Mulitifamily Dryer Loads

[reb707]

I am studying for a contractors test and one of the questions has me stumped. Here it is

We have an 8 unit apartment complex, with a 120/208, 3 phase, 4 wire service. Each Apartment has a 4.5kw single phase 120/208 volt dryer. The question is what would be the minimum service demand be for these dryers. My calculations are below, but my answer is not one of the options.

8 / 3 = 2.6, round up to 3 then 3 * 2 = 6
8 - 6 = 2

5000 * 6 = 30,000 kva
5000 * 2 = 10,000 kva

40000kva * .6 (60%) = 24 kva But 24 is not an option

The options are 21, 27, 36 and 40 kw

 

[charlie b]

I get 18 KVA. You will have 3 dryers on A-B, 3 dryers on A-C, and 2 dryers on B-C. So the most between two phases is 3. The total load is therefore 5000 times (3x2), or 30 KVA. You reached this same point in your approach. I don't know why you are adding 10 KVA more. The code says that the total load is twice the load between any two phases. That is the 30 KVA. All that remains is to apply the 60% demand factor to the 30 KVA, and that is how I reached an answer of 18 KVA.

Now perhaps someone can tell me if my approach is flawed.

 

[Dennis Alwon]

Charlie I think you have to use 8 dryer and use 75%. I get 33,750kw

 

[Dennis Alwon]

I agree with the 3 max per phase. 3 x 2= 6
6 dryer in Table 220.54 = 75%

6x 5000 = 3000 *.75 = 22500

22500 x 3/2= 33750kw

 

[charlie b]

Charlie I think you have to use 8 dryer and use 75%. I get 33,750kw

There is room for confusion here. My interpretation is that the total load could have been set at 8 times 5000. But it did not have to. We are allowed to count the total load as twice the load seen by the highest number of dryers connected between two phases. Balancing the load as best as possible puts no more than 3 dryers on any pair of phases. Twice that is 6, and 6 x 5000 is 30,000. So I will stick with that.

I did use the demand factor for 8 dryers. That is 60%, not 75%.

 

[charlie b]

I agree with the 3 max per phase. 3 x 2= 6. 6 dryer in Table 220.54 = 75%

Here again, I think the 3 x 2 is supposed to be used to determine how much VA we have as a total load.

22500 x 3/2= 33750kw

I have no idea where the 3/2 comes from here. Please clarify.

 

[Dennis Alwon]

Look at the example D(5a) in the back of the book in annex D

 

[Smart $]

Look at the example D(5a) in the back of the book in annex D

That shows the calculation for ranges but the approach is nearly identical...

Dryers: Maximum number between any two phase legs = 3
3 per phase x 2 = 6 per leg
6@5,000VA ea. = 30,000VA
Table 220.54 demand = 30,000VA × 6@75% = 22,500VA
Per phase demand = 22,500 VA ÷ 2 = 11,250 VA
Equivalent 3-phase load = 3 legs @ 11,250 VA ea. = 33,750VA

The last two lines is where the 3/2 multiplier comes in.

 

[Dennis Alwon]

That shows the calculation for ranges but the approach is nearly identical...

Dryers: Maximum number between any two phase legs = 3
3 per phase x 2 = 6 per leg
6@5,000VA ea. = 30,000VA
Table 220.54 demand = 30,000VA × 6@75% = 22,500VA
Per phase demand = 22,500 VA ÷ 2 = 11,250 VA
Equivalent 3-phase load = 3 legs @ 11,250 VA ea. = 33,750VA

The last two lines is where the 3/2 multiplier comes in.

Is that not the answer I had? I know it is for ranges but like you said it is the same calc.

 

[Smart $]

Is that not the answer I had? I know it is for ranges but like you said it is the same calc.

The answer in my post—a slightly revised version of the Annex D Example—is exactly the same as your answer.

 

[Dennis Alwon]

The answer in my post—a slightly revised version of the Annex D Example—is exactly the same as your answer.

I gotcha. I just combined the division and multiplication to get 3/2- I guess I could have spelled it out but I figured if Charlie saw the example in the book he would see where it came from... Thanks

 

[kwired]

I am studying for a contractors test and one of the questions has me stumped. Here it is

We have an 8 unit apartment complex, with a 120/208, 3 phase, 4 wire service. Each Apartment has a 4.5kw single phase 120/208 volt dryer. The question is what would be the minimum service demand be for these dryers. My calculations are below, but my answer is not one of the options.

8 / 3 = 2.6, round up to 3 then 3 * 2 = 6
8 - 6 = 2

5000 * 6 = 30,000 kva
5000 * 2 = 10,000 kva

40000kva * .6 (60%) = 24 kva But 24 is not an option

The options are 21, 27, 36 and 40 kw

No one that has replied has come up with any of those answers, possible the person that wrote the question doesn't know the answer either?

 

[Dennis Alwon]

My guess is they did exactly what Smart$ did only they used 60% instead of 75% (which is incorrect, IMO) and that comes to 27kw