apartment complex - cloths dryer load calculation?

[HeyItsMe]

I don?t remember the exact question, but I am trying to to figure out a question I got on a test I am going to be retaking soon

The question was something like this:

41 unit apartment building and an electric clothes dryer will be installed in each unit, what is the calculated load on all the dryers and what is the load on the neutral.

There was 4 answers but I forget them, all I remember is 2 were 00,000 long and 2 were 000,000


5,000 * 41 = 205,000

205,000 * 35% = 71750

41 - 23 = 18

* 0.009 (0.5%)

i am lost been trying to figure this out for way to long

i do know what to say about the

"2 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 max number connected between any 2 phases"

none of that is mentioned in the question at all, unless something is assumed that i dont understand.

 

[newservice]

I have a few questins, but Ill wait and let someone else chime in.

 

[HeyItsMe]

I am not sure that article 220.54


"2 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 max number connected between any 2 phases"


is applied at all because that information was not supplied in the question, unless its assumed

And the rest of the information I got was from table 220.54


24-42 = 35% minus 0.5% for each dryer exceeding 23

So I started with 41 dryers * 5000 watts (volt amperes)
205,000 * 35% - 71750
Then 41 (units/dryers) - each dryer exceeding 23 = 18
18 * 0.5% = 0.009

Is my work right to this point? I am kind of lost here??

Then what ever number I come up with I multiply times 70% for the neutral, right?

The answers given were like this
00,000 ? 00,000
00,000 ? 00,000
000,000 ? 00,000
000,000 ? 00,000

 

[Dennis Alwon]

There is an example for ranges - which would be similar- in Annex D of the NEC-- Example D5(a) -- I am cooking and don't have time to work it out.

 

[Dennis Alwon]

In case you don't have the book with you

Example D5(a) Multifamily Dwelling Served
at 208Y/120 Volts, Three Phase
All conditions and calculations are the same as for the multifamily dwelling
[Example D4(a)] served at 120/240 V, single phase except as follows:
Service to each dwelling unit would be two phase legs and neutral.
Minimum Number of Branch Circuits Required for Each Dwelling
Unit (see 210.11)
Range Circuit: 8000 VA ? 208 V = 38 A or a circuit of two 8 AWG
conductors and one 10 AWG conductor in accordance with 210.19(A)(3)
Minimum Size Feeder Required for Each Dwelling Unit (see 215.2)
For 120/208-V, 3-wire system (without ranges),
Net calculated load of 3882 VA ? 2 legs ? 120 V/leg = 16 A
For 120/208-V, 3-wire system (with ranges),
Net calculated load (range) of 8000 VA ? 208 V = 39 A
Total load (range + lighting) = 38.5 A + 16.2 A = 54.7 A
Feeder neutral: (range) of 8000 VA ? 70% = 5600 VA ? 208 V = 27 A
Total load: (range + lighting) = 27 A + 16 A = 43 A
Minimum Size Feeders Required from Service Equipment to Meter
Bank (For 20 Dwelling Units ? 10 with Ranges)
For 208Y/120-V, 3-phase, 4-wire system,
Ranges: Maximum number between any two phase legs = 4
2 ? 4 = 8.
Table 220.55 demand = 23,000 VA
Per phase demand = 23,000 VA ? 2 = 11,500 VA
Equivalent 3-phase load = 34,500 VA
Net Calculated Load (total):
40,590 VA + 34,500 VA = 75,090 VA
75,090 VA ? (208 V)(1.732) = 208 A

 

[Smart $]

I am not sure that article 220.54

...
24-42 = 35% minus 0.5% for each dryer exceeding 23
...

You are applying the demand factor incorrectly.

For each unit over 23, the demand factor decreases by 0.5%...

24 units at 34.5%
25 units at 34%
26 units at 33.5%
27 units at 33%
and so forth.

For 41 dryers, it is...

5,000kVA * 41 units * (35% - (41-23)*0.5%) =
5,000kVA * 41 units * (35% - 9%) =
5,000kVA * 41 units * 26% = 53.3kVA

 

[Smart $]

...
"2 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 max number connected between any 2 phases"
...

If your service is 208/120V 3?, and each apartment is supplied 120/208 1?... and divided equal as possible...

41 ? 3 = 14 rounded up

5,000kVA * 14?2 * (35% - (14?2-23)*0.5%) =
5,000kVA * 28 * (35% - 2.5%)
5,000kVA * 28 * 32.5% = 45.5kVA

 

[Smart $]

?
Then what ever number I come up with I multiply times 70% for the neutral, right?
...

If the supply is split phase, such as 120/240, you can.

If it is 120/208V 1? to each apartment, you cannot apply the 70% demand factor for the individual apartment feeder neutral [ref: 220.61(C)(1)]. On the supply side of the 208/120V 3? to 120/208V 1? distribution panel you can.

