# NCEES exam problem 129. I am now freaking out about voltage drop



## Sparky Bill PE (Dec 1, 2020)

It says "You can use the other formula and get same answer, and I'm not getting same answer. I'm getting like 473.8 (of course just enough to make you chose wrong answer). What am I missing here? 

I don't see anywhere in reference handbook, or NEC telling me to use this method of keeping the resistance/reactance or converting to impedance. If I can use this formula, then why would we ever convert to effective Z?

Really gonna suck when this problem is on exam and its a coin flip on which formula to use to get the problem correct.


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## Ampera18 PE (Dec 1, 2020)

The one in the Reference manual just looks like an approximation. I'm getting the same as you, and i don't know what's wrong, except that the reference manual says it's an approximation, and this solution is basically V=IR, but the R is the cable. Seems more intuitive.


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## Sparky Bill PE (Dec 1, 2020)

Ampera18 said:


> The one in the Reference manual just looks like an approximation. I'm getting the same as you, and i don't know what's wrong, except that the reference manual says it's an approximation, and this solution is basically V=IR, but the R is the cable. Seems more intuitive.


I have a lot of experience with the FE, and a lot of it was more of "trying to figure out the cute ncees way" of solving things instead of actually solving it like an engineer. I'm hoping I'm wrong on this. Otherwise, I know I'm going to be super pissed as soon as I see a voltage drop question and try to mind read how they want it.


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## Cuseman17 (Dec 1, 2020)

From the alternate method, I am getting Vd of 5.2V. Now given Vline is 480 and the Vd is per phase, so converting it to Vphase, we get V=277.128.
Vphase - Vd= 277.128 - 5.2 = 271.92. Converting it back to line value, Vline=sq.rt 3 * 271.92 = 470.99

Let me know if there is something wrong with this method.


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## Sparky Bill PE (Dec 1, 2020)

Cuseman17 said:


> From the alternate method, I am getting Vd of 5.2V. Now given Vline is 480 and the Vd is per phase, so converting it to Vphase, we get V=277.128.
> Vphase - Vd= 277.128 - 5.2 = 271.92. Converting it back to line value, Vline=sq.rt 3 * 271.92 = 470.99
> 
> Let me know if there is something wrong with this method.


What is "alternate method"? Is that you calculating Z effective, then multiply that by current? Or is that you NOT finding Z effective and just using resistance/reactance?


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## Cuseman17 (Dec 1, 2020)

Alternate method being the alternate solution they mentioned above. I calculated the Zeff based on the formula on NEC chapter 9 Table 9. Then multiplied it by the current. The same formula is also given in the handbook.

NEC - 


 

Handbook -


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## Sparky Bill PE (Dec 1, 2020)

Cuseman17 said:


> Alternate method being the alternate solution they mentioned above. I calculated the Zeff based on the formula on NEC chapter 9 Table 9. Then multiplied it by the current. The same formula is also given in the handbook.
> 
> NEC -
> View attachment 19686
> ...


I didn't get 471, I'll post my work tomorrow and we can compare to what I'm doing wrong getting a different answer.


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## Cuseman17 (Dec 2, 2020)

This is from my notes when I solved it in the exam. Its a bit messy so let me know if you cant read it and I will try to write it up on a clean sheet.


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## Sparky Bill PE (Dec 2, 2020)

Cuseman17 said:


> This is from my notes when I solved it in the exam. Its a bit messy so let me know if you cant read it and I will try to write it up on a clean sheet.
> 
> View attachment 19695


Your VD is I * Zeff, why didn't you use the angel with I instead of just its magnitude?


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## Cuseman17 (Dec 2, 2020)

In my opinion, the angle is already taken care of when using the Zeff formula. We can either use the pf in Zeff calculation and mulitply with I mag *OR* use (R+jX) and then use I at an angle.


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## Sparky Bill PE (Dec 2, 2020)

Cuseman17 said:


> In my opinion, the angle is already taken care of when using the Zeff formula. We can either use the pf in Zeff calculation and mulitply with I mag *OR* use (R+jX) and then use I at an angle.


I agree angle in Z eff is taken care of, in your math there you have I * Z = 5.2. You're not using the angle of your current. Am I not supposed to be using the angle of my current provided by the power factor? My calculation is 400 at the angle of -36.9.


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## Byk (Dec 3, 2020)

SparkyBill said:


> I agree angle in Z eff is taken care of, in your math there you have I * Z = 5.2. You're not using the angle of your current. Am I not supposed to be using the angle of my current provided by the power factor? My calculation is 400 at the angle of -36.9.


