erich Posted September 2, 2006 Report Posted September 2, 2006 This insulation company have made claims of R-15 per 5mil thickness for their coating product. I feel they are playing fast and loose with the R-value claim. Ceramic Space shuttle tiles are R-50 / Inch. At R-15 / 5mil that's R-75 / Inch, They quote the cost for 5 Mil at .35 cents / Sqft, that would be $1.75 / Inch ??...... I thought That these low resistance numbers were primarily from the vacuum spaces, I can't see how the molecular vibrations responsible for conduction are less in a paint binder than the matrices of the Shuttle's ceramic tiles?? They claim that their low conductance is due to the The Knudsen effect ....that was new to me.... but makes sense. http://www.industrial-nanotech.com/howitworks.htm , But they still obfuscate about publishing U or R values. Why can't this company have these numbers, U and R-values, verified by the government labs or construction product testing institutions? That would end this discussion. They claim That Aerogels are the only material that has a lower conductance, but Aerogel Panels http://www.kalwall.com/nano1.htm are R-20 @ about 2 inch thickness. However, This other Nano insulation company has no problem claiming a r-value of R-40: http://www.nanopore.com/thermal.html Please, I need some thermo dynamic help here. Erich J. Knight Quote
erich Posted October 15, 2006 Author Report Posted October 15, 2006 Well I have been told off By the inventor of Nansulate coating (Hydro-NM-Oxide), Stuart Burchill. http://www.industrial-nanotech.com/howitworks.htm I understand my mistaken assumption, exponential, not arithmetical. I still don't understand what precludes them from stating a properly exponentially calculated R-value per the suggested 5 mil thickness from their data : Nansulate: Test Method: ASTM C-518 Sample Thickness = 0.048” Thermal Conductivity = 0.058 Btu/hr ft °F R-Value = .048/12(.058) = 0.069 hr ft² °F/Btu t-mean = 170 °F "Dear Erich, Regarding your post of poorly interpreted data, scientifically incorrect conclusions, and the rapid further incorrect extrapolations. (Excerpts from Erich's Post to message board:) “No wonder they don't post this on their web page! This puts the R-Value of their suggested application rate of 5 mils at R-0.28 or R- 1.4 per inch. Yet they do post the R-10 value/inch of the pure nano material (Hydro-NM-Oxide).which they use in the coating formulation, At best this is misleading. In her email I was warned not to repost this information, I called her back and got permission, and am now waiting for a call from Mr. Stuart Burchill (inventor/CEO) to explain to me this playing fast and loose with how they present the product.” A rudimentary understanding of thermal conductivity is clearly not integrated into the thought processes of your posts. 1) Basic Fact of Thermal Dynamics 101: The thermal conductivity (K value) of any given material increases with temperature. 2) Therefore… the thermal conductivity of the section of insulation material closest to the heat source is higher than the thermal conductivity of the section of the material furthest away from the heat source. As the heat energy moves through the insulation, it is reduced, therefore each subsequent section is “dealing” with a lower heat energy, or temperature, and therefore the thermal conductivity of that section is lower (per fact 1) and therefore even more effective and reducing the rate of heat transfer as the heat energy moves to the next section which then deals with an even lower heat energy (temperature) and therefore is dealing with it with an even lower thermal conductivity (lower rate of heat energy transfer) and the process of heat reduction goes on exponentially, not arithmetically, as it moves through 3) Therefore… the R-Value, which is a factor of distance (thickness) divided by K (thermal conductivity) is actually the distance divided by the average sum total of the decreasing K values which occur as you move from the side facing the heat source to the other side of the insulation. 4) Therefore you cannot correctly extrapolate a comparative R-Value against insulation that has to be applied more thickly from a thin film thickness. If you just take a moment to think it through logically… how much insulation effect would you get from a 5/1000 inch thick section of R13 fiberglass batting or polyurethane foam? Try it, if you can even slice it that thin. 5) In addition, your calculations do not take into consideration other factors of the physics of heat transfer including emissivity. 6) Bottom line… you don’t understand how we do what we do and that is understandable. Most people don’t. What is not understandable is that you take data and a field of science that you don’t understand and draw incorrect conclusions and then post them as fact. In addition, the test data you were provided is from one of my original formulations and is dated September 21st, 2004. I have not been sitting idle since then and have continued to improve performance. I tried to call you, but got your home answering machine. Instead of trying to look for problems and creating them where there are none with your misinformation dissemination, why don’t you tell a friend or a neighbor about our great products THAT HAVE STOOD THE TEST OF TIME AND PROVEN THEIR PERFORMANCE IN THE LABORATORIES AND ON THE PRODUCTS AND FACILITES AND RESIDENCES OF GREAT COMPANIES AND GREAT PEOPLE WORLDWIDE AGAIN AND AGAIN AND AGAIN? Stuart Burchill Industrial Nanotech, Inc. " Erich J. Knight Quote
