Author Topic: I need help!  (Read 426 times)
lightinglover8902
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Power distributor: CenterPoint Energy. 120V 60Hz


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I need help! « on: June 22, 2018, 03:09:22 PM » Author: lightinglover8902
I'm working on a Penn Foster project which is based on the NEC, and I need a little bit of help.


Example 1: Suppose that you’re working in a home that has
a 15 kW oven that operates on 240 V. The oven is on a branch
circuit by itself, as shown in Figure 4.

Question 35: What is the demand load for this circuit? (Show
all of your calculations on the calculation sheet at the end of
Part 3.)

Question 36: What size TW copper conductor should be used
for the branch circuit? (Show all of your calculations on the
calculation sheet at the end of Part 3.)

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Save the Cooper OVWs!! Don't them down by crap LED fixtures!!!

Ash
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Re: I need help! « Reply #1 on: June 22, 2018, 06:31:46 PM » Author: Ash
Demand, in the sense of demand coefficient, relates to the fact that not all appliances on the premises are switched on at the same time. This allows to undersize the main feeders, and undersize more the feeders feeding them, and so on - which means that they could be overloaded, but the overload condition is unlikely to happen

Why this undersizing is good ?

Let's say you size the feeder for the house according to the xum of all connected appliances, switched on at the same time

Then the main feeder feeding the row of houses will have to be sized for all of them having everything on at the same time

Then the distribution grid branch will have to be sized like that too

And the electrical grid as a whole..

And the power plants

This will require building huge power plants, designed for the slimmest chance that never happens, that everything in the entire country is on at the same time. In reality, those huge plants will never ever have to be powered on, or will be permanently powered at nearly no load, burning fuel for nothing. And require very thick wiring and higher rated transformers throughout the grid, that will be nothing but waste of resources

If only the power plants are undersized according to the realistic loads present, then in the unlikely event that those loads will be exceeded, the entire grid will come down, because the power plants are overloaded

So, the ampacity reduction so to speak is distributed downstream, making it more likely to trip as little of the system as possible when users are overloading :

 - If you overload one circuit you trip that circuit

 - If you don't overload any particular circuit, but do overload the feeder for the house as a whole, you trip the main for the house

 - If you don't overload in any particular house, but do overload the feeder for the row of houses as a whole, you trip the main for the row

..
..

This means that the most likely overloading and resulting outages are moved away from the plants, big parts of the grid, and so on, towards small local outages down to the user's circuits or feeder, and enable building the power plants, grids, distribution, feeders, and such with only as much resources as they actually need to work "most of the time", yet safe enough to work for decades without the overload happening



To estimate how much a feeder can be downsized compared to the sum of things it powers, there is a statistical evaluation : "Users at home use in reality at most X of the max power they could use in theory". This evaluation had been done for every place where the downsizing is done :

For homes

For commercial loads

For feeders that feed X, Y, Z

etc..

The result of the evaluation, which you can use, is the demand factor. It's a number less than 1

So :

For example you have 10 20A circuits coming from a panel. Calculate the required size of feeder for this panel. Max load = 200A. If say the demand coefficient for this sort of place is 0.7, then it requires 0.7 * 200 = 140A. Round that up to the nearest standard size, that would be for example 150A

The coefficeints are provided in books, and for the general cases you could use them. But do your judgement : If you know the way how the circuit will be used, and you think it justifies a different demand coefficient, use your own one



For one appliance, this is not applicable : If the appliance is switched on, it will really draw up to it's full power. So the circuit for the 15kW heater must be designed to handle 15kW normal load, there is nothing that could be undersized in this case

(for some appliances you still could undersize based on the assumption that they don't run for long at a time, but it's not the case here either)

15kW * 5/4 = 18 3/4 kW (apply safety margin - don't design circuits to be loaded above 4/5s of their capacity in normal use)

18 3/4 kW / 240V = 78A

Choose the breaker that is next above 78A (well, in this case i dont think you are doing anything wrong if you'd choose 75A, if that size exists - It's still plenty of safety margin, even if a little short of the standard 4/5ths)

Choose the cable according to :

 - The breaker current rating you chose (for example, if the load is 78A but the closest next available breaker size is 100A, then the cable is to be chosen for 100A)

 - The cable installation method/conditions (in wall, in cable raceways, underground, type of cable, etc)

 - The ambient temp at the place of installation (apply derating if ambient temp is high)

 - The number of heating conductors going together. For single phase load it's 2. For 3 phase load it's 3. (even if Neutral is connected, it does not carru current, so is not counted as a conductor that contributes to the heat of the cable)

 - If the cable run is very long, you might choose bigger cable size to reduce voltage drop or avoid excessive short circuit loop impedance



In reality at this size you have good reason to use Aluminum over Copper wire for the circuit



Not NEC compliant answer here, just plain considerations..
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