3 ms·

The ‘sand battery’ explained: Insulate a lot of sand and heat it with an electrical element (300-500°C). When you need the energy back, pump air into it and you

The ‘sand battery’ explained: Insulate a lot of sand and heat it with an electrical element (300-500°C). When you need the energy back, pump air into it and you get hot air to power generators and heat buildings.

koheripbal 4y ago
It's important to remember with most of these thermal batteries is that they are limited in efficiency by the laws of thermodynamics, both when putting the energy in and then again pulling it back out. The net round trip total power loss is usually well above 50%.
- einpoklum 4y ago
  I'm no thermodynamicist, so I can only take you at your word about the power loss. But still: * Suppose you can effectively and consistently get 40% of the energy you put into this kind of battery. * Suppose you can make enough sand batteries to cover your overall power consumption (this is a big ask, but for the sake of discussion let's assume we can achieve this). * Suppose that you have an energy source that's non-uniform across the day. For simplicity suppose that it produces 2N MegaWatts for 12 hours and 0 MegaWatts for 12 hours, i.e. N MegaWatts on average. * Assume that electric power consumption, not including the batteries, is uniform across day and night (It isn't; it would actually be lower when power is generated if we're talking about solar). * suppose we want a uniform power source when considering the batteries. ------ Now, let delta be the fraction of our power output which we divert towards batteries, which are only charging during the 12 hours in which that is possible. Solving for uniformity, fraction not diverted to batteries = fraction recovered from batteries (1 - delta) = 40% \* delta 1 = 1.4 \* delta delta = 1/1.4 = 5/7 ~= 0.7143 1-delta ~= 0.285 So, you get a power source that's uniform over the entire day at 0.285 * 2N = 0.55 * N MegaWatt, i.e. over half of your average power output of the unstable source. That's not great, but it's pretty damn good!
- ktzar 4y ago
  From other articles, the key is that, if the energy recovery happens during cold months and the residual energy in heat form is used to feed municipal heating networks, efficiency can be up to 90%.
- ClumsyPilot 4y ago
  Thats for electrical power, if all you want to do is extract heat, you only major loss is thermal cooling over time and in the pipes. You should be able to get most of your energy back from a well-insulated system
  - s1artibartfast 4y ago
    Exactly, for just storing heat the theoretical limit is 100%. Imagine putting hot water in a insulated thermos when you have some, waiting 6 months, then pouring it in your bathtub when you do want it. Nothing is lost

koheripbal 4y ago

It's important to remember with most of these thermal batteries is that they are limited in efficiency by the laws of thermodynamics, both when putting the energy in and then again pulling it back out. The net round trip total power loss is usually well above 50%.

einpoklum 4y ago
I'm no thermodynamicist, so I can only take you at your word about the power loss. But still: * Suppose you can effectively and consistently get 40% of the energy you put into this kind of battery. * Suppose you can make enough sand batteries to cover your overall power consumption (this is a big ask, but for the sake of discussion let's assume we can achieve this). * Suppose that you have an energy source that's non-uniform across the day. For simplicity suppose that it produces 2N MegaWatts for 12 hours and 0 MegaWatts for 12 hours, i.e. N MegaWatts on average. * Assume that electric power consumption, not including the batteries, is uniform across day and night (It isn't; it would actually be lower when power is generated if we're talking about solar). * suppose we want a uniform power source when considering the batteries. ------ Now, let delta be the fraction of our power output which we divert towards batteries, which are only charging during the 12 hours in which that is possible. Solving for uniformity, fraction not diverted to batteries = fraction recovered from batteries (1 - delta) = 40% \* delta 1 = 1.4 \* delta delta = 1/1.4 = 5/7 ~= 0.7143 1-delta ~= 0.285 So, you get a power source that's uniform over the entire day at 0.285 * 2N = 0.55 * N MegaWatt, i.e. over half of your average power output of the unstable source. That's not great, but it's pretty damn good!
ktzar 4y ago
From other articles, the key is that, if the energy recovery happens during cold months and the residual energy in heat form is used to feed municipal heating networks, efficiency can be up to 90%.
ClumsyPilot 4y ago
Thats for electrical power, if all you want to do is extract heat, you only major loss is thermal cooling over time and in the pipes. You should be able to get most of your energy back from a well-insulated system
- s1artibartfast 4y ago
  Exactly, for just storing heat the theoretical limit is 100%. Imagine putting hot water in a insulated thermos when you have some, waiting 6 months, then pouring it in your bathtub when you do want it. Nothing is lost