Just took a state test in oregon

Status
Not open for further replies.
kwired said:
I think we both did the same thing just in a little different way.

What doesn't seem quite right is that two or three dryers (balanced across the phases) comes out to 10 kVA. But then if you consider the dryer is likely rated 5000 to 5500 watts at 240 volts, and you put it on a 208 volt system it will probably only operate at 3750 to 4100, though you are using 5000 in your calculation. Plus the fact you need more then 4 before a demand factor less then 100% applies.

I really have little concern that you would be short on supply capacity with that calculation method.
Click to expand...
That's because your calculation isn't right. It comes out as 15kVA for 2 or 3 dryers distributed as best possible.

You are correct that Code makes no allowance for operating a 240V-only-rated dryer on a 208V system. Another oddity is no compensation is required for one dryer on a 3Ø system. You go from 5kVA for 1 dryer, to 15kVA for 2 and 3. Look what happens after that...

1 5,000
2 15,000
3 15,000
4 30,000
5 25,500
6 22,500
7 29,250
8 27,000
9 24,750
10 30,000
11 28,200
12 27,600
13 33,750
14 33,000
15 32,250
 
Smart $ said:
That's because the calculation is not finished.

  1. Determine maximum number connected between any two phases (aka lines or legs)
  2. Multiply load in kVA by 2.
  3. Determine demand load per Table 220.54.
  4. Divide the above result by 2 and multiply by 3 to determine overall demand load.
Note you get the same result with two or three dryers and a maximum of one connected between any two legs.
Click to expand...
I am not seeing why you go through the steps you went through.

Where two or more single-phase dryers are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases.
Click to expand...

That says to me if I have two or three dryers there is maximum number connected between any two phases is two. If we are sticking with 5kVA as the load of each unit, twice that is 10kVA. Demand factor for 2 or 3 units is 100% - this tells me the load is 10 kVA, which I agree don't make sense as we all know there is 15 kVA there, but where in between the lines does it say to divide by 2 and then multiply by 3?

I will admit to never having to do this calculation that I can recall. Why do they even mention twice the current other then maybe they are rounding up the square root of three?

There has to be something I am missing in there.
 
kwired said:
I am not seeing why you go through the steps you went through.



That says to me if I have two or three dryers there is maximum number connected between any two phases is two. If we are sticking with 5kVA as the load of each unit, twice that is 10kVA. Demand factor for 2 or 3 units is 100% - this tells me the load is 10 kVA, which I agree don't make sense as we all know there is 15 kVA there, but where in between the lines does it say to divide by 2 and then multiply by 3?

I will admit to never having to do this calculation that I can recall. Why do they even mention twice the current other then maybe they are rounding up the square root of three?

There has to be something I am missing in there.
Click to expand...
Well for confirmation, look at Annex D Example D5(a) calculations for ranges in a multifamily dwelling. That's the only place there's any indication of dividing the demand by 2 and multiplying by three.

Heck. For all I know, I may be getting it wrong. :blink:
 
I'm still confused on this ...so if I have 10 units..the max on one phase is 4 units I multiply this X 2..that gives me 8..so how do I apply that to the table..which is in %..
 
Smart $ said:
Well for confirmation, look at Annex D Example D5(a) calculations for ranges in a multifamily dwelling. That's the only place there's any indication of dividing the demand by 2 and multiplying by three.

Heck. For all I know, I may be getting it wrong. :blink:
Click to expand...
Looked at the example you mentioned - still confused. I sort of get what they did, yet don't get why.

For ten 8 kVA ranges they come up with a feeder load after demand factors of 34.5kVA on a three phase 208 wye system.

Now go with a 240 single phase system and 220.55 only says we need 25 kVA for the same ten ranges.
 
morg123452000 said:
I'm still confused on this ...so if I have 10 units..the max on one phase is 4 units I multiply this X 2..that gives me 8..so how do I apply that to the table..which is in %..
Click to expand...
I agree the dryers is even more confusing, do we apply demand factor to 4 units, 8 units, VA, VA per leg (before or after doubling the maximum connected to one phase), maybe we could double 220.55 and go to 440.110 and see what it says there.
 
You use the demand factor percentage for 2X units but you multiply that percentage by 1X wattage.
Then since that is the kW for one phase you multiply by three.
Couldn't be simpler, right?

Sent from my XT1585 using Tapatalk
 
kwired said:
Looked at the example you mentioned - still confused. I sort of get what they did, yet don't get why.
...
Click to expand...
The why is they are padding the demand load so when you divide by voltage and root of three to determine service or feeder conductor ampacity it will support a full compliment of balanced units... or something close to that. For example, with ten units, the calculated demand load will support twelve units (the next greater multiple of three).
 
GoldDigger said:
You use the demand factor percentage for 2X units but you multiply that percentage by 1X wattage.
Then since that is the kW for one phase you multiply by three.
Couldn't be simpler, right?

Sent from my XT1585 using Tapatalk
Click to expand...
I guess if you are used to doing load calculations for each line it makes sense.

VA is VA though, it just happens to be distrubuted differently with a three phase system. Still doesn't seem right that the three phase example in the annex comes up with 34.5 kVA for ten ranges yet if it were a single phase source the demand for the same load should be 25 kVA. Only thing that should change IMO is measured amps on an individual conductor, VA should be the same either way.
 
When I take the test again in oregon I'm thinking of Calc just the stoves for the 3 phase system and do the dryers normal way..it's to confusing and I don't beleave the writer of the test would include that..the only examples for single phase appliances on a 3 phase feed is for the stoves I've never seen one for dryers even though the code mentions it in 220-54....any thoughts on that?
 
kwired said:
I guess if you are used to doing load calculations for each line it makes sense.

VA is VA though, it just happens to be distrubuted differently with a three phase system. Still doesn't seem right that the three phase example in the annex comes up with 34.5 kVA for ten ranges yet if it were a single phase source the demand for the same load should be 25 kVA. Only thing that should change IMO is measured amps on an individual conductor, VA should be the same either way.
Click to expand...
I wouldn't go as far as to say that. IMO it is a poor attempt at compensating for 3Ø load distribution. I could probably come up with a better method.
 
morg123452000 said:
When I take the test again in oregon I'm thinking of Calc just the stoves for the 3 phase system and do the dryers normal way..it's to confusing and I don't beleave the writer of the test would include that..the only examples for single phase appliances on a 3 phase feed is for the stoves I've never seen one for dryers even though the code mentions it in 220-54....any thoughts on that?
Click to expand...
Method for both is very similar, main difference is the ranges come in a wider variety of kW ratings, where dryers are always in the range of 5- 5.75 kW. That results in the demand factor table for dryers being just a percentage where the range table gives you a specific demand based on kW rating of the appliances. If you want to make it more confusing mix and match kW ratings of the ranges, cooktops, ovens. Several apts with same range rating is the simpler method.
 
Smart $ said:
IMO it is a poor attempt at compensating for 3Ø load distribution.
Click to expand...

Was kind of my thinking from the start. But since it involves demand factors we also need to know why they chose what they did before shooting it down too far. It definitely is not the way to determine actual load, but demand factor loading is different. I still don't see why applying 50, 60, 80% to a certain number of units works for single phase but not three phase. But they also disregard the fact that most of these appliances are rated at 120/240 volts, yet when connected to a three phase system they are likely only going to operate at 208 volts. Maybe they have considered that they will run longer heating cycles on 208 then on 240 so it impacts necessary demand factor some?

Regardless I think the process could somehow be explained a little better in the code requirement.
 
Status
Not open for further replies.
Top