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    Who are the NERD fund donors Mr Snyder?

    Raise the curtain.

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    This is odd (none / 0) (#23)
    by Seth9 on Fri Sep 10, 2010 at 01:29:29 PM EST
    As you've shown, my original thought was probably incorrect (I was just trying to guess what the paper's number actually meant based on similar articles I've read). I've started running the numbers but I have to get to class. I'll get back to you on this later today.

    Also, I looked up my numbers from a project for a class last year which involved solar panels. We found that it cost somewhere in the neighborhood of $350-$400/m^2 per solar panel (lasts 25 years), so the solar panels themselves cost around $8750-$10000 at 5 m^2 per panel. Then they'd have to buy an inverter and do some wiring and stuff (the cost of this shouldn't be too high though). So I'm guessing that the turbines produces the bulk of the electricity because they invested a lot more money on it. That, or the people doing this are clinically dumb.


    Calculations and conclusions (none / 0) (#26)
    by Seth9 on Fri Sep 10, 2010 at 07:00:15 PM EST
    I'm going to assume that they get power for the same price that you do. There's no reason why they wouldn't. I'm also going to treat the month as 24 days (they don't use the building on weekends, but they also probably don't shut everything off). Also, Michigan gets about 1 KW/m^2 of direct sunlight on average, not 1.4, if I remember correctly (I can't find a confirming source, so I'm going with what I remember), due to the high levels of cloud cover. However, by angling the panels to the south, this will jump somewhat because the sun is always in the southern part of the sky. I'll just go with 1.2 KW/m^2, but this could be off because I'm going on memory and guesswork.

    Now, using these revised figures, we come up with this:

    $14000/month ÷ $.131835/month = apx. 106193 KWh/month
    106193 KWh/month ÷ 24 days * 20% (conv. factor) = 884.9 KWh/day (by system)
    884.9 KWh ÷ 8 hours (workday is 9-5) = 110.6 KW

    This is clearly lower than 262 KW, implying that either my calculations are flawed due to a faulty assumptions, one of their figures are flawed, or there is a wide variance in the amount of power this system produces. I'm guessing that its a combination of the three, as I'm making guesses here, the guy from the article was providing numbers that look like they were rounded up, and variance in power output occurs in both solar and wind systems.

    Now, with regard to the solar cells, here's my calculations. Please note that I changed several of your assumptions to fit values that I actually have a source for costs on. :

    1.2 KW/m^2 * .135 (eff factor) = .162 KW/m^2
    .162 KW/m^2 * 5 m^2 * 5 panels = 4.05 KW
    0.81 KW * 6 hours (8 seemed high to me) = 24.3 KWh/day

    Now, last year when I acquired these numbers for a class assignment, it cost $360/m^2 for a solar panel with 13.5% efficiency. This comes out to $9000 for 25 m^2 of panels. These panels are supposed to last 25-30 years, but I'm betting that with wear and tear, they only produce 20 years worth of a full load. We can also tack on about $1000 to install the panels, making the solar system cost about $10000.

    So, taking these figures, we can estimate the cost of electricity for the solar system:

    24.3 KWh * 365 days/year * 20 years = 177390 KWh
    $10000/177390 KWh = $.05637, less than half of your rate for electricity. Now, my calculations employ numerous assumptions and ignored several factors that could mean that this cost would be higher or lower. But even if it actually comes out to $.10/KWh, it is still a significant savings over what you pay. Of course, this only accounts for the savings from solar power. I have no idea what the figures are for the wind turbines, but I'm guessing that they have lower long-term savings because they require much more maintenance as they have moving parts.


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