erich Posted October 27, 2006 Author Report Posted October 27, 2006 Here is my tentative reply to Dr. Burchill, please comment if I'm sticking my foot in it again. "Dear Dr. Burchill: When I posted to Ragging Bull (RB) I was frustrated by several contradictory replies to my questioning of the veracity of post # 2412 concerning R-15 claim for a 5mil thickness By: thatsmrnerd2u 18 Aug 2006, 11:59 AM EDT Msg. 2412 of 2417Jump to msg. # Just got off the phone with INTK's local distributor and I must say, I'm impressed. I get the impression that there are big things on the near horizon and judging by the increase in performance in the racecars that have used their product are getting, I think this BMW thing is going to be huge. Also, he didn't have an exact number but said a 5 mil coating would give about an R15 insulating value. That's the equivalent of 3" of polystyrene board insulation. I can tell you that composite aluminum manufactures (Reynolds - owned by Alcoa, Alucabond) hate having to have a deep cavity to make room for insulation but it's nec. to meet energy code requirements. Mr. Nerd" This lead me to Ms Crowlley, ........I explained my concerns about your stating on your web page that your coating is 70% Hydro-NM-Oxide, then right below that showing pure Hydro-NM-Oxide at R-10 to 13 per inch, how this can be misleading.After she sent the ASTM C-518 test data, I emailed her, restated my concerns, and my Incorrect Extrapolations, and was referred to ZA Consulting for any more information. They told me you would call. Four days later I made the post to RB stock board to which you replied. I'm sorry I missed your call, I think the very day I posted (10/9). I did return it. To your reply on RB: 1) "K value increases with temperature" :Yes , but the temperature differences we are talking about, in a buildings thermal envelope, are just the climatic range either side of 70 degrees 2) "exponentially, not arithmetically": Yes , however it is common in the building trade, it is marginally accurate in this application : http://en.wikipedia.org/wiki/Thermal_conductivity :Second definition (buildings)When dealing with buildings, thermal resistance or R-value means what is described above as thermal insulance, and thermal conductance means the reciprocal. For materials in series, these thermal resistances (unlike conductances) can simply be added to give a thermal resistance for the whole. A third term, thermal transmittance, incorporates the thermal conductance of a structure along with heat transfer due to convection and radiation. It is measured in the same units as thermal conductance and is sometimes known as the composite thermal conductance. The term U-value is another synonym." 3)"Sum total of decreasing K values": So can you publish this proper, exponentially calculated, number for 5 mil Nansulate? How should designers and architects evaluate the composite thermal performance of a building's thermal envelop to establish heating and cooling loads with your product as a component? 4)"Comparative R-Values": Aerogels at 5 mils thickness would be marginally R2. Does the Knudsen effect in particular take your product exponentially beyond the marginal differences in the linear addition of the R values of conventional insulation materials? 5) " Heat transport and emmisivity" I could not find your Emittance and Reflectance numbers on the web site. 6) I thank you for explaining my miscalculation. I apologise for my stupid "fast and loose" remark concerning your presentation. I have sent many posts about your technology to a wide range of people, including all the new space explorers - Branson, Scaled Composites, Bezos, etc-, my local Coors brewery, and engineers in the thermal solar ( OEGY.OB ) and oil industries ( EDNE.OB ). Also Bio & solar H2 companies NNLX.PK and http://www.hydrogensolar.com/index.html Thanks for your attention, Erich" Quote
erich Posted November 5, 2006 Author Report Posted November 5, 2006 Well, I sent my post above ,unaltered, to INTK last week with no reply.So I posted it to Ragging Bull, a stock message board today, to see if they will reply to my questions there. http://ragingbull.quote.com/mboard/b...cgi?board=INTK Over the past weeks I've posted it to several physics forums. The comments have been all supportive, and that my questions were valid. Erich Quote
HydrogenBond Posted November 5, 2006 Report Posted November 5, 2006 An interesting insulation material is mylar. It is thin plastic sheeting coated with a highly reflective silver or gold mirror finish. The mylar reflects heat allowing less heat to conduct through the thin film and into insulation. If anyone goes camping a similar affect can be made with aluminum foil. First place your hand near the flame, it gets hot due to radiation. Next put your hand at the same point with a piece of alumimum foil in front of your hand to reflect the thermal radiation from the fire. Your hand will feel cool. Quote
Jay-qu Posted November 5, 2006 Report Posted November 5, 2006 I cant say that you should expect them to admit they are wrong. People in these big companies dont like to get told they are wrong, even with direct and clear evidence.. Quote
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.