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The impression I get is that the propellant production starts only after humans arrive. But I couldn't give you a citation for that!
I am still trying to figure out what plan is being used....
A. 2 cargo that land and start manufacturing enough fuel for a starship that will come on the next launch cycle
B. 2 cargo that land and start manufacturing enough fuel with 2 more cargo and a crewed starship that will come on the next launch cycle
C. 2 cargo that land and start manufacturing enough propellant for a crew that lands with them.Each has different requirements for power, fuel creation time and risk mitigation safety requirements.
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Have you got a citation? Because I've previously cited academic papers that don't support that contention.
To add: NASA's statement vary in credibility...oftentimes they are just creating very simple accounts for the general public or schools. You have to see what NASA sponsored academic papers say I think.
I've previously cited excerpts from various technical documents and measurements taken from actual spacecraft with actual solar and battery power systems that have actually been to Mars. You ignored all of those. You'll ignore this, too, but I'll continue to counter all assertions that disagree with recorded measurement and performance data taken from spacecraft that have been to Mars.
Google "MERB Notebook" - Opportunity (MERB) Analyst's Notebook - for actual Tau measurement data received from Opportunity
When Opportunity died, its solar panels went from producing 650Wh/day to 22Wh/day. The mission clock aboard the rover required 1W of power, or 24Wh/day. If the panel was producing 850Wh/day, as it did when new, that amounts to a maximum of 29Wh/day. A panel that produces 850Wh/day when clean and new produces just 3% of that power output during a major dust storm.
Based upon opacity measurements (top down from MRO and bottom up from MER-A / MER-B / MSL) and the actual recorded energy storage provided by telemetry data from MER-A and MER-B, major dust storms reduce insolation to as low as 100Wh/m^2/day. At locations significantly above or below the equator, yearly insolation can be half of what it is at the equator to begin with. The absolute maximum amount of power a completely clean and new photovoltaic array with a 35% efficiency could produce during a Martian dust storm amounts to 35Wh/m^2/day. Juno, the PV powered probe NASA sent to Jupiter, was still producing 144Wh/m^2/day, at Jupiter. That's more than 4 times as much power as a Mars surface PV panel could produce during a major dust storm. Therefore, a spacecraft at least as far away as Jupiter is much better off than a spacecraft on the surface of Mars during a major dust storm.
Google "Mars Surface Solar Arrays" for an actual graph of the surface insolation level using measurements taken before and during dust storms, which includes the reference AM0 measurement. Review slides 26, 27, 29, 30, and 31 for actual measurements. Note that on slide 35, NASA has experimented with everything they could think of that could clean the panels, including compressed CO2 (requires more power and has more potential points of failure) / wiper blades (which broke during testing) / ESD (works well, passed testing, and went to Mars aboard Spirit and Opportunity, where it was successfully used when there was power available to use it). PV panels attract dust because the dust particles and panels are oppositely charged. Around 38Wh/m^2 is required to completely clean the panels. During a major storm, there's insufficient power to clean the panels, so the energy reserve to clean the panels exceeds the panels' potential output.
Provided that you could keep a PV panel perfectly clean during a dust storm, which is obviously impossible, then 1m^2 of a brand new 35% efficient PV panel would produce 35Wh/day. The minimum array size to continue producing 1MWe using ATK panels is 28,571m^2, or 6.4 football fields. The best thin film arrays I'm aware of are about 12% efficient. The minimum array size to produce 1MWe with thin film is 83,333m^2, or 18.7 football fields. If the arrays aren't at least that large, then you'll run out of available stored power in less than a week. You can forget about propellant production and start thinking about how to avoid killing everyone from running out of power.
Further notes on batteries suitable for use on Mars:
The small cylindrical Lithium-ion batteries that Tesla uses in vehicles, which have better energy density than prismatic cells, also have poor cold temperature charge / discharge performance in comparison to Yardney's prismatic cells, which is why NASA doesn't use them for any of their robotic missions to Mars. None of the cells with the chemistry used by Panasonic and Tesla have passed testing thus far, as they'd be unable to charge or discharge at a sufficient rate to store enough of the power output from a solar array without first expending more energy to heat the cells to a temperature within their operational temperature design limitations. After more than 14 years of operation on Mars, the prismatic cells aboard MER-B had more than 85% of their original capacity remaining, so prismatic cells with that particular cell chemistry have energy densities that's about as good as it gets using current technology (118.4Wh/kg for the cells alone and 80.4Wh/kg for the complete Orion MPCV battery pack, which passes all operational tests required for human space flight in deep space; every year this goes up a little as Yardney refines their technology). Since Starship is a human mission, only human-rated systems will be going along for the ride. Until batteries with triple the capacity of existing cells appear, that can actually operate over the temperature ranges required, the viability of a solar powered mission is essentially a question of how long a dust storm lasts until you can start producing enough power to clean the arrays and resume storing enough energy to make it to the next sunrise. Any storm that lasts for more than a month is still a mission killer.
Either way, 1m^2 of all known PV panel types can't produce enough energy through the course of an entire day to actually clean 1m^2 of PV in a severe dust storm, so any available power output will inevitably drop to zero. If the dust storm reduces reduces output to zero for just 1 week, then a constant 50kWe power requirement to keep the people and electronics from dying, an absurdly low figure given the amount of electronics you plan on bringing to Mars, is 8.4MWh worth of power (3.4MWh more than the maximum you thought you'd have to bring). That's equivalent to 104.5t of batteries that have actually been to Mars. If the storm lasts for 2 weeks, then that's 209t of batteries, which is 50t more than the weight of 100 KP reactors and all power cables, even with a power cable for each reactor that goes all the way back to whatever it's powering from 2km away (worst case scenario for weight, but best case for redundancy). Any storm that kills PV array output for longer than 2 weeks is a complete mission killer when using solar power, full stop.
This would be exactly why NASA is pursuing nuclear power to enhance our space exploration capabilities. NASA has engineers who can do basic math, despite all the aspersions cast upon their technical capabilities. If there's a 1MWe power requirement, or even a comparatively modest 50kWe constant power requirement to keep the lights on, along with a 500t delivered tonnage limitation for a single mission, then we can't throw enough tonnage of PV and batteries at this problem to actually solve it.
