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

[DarkLegion PE]

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

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

 

[rburns18 PE]

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.

 

[DarkLegion PE]

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

 

[DarkLegion PE]

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?

 

[Ampera18 PE]

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