Thank you for the help. It is much appreciated. ]]>

Story here:

http://m.seattlepi.com/news/us/artic...rt-5704156.php ]]>

Also does this mean to watch 4 cameras at once using the DVR card software I need 4 antennas on the receiver. (To pick up 4 signals at once)? Any advice...........?

Thank you! ]]>

Should the 20A circuit breaker have tripped?

In doing checks I noted that the hot and neutral were revered in the outlet and a GFCI outlet was not used, could this have played a factoring the electrocution?

The plug for the pump also had a short between hot and neutral, how would this have played a factor. ]]>

Hi guys. I am a long time reader and this is my very first post. I'm an EE out of Auburn University, graduating in 2006. I know this post is long but I need help and I hope you all don't think this is a stupid question/post. Hopefully this will help someone years down the road. I am trying to understand the concept of doing arc flash calculations using IEEE 1584 equations. I saw NFPA’s equation and it was very simple once you had the data to input, but it’s only limited from 16-50kA. Before I attempt to use an actual arc flash program, I want to understand the engineering behind the calculator. Pleases keep in mind that I will not do any arc flash analysis without shadowing an experience mentor and going to a few training classes.

PLEASE help me and I’m sorry if this post is redundant. I did a lot of searching and reading last night and I could never find an answer, so my head is hurting this morning lol. I would like to attempt to calculate an area at one of our facilities in the field. This is an example run, I’m not inputting anything into any model. I want to calculate what the estimated category rating is at the 3 phase 200A disconnect at the pole downstream of a 3-25kVA, 1 Phase transformer bank (75kV x-former bank), with Delta-Wye grounded. This disconnect is located on the same pole as the x-formers bank.

**Equation Factors**

Utility is 12.47kV ACSR

%Z impedance – 2.3%

X-former is 12.47kVA – 480V

Short circuit/bolt fault current – 2000A (an assumption for example sake)

3/0 copper feeding line side of 200A disconnect

.01 clearing time

**IEEE 1584 Equations**

*For systems between 0.208 and 1 kV:*

lg Ia = K + 0.662(lg Ibf) + 0.0966(V) + 0.000526(G) + 0.5588(V)(lg Ibf) - 0.00304(G)(lg Ibf)

**Based on factors above: Determine arc fault currents**

lg[Ia] = -0.097 + 0.662 (lg(2kA) + 0.0966(.480) + .000526(25) + .5588(.480)(lg(2kA) - .00304(25)(lg(2kA))

= -0.097 + 0.662 (0.3010) + 0.0464 + .0132 + .0807 - .0229

= 0.2197

**Ia = log (.2197) = -.6581**

This is my first issue guys. I highly doubt this number should be negative once completed. Maybe I messed up a decimal point somewhere, maybe miscalculated something? Can someone please confirm my math? If the top part is incorrect, of course everything else will the wrong going forward. That’s ok, as long as I can get assistance with the part that’s incorrect, it’s simple to fix my math ?. Let’s keep moving forward shall we…..

**Based on factors above: Determine incident energy**

lg En = K1 + K2 + 1.081(lg Ia) + 0.0011(G)

lg[En] = -0.555 + (-.113) + 1.081 (lg (-.6581) + .0011(25)

= -.668 + 1.081(ERROR) + .0275

En = lg(unknown)?

This is my 2nd problem. Since I have the incorrect arc fault current, it’s giving me an “error” once I try to compute it in the equation. The log of the negative Ia is giving me an error. I will still continue to the follow through on the following equations just to see if my “theory”. Please forgive me guys, I’m learning, I hate to ask stupid questions. Your feedback is greatly appreciated.

** (1) E = 4.184(Cf)(En)(t/0.2)(610x/Dx) **

E = 4.184(1.5)(unknown)(0.1/0.2)(610^1.641/455^1.641)

= 6.276(unknown)(.5)(37215/23003)

= 6.276(unknown)(.8089)

= and this is where I’m stuck again.

Based on what I read, this Incident energy is in joule/cm^2, so if I had everything correct above, I would take the E answer and then multiply by 0.239 cal/cm^2 to get a estimated category rating. Is this correct? Once again please help me. I now report to a boss who isn’t an EE and he thinks that you can simply plug in an equation to get the answer. I’ve been trying to tell him that there are SO MANY variables to determine the true value, thus the need for the arc flash program, but he still wants me to calculate an “example” by hand.

