Computing stuff tied to the physical world

Solar Railway Station

In Hardware on Oct 19, 2012 at 00:01

On a recent trip to Germany, we visited Uelzen, about 100 km north of Braunschweig. Its railway station was “upgraded” (pimped?) some 12 years ago by an Austrian artist called Friedensreich Hundertwasser in what looks very Gaudi-like in style and appearance:


He was interested in organic shapes, creating round and uneven fairy-tale like interiors:


But the reason I’m mentioning all this, is that the station also includes this display panel:


The roof is covered with solar panels, and has been generating electricity for the community since 1997. Almost half a megigawatt-hour produced so far, and still generating just under 8 Kilowatt on a half-cloudy October day.

Way to go! So gibt mann das gute Vorbild!

PS. Calculating back from the figures shown, this would appear to be just 42 m2 of panels, which seems off by an order of magnitude. Oh, wait… that’s not accounting for the panel’s conversion efficiency, I’m guessing that to be somewhere in the 10% range for those panels.

  1. Don’t trust these displays. In a local school I’ve seen the numbers in a similar display count only 1/10th of the kWh. Wrong dip switch setting. Oops. Let’s do the math: Bestrahlungsstarke is 189 W/m2 currently; that’s 1/5th of a bright summer’s day (1000W/m2). That means the installation can yield 3.5 kWp; and with 7.5 kW on the sunny day-with-some-clouds you were there around noon (don’t see a clock, but the shadows seem to indicate that), that fits. At 950 kWh/kWp, 3.5 kWp yield 3350 kWh/y; in 15 years that is 49875 kWh. The correction factor of 0.85 with the actual reading suggests a panel inclination of 10 degrees from level, yielding 85% of max production. Isn’t there a decimal point in that display we can’t see? Or a wrong setting of dip switches?

    • I think everything is alright: “Bestrahlungsstarke is 189 W/m2 currently; that’s 1/5th of a bright summer’s day (1000W/m2)” Therefore we have to multiply the currently produced power by 5 to get the peak power (im kWp): 7.3 kW * 5 = 36.5 kWp. With your assumed 950 kWh/kWp (per year) thats 520125 kWh in total which is not that far away from the claimed 425244 kWh

  2. “Almost half a megawatt-hour produced so far”

    425 244 kWh = 425,244 MWh = 0,425244 GWh

    if the display is correct of course.

  3. I guess they have actually decreased the usage of PV generated electricity and minimized the need for batteries in off-grid system by implementing additional solar panels for direct air heating (hot-air heating system). Actually such solar heater panels can be build very cheap, especially if recycled parts are used (for example pop-cans). Detailed do-it-yourself instructions can be found here:

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