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| arc flash calculations - worksheet b page 19 of free guide https://brainfiller.com/arcflashforum/viewtopic.php?f=23&t=2490 |
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| Author: | clive rayner [ Wed Oct 17, 2012 2:20 am ] |
| Post subject: | arc flash calculations - worksheet b page 19 of free guide |
Hi I am struggling to understand where the figures in the above sample calculation step 1 are derived. I can see that 610/457.2 ^1.641 = 1.60506 but 12,670/8286^1.641= 2.009. Can anyone help me understand where these figures are derived thanks |
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| Author: | Jim Phillips (brainfiller) [ Wed Oct 17, 2012 3:04 am ] |
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clive rayner wrote: Hi I am struggling to understand where the figures in the above sample calculation step 1 are derived. I can see that 610/457.2 ^1.641 = 1.60506 but 12,670/8286^1.641= 2.009. Can anyone help me understand where these figures are derived thanks The information shown is for adjusting the incident energy from the normalized working distance of 24 inches (610 mm) to your distance of 18 inches (457) mm. The adjustment is based on the ratio of the two distances raised to the exponent of 1.641 The theoretical exponent was traditionally “2” but IEEE testing shows that this value varies depending on equipment type. 1.641 is for a panel. The basic concept is the incident energy decreases exponentially with increasing distance. It also increases exponentially as the distance decreases. As far as the calculations, here is what I have: (610/457.2)^1.641 = 1.605 Or as IEEE has it in their equations: (610^1.641) / ( 457.2^1.641) 37215.7 / 23,186.5 = 1.605 The result of 1.605 means that the incident energy will increase by a factor of 1.605 when a person moves from the normalized distance of 24 inches to a working distance of 18 inches. Hope this helps. |
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| Author: | clive rayner [ Thu Oct 18, 2012 4:00 am ] |
| Post subject: | |
brainfiller wrote: The information shown is for adjusting the incident energy from the normalized working distance of 24 inches (610 mm) to your distance of 18 inches (457) mm. The adjustment is based on the ratio of the two distances raised to the exponent of 1.641 The theoretical exponent was traditionally “2” but IEEE testing shows that this value varies depending on equipment type. 1.641 is for a panel. The basic concept is the incident energy decreases exponentially with increasing distance. It also increases exponentially as the distance decreases. As far as the calculations, here is what I have: (610/457.2)^1.641 = 1.605 Or as IEEE has it in their equations: (610^1.641) / ( 457.2^1.641) 37215.7 / 23,186.5 = 1.605 The result of 1.605 means that the incident energy will increase by a factor of 1.605 when a person moves from the normalized distance of 24 inches to a working distance of 18 inches. Hope this helps. Hi Jim Good to hear from you again, we met last week at Haydock Park in the U.K. Thanks for the reply. thanks |
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