6 ms·

Let's do out the math on this... A subway tunnel might have a diameter of about 6 meters, so cross section = 3 * 3 * pi = 28 square meters. Digging subway tunn

Let's do out the math on this... A subway tunnel might have a diameter of about 6 meters, so cross section = 3 * 3 * pi = 28 square meters. Digging subway tunnel through rock costs about $100M per kilometer. On the one hand, these holes would be vertical, which is harder than horizontal; on the other hand, they wouldn't need ventilation and train tracks and stuff. Let's handwave and say it's $100M for a 1 km deep hole. Now, you can't fill the whole tunnel perfectly or the air can't escape, so our total volume of mass will be about 25 m^2 * 1000 m = 25,000 cubic meters. If the weights are made from lead, that's a total mass of ~280,000 tons or 2.8 * 10^8 kg, at a mean depth of 500 meters, so our total potential energy is 2.8 * 10^8 * 500 * 9.8 = 1.4 TJ, or 1.3 TJ net assuming you get the efficiency they claim. 1 kWh is 3.6 MJ, so you can store ~400,000 kWh at $100M capital cost (ignoring for the moment the cost of weights, generators, etc.), which is $250 per kWh installed capacity. That's pretty good... but you also have to pay for weights and a bunch of other stuff. Bulk lead costs about $2,000 per ton on the current market, so that's $560M for the weights, which puts you back in the $2,000 per kWh range which doesn't beat lithium batteries. So you have to use iron or some cheaper material... but then you don't have as much storage capacity because the density is lower, and even with iron you're paying $400 per metric ton or $80M for all your weights. So this isn't obviously impossible like Solar Roadways, but even in the best case it won't make storage dramatically cheaper.

