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I'd like to power a small, headless, linux box in the desert for 10+ hours a day. It draws the lower end of 30-40 watts. I'd like to not use a generator and go the battery + solar route. Can you tell me the best setup I should use in terms of battery size vs solar panel output?

EDIT: Adding additional info: The desert where this unit will be for a week will have about 5 hours of useable sunlight each day (Black Rock Desert, early September). The computer is a standard shuttle (mini-pc) with SSD and wifi (more details in the comments). I know I can get lower than 30-watts PC but for the application and user load I expect, I'd like to stick with what I have now for the initial version (though hardware suggestions are always welcome).

I will be charging the battery up fully before the week long stay, which would allot me a little time with no sun / solar panel setup. Thanks.

Update! I took some advice below and got an old NSLU2 off of Craigslist for $45. Now I'm running a Debian web server at 2.5 watts! (6.4 watts with two usb hard drives and while transferring files.) I may be able to run the whole setup off a battery all week without the need for a charge.

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    \$\begingroup\$ Is there any reason why it draws so much? What kind of hardware are you running and what task is it meant to do...? I ask because there are a variety of fairly capable embedded linux platforms which draw anywhere between 1.5 Watts and 5 watts given their computational power. \$\endgroup\$ – Jon L Sep 20 '11 at 5:19
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    \$\begingroup\$ Is it a fixed setup, like in a house, or mobile, like in a truck? Will it be a single prototype, or built multiple times? Should it be networked, cheap, reliable, rugged, fancy? \$\endgroup\$ – posipiet Sep 20 '11 at 8:48
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    \$\begingroup\$ The Linksys NSLU2 draws about 3.5W (with no HD) and can run from a flash stick. Unless you have a fixed requirement for the box you have, I'd consider something lower power. \$\endgroup\$ – Toby Jaffey Sep 20 '11 at 14:57
  • \$\begingroup\$ @Jon, the machine draws so much power because I built it as cheaply as possible ($220). Here's the stats: Foxconn R10-G3 Supports 65W for Intel Core 2 Quad, Core 2 Duo processors Intel Socket T(LGA775) Intel G31 Intel GMA 3100 ... Item #: N82E16856119018 Rendition by Crucial 2GB 240-Pin DDR2 SDRAM DDR2 800 (PC2 6400) Desktop Memory Model RM25664AA800 Item #: N82E16820148235 Intel Celeron 430 Conroe-L 1.8GHz LGA 775 35W Single-Core Processor BX80557430 Item #: N82E16819116039 Patriot Torqx 2 PT232GS25SSDR 2.5" 32GB SATA II Internal Solid State Drive (SSD) Item #: N82E16820220580 \$\endgroup\$ – Mauvis Ledford Sep 20 '11 at 16:27
  • \$\begingroup\$ @posipiet: This is a standard shuttle computer with regular AC input. It won't be attached to a car, probably will be inside of a tent at all times. Has a built in wifi card and a 32GB SSD to host a text-only version of wikipedia and messaging system among other things. It's an fun project for Burning Man. Anticipated max load on the server is 10 people at once connected via Androids and iPhones. If it's successful, I'll offer up the blueprint and software to anyone that wants it. \$\endgroup\$ – Mauvis Ledford Sep 20 '11 at 16:36
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You want a solar powered power supply that provides 40 W for 10 hours every day, for 400 W-h per day. Obviously all this power originally comes into the system via the solar panel, so it must be sized accordingly. Let's say the switching power supplies in the system are a total of 70% efficient. Then there is power lost in storing and later retrieving it from the battery. Let's say that's another 70%. Combining those two, you have about 50% efficiency from solar panel output to ultimate load.

Now you know the solar panel has to produce about 800 W-h per day. With a very large battery, it only needs to produce this averaged over a long time. The smaller the battery, the smaller the averaging window where the panel still has to produce this power. How much is reasonable depends on factors you haven't told us.

Let's say you've sized the system so that you need the 800 W-h/day average over a few days. You left out a lot of information, like what latitude this is at and therefore what the minimum length of sunlight in the winter is, what probability of failure you can tolerate, what minimum percentage full sun your location expects averaged over a few days, etc. For example, if you conclude that worst case over a few days you can only count on the equivalent of 1 hour full sun per day, then the panel needs to be able to put out 800 W in full sunlight.

The next question is the battery. From the previous example, it looks like the battery should be able to run the system without any input power for at least a full day usage, which is 400 W-h into the load. Let's say half the total switching power supply loss of 70% assumed above is between the battery and the load, which means from battery to load is 84% efficient. 400 W-h / 84% = 480 W-h, which is what the battery has to be able to produce without input power and without it being exceptional and therefore significantly degrading the battery.

