sub pannel

[johnsmith09800]

Working on a residential remodel and going to install a new 200a panel and 125a sub panel feed by 80a breaker with #2, (boss mans instructions!) There is gona be roughly 50 circuits total. I prefer to load the sub panel with as many 120v circuits as allowed to save space in the main panel. With the load unknown, how do i determine exactly how many circuits I can put in the sub with 80a main?

 

[kwired]

Working on a residential remodel and going to install a new 200a panel and 125a sub panel feed by 80a breaker with #2, (boss mans instructions!) There is gona be roughly 50 circuits total. I prefer to load the sub panel with as many 120v circuits as allowed to save space in the main panel. With the load unknown, how do i determine exactly how many circuits I can put in the sub with 80a main?

When I do this I try to put all the bedrooms, living room, other areas that typically are not loaded heavily in the sub panel, and try to put some of the more continuous or frequently operating heavier loads in the main panel. I will say it is not always easy to determine exactly what circuits are heavier loads, but maybe if you know much about the owner and their family it may give you some hints of what tendencies may be.

A bathroom requires 20 amp circuit for receptacles, and yes if they use hairdryers it will have heavy loading, but maybe for only 10 minutes a day, it goes in the subpanel most of the time for me.

 

[johnsmith09800]

That is very good advice to consider (TY), most of the circuits are dedicated for specific use's (dish, disposal, micro, x2 fridge, x2 kitchen app, bath gfi, exterior gfi, jucuzzi,

Im thinking I should put Jucuzzi(motor) Dish(motor/heat) disposal(motor) bath gfi(heat) kit app(unknown/heat) washer(motor) and all 220v circuits into main panel (GIVEN)

I really want to know is how can I mathitamicaly calculate the max load I can put on the 80a sub main from the standard unknown loads of the remaining circuits of general purpose?
If the sub main is 80a and i get 80% of that to use is 64a, I load my sub panel with 6 - 20a breakers = 6x20=120a

this is where I get lost, I know my pannel wouldnt be overloaded, but I cant figure out what im missing here

 

[suemarkp]

The "code" answer is take the rooms served by the circuits you put in the subpanel and treat them as a 3 VA per square foot load. If you put 2000 sq ft of general purpose lights and receptacles in this panel, that is 6 KVA of load. It doesn't matter whether this is 3 circuits or 10 circuits -- the load is the same based on the floor area they cover.

Strangely, the bathroom receptacle and outside receptacle have no load in a calculation -- they are zero. As a practical matter I'd leave about 15A of headroom to cover all these unknowns.

Kitchen and laundry small appliance circuits count as 1500VA of your feeder for each one you include.

A circuit for a specific appliance (dishwasher, disposal, etc) uses the nameplate for its load value.

I think I'd rather see a bathroom circuit in the subpanel before a hot tub or electric range -- they eat up so much of your feeder and may affect the subpanel circuits if your feeder is marginally sized. But you have plenty of wire size for this feeder (you could easily go to a 100A feeder if the #2 is copper as many times a 100A breaker is cheaper than a 70, 80, or 90).

Remember that you don't just add up all these VA values and stop. Demand factors are allowed to be applied per the standard load calculation (a 35% factor applies after the first 3 KVA for general lighting and small appliance/laundry loads, a 75% factor on fixed appliances if you have 4 or more, etc).

In a residence, I don't think you'll find many if any continuous loads. So you can use all 80A (19.2 KVA on a 240V feeder). In your example of 6 20A breakers, remember you have two phases. So you'd have 60A on each leg of the feeder if you put them balanced between the two legs. So you'd be using at worst case 60A of your 80A feeder. But you'll hardly ever treat a 20A circuit as a 20A load -- use the values above based on whether it is general purpose (square footage based), equipment specific, or one of the 1.5 KVA required circuits.

 

[kwired]

The "code" answer is take the rooms served by the circuits you put in the subpanel and treat them as a 3 VA per square foot load. If you put 2000 sq ft of general purpose lights and receptacles in this panel, that is 6 KVA of load. It doesn't matter whether this is 3 circuits or 10 circuits -- the load is the same based on the floor area they cover.

Strangely, the bathroom receptacle and outside receptacle have no load in a calculation -- they are zero. As a practical matter I'd leave about 15A of headroom to cover all these unknowns.

Kitchen and laundry small appliance circuits count as 1500VA of your feeder for each one you include.

A circuit for a specific appliance (dishwasher, disposal, etc) uses the nameplate for its load value.

I think I'd rather see a bathroom circuit in the subpanel before a hot tub or electric range -- they eat up so much of your feeder and may affect the subpanel circuits if your feeder is marginally sized. But you have plenty of wire size for this feeder (you could easily go to a 100A feeder if the #2 is copper as many times a 100A breaker is cheaper than a 70, 80, or 90).

Remember that you don't just add up all these VA values and stop. Demand factors are allowed to be applied per the standard load calculation (a 35% factor applies after the first 3 KVA for general lighting and small appliance/laundry loads, a 75% factor on fixed appliances if you have 4 or more, etc).

In a residence, I don't think you'll find many if any continuous loads. So you can use all 80A (19.2 KVA on a 240V feeder). In your example of 6 20A breakers, remember you have two phases. So you'd have 60A on each leg of the feeder if you put them balanced between the two legs. So you'd be using at worst case 60A of your 80A feeder. But you'll hardly ever treat a 20A circuit as a 20A load -- use the values above based on whether it is general purpose (square footage based), equipment specific, or one of the 1.5 KVA required circuits.

