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Technology Stocks : Energy Conversion Devices

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To: Michael Latas who wrote (3383)3/23/1999 8:17:00 AM
From: fred whitridge  Read Replies (2) of 8393
 
Michael-- There's a PV rule of thumb that there is 1000 watts of solar insolation falling on each square meter at noon at the equator, or thereabouts. So if our panels are 10% efficient they are snagging 100 watts per meter. another rule of thumb (very latitude dependent) is that the "solar day" might be 4 peak hours. So the (approx.) 30 square feet in a meter would yield (maybe) 100 watts for (about) 4 hours, or 1.2 kilowatt hours in a day. Now pull out your electric bill and see how much you use in an average day and then you'll have an idea how many square feet of your roof will be devoted to PV.

i'm very curious about the space PV. I have no idea what the power drain is on a LEO satellite. But I like the fact that our process may be the only one that can lay down multiple layers on a flexible substrate and save a huge amount of weight. And that is the key: if it truly costs $10,000 for each pound we can lob into orbit, then we can charge a big premium on those pounds saved. if we can really pull in $30 a watt for space PV then our 5MW machine (assuming we can build the space PV on that machine) would be PRODUCING $150,000,000 OF PRODUCT REVENUE PER YEAR. So how big is the market for space PV and what might our margins be and can we make it on existing machines?
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