Let's see how the numbers work out for a 12V lead-acid battery. 48W / 12V = 4A drain when the load is powered. Since the load needs to run at this power level for 10 hours, that represents 40 A-h capacity. However, that needs to be significantly derated. A new lead-acid 40 A-h battery can do this once perhaps at the right temperature, but running it down to empty will kill it. For lead-acid you want a "deep cycle" battery but still derate significantly. Something like a 80 A-h "marine" battery might do it. Other battery technologies have different tradeoffs with how fully they can be discharged, operating temperature range, life time, life cycles, cost, availability, etc, etc.

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  • \$\begingroup\$ Thanks Olin. As you've said, I'm happy to have two unique answers give similar results. I'm accepting Russel's answer since he gave it first, but I found yours equally as valuable. \$\endgroup\$ – Mauvis Ledford Sep 20 '11 at 17:58
  • \$\begingroup\$ @MauvisLedford - Olin answered before Russel: 14:02 vs 15:25. Hover over the "X hours ago" to get the exact post time. You can change which answer you've accepted if you wish. \$\endgroup\$ – Kevin Vermeer Sep 20 '11 at 18:39
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Parameters:

  • Define "sunshine hour" as 1 hour of full sunlight (1000 W/m^2) or an equivalent amount of light at a lesser level delivered over more than 1 hour.
    Typical sunshine hours per day worldwide in summer is 4 to 5 hours with less or much less in winter.

    A superb resource is www.gaisma.com which provides detailed insolation (sunshine) and related matter for numerous locations worldwide. As Mauvis is shown as being in San Francisco USA see http://www.gaisma.com/en/location/san-francisco-california.html

Average sunshine hours per day each month for January to December are shown there as

  • 2.05 3.05 4.49 5.93 7.06 7.72
    7.50 6.69 5.38 3.85 2.50 1.85

So highest insolation is a massive 7.7 sunshine hours per day average in June and lowest is 1.85 sunshine hours per day average in December.

For comparison, Nairobi in Kenya has only 6.3 sunshine hours per day average max (in February) BUT a worst case month of 4.4 sunshine hours / day in July. Solar panel requirements in Nairobi would be less than half those in SF.

  • A modern silicon on glass laminated PV panel will deliver about 130 Watt's / m^2 of area.

  • If you have an MPPT tracking controller you'll get perhaps 95% of this into the battery. Without MPPT you may get 70%-80% depending on conditions. Maybe more.
    Say 75% for initial calculations.

  • Lead acid battery will deliver 80%+ of energy stored into it.
    LiFePo4 battery will deliver 90%+ of energy stored into it. Both have adequately low self discharge rates.

SO

Energy available from a PV (Photovoltaic panel / solar panel) saved to battery and then recovered is about:

  • 130 W x 75% x 80% =~ 80 Watts per square metre in FULL sunshine.

If this battery capacity is to be used over 10 hours then the Wattage supported per square metre is 80/10 = 8 Watts equipment load per metre^2 of panel per sunshine hour.

If you want the system to run for N days with no sun (sand storm ? :-) ) you need N metre^2 of panel per 8 Watts or you can power 8/N Watts of equipment per square meter per sunshine hour.

Using the 1.85 sunshine hours per Day December figure you can power 8W x 1.85 =~ 15 Watts of equipment for 10 hours from an average December days sun per square metre of panel.

So, to run your 40 W of equipment safely in December you'll need 40/15 =~ 2.66 m^2 of panels or about 2.66 x 130W = 350 Watts of solar panels. Note that this is to provide one days operation of 10 hours from 1.85 hours of full sun equivalent.

If you want to be able to withstand 2 sunless days you need to double that to 700 Watts of panel.


The battery needs to be sized to handle this amount of energy. The above was calculated on 75% of panel energy being used to charge the battery, so energy in is

350W x 1.85hr x 75% =~ 480 Watt-hours.
At 12V that's 480/12 = 40 Amp hours of battery capacity.

A 100 Ah deep cycle battery is liable to suffice.

The above requirement will be reduced by

  • MPPT controller - moderate

  • LiFePO4 battery - moderate

  • Summer rather than winter insolation - massive - 300%+ more sun.

  • Lower powered equipment - potentially very significant.


FWIW: I started this reply hours ago but didn't finish it. I now see Olin has now also provided a longish answer. I would not have gone to such length if his answer had been there when I started.


Gaisma information:

Burning Man is in the Black Rock Desert in Nevada, 120 miles north of Reno.
The following Reno information should be reasonably applicable.