tempestn 12y ago
One major advantage this would have over chemical batteries (Lithium-Ion or otherwise) is that it should allow for many, many more cycles. You would have to replace cables and motors and such at some point, but the expensive components - the hole and the weight - should last near-indefinitely. It certainly appears more expensive than pumped storage, but as they say, it doesn't require a convenient mountain and lake. That makes for much more flexibility in placement, and the closer you can place the storage to the demand, the less is wasted in transmission, and the less storage you ultimately need. That said, the pumped heat storage design[1] that hit the homepage yesterday would have all those same advantages, and although the efficiency would be slightly lower, it looks much cheaper to build. [1]: http://www.windpowerengineering.com/featured/business-news-projects/didnt-think-electricity-stored-temperature-difference/ http://www.windpowerengineering.com/featured/business-news-p...
- marcosdumay 12y ago
  Now that you mentioned a mountain... Why again is a hole needed? Edit: I mean that an inclined rail could be just as good as a hole, and way cheaper.
  - tempestn 12y ago
    For the gravity-based design, you need to lower a weight directly up and down in a controlled manner (ie without swinging etc.) That basically requires a shaft in the ground. If you have a mountain and a lake handy, pumped storage would be more economical, but the advantage of this design - as I understand it - is it can be deployed in places that don't have those things.
    - ChuckMcM 12y ago
      Actually an inclined plane and a hole in the ground are identical in terms of their energy storage, both are a function of their highest and lowest points. Which suggests a very simple implementation which is to run a pair of rails up the side of a mountain and but the winder/generator at the top and tie it off to the end of a rail car filled with rocks. Sure you would give up some efficiency with the friction of rail on steel but it would be pretty cheap to implement. On some of the alluvial fans in the Mohave it would make for an interested 'excess energy' storage mechanism for solar. That said, how are those gravity fed porch lights doing? That seems like it would be an interesting proxy here.
      - pm90 12y ago
        I think the important factor here is the possibility of placing these things anywhere one wants. If mountains are available, I don't see why not (except maybe they would ruin the natural beauty and hence tourism...but I don't know how much impact that has). But if they are not, you can always just dig a big hole.
      - mmcconnell1618 12y ago
        External rails on a mountain don't allow for the optional compressed air part of the system that (according to the page) can double the efficiency of the system at less than double the cost.
the_cat_kittles 12y ago
wouldn't this last a lot longer than lithium batteries?
- msandford 12y ago
  Maybe? But what if it cost way, way more? You actually have to look at the time value of money for anything that takes more than a year or two to pay off to be sure that there isn't some kind of false economy. If you assume that the thing lasts forever, well, you divide $x/infinity so it's free. EXCEPT that you've got $x tied up that could be earning interest instead. So it's not $x/infinity, it's the opportunity cost on $x versus anything else out there like batteries, the grid, etc.
Gurkenglas 12y ago
If the capacity is proportional to the depth of the tunnel and the mass of the weights, and the maximal mass of the weights is proportional to the depth of the tunnel, then wouldn't the capacity grow quadratically with the invested money? Edit: You assumed that they can both fill most of the tunnel up with lead and move that up or down 500 meters. The volume that they can fill with that is thus only half of what you say.
- msandford 12y ago
  Why is the maximal mass of weights proportional to depth? It's more proportional to diameter than anything else. There's a practical limit to how much weight you can hang. Sure if you make a 10km deep hole and a 3km deep weight you could store a lot of energy. But a 3km deep weight might not hold itself together.
  - Gurkenglas 12y ago
    If we have enough volume that we can't use the most dense possible weight for all of it, we can find cheaper weight materials by dropping the density constraint. Couldn't you have the weight grip the sides of the tunnel with gears attached to a generator/motor (inside the weight) so the weight wouldn't rip itself apart? It would be the same machinery that ordinarily would operate the pulley at the top, just moved down into the weights (cause with this, you would need no more pulley). (To illustrate why it wouldn't rip itself apart, imagine gaps on the weight every 100 meters.)
    - msandford 12y ago
