pet peeves

Plugs in a Kitchen Why can't they be adjacent?....This one was fixed in the 2006 Code!!!

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Smaller Ground wire is fine for the rest of Canada, And all of the United States, But here in B.C. it is too small for some reason (ie: #6 for 200 amp service is good everywhere but B.C.).....Hmmmmm?

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Service Masts....When we upgrade a service (usually from 100 to 125 or 200 etc).. Why in hell can't we use the old 2" mast that was there for 40 years. Yeah there are some Grow ops that take advantage of in the wall services, but why does everyone have to suffer and have a ridiculous expense and make their house look ugly with pipes on the outside. Why not make peanalties for the abusers (Like "No power for 5 years") and let the rest of the honest people keep their costs down and homes look good.

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Voltage drop? What is the maximum distance blah blah blah!!!

Dont get me wrong...I know it is very important to consider voltage drop. What I dont get is who is the rocket scientist that developed the method of finding it? And even worse, they convinced our electrical instructors to teach our apprentices this way!!

What is wrong with these people? Anybody out there ever made the statement " O.K Bob This circuit would be perfect for a 32 circular mill #13 cable. Go Special order it"? Not Likely!

We only use a few wire sizes!!! Its either Copper or Aluminum!

Why not have a table like this in the code book, and teach it like this. Just Ohms Law!:

Copper wire resistance per foot. (Could also be meters)

#4   .000292
#6   .000465
#8   .000739
#10 .00118
#12 .00187
#14 .00297
#16 .00473
#18 .00751
#20 .0119
#22 .0190
#24 .0302
#26 .0480
#28 .0764

For those of us who forget how it is currently written in the code and how we are taught, here is an example!

This calculation was done by a J/M who just finished passing his final exams and IP.
Find the wire size needed for a cable to carry 57 amps 256 feet with a 3% voltage drop.

1     Convert to meters .256' / 3.281 = 78.025m
2.    Basic meters (from Tab D3 sub (3) * Voltage Drop * 1 * (Volts used/120) * = Distance
3.    78.025 / ( 3 * ~1 * 3.142 ) = 8.2776(Basic Meters)
4.    Basic Meters = 9.9 First we assume a #4 might work (#4awg)
5.    57 / 85 = of allowable ampacity
6.    67.05 (70) % @ 90 degrees (rating of wire) = 1
7.    9.9 * 3 * .95 * 3.142 = 88.65 meters
8.    88.65 meters * 3.281 = 290.87'
9.    check #6 awg to see if it will work
10. 6.2 * 3 * 3.142 * .91 = 53.18 meters
11. 53.18 meters * 3.281 = 174' so #6 is too small

Here it is again doing it with the wire resistance. All you need is ohms law!

Find wire size needed to run 256 feet carrying 57 amps with a 3% voltage drop.

256 feet x 2 for there and back wire = 512 feet of wire total
3% of 377 volts is .03 x 377 = 11.31 volts max voltage drop
11.31 max drop divided by 57 amps is max resistance = .19842 ohms ( E / I = R )
.19482 divided by 512 is max ohms per foot. = .0003875
Look at the table and use #4

I remember in Trade school all the confusion and time I spent learning this and then helping others learn how to do this. I was lucky and understood it after trying it a few times. It took the instructor a lot of time and patience to get through to everybody. Time that could have been spent teaching something else. I also know that I had forgotten it all and had to look it all up again to try and do it for this page. The fresh Journeyman who did the above calculation also had to look it all up again too after being out of school for 4 months!! This seems to be a huge problem as almost every electrical site on the web has a table or calculator for voltage drop.