If new solar panel and battery technology becomes available, then we'll revisit this topic. Until then, I'm done responding to assertions that completely ignore actual scientific measurements of the actual surface environmental conditions on Mars. NASA has gone to the trouble of creating pretty color graphs of their scientific data that are very easy to understand. I've checked everything I have from NASA on this and there are no significant disagreements about the surface insolation conditions. Just as GW stated previously and repeatedly, we've known what those surface insolation conditions were like from the very first probes we sent there. Every subsequent mission has only added more data points to the pile of data that we already have. If the scientists from NASA and JPL can't speak authoritatively on this matter, then nobody else can.
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I know I said I was done addressing this egregiously bad solar power idea in my prior post, but the categorical refusal of certain parties to employ any basic math to this energy provisioning problem has motivated me to illustrate just how bad solar power is under optimal conditions on Mars in comparison to nuclear power. I'm sick of having the scientific data acquired by NASA / JPL questioned at every turn, despite the fact that the only reason a mission to Mars is possible at all is because of the decades of hard work by our space exploration agency. Feel free to continue to question all the politically-motivated make-work projects distracting us from our ultimate goal, or at least the goal that everyone says they agree upon.
NASA's avg. insolation per sol over 688 sols without dust storms
50 deg N: 277.32Wh/m^2
0 deg: 492.14Wh/m^2
30 deg S: 412.32Wh/m^2
The figures shown above are what you get to work with, not what you actually receive from less than 100% efficient solar panels.
At the equator, that's 147.642Wh/sol for a 30% efficient panel or 172.249Wh/sol for a 35% efficient panel. The lowest credible mass figure I've seen using a 1,000m^2 Vectran inflatable deployable array comes in at around 700kg and produces 50.79MWh over 688 sols, or 73.8kWh/sol or 3,075.9W of constant power per sol. The mass associated with ATK fan technology is completely unworkable at the output level required for propellant production. Therefore, a solar array that produces equivalent power to a 10kWe KP reactor would weigh at least 2,275kg if it's operating at the equator under 100% clear sky conditions. This tracks precisely with NASA's assertion that KP would beat PV by at least a factor of 2, even at locations that are favorable to solar power.
If the power requirement is 1MWe constant power for propellant production, then the mass of 30% efficient solar panels alone is 227.5t. If we add 209t of batteries to survive for 2 weeks without power with a 50kWe constant power requirement, we've arrived at 436t. The PV arrays would only fill every square meter of the 825m^3 cargo volume of 2 Starships, so I guess that's a bonus, right? There's only 440 PV arrays that each weigh more than the 100 KP reactors that have to be deployed, along with 1,400 batteries. At least now I know why SpaceX has to land 5 Starships. There's no way they could ever make this work with fewer Starships. Since this new Starship is "Super Heavy", just like any PV power supply will be, I guess this new propellant plant and drilling equipment must be "Super Light".
Nukology for Non-Nukists
Anyone interested in understanding why NASA chose nuclear power for MSL and the Mars 2020 rovers should Google and read the 317 page "Final Environmental Impact Statement for the Mars 2020 Mission" report. Incidentally, this is what professional quality work, as opposed to PowerPoint presentations, actually looks like. Those people would then go directly to Page 2-39 and read through Page 2-50 to understand why nuclear power was selected over solar power. NASA doesn't make mission planning decisions by flipping coins. It wasn't "nukists" (whatever those are) saying "Man, it'd be super cool if we put some Plutonium on our Mars Mini Cooper! Hold my beer and watch this!". It certainly wasn't to offend the fragile sensibilities of our "green energy" religious zealots. It just might have been the fact that it was the only energy source that would actually get the job done without running smack into mass and volume constraints.
As it turns out, nuclear power permits 100% functionality year round because nuclear power provides 24/7/365 power. Photovoltaic arrays with 35% BoL efficiency, photovoltaics combined with Radioisotope Heater Units (RHU's), and Radioisotope Thermoelectric Generators (RTG's) were all considered as candidate power sources for the Mars 2020 rover. Of all available options, even with the rover's miserly daily power requirement of 100Wh to 600Wh for mobility and science operations, only nuclear power could provide the energy required at all times at latitudes between the equator 30 deg N or 30 deg S.
A guaranteed operational lifetime of 1 calendar year using solar power was assessed as 50% or less (with significant impacts to available power during certain times of the year for any science to actually occur), irrespective of latitude, whereas the operational lifetime of 1 calendar year using nuclear power was assessed as 100%, irrespective of the latitude of the exploration sites under consideration for the mission. That means the rover could land anywhere between the equator and 30 deg N or 30 deg S and still meet 100% of its power demands to survive the cold and still perform the science experiments that American tax payers paid billions of dollars to execute. Those beautiful color graphs (Figure 2-18 on Page 2-40 and Figure 2-20 on Page 2-46) indicate seasons of the year and latitudes where mere survival of a solar or solar + RHU rover was not possible, given the rover's power requirements.
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It's a convention of debates in internet forums I would say that if you give a citation, you give a link...you don't ask someone to go Googling.
I don't accept that the experience of tiny, vulnerable solar-powered robots on the Mars surface is particularly relevant to the issue of how much insolation is available during a dust storm or could be captured by a six person mission with 500 tons of equipment backing them up. The two cases are really not comparable. That said, I have never read anything that implies the rovers had ceased generating PV power, just that it was better for the survival of the robots to go into "hibernation".
So you are claiming that Opportunity was producing only about 3.4% of normal power. Well if that was applied to a PV system generating an average of 2000 Kwes average, that would be an average of 68Kwes being produced. So even without any active cleaning of the panels and without using any battery storage or methane fuel storage, you would be able to operate the base and keep people alive.
That said, I think I need to look into that figure of 22 Whs in one day because it sounds at variance with what I have read. Was that a complete day?
Remember, as well the peak of such severe dust storms lasts for only a few sols at most.
louis wrote:Have you got a citation? Because I've previously cited academic papers that don't support that contention.