      Technically that would work fine. Economically, it's a disaster. McMaster prices are at least 2x what's available if you buy direct from a source, and if you were buying in bulk it'd probably get cheaper still. But this is cheap, cheap stuff for not that much weight. You'd need something I dunno, 100x-10000x stronger to do what you're talking about: http://www.mcmaster.com/#racks-and-pinions/=upn9xl http://www.mcmaster.com/#racks-and-pinions/=upn9xl The appeal of the cables is that they're cheap. And they're cheap because they're easy to make (all things considered). Once you start applying sophistication to the design the price gets higher, and it's already too high from having to drill the hole.
cperciva 12y ago
Digging subway tunnel through rock costs about $100M per kilometer. On the one hand, these holes would be vertical, which is harder than horizontal I would have thought that a better comparison would be oil wells, which cost about $500 per ft of depth, or $1.5M per km. our total volume of mass will be about 25 m^2 1000 m = 25,000 cubic meters. If the weights are made from lead, that's a total mass of ~280,000 tons* You seem to be assuming that the entire depth of the shaft is filled by weights. My impression was that the weight was much smaller than the shaft it fell down.
- msandford 12y ago
  Oil wells are also only a foot in diameter. Makes the economics worse, really.
  - deleted 12y ago
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- darkmighty 12y ago
  Actually 1/2 the shaft filled by weights is optimal independently of material, considering only raw energy. (E=p A h(H-h), respectively density, area, payload height and shaft height, Eopt=p A H^2/2 ). Pretty surprising to me that the energy increases quadratically with depth.
  - cperciva 12y ago
    Yes, if you're considering only energy. But in practice there are other considerations, such as the cost of the weight, the strength (and thus cost) of the cables, the amount of torque which can be produced for lifting the weight, etc.
    - deleted 12y ago
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ortusdux 12y ago
I feel this idea would mesh well with the mining industry. Dig straight down, reinforce the walls on the way, separate out the desired minerals, pull up the drill, put in a pendulum and move the rig over and start again. You already need to setup giant motors to move the pendulum, so getting the ore up would be easy. An ideal site would have an abundance of lead, which would cut down on the pendulum cost. Unused rock could be mixed into cement for the walls. The small footprint combined with the benefits of being able to store solar or wind energy for long periods of time could make it the first form of mining that environmentalists approve of.
malandrew 12y ago
What's the math look like for the side of a mountain or hill, where you pour a concrete channel and put wheels on the weight so it can be rolled up and down the hill. I'm just wondering if there are places like in the Rockies where this idea would be viable.
- fractalsea 12y ago
  I don't understand why they need to dig holes to do this. Why not build something equivalent onto the side of a cliff?
  - jstanek 12y ago
    The depth requirement is 500m to 1500m. Though cliffs can be much taller than this, they usually are not totally vertical and cannot accomodate this much depth. Disclaimer, not a geologist.
    - fractalsea 12y ago
      True, although I wonder if doing it over even a 100m cliff, but with a much larger weight, would be cheaper than the drilling operation.
  - dools 12y ago
    It looks as though compressed air is a key component.
    - dgemm 12y ago
      It's not.
      - tempestn 12y ago
        It's not a key component, but it is a significant benefit of their system. The main reason though is that the primary benefit of this system is flexibility of placement. By locating grid storage close to where it's used most, you save significantly on transmission losses (and infrastructure). Finding sufficient vertical cliffs would be even more difficult than finding suitable locations for pumped hydro. Also if you expose the weight to open air and wind the whole system would be much trickier than using an enclosed shaft.
tomp 12y ago
How are you going to hang a 280 000 ton weight, and more importantly, how are you going to move it up and down?
- pm90 12y ago
  We can use steel cables. I can't find concrete reference atm, but I'm sure it can be done
- PhasmaFelis 12y ago
  You're not. Apsec's math is based on the weight being as tall as the hole is deep, which obviously doesn't make sense.
malambic 12y ago
Can we get a source for the $100M per kilometer figure? The NYC subway system has 373 km of tunnels, 60% of which is underground, which would translate to roughly 22 billion dollars just for the digging... that just seems way too high to me.
- deleted 12y ago
  [deleted]
- Anechoic 12y ago
  2nd Avenue subway [0] requires about 8.5 miles worth of tunnels (several hundred feet of which were already dug back in the 70's) and it's budget is $17 billion. Of course that whole cost isn't just for digging tunnels, but it's a clear indication that digging tunnels (especially in dense urban areas) can be extremely expensive. [0] http://en.wikipedia.org/wiki/Second_Avenue_Subway http://en.wikipedia.org/wiki/Second_Avenue_Subway
