dryer

[treyp3]

what is the 3 phase demand load in kw for 40 1 phase 5 kw dryers supplied by a 120/208v 3 phase 4 wire system.

I get 68.25kw like this: 40/3=14 dryers x 2 per phase = 28 dryers. 35-2.5=32.5%
28x5kw=140kw x 32.5%=45.5kw. 45.5/2=22.75 x 3 phases =68.25

The test book gives nearly three times this amount as an answer. Am I way off?

 

[squaredan]

Dryer

Dryer

hey treyp3, I have been studying to take the test and doing alot of calculations and that is how i would do it...maybe we are both missing something...Dan did the question say anything about dwelling unit or commerical dyers cuz 220.54 is for dwelling units, same with Table 220.54 household dyers

 

[squaredan]

Dryer

Dryer

What was the answer?

 

[Dennis Alwon]

what is the 3 phase demand load in kw for 40 1 phase 5 kw dryers supplied by a 120/208v 3 phase 4 wire system.

I get 68.25kw like this: 40/3=14 dryers x 2 per phase = 28 dryers. 35-2.5=32.5%
28x5kw=140kw x 32.5%=45.5kw. 45.5/2=22.75 x 3 phases =68.25

The test book gives nearly three times this amount as an answer. Am I way off?

How could it be 3 times your answer. If you just took 5000 * 40 = 200kw so it can't be correct.

I believe you are correct. I edited my work.

 

[George Stolz]

I agree with Dennis.

Going out on a limb, now, I'm going to say that if I were to do this problem I would do this:

40 dryers x 5kW = 200 kW
(now demand)
40 dryers = 23 + 17
17 x 0.5% = 8.5%
35% - 8.5% = 26.5%
200 kW x 26.5% = 53 kW

Note that I made no allowance for some dryers being on A*B, some B*C, and some C*A. My reason is because I do not see a panel schedule attached to the question, and I have no assurance that the load is evenly balanced. IMO, the "maximum load connected between any two phases" means exactly that - every dryer connected A*B, no balancing of dryers.

 

[bob]

what is the 3 phase demand load in kw for 40 1 phase 5 kw dryers supplied by a 120/208v 3 phase 4 wire system.

I get 68.25kw like this: 40/3=14 dryers x 2 per phase = 28 dryers. 35-2.5=32.5%
28x5kw=140kw x 32.5%=45.5kw. 45.5/2=22.75 x 3 phases =68.25

The test book gives nearly three times this amount as an answer. Am I way off?

The 45.5 kw is the demand for the 40 dryers. No need to go any further.
45.5 kw/(0.208 x 1.73) = 126 amps.

 

[bob]

Correction to my post. Check this article

http://www.ecmag.com/?fa=article&articleID=8393

 

[George Stolz]

Seems like a good article, just skimmed it. I didn't see where he said how to cope with not knowing the number of dryers per phase, however.

 

[Dennis Alwon]

I agree with Dennis.

Going out on a limb, now, I'm going to say that if I were to do this problem I would do this:

40 dryers x 5kW = 200 kW
(now demand)
40 dryers = 23 + 17
17 x 0.5% = 8.5%
35% - 8.5% = 26.5%
200 kW x 26.5% = 53 kW

Note that I made no allowance for some dryers being on A*B, some B*C, and some C*A. My reason is because I do not see a panel schedule attached to the question, and I have no assurance that the load is evenly balanced. IMO, the "maximum load connected between any two phases" means exactly that - every dryer connected A*B, no balancing of dryers.

George here is the issue I have with your calculation as I did the same thing. .5% should be taken on the total dryers used in the demand not the actual number (40) . We have a demand of 28 Dryers so we should be taking .5% of (28-23).

.5% of 5 not 17

 

[Little Bill]

George here is the issue I have with your calculation as I did the same thing. .5% should be taken on the total dryers used in the demand not the actual number (40) . We have a demand of 28 Dryers so we should be taking .5% of (28-23).

.5% of 5 not 17

Dennis, are you (and George) taking into account the last sentence in 220.54?

 

[Dennis Alwon]

Dennis, are you (and George) taking into account the last sentence in 220.54?

Not sure that would affect anything.

 

[jumper]

Seems like a good article, just skimmed it. I didn't see where he said how to cope with not knowing the number of dryers per phase, however.

In general, wouldn't someone try and balance them according to the total amount to be connected?

 

[jumper]

George here is the issue I have with your calculation as I did the same thing. .5% should be taken on the total dryers used in the demand not the actual number (40) . We have a demand of 28 Dryers so we should be taking .5% of (28-23).

.5% of 5 not 17

I am not seeing as how you came to this conclusion. I may be wrong, but I do not understand.

 

[Dennis Alwon]

I am not seeing as how you came to this conclusion. I may be wrong, but I do not understand.

Well article 220.54 tells us how to get the demand. there are 14 connections to 2 phases so the total number of dryers is based on 2 times 14 or 28.

We look in T. 220.54 and we see 35% minus .5% for each dryer over 23. If the demand is 28 dryers then we have 28-23= 5. We need to subtract .5% * 5 or 2.5% from (35% of the 200kw)

It seems if we are using that table at 28 dryers then why we we use that table and subtract 23 from the 40 dryers.

 

[George Stolz]

Dennis, I'm on the phone so this is abrupt: you're wrong.

There are 40 dryers, I really can't see where you got 28...?

 

[Little Bill]

Well article 220.54 tells us how to get the demand. there are 14 connections to 2 phases so the total number of dryers is based on 2 times 14 or 28.

We look in T. 220.54 and we see 35% minus .5% for each dryer over 23. If the demand is 28 dryers then we have 28-23= 5. We need to subtract .5% * 5 or 2.5% from (35% of the 200kw)

It seems if we are using that table at 28 dryers then why we we use that table and subtract 23 from the 40 dryers.

George's math seems to be what table 220.54 says to do. 35%-0.5 x (40 dryers -23 = 17)=8.5. So 35-8.5=26.5%, then 26.5%x200 kva = 53 kva. You could also use each dryer at 5 kva. 5 kva x 26.5%=1.33 kva x 40 dryers = 53.20 kva. I'm not sure how the last sentence in 220.54 is used.

 

[kwired]

Another thing to consider is the 5000VA rating of the dryer. This is likely the rating for 240 volts. If so, same unit operating at 208 volts is only about 3760 VA.

 

[jumper]

Another thing to consider is the 5000VA rating of the dryer. This is likely the rating for 240 volts. If so, same unit operating at 208 volts is only about 3760 VA.

A dryer is not a purely resistive load, so are sure this is correct?

 

[jumper]

Well article 220.54 tells us how to get the demand. there are 14 connections to 2 phases so the total number of dryers is based on 2 times 14 or 28.

We look in T. 220.54 and we see 35% minus .5% for each dryer over 23. If the demand is 28 dryers then we have 28-23= 5. We need to subtract .5% * 5 or 2.5% from (35% of the 200kw)

It seems if we are using that table at 28 dryers then why we we use that table and subtract 23 from the 40 dryers.

I agree with George, NOPE.

 

[kwired]

A dryer is not a purely resistive load, so are sure this is correct?

Inductive load is likely 1/3 - 1/2 hp so we are looking at what 4-600 VA for a motor and the other 90% or so of the load is resistive. Plus I said 'about 3670' and not 'is 3670'.