In addition to my sibling comment: The cost of the panels is a rather small fraction of the total cost of a typical installation. Most of that cost ist labor, some regulatory requirements and the inverter. Whether you pay a factor of 2 for the panels or not typically doesn't matter. In other words: Reusing used panels will only ever be able to safe you a minuscule amount.
IDK sounds like you got ripped off. I diy'd and panels were cheap of course, but fittings were perhaps 3-5x cheaper. Inverter is typically same as your panels (hybrid, grid-tied are quite a bit cheaper).
For all of your context/reference, if you buy whole pallets from a central European port warehouse, glass-glass modules run around $0.11/Wp plus shipping.
Unless you're just bolting them to the floor or to an uninsulated wall, mounting will (sadly) run you a sizable fraction of that cost in the best case.
Maybe, but these aren’t fittings, they’re ground mounts with large screws that screw into the ground to hold the entire array down, including under high wind (and have to come with PE stamped system-level engineering drawings talking about things like rated wind load of the whole array to pass building inspections).
But yeah, at the end of the day, just bent bars of aluminum with ground screws and bolts to hold the corners of the panels, versus the technological marvels of the solar panels they hold.
These days it’s a stack of microinverters. Which are not cheaper but do improve array efficiency outside of idea conditions. But that’s another up front cost.
The low cost of the modules themselves has led to the suggestion of cost optimized DC-coupled PV systems being used to directly drive resistive heaters. The cost per unit of thermal energy in a cost optimized system moderate scale system (> residential, < utility scale) may be in the range of $3-5/GJ, very competitive with natural gas. Low cost maximum power point trackers would be useful; inverters would not be needed.
Low cost modules allow one to do away with things like optimally tilted modules and single axis tracking. The modules can also be tightly packed, reducing mounting and wiring costs.
What's the proposed system design? For example, in January, I get about 9 hours of sunlight and have an average daily high of 25 F. I'm gonna need to store heat somehow or another.
The place I saw this most clearly described was in Standard Thermal's concept, which will store the heat in huge piles of dirt heated to 600 C. The thermal time constant of such piles can be many years.
The surface will always be only slightly hot. Heat will be stored inside, insulated by overlying dirt. Dirt isn't the best insulator by thickness, but it's a very good insulator by $.
If we assume the delta of 550 degrees (600 down to 50), you'll need: 7.913×10^10 J / (550K * 1000Jkg^-1K^-1) = 143,872,727 kg of material in your pile. This is a ridiculously stupid number. And I don't see any obvious mistakes?
I use units(1), which also helps me avoid dimensional errors (dividing when I should have multiplied, etc.):
You have: 7.913e10 J / 550K / (1J/g/K)
You want: kg
* 143872.73
/ 6.9505876e-06
maxerickson says, "Still big number," and 144 tonnes would typically be an unwieldy quantity of material if you had to buy it. But Standard Thermal's intention is not to buy dirt, just pile up already-on-site dirt with a bulldozer or excavator. If we assume 1.3 tonnes/m³, that's 110m³, or, in medieval units, 144 cubic yards. https://www.eaglepowerandequipment.com/blog/2022/03/how-much... tells us:
> An excavator could be used to dig anywhere from 350 to 1,000 cubic yards per day, depending on a number of factors including bucket capacity, type of ground, operator skill and efficiency level, and more. (...)
> One of the biggest factors that impact how much an excavator can dig in one day is the unit’s bucket size, which typically ranges from 0.5 to 1.5 cubic yards of bucket capacity. Most common regular-size excavators have a 1 cubic yard bucket capacity, and mini excavators are closer to the 0.5 cubic yard capacity.
So, with this number, we're talking about a few hours of work for a "mini excavator". https://www.bigrentz.com/rental-locations/pennsylvania/pitts... tells us that a "4,000 lb. mini excavator" rents for US$197 per day. So the expense of moving the dirt is not really significant, compared to other household projects such as replacing the roof, insulating the walls, or repainting the exterior.
Standard Thermal mentions that they are in effect firing the clay in the ground, that they've had significant trouble with resistance-heater reliability, and that their objective is to power steam-turbine power stations with the stored heat. These three facts lead me to believe that they're targeting a temperature closer to 1000° than to 600°.
600 C is about what a coal fired power plant would use. And 600 C is around the maximum that you want if you're using cheap steel for the pipes. Much beyond that and creep becomes a problem. So I don't think 1000 C is their target.