  - Panoramix 12y ago
    There is a very large difference between building complex tunnels and a simple hole in the ground. The cost for drilling a straight exploratory/mining hole was 10 Million/km in the 70's and 80's[0]. I can easily imagine this being cheaper by a factor of 2-5 with current advances. Not to mention this concept has a lot of synergy with mining companies. [0] http://facstaff.gpc.edu/~pgore/geology/geo101/interior.htm http://facstaff.gpc.edu/~pgore/geology/geo101/interior.htm
    - Anechoic 12y ago
      There is a difference, but it's not necessarily large for deep drilling. Once you start drilling large diameter holes deep into the ground, you have to shore up that hole to prevent it from collapsing in on itself. For wells or expiatory holes, water or slurry is used, but that's not an option in this case, so there will need to be a casing involve. Installing that casing so that the pieces line line up perfectly is a complex endeavor (not impossible, it's been done, just not straight forward as "drill a hole")
PhasmaFelis 12y ago
> Now, you can't fill the whole tunnel perfectly or the air can't escape, so our total volume of mass will be about 25 m^2 1000 m = 25,000 cubic meters.* You seem to have accidentally assumed that the weight will stretch the full height of the hole. Obviously this would leave it unable to move vertically.
bluthru 12y ago
>On the one hand, these holes would be vertical, which is harder than horizontal I don't think this is the case. We dig vertical holes all the time for wells, geothermal, and pilings for buildings: http://www.bdonline.co.uk/Pictures/web/r/r/g/CCTV_17_foundation_ready.jpg http://www.bdonline.co.uk/Pictures/web/r/r/g/CCTV_17_foundat... Look at all of them! For one building!
sargun 12y ago
What about depleted Uranium? 19G / cm^3, 1.7 the density of lead. No idea what the cost is, but I can't imagine that waste material would be that costly.
nl 12y ago
This is wrong. 1. Digging mine shafts is dramatically cheaper than structal tunnels. They dig exploratory shafts frequently to roughly these depths for well under $1m 2. Using a disused mineshaft would have a negative cost associated with it because the security associated with keeping it safe is non zero.
throwaway_yy2Di 12y ago
Actually, the majority of the weight would be in steel rope. For one type [0], the safe limit is a tensile load of about 134-150 MN/m^2 (= MPa) [1]. At a density of 8 g/cm^3, the limiting length of a uniform cable is ~1.7 - 1.9 kilometers. If you have a stationary, suspended cable of this length, its own weight puts it at its maximum load; it can't lift anything else. Steel rope on McMaster is around $10,000/ton [of rope]. So, these assumptions are a dead end. You can't solve this by using shorter cables, because that decreases your energy capacity at the same rate. If you use cables of 1/10th the length (~100 m), you get only 1/10th the potential energy storage per ton. The cable thickness per lifted ton is constant. [0] http://www.engineeringtoolbox.com/wire-rope-strength-d_1518.html http://www.engineeringtoolbox.com/wire-rope-strength-d_1518.... [1] The breaking limit of rope is far higher (~700 MPa), and the breaking limit of a single wire strand -- the tensile strength -- is higher still (1,770 MPa according to [2]) For the cross section area, I'm assuming a circular rope (not accurate). [2] http://www.gabaswire.com/en/overview/grades-of-wire-rope.html http://www.gabaswire.com/en/overview/grades-of-wire-rope.htm...
- throwaway_yy2Di 12y ago
  "You can't solve this by using shorter cables, because that decreases your energy capacity at the same rate. If you use cables of 1/10th the length (~100 m), you get only 1/10th the potential energy storage per ton. The cable thickness per lifted ton is constant." This is a simple "figure of merit" for cable in this problem, cost / (length * load capacity (N)) The is the same as the cost / energy stored. The denominator is simply the work equation (distance * force) -- the mechanical work the cable can do before it runs out of length. = cost / energy This is actually sort-of constant, since the denominator is ~proportional to the cable volume. (The load capacity is ~ the cross sectional area d^2). For steel rope from [0], it looks like a lower bound of about $1,200/kWh. [0] http://www.mcmaster.com/#standard-wire-rope/=upui3o http://www.mcmaster.com/#standard-wire-rope/=upui3o (E.g. item "3440T68", 5/8" plain steel, $5.16/foot for 9,080 lbf lifting capacity; $5.16 / (9,080 lbf * 1 foot) = $1,509/kWh)
DennisP 12y ago
Here's a related idea: take a piece of land made of granite. Cut a 1km diameter plug of granite, one km deep. Store power by pumping water under the plug, raising the whole thing half a kilometer in the air. It would store enough energy to power Germany for a day. http://www.solarserver.com/solar-magazine/solar-energy-system-of-the-month/hydraulic-hydro-storage.html http://www.solarserver.com/solar-magazine/solar-energy-syste...