To add: NASA's statement vary in credibility...oftentimes they are just creating very simple accounts for the general public or schools. You have to see what NASA sponsored academic papers say I think.
I've previously cited excerpts from various technical documents and measurements taken from actual spacecraft with actual solar and battery power systems that have actually been to Mars. You ignored all of those. You'll ignore this, too, but I'll continue to counter all assertions that disagree with recorded measurement and performance data taken from spacecraft that have been to Mars.
Google "MERB Notebook" - Opportunity (MERB) Analyst's Notebook - for actual Tau measurement data received from Opportunity
When Opportunity died, its solar panels went from producing 650Wh/day to 22Wh/day. The mission clock aboard the rover required 1W of power, or 24Wh/day. If the panel was producing 850Wh/day, as it did when new, that amounts to a maximum of 29Wh/day. A panel that produces 850Wh/day when clean and new produces just 3% of that power output during a major dust storm.
Based upon opacity measurements (top down from MRO and bottom up from MER-A / MER-B / MSL) and the actual recorded energy storage provided by telemetry data from MER-A and MER-B, major dust storms reduce insolation to as low as 100Wh/m^2/day. At locations significantly above or below the equator, yearly insolation can be half of what it is at the equator to begin with. The absolute maximum amount of power a completely clean and new photovoltaic array with a 35% efficiency could produce during a Martian dust storm amounts to 35Wh/m^2/day. Juno, the PV powered probe NASA sent to Jupiter, was still producing 144Wh/m^2/day, at Jupiter. That's more than 4 times as much power as a Mars surface PV panel could produce during a major dust storm. Therefore, a spacecraft at least as far away as Jupiter is much better off than a spacecraft on the surface of Mars during a major dust storm.
Google "Mars Surface Solar Arrays" for an actual graph of the surface insolation level using measurements taken before and during dust storms, which includes the reference AM0 measurement. Review slides 26, 27, 29, 30, and 31 for actual measurements. Note that on slide 35, NASA has experimented with everything they could think of that could clean the panels, including compressed CO2 (requires more power and has more potential points of failure) / wiper blades (which broke during testing) / ESD (works well, passed testing, and went to Mars aboard Spirit and Opportunity, where it was successfully used when there was power available to use it). PV panels attract dust because the dust particles and panels are oppositely charged. Around 38Wh/m^2 is required to completely clean the panels. During a major storm, there's insufficient power to clean the panels, so the energy reserve to clean the panels exceeds the panels' potential output.
Provided that you could keep a PV panel perfectly clean during a dust storm, which is obviously impossible, then 1m^2 of a brand new 35% efficient PV panel would produce 35Wh/day. The minimum array size to continue producing 1MWe using ATK panels is 28,571m^2, or 6.4 football fields. The best thin film arrays I'm aware of are about 12% efficient. The minimum array size to produce 1MWe with thin film is 83,333m^2, or 18.7 football fields. If the arrays aren't at least that large, then you'll run out of available stored power in less than a week. You can forget about propellant production and start thinking about how to avoid killing everyone from running out of power.
Further notes on batteries suitable for use on Mars:
The small cylindrical Lithium-ion batteries that Tesla uses in vehicles, which have better energy density than prismatic cells, also have poor cold temperature charge / discharge performance in comparison to Yardney's prismatic cells, which is why NASA doesn't use them for any of their robotic missions to Mars. None of the cells with the chemistry used by Panasonic and Tesla have passed testing thus far, as they'd be unable to charge or discharge at a sufficient rate to store enough of the power output from a solar array without first expending more energy to heat the cells to a temperature within their operational temperature design limitations. After more than 14 years of operation on Mars, the prismatic cells aboard MER-B had more than 85% of their original capacity remaining, so prismatic cells with that particular cell chemistry have energy densities that's about as good as it gets using current technology (118.4Wh/kg for the cells alone and 80.4Wh/kg for the complete Orion MPCV battery pack, which passes all operational tests required for human space flight in deep space; every year this goes up a little as Yardney refines their technology). Since Starship is a human mission, only human-rated systems will be going along for the ride. Until batteries with triple the capacity of existing cells appear, that can actually operate over the temperature ranges required, the viability of a solar powered mission is essentially a question of how long a dust storm lasts until you can start producing enough power to clean the arrays and resume storing enough energy to make it to the next sunrise. Any storm that lasts for more than a month is still a mission killer.
Either way, 1m^2 of all known PV panel types can't produce enough energy through the course of an entire day to actually clean 1m^2 of PV in a severe dust storm, so any available power output will inevitably drop to zero. If the dust storm reduces reduces output to zero for just 1 week, then a constant 50kWe power requirement to keep the people and electronics from dying, an absurdly low figure given the amount of electronics you plan on bringing to Mars, is 8.4MWh worth of power (3.4MWh more than the maximum you thought you'd have to bring). That's equivalent to 104.5t of batteries that have actually been to Mars. If the storm lasts for 2 weeks, then that's 209t of batteries, which is 50t more than the weight of 100 KP reactors and all power cables, even with a power cable for each reactor that goes all the way back to whatever it's powering from 2km away (worst case scenario for weight, but best case for redundancy). Any storm that kills PV array output for longer than 2 weeks is a complete mission killer when using solar power, full stop.
This would be exactly why NASA is pursuing nuclear power to enhance our space exploration capabilities. NASA has engineers who can do basic math, despite all the aspersions cast upon their technical capabilities. If there's a 1MWe power requirement, or even a comparatively modest 50kWe constant power requirement to keep the lights on, along with a 500t delivered tonnage limitation for a single mission, then we can't throw enough tonnage of PV and batteries at this problem to actually solve it.
If new solar panel and battery technology becomes available, then we'll revisit this topic. Until then, I'm done responding to assertions that completely ignore actual scientific measurements of the actual surface environmental conditions on Mars. NASA has gone to the trouble of creating pretty color graphs of their scientific data that are very easy to understand. I've checked everything I have from NASA on this and there are no significant disagreements about the surface insolation conditions. Just as GW stated previously and repeatedly, we've known what those surface insolation conditions were like from the very first probes we sent there. Every subsequent mission has only added more data points to the pile of data that we already have. If the scientists from NASA and JPL can't speak authoritatively on this matter, then nobody else can.