Please help guys. Thanks.

]]>PLEASE help me and I’m sorry if this post is redundant. I did a lot of searching and reading last night and I could never find an answer, so my head is hurting this morning lol. I would like to attempt to calculate an area at one of our facilities in the field. This is an example run, I’m not inputting anything into any model. I want to calculate what the estimated category rating is at the 3 phase 200A disconnect at the pole downstream of a 3-25kVA, 1 Phase transformer bank (75kV x-former bank), with Delta-Wye grounded. This disconnect is located on the same pole as the x-formers bank.

Utility is 12.47kV ACSR

%Z impedance – 2.3%

X-former is 12.47kVA – 480V

Short circuit/bolt fault current – 2000A (an assumption for example sake)

3/0 copper feeding line side of 200A disconnect

.01 clearing time

lg Ia = K + 0.662(lg Ibf) + 0.0966(V) + 0.000526(G) + 0.5588(V)(lg Ibf) - 0.00304(G)(lg Ibf)

lg[Ia] = -0.097 + 0.662 (lg(2kA) + 0.0966(.480) + .000526(25) + .5588(.480)(lg(2kA) - .00304(25)(lg(2kA))

= -0.097 + 0.662 (0.3010) + 0.0464 + .0132 + .0807 - .0229

= 0.2197

This is my first issue guys. I highly doubt this number should be negative once completed. Maybe I messed up a decimal point somewhere, maybe miscalculated something? Can someone please confirm my math? If the top part is incorrect, of course everything else will the wrong going forward. That’s ok, as long as I can get assistance with the part that’s incorrect, it’s simple to fix my math ?. Let’s keep moving forward shall we…..

lg En = K1 + K2 + 1.081(lg Ia) + 0.0011(G)

lg[En] = -0.555 + (-.113) + 1.081 (lg (-.6581) + .0011(25)

= -.668 + 1.081(ERROR) + .0275

En = lg(unknown)?

This is my 2nd problem. Since I have the incorrect arc fault current, it’s giving me an “error” once I try to compute it in the equation. The log of the negative Ia is giving me an error. I will still continue to the follow through on the following equations just to see if my “theory”. Please forgive me guys, I’m learning, I hate to ask stupid questions. Your feedback is greatly appreciated.

E = 4.184(1.5)(unknown)(0.1/0.2)(610^1.641/455^1.641)

= 6.276(unknown)(.5)(37215/23003)

= 6.276(unknown)(.8089)

= and this is where I’m stuck again.

Based on what I read, this Incident energy is in joule/cm^2, so if I had everything correct above, I would take the E answer and then multiply by 0.239 cal/cm^2 to get a estimated category rating. Is this correct? Once again please help me. I now report to a boss who isn’t an EE and he thinks that you can simply plug in an equation to get the answer. I’ve been trying to tell him that there are SO MANY variables to determine the true value, thus the need for the arc flash program, but he still wants me to calculate an “example” by hand.

Please help guys. Thanks.

Any opinions are appreciated on the safety of moving this cable while energized. ]]>

1) With an arc flash suit I can do any kind of hot work.

2) If the hot work is allowed I must wear the correct type of safety gear.

Just a bit bored tonight so I figured this could be a good topic.

Just wondering if anyone else deals with this. ]]>

Outside disconnect and main 'sub' panel properly wired.

GEC from the disco to the copper water lines entering the home. Not within the 5' rule but even if it was I would still see objectionable current at the test point.

The GEC run through a chase in the house and 'may' be the reason for the high EMF readings.

Assumption on my part: I believe that this bare wire may be coming in contact with the interior water lines that are also in this overhead chase prior to it's proper termination. Giving multiple parallel paths back to the source.

4.1 Amps measured at the pipe in the picture.

Now going outside, more than 10', and isolating the structure plumbing from the city plumbing should get rid of the reading.

My question is - Shouldn't there still be an objectionable current 'touch potential' present at this or any point on the plumbing system?

If so how does one solve this? Can't just not bond the interior metal water lines. Correct?