Hmm! Interesting! I would have thought that 600° would be close to the minimum for producing supercritical steam, so any energy stored up to 600° would be "overhead" that couldn't be effectively recovered—only the heating above that. And I assumed they would have to use cheap ceramic for the pipes, because oxidation is usually a problem for cheap steel even below 600°.
Oh, apparently because of "dramatic improvements in power plant performance":
> Starting with the
traditional 2400 psi / 1000 F (165 bar / 538 C)
single-reheat cycle, dramatic improvements in
power plant performance can be achieved by
raising inlet steam conditions to levels up to
4500 psi/310 bar and temperatures to levels in
excess of 1112 F/600 C. It has become industry
practice to refer to such steam conditions, and
in fact any supercritical conditions where the
throttle and/or reheat steam temperatures
exceed 1050 F/566 C, as “ultrasupercritical”.
Anyway, those are the plants that Standard Thermal wants to sell their product/service to. And once the hot dirt falls below 600°, it can no longer heat the water to 600°. So I think they have to be aiming far above that temperature, which is also why heating element reliability is a challenge and why the clays in the soil are firing (a phenomenon which only happens at 600° for the lowest-firing terra-cotta clays, more typically requiring 1000°–1400°).
I haven't seen pfdietz's proposed system design, but a so-called "sand battery," consisting of a box of sand with a heating element running through it, should work fine. You can PWM the heating element with a power MOSFET to keep it from overheating; you can measure its temperature with its own resistance, but also want additional thermocouple probes for the sand and to measure the surface of the box. A fan can blow air over or through the sand to control the output power within limits.
I'll work out some rough figures.
Let's say your house is pretty big and badly insulated, so we want an average of 5000 watts of heating around the clock with a time constant on the order of 10 hours, and we don't want our heating element to go over 700°. (Honest-to-God degrees, not those pathetic little Fahrenheit ones.) That way we don't have to deal with the ridiculous engineering issues Standard Thermal is battling. There's a thermal gradient through the sand down to room temperature (20°) at the surface. Suppose the sand is in the form of a flat slab with the heating element just heating the center of it, which is kind of a worst case for amount of sand needed but is clearly feasible. Then, when the element is running at a 100% duty cycle, the average sand temperature is 360°. Let's say we need to store about 40 hours of our 5000W. Quartz (cheap construction sand) is 0.73J/g/K, so our 720MJ at ΔT averaging 340K is 2900kg, a bit over a cubic meter of sand. This costs about US$100 depending mostly on delivery costs.
The time constant is mostly determined by the thickness of the sand (relative to its thermal diffusivity), although you can vary it with the fan. The heating element needs to be closely enough spaced that it can heat up the sand in the few hours that it's powered. In practice I am guessing that this will be about 100mm, so 1.5 cubic meters of sand can be in a box that's 200mm × 2.7m × 2.7m. You can probably build the box mostly out of 15m² of ceramic tiles, deducting their thermal mass from the sand required. In theory thin drywall should be fine instead of ceramic if your fan never breaks, but a fan failure could let the surface get hot enough to damage drywall. Or portland cement, although lime or calcium aluminate cement should be fine. You can use the cement to support the ceramic tiles on an angle iron frame and grout between them if necessary.
7.5m² of central plane with wires 100mm apart requires roughly 27 2.7m wires, 75m, probably dozens of broken hair dryers if you want to recycle nichrome, though I suspect that at 700° you could just use baling wire, especially if you mix in a little charcoal with the sand to maintain a reducing atmosphere in the sand pore spaces. (But then if it gets wet you could get carbon monoxide until you dry it out.) We're going to be dumping the whole 720MJ thermal charge in in under 9 hours, say 5 hours when the sunshine is at its peak, so we're talking about maybe 40kW peak power here. This is 533 watts per meter of wire, which is an extremely reasonable number for a wire heating element, even a fairly fine wire in air without forced-air cooling.
If we believe https://www.nature.com/articles/s41598-025-93054-w/tables/1 the thermal conductivity of dry sand ranges from 0.18 W/m/K to 0.34 W/m/K. So if we have a linear thermal gradient from our peak design temperature of 700° to 20° over 100mm, which is 6800K/m, we should get a heat flux of 1200–2300W/m² over our 15m² of ceramic tiles, so at least 18kW, which is more than we need, but only about 3×, so 200mm thickness is in the ballpark even without air blowing through the sand itself. (As the core temperature falls, the heat gradient also falls, and so does the heat flux. 720MJ/18kW I think gives us our time constant, and that works out to 11 hours, but it isn't exactly an exponential decay.) Maybe 350mm would be better, with corresponding increases in heating-element spacing and decreases in wire length and box surface area and footprint.