Last edited by louis (2019-10-20 06:46:22)
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This article is helpful, referencing an MIT study that backs solar power as best for a Mars Mission:
https://www.universetoday.com/21293/des … -colonies/
Your figures for average insolation are meaningless. If I was an alien planning a mission to Earth that needed maximum solar power why would I plan to land on Rhode Island rather than in Arizona? Average figures are misleading as to what is possible. We will go where insolation is higher than average. That's clearly what Space X are planning to do. I note also that you do give figures for the optimal northern latitudes. The article above states: " Southern latitudes have much less solar energy available for most of the year."
Fig 3 in the following scientific paper gives Whs of insolation per sol for various latitudes (equator and mid latitudes) on Mars:
http://systemarchitect.mit.edu/docs/cooper10.pdf
This appears to vary between 4,500 Whs and about 10,000 Whs per sq metre per sol over a Martian year. So I am struggling to see where you get your figures of between 227 Whs and 492 Whs. Clearly someone is wrong. You don't give a link to your figures - I've given a link so all can see the reference.
The average from that scientific paper looks to be about 7,500 Whs per sol. If we capture 15% of that, that's 1.125 Kwhs per sq metre per day. So a 100 x 100 metre array would generate 11.25 MwHes per sol.
Mars 2020 is a another little rover! - just over a ton and consuming 110Wes. You're telling us that has something to teach us about a 500 ton human-crew mission to Mars that needs 1 Mwe of power? Not in any way plausible I'm afraid.
I know I said I was done addressing this egregiously bad solar power idea in my prior post, but the categorical refusal of certain parties to employ any basic math to this energy provisioning problem has motivated me to illustrate just how bad solar power is under optimal conditions on Mars in comparison to nuclear power. I'm sick of having the scientific data acquired by NASA / JPL questioned at every turn, despite the fact that the only reason a mission to Mars is possible at all is because of the decades of hard work by our space exploration agency. Feel free to continue to question all the politically-motivated make-work projects distracting us from our ultimate goal, or at least the goal that everyone says they agree upon.
NASA's avg. insolation per sol over 688 sols without dust storms
50 deg N: 277.32Wh/m^2
0 deg: 492.14Wh/m^2
30 deg S: 412.32Wh/m^2The figures shown above are what you get to work with, not what you actually receive from less than 100% efficient solar panels.
At the equator, that's 147.642Wh/sol for a 30% efficient panel or 172.249Wh/sol for a 35% efficient panel. The lowest credible mass figure I've seen using a 1,000m^2 Vectran inflatable deployable array comes in at around 700kg and produces 50.79MWh over 688 sols, or 73.8kWh/sol or 3,075.9W of constant power per sol. The mass associated with ATK fan technology is completely unworkable at the output level required for propellant production. Therefore, a solar array that produces equivalent power to a 10kWe KP reactor would weigh at least 2,275kg if it's operating at the equator under 100% clear sky conditions. This tracks precisely with NASA's assertion that KP would beat PV by at least a factor of 2, even at locations that are favorable to solar power.
If the power requirement is 1MWe constant power for propellant production, then the mass of 30% efficient solar panels alone is 227.5t. If we add 209t of batteries to survive for 2 weeks without power with a 50kWe constant power requirement, we've arrived at 436t. The PV arrays would only fill every square meter of the 825m^3 cargo volume of 2 Starships, so I guess that's a bonus, right? There's only 440 PV arrays that each weigh more than the 100 KP reactors that have to be deployed, along with 1,400 batteries. At least now I know why SpaceX has to land 5 Starships. There's no way they could ever make this work with fewer Starships. Since this new Starship is "Super Heavy", just like any PV power supply will be, I guess this new propellant plant and drilling equipment must be "Super Light".
Nukology for Non-Nukists
Anyone interested in understanding why NASA chose nuclear power for MSL and the Mars 2020 rovers should Google and read the 317 page "Final Environmental Impact Statement for the Mars 2020 Mission" report. Incidentally, this is what professional quality work, as opposed to PowerPoint presentations, actually looks like. Those people would then go directly to Page 2-39 and read through Page 2-50 to understand why nuclear power was selected over solar power. NASA doesn't make mission planning decisions by flipping coins. It wasn't "nukists" (whatever those are) saying "Man, it'd be super cool if we put some Plutonium on our Mars Mini Cooper! Hold my beer and watch this!". It certainly wasn't to offend the fragile sensibilities of our "green energy" religious zealots. It just might have been the fact that it was the only energy source that would actually get the job done without running smack into mass and volume constraints.As it turns out, nuclear power permits 100% functionality year round because nuclear power provides 24/7/365 power. Photovoltaic arrays with 35% BoL efficiency, photovoltaics combined with Radioisotope Heater Units (RHU's), and Radioisotope Thermoelectric Generators (RTG's) were all considered as candidate power sources for the Mars 2020 rover. Of all available options, even with the rover's miserly daily power requirement of 100Wh to 600Wh for mobility and science operations, only nuclear power could provide the energy required at all times at latitudes between the equator 30 deg N or 30 deg S.
A guaranteed operational lifetime of 1 calendar year using solar power was assessed as 50% or less (with significant impacts to available power during certain times of the year for any science to actually occur), irrespective of latitude, whereas the operational lifetime of 1 calendar year using nuclear power was assessed as 100%, irrespective of the latitude of the exploration sites under consideration for the mission. That means the rover could land anywhere between the equator and 30 deg N or 30 deg S and still meet 100% of its power demands to survive the cold and still perform the science experiments that American tax payers paid billions of dollars to execute. Those beautiful color graphs (Figure 2-18 on Page 2-40 and Figure 2-20 on Page 2-46) indicate seasons of the year and latitudes where mere survival of a solar or solar + RHU rover was not possible, given the rover's power requirements.
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sine function of angle to location from reference point of the equator gets the degree location on orbit reduction...math again
sorry you had to search for the reference but thanks for those you listed.
