Can someone please help me with this dryer calculation please?

[kefox81]

Hello again everyone. I need help with this dryer calculation on my practice test. Here it is.
An eight unit apartement complex is served by a 120/208 volt 3 phase 4 wire system. Each apartment has a 4.5 kw single phase 120/208 volt dryer. What would be the minimum service demand for these dryers?

a. 40kw
b. 36kw
c. 27 kw
d. 21 kw.

The practice test says the correct answer is 27kw but I cant figure it out. This is what I can do so far. I turn to article 220.54 and use a rating of 5000 watts per unit per 200.54.
8 dryers x 5000 watts = 40,000.
I then go to table 220.54 and see that 8 dryers use a demand factor of 60%. I then do
40,000 x 60% =24,000.
That's as far as I can get. Can anyone help me please?

 

[Danelectron]

Hello again everyone. I need help with this dryer calculation on my practice test. Here it is.
An eight unit apartement complex is served by a 120/208 volt 3 phase 4 wire system. Each apartment has a 4.5 kw single phase 120/208 volt dryer. What would be the minimum service demand for these dryers?

a. 40kw
b. 36kw
c. 27 kw
d. 21 kw.

The practice test says the correct answer is 27kw but I cant figure it out. This is what I can do so far. I turn to article 220.54 and use a rating of 5000 watts per unit per 200.54.
8 dryers x 5000 watts = 40,000.
I then go to table 220.54 and see that 8 dryers use a demand factor of 60%. I then do
40,000 x 60% =24,000.
That's as far as I can get. Can anyone help me please?

Halfway through 220.54 it states that for two or more single phase dryers that are supplied by a 3-phase, 4-wire service the total load will be calculated on the basis of twice the maximum number connected between any two phases. So the max number of dryers across any two phases in a properly distributed panel is 3.
Worked out using 4.5kW you have 4.5kW x 3(dryers) x 2= 27kW.
I probably would have used 5kW for each dryer load (per the first sentence in 220.54) resulting in 30kW and scratched my head as well.

 

[Dennis Alwon]

I think what they did was used 6 dryersvat 5000 = 30000

But then they took 8 Dryers demand = 60%

60% x 30000 = 18000 / 2= 9000 x # of phases 3 = 27,000

I don't think that is correct, they should have used 75% which is the demand for 6 dryers

Look at the example for ranges in Annex D(5)(a)

 

[Santa49]

Santa49

Santa49

I think what they did was used 6 dryersvat 5000 = 30000

But then they took 8 Dryers demand = 60%

60% x 30000 = 18000 / 2= 9000 x # of phases 3 = 27,000

I don't think that is correct, they should have used 75% which is the demand for 6 dryers

Look at the example for ranges in Annex D(5)(a)

The nearest I can get to the given answer is the following.
Article 220.54 states where 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 maximum number connected between any 2 phases. Based on a balanced system there are 3 dryers maximum between phases. Times 2=6 dryers.
6 dryers at 4.5kw=27kw.
Someone please correct me if that is not correct.

 

[Dennis Alwon]

The nearest I can get to the given answer is the following.
Article 220.54 states where 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 maximum number connected between any 2 phases. Based on a balanced system there are 3 dryers maximum between phases. Times 2=6 dryers.
6 dryers at 4.5kw=27kw.
Someone please correct me if that is not correct.

I don't believe that is correct because in that same article it states that you must 5000 watts or higher so that eliminates 4500 watts. Also, that is not how they did it in annex d

 

[JFletcher]

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