To limit heat loss when the fan is off, instead of a single humongous wall, you can split the beast into 3–6 parallel walls with a little airspace between them, so they're radiating their heat at each other instead of you, and cement some aluminum foil on the outside surfaces to reduce infrared emissivity. The amount of air the fan blows between the walls can then regulate the heat output over at least an order of magnitude. (In the summer you'll probably want to leave the heating element off.)
The sand, baling wire, aluminum foil, lime cement, angle irons, charcoal, thermocouples, power MOSFETs, microcontroller, fans, and ceramic tiles all together might work out to US$500. But the 40kW of solar panels required are about US$4000 wholesale, before you screw them to your siding or whatever. At US prices they'd apparently be US$10k.
720MJ is 200kWh in cursed units, so this is about US$2.50/kWh. Batteries are about US$80/kWh on the Shanghai Metals Market.
A thing I forgot to calculate: with 75m of wire dissipating 533 watts per meter, how thick should the wire be? Suppose we divide it into three 25m circuits so that we still have most of our heat if a wire burns out, and suppose we're using 48Vdc. So E²/R = 13.3kW, R = E²/13.3kW = 0.173Ω, and each of those elements is carrying an astonishing 277 amps. So we want 7 milliohms per meter. It turns out that that's about 12-gauge copper wire, nominally 5 milliohms per meter. 2 millimeters across. A higher-resistivity metal like iron or nichrome would have to be even thicker.
Better idea: put 9 2.7-meter wires in parallel on each of the three circuits, so each wire can have 9×0.173Ω = 1.56 Ω = 0.58Ω/m. That's 32-gauge copper magnet wire, 0.2mm diameter, 0.54Ω/m; or its thicker equivalent in other metals. Iron's resistivity is 5.7 times copper's, so you need a 5.7 times thicker wire: 0.5mm, 24-gauge. Nichrome is 11 times the resistivity of iron, so you'd need 1.6-mm-diameter nichrome.
I don't know, I think the copper would probably melt faster than the sand could conduct the heat away from it, and the nichrome would definitely be fine, but too expensive. But you can extrapolate from this how to solve the problem: by shortening the distance along the heating wires to low-resistance busbars (possibly made of rebar or leftover angle iron) and thus increasing the number of parallel paths, you allow the use of higher-resistance-per-unit-length and thus cheaper and more workable heating elements; the limit of this lightweighting is that the wires' surface area in contact with the sand must cool them enough to prevent melting. By this method you can use a small amount of a conductor of any resistivity at all, limited mainly by the temperature.
All these metals are fine at 700°, or for that matter 1000°. Copper will have less of a tendency to oxidize than iron, which would require a reducing atmosphere, and nichrome will oxidize but remain protected by its oxidation. (A reducing atmosphere will destroy nichrome.) But, at a lower temperature still, like 600°, you could use 10μm thick household aluminum foil, which is much easier to work with than any kind of 20μm wire, but has a similar ratio of surface area to volume. It has 54% more resistivity than copper, so a 10μm × 1mm strip is 2.7 ohms per meter. Our previous objective of 0.58Ω/m is a 4.6mm-wide-strip, which transfers heat to the sand along its 9.2mm perimeter, like a 10-gauge wire. 75m × 4.6mm is the size of about 5 or 6 pages of A4 paper cut into strips.
Austin Vernon claims they have a very cheap resistor material for Standard Thermal but hasn't said what it is. I look forward to hearing that detail when it leaks out. A good chunk of their work while in stealth was on the resistors, I understand.
I think I've shown above that you can make the resistor material itself almost arbitrarily cheap, calculating for example how you can get 40 kilowatts out of 9.3 grams of aluminum foil, and showing that with more busbars you can use even less resistor material than that. Aluminum itself wouldn't work for Standard Thermal's target temperatures, but you can make an arbitrarily thin foil out of any metal, supporting it as a thin film on an insulating ceramic such as porcelain if necessary. Copper, gold, silver, mild steel, nickel, nichrome, other stainless, titanium, platinum, and platinum/iridium, could all be made to work, and in no case would the material cost be significant. Metal film resistors supported on ceramic are being used to convert electrical energy into heat in probably every electronic device in your house.