Here is a link for the tau on the last day:
https://space.stackexchange.com/questio … efined-and
Vacuum would have a τ of zero. An opacity of τ means that the atmosphere is reducing the direct intensity of light from the Sun, if it were directly overhead, by a factor of e−τ
It was measured by the rovers every sol by pointing the PanCam at the Sun, or where the Sun is supposed to be, measuring the intensity, correcting for the slant angle to the Sun, dividing by the intensity of the sunlight at Mars' current distance from the Sun, and taking the negative natural log of that. The reality is more complicated, but that's the general idea.
A typical τ on Mars at a good time of year is 0.5. It is not unusual for it to get to 1. At τ=2, you're seeing some dust storm activity. At τ=3, it's getting pretty bad. τ=4 or τ=5 is crazy bad.
A τ of 10.8 is almost non-sensical.
Emily Lakdawalla made a nice animation and table of what the Sun looks like as τ went up at one point for Spirit:
http://www.planetary.org/blogs/emily-la … /1053.htmlplot of the τ measured by Opportunity over its entire mission
https://www.lpl.arizona.edu/~lemmon/mars-tau-b.html
So an average t of 1 to 2 should be expected for every hour of a mars sol...
https://www.ssc.wisc.edu/~jfreese/soc75 … asures.pdf
Tau-equivalent measures
keep in mind wattage is not hourly based that is the collected and expended values with respect to time with regards to Wh/m^2 or in our case KWh/m^2 or MWh/m^2 so the
appears to vary between 4,500 Whs and about 10,000 Whs per sq metre per sol over a Martian year
so what is a Whs; as s is seconds. Then what is per sq meters per sol which is 25 hrs and not the days energy that can be recieved during the flat recieving condition or angled value or tracked values...
According to Hofstetter, a Mars mission should be able to transport several 2 metre-wide rolls of thin-film solar panel arrays. Rolling out an array of these thin-film rolls could supply ample energy to a colony. For example, if the array is positioned at 25° north, measuring 100×100 metres, 100 kilowatts can be generated.
notice the array is wattage and not KWhrs which is the high noon verticle overhead value.
Edit.. sorry I made an error with panel count:
working to the efficiency and input isolation for an input of 5 hrs day recieved energy since we are not at the equator where that number would be 6hrs. They said an array with ouptut of 100kw / 10,000 panels is giving you an output of 10 w or 0.01kw output only for the 1 hour time period for each panel. The panels efficiency looks like 3% instead of the 35% were on the rovers for a 128 w for the thin film flexible array which also means no dust storm or dust on the panels. Thats not mars at all and we know there is lots of dust that effect even the high noon hour recieving values as well as the daily whr that we can recieve..
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The exoatmospheric solar constant at Earth is 1353 W/sq.m, a value accepted as true and accurate for a long time now, by many authorities. It is just about 2.2 times less at Mars, or near 615 W/sq.m. An exoatmospheric solar panel pointed directly at the sun at Mars collects energy at the 615 W/sq.m rate. That would be incident energy upon the panel collected over a 4-hour period of about 2460 W-hours per sq.m. More below why I selected a 4-hour period.
If the solar panel were 20% efficient converting sunlight to electricity, that would be 492 W-hrs per sq.m of electric power collected over that 4-hour interval. Mars's atmosphere is both unclouded, and very thin. So a solar panel on the surface pointed directly at the sun at mid-day could produce something like 480-490 W-hrs per sq.m over that same 4-hour interval. But only while the sun is high in the sky, and with the correct inclination for a non-tracking panel.
I picked a 4 hour interval because solar has best recovery from about 10 AM to about 2 PM solar hour angle time. From dawn to 10 AM, and from 2 PM to sunset, recovery of power is much reduced because of adverse incidence angles. And it recovers zero at night. The only way around that adverse angle problem is the tracking collector, which is heavy, expensive, and difficult to maintain in a dusty environment.
What all that says is that you can collect at most something like 500-600-700 W-hours of electricity per sq.m per day, with fixed collectors on Mars, inclined correctly, at pretty much any latitude not polar. Hundreds of W-hrs, not 1000's or 10,000's, Louis!
You can find papers to cite that still say the Earth is flat. Just because some idiot at MIT or anywhere else published something doesn't make it true. You have to understand what the guy was really doing before misquoting his numbers.
Let's just say I am less than impressed, just because of some published-paper citation. I trust the fundamental numbers that I can actually verify. Kbd512 has them for solar on Mars, Louis does not.
GW
Last edited by GW Johnson (2019-10-20 10:36:30)
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https://www.planetary.org/blogs/emily-l … /1053.html
https://en.wikipedia.org/wiki/Tau
Tau-equivalent measures
https://mars.jpl.nasa.gov/msl/mission-u … asurements
https://mars.nasa.gov/resources/21921/o … is-energy/
This is the wind blowing the sand on the surface of mars
Seems that the daily tau is just under 1
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Mit edu are college students that want to be engineers with there publishing of papers which may or may not be riddled with errors just like my spelling and punctuation, so take them with a grain of salt....
Here is the seasonal rotation of mars for northern summer which is clear in the red color
A12 month period is used to mark the rotation of which the low dust months are 1-6 with 7-12 being the high dust months.
here is the energy levels for earth
Why we have different levels of energy as we move away from the equator where we will recieve 98% of the energy from orbit has to due with the atmospher thickness
When we angle the panels toward the sun we will recieve the total amount of energy which is per meter square but when we layu a panel flat on the ground the suns rays that would hit a 1 meter square misses most of the target panel.
This represents the level of energy for the earth and with mars the levels are lower such that we get what kdb512 indicated for the distance and smaller radiaus of mars
Page 4 of this document shows what mars will recieve
https://deepblue.lib.umich.edu/bitstrea … sequence=1
Solar Radiation Incident on Mars and the Outer Planets: Latitudinal, Seasonal, and Atmospheric Effects
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Sorry, hadn't spotted that solar radiation figure related to in Fig 3 related to a tracking array. That explains the v. high daily figures which aren't relevant since I don't think anyone is seriously proposing tracking arrays for Mission One.
So ignore that!
