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		<title>QuantumHeat.org</title>
		<description>Discuss QuantumHeat.org</description>
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			<title>bob says:</title>
			<link>http://www.quantumheat.org#comment-4437</link>
			<description><![CDATA[@123star I agree with item 3 completely. I would add that we should switch to some form of heat collecting/inte grating calorimeter. The closest we've come so far is the water bucket test. Water is a pretty good heat collector/integ rator. This serves to eliminate any longitudinal dependance for temperature sensing. Water is good but to simplify we might even get away with air as the integrating fluid. eg. empty bucket insulated top and bottom with small fan. While we are at it we should add one extra temperature sensor on the outside of the bucket and plot heat flux vs delta temp across the bucket wall. My prediction is that we get a good corelation.]]></description>
			<dc:creator>bob</dc:creator>
			<pubDate>Tue, 24 Sep 2013 13:18:23 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4437</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-4436</link>
			<description><![CDATA[Hi there! I am still following your posts even if I am not writing anymore. I think I could say "I told you so" but, we have still to quantify this effect to know if it can be neglected or not. I'd say that this effect could account for a output heat overestimate of about 5% (ballpark figure). My suggestions are: (1) Do a thorough temperature sweep test with supposedly inert wires (NiCr). While keeping the same input power, vary the diameter and/or the number of wires in the calorimeter (i.e. you have to vary the radiating area, which is the total wire surface). Plot T_ext vs "total wire surface radiating area". (2) Paint one of the copper bands that hold the thermocouples with an IR absorbing paint (just paint the internal-facing side), and see if this worsen the "temperature dependence behavior" of the thermocouple being held by the painted copper band. (3) Switch to a 100% opaque calorimeter as soon as possible.]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Tue, 24 Sep 2013 08:40:05 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4436</guid>
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			<title>Paul says:</title>
			<link>http://www.quantumheat.org#comment-4435</link>
			<description><![CDATA[Ecco has been asking for the foil wrap for quite awhile. I think it's time to give it a try.]]></description>
			<dc:creator>Paul</dc:creator>
			<pubDate>Tue, 24 Sep 2013 02:41:14 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4435</guid>
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			<title>bob says:</title>
			<link>http://www.quantumheat.org#comment-4433</link>
			<description><![CDATA[One very easy way to " check" if IR transmission through the glass is a significant contributor to heat transfer is to wrap the tube in foil. If the interior temperature is significantly higher for given input power with foil then IR loss is significant.]]></description>
			<dc:creator>bob</dc:creator>
			<pubDate>Tue, 24 Sep 2013 00:19:51 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4433</guid>
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			<title>Ecco says:</title>
			<link>http://www.quantumheat.org#comment-4427</link>
			<description><![CDATA[@Ryan Hunt: below a wavelength of about 2.7 um, borosilicate glass becomes mostly transparent, depending on its thickness and other technical characteristics . Its transmittance also briefly peaks at 3 um and 4 um, to a lower extent. http://www.ljstar.com/technical/images/sightglass_image_2.jpg Although it's true that in general borosilicate glass is rather opaque to IR, it still allows a relevant amount of energy to leave the cell though that pathway, especially as temperatures get increased. Furthermore, the fact that the wires and the interior of the cell aren't ideal blackbodies adds much uncertainty to exactly how much emitted energy leaves the cell. I think this is a good reason to use 100% opaque tubes. Here's a spectral radiance chart calculator, by the way: http://www.wolframalpha.com/input/?i=Calculator+of+Wien%27s+Displacement+Law&a=*FS-_**WiensDisplacementLaw.lambda-.*WiensDisplacementLaw.T--&f2=450+C&f=WiensDisplacementLaw.T_450+C&a=*FVarOpt.1-_***WiensDisplacementLaw.lambda--.***WiensDisplacementLaw.nu---.*--]]></description>
			<dc:creator>Ecco</dc:creator>
			<pubDate>Mon, 23 Sep 2013 16:59:37 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4427</guid>
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		<item>
			<title>Ryan Hunt says:</title>
			<link>http://www.quantumheat.org#comment-4425</link>
			<description><![CDATA[The difference between the glass cells and the multi-wire test int he CTC is that the wire can be heated directly in the glass cell. We do not have that ability on the multi-wire test without some significant re-wiring. The borosilicate glass is very opaque to IR. I do not think much of the heat radiates out. It also is a rather good thermal conductor, so it will tent to lose heat that way pretty well.]]></description>
			<dc:creator>Ryan Hunt</dc:creator>
			<pubDate>Mon, 23 Sep 2013 16:27:18 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4425</guid>
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			<title>Ecco says:</title>
			<link>http://www.quantumheat.org#comment-4423</link>
			<description><![CDATA[I told you so! The temperature/exc ess power behavior (yet to be confirmed - I have to stress this - but very promising) of the latest multi-wire CTC cell experiment indeed seems to point to this. This is why I've been telling to crank up input power to make temperatures increase to 500°C and beyond. The excess power rise with temperature seems exponential, but at lower temperatures you're basically playing with hard to measure sub-watt excess heat. Unfortunately due to technical limitations increasing temperatures substantially just isn't possible in standard glass tube cells. Also, the very fact that they're transparent means that much input energy leaves the cell without heating the wires and that therefore it's wasted.]]></description>
			<dc:creator>Ecco</dc:creator>
			<pubDate>Mon, 23 Sep 2013 16:18:02 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-4423</guid>
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