Interseasonal heat storage / sand battery insulation?

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  • #1
PaulMuadDib
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Hi there,
I am not a thermal sciences specialist, and I try go get an idea of the insulation required to retain heat in a sand battery for a few months.
I have a small pool (4 x 2 x 1.5 m) and I am questionning myself about two options: either converting this pool into a water storage or into a heat storage (heating sand during summer).

I have tried a very simple modelling (differential equation) to get an idea of the needed insulation for this interseasonal heat storage:

dQ/dt = m [ Cp + (T-Tex) dCp/dT ] dT/dt = - S phi = - S/R (T - Text)

where m is the mass (kg) of sand, Cp is the heat capacity (J/kg.°K) of sand (varying with temperature, capturing the quartz inversion), S is the surface of the sand volume (to be insulated), R is the thermal resistance (R value in °K.m²/W) of the insulator (I take one of the best: lambda = 0.025 °K.m²/W for polyurethane).

I am very surprised to see that I would need 75 cm thick polyuretane walls (R-value of 30 !) to get enough heat remaining after 6 months (meaning 400°C left or 35 % of the thermal energy left, starting from 1000°C), knowing these walls would not stand these high temps: is this realistic ? (for the R-value & the modelling) If not, how should I model this ?

1712508427519.png
 

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  • #2
I did not check your analysis, but your calculations seem right. I have done similar calculations with different assumptions. My assumptions generally had heat energy stored in a volume underneath a house, where the volume had the dimensions of the house and 2 to 3 meters deep. Even with extremely good insulation in the house, there was not enough heat stored to be practical.

On a separate project, I did the calculations for water heat storage for a friend. He had a four zone hydronic system with a gas boiler and a wood boiler. He wanted enough heat storage to be able to build a fire in the evening, and keep all four zones at the proper temperature until the following evening without using the gas boiler. With an 1100 gallon water tank, the system did just that down to outside temperature of 0 degrees F. At lower temperatures, a small fire in the morning added just enough heat to keep it warm until evening.

A more complete analysis will estimate the cost of your proposed heat storage system, plus estimate the value of the stored heat available to heat the house. Then take that estimated cost, put it into improved insulation in the house, and estimate the value of the heat saved. Then compare.
 
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  • #3
PaulMuadDib said:
starting from 1000°C
Is there a particular reason you chose this temperature? It seems unreasonably hot to me for what appears to be personal thought experiment. I only say this as, not only will your "polyuretane" walls melt, but so will many metals in this list by engineering toolbox.

Also, how would you heat your 4 cubic meters of sand up to 1000°C? With some type of piping filled with molten salts? Or are you just using hot air?
 
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  • #4
Hello. These 1000°C are required to reach a sufficient heat amount (1800 kWh) initially, to get a minimum of 600 kWh 6 months later, but this will not be praticable (heat resistance of the tubular heat exchanger / TBE within the sand, dilatation, volume of the insulator, price,...). I don't know what is the Polar night heat exchanger which can operate well over these 600°C, but I cannot afford it. The polyurethane would have been the outer (and best) insulating layer, behind a rockwool layer, itself behind a protective fire brick wall (I should compute the temperature profile within this 3 layered insulator). The same TBE would both circulate super hot air (to heat the sand reservoir), heated in closed loop by a Serpentine (Tutco) tubular resistance heater and would also circulate fresh air to heat it up for the house. If my calculations are correct (not sure, not my field) then obviously, I have to reduce my goals. I can store a sufficient heat (here it was ~1.8 MWh) in may, june, july and maintain it in august, september and october with the remaining solar excess. If I need this heat in december, january, it's more like 3 months later (not 6). If I heat those 4 m^3 of sand to 600°C (~1 MWh), 3 months later I still have 460 kWh of heat with a R-value of ~20 with a 3 layer insulation of 10 cm, 25 cm and 25 cm of fire brick, rockwool and polyurethane (still meaning 0.8, 2.5 and 3.25 m^3 of these materials !)
1712682869610.png
 
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  • #5
I think a much simpler method would be to drill a 30 meter deep well for seasonal thermal storage and use your pool filled with water for daily thermal storage, similar to the way the Drake Landing system is designed. Since they use a glycol/water solution to transport heat, I'm guessing their maximum operating temperature is not too much above 100°C.
 