 You don't need to use the angle for your calcs. If you have NEC Handbook I highly recommend looking at the examples in chapter 9, right after table 9.


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## Cuseman17 (Dec 3, 2020)

Yes that is correct. As far as I understand, when we use the Zeff formula, Zeff is already at the pf angle, -36.86. If we again multiply it by [email protected] degrees, the Vd will be at an angle of - 72.73 which will be incorrect. 

May be my understanding is incorrect, but that's how I thought it works.


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## Sparky Bill PE (Dec 3, 2020)

Byk said:


> You don't need to use the angle for your calcs. If you have NEC Handbook I highly recommend looking at the examples in chapter 9, right after table 9.


Maybe that's what I'm doing wrong, so when doing voltage drop and using Z eff  you always take the current magnitude, not the complex current?


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## Byk (Dec 3, 2020)

SparkyBill said:


> Maybe that's what I'm doing wrong, so when doing voltage drop and using Z eff  you always take the current magnitude, not the complex current?


That is correct, magnitude only because voltage drop of a circuit is in direct proportion to the resistance of the conductor and the magnitude of the current


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## DJMB (Dec 3, 2020)

NEC - Table 9 - Note 2

Ze (for other than .85 PF) = R*PF+Xl*sin(cos^-1(PF))

R(1000ft)=.029

R(250ft)=.029*(250/1000) = .00725

X(1000ft)=.048

X(250ft)=.048*(250/1000)=.012

PF=.8

Ze = .00725*.8+.012*sin(cos^-1(.8))

Ze = .013

Vd = I*Ze = 400*.013 = 5.2V

V = Vs(L-N)-Vd

V=277-5.2 = 271.8

V(LL) = V(LN)*1.732

V(LL) = 271.8*1.732 = 470.77 ≈ 471V


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## Art (Dec 13, 2020)

The pf drop from 0.85 to 0.8 has little effect. Basically sqrt(ratio): if pf goes down, Z goes up (Vdrop goes up), and the inverse. From table Z = 0.057

Vdrop = 400 x Sqrt3 x 250/1000 x 0.057 = 9.87, net 470.13

adjust Z for pf ~ 0.057 x sqrt(0.85/0.8)

= 0.0588

Vdrop 10.18, net 469.82

470 either way


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## Sparky Bill PE (Dec 13, 2020)

Art said:


> The pf drop from 0.85 to 0.8 has little effect. Basically sqrt(ratio): if pf goes down, Z goes up (Vdrop goes up), and the inverse. From table Z = 0.057
> 
> Vdrop = 400 x Sqrt3 x 250/1000 x 0.057 = 9.87, net 470.13
> 
> ...


Yeah i found out my problem is I was using the complex current instead of just using the magnitude of current.


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## Ampera18 PE (Dec 14, 2020)

Sparky Bill said:


> Yeah i found out my problem is I was using the complex current instead of just using the magnitude of current.


i think it should use complex current. this problem is bad.


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## Sparky Bill PE (Dec 14, 2020)

Ampera18 said:


> i think it should use complex current. this problem is bad.


Exactly. I quit trying to explain myself its kind of a joke lol


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## DarkLegion PE (Dec 16, 2020)

DJMB said:


> NEC - Table 9 - Note 2
> 
> Ze (for other than .85 PF) = R*PF+Xl*sin(cos^-1(PF))
> 
> ...


For the bolded part why did you drop to i imaginary term for .012*sin(cos^-1(.8))? with it I get a completely different answer


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## rburns18 PE (Dec 16, 2020)

DarkLegion said:


> For the bolded part why did you drop to i imaginary term for .012*sin(cos^-1(.8))? with it I get a completely different answer


The formula at the bottom of the table in the NEC and in the NCEES Handbook don't have the j term. The X is just the value of the reactance. You don't include jX, just X. In the end, you just get a number.


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## DarkLegion PE (Dec 16, 2020)

Rburns18 said:


> The formula at the bottom of the table in the NEC and in the NCEES Handbook don't have the j term. The X is just the value of the reactance. You don't include jX, just X. In the end, you just get a number.


Thanks, all this time the numbers have been working out for me except for this question. I feel lucky I caught this mistake before it was too late lol


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## DarkLegion PE (Jan 1, 2021)

Sorry to revive an older topic but looking at the NEC handbook example we also do not use the complex values  if the PF=.85, is this correct?

So just use magnitudes when doing voltage drops for either .85 PF or =/= .85 PF?


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## Ampera18 PE (Jan 1, 2021)

yes i believe one should just use magnitude in the impedance and angle in the current


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