Don't forget the 70% demand factor only applies to the ranges, ovens, counter-mounted cooking units, and electric clothes dryer portion of the maximum unbalanced load... not the entire load, and not the entire maximum unbalanced load.

 

[HeyItsMe]

Smart $ and everyone else, thanks! I now remember that one of the answers were 53.3, and I can see what I was doing wrong with my math. Thank you very much everyone here for helping with this question! Maybe one of these days I will be able to answers questions on here too.


long time reader, first time posting......



edit: spelling

 

[JDBrown]

You are applying the demand factor incorrectly.

For each unit over 23, the demand factor decreases by 0.5%...

24 units at 34.5%
25 units at 34%
26 units at 33.5%
27 units at 33%
and so forth.

For 41 dryers, it is...

5,000kVA * 41 units * (35% - (41-23)*0.5%) =
5,000kVA * 41 units * (35% - 9%) =
5,000kVA * 41 units * 26% = 53.3kVA

Wait a minute ... so, if there were 93 dryers ...

5,000kVA * 93 units * (35% - (93-23)*0.5%) =
5,000kVA * 93 units * (35% - 35%) =
5,000kVA * 93 units * 0% = 0kVA

If there are 93 dryers, we don't have to count the load from any of them?

And if there are more than 93 dryers we get a load credit, as it were?

I think my head just exploded. :blink:

 

[Ragin Cajun]

Typical dryer has a 5000w element at 240V PLUS the motor of ~~500w (can't forget the motor)

At 208V that 5000w drops down to ~3750w plus motor for ~~4250w.

Thus 4250 * 41 * .26 = 45.3 kw?


RC

 

[david luchini]

At 208V that 5000w drops down to ~3750w plus motor for ~~4250w.

220.54 says the load for each dryer shall be 5000w or the nameplate rating, whichever is greater.

 

[GoldDigger]

Wait a minute ... so, if there were 93 dryers ...

5,000kVA * 93 units * (35% - (93-23)*0.5%) =
5,000kVA * 93 units * (35% - 35%) =
5,000kVA * 93 units * 0% = 0kVA

If there are 93 dryers, we don't have to count the load from any of them?

And if there are more than 93 dryers we get a load credit, as it were?

I think my head just exploded. :blink:

43 and over 25%

(by the formula for 24-42, we see that 42 will be 25.5%)
Pick up your head and set it back in place gently this time.

 

[Smart $]

Typical dryer has a 5000w element at 240V PLUS the motor of ~~500w (can't forget the motor)

At 208V that 5000w drops down to ~3750w plus motor for ~~4250w.

Thus 4250 * 41 * .26 = 45.3 kw?


RC

220.54 says the load for each dryer shall be 5000w or the nameplate rating, whichever is greater.

Also note the example I worked out for a 208/120V 3? 4W service with 120/208V 1? 3W distribution. That condition results in less calculated load for the same number of dryers.

 

[david luchini]

Also note the example I worked out for a 208/120V 3? 4W service with 120/208V 1? 3W distribution. That condition results in less calculated load for the same number of dryers.

Wouldn't you have to multiply by 1.5 to get the equivalent 3 phase load, as suggested by Example D5(a)? Or 68.25kVA (33.3% demand?)

 

[JDBrown]

(by the formula for 24-42, we see that 42 will be 25.5%)
Pick up your head and set it back in place gently this time.

:slaphead: That's what I get for just reading the thread and not looking at the actual Code section before posting.

 

[Smart $]

Wouldn't you have to multiply by 1.5 to get the equivalent 3 phase load, as suggested by Example D5(a)? Or 68.25kVA (33.3% demand?)

That's a good question... for which I have no definitive answer. The example noted demonstrates the calculation regarding ranges, not dryers. However, 220.55 has an identical clause to 220.54 for calculating the total load... "Where two or more single-phase dryers are supplied by a 3-phase, 4-wire feeder or service..."

The only difference is 220.55 has an IN that says see the example... and we both know IN's are not enforceable. Neither clause actually says to divide the result by two then multiply by three, as in the example (or in effect, multiply by 1.5 as you stated). I suppose that is what "on the basis" means... but "on the basis" is far from an explicit method to derive a definitive equation.

It would be better worded if it stated, rather than...

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.

...as...

Where two or more single-phase dryers are supplied by a 3-phase, 4-wire feeder or service, the demand factor shall be determined on the basis of twice the maximum number connected between any two phases.

Using the latter and revised clause, it'd be more obvious to calculate as such...

41 ? 3 = 14 rounded up
14 ? 2 = 28

5,000kVA ? 41 units ? (35% - (28-23)*0.5%) =
5,000kVA ? 41 units ? (35% - 2.5%)
5,000kVA ? 41 units ? 32.5% = 66,625VA

 

[GoldDigger]

Wouldn't you have to multiply by 1.5 to get the equivalent 3 phase load, as suggested by Example D5(a)? Or 68.25kVA (33.3% demand?)

