7

Good morning everyone,

I am working on a project for my university where we need to perform a PrePhase A study for a Mars Infrastructure for Propellant production as a means of support for future missions.

Summarizing, the power requirements for the propellant production are 10.1 kW (constantly during day and night). I am trying to size the Solar Array with the following model for the Solar Flux.

SolarFluxOnMArs

When dimensioning the Solar Array, I am using this basic formula:

Formula

Then, I integrate the power of the Array during all day and make sure that the total energy [Wh] produced is equal to the total energy required [10.1 kW * 24.65h]. If the power required is higher than the Solar Array power, batteries will be in charge of providing it.

What I am missing right now is a value for the weight of a solar panel. I currently have in mind 2 models:

  1. UltraFlex Solar Array
  2. Azur Space Triple Junction Solar Cell 3G30C-Advanced

Apparently the first one offers a better weight performance, while the second one has a better efficiency. However, I cannot find a value for the weight per square meter of both models.

Would you have any information on that or do you have any input on usual weight of space solar panels as a function of panel surface?

Also, what is your critisicm on my sizing procedure? This is the first time doing this, so I am not sure if I am following the right process.

Thank you very much.

  • what research have you done? https://www.orbitalatk.com/space-systems/space-components/solar-arrays/docs/UltraFlex_Factsheet.pdf suggests 150W/kg at beggining of life –  Apr 09 '18 at 08:33
  • Yes, but under what conditions are those 150W/kg. Do they provide the same power output with independence of solar conditions? Solar Flux on Earth is about 1360 W/m2, whereas for Mars is less than half of that value. Sorry for the ignorance, but I don't know many things about solar energy, so I don't know what it means when they specify 150W/kg. – Roger Pedrós Bòria Apr 09 '18 at 08:36
  • is this not answered in https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19890018252.pdf –  Apr 09 '18 at 08:45
  • The paper is related to Solar Irradiance on Mars. My question was about under what conditions this 150W/kg are specified. – Roger Pedrós Bòria Apr 09 '18 at 09:22
  • Note that PV efficiency goes up as the temperature goes down. Then, if you do not reduce the number of panels you must upgrade wiring, inverters, and so on. – MBM Apr 09 '18 at 17:53
  • 5
    +1 This is a model first question for a new user. What I see is the time and effort taken to explain the problem, to outline the OP's current level of understanding of the topic, to then outline the steps already taken by the OP, and then to ask specific, well-defined questions and to invite comment on ways to improve the approach. – uhoh Apr 10 '18 at 02:14
  • @MBM is right. See for example this answer to the question Temperature of photovoltaic cells vs efficiency - is it ever actively controlled? – uhoh Apr 10 '18 at 02:26
  • @uhoh so, the question doesn't maky any sense to you? – Roger Pedrós Bòria Apr 10 '18 at 10:12
  • @MBM I am aware of this fact, but what does it have to do with the more general question that I am having? (consider this is a PrePhase A study) – Roger Pedrós Bòria Apr 10 '18 at 10:13
  • @RogerPedrósBòria your question makes perfect sense to me. There's one "unclear what you're asking" vote to close your question that perhaps you can't see yet (may need more reputation points for this privilege) so I'm just saying that "it certainly IS clear what you are asking!" If nobody posts an answer in a day or two I will add one myself. – uhoh Apr 10 '18 at 10:22
  • @uhoh any more insight? – Roger Pedrós Bòria Apr 23 '18 at 13:55
  • 1
    @RogerPedrósBòria I'll take a look in the next day or so, thanks for the ping! – uhoh Apr 23 '18 at 14:48
  • Remember if you need the power 24.5/7, you also need to budget for the power conversion efficiency, storage loss, re-conversion from storage efficiency. Sizing: Do you plan on stationary flat panels, or sun-tracking? Consider the issues of dust accumulation. Consider the possible issues of varying performance in differing thermal environments(mars gets cold overnight) – CuteKItty_pleaseStopBArking Sep 25 '21 at 17:28
  • Multiply by 37 and then decide to use Kilopower instead. – Chris B. Behrens Nov 15 '21 at 16:23

3 Answers3

4

It is usually easier to first determine the energy required by your system. You have done that already - assuming that the ISRU plant needs to continuously operate at 10.1 kW per sol:

$$ 10.1 \ [kW] * 24.65 \ [h] \approx 249 \ [kWh] $$

The solar arrays will need to be able to generate this much energy every sol to keep propellant production running continuously. You need to estimate the area of solar arrays needed to produce this energy quota.