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  • #6
PaulMuadDib said:
600 kWh 6 months later
Is this just a fun exercise or are you planning on building the system? It seems impractical to me, in the sense that in addition to the equipment cost, you need to put 1800 kWh in (to meet your stated:
35 % of the thermal energy left

Where I live, my electric service is $0.115/kWh, so 1800 kWh costs me $210; while I could just wait and buy the 600 kWh later for $70. Or, are you somehow able to input that 1800 kWh off-grid from another source? I can't imagine what you will use to heat the sand to 1000C.
 
  • #7
Your heat balance equation is incorrect. It should just have ##mC_p(T)\frac{dT}{dt}##. Enthalpy is $$H=\int_{T_0}^T{C_p(T')dT'}$$where T is a dummy variable of integration. The time derivative of this is $$\frac{dH}{dt}=C_p(
T)\frac{dT}{dt}$$
 
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  • #8
A lot of heat is lost by radiation, which is not taken into account.
 
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  • #9
@Chestermiller Thank you for the correction of the time derivative !
@Demystifier I would have to add a radiative term $$\sigma e S (T^4 - T_{ext}^4)$$ where T and e are the temperature and the emissivity of the outer layer of the insulator ? (polyurethane). It seems indeed huge: for S = 15 m² and T = 40°C ; T_ext = 10°C, and sigma = 0.88, this radiative loss would be 2.4 kW ?! :oops:
 
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  • #10
gmax137 said:
Is this just a fun exercise or are you planning on building the system? It seems impractical to me, in the sense that in addition to the equipment cost, you need to put 1800 kWh in (to meet your stated:


Where I live, my electric service is $0.115/kWh, so 1800 kWh costs me $210; while I could just wait and buy the 600 kWh later for $70. Or, are you somehow able to input that 1800 kWh off-grid from another source? I can't imagine what you will use to heat the sand to 1000C.

This an exploration of feasability (technical, economical) of a basic thermal solution storage, but I am not a specialist in this area !
I have a small pool I will no longer fill (water becomes scarce here, I am at 15 min from the sea).
In winter, I use a heat pump and a small wood stove (december, january, part of febuary) but I would like to get rid off this last source of CO2 in my house energy.
Being autonomous in energy (PV +LFP storage), I also have a large PV surplus (3 MWh / year) starting from march to october) and I cannot inject/sell on the grid (my installation is off grid).
Hence my questionning about storing these MWh in excess in my piscinetta (4 x 2 x 1.5 m) for the heat of the winter.
I am also looking to paraffin / wax (for its latent heat energy stored in the solid <--> liquid phase transition) to operate at lower temperatures, but this is more costly than the sand or the water ! :rolleyes:
 
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  • #11
PaulMuadDib said:
@Chestermiller Thank you for the correction of the time derivative !
@Demystifier I would have to add a radiative term $$\sigma e S (T^4 - T_{ext}^4)$$ where T and e are the temperature and the emissivity of the outer layer of the insulator ? (polyurethane). It seems indeed huge: for S = 15 m² and T = 40°C ; T_ext = 10°C, and sigma = 0.88, this radiative loss would be 2.4 kW ?! :oops:
It has to be huge, that's how radiators make our houses warm.
 
  • #12
Demystifier said:
It has to be huge, that's how radiators make our houses warm.
The radiation is in series with the insulation, not in parallel. In my judgment, the convective heat loss will be much higher than the radiative heat loss.
 
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  • #13
Chestermiller said:
The radiation is in series with the insulation, not in parallel. In my judgment, the convective heat loss will be much higher than the radiative heat loss.
If the heat storage is underground, can I at least hope that convective heat losses are limited ? 🫣
The radiative heat loss already kills my hope in fact... I'm gonna heat the ground even with a good thermal insulator 🙄 Unless I can shield radiative losses in an economical manner ? In the same way I have insulated my water heater tank ? (metallized foils separated by a layer of polypropylene). Do the emissivities multiply, considering a serie of metallized foils ?
 
  • #14
PaulMuadDib said:
Hello. These 1000°C are required to reach a sufficient heat amount (1800 kWh) initiall
With a half meter wall of polyurethane completely encasing, and a 100 kW constant power source to heat the system, I'm getting at least 5000 kWh. The less input power you use the worse that gets. This doesn't include the thermal energy absorbed and stored by the ( rather massive) wall. I assume linear thermal gradients across the wall at all times ##t##, and unform temperature of sand.

I agree with @Chestermiller that radiation is going to be very small.