Insolation = Sunshine-hours = 4.95 average for September
and 5.92 per day for August.
As BM is in early September use say 5 hours equivalent full sun per day.
There are about 2 wet days per month around this time - hope they are jot during BM :-).

enter image description here

I'll leave readers to extract the fine details from the following wonderful diagram below. I can comment if anything can't be understood (also see gaisma help page).
The BM line will be slightly above the orange day line which is for late September.
Sunrise about 6:40am and sunset about 7pm.
Sun angle at midday about 50 degrees above horizon.
9am to 3pm sun angles 20 degrees or higher above horizon.

Sun swings from about 110 degrees to 230 degrees 9am to 3pm = +/- 60 degrees
Sine of 60 degrees is 0.87 so pointing panels at midday sun position would lose about 13% of available energy at 3pm and 9am positions. So moving panels once or twice in day manually would produce modest gains.

enter image description here

Angle change above horizon during peak sun periods is (50-20) = 30 = +/- 15 degrees so vertical angle change not worthwhile across day.

Note that sun is at maximum height at about 1pm. Daylight saving. Adjusting my 9am and 3pm comments to true times (10am to 4pm) would better centre results on true noon peak BUT results will not vary much.

Note that at sunrise and sunset on day this graph was plotted (orange line) the sun rises and sets at about +/- 90 degrees from midday angle. For earlier dates back as far as June 21 the sun sets and rises progressively greater distances past 90 degrees from midday so if you wanted a panel to get all sunlight it would need to point "behind" it's normal midday pointing position. ie sun rises and sets "over your shoulder" in summer months.


12VDC to PC-power power supply

This question relating to PC's powered from 12VDC was asked in September 2011.

The user bought a 12V to micro-At power supply from ebay.
It looks potentially useful in your application and shows what is available and slo, usefully, the level of complexity required in 'rolling your own'.

Bought from here

And looked like this:

enter image description here

enter image description here

PW-200-M 200W micro-ATX DC/DC Mini ITX Power Supply PSU

They say:

  • Power any Pentium 4 motherboard with this super small, cable-free PW-200-M 200W micro-ATX DC to DC power supply which works with the full range of mini-ITX motherboards.

    Featuring noise-free, low-heat operation, this power supply connects directly to your motherboard ATX connector providing a fast, compact and convenient power solution.

    Benefits:
    The only cable-free micro-ATX DC to DC supply that is compliant with the full range of mini-ITX motherboards Supports Pentium 4 and powers most motherboards up to 3.0GHz Power your PC and peripherals from a single 12V power supply Total noise-free operation Connects directly to motherboard ATX connector Provides up to 200W from a single 12V supply 200,000 hour life span Compact size saves you space: 57 x 61mm The PW-200-M 200W micro-ATX DC to DC power supply is brand new and unused.

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  • \$\begingroup\$ I don't see a problem with another detailed answer whether it was after mine or not. It's interesting to note the different ways to attack the problem and different numbers we picked for examples, but how we came up with remarkably similar answers. You said 700W panel and I 800W, you said 100 A-h deep cycle lead-acid 12V battery and I said 80 A-h. Those are basically identical given the sparse information in the question. \$\endgroup\$ – Olin Lathrop Sep 20 '11 at 16:19
  • \$\begingroup\$ The only nit I have with your answer is that you can't use average daily insolation if you're only averaging for a few days. Any few consecutive days can be considerably more cloudy than the average. \$\endgroup\$ – Olin Lathrop Sep 20 '11 at 16:22
  • \$\begingroup\$ @Olin Lathrop - Yes. Have to stop somewhere, alas ;-). I noted "If you want the system to run for N days with no sun (sand storm ? :-) ) ... you can power 8/N Watts of equipment per square meter per sunshine hour". Knowing what value is safe to use for N is problematical - having some idea of the actual environment helps. I could hazard a guess for AK, NZ, but may be well out for SF, USA. \$\endgroup\$ – Russell McMahon Sep 20 '11 at 16:47
  • \$\begingroup\$ Thanks greatly for your detailed response and apologies for any valuable details I didn't give. The desert where this unit will be for a week will have about 5 hours of useable sunlight each day (Black Rock Desert, early September). So can I decrease the amount of solar panel in your projection by 270%? Re: the lower powered equipment, I'd like to have a machine that handles at least 10 simultaneous users at once with the software I'm building, but am definitely looking into it. \$\endgroup\$ – Mauvis Ledford Sep 20 '11 at 17:28
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    \$\begingroup\$ Russell, jeezum, thanks for the great info! Yes it is. I'm starting early. :) \$\endgroup\$ – Mauvis Ledford Sep 22 '11 at 5:35

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