NCEES #129

[cruzy]

How can you use the effective Z to find the voltage drop and still get the same answer? it specifically says that it is an approximation of the answer, but when I do it (400 amps with angle corresponding to 0.8, times the effective Z), I get 473.xxx, which is closer to the 475 answer than the 471.

 

[cableguy]

Did you account for going from phase voltage to line voltage? The voltage drop formula gives you a Vphase drop, you have to multiply by sqrt(3) to get the line drop.

 

[cruzy]

Ya I did...I still come up with 472.7V, which is still closer to 475 than to 470.

Anyone done it using the effective Z method and have step by step they want to share?

 

[knight1fox3]

Ya I did...I still come up with 472.7V, which is still closer to 475 than to 470.
Anyone done it using the effective Z method and have step by step they want to share?

Here is the way the GA Tech course instructor solved the problem:

Using Ze=R*pf + X*[sin(arccos(0.8))] where R=0.029 and X=0.048

Working the math Ze=0.0232+0.0288=0.052 Ohm-N

VdropL-N=(0.052)*(250/1000)*400=5.2V

VdropL-L=5.2*sqrt(3)=9V

Therefore VpanelA=480-9=471V --> answer B

 

[cruzy]

Ya I did...I still come up with 472.7V, which is still closer to 475 than to 470.
Anyone done it using the effective Z method and have step by step they want to share?

Here is the way the GA Tech course instructor solved the problem:

Using Ze=R*pf + X*[sin(arccos(0.8))] where R=0.029 and X=0.048

Working the math Ze=0.0232+0.0288=0.052 Ohm-N

VdropL-N=(0.052)*(250/1000)*400=5.2V

VdropL-L=5.2*sqrt(3)=9V

Therefore VpanelA=480-9=471V --> answer B

I guess you have to leave out the angle difference when multiplying by 400? When I include the 0.8 pf lag in the 400 amps and carry out the calcs, I get the 9V angle (-36.8), but then when I subtract from 480 I get 472.8. So do you leave the angle out of the VD calc method or is my calculator just wrong (using a TI-83 and it has given me misleading answer sometimes because it doesn't tell the difference between complex quadrants when doing angle calcs)??

 

[GabeM]

knight1fox3's solution is correct.

You mentioned you have the IEEE books in another post. Check the IEEE red book page 97. That phasor diagram was the only way I could understand that equation, even after reading many posts on this forum trying to explain it.

 

[skwo]

If the conduit runs 250 ft, doesn't that imply that the total circuit distance is 500 ft? Obviously the solution uses 250 ft, but how can I be sure not to make that mistake on the exam? Does anyone have any pointers?

 

[Dolphin P.E.]

If the conduit runs 250 ft, doesn't that imply that the total circuit distance is 500 ft? Obviously the solution uses 250 ft, but how can I be sure not to make that mistake on the exam? Does anyone have any pointers?

The problem states 3 Phase System, Then you use Square root of 3.

 

[skwo]

Thanks. Brain fart!

 

[ros]

Ya I did...I still come up with 472.7V, which is still closer to 475 than to 470.
Anyone done it using the effective Z method and have step by step they want to share?

Here is the way the GA Tech course instructor solved the problem:

Using Ze=R*pf + X*[sin(arccos(0.8))] where R=0.029 and X=0.048

Working the math Ze=0.0232+0.0288=0.052 Ohm-N

VdropL-N=(0.052)*(250/1000)*400=5.2V

VdropL-L=5.2*sqrt(3)=9V

Therefore VpanelA=480-9=471V --> answer B

I am confuse now....

what i did is .... (480/sqt3)- (400<-36.86)*0.013 which is phase voltage ... so i multiply answer by sqt 3 and get 472.7 V ... no what did i do wrong here....

i got 0.013 = 0.052)*(250/1000)

 

[Dolphin P.E.]

Ya I did...I still come up with 472.7V, which is still closer to 475 than to 470.
Anyone done it using the effective Z method and have step by step they want to share?

Here is the way the GA Tech course instructor solved the problem:

Using Ze=R*pf + X*[sin(arccos(0.8))] where R=0.029 and X=0.048

Working the math Ze=0.0232+0.0288=0.052 Ohm-N

VdropL-N=(0.052)*(250/1000)*400=5.2V

VdropL-L=5.2*sqrt(3)=9V

Therefore VpanelA=480-9=471V --> answer B

I am confuse now....

what i did is .... (480/sqt3)- (400<-36.86)*0.013 which is phase voltage ... so i multiply answer by sqt 3 and get 472.7 V ... no what did i do wrong here....

i got 0.013 = 0.052)*(250/1000)

Do this: (480/sqt3)-(400*.013)=271.9V

271.9*sqt3=470.99V

you already accounted for the power factor when you got the equivalent impedance. Don't account for it again in the form of using the current in Phasor form. Just use the magnitude of the current.

 

[ros]

Thanks.... Another question similar to this is prob #512

Here in this problem why they divide by 2 instead of multiplying by 2?

They way i understood this prob is that, there are two set of cable so multiply with 2.... Correct?

 

[ros]

Thanks.... Another question similar to this is prob #512
Here in this problem why they divide by 2 instead of multiplying by 2?

They way i understood this prob is that, there are two set of cable so multiply with 2.... Correct?

Ooooooo i get it... Its bcoz they r parallel...

 

[Dolphin P.E.]

Thanks.... Another question similar to this is prob #512
Here in this problem why they divide by 2 instead of multiplying by 2?

They way i understood this prob is that, there are two set of cable so multiply with 2.... Correct?

No, if you have 2 sets of cables (assuming they have the same propertis) then treat them as resistors in parallel. R//R = R/2.

 

[HornTootinEE]

I used the NEC method with the Effective Z corrected for power factor, since I think that problem flat out states NEC 2008 in it.

 

[slowhokie]

Digging this topic back up, because I still can't make sense of these!

For 129, the method posted above with Ze=R*pf + X*[sin(arccos(0.8))] I can understand, but then isn't the Ze=0.0232+j 0.0288? Not seeing how to get 0.052 Ohm-N...

Also the NEC and previous poster note that the R and X values from Table 9 need to be corrected for power factor, by Ze=R*pf + X*[sin(arccos(pf))]. Why do the NCEES solutions not do this for 129 or 512?