Feeder to Work shop Calculation.

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AC\DC

Senior Member
Location
Florence,Oregon,Lane
Occupation
EC
Hello, doing a bid for a shop that the owner is adding some decent loads. With some possible in future. Ill state all the info then my question.
ALL NEW
Car hoist
10-Hp (FLC 28)
230 volt

Compressor
2-HP (FLC 12)
230 volt

Welder
20 Amp
240 volt

RV With 50/30/20 setup

WHAT IS CURRENTLY ON PANEL
Currently Feeding a HW tank 4500va
General recep and lighting for shop.

Service is out on the yard with one breaker feeding the house and the current 125 amp feeding this shop that is attached to the house.

Trying to find any reduction i can do to make the feeder smaller. Don't seem to many.

So going through 430.24. 125% of Largest FLC plus 100% of the other FLC
430..26 states I can get AHJ allow smaller ampacity for feeder than allowed in 430.24
Would rather not use that if I don't have to.

So for Welder i could not find anything in 630. so take it at 100%

In 551.73 States Not less than 12,000 va for 50 amp then in last sentence states I only use the highest rated receptacle. So I don't have to include the 30 amp or 20 amp.

Then Just 4500va for water heater then 3500 va for general lighting rounded up for easy. So if everything looks good so far, here is my questions

1: I can't use part IV of 220.83 since is does not serve the TOTAL connected load of the Service. Correct. Second guessing my self.
2: I Don't see any reductions beside 430.26
3: will be supplying this for Rv setup. Its just one RV location its rated for 100 amps But i was just debating on reducing that since they are either going to use the 50 or the 30. They don't have the RV yet.

With no reduction It will just be adding everything up and sizing off that.

Then to figure out if current service is sized right for this will follow 220.84 and then add these loads I specified and then I can use optional method.( I don't see this being a problem but would rather be safe)

Thank you guys for any input on this. If i don't get a reply Ill assume I am correct ;)
 
Take the square root of the duty cycle of your welder and use that as your effective amps factor. Say your duty cycle is 60%, your reduction factor will be 20A X (0.6)^0.5 = 15.5 amperes! (0.77 reduction)!
 
Is it a one man shop so you could argue certain things cannot be used at once? Even with that I don't see much you could drop other than you can't lift a car and weld at the same time (so drop the lower load). Everything else seems to be automatic so can't be eliminated.
 
AC\DC said:
His is a Ark welder. I believe that is for Resistance welders
Click to expand...
I’m almost certain there’s a duty cycle chart for arc welder.(assuming they would be considered the generator or transformer type welders listed in the NEC.) don’t have my book handy or I would look but I also think it says you have to use the current input listed on the welder, and if it doesnt have such you can use the primary current rating and multiply with the duty cycle chart. If I a wrong someone correct me , thanks
 
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As mentioned, if one man shop, regardless what NEC might say many hand held or direct use tools would never be used at same time. Air compressor is automatic and could run at any time. Dust collectors or exhaust collectors and other similar items also could run while other items are in use. The water heater you would definitely need to account for - it could run at any time.

The hoist, the welder, a lathe or drill press - all are kind of things that you generally would use only one at a time. Take the largest of those and can sort of ignore the others in most cases. Welder does have duty cycle as mentioned. The hoist kind of does just by nature of how they are used. You would seldom have that motor run for more than 20-30 seconds.
 
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