VD for MC Cable

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Alwayslearningelec

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NJ
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Estimator
Trying to figure out calculation for VD on 12/2 MC cable with steel armor.
120V. It's a 20A breaker feeding 5 receptacles.
How would one determine the max length this MC could run to keep VD under 3%?
I've always used VD calculators for wire and not a cable.
Thanks.
 
I would not imagine there would be much if any difference between MC & metal conduit.
Your biggest challenge would be deciding on what load to use. :)

If the load is actually 20 amps for a 3% or less drop you are looking at around 50 ft.
 
augie47 said:
I would not imagine there would be much if any difference between MC & metal conduit.
Your biggest challenge would be deciding on what load to use. :)

If the load is actually 20 amps for a 3% or less drop you are looking at around 50 ft.
Click to expand...
No so i don't know the actual load of the receptacle or what's plugged into it. It's a convenience outlet so various things will be plugged in. They are 20A receptacles. I'd have to imagine they wouldn't draw more than a few amps each.
So how would one approach a VD calc in this situation? Thanks
 
horsegoer said:
No so i don't know the actual load of the receptacle or what's plugged into it. It's a convenience outlet so various things will be plugged in. They are 20A receptacles. I'd have to imagine they wouldn't draw more than a few amps each.
So how would one approach a VD calc in this situation? Thanks
Click to expand...
Since you do not know what will be plugged in, there is no way to know what the load is going to be.
 
As a plans reviewer I would put the monkey back on the engineers back.........
Most of the plans I see have a maximum distance for #12
 
As Bob said there is no way to know what amp number to use in the calculation. I would go with the rule of thumb of up to 100' using #12. After that #10.
 
horsegoer said:
Trying to figure out calculation for VD on 12/2 MC cable with steel armor.
120V. It's a 20A breaker feeding 5 receptacles.
How would one determine the max length this MC could run to keep VD under 3%?
I've always used VD calculators for wire and not a cable.
Thanks.
Click to expand...
why 3%?

it's a good design criteria, but most state building codes have been updated to allow for 5% total branch and feeder, not 2% feeder and 3% branch as it used to be. Florida for example.
 
One thing I do is look at the panel sch. It will have a connected load in kva. If none listed and their convenient recpt. Use 180 VA per.
 
From a design perspective, (5) 20A outlets are calculated at a demand of 0.9kva (5*0.18kva per).

If the receptacle isn't for a designated appliance or device then 180va is the demand attributed to the receptacle - this is why you'll never see more than 10 convenience duplex on the same circuit - it theoretically would be drawing 1800va. Divide that by 120v and your at your 15A (max amps) for the 20A circuit, add an additional receptacle and your at 16.5A, multiply by 125% and your exceeding 20A ocpd.

Technically speaking the load with 5 duplex is 7.5A @ 120v line to neutral.

7.5A "could" be your amperage from the 12 Gauge conductors and "should" be under 3%VD up to 138'.



In a perfect world...
 
NEClayR said:
From a design perspective, (5) 20A outlets are calculated at a demand of 0.9kva (5*0.18kva per).

If the receptacle isn't for a designated appliance or device then 180va is the demand attributed to the receptacle - this is why you'll never see more than 10 convenience duplex on the same circuit - it theoretically would be drawing 1800va. Divide that by 120v and your at your 15A (max amps) for the 20A circuit, add an additional receptacle and your at 16.5A, multiply by 125% and your exceeding 20A ocpd.

Technically speaking the load with 5 duplex is 7.5A @ 120v line to neutral.

7.5A "could" be your amperage from the 12 Gauge conductors and "should" be under 3%VD up to 138'.



In a perfect world...
Click to expand...

One does not include 125% on receptacle loads so you can have 13 receptacles on a 20 amp cir

180 x 13= 2340

20 x 120 = 2400 allowed. This is for load calculation as their are some who believe their is no limit to the number of receptacles on a circuit... We won't get into that here
 
Some ahj have amendments limiting the number of recpt. on a 15 and 20 amp branch circuit. In Tulsa they use a poor choice of word and use the word "outlets".
I just smile and move on.
Title 52 here (pic).

In this case for us it's 180*8 or 1440 va for a 20 amp recpt branch circuit. So for load on the VD question we would use 1440/120 or 12 amps.
One issue with this is you have to figure VD as you go and drop 180 VA (1.5 amps) at each recpt outlet location for total length. In our case just use 12 amps.
Simple I,E,R VD= IxR
To solve for distance. R= VDp/I
VDp is voltage drop permitted.
120*.03= 3.6 VDp
#12 standard 1.98 per 1000
Or .00198 per foot.
3.6 / 12= .3 R
.3/.00198=151.51 total feet of wire

Then you have what most use
For D
CM x VDp/ 2xKxI
6530x3.6= 23508
2x12.9x12= 309.6
23508/309.6= 75.93 feet one way.
75.93x2= 151.86 feet of wire.
So, have fun with that.
I use the simple VD= I x R in the field for teaching basic VD based on ohms law triangle for apprentices. They only need to remember R for #12 (1.98) and #10 (1.24) stranded. These are the DC values in table 8. It puts the triangle to work before moving into the ohms law circle.
An example, #12 and a 150 foot run
.00198 x 150 (of wire) = .297 R or R total
.297x 12 (I) = 3.564. VD.
Supper easy for field work fun.
I have them apply this to every circuit they pipe and pull. So when they grab 200 feet of 1/2 for recpt circuit I go Ummmmm. That's a minimum of 400' of wire.
(.00198*400)x12= (9.504/120)*100=. 7.92 %
It's an eye opener for most and they one that go so what, I watch real close.
Little long ,sorry.
For fun use #10 for the 400' and see what you get for %
 

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