3 ms·

You need a part of the system anyway so you can't count those parts (water tanks, hydrogen tanks, oxygen tanks, electrolyzer) into any of the power options. The

by jhgb 5y ago

You need a part of the system anyway so you can't count those parts (water tanks, hydrogen tanks, oxygen tanks, electrolyzer) into any of the power options. They will be there regardless of whether you want them or not. Also I've noticed that despite of what ben_w claims the article says, the article still assumes 100 kWh/t for a battery system and 1 MWh/t for a fuel cell system. So it doesn't really say that "the dry mass of the fuel cells would mass similar to LiIon of the same energy storage capacity", otherwise the numbers would be in the opposite order.

Retric 5y ago
Due to shipping and redundancy concerns redundant tanks are used to expand storage. So, the dry mass of tanks should scale nearly linearly with the total amount of water, hydrogen, and air stored.
- jhgb 5y ago
  Nobody said they don't. But 1) that doesn't make the fuel cells bigger (so the "similar dry mass of the fuel cells to batteries of a given capacity" is a category error), and 2) the additional tank mass is very mild. We can for example store 20.8 tonnes of hydrogen/oxygen mixture in a ~2 tonne tank. We know this because the Centaur stage does it. And those two tonnes of tankage correspond to ~50 MWh of electricity generated by a fuel cell. That's almost 300 tonnes worth of batteries in the form of 23 tonnes of mass, 21 of which might even not need transportation from Earth. Potentially up to a 100:1 win for Team Fuel Cells! ("Only" ~10:1 if you do carry water from Earth, but that's still substantial.)
  - Retric 5y ago
    Sure, dry mass of fuel cells vs dry mass for fuel cells vs total mass need for fuel cells. Also, that’s just the start. Team fuel cells are also dramatically lower efficiency so you need a extra solar panels and all the associated equipment with that. Which again scales linearly with energy demands.
    - jhgb 5y ago
      Total mass for a fuel cell system still wins against any battery system. And solar panels go up to 200 W/kg these days (UltraFlex/MegaFlex design), with up to 300 W/kg in labs. They're the least of your worries.
      - Retric 5y ago
        For non tracking panels at 100% efficiency and 200W/kg assuming an optimistic 1/2 of total weight for solar collection is panels runs around 50 kg per MWh per month on the moon. Batteries at ~90% bump that to 55kg/MWH. Fuel cells at ~30% bump that to ~170 kg or another 110kg /MWh, not that significant but still important.
        jhgb 5y ago
        How does panel efficiency go into this? Those 200 W/kg are post-conversion, so whatever efficiency they have is already included. > assuming an optimistic 1/2 of total weight for solar collection is panels What does that even mean? 200 W/kg is system efficiency, so not just cells but all the structures involved. > runs around 50 kg per MWh per month on the moon. I don't get it. "MWh per month" is 114 watts. A 50 kg array at 200 W/kg has 10 kW peak. At the south pole, a non-tracking unifacial horizontally mounted and elevated array will have at least a ~30% capacity factor, a bifacial array will have a ~60% capacity factor. That's 3 kW or 6 kW on average, respectively, far from your 114 watts. EDIT: Should have really been 1.4 kW instead of 114 watts - I calculated with 1 MWh per year by a mistake. Still a substantial difference from 3/6 kWh. > Fuel cells at ~30% A round-trip from a fuel-cell system will be at ~40%. Maybe ~50% if you use top equipment. But even bog-standard fuel cells are at ~50-55% (LHV) efficiency and bog-standard PEM electrolyzers are at ~80% efficiency, which should give you no less than a 40% round-trip.
        Retric 5y ago
        > How does panel efficiency go into this? It doesn’t but a measure of output from solar panels ignores the storage efficiency which is what we care about here. > 200w/kg is system efficiency. You listed that as the panels in your post. First actual array output falls over time. Whatever it would be new is irrelevant you scale for how long it’s in use. Panels on the moon need to be mounted at an appropriate angle for maximum efficiency, cooling, they need wires to move electric from the panel to your power conversion, electronics to handle that power etc. You also get losses from lunar dust, etc. Simply saying all that doubles weight is a reasonable ballpark. Don’t forget the moon is dealing with days of full sun panels need to maintain low temperatures for efficiency and you can’t just dump all that heat into the moon. > I don’t get it A 50kg array consisting of 25kg of solar panels, at 30% efficiency produces: 25 different 0.2kw panels for 30 percent of the time over 28 days times 24 hours = 25 * 0.200kw * 0.30 * 28 * 24h = 1008 kWh or 1.008 MWh. Edit: 1.4kW * 28 days * 24h/day = 940 kWh or 0.94 MWh. > A round-trip from a fuel-cell system will be at ~40%. I have yet to read about an actual working system over 30% water to hydrogen to water. Do you have any citations or is this assuming some unknown breakthrough?
ben_w 5y ago
Right next to where it suggests 1 MWh/t “For the sake of argument”, it says: """Even using Lunar water, the mass of the cells, condensers, electrolysers, power electronics, storage tanks, and heat exchangers is comparable a Lithium-ion battery, while the round trip efficiency is much lower. And that’s not even including the mass overhead for mining lunar water!"""
- jhgb 5y ago
  Yes, but that makes no sense. Partly because as I said, lots of those components will be mandatory anyway so they should count into the fuel cell system at best only partially, and partly because pretty much the only part that scales with capacity is the tanks anyway. The rest scales overwhelmingly with average power instead.