It's really a simplified model and its already too complex.

$$ q - h A_s ( T_s - T_{\infty} ) = M c_p \frac{dT}{dt} $$

##q## heat rate input ( kW)
##h## convection coefficient of surroundings air
##A_s## exterior wall surface area
##T_s## wall surface temp.
##T_{\infty}## enviroment/surroundings temp.
##T## is the instantaneous sand temp.
##M## mass of sand
##c_p## specific heat
##L## is wall thickness
##k## thermal conductivity of wall matherial
##\beta## is a time constant ## = \frac{h A_s}{ M c_p \left( \frac{Lh}{k} + 1 \right) }##


Solution (under stated assumptions) and 100 kW input:

1712853923454.png



Here are the parameter values:
1712854571107.png
 
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  • #15
PaulMuadDib said:
This an exploration of feasability (technical, economical) of a basic thermal solution storage, but I am not a specialist in this area !
I have a small pool I will no longer fill (water becomes scarce here, I am at 15 min from the sea).
In winter, I use a heat pump and a small wood stove (december, january, part of febuary) but I would like to get rid off this last source of CO2 in my house energy.
Being autonomous in energy (PV +LFP storage), I also have a large PV surplus (3 MWh / year) starting from march to october) and I cannot inject/sell on the grid (my installation is off grid).
Hence my questionning about storing these MWh in excess in my piscinetta (4 x 2 x 1.5 m) for the heat of the winter.
I am also looking to paraffin / wax (for its latent heat energy stored in the solid <--> liquid phase transition) to operate at lower temperatures, but this is more costly than the sand or the water ! :rolleyes:
So stop me If I've figured something wrong but you have a surplus of 3MWh over the span of 7 month ( 210 days). That means you have an average available input power of:

$$\bar{q} = \frac{ 3 \cdot 10^6 ~\text{W h}}{210~ \text{day}}\cdot \frac{1 ~\text{day}}{ 24 ~\text{h}} \approx 0.6 ~\text{kW} $$

Here is what the graph looks like for 0.6 kW.

1712868936423.png


Notice the asymptote at about 235 C. I wouldn't start digging yet...
 
  • #16
PaulMuadDib said:
Thank you for your simulation @erobz ! My own simulation was rather with 4 m^3 of sand (and the quantity of insulation for such a volume is already dissuasive !). No conduction loss in your differential model ? (although some conduction terms like L and k appear in your solution & time constant beta). If it is an underground storage, is this convection part significant versus the others ? How did you choose the h value ? and what about the T_ value ?
I have estimated the temperature profile within the 3 layers insulation (completely following the electrical analogy): I compute the heat flux ##\phi=\frac {A}{R}(T-T_{ext})##, where R is the thermal resistance of the whole insulation: ##R=\frac {L_1}{\lambda_1}+\frac {L_2}{\lambda_2}+\frac {L_3}{\lambda_3}## (the 3 resistances in serie), and then, assuming the heat flux is constant through the 3 layers, for the first layer (fire brick) ##T(x)=T-\frac {R_1\phi}{A L_1}x## giving ##T_1=T-\frac {R_1 \phi}{A}## (at the interface of layer 1 and 2), then for the 2nd layer (rock wool): ##T(x)=T_1-\frac {R_2\phi}{A L_2}(x-L_1)## and ##T_2=T_1-\frac {R_2 \phi}{A}##... And the same for the last layer (polyurethane). I get this temperature profile (conduction only):

View attachment 343238

Doing so, I naturally get a surface temperature ##T_3## (##T_S## in @erobz 's model) equal to ##T_{ext}=10°C## the underground temperature (##T_{\infty}## in @erobz 's), such that the radiation and convection terms are both zeroed.
Should I do otherwise to compute the temperature profile in my 3 layers insulation ? Is it normal to assume that the exterior surface is at equilibrium (##T_3=T_{ext}##) with the environment ?
 
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  • #17
First simulation (conduction only) for paraffine/wax. With 4 ##m^3## of wax heated to 150°C, I initially store 550 kWh, I get 350 kWh 3 months later (63 % remaining, thanks to 210 kWh stored as latent heat) with a much more reasonnable insulation (no fire brick, 15 cm of rock wool and 15 cm of polyurethane). I had to model the wax apparent ##C_P## as $$C_P(T)=aT+ b + \frac{c}{d(T-T_{melt})^2+1}$$ (with the values of a, b, c and d as on the excel sheet below). This would amount to 200 kJ/kg of latent heat. And this would be much more encouraging than the sand (lower temperatures, far less insulation, plateau in the cooling thanks to the latent heat)

1712928144046.png
 

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  • #18
Thank you for your simulation @erobz ! My own simulation was rather with 4 m^3 of sand (and the quantity of insulation for such a volume is already dissuasive !). No conduction loss in your differential model ?

