NCEES POWER #129 NEW

Forgive me if this is answered somewhere else, I wish that I could search by "129" in this site....

BUT

for this problem, the NCEES answer uses the effective Z from Table 9...which is for a pf of .85...but the pf for this problem is at .8...so shouldn't Z be calculated from the values given in the table and Z=Rpf+sin[arccos(pf)]?

When you do so, I get Z=.0058+j.007, which creates a voltage drop of 3.64V (this also takes into account the 250ft of length based off of the Table values R=.029 and X=.048)

SO my answer was A since (277-3.64)*root(3)=473.47V....

But the correct answers by the answers is B, using the Z effective from Table 9, which equates to a higher impedance and a higher voltage drop.

Help me feel less insane, or if I missed something, whelp, time to learn...

snerts50 said:

Forgive me if this is answered somewhere else, I wish that I could search by "129" in this site....
BUT

for this problem, the NCEES answer uses the effective Z from Table 9...which is for a pf of .85...but the pf for this problem is at .8...so shouldn't Z be calculated from the values given in the table and Z=Rpf+sin[arccos(pf)]?

When you do so, I get Z=.0058+j.007, which creates a voltage drop of 3.64V (this also takes into account the 250ft of length based off of the Table values R=.029 and X=.048)

SO my answer was A since (277-3.64)*root(3)=473.47V....

But the correct answers by the answers is B, using the Z effective from Table 9, which equates to a higher impedance and a higher voltage drop.

Help me feel less insane, or if I missed something, whelp, time to learn...

Click to expand...

Try Z=(.029/4)cos(36.9)+(.048/4)sin(36.9), this will get you the right answer with a voltage drop of 5.198, which is subtracted from 277, then multiplied by sqrt 3 for a final answer of 470.8V. This comes from Table 9 Note 2.

This is a side question from the above and not really the intent of question #129. I am just looking at each question from many angles. Question #129 mentions that there are 3 copper conductors and a neutral. Does the neutral count as a current-carrying conductor and makes the total conductors in the conduit 4? Therefore subject to an adjustment factor according to Table 310.15B(2)(a)?

EEVA said:

Try Z=(.029/4)cos(36.9)+(.048/4)sin(36.9), this will get you the right answer with a voltage drop of 5.198, which is subtracted from 277, then multiplied by sqrt 3 for a final answer of 470.8V. This comes from Table 9 Note 2.

Click to expand...

This still gives Z = 0.0058 + j0.0072 and a voltage drop of 3.7. If you don't use the effective Z option and apply the power factor to the current, i.e. ([email protected]º)*(0.00725 +j0.012), then you get a voltage drop of 5.6.

I still can't figure out why myself why it doesn't work the other way, unless my calculator is lying to me...

ktown said:

EEVA said:

Try Z=(.029/4)cos(36.9)+(.048/4)sin(36.9), this will get you the right answer with a voltage drop of 5.198, which is subtracted from 277, then multiplied by sqrt 3 for a final answer of 470.8V. This comes from Table 9 Note 2.

Click to expand...

This still gives Z = 0.0058 + j0.0072 and a voltage drop of 3.7. If you don't use the effective Z option and apply the power factor to the current, i.e. ([email protected]º)*(0.00725 +j0.012), then you get a voltage drop of 5.6.

I still can't figure out why myself why it doesn't work the other way, unless my calculator is lying to me...

Click to expand...

Hmmm.....(Maybe I am missing something, but) Effective Z is .01299 which has the .8 pf included. Then 400A X .01299 = 5.196. Then 277 - 5.196 = 271.8, then times by sqrt 3 = 470.8V.

EEVA said:

This is a side question from the above and not really the intent of question #129. I am just looking at each question from many angles. Question #129 mentions that there are 3 copper conductors and a neutral. Does the neutral count as a current-carrying conductor and makes the total conductors in the conduit 4? Therefore subject to an adjustment factor according to Table 310.15B(2)(a)?

Click to expand...

Under normal conditions (balanced 3-phase load) the neutral doesn't carry any current and therefore doesn't contribute any heat. In reality many times a load isn't perfectly balanced across the 3 phases so there is some current on the neutral. It typically is not treated as a current-carrying conductor by the NEC, unless there are significant harmonics. Hope this helps.

ElecPwrPEOct11 said:

EEVA said:

This is a side question from the above and not really the intent of question #129. I am just looking at each question from many angles. Question #129 mentions that there are 3 copper conductors and a neutral. Does the neutral count as a current-carrying conductor and makes the total conductors in the conduit 4? Therefore subject to an adjustment factor according to Table 310.15B(2)(a)?

Click to expand...

Under normal conditions (balanced 3-phase load) the neutral doesn't carry any current and therefore doesn't contribute any heat. In reality many times a load isn't perfectly balanced across the 3 phases so there is some current on the neutral. It typically is not treated as a current-carrying conductor by the NEC, unless there are significant harmonics. Hope this helps.

Click to expand...

Thanks, I see it mentioned in NEC 310.15B 4C

EEVA said:

ktown said:

EEVA said:

Try Z=(.029/4)cos(36.9)+(.048/4)sin(36.9), this will get you the right answer with a voltage drop of 5.198, which is subtracted from 277, then multiplied by sqrt 3 for a final answer of 470.8V. This comes from Table 9 Note 2.

Click to expand...

This still gives Z = 0.0058 + j0.0072 and a voltage drop of 3.7. If you don't use the effective Z option and apply the power factor to the current, i.e. ([email protected]º)*(0.00725 +j0.012), then you get a voltage drop of 5.6.

I still can't figure out why myself why it doesn't work the other way, unless my calculator is lying to me...

Click to expand...

Hmmm.....(Maybe I am missing something, but) Effective Z is .01299 which has the .8 pf included. Then 400A X .01299 = 5.196. Then 277 - 5.196 = 271.8, then times by sqrt 3 = 470.8V.

Click to expand...

Looks like I got it wrong with effective Z, it does not give you the right answer. I got 472.8, which leads one to pick (A)475V not 470V. I never included the pf with the current which lead to the answer in the book.

Maybe this is a last minute panic attack, since I already walked away from this one before satisfied that had it down, but...

Why doesn't the abbreviated version as provided in Ugly's or Ferm's work for this problem?

Their shortcut version states that

Vd=(Sq rt 3*K (for copper)*L*I)/cm

Vd=(1.732*12.9*250'*400A)/500,000=4.468v???

Even if, as described above, you say 277-4.468=272.53*1.732=472.02v.

While yes, it does technically push you to the correct answer, it's very close to putting you on the fence to decide between 470 and 475.

I'm fine with the Chapter 9 Table 9 method-I just don't like the six minutes that would it would have to fit neatly into. Words of wisdom?

Audienceof1 said:

Maybe this is a last minute panic attack, since I already walked away from this one before satisfied that had it down, but...

Why doesn't the abbreviated version as provided in Ugly's or Ferm's work for this problem?

Their shortcut version states that

Vd=(Sq rt 3*K (for copper)*L*I)/cm

I'm fine with the Chapter 9 Table 9 method-I just don't like the six minutes that would it would have to fit neatly into. Words of wisdom?

Click to expand...

just glancing quickly but it looks like this equation is ignoring the inductive reactance factor of the cable which is probably one of the points of the problem. It also appears it is giving you a line-line drop so no further multiplication by √3 is needed and you are subtracting it from a line-neutral voltage. If the conductor size is large, the reactance term is not negligible.