Sizing 3 phase service conductors

[Jpflex]

I’m trying to understand what is wrong here

Assuming we did a load calculation for a commercial variety store and were trying to calculate service wire size for a 3 phase 4 wire (neutral) 208 volts system, ambient temperature standard 30 degrees Celsius we have:

48,120 VA noncontinuous loads

36,200 VA continuous load

Neutral load 47,602 VA


1. Calculate continuous load 36,200 x 125% = 45,200 VA

2. For total ungrounded conductor load, add continuous load plus non continuous load for ungrounded conductor total 45,250 VA + 48,120 VA = 93,370 VA

3. Neutral load remains the same 47,602 VA


We then must derate the ungrounded service wire ampacity to 80% because this 3 phase system has more than 3 current carrying conductors with a derived neutral

Therefore

Ungrounded conductor 93,370 VA / 208 x 1.73 = 259i amperes required

With wires at 80% ampacity we have 259i / 0.80 = 324i actual conductor ampacity needed

Table NEC 310.15 B 16
350 kcmil = 350i amperes at 90 degrees Celcius, 350i x 0.80 = 280i max
This is also good for 259i at 75 degrees Celcius

My answer 350 kcmil THHN ungrounded service conductor

However book says the service conductor should be a 400 kcmil THWN good for target 324i amperes

The book did not specify the type of wire to be used or type of insulation rating so based on THHN wire would my calculated size of 350 kcmil work?

 

[augie47]

My answer would have been 400AL or 300 Cu so I'm no help.

 

[david luchini]

I don't see why 300kcmil wouldn't do the job.

Edit: Gus types faster.

 

[infinity]

We then must derate the ungrounded service wire ampacity to 80% because this 3 phase system has more than 3 current carrying conductors with a derived neutral

Are you sure that more than 50% of the load is non-linear? I'm guessing that no derating is required.

 

[Jpflex]

My answer would have been 400AL or 300 Cu so I'm no help.

But 300 kcmil cu THHN is limited to 320i amperes at 90 degree Celcius insulation. 320 I x 0.80 = 256i which is only 3i amperes smaller than required which is probably why you mention using this. That extra 3 amperes insufficient is why I selected the next size up to handle at least 256i amperes

 

[Jpflex]

Are you sure that more than 50% of the load is non-linear? I'm guessing that no derating is required.

This is straight off the book

 

[Jpflex]

But 300 kcmil cu THHN is limited to 320i amperes at 90 degree Celcius insulation. 320 I x 0.80 = 256i which is only 3i amperes smaller than required which is probably why you mention using this. That extra 3 amperes insufficient is why I selected the next size up to handle at least 256i amperes

Correction

 

[david luchini]

But 300 kcmil cu THHN is limited to 320i amperes at 90 degree Celcius insulation. 320 I x 0.80 = 256i which is only 3i amperes smaller than required which is probably why you mention using this. That extra 3 amperes insufficient is why I selected the next size up to handle at least 256i amperes

234A is required for the load.

250.5A is required for the OCPD (assuming the service OCPD is 300A

 

[Jpflex]

Miss type I selected the next size up to handle at least 259 amperes not 256i

 

[david luchini]

This is straight off the book

You should probably toss that book

 

[Jpflex]

234A is required for the load.

250.5A is required for the OCPD (assuming the service OCPD is 300A

259 is required for the load after 80% was applied to the conductor original ampacity

So a conductor of 324i is needed but 300 kcmil only shows ampacity of 300i

 

[Jpflex]

You should probably toss that book

Really? Is it wrong?

 

[augie47]

Sine the book states a majority of the load is non-linear (fact committed from the OP) then the 80% would come into play and you would need a 400Cu to meet the 324amp load.

 

[Jpflex]

Sine the book states a majority of the load is non-linear (fact committed from the OP) then the 80% would come into play and you would need a 400Cu to meet the 324amp load.

Ok I get that because this will require the 80% conductor bundling ampacity adjustment for a three phase service with non linear loads.

However I still think the mistake in the book was to not specify what type of wire to use such as THWN or THHN

If I used THHN I can use a 350 kcmil at 350i x 0.80 = 280i amperes

Or

If I used THWN I get 400 kcmil 335i x 0.80 = 268i amperes

Either wire type works? Does this look correct?

 

[infinity]

Since the book states a majority of the load is non-linear (fact omitted from the OP) then the 80% would come into play and you would need a 400Cu to meet the 324amp load.

Yes minute details are important when trying to give the correct answer.

 

[augie47]

Ok I get that because this will require the 80% conductor bundling ampacity adjustment for a three phase service with non linear loads.