by jhgb 5y ago

Retric 5y ago

Due to shipping and redundancy concerns redundant tanks are used to expand storage. So, the dry mass of tanks should scale nearly linearly with the total amount of water, hydrogen, and air stored.

jhgb 5y ago
Nobody said they don't. But 1) that doesn't make the fuel cells bigger (so the "similar dry mass of the fuel cells to batteries of a given capacity" is a category error), and 2) the additional tank mass is very mild. We can for example store 20.8 tonnes of hydrogen/oxygen mixture in a ~2 tonne tank. We know this because the Centaur stage does it. And those two tonnes of tankage correspond to ~50 MWh of electricity generated by a fuel cell. That's almost 300 tonnes worth of batteries in the form of 23 tonnes of mass, 21 of which might even not need transportation from Earth. Potentially up to a 100:1 win for Team Fuel Cells! ("Only" ~10:1 if you do carry water from Earth, but that's still substantial.)
- Retric 5y ago
  Sure, dry mass of fuel cells vs dry mass for fuel cells vs total mass need for fuel cells. Also, that’s just the start. Team fuel cells are also dramatically lower efficiency so you need a extra solar panels and all the associated equipment with that. Which again scales linearly with energy demands.
  - jhgb 5y ago
    Total mass for a fuel cell system still wins against any battery system. And solar panels go up to 200 W/kg these days (UltraFlex/MegaFlex design), with up to 300 W/kg in labs. They're the least of your worries.
    - Retric 5y ago
      For non tracking panels at 100% efficiency and 200W/kg assuming an optimistic 1/2 of total weight for solar collection is panels runs around 50 kg per MWh per month on the moon. Batteries at ~90% bump that to 55kg/MWH. Fuel cells at ~30% bump that to ~170 kg or another 110kg /MWh, not that significant but still important.
      - jhgb 5y ago
        How does panel efficiency go into this? Those 200 W/kg are post-conversion, so whatever efficiency they have is already included. > assuming an optimistic 1/2 of total weight for solar collection is panels What does that even mean? 200 W/kg is system efficiency, so not just cells but all the structures involved. > runs around 50 kg per MWh per month on the moon. I don't get it. "MWh per month" is 114 watts. A 50 kg array at 200 W/kg has 10 kW peak. At the south pole, a non-tracking unifacial horizontally mounted and elevated array will have at least a ~30% capacity factor, a bifacial array will have a ~60% capacity factor. That's 3 kW or 6 kW on average, respectively, far from your 114 watts. EDIT: Should have really been 1.4 kW instead of 114 watts - I calculated with 1 MWh per year by a mistake. Still a substantial difference from 3/6 kWh. > Fuel cells at ~30% A round-trip from a fuel-cell system will be at ~40%. Maybe ~50% if you use top equipment. But even bog-standard fuel cells are at ~50-55% (LHV) efficiency and bog-standard PEM electrolyzers are at ~80% efficiency, which should give you no less than a 40% round-trip.
        Retric 5y ago
        > How does panel efficiency go into this? It doesn’t but a measure of output from solar panels ignores the storage efficiency which is what we care about here. > 200w/kg is system efficiency. You listed that as the panels in your post. First actual array output falls over time. Whatever it would be new is irrelevant you scale for how long it’s in use. Panels on the moon need to be mounted at an appropriate angle for maximum efficiency, cooling, they need wires to move electric from the panel to your power conversion, electronics to handle that power etc. You also get losses from lunar dust, etc. Simply saying all that doubles weight is a reasonable ballpark. Don’t forget the moon is dealing with days of full sun panels need to maintain low temperatures for efficiency and you can’t just dump all that heat into the moon. > I don’t get it A 50kg array consisting of 25kg of solar panels, at 30% efficiency produces: 25 different 0.2kw panels for 30 percent of the time over 28 days times 24 hours = 25 * 0.200kw * 0.30 * 28 * 24h = 1008 kWh or 1.008 MWh. Edit: 1.4kW * 28 days * 24h/day = 940 kWh or 0.94 MWh. > A round-trip from a fuel-cell system will be at ~40%. I have yet to read about an actual working system over 30% water to hydrogen to water. Do you have any citations or is this assuming some unknown breakthrough?

ben_w 5y ago

Right next to where it suggests 1 MWh/t “For the sake of argument”, it says: """Even using Lunar water, the mass of the cells, condensers, electrolysers, power electronics, storage tanks, and heat exchangers is comparable a Lithium-ion battery, while the round trip efficiency is much lower. And that’s not even including the mass overhead for mining lunar water!"""

jhgb 5y ago
Yes, but that makes no sense. Partly because as I said, lots of those components will be mandatory anyway so they should count into the fuel cell system at best only partially, and partly because pretty much the only part that scales with capacity is the tanks anyway. The rest scales overwhelmingly with average power instead.