As I read it, the calculation for single phase dryer loads is calculated ONLY via the two times the largest number on any single phase, with no application of any of the other demand calculations.
My explanation, to myself and anyone who wants to listen, on why this is done is as follows:

1. For three phase loads, you can assume, to a first approximation, that each of the dryers will present a balanced load across all three phases. (To the extent that there is a single phase motor or control load, that will be a small percentage and can still be distributed fairly well by rotating the phase that the motor is connected to across different dryers.)
So, the laws of probability can be based on the diversity of the total number of dryers. Since each dryer is used by a single tenant, the odds of a whole bunch being used at the same time are low. (Events like a pool party for the whole complex might change that. )
2. The load will be distributed over all phases, so the calculation of the size (amperage) of the individual phase wires can be estimated perfectly well by looking at the total power consumption. 1. and 2 are taken into account in the formulas and tables.
3. When you have purely single phase loads, the diversity factor affecting each phase is lower. Instead of the diversity of 33 dryers, you now have, for example, three groups of 11 dryers each. And the diversity that you can apply to any single phase will be that associated statistically with 11 dryers, not 33. The chances of overload any one phase are now what is important, not the total power consumption. That fits in with Smart $'s suggestion.
4. But instead of doing that, the Code, IMHO, takes the easy way out and just flat out says that you take twice the power drawn by the number of units on the heaviest loaded phase and that is the final answer. Then you assume that that power number is supplied equally by all three phases. (Or at least that 1/3 of that number should be used to size the individual phase conductors, feeders, etc.)

 

[david luchini]

As I read it, the calculation for single phase dryer loads is calculated ONLY via the two times the largest number on any single phase, with no application of any of the other demand calculations.

I don't believe that is what the Code reads. Nor would it make much sense. To read it that way, 60 dryers on a single phase system would have a demand of 15 dryers (75.0kW), but the same 60 dryers on a 3 phase system would have a demand of 40 dryers(200kW.) For 90 dryers the demand would be 22.5 (112.5kW) on single phase and 60 (300kW) on three phase. Why would the demand be so much higher just because they are on a 3 phase system?

1. For three phase loads, you can assume, to a first approximation, that each of the dryers will present a balanced load across all three phases. (To the extent that there is a single phase motor or control load, that will be a small percentage and can still be distributed fairly well by rotating the phase that the motor is connected to across different dryers.)
So, the laws of probability can be based on the diversity of the total number of dryers. Since each dryer is used by a single tenant, the odds of a whole bunch being used at the same time are low.

That's the point of the demand factors.

3. When you have purely single phase loads, the diversity factor affecting each phase is lower. Instead of the diversity of 33 dryers, you now have, for example, three groups of 11 dryers each. And the diversity that you can apply to any single phase will be that associated statistically with 11 dryers, not 33. The chances of overload any one phase are now what is important, not the total power consumption. That fits in with Smart $'s suggestion.

The problem with this is that there are not 11 dryers associated with any single phase, but 22. If there are 11 dryers between A-B and 11 between B-C, then phase B has 22 dryers associated with it. But as the system is 3 phase and the A-B and B-C currents are not directly additive you could say that each phase has 18 dryers associated with it (11*1.732.)

4. But instead of doing that, the Code, IMHO, takes the easy way out and just flat out says that you take twice the power drawn by the number of units on the heaviest loaded phase and that is the final answer. Then you assume that that power number is supplied equally by all three phases. (Or at least that 1/3 of that number should be used to size the individual phase conductors, feeders, etc.)

By applying the demand factors to twice the highest number connected between any two phases and then applying the equivalent 3 phase (as shown in example D5(a)) you will find that the demand is slightly larger for the same number of dryers (starting with a low quantity) on a 3 phase system, but that as more dryers are added, the difference between 3 phase and single phase demand is reduced, until you eventually get the same demand for both. This addresses the lower diversity for 3 groups that you note in your Point #3.

For instance, with 33 dryers the single phase demand is 49.5kW and the 3 phase demand is 59.4kW. (Compare to 110.0kW for the two times the largest with no demand)

With 60 dryers, the single phase demand is 75.0kW and the 3 phase demand is 79.5kW. (Compare to 200.0kW)

With 90 dryers, the single phase demand is 112.5kW and the 3 phase demand is 112.5kW. (Compare to 300.0kW)

 

[Smart $]

I don't believe that is what the Code reads. Nor would it make much sense. ...

IMO the Code as written doesn't make sense. First, we have no empirical data upon which to base any demand factors for a single phase system let alone determine different demand factoring where the same number of identical single-phase dryers are supplied by a 3-phase, 4-wire feeder or service.

Without supporting evidence, the demand for multiple single phase dryers on a two-conductor system should be the same as on a three-conductor system. The only difference that should actually be accounted for is the issue brought up by Ragin Cajun (i.e. kW at 240V vs. 208V).

To say the demand factor for the same number of identical dryers on a three phase system is higher than on a single phase system is ludicrous from the perspective of statistical analysis.