Voltage drop, power factor question

[Sparky Bill PE]

when working this problem, I took .85 as face value for power factor and used Z chart, obviously I'm wrong but I don't know why. 

 

[BebeshKing PE]

when working this problem, I took .85 as face value for power factor and used Z chart, obviously I'm wrong but I don't know why. 

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I got it right,though. I used Zeff=0.08ohm/1000ft for p.f=0.85 column on Table 9 chapter 9.

Vdrop-3phase = (sqrt of 3)(Iline)(Zeff)(length)=(Sqrt of 3)(566)(0.08/1000)(100)=7.84 Volts (answer B)

 

[Sparky Bill PE]

Do you transform one of these equations or do you memorize that for the test? I'm doing the best I can to stick to the reference book I don't use any of my books or Zachs printed notes. I have a damn 12 month catalog of references that are useless.

 

[BebeshKing PE]

Do you transform one of these equations or do you memorize that for the test? I'm doing the best I can to stick to the reference book I don't use any of my books or Zachs printed notes. I have a damn 12 month catalog of references that are useless.

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I just used the reference handbook as my guide but I don't used most of their formulas exactly the way they show it.

Just for example that Voltage drop formula. It say that if it is 3 phase you will need to use K=2/sqrt of 3. But in my mind, the 2 will just cancel to the formula and it will end up with(sqrt of 3)(L)(R)(I)/1000 . and we know that the R/1000 is the effective impedance. 

That 1000 in the denominator should not be there since you will already divide the effective impedance by 1000.

 

[Sparky Bill PE]

I just used the reference handbook as my guide but I don't used most of their formulas exactly the way they show it.

Just for example that Voltage drop formula. It say that if it is 3 phase you will need to use K=2/sqrt of 3. But in my mind, the 2 will just cancel to the formula and it will end up with(sqrt of 3)(L)(R)(I)/1000 . and we know that the R/1000 is the effective impedance. 

That 1000 in the denominator should not be there since you will already divide the effective impedance by 1000.

Thanks for explaining the 2/root 3

 

[Zach Stone P.E.]

Do you transform one of these equations or do you memorize that for the test? I'm doing the best I can to stick to the reference book I don't use any of my books or Zachs printed notes. I have a damn 12 month catalog of references that are useless.

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Hi @SparkyBill

Take a look at the live class recording for voltage drop from last semester and it will clear this up for you and demystify the handbook formula. It's really just using the nec effective impedance approach for approximating voltage drop using magnitudes only. If you have any issues, be sure to post in the live class discussion board so we can clear it up