4 ms·

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 co

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.

maxerickson 1y ago
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.
- pfdietz 1y ago
  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. https://www.orcasciences.com/articles/standard-thermal-copy https://www.orcasciences.com/articles/standard-thermal-copy https://austinvernon.substack.com/p/building-ultra-cheap-energy-storage https://austinvernon.substack.com/p/building-ultra-cheap-ene... https://news.ycombinator.com/item?id=45012942 https://news.ycombinator.com/item?id=45012942
  - mjevans 1y ago
    I'm going to want that pile hot enough to kill all the bugs and pets that want to get near it.
    - pfdietz 1y ago
      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 $.
  - cyberax 1y ago
    I ran the numbers on that, and it just doesn't work. Stone has rather lousy specific heat capacity (less than 1kJ/kg/K, compared to 4.2kJ for water). A typical house in Midwest needs around 22,000kWh (7.913×10^10 J) over the winter (75 million BTU - https://www.eia.gov/todayinenergy/detail.php?id=57321 https://www.eia.gov/todayinenergy/detail.php?id=57321 ). 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?
    - pfdietz 1y ago
      Your decimal point slipped three places in that last calculation; the result is too high by a factor of 1000. A more worthy criticism is that the pile for just a single house is too small and would cool off too quickly.
      - cyberax 1y ago
        I don't believe it did? Delta of 550 degrees Kelvin multiplied by 1000J per kg per Kelvin.
        pfdietz 1y ago
        7.913e10 / ( 5.5e2 * 1.0e3 ) = 1.438e5, not 1.438e8 When doing calculations like this I just fire up a lisp and enter the thing to be calculated as lisp form.
        kragen 1y ago
        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-can-an-excavator-dig-in-a-day-includes-formula/ 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/pittsburgh/mini-excavators 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°.
- kragen 1y ago
  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 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. What do you think?
alvah 1y ago
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.
- bluGill 1y ago
  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.
  - pfdietz 1y ago
    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.
    - bluGill 1y ago
      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.
- kragen 1y ago
  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.
jopsen 1y ago
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.
- pfdietz 1y ago
  Agriculture is one of the target markets for Standard Thermal. For example, grain drying is a large consumer of propane, which is expensive.

maxerickson 1y ago

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.

pfdietz 1y ago
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. https://www.orcasciences.com/articles/standard-thermal-copy https://www.orcasciences.com/articles/standard-thermal-copy https://austinvernon.substack.com/p/building-ultra-cheap-energy-storage https://austinvernon.substack.com/p/building-ultra-cheap-ene... https://news.ycombinator.com/item?id=45012942 https://news.ycombinator.com/item?id=45012942
- mjevans 1y ago
  I'm going to want that pile hot enough to kill all the bugs and pets that want to get near it.
  - pfdietz 1y ago
    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 $.
- cyberax 1y ago
  I ran the numbers on that, and it just doesn't work. Stone has rather lousy specific heat capacity (less than 1kJ/kg/K, compared to 4.2kJ for water). A typical house in Midwest needs around 22,000kWh (7.913×10^10 J) over the winter (75 million BTU - https://www.eia.gov/todayinenergy/detail.php?id=57321 https://www.eia.gov/todayinenergy/detail.php?id=57321 ). 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?
  - pfdietz 1y ago
    Your decimal point slipped three places in that last calculation; the result is too high by a factor of 1000. A more worthy criticism is that the pile for just a single house is too small and would cool off too quickly.
    - cyberax 1y ago
      I don't believe it did? Delta of 550 degrees Kelvin multiplied by 1000J per kg per Kelvin.
      - pfdietz 1y ago
        7.913e10 / ( 5.5e2 * 1.0e3 ) = 1.438e5, not 1.438e8 When doing calculations like this I just fire up a lisp and enter the thing to be calculated as lisp form.
        kragen 1y ago
        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-can-an-excavator-dig-in-a-day-includes-formula/ 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/pittsburgh/mini-excavators 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°.
kragen 1y ago
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 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. What do you think?

alvah 1y ago

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.

bluGill 1y ago
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.
- pfdietz 1y ago
  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.
  - bluGill 1y ago
    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.
kragen 1y ago
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.

jopsen 1y ago

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.

pfdietz 1y ago
Agriculture is one of the target markets for Standard Thermal. For example, grain drying is a large consumer of propane, which is expensive.