The effects of conduction are hidden in ##T_s##. Because I'm ignoring the heat absorption by the wall itself, conduction is just an intermediary step to convection. My surface boundary condition (power balance) is given by:

$$ h A_s ( T_s - T_{\infty}) = k A_s \frac{T - T_s}{L} $$

That is solved for ##T_s## in terms of the sand temp ##T##, and subbed into the ODE.

If it is an underground storage, is this convection part significant versus the others ?
I'm not sure computationally how to handle underground burial exactly, but one thing is for sure it would certainly lower the convection coefficient. and better shielded from most losses. You will still inevitably heat up the surrounding ground, which will be conducted to the surface at a much slower rate.

How did you choose the h value ?

It's a typical value for convection to air with a mix of free and forced convection. It's not precise, its practical.

and what about the T_ value ?

##T_{\infty}## is ambient temperature of surroundings at the time of heat input( when you are heat up the sand ). My equation does not describe the slow release of heat over time once its already hot. I shared it to point out some of the practical limitations with trying to heat this sand to ##1000 C## with ## 0.6 kW## surplus you have available (at least above ground).

As for the draining phase, you might be tempted to set ## q = 0 ##, but you will arrive at ##T(t) = T_{\infty}## if you do. The draining phase will be a different result that needs its own analysis. Furthermore I was only trying to look for analytical results. If you want to pursue the variable ##c_p(T)## , you are going to need numeral solution.

Is it normal to assume that the exterior surface is at equilibrium (##T_3=T_{ext}##) with the environment?

No if they are equal, the system and its surroundings are in thermal equilibrium. Its ideal, but it unrealistic (there would be no loss of heat to the surroundings). The convection/conduction coefficients ##h,k## drives the temperature difference between the exterior wall and its surroundings.
 
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  • #19
erobz said:
The effects of conduction are hidden in ##T_s##. Because I'm ignoring the heat absorption by the wall itself, conduction is just an intermediary step to convection. My surface boundary condition (power balance) is given by:

$$ h A_s ( T_s - T_{\infty}) = k A_s \frac{T - T_s}{L} $$

That is solved for ##T_s## in terms of the sand temp ##T##, and subbed into the ODE.


I'm not sure computationally how to handle underground burial exactly, but one thing is for sure it would certainly lower the convection coefficient. and better shielded from most losses. You will still inevitably heat up the surrounding ground, which will be conducted to the surface at a much slower rate.



It's a typical value for convection to air with a mix of free and forced convection. It's not precise, its practical.



##T_{\infty}## is ambient temperature of surroundings at the time of heat input( when you are heat up the sand ). My equation does not describe the slow release of heat over time once its already hot. I shared it to point out some of the practical limitations with trying to heat this sand to ##1000 C## with ## 0.6 kW## surplus you have available (at least above ground).

As for the draining phase, you might be tempted to set ## q = 0 ##, but you will arrive at ##T(t) = T_{\infty}## if you do. The draining phase will be a different result that needs its own analysis. Furthermore I was only trying to look for analytical results. If you want to pursue the variable ##c_p(T)## , you are going to need numeral solution.



No if they are equal, the system and its surroundings are in thermal equilibrium. Its ideal, but it unrealistic (there would be no loss of heat to the surroundings). The convection/conduction coefficients ##h,k## drives the temperature difference between the exterior wall and its surroundings.
I understand your model with the surface boundary condition, thank you @erobz. Solving it for ##T_s##, I get ##T_s=\frac {T + R.h\ T_{\infty}}{R.h + 1}## and if the product ##R.h## (##=\frac{L h}{k}##) is large (with your values ##R.h=2500##) then ##T_s - T_{\infty} \approx \frac{T}{R.h} = 0.4°C## ? (@##T=1000°C##).
If ##T_{\infty}=10°C## and ##T_S \approx 10.4°C## then the loss by conduction/convection would be ##\frac{A_s}{R}(T-T_s) \approx 590\ W## and the loss by radiation ##\sigma\ e\ A_s(T_s^4-T_{\infty}^4) \approx 30\ W## ?
 
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  • #20
PaulMuadDib said:
I understand your model with the surface boundary condition, thank you @erobz. Solving it for ##T_s##, I get ##T_s=\frac {T + R.h\ T_{\infty}}{R.h + 1}## and if the product ##R.h## (##=\frac{L h}{k}##) is large (with your values ##R.h=2500##) then ##T_s - T_{\infty} \approx \frac{T}{R.h} = 0.4°C## ? (@##T=1000°C##).
If ##T_{\infty}=10°C## and ##T_S \approx 10.4°C## then the loss by conduction/convection would be ##\frac{A_s}{R}(T-T_s) \approx 590\ W## and the loss by radiation ##\sigma\ e\ A_s(T_s^4-T_{\infty}^4) \approx 30\ W## ?
Seems reasonable to me. Just note that if you’re including radiation it would be a part of the surface balance, and a part of the ODE. However, it serves to compare order of magnitude difference between convection and radiation.
 