tempestn 12y ago

One major advantage this would have over chemical batteries (Lithium-Ion or otherwise) is that it should allow for many, many more cycles. You would have to replace cables and motors and such at some point, but the expensive components - the hole and the weight - should last near-indefinitely. It certainly appears more expensive than pumped storage, but as they say, it doesn't require a convenient mountain and lake. That makes for much more flexibility in placement, and the closer you can place the storage to the demand, the less is wasted in transmission, and the less storage you ultimately need. That said, the pumped heat storage design[1] that hit the homepage yesterday would have all those same advantages, and although the efficiency would be slightly lower, it looks much cheaper to build. [1]: http://www.windpowerengineering.com/featured/business-news-projects/didnt-think-electricity-stored-temperature-difference/ http://www.windpowerengineering.com/featured/business-news-p...

marcosdumay 12y ago
Now that you mentioned a mountain... Why again is a hole needed? Edit: I mean that an inclined rail could be just as good as a hole, and way cheaper.
- tempestn 12y ago
  For the gravity-based design, you need to lower a weight directly up and down in a controlled manner (ie without swinging etc.) That basically requires a shaft in the ground. If you have a mountain and a lake handy, pumped storage would be more economical, but the advantage of this design - as I understand it - is it can be deployed in places that don't have those things.
  - ChuckMcM 12y ago
    Actually an inclined plane and a hole in the ground are identical in terms of their energy storage, both are a function of their highest and lowest points. Which suggests a very simple implementation which is to run a pair of rails up the side of a mountain and but the winder/generator at the top and tie it off to the end of a rail car filled with rocks. Sure you would give up some efficiency with the friction of rail on steel but it would be pretty cheap to implement. On some of the alluvial fans in the Mohave it would make for an interested 'excess energy' storage mechanism for solar. That said, how are those gravity fed porch lights doing? That seems like it would be an interesting proxy here.
    - pm90 12y ago
      I think the important factor here is the possibility of placing these things anywhere one wants. If mountains are available, I don't see why not (except maybe they would ruin the natural beauty and hence tourism...but I don't know how much impact that has). But if they are not, you can always just dig a big hole.
    - mmcconnell1618 12y ago
      External rails on a mountain don't allow for the optional compressed air part of the system that (according to the page) can double the efficiency of the system at less than double the cost.

the_cat_kittles 12y ago

wouldn't this last a lot longer than lithium batteries?

msandford 12y ago
Maybe? But what if it cost way, way more? You actually have to look at the time value of money for anything that takes more than a year or two to pay off to be sure that there isn't some kind of false economy. If you assume that the thing lasts forever, well, you divide $x/infinity so it's free. EXCEPT that you've got $x tied up that could be earning interest instead. So it's not $x/infinity, it's the opportunity cost on $x versus anything else out there like batteries, the grid, etc.

Gurkenglas 12y ago

If the capacity is proportional to the depth of the tunnel and the mass of the weights, and the maximal mass of the weights is proportional to the depth of the tunnel, then wouldn't the capacity grow quadratically with the invested money? Edit: You assumed that they can both fill most of the tunnel up with lead and move that up or down 500 meters. The volume that they can fill with that is thus only half of what you say.

msandford 12y ago
Why is the maximal mass of weights proportional to depth? It's more proportional to diameter than anything else. There's a practical limit to how much weight you can hang. Sure if you make a 10km deep hole and a 3km deep weight you could store a lot of energy. But a 3km deep weight might not hold itself together.
- Gurkenglas 12y ago
  If we have enough volume that we can't use the most dense possible weight for all of it, we can find cheaper weight materials by dropping the density constraint. Couldn't you have the weight grip the sides of the tunnel with gears attached to a generator/motor (inside the weight) so the weight wouldn't rip itself apart? It would be the same machinery that ordinarily would operate the pulley at the top, just moved down into the weights (cause with this, you would need no more pulley). (To illustrate why it wouldn't rip itself apart, imagine gaps on the weight every 100 meters.)
  - msandford 12y ago
    Technically that would work fine. Economically, it's a disaster. McMaster prices are at least 2x what's available if you buy direct from a source, and if you were buying in bulk it'd probably get cheaper still. But this is cheap, cheap stuff for not that much weight. You'd need something I dunno, 100x-10000x stronger to do what you're talking about: http://www.mcmaster.com/#racks-and-pinions/=upn9xl http://www.mcmaster.com/#racks-and-pinions/=upn9xl The appeal of the cables is that they're cheap. And they're cheap because they're easy to make (all things considered). Once you start applying sophistication to the design the price gets higher, and it's already too high from having to drill the hole.

cperciva 12y ago

Digging subway tunnel through rock costs about $100M per kilometer. On the one hand, these holes would be vertical, which is harder than horizontal I would have thought that a better comparison would be oil wells, which cost about $500 per ft of depth, or $1.5M per km. our total volume of mass will be about 25 m^2 1000 m = 25,000 cubic meters. If the weights are made from lead, that's a total mass of ~280,000 tons* You seem to be assuming that the entire depth of the shaft is filled by weights. My impression was that the weight was much smaller than the shaft it fell down.

