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| 40 cal/cm^2 equals what temperature? https://brainfiller.com/arcflashforum/viewtopic.php?f=2&t=2994 |
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| Author: | WBurtt [ Mon Sep 23, 2013 11:56 am ] |
| Post subject: | 40 cal/cm^2 equals what temperature? |
Hello Folks; I have been able to find mention in various places that 1.2 cal/cm^2 is about 174F and 8 cal/cm^2 is about 199F. I have not been able to find any other similar data for other cal/cm^2, specifically 40 cal/cm^2. Now, I have seen the literature that states arc flash events can result in temperatures of 35,000F but they don't give the incident energy of the event that would cause that temperature. So, does the 35,000F coincide with a 40 cal/cm^2 event? While on the topic, might as well flesh out the chart, what does 25 cal/cm^2 give for a temperature? Thanks. Bill |
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| Author: | PaulEngr [ Mon Sep 23, 2013 5:02 pm ] |
| Post subject: | |
Incorrect altogether. Temperature and heat flux are two totally different things. At 1.2 cal/cm^2 for 1 second skin surface temperature will rise to 140 F but if the time interval is different, so is the resulting surface temperature. The temperature in the arc column can be estimated but doesn't really mean a whole lot since at that point everything is all plasma (electrons are truly just a cloud) and we don't really have standards to calibrate against above around 4,000F. |
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| Author: | WBurtt [ Tue Sep 24, 2013 10:06 am ] |
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Paul; Fair enough. I think I was just asking the incorrect question. During the original arc flash experiments the heat had to be measured in some manner so that the incident energy could be calculated. So, I searched for arc flash testing of PPE and found several papers that referenced the use of copper calorimeters. The papers listed the formula to determine the incident energy from the rise in temperature. They included a statement that sensor absorptivity measurements have determined that absorbed energy is equal to or greater than 90% of incident energy for copper calorimeters. Incident Energy = 0.135 cal/cm^2/C So I worked the formula backwards and solved for “C” for a given "Incident Energy". I also converted C to F. 1.2 cal/cm^2 = 48F temperature rise 4.0 cal/cm^2 = 85F temperature rise 8.0 cal/cm^2 = 138F temperature rise 25.0 cal/cm^2 = 365F temperature rise 40.0 cal/cm^2 = 565F temperature rise Bill |
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| Author: | PaulEngr [ Tue Sep 24, 2013 7:50 pm ] |
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Back when Alicia Stoll did her experiments to measure the threshold for a second degree burn, she compared the heat source used to calorimetry measurements with a copper calorimeter at different exposure times to a heat source. This gives a way to then use the energy measured by a copper calorimeter instead of testing on humans. The curve is called the Stoll curve. Temperatures of the copper calorimeter have no relationship to human skin other than the Stoll curve. |
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| Author: | jghrist [ Mon Sep 30, 2013 11:28 am ] |
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Bill, You are still missing the time parameter. You've also injected temperature rise. Consider the 1.2 cal/cm² case. This is the limit of second-degree burns. If the ambient were 32°F, this would mean exposure to 80°F would cause second degree burns. |
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| Author: | PaulEngr [ Mon Sep 30, 2013 4:17 pm ] |
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Not only that but there are absorption/cooling considerations. There are substantial measured differences between ablative laser experiments and the Stoll curve. |
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| Author: | Flash [ Mon Oct 07, 2013 1:29 pm ] |
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Take a look at the first IEEE video, I believe from the 90's. This is the one that Dupont was selling. Those tests actually had temp sensors on the dummy. |
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