And the old standby for resistive heating of giant piles of dirt, for example to bake it into carborundum, isn't a metal at all—it's plain old carbon, which you can if necessary bake in situ. Carborundum itself can also work, though it's not malleable, and controlling its resistivity can be tricky.
MIG welding wire is an interesting possibility.
The main potential obstacle, I think, is the manufacturing cost, and as sandy234590 was saying, potentially durability in use. Vernon said resistor durability had been one of their major problems; I'd think that sand would impose less stress on the resistors than generic dirt, but, with quartz in particular, you could greatly reduce the risk by not crossing the quartz dunting temperature at 573°: https://digitalfire.com/glossary/quartz+inversion That obviously isn't an option for Standard Thermal, but it would be completely viable for household climate control, just requiring somewhat more sand.
Sandy points out, implicitly, that mild steel such as the baling wire I suggested typically does not last long at high temperatures. But that's because it oxidizes. The same vulnerability is present in most metals, though not silver, gold, platinum, and platinum/iridium alloys, and only to a limited extent for nickel, nichrome, and other stainlesses. That oxidation can only happen in an oxidizing atmosphere; the thin iron ballast wires in Nernst lamps last indefinitely because they're sealed in a reducing (hydrogen) atmosphere. As I said, I think you can maintain a reducing atmosphere in the sand pore space by just including a little charcoal, which will scavenge any oxygen that gets close to the heating elements when they're hot, and may even be able to reduce any oxide that does form, at the cost of carbon monoxide emission.
If the atmosphere inside the sand is oxidizing, you'd probably want to either use something that won't be damaged by oxidization, such as gold or nichrome, or use a very thick heating element such as carbon so that it will have an adequate service life despite the oxidation. Most stainless steels will start to oxidize at a few hundred degrees, even though they're fine at room temperature.
(The main heating element in Nernst lamps, cubic zirconia, was also immune to oxidation, but it had some other drawbacks; for example, it needed to be preheated into its conductive range with a platinum preheat wire, and its rather aggressive negative temperature coefficient of resistance made it prone to thermal runaway when operated on a constant-voltage source—thus the iron ballast wire.)
I would instead say that, familiar with many designs from millennia of history of using thermal mass for indoor climate control, I outlined a design of an electric night storage heater that is especially cheap and convenient. Or an "electric day storage heater", I guess, since the day is when it stores heat.
Sand batteries have a much higher cost per unit of energy storage capacity, so they are in more direct competition with batteries for shorter term storage. It's hard to compete with a storage material you just dig out of a local hole. The economics pushes toward crude and very cheap.
Having said that: a good design for sand batteries would use insulated silos, pushing/dropping sand into a fluidized bed heat exchanger where some heat transfer gas is intimately mixed with it. This is the NREL concept that Babcock and Wilcox was (still is?) exploring for grid storage, with a round trip efficiency back to electricity of 54% (estimated) using a gas turbine. Having a separate heat exchanger means the silos don't have to be plumbed for the heat exchange fluid or have to contain its pressure.
Getting the sand back to the top (where it will be heated and dropping into silos) is a problem that could be solved with Olds Elevators, which were only recently invented (amazingly).
(I completed my parent comment since you wrote your response, which may make it confusing to read your response; sorry about that.)
I agree that local dirt is much cheaper than trucked-in construction sand, but I think my design sketch above shows that a "sand battery" whose only moving parts are fans will be about 30× cheaper than a real battery at household scale, even though the sand is still most of the estimated cost. A "sand battery" designed to power a steam turbine is a much more difficult problem to solve, but in this case the stated problem is just that it's 24°F (-3°) outside, so I think much cheaper solutions are fine, with no pressure vessels, stainless steel, insulated silos, sand conveyors, or heat transfer fluids other than garden-variety air.
Do you have a good handle on the pressure (and therefore power) requirements for getting air to flow upward through sand? I feel like you ought to be able to get a pretty decent amount of thermal power out of half a tonne of sand with a really minimal amount of pumping, but that's only a gut feeling. Definitely as you go to graded-granulometry gravel the required head drops off to almost nothing.
Thanks for the link to the Olds device! That's utterly astounding. Archimedes could have used it for raising sand, although making a sturdy enough tube out of wood might have been a bit of a chore.