msandford 12y ago
Oil wells are also only a foot in diameter. Makes the economics worse, really.
- deleted 12y ago
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darkmighty 12y ago
Actually 1/2 the shaft filled by weights is optimal independently of material, considering only raw energy. (E=p A h(H-h), respectively density, area, payload height and shaft height, Eopt=p A H^2/2 ). Pretty surprising to me that the energy increases quadratically with depth.
- cperciva 12y ago
  Yes, if you're considering only energy. But in practice there are other considerations, such as the cost of the weight, the strength (and thus cost) of the cables, the amount of torque which can be produced for lifting the weight, etc.
  - deleted 12y ago
    [deleted]

ortusdux 12y ago

I feel this idea would mesh well with the mining industry. Dig straight down, reinforce the walls on the way, separate out the desired minerals, pull up the drill, put in a pendulum and move the rig over and start again. You already need to setup giant motors to move the pendulum, so getting the ore up would be easy. An ideal site would have an abundance of lead, which would cut down on the pendulum cost. Unused rock could be mixed into cement for the walls. The small footprint combined with the benefits of being able to store solar or wind energy for long periods of time could make it the first form of mining that environmentalists approve of.

malandrew 12y ago

What's the math look like for the side of a mountain or hill, where you pour a concrete channel and put wheels on the weight so it can be rolled up and down the hill. I'm just wondering if there are places like in the Rockies where this idea would be viable.

fractalsea 12y ago
I don't understand why they need to dig holes to do this. Why not build something equivalent onto the side of a cliff?
- jstanek 12y ago
  The depth requirement is 500m to 1500m. Though cliffs can be much taller than this, they usually are not totally vertical and cannot accomodate this much depth. Disclaimer, not a geologist.
  - fractalsea 12y ago
    True, although I wonder if doing it over even a 100m cliff, but with a much larger weight, would be cheaper than the drilling operation.
- dools 12y ago
  It looks as though compressed air is a key component.
  - dgemm 12y ago
    It's not.
    - tempestn 12y ago
      It's not a key component, but it is a significant benefit of their system. The main reason though is that the primary benefit of this system is flexibility of placement. By locating grid storage close to where it's used most, you save significantly on transmission losses (and infrastructure). Finding sufficient vertical cliffs would be even more difficult than finding suitable locations for pumped hydro. Also if you expose the weight to open air and wind the whole system would be much trickier than using an enclosed shaft.

tomp 12y ago

How are you going to hang a 280 000 ton weight, and more importantly, how are you going to move it up and down?

pm90 12y ago
We can use steel cables. I can't find concrete reference atm, but I'm sure it can be done
PhasmaFelis 12y ago
You're not. Apsec's math is based on the weight being as tall as the hole is deep, which obviously doesn't make sense.

malambic 12y ago

Can we get a source for the $100M per kilometer figure? The NYC subway system has 373 km of tunnels, 60% of which is underground, which would translate to roughly 22 billion dollars just for the digging... that just seems way too high to me.

deleted 12y ago
[deleted]
Anechoic 12y ago
2nd Avenue subway [0] requires about 8.5 miles worth of tunnels (several hundred feet of which were already dug back in the 70's) and it's budget is $17 billion. Of course that whole cost isn't just for digging tunnels, but it's a clear indication that digging tunnels (especially in dense urban areas) can be extremely expensive. [0] http://en.wikipedia.org/wiki/Second_Avenue_Subway http://en.wikipedia.org/wiki/Second_Avenue_Subway
- Panoramix 12y ago
  There is a very large difference between building complex tunnels and a simple hole in the ground. The cost for drilling a straight exploratory/mining hole was 10 Million/km in the 70's and 80's[0]. I can easily imagine this being cheaper by a factor of 2-5 with current advances. Not to mention this concept has a lot of synergy with mining companies. [0] http://facstaff.gpc.edu/~pgore/geology/geo101/interior.htm http://facstaff.gpc.edu/~pgore/geology/geo101/interior.htm
  - Anechoic 12y ago
    There is a difference, but it's not necessarily large for deep drilling. Once you start drilling large diameter holes deep into the ground, you have to shore up that hole to prevent it from collapsing in on itself. For wells or expiatory holes, water or slurry is used, but that's not an option in this case, so there will need to be a casing involve. Installing that casing so that the pieces line line up perfectly is a complex endeavor (not impossible, it's been done, just not straight forward as "drill a hole")

