Feeder sizing

Status
Not open for further replies.
B

bigpapa400lb

Member
A 120/208 3-phase 4-wire feeder supplies a distribution panelboard. The 32,000 VA load supplied by the panelboard is 80% nonlinear, and 100% continuous. What is the minimum required amperage rating, and THW copper conductior size required for this feeder if it passes through a 45 degree C ambient temperature?

My Calculations:

32000/208 x 1.732= 266A x .82 x 80%= 174A 2/0 CU
Answer key says:169A 2/0 CU
My Conductor size is right but the calculation is wrong?
 
The way I did the calculation was as follows:
32000/208x1.732=88.88 amps
Nonlinear brings it to 111.11 (88.88 x 1.25)
Conrinuous brings it to 138.88 (111.11 x 1.25)
45? C for THW provides for a .82 multiplier 111.11/.82=169.37
 
augie47 said:
The way I did the calculation was as follows:
32000/208x1.732=88.88 amps
Nonlinear brings it to 111.11 (88.88 x 1.25)
Conrinuous brings it to 138.88 (111.11 x 1.25)
45? C for THW provides for a .82 multiplier 111.11/.82=169.37
Click to expand...

why did your .82 provided for ambient apply to 111.11and not 138.88?

1.11/ .82 does not equal 169.37
I cannot follow what you did
 
typo on my part..the .82 did apply to the 138...
138/,82 = 169

thanks
 
augie47 said:
The way I did the calculation was as follows:
32000/208x1.732=88.88 amps
Nonlinear brings it to 111.11 (88.88 x 1.25)
Conrinuous brings it to 138.88 (111.11 x 1.25)
45? C for THW provides for a .82 multiplier 111.11/.82=169.37
Click to expand...

Where are you getting the 1.25 factor (125%) for non-linear loads?
 
bigpapa400lb said:
A 120/208 3-phase 4-wire feeder supplies a distribution panelboard. The 32,000 VA load supplied by the panelboard is 80% nonlinear, and 100% continuous. What is the minimum required amperage rating, and THW copper conductior size required for this feeder if it passes through a 45 degree C ambient temperature?

Answer key says:169A 2/0 CU
Click to expand...

Is this a test or test-prep question?

It appears the answer key is wrong.
 
Smart $ said:
Where are you getting the 1.25 factor (125%) for non-linear loads?
Click to expand...

He stated a feeder using THW so it had to be a raceway system. 3 phase, 4 wire would be 4 current carrying conductors (neutral counts on major non-linear load per 310.15(A)(4).
310.15(B)(2) has 80% adjustment for 4-6 current carrying (or 1.25 multiplier)
 
augie47 said:
He stated a feeder using THW so it had to be a raceway system. 3 phase, 4 wire would be 4 current carrying conductors (neutral counts on major non-linear load per 310.15(A)(4).
310.15(B)(2) has 80% adjustment for 4-6 current carrying (or 1.25 multiplier)
Click to expand...
Ahh!! Would have made more sense if you had labeled it 4 CCC derating. It's not that you labeled it wrong... just my mind didn't make the implied connection.

Smart $ said:
It appears the answer key is wrong.
Click to expand...
Ignore the quoted remark.
 
augie47 said:
The way I did the calculation was as follows:
32000/208x1.732=88.88 amps
Nonlinear brings it to 111.11 (88.88 x 1.25)
Conrinuous brings it to 138.88 (111.11 x 1.25)
45? C for THW provides for a .82 multiplier 111.11/.82=169.37
Click to expand...

So you have to multiply the nonlinear and continuous loads by 1.25
 
The original post stated the load was 100% continuous so that would require a 1.25 multiplier.
The original post says the load is 80% non-linear which requires that the neutral be counted as a current carrying conductor. Since the neutral is a CCC, the feeder has 4 current carrying conductors which requires a 80% derating (or 1.25 multiplier)_
 
Different question same type:

A 120/208v, four wire THW feeder is installed in a raceway and supplies a 10,000VA nonlinear load which will operate for three or more hours of continuous operation. The raceway containing the feeder will pass through an ambient temperature of 25 degrees C. What is the allowable ampacity of this feeder under this condition of use?

My calculation:
10000/360= 28A x 1.25(continuous duty) = 35A x 1.05(ambient temp)= 37A
37A x .80 = 30A

But the answer is different in the answer key. What am I doing wrong?
 
bigpapa400lb said:
My calculation:
10000/360= 28A x 1.25(continuous duty) = 35A x 1.05(ambient temp)= 37A
37A x .80 = 30A

But the answer is different in the answer key. What am I doing wrong?
Click to expand...
When you use wire ampacity adjustments to adjust loads or their minimum circuit amapacity rating to determine wire size and type, you use the inverse operation or value...
10000(load)/360≈28A(load/line)
...?125%(adjust load/line to minimum circuit ampacity)≈35A
...?1.05(adjust minimum circuit ampacity to minimum wire ampacity)≈33A
...?80%(adjust minimum circuit ampacity to minimum wire ampacity)=41.3A
 
Last edited:
Samrt$, We are probbaly working on homework :), but I am concerned about the 1.05 multiplier. The OP states the feeder "passes thru" the 25? C area and, I agree, the 41.3 would be applicable while passing thru and since we don't have data on the other area we can only summise the ambient.
 
augie47 said:
Samrt$, We are probbaly working on homework :), but I am concerned about the 1.05 multiplier. The OP states the feeder "passes thru" the 25? C area and, I agree, the 41.3 would be applicable while passing thru and since we don't have data on the other area we can only summise the ambient.
Click to expand...
You didn't seem to have this area doubt for the first question ;)

...but you are quite right in that the pass-through area (in both cases) could be on either side of the 10% or 10' rule.

Should we assume one, or both, is a trick question?
 
Last edited:
In the first question, it was a "limiting" factor and again I was assuming that was the highest (plus you gave us the answer :) )
In the second the ambient is an "enhancing" value. The "inspector" in me looks for the more restrictive number :)
Since they gave you that number I would ass-u-me that they want it figured into the answer, so I would side with Smart$.
Do you know the desired answer ?
 
Status
Not open for further replies.
Top