That answer is worth some consideration, but I don't really know you can go directly from art 220 calculations and demand factors either. Like you said the bath and outdoor receptacles have no real value to plug into the calculation, so would unfinished basements, garages, or other similar areas. NEC kind of figures these will be covered by the 3VA/SF general lighting load when supplying a feeder or service that supplies the entire dwelling, but has no real way to calculate the load on these circuits when not with the remainder of the dwelling circuits.

You mentioned the disposer and its nameplate value. Though that nameplate value is important for calculating the branch circuit, it has very little meaning on a service or feeder as this is an item that is only loaded for maybe 15-30 seconds most of the time and in cases where it is used a lot, that still only ends up being maybe 4-6 times a day. Put in a separate meter for the disposer and you will pay more for minimum monthly charge on the meter than you will pay for energy charges in most cases.

 

[suemarkp]

While I agree that the burden on a service or feeder of intermittent motors (e.g. garage door openers, garbage disposals, trash compactors, etc) is minimal, I don't see the NEC as allowing us to ignore them. That would be a good code proposal in my opinion, since the dwelling calculations seem to always cause a much larger feeder or service than is really necessary.

I don't see anything in the calculation that says you can't break up a dwelling into multiple sections with separate feeders serving each and doing the calculations as if it were a full dwelling (including only the relevant parts served by that feeder). How else would you propose to do a 400A service where two 200A panels feed a dwelling?

 

[kwired]

While I agree that the burden on a service or feeder of intermittent motors (e.g. garage door openers, garbage disposals, trash compactors, etc) is minimal, I don't see the NEC as allowing us to ignore them. That would be a good code proposal in my opinion, since the dwelling calculations seem to always cause a much larger feeder or service than is really necessary.

I don't see anything in the calculation that says you can't break up a dwelling into multiple sections with separate feeders serving each and doing the calculations as if it were a full dwelling (including only the relevant parts served by that feeder). How else would you propose to do a 400A service where two 200A panels feed a dwelling?

There is no demand factors to apply to a garage door opener, waste disposers, etc. and no real way of determining the real load.

Sure one can add the nameplates of these items to come up with a feeder size, but in reality you could run a fairly small feeder to serve all these kinds of loads and never trip the feeder overcurrent device.

 

[Volta]

There is no demand factors to apply to a garage door opener, waste disposers, etc. and no real way of determining the real load.

Sure one can add the nameplates of these items to come up with a feeder size, but in reality you could run a fairly small feeder to serve all these kinds of loads and never trip the feeder overcurrent device.

I think you could use a demand factor of 75% for many of those types of loads via 220.53.

 

[kwired]

I think you could use a demand factor of 75% for many of those types of loads via 220.53.

I suppose you could, I also think the demand factor should be even less than that myself for things like a garage door opener, or a disposer. The duty rating of some of this equipment doesn't even allow it to run for very long in some cases. I was servicing a garage door opener once and the frequent cycling while working on it got it hot enough the internal overload opened and we had to wait for it to cool to continue. I think I was having issues with setting the stop limit switches on that one and we did cycle it quite a bit in a short time.

So if you have a 100 amp feeder with actual 100 amps of steady load, and you add a 6 amp garage door opener to it for maybe 10-15 seconds, my guess is you never trip the 100 amp feeder breaker, if you do it is like one time in several thousand tries.

 

[kbsparky]

....
If the sub main is 80a and i get 80% of that to use is 64a, I load my sub panel with 6 - 20a breakers = 6x20=120a

this is where I get lost, I know my pannel wouldnt be overloaded, but I cant figure out what im missing here

First, you are not required to use 80% when figuring the demand loads for the sub-panel. The so-called 80% rule only applies to continuous loads, such as electric resistance heat, or electric water heaters. If none of those are connected to the sub-panel, then eliminate any use of this part of the equation. :huh:

Second, your calculation of 6x20 Amp breakers fails to recognize that your sub-panel is fed with a 120/240 feeder, which has 2 hots. Your 6x20 Amp breaker total only takes 120 Volt loads into consideration. :?

You have to divide that total by 2 to be able to calculate the loads across both hot legs..... BUT ....

The third thing you are doing here is taking the overcurrent value of the breakers, and adding them up as if they will be loaded to 100% of their rating all the time. This is not the proper way to perform load calculations. :blink:

Start over by studying section 220 of the NEC ....

 

[jxofaltrds]

Try using this first:

http://www.mikeholt.com/documents/calculations/formulas/ResidentialLoadCalculations.xls

 

[kwired]

First, you are not required to use 80% when figuring the demand loads for the sub-panel. The so-called 80% rule only applies to continuous loads, such as electric resistance heat, or electric water heaters. If none of those are connected to the sub-panel, then eliminate any use of this part of the equation. :huh:

Second, your calculation of 6x20 Amp breakers fails to recognize that your sub-panel is fed with a 120/240 feeder, which has 2 hots. Your 6x20 Amp breaker total only takes 120 Volt loads into consideration. :?

You have to divide that total by 2 to be able to calculate the loads across both hot legs..... BUT ....

The third thing you are doing here is taking the overcurrent value of the breakers, and adding them up as if they will be loaded to 100% of their rating all the time. This is not the proper way to perform load calculations. :blink:

Start over by studying section 220 of the NEC ....

To add to what you have already said: OP has 80 amp subpanel @ 120/240 volts. If he is drawing 80 amps on both legs he is using 19200 VA. Divide that by general lighting load of 3 VA per SF and he can supply the general lighting load for 6400 Square feet from this panel. That is 80 x 80 feet, a pretty big house in most places.

 

[johnsmith09800]

Thanks for the helpful replies. After considering your responses, I think im worrying over nothing. I can estimate what most of the loads will be, so if they give me .. say a 20 circuit sub, I can probably use all 20 and still be no where close to being overloaded.