PhasmaFelis 12y ago

> Now, you can't fill the whole tunnel perfectly or the air can't escape, so our total volume of mass will be about 25 m^2 1000 m = 25,000 cubic meters.* You seem to have accidentally assumed that the weight will stretch the full height of the hole. Obviously this would leave it unable to move vertically.

bluthru 12y ago

>On the one hand, these holes would be vertical, which is harder than horizontal I don't think this is the case. We dig vertical holes all the time for wells, geothermal, and pilings for buildings: http://www.bdonline.co.uk/Pictures/web/r/r/g/CCTV_17_foundation_ready.jpg http://www.bdonline.co.uk/Pictures/web/r/r/g/CCTV_17_foundat... Look at all of them! For one building!

sargun 12y ago

What about depleted Uranium? 19G / cm^3, 1.7 the density of lead. No idea what the cost is, but I can't imagine that waste material would be that costly.

nl 12y ago

This is wrong. 1. Digging mine shafts is dramatically cheaper than structal tunnels. They dig exploratory shafts frequently to roughly these depths for well under $1m 2. Using a disused mineshaft would have a negative cost associated with it because the security associated with keeping it safe is non zero.

throwaway_yy2Di 12y ago

Actually, the majority of the weight would be in steel rope. For one type [0], the safe limit is a tensile load of about 134-150 MN/m^2 (= MPa) [1]. At a density of 8 g/cm^3, the limiting length of a uniform cable is ~1.7 - 1.9 kilometers. If you have a stationary, suspended cable of this length, its own weight puts it at its maximum load; it can't lift anything else. Steel rope on McMaster is around $10,000/ton [of rope]. So, these assumptions are a dead end. You can't solve this by using shorter cables, because that decreases your energy capacity at the same rate. If you use cables of 1/10th the length (~100 m), you get only 1/10th the potential energy storage per ton. The cable thickness per lifted ton is constant. [0] http://www.engineeringtoolbox.com/wire-rope-strength-d_1518.html http://www.engineeringtoolbox.com/wire-rope-strength-d_1518.... [1] The breaking limit of rope is far higher (~700 MPa), and the breaking limit of a single wire strand -- the tensile strength -- is higher still (1,770 MPa according to [2]) For the cross section area, I'm assuming a circular rope (not accurate). [2] http://www.gabaswire.com/en/overview/grades-of-wire-rope.html http://www.gabaswire.com/en/overview/grades-of-wire-rope.htm...

throwaway_yy2Di 12y ago
"You can't solve this by using shorter cables, because that decreases your energy capacity at the same rate. If you use cables of 1/10th the length (~100 m), you get only 1/10th the potential energy storage per ton. The cable thickness per lifted ton is constant." This is a simple "figure of merit" for cable in this problem, cost / (length * load capacity (N)) The is the same as the cost / energy stored. The denominator is simply the work equation (distance * force) -- the mechanical work the cable can do before it runs out of length. = cost / energy This is actually sort-of constant, since the denominator is ~proportional to the cable volume. (The load capacity is ~ the cross sectional area d^2). For steel rope from [0], it looks like a lower bound of about $1,200/kWh. [0] http://www.mcmaster.com/#standard-wire-rope/=upui3o http://www.mcmaster.com/#standard-wire-rope/=upui3o (E.g. item "3440T68", 5/8" plain steel, $5.16/foot for 9,080 lbf lifting capacity; $5.16 / (9,080 lbf * 1 foot) = $1,509/kWh)

DennisP 12y ago

Here's a related idea: take a piece of land made of granite. Cut a 1km diameter plug of granite, one km deep. Store power by pumping water under the plug, raising the whole thing half a kilometer in the air. It would store enough energy to power Germany for a day. http://www.solarserver.com/solar-magazine/solar-energy-system-of-the-month/hydraulic-hydro-storage.html http://www.solarserver.com/solar-magazine/solar-energy-syste...