However I still think the mistake in the book was to not specify what type of wire to use such as THWN or THHN

If I used THHN I can use a 350 kcmil at 350i x 0.80 = 280i amperes

Or

If I used THWN I get 400 kcmil 335i x 0.80 = 268i amperes

Either wire type works? Does this look correct?

Your terminations are likely 75°C so the you would need a conductor that has an ampacity of 324 in the 75° column. ie: 400

 

[Tulsa Electrician]

Copy from the text of the question. On the four conductors

Ok I get that because this will require the 80% conductor bundling ampacity adjustment for a three phase service with non linear loads.

However I still think the mistake in the book was to not specify what type of wire to use such as THWN or THHN

If I used THHN I can use a 350 kcmil at 350i x 0.80 = 280i amperes

Or

If I used THWN I get 400 kcmil 335i x 0.80 = 268i amperes

Either wire type works? Does this look correct?

Keep up n mind it's over 100 amp so use 75 c as default for conductor termination.
So the choice is yours Al or CU. Since you are derating use a 90 c rated conductor with final termination of @75 c for the calculated load. You have to meet both ( derated ampacity and calculated load) with your wire choice

Book showed 259 amps.
Since your derating a THHN/THWN-2 conductor would be a good choice Since these are service conductors.

Take the 259*1.25 and choose.
259*1.25= 323.75 or 259/.80
Now look at the 90 c column
350mcm CU is 350 and at the 75 c 310 amps.
300 mcm Cu is to small.
320*.80= 256, must meet 259 which is the minimum for derate.
500 mcm AL is 350 amps and at 75c 310 amps.

This the difference between reality and testing.

 

[wwhitney]

Your terminations are likely 75°C so the you would need a conductor that has an ampacity of 324 in the 75° column. ie: 400

I don't believe that's correct. When checking the terminations at the termination temperature of 75C, we do not need to simultaneously apply correction or adjustment factors. So with 75C terminations the wire selected needs to have a 75C unadjusted ampacity (from the table) of at least 259A, and our OCPD also needs to be at least 259A (which means 300A).

Then for the run of the conductor, we do need to apply the correction and adjustment factors, but don't need to apply the 125% continuous use factor. So for that check we need a wire with an ampacity of at least (48,120+36,000)/3/120 = 233A. Except the OCPD needs to be at least 300A to exceed 259A, so we also need a wire with an ampacity of at least 251A per 240.4(B). So for 90C insulation and a 0.8 factor for 4 CCCs, we need a wire with a 90C unadjusted ampacity (from the table) of at least 251/0.8 = 314A.

Note I didn't click through to look at the screenshots, so hopefully I've extracted the correct problem values just from the text in this thread.

Cheers, Wayne

 

[Jpflex]

Copy from the text of the question. On the four conductors

Keep up n mind it's over 100 amp so use 75 c as default for conductor termination.
So the choice is yours Al or CU. Since you are derating use a 90 c rated conductor with final termination of @75 c for the calculated load. You have to meet both ( derated ampacity and calculated load) with your wire choice

Book showed 259 amps.
Since your derating a THHN/THWN-2 conductor would be a good choice Since these are service conductors.

Take the 259*1.25 and choose.
259*1.25= 323.75 or 259/.80
Now look at the 90 c column
350mcm CU is 350 and at the 75 c 310 amps.
300 mcm Cu is to small.
320*.80= 256, must meet 259 which is the minimum for derate.
500 mcm AL is 350 amps and at 75c 310 amps.

This the difference between reality and testing.

When the loads were calculated 125% was already multiplied by continuous loads total and added to non continuous loads to obtain 93,370 KVA total service demand.

93,370 / 208 x 1.732 = 259 i amperes ungrounded conductor ampacity after adjustment

Should we really be applying an additional 125% to this?

 

[Jpflex]

Copy from the text of the question. On the four conductors

Keep up n mind it's over 100 amp so use 75 c as default for conductor termination.
So the choice is yours Al or CU. Since you are derating use a 90 c rated conductor with final termination of @75 c for the calculated load. You have to meet both ( derated ampacity and calculated load) with your wire choice

Book showed 259 amps.
Since your derating a THHN/THWN-2 conductor would be a good choice Since these are service conductors.

Take the 259*1.25 and choose.
259*1.25= 323.75 or 259/.80
Now look at the 90 c column
350mcm CU is 350 and at the 75 c 310 amps.
300 mcm Cu is to small.
320*.80= 256, must meet 259 which is the minimum for derate.
500 mcm AL is 350 amps and at 75c 310 amps.

This the difference between reality and testing.

You took 259 amperes from book and multiplied by 125% but Book separated loads from continuous and non continuous loads and then multiplied the continuous loads by 125% and added this to the non continuous loads at 100% to obtain total KVA

But on your example you took 259 I ampere and multiplied by 125% is this what you intended?