This map gives solar radiance on Earth. The areas with highest insolation reach just over 9 KwH per sq. metre daily average .
https://upload.wikimedia.org/wikipedia/ … ion_01.png
At a minimum, taking 43% of that figure, I would expect then there would be places on Mars with figures at least around 3.87 KwHs per sq metre. But Mars's seasonal wobble and greater levels of indirect insolation might raise that higher possibly...
3870 Whs per sol is way better than the best figure quoted by kbd, of 492 Whs per sol. So something is awry. At 15% efficiency that would give you 0.58 Kwh per sq. metre so a 100 x 100 metre array (10,000 sq. metres) would give 5800 KwHes. At 30% efficiency you would get 11600 KwHes.
This ESA paper indicates the proposed rover will achieve 607Whe/m2 per Sol for its solar array.
https://sci.esa.int/documents/34923/361 … report.pdf
Last edited by louis (2019-10-20 15:27:52)
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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See my post above - I misrepresented what the MIT guys were saying - they were talking about what a tracking solar array would receive...so not relevant to Mars Mission One I think. It's an odd way to approach the problem...
However, I still think kbd's figures are not indicative of best insolation sites on Mars. The ESA solar panels must be receiving insolation of around 1,700 Whs per sol if the solar panels have v. high efficiency of 35%. That's about four times more than kbd highest figure.
Mit edu are college students that want to be engineers with there publishing of papers which may or may not be riddled with errors just like my spelling and punctuation, so take them with a grain of salt....
Here is the seasonal rotation of mars for northern summer which is clear in the red color
http://www-mars.lmd.jussieu.fr/mars/time/orbit.pngA12 month period is used to mark the rotation of which the low dust months are 1-6 with 7-12 being the high dust months.
here is the energy levels for earth
https://www.altenergymag.com/articles/0 … z/fig8.gifWhy we have different levels of energy as we move away from the equator where we will recieve 98% of the energy from orbit has to due with the atmospher thickness
https://www.altenergymag.com/articles/0 … z/fig1.gif
When we angle the panels toward the sun we will recieve the total amount of energy which is per meter square but when we layu a panel flat on the ground the suns rays that would hit a 1 meter square misses most of the target panel.https://www.altenergymag.com/articles/0 … z/fig6.gif
This represents the level of energy for the earth and with mars the levels are lower such that we get what kdb512 indicated for the distance and smaller radiaus of mars
https://www.altenergymag.com/articles/0 … z/fig7.gif
Page 4 of this document shows what mars will recieve
https://deepblue.lib.umich.edu/bitstrea … sequence=1
Solar Radiation Incident on Mars and the Outer Planets: Latitudinal, Seasonal, and Atmospheric Effects
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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I see why the error now. I also see the tracking as well which will raise the ouytput values by 30% more for the sol number.
The earth image gives the recieved energy on the surface, it is not what a solar cell will capture and not from orbit as we are showing for mars for kbd512 numbers. That is due to the level of dust not being all that consistent.
The area you quoted is also at the tops of mountians which is not the same as we would see in the valleys which would be lower. The mountians are closer to the sun so less solar energy spread occurs for each meter which makes it higher. This would also hold true for mars as we can only land in the valleys where the energy levels will be lower as well.
Earth equator is 1100 w with mars equator being 430 w each at a sq. meters measurement for the hour period. That is why there is such difference for mars on its surface.
I will add in the insight numbers in a bit as these are the target application for an array which is high grade solar on a hinged panel that makes it fan out. Its not a flexible thin film array.
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This NASA paper gives a high of 3882 Whs for insolation per sq metre per sol. (Figure IV on page 18).
https://ntrs.nasa.gov/archive/nasa/casi … 018252.pdf
The highest direct beam insolation is 2553 Whs per sq metre per sol.
Harvesting 20% of that would give you just over 0.5 KwHes per sq. metre.
80,000 sq. metres (282 x 282 metres), producing 40 MwHes per sol should provide more than enough cover for storage and the negative impact of very severe and prolonged dust storms.
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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Each of InSight's two solar wings are 7 feet (2.2 meters) wide consisting of SolAero ZTJ triple-junction solar cells made of InGaP/InGaAs/Ge arranged on Orbital ATK UltraFlex arrays. Solar panels yielded 4.6 kilowatt-hours. for a mars sol of 25 hrs.
They only expect 3.0 kwhrs or 3,000 whrs typical so this is a dusty panel number before going lower..
a=pi * r^2 turns out its 3.8 m^2 for each panel or a total of 7.6 m^2 to recieve the solar energy on mars.
https://www.jpl.nasa.gov/news/press_kit … esskit.pdf
4.6 kwhrs / 7.6 m ^2 = 605 whrs for each sq meter of panels
430 w * 35% = 150.5 w
605 whrs / 150.5 = 4 hrs is solar collection time....
A pair of rechargeable, 25 amp-hour lithium-ion batteries located on the lander will provide energy storage. The lithium-ion batteries are from the Yardney Division of EaglePicher Technologies, East Greenwich, Rhode Island
notice the amp hr rating matches the mars solar day....which makes the bus voltage 48v
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Louis, To get the amount of energy of 40 MWhes you are going to deploy 8,700 insight units....which are fully trackable.
https://www.jpl.nasa.gov/news/press_kit … pacecraft/
https://ntrs.nasa.gov/archive/nasa/casi … 112621.pdf
https://ntrs.nasa.gov/archive/nasa/casi … 015793.pdf
energy storage technologies
https://www.lpi.usra.edu/opag/meetings/ … ampudi.pdf
https://www.eaglepicher.com/resources/n … ht-lander/
Two of EaglePicher’s 8-cell, 28 volt, 30 amp-hour batteries are mounted to InSight’s baseplate and will be used to store solar power that the lander collects. Thanks to a new, remarkable engineering development of electrolyte formulation, the batteries will continue to operate down to -35°C (-31°F) with no lithium plating. This will enable the battery to endure all Martian seasons. EaglePicher batteries were used in the Mars rovers Spirit, Opportunity and Curiosity.
The higher amp hr and voltage allows for a higher level of storage in summer.