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  • #21
For the cooling phase what are the average temperatures the unburied sand battery would experience?
 
  • #22
erobz said:
For the cooling phase what are the average temperatures the unburied sand battery would experience?
Over the last year, the annual average was 16°C (in a shaded box outside)
1713098711799.png
 
  • #23
PaulMuadDib said:
Over the last year, the annual average was 16°C (in a shaded box outside) View attachment 343381
You are trying to store heat over the winter months though? Visually maybe 5 C?
 
  • #24
erobz said:
You are trying to store heat over the winter months though? Visually maybe 5 C?
Sorry @erobz, I misunderstood your question, the cooling phase will start from end of october to early febuary: mean temperature is 9.9°C during this period (while my underground garage stays around 13°C in a very stable manner from november to febuary)

1713107243447.png
 
  • #25
@erobz I have added the convective model (with h = 100 W/m²K, trying to find litterature about underground convection) in my excel sheet... it basically creates a small difference between the outward wall surface and the underground environment, and enables a small loss by radiation (significant though, on several months). If my model of latent heat for paraffin is correct, it seems far more interesting to melt wax in the period of solar excess ! (lower operationnal temperatures, less losses, far less insulation, better ratio of remaining energy than sand thanks to the latent heat)
 
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  • #26
A nice article on seasonal heat energy storage that is actually being built: https://cleantechnica.com/2024/04/1...rmal-battery-for-district-heating-in-finland/. It is presently a design on paper, but according to the article: Construction of the storage facility’s entrance is expected to start in summer 2024, and the seasonal thermal energy storage facility could be operational as early as 2028.

The scale of the project is enormous. There will be three chambers buried 100 meters (330 feet) underground. Each chamber will measure 300 meters (984 feet) in length, 40 meters (131 feet) high, and 20 meters (66 feet) wide. The combined volume of all three chambers, including processing facilities, will be 1.1 million cubic meters (39 million cubic feet). When completed, Veranto will be the largest thermal energy storage system in the world.

The hot water will be stored in three massive underground chambers buried deep underground. Just like the cooling system in a conventional car, the system will be pressurized to allow the temperature of the water to reach 140º C (284º F). Vantaa Energy says the completed system will store up to 90 MWh of energy — enough to heat a medium size city like Vantaa for a year and making it the largest thermal energy storage system in the world.

The temperature of the water going to buildings is between 80 and 115° C (175 to 240º F), depending on the weather. In the return pipe, the water that has released its thermal energy is typically around 35 to 50° C (95 to 112º F).


Some of the numbers make me wonder. When I calculate the thermal energy in water with the temperature differences in the article, I get a lot more than 90 MWH.

The discussion in the thread so far has shown that a small scale system has large heat loss compared to the amount of usable heat. The ratio of heat loss to usable heat reduces as the system gets larger because the heat stored is proportional to the cube of the system dimensions, while heat loss is proportional to the square of the system dimensions. This is an example of the cube-square law.
 
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  • #27
jrmichler said:
Some of the numbers make me wonder. When I calculate the thermal energy in water with the temperature differences in the article, I get a lot more than 90 MWH.
Yeah, me too. 39 million ft3 is about 2 billion pounds of water. Taking the averages of the stated supply and return temperatures I get a change of 104F. That works out to about 60,000 MWh.

Plus, I don't think the stated 90 MWh is enough to keep a city warm.

Also, 300 x 40 x 20 x 3 is 720,000 m3 (25 million ft3).

My wife makes fun of me 'cause I read/watch the news with a calculator in hand.
 
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  • #28
The very first comment in the comments section of the cleantechnica article was someone pointing out about 90MWh being wrong. The writer of the article admitted to the error, and the article now lists the corrected value:

Vantaa Energy says the completed system will store up to 90 GWh of energy...

Apparently, it was a simple transcription error from the webpage of the company building the battery: https://www.vantaanenergia.fi/en/carbon-negativity-2030/heatstorage/


gmax137 said:
That works out to about 60,000 MWh.

I'm guessing the extra 30,000 MWh comes from the surrounding stone.
 
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