I've heard of farmers doing this, well I think they actually had an inverter. But limits on how much they could dump into the grid, meant that they had lots of surplus electricity and installing resistive heating was very cheap.
Even if they don't have surplus electricity all the time.
Is it worth using heat pumps in this setup (in addition to resistive elements)? I understand they can't reach the absolute temperature of resistive heating, but from an efficiency POV for the first few tens of degrees they are much more efficient.
Depends - the problem with heat pumps is when you need them the most they don't work. If it never gets below -10c (exact temperature needs more study, could be as low as -25) where you live they are fine - but that implies you live in an area where you don't get many cold days and so the expense isn't worth it (it also implies you live where it gets hot in sumner so you want ac anyway and the marginal additional cost makes it worth it again). If you live in an area where it gets colder you need additonal backup heat that can cover those really cold days and so you may as well run that system only.
I think unless you're in an area dominated by cooling needs, an optimally sized heat pump system will not cover 100% of heating needs. It would make sense to make it smaller and use a backup resistive heater for rare very cold events.
cooling is more important but not by much. however the real problem is temperature delta: 100f-70f is 30 degrees, 70f-10f is 60 degrees. If you size a system for cooling it can't make up.
of course things arenot actually linear on temperature but as a rough estimate it gets the point across.
Efficiency allows you to use less solar panels, but more solar panels are cheaper than a heat pump. I think the ratio is about 5:1 at this point and widening.
To be concrete, I'm told that recently in the US a certain 34000btu/hour (10kW) output heat pump consuming up to 14A at 220V at the compressor (3kW) cost US$2700 installed, which is 27¢ per peak watt of output. But https://www.solarserver.de/photovoltaik-preis-pv-modul-preis... gives a price of €0.055 per peak watt (US$0.065/Wp) for low-cost solar panels. So the heat pump costs, in some sense, 4.2 times as much as the solar panels.
But the heat pump doesn't save you 10kW over resistive heating when it's running full-tilt. It saves you 10-3 = 7kW. So it costs 39¢ per watt of saved energy, which is 6 times as much as the solar panels.
In some simplified theoretical sense, if you decide you need another 10kW of heating for your house, you could spend US$2700 on this heat pump, and also buy 3000 Wp of solar panels to power it, costing US$194, for a total cost of US$2894. Or you could buy 10000 Wp of solar panels, costing US$645, and a resistive wire, costing US$10, for a total cost of US$655. US$655 is almost five times cheaper than US$2894. (4.4 times cheaper.)
There are a lot of factors that this simplified cost estimate overlooks; for example:
• Maybe you need to run the heater 16 hours a day but you only get sunlight 7 hours a day, either because it's winter in Norway, or because there are tall pine trees that shade your property most of the day, and you can't put the panels up on the trees. So maybe in some sense one watt of peak heater output is worth 2.3 watts of peak solar panel output. Or maybe it's the other way around, where your house only needs active heating during a few hours at night, so one watt of peak heater output is only worth 0.43 watts of peak solar panel output.
• The prices are in different countries. Solar panels are more expensive in the US, even wholesale.
• US$2700 is the retail price of the heat pump, including installation and warranty, and 6.5¢/Wp is the wholesale price of low-cost solar panels with no warranty ("Minderleistungs-Solarmodule, B-Ware, Insolvenzware, Gebrauchtmodule, PV-Module mit eingeschränkter oder ohne Garantie, die in der Regel auch keine Bankability besitzen.") Even in Europe the retail price of solar panels is three or four times this.
• Driving a resistive heating element from solar panels is considerably easier than driving a heat pump from solar panels; adapting a heating element to run on lower voltage is just a matter of connecting more wires to the middle of it, while adapting a heat pump to run on lower voltage may involve redesigning the whole power supply board or even rewinding the motor. Which is in a hermetically sealed refrigerant circuit, by the way, which you'd have to reseal. In practice, you'd just buy an inverter, but a 3000-watt inverter is expensive.
• As you said, for sensible-heat thermal storage, the heat pump craps out at about 50° or 60°, while any garden-variety resistive heating element (plus a lot of crappy improvised ones) will be just fine at 600° or 700°. That means you need ten times as much thermal mass for the same amount of storage. Sand is dirt cheap, but once you get into the tens of tonnes, even dirt isn't really cheap.
Despite such complications, I still think that pair of numbers is a useful summary of the situation: the heat pump costs 39¢ per watt saved, while the solar panel costs 6.5¢ per watt produced.
