220.53 Fixed Appliance Loads Demand Factors

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Joequinones

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I'm attending the Residential Wiring course at Carteret Commuinty College in Morehead City NC. I'm doing a project on Calculations, my question is: 220.53 allows a demand factor of 75% for 4 or more fixed appliances. Does the 75% get added to the nameplate load demends increasing the total load demands of the 4 or more appliances by an added amount of 75%? Secondly, since Fixed Appliances are part of the General Lighting Load, do I adjust this demand load under 220.42, and Table 220.42? Lastly, what if the Fixed Appliances total less than 4, do I use nameplate loads and then apply 220.42?
 
You have a few basic misconceptions going on here. That’s OK, as that is the reason we go to classes.

The first basic principle is that we need to provide enough power to a building to handle its loads. We can certainly do that by adding up every single thing that has a wire attached. That would be called the “total connected load.” But the code recognizes that it is not likely that every single thing in the building will be running at full power at the same time. The purpose of “demand factors” is to let us provide a lesser amount of power than the total connected load.

In your first question, if you have a lot of appliances (e.g., dishwasher, trash compactor, garbage disposal), the owner is not likely to be running them at the same time. So you take the total load associated with these appliances, and multiply that value by 75%. This gives you a smaller number, not a larger one. It is not often that a single family dwelling unit will have four or more appliances that count under this article. We use the 75% factor more often on an apartment or condo building. When you calculate the service to the whole building, you get to say that the 15 units in the building each have a dishwasher and a disposal, so that is 30 fixed appliances, and you can take the 75% factor on those 30.

The next principle that often comes into load calculations (but not in your specific example, by the way) is that we need to increase the calculated amount by adding 25% for certain specific loads. The largest motor in the building gets 25% extra added to its calculated load. Any load that is likely to be running continuously for over 3 hours (i.e., a “continuous load”) also gets an extra 25% added. There are other examples, but as I said your question does not involve any of these.

The third principle that your question brings to my mind is that the appliances you are discussing are not associated with lighting at all. Table 220.42 does not come into play, when you add up appliance loads.

Welcome to the forum.
 
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Thank you Charlie B. for the reply, but I'm reading the NEC Handbook shaded area at the end of the Solution to the problem example they give. This is what is giving me a problem. The instructor did tell me to use the smaller amount using the 75%, but the book is telling me otherwise, or am I reading the NEC text incorrectly?
 
Help me understand. Which version of the handbook are you using, and where is the sample problem that you are discussing? I need to read their solution, in order to figure out what they are saying.
 
OK. I borrowed my cubicle neighbor’s 2008 handbook, and I found a sample calculation in the shaded area right after 220.53. So I will presume that is what you are talking about.
Joequinones said:
The instructor did tell me to use the smaller amount using the 75%, but the book is telling me otherwise, or am I reading the NEC text incorrectly?
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You are not reading the text correctly.

As I said earlier, you can always size a service or feeder by adding up 100% of everything that is connected to the power system. But you don’t need to do that, since the NEC gives demand factors for certain cases.

In the Handbook example, you have to include, in the feeder load calculation, the load associated with six fixed-in-place appliances. Their “connected load” adds up to 8944 VA. You could stop right there, and just use 8944 VA as the load for those six items. But you can also, if you wish, apply the 75% demand factor. Thus, instead of including 8944 VA, you need only include 6708 VA. You do have to include something, since these things do require power. You could add the "big number something," or a "smaller number something," but you must account for these loads. I think that is where you are getting confused.

Again, please note that this type of load is not included with the general lighting load, and the demand factors of Table 220.42 cannot be applied to them.
 
charlie b said:
Help me understand. Which version of the handbook are you using, and where is the sample problem that you are discussing? I need to read their solution, in order to figure out what they are saying.
Click to expand...

Don't mean to butt in but this is the example shown in the 2008 NECH

220.53 Appliance Load ? Dwelling Unit(s).
It shall be permissible to apply a demand factor of 75 percent to the nameplate rating load of four or more appliances fastened in place, other than electric ranges, clothes dryers, space-heating equipment, or air-conditioning equipment, that are served by the same feeder or service in a one-family, two-family, or multifamily dwelling.

For appliances fastened in place (other than ranges, clothes dryers, and space-heating and air-conditioning equipment), feeder capacity must be provided for the sum of these loads; for a total load of four or more such appliances, a demand factor of 75 percent may be applied. See Table 430.248 for the full-load current, in amperes, for single-phase ac motors, in accordance with 220.50.

Calculation Example
Determine the feeder capacity needed for a 120/240-volt fastened-in-place appliance load in a dwelling unit for the following:

Appliance Rating Load
Water heater 4000 W, 240 V 4000 VA
Kitchen disposal 1/2 hp, 120 V 1176 VA
Dishwasher 1200 W, 120 V 1200 VA
Furnace motor 1/4 hp, 120 V 696 VA
Attic fan 1/4 hp, 120 V 696 VA
Water pump 1/2 hp, 240 V 1176 VA

Solution
STEP 1.
Calculate the total of the six fastened-in-place appliances:

Total load = 4000VA + 1176 Va + 1200 VA
+696 VA + 696 VA + 1176 VA
=8944 VA

STEP 2.
Because the load is for more than four appliances, apply a demand factor of 75 percent:

8944 VA X 0.75 = 6708 VA

Thus, 6708 volt-amperes is the load to be added to the other determined loads for calculating the size of service and feeder conductors.
 
220.53

220.53

To all who made this NEC issue clear to me, thank you all. We had quite a discussion in class about the calculations. This forum is going to be a big help for all of us in the future. Most of the students, I'm sure will be signing up, this is an great tool.

Joe
 
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