Example 54.3 from Power Reference Manual

[Dodgeviper1017]

A 480Y/277, 80 kW noncontinuous load with a 0.8 pf consists of less than 50% ballast-type lighting. The feeder length is 110m. Applying the 3% fine print note No. 2 recommendation from Sec. 215.2(A)(3), what is the maximum recommended voltage drop for every 1000 Ampere-Foot? 

Here is what they did:

I = P/(sqrt(3)*V*pf)

Ampere-Feet = I*length*3.28 

Vdrop = .03*277

then Vdrop/Ampere-Feet

Which gives 0.192/1000 Ampere-Feet

What I don't understand is why they didn't multiply the length by 2? I also don't understand a little extra information they provided in solution it reads: "The reason for specifying noncontinuous is to emphasize that the ampacity is to be calculated at 100% not 125%. Regardless of the load, continuous or noncontinuous, the voltage drop is calculated using the 100% ampacity. Nevertheless, the conductor must be sized to handle 125% ampacity according to the Code standards. The OCPD must be set at 125% ampacity as well." I don't understand this part why would I size a noncontinuous load at 125% if it is noncontinuous would sizing it in this manner not convert it to continuous in a sense?

 

[Phatso86]

I don't remember coming across this problem, but what I do know is that the PPI books are ridiculous with their errors.

I've seen two identical problems. Literally clones with different values. One problem in the text, on in the workbook and the answers were vastly different.

A lot of the errata for volume two were submitted by me

 

[Dodgeviper1017]

I know man this is what scares me about this test though. How do i know what is right? What would be your opinion on how to solve it even though you haven't seen based on what I have posted?

 

[Ken PE 3.1]

A 480Y/277, 80 kW noncontinuous load with a 0.8 pf consists of less than 50% ballast-type lighting. The feeder length is 110m. Applying the 3% fine print note No. 2 recommendation from Sec. 215.2(A)(3), what is the maximum recommended voltage drop for every 1000 Ampere-Foot? Here is what they did:

I = P/(sqrt(3)*V*pf)

Ampere-Feet = I*length*3.28 

Vdrop = .03*277

then Vdrop/Ampere-Feet

Which gives 0.192/1000 Ampere-Feet

What I don't understand is why they didn't multiply the length by 2? I also don't understand a little extra information they provided in solution it reads: "The reason for specifying noncontinuous is to emphasize that the ampacity is to be calculated at 100% not 125%. Regardless of the load, continuous or noncontinuous, the voltage drop is calculated using the 100% ampacity. Nevertheless, the conductor must be sized to handle 125% ampacity according to the Code standards. The OCPD must be set at 125% ampacity as well." I don't understand this part why would I size a noncontinuous load at 125% if it is noncontinuous would sizing it in this manner not convert it to continuous in a sense?

3 phase (480Y/277) is calculated as the length. Single phase runs are doubled. Be mindful of a 240/120 run to a house, still single phase.

 

[Ultrafault]

You are correct neither the conductor or the OCPD must be sized 125%. I also agree that you would not in this case multipy by 2 as it is 3 phase. There is such a beast as a high leg delta that is 240\120 that is 3 phase. Another interesting fact is that the propper way to designate this system as a Y is to list the voltage a 277\480.