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Louis,
Here's what I'm trying to point out. Let's use your figures since we're both using the same source for a change. The total daily power requirement is 24MWh. So, let's say we have 2,553Wh/m^2/sol of total power. A 35% efficient array provides 893.55Wh/m^2/sol.
893.55Wh/sol / 24.65hrs/sol = 36.25W/m^2 of constant time-averaged power
In LEO we'd get 695.5W of constant time-averaged power
This power could be used in 1 of 2 ways:
1. Try to use all of the power whenever it's available
The array would be dramatically oversized to produce every bit of the 24MWh power requirement during the day, leading to insanely high power fluctuations measured in megawatts. Whenever power grids here on Earth have power fluctuations of that scale, they also tend to have brownouts or blackouts or blown transformers. Apart from rail guns and particle physics experiments, we just don't design power systems that way. If you've ever seen the power cables and switches used by rail guns, it'd be fairly obvious why not. The power cables closely resemble small tree branches. If the power conversion equipment can't reconcile the load with the source, then the load drops or the power goes to ground or the power conversion equipment gets hot enough to start melting things.
2. Store some of that power and discharge it later
If the propellant plant won't shut down at night or won't handle that kind of power fluctuation, which seems a dubious proposition at best, then we're obviously talking about storing some of that power. It seems like some kind of serious power buffer is required. That would typically mean batteries and lots of them. If only half of the power produced by the solar array is stored, then 8.4MWh provides 525kWh worth of power for 16 hours. That's pretty close to half-rate production during the hours when maximum solar power isn't available. That kind of discharge rate seems feasible and they'd be recharged during peak hours.
Caveats:
If there's a constant power requirement for 50kWe of keep-alive power, then one silly little week of going without power represents 8.4MWh worth of power. I don't think you could retain all the propellants during the week using battery power alone, but you could at least reduce boil-off by using IVF technology.
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I'd refer you to the Reddit post from Blake I linked to on the other thread:
https://www.reddit.com/r/spacex/comment … ant_plant/
1. As for managing the system, I presume you would have some sort of battery buffer. In my proposal, I allocated 30 tons for battery storage which could store 6 MwHes perhaps, in addition to whatever battery storage is available on the Starships (substantial - probably around 3 Mwhe). Mars is not an environment like Earth where it can be cloudy, low light one moment and brilliant sunshine the next. So I don't think there will be wild fluctuations in power being inputted. So, it's not like trying to handle extreme fluctuations that you get when you are managing wind and solar on Earth (but of course, even here, 99.9% of the time we manage those successfully). I normally allow 30% for additional equipment required to run the system. That's pretty generous, I think.
2. Blake's gone into this in some detail. He thinks a day-night storage of 1.6 Mwhs would be sufficient to protect the propellant plant facility (100 Kwhe) and has given mass for a PV powered propellant plant on the assumption of a peak of 4Mwes passing through. On my proposal you would have chemical batteries with 9 Mwhes and a store of methane of 3 tons. Even if somehow (I don't believe possible) there was the severest ever dust storm which reduced PV output to zero, you'd be able to keep going at 100 Kwe for over 30 sols (at a minimum as that's assuming you haven't already produced any methane and oxygen through the propellant plant). Working at 100 Kwe would in those circumstances require more 10Kwe methane generators at maybe 500 Kg a time...so 5 tons in total. However, looking at this sensibly, you can probably shut down the propellant plant temporarily if necessary without disabling it. That would make more sense.
We need to distinguish between "day-night storage" and "major shortfall" storage.
3. I think Blake's analysis is good but would criticise it on two grounds: (a) they are too optimistic about the PV power-mass ratio I think - you could probably double his estimate of 20 tons for the PV film itself and (b) they seem not to have accounted for severe dust storms - ie reduction in PV power, meaning you have to have a large PV facility and a bigger propellant plant facility.
Louis,
Here's what I'm trying to point out. Let's use your figures since we're both using the same source for a change. The total daily power requirement is 24MWh. So, let's say we have 2,553Wh/m^2/sol of total power. A 35% efficient array provides 893.55Wh/m^2/sol.
893.55Wh/sol / 24.65hrs/sol = 36.25W/m^2 of constant time-averaged power
In LEO we'd get 695.5W of constant time-averaged power
This power could be used in 1 of 2 ways:
1. Try to use all of the power whenever it's available
The array would be dramatically oversized to produce every bit of the 24MWh power requirement during the day, leading to insanely high power fluctuations measured in megawatts. Whenever power grids here on Earth have power fluctuations of that scale, they also tend to have brownouts or blackouts or blown transformers. Apart from rail guns and particle physics experiments, we just don't design power systems that way. If you've ever seen the power cables and switches used by rail guns, it'd be fairly obvious why not. The power cables closely resemble small tree branches. If the power conversion equipment can't reconcile the load with the source, then the load drops or the power goes to ground or the power conversion equipment gets hot enough to start melting things.
2. Store some of that power and discharge it later
If the propellant plant won't shut down at night or won't handle that kind of power fluctuation, which seems a dubious proposition at best, then we're obviously talking about storing some of that power. It seems like some kind of serious power buffer is required. That would typically mean batteries and lots of them. If only half of the power produced by the solar array is stored, then 8.4MWh provides 525kWh worth of power for 16 hours. That's pretty close to half-rate production during the hours when maximum solar power isn't available. That kind of discharge rate seems feasible and they'd be recharged during peak hours.
Caveats:
If there's a constant power requirement for 50kWe of keep-alive power, then one silly little week of going without power represents 8.4MWh worth of power. I don't think you could retain all the propellants during the week using battery power alone, but you could at least reduce boil-off by using IVF technology.
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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When solar is collecting energy the charge convertor outputs it to batteries depending on the charge on them until they are full at which time they switch into a trickle mode charging where the excess power is disapated into a load such as a hot water tank or some other method to make use of that output power from the solar panel.
Panel power of collection max must match the storing power capacity of the batteries. Any amount to excessive for storage means batteries that are always dead and to little means over heated circuits and damaged batteries.
