<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
		>
<channel>
	<title>Comments on: How do you transport solar power tower energy to a building that is 4 kms away?</title>
	<atom:link href="http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/feed/" rel="self" type="application/rss+xml" />
	<link>http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/</link>
	<description>the homes of the future!</description>
	<lastBuildDate>Thu, 02 Feb 2012 01:23:41 +0200</lastBuildDate>
	<generator>http://wordpress.org/?v=2.8.2</generator>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
		<item>
		<title>By: Frank N</title>
		<link>http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/comment-page-1/#comment-19</link>
		<dc:creator>Frank N</dc:creator>
		<pubDate>Fri, 28 Nov 2008 17:36:21 +0000</pubDate>
		<guid isPermaLink="false">http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/#comment-19</guid>
		<description>If you want the heat energy, consider tracking mirrors at the site and solar collectors on the houses.  If you want electric energy, that area is enough to support a traditional steam turbine, much more cost-effective than photovoltaic.  That area is about 15 times the area that Solar One used to generate 10 megawatts in the Mohave Desert, so it could supply a theoretical maximum of 150 megawatts during peak sunshine.  That&#039;s probably enough to match the demand of about 25,000 homes.  Best bet would be to feed it to the electric power grid.</description>
		<content:encoded><![CDATA[<p>If you want the heat energy, consider tracking mirrors at the site and solar collectors on the houses.  If you want electric energy, that area is enough to support a traditional steam turbine, much more cost-effective than photovoltaic.  That area is about 15 times the area that Solar One used to generate 10 megawatts in the Mohave Desert, so it could supply a theoretical maximum of 150 megawatts during peak sunshine.  That&#8217;s probably enough to match the demand of about 25,000 homes.  Best bet would be to feed it to the electric power grid.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Helmut</title>
		<link>http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/comment-page-1/#comment-18</link>
		<dc:creator>Helmut</dc:creator>
		<pubDate>Fri, 28 Nov 2008 09:05:55 +0000</pubDate>
		<guid isPermaLink="false">http://solarpower-homes.com/solar-power/how-do-you-transport-solar-power-tower-energy-to-a-building-that-is-4-kms-away/35/#comment-18</guid>
		<description>At 40 % efficiency and 1252 W/m^2 the maximum available energy for a square kilometer is
(0.4)(1,252 W/m^2)(10^6 m^2) = 500.8 MW if the entire array is normal to the incident sunlight on a clear day.

The average dwelling has electrical service of 100A at 240 VAC, or 24 KW, so 5 houses would need about 120 KW.

For energy transmission you need a (very large) capacitor or battery bank for energy storage, a motor-generator set (90% efficient as opposed to &lt; 50% efficient for an inverter), a step-up transformer and step-down transformers.

Note that atmospheric attenuation of sunlight increases with both latitude and time from local Noon.  Even on a clear day sunlight is available for at most 12 hr/da.  It is also extremely expensive to &quot;track&quot; the Sun (keep the array perpendicular to the sunlight), especially so for such a large array.</description>
		<content:encoded><![CDATA[<p>At 40 % efficiency and 1252 W/m^2 the maximum available energy for a square kilometer is<br />
(0.4)(1,252 W/m^2)(10^6 m^2) = 500.8 MW if the entire array is normal to the incident sunlight on a clear day.</p>
<p>The average dwelling has electrical service of 100A at 240 VAC, or 24 KW, so 5 houses would need about 120 KW.</p>
<p>For energy transmission you need a (very large) capacitor or battery bank for energy storage, a motor-generator set (90% efficient as opposed to < 50% efficient for an inverter), a step-up transformer and step-down transformers.</p>
<p>Note that atmospheric attenuation of sunlight increases with both latitude and time from local Noon.  Even on a clear day sunlight is available for at most 12 hr/da.  It is also extremely expensive to &#8220;track&#8221; the Sun (keep the array perpendicular to the sunlight), especially so for such a large array.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

