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I am trying to size (weight and surface) a Solar Array for a Mars Base (university project) and the main specification I get is Specific Power [W/kg]. However, I need to relate both weight and surface, so a figure like [kg/m2] would be suitable for me (can't find it anywhere).

Obviously, specific power is a measure of power per kilogram of Solar Array. But under what conditions is this value usually given? Surely it is not the same to have the Solar Array in a GEO orbit, on Mars surface, during peak Solar Flux, at sunrise...

Could you clarify this for me please?

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  • $\begingroup$ Can you add links to one or two specific examples of where you are seeing specific power quoted but no specific conditions are included in an explanation? $\endgroup$ – uhoh Apr 10 '18 at 10:50
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    $\begingroup$ If you’re seeing W/kg figures in the context of spacecraft power supply, I would expect it to be referring to average solar flux at 1 AU from the sun, unless it’s in a comparison of options for a specific interplanetary mission. $\endgroup$ – Russell Borogove Apr 10 '18 at 11:27
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    $\begingroup$ @uhoh orbitalatk.com/space-systems/space-components/solar-arrays/docs/… specifies a power output of 150W/kg (I assume it's 1AU). Then on the second page it gives a power output of >103 W/kg for the Mars Phoenix Lander Mission. I am aware of the fact that the cells may not be exactly the same, but the difference just seems too big considering that the Solar Flux on Mars surface is much less when compared to Earth. $\endgroup$ – Roger Pedrós Bòria Apr 10 '18 at 13:05
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    $\begingroup$ @RogerPedrósBòria I suspect the 103 W/kg to 150 W/kg difference is due to 10+ years of development since Mars Phoenix Lander, not due to Earth-Mars differences. $\endgroup$ – Russell Borogove Apr 10 '18 at 17:45
  • $\begingroup$ Be careful with the data - solar flux on Mars surface is quite similar to that on Earth surface. Earth atmosphere absorbs a lot. If that number is used instead of LEO value it all becomes apples to oranges. $\endgroup$ – SF. Apr 26 '18 at 0:13
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W/kg is just that: relation of generated power per kg of mass. If nothing else is quoted, you can assume that this is given for an earth orbit, where you have as input to he solar arry one solarconstant of radiation.

As to your comment on solar flux: Don't forget that the flux reduces with the square of the distance, as for every radiation source. Have you done the math?

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  • $\begingroup$ Let's add Watt/kg is an exceptionally important factor for ion engine powered craft. With very low thrust and huge energy consumption of these engines, the panels contribute a lot to mass of the craft - the thrust to weight ratio may become prohibitively low, resulting in unacceptably long burns. In case of Mars this parameter isn't as important as it only influences payload cost/value for transporting them to Mars. $\endgroup$ – SF. Apr 25 '18 at 23:32
  • $\begingroup$ You may find this parameter being named "alpha" as well. Lack of energy sources with sufficiently good alpha is the primary reason why VASIMR engines are not practically viable currently. $\endgroup$ – SF. Apr 25 '18 at 23:35
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    $\begingroup$ "Have you done the math?" is not part of a good SE answer, which should be written in a way that is useful to future readers as well. If you would like to address the OP directly, it's best to do as a comment, ideally in a way that asks for clarification. In the question the OP gives examples of both different distances, and different incident angles, and is asking for an answer which comprehensively addresses these issues, so I don't see how this is actually an answer to the question as asked. $\endgroup$ – uhoh Apr 26 '18 at 0:34
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A solar array consists mainly of solar cells, a supporting structure and an electric harness.

The mass of the solar cells and their efficiency can be found in datasheets online; look for example for the azurspace website. Multiplying the efficiency with the incoming solar flux on Mars surface, will give you the power per square meter of solar cell generated. There are some additional losses in the power system and due to thermal effects, but this will give you a good first estimate. So: P[W/m^2]=efficiency * mars flux [W/m^2]

The solar cells need to be supported. In satellites this currently is done with honeycomb panels. However for your Mars base you might want a specific design. It's probably best if you make an initial structure, and estimate the mass of that structure (per unit area).

If the mass of the electrical harness is ignored you can calculate the mass of a square meter of cells, and add that to the mass of a square meter of your structure.

EDIT Based on Nathan Tuggy's comment, let me clarify some things. I might have misunderstood the question. I don't think it's possible to get the area from the specific power (per unit mass) alone. You would indeed need to know the mass per area of the panels in order to get the specific power per unit area.

The specific power should be given within some context to be useful. Power generated is dependent on incoming flux, efficiency and temperature as well.

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  • $\begingroup$ This seems to be coming at things from the other end, calculating specific power from mass, flux, and area, rather than area from specific power and power needed. $\endgroup$ – Nathan Tuggy Jun 23 '18 at 8:37
  • $\begingroup$ I've updated the answer. I've given that approach, since I don't think it's possible to get the area from this variables alone. $\endgroup$ – NvdP Jun 23 '18 at 8:54

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