First integrate the solar flux profile to get the solar irradiance per sol (measured in units Wh/m2/sol). Then multiply this value by the expected solar cell conversion efficiency factor to get the solar energy generated per unit area by the solar arrays. (Note that the cell conversion efficiency is not trivial to calculate as it will be dependent on temperature, attenuation of light in the atmosphere, optical depth, Mars solar longitude etc. but for a first order guess ~20% might suffice if considering 3G30C cells). You can now use this value to work out the area of solar array needed to provide ~249 kWh per sol.

To estimate the mass of this system you can make some assumptions about ultra-flex arrays using information online:

Using these values you can work out the approximate power an array will produce at 1AU. Assuming a solar flux of ~1360 W/kg, and a cell efficiency of ~30% would indicate that the fold out arrays may have an areal density between 2.7 - 5.1 kg/m2 (depending on the specific power assumed). You can then use this value with your computed total solar array area to estimate the mass.

To size the batteries work out how much of the 249 kWh total energy needs to be provided outside of sunlight hours (i.e. multiple the 10.1 kW by time the battery needs to power operations). Note that there will be some conversion losses associated with storing energy in the battery during sunlight hours. This would increase the total energy requirement calculated in the first step.

For extra credit you may want to also consider the impact of any of the following:

  • Dust accumulation on the solar arrays with mission duration
  • Variations in the solar flux profile as a function of latitude, solar longitude and optical depth (i.e. where on the surface of Mars the plant is located, Mars season, duration of sunlight on a given day)

These will impact the amount of power the solar arrays will be able to collect on a given day. You can find some inputs on how to assess this here: https://ntrs.nasa.gov/api/citations/19890018252/downloads/19890018252.pdf

Though, given this problem is for a pre-phase A assessment, your constant solar flux profile is probably representative enough to assess feasibility.

Fred
  • 13,060
  • 4
  • 41
  • 78
Armadillo
  • 1,426
  • 6
  • 17
2

I like this question and the broader ISRU return fuel for Mars in general. So I thought I'd throw my two cents worth even though I'm not qualified ... So lets start by sizing an Earth based 10kw solar panel system, and scale up from there. https://www.sunpowerbythesolarquote.com/post/10 kW-solar-system

...This means that to install the panels for this system, you will need about of 660 square feet of roof space. A 10 kW system is comprised of 30 to 40 panels total depending on the efficiency of the panels you choose ... In total, this system generates about 10,000 watts of electricity per hour as defined by laboratory Standard Test ... This breaks down to an average of between 29 and 46 kWh per day.

So lets say we're near the Mars equator and use the 45 kWh per day multiplied by 0.59 since Mars gets less Solar energy than Earth.

I'm using a 24 hour day, but that shouldn't effect the average energy required, just increase the batteries a little for the extra hour per day. This 30 panel system if on Mars would generate: $$ 45\text{kWh} * 0.59 = 26.5\text{kWh}/\text{day} $$ We want $10\text{kW}*24=240\text{kWh}/\text{day}$ so we divide 240kWh/26.5kWh and then scale up the number of panels by a factor of ~10x.

So a 300 panel Mars system might generate 10 kW continuously with battery storage and with a peak power of 59 kW with each panel about 1.6 m2 in area, so the Mars 10 kW continuous solar farm might be 25 meters x 20 meters in total.

Fred
  • 13,060
  • 4
  • 41
  • 78
Sheldon
  • 555
  • 4
  • 17
2

Your two models are not really comparable. A solar array is constructed of the cells (plus wiring, etc) and the substrate (i.e. what the cells are bonded to). The UltraFlex Solar Array is the whole thing whereas the Azur cells are just that - the cells alone. Clearly, the substrate is quite a significant contributor to the mass. If you have mechanisms to unfold the panels, they need to be included, too. Without a detailed look, it is probably not too much of a stretch to assume you could couple aspects of the UltraFlex with the better cells.

There is also a trade-off with battery size to consider. The battery will won't be small, so if you are worried about mass, you would probably need to include that in the overall power decision.

Paul Norridge
  • 381
  • 1
  • 7