So far we have the average panel for a total mars but the seasons make the energy in summer go up for whrs as the period of day hours of collection goes up while in winter it goes down for the same panels in terms of whr collected.
That said to get a constant level of energy you will need to play with angles of the panels or shading to compensate for the over power creation as well as for the lower winter levels.
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Solar inherently requires batteries. Intermittency. The longer the darkness, the bigger the battery. Or else the stronger the requirement for "something else" that works in the dark. Or most likely, both.
That's just a dirty little fact of life. Don't need any citations to know that! Learned about it in engineering school. Decades and decades ago.
GW
GW Johnson
McGregor, Texas
"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"
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In relation to propellant production, the issue is how far you need to maintain constant power at x figure. I doubt that is an essential requirement. So then it all becomes a question of balancing PV mass, battery mass, other energy storage mass and the propellant plant mass (which would have to be larger for PV compared with nuclear).
You then need to compare all that mass with nuclear power mass (which certainly appears on the face of it to be larger) and address questions of ease of deployment and maintenance.
Solar inherently requires batteries. Intermittency. The longer the darkness, the bigger the battery. Or else the stronger the requirement for "something else" that works in the dark. Or most likely, both.
That's just a dirty little fact of life. Don't need any citations to know that! Learned about it in engineering school. Decades and decades ago.
GW
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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Think of solar as if you were charging the battery with and alternator in you car. The 12 lead acid battery will only charge when the alternator is providing at least 0.1 v above the disconected battery voltage but sine we are loading the battery when that happens the input must be even greater to allow for it to actually charge at all. The alternator charges at its peak value of roughly 14v which conisides with the high noon value for solar. Once the voltage to charge the battery falls just 0.1 vol below the battery value its now in a discharge value. In this case a dead lead acid voltage is about 11v even thou it can still give power for lower voltaged loading electrical items. Lithium is different in that once you cross that low value you can damage the internals of the battery and is why fast charging is bad when its down so low.
Solar has one more block to connect between it and the battery thats call the charge convertor which controls the battery charging action. This circuity turns on only after you are above the voltage of the battery to charge it even if its dust covered values is the same as the batteries at high noon it will not be charging the battery at all. Unloading the battery loads will not help to make that battery charge under that condition and if you do load the battery down the input power from the panel will not cause charging as its not getting all that much power coming out of the panel as as to charge the battery even thou the condition of the voltage bing lower than the panles voltage on the convertor.
Solar does not directly connect from the panel to the battery in any design or use.
Now back to the power wall diagram of solar charging and outputing for the night from it. The charging curve is higher than the battery but to allow the most energy to go into the battery you need to be below the whr for where it turns on for charging in order to have the most energy for the night hours for the enrgy you are using from the day times solar energy collection.
I have meantioned it before that the turn on and off occurs around 70% of the high noon value but its more like 60 to 55 % of what you can load the batteries voltage that matter the most so as to be able to charge the battery for its total whr use for the night.
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I get it SpaceNut - no PV panel can ever usefully charge a battery that can in turn usefully discharge that electricity into a circuit. Thanks for the technical advice.
Meanwhile, in South Australia, they take a slightly different view:
https://www.sciencealert.com/remember-t … first-year
Think of solar as if you were charging the battery with and alternator in you car. The 12 lead acid battery will only charge when the alternator is providing at least 0.1 v above the disconected battery voltage but sine we are loading the battery when that happens the input must be even greater to allow for it to actually charge at all. The alternator charges at its peak value of roughly 14v which conisides with the high noon value for solar. Once the voltage to charge the battery falls just 0.1 vol below the battery value its now in a discharge value. In this case a dead lead acid voltage is about 11v even thou it can still give power for lower voltaged loading electrical items. Lithium is different in that once you cross that low value you can damage the internals of the battery and is why fast charging is bad when its down so low.
Solar has one more block to connect between it and the battery thats call the charge convertor which controls the battery charging action. This circuity turns on only after you are above the voltage of the battery to charge it even if its dust covered values is the same as the batteries at high noon it will not be charging the battery at all. Unloading the battery loads will not help to make that battery charge under that condition and if you do load the battery down the input power from the panel will not cause charging as its not getting all that much power coming out of the panel as as to charge the battery even thou the condition of the voltage bing lower than the panles voltage on the convertor.
Solar does not directly connect from the panel to the battery in any design or use.
Now back to the power wall diagram of solar charging and outputing for the night from it. The charging curve is higher than the battery but to allow the most energy to go into the battery you need to be below the whr for where it turns on for charging in order to have the most energy for the night hours for the enrgy you are using from the day times solar energy collection.
I have meantioned it before that the turn on and off occurs around 70% of the high noon value but its more like 60 to 55 % of what you can load the batteries voltage that matter the most so as to be able to charge the battery for its total whr use for the night.
Let's Go to Mars...Google on: Fast Track to Mars blogspot.com
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Thanks for the link.
For solar which is DC you cannot connect the units directly without changing it to the the matching frequency of the power grid in AC form (dc to ac invertor). This is for net metering for home users as you are grid tieing in the system to offset money spent on pwer from the power companies. These units take the excess energy in from the AC grid and store it once its converted back to dc in the batteries with a charge convertor, to deliver it back out from the storage batteries it then goes through an DC to AC convertor or invertor back to the grid for use. What this does is make use of more solar and off loads the running of power plants that use fossil fuels in all forms. This works just like an UPS or universal power source which is ac in and ac comes out when the power fails via opening and closing relays inside.
edit
https://www.carboncommentary.com/blog/2 … -and-reuse
The 90 kW PV installation at the Augsburg building will produce an average of around 10 kW of power over the course of a year. (The panels have to be laid flat, reducing the yield).
Sure its on earth but even with the mars dust difusing the light still will be not as good as this would indicate...
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For the topic "Nuclear power is safe" the article at the link below should be a good fit.
It should also drive Louis bonkers << grin >>
The developers of the system described even include solar panels in their design, to show a positive attitude toward renewable energy.
https://www.yahoo.com/finance/news/oklo … 00370.html
(th)
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The fuel type sounds like that which is being used by the kilowatt reactors for mars with the heat pipes....
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