentage of Carnot efficiency for this temperature range.

6) Show the results in a table or tables - temperatures, flow rates, pipe diameters, heat exchanger areas, pumping power, engine power at output shaft, and net output power.

7) Enjoy - it looks like a fun project.