# NCEES #111



## VoltageDrop (Oct 22, 2012)

When they convert the load voltage VAB to Van, what relationship causes it to be shifted by -30 degrees?


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## soma (Oct 22, 2012)

It's a delta connected load, so voltages do not shift nor do they change magnitude. Just use Vab = VAB + IaA * Z line. Though the answer is correct, the method used in the solution is wrong.


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## Insaf (Oct 22, 2012)

When they convert the load voltage VAB to Van, what relationship causes it to be shifted by -30 degrees?

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In balanced three phase system Van, Vbn, Vcn are 120 degree apart from each other. Suppose we like to have phase relationship between Vab and Van:

Vab= Van+Vnb= Van - Vbn; If we draw geometrically, we will see that Vab is leading Van by 30 degree.

If we consider a numerical example, Van=1&lt;0 deg, Vbn= 1&lt;-120 then Vab= 1.732&lt;30 deg = 1.73* 1&lt;30 degree.

Please note that delta connection can be solved in same way as Y connection, considering a virtual neutral point. Bottom line is that irrespective of delta or wye, phase voltage always lag line voltage by 30 degree.


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## da_souljah (Oct 22, 2012)

Can someone provide an alternate way to reach the solution?


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## thammond (Oct 22, 2012)

Why do they not use 2 time R+jX because where the terminals Vab are you would have the impedance of b as well. Any help would be appreciated

Thanks


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## katag (Oct 22, 2012)

I used Voltage Loop to solve this problem. We know that Vab = (R+jX)*IaA + VAB + (R+jX)*-IbB

Vab = (5+10i) * 70&lt;-20 + 12500 + (5+10i) *-70&lt;-140 = 12954.6 which is closest to 13.0kV answer C. You just have to remember when you go around the loop that IbB is going the opposite direction of IaA and that it is 120 deg shifted from Phase A.

Hope this helps!


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## soma (Oct 22, 2012)

thammond said:


> Why do they not use 2 time R+jX because where the terminals Vab are you would have the impedance of b as well. Any help would be appreciated
> 
> Thanks


This is the alternate way I posted above *Vab = V**AB +** IaA * Z line*

two times impedance for calculating voltage drop should be only used for single-phase two-wire circuits.


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## VoltageDrop (Oct 22, 2012)

Thanks, everybody! Katag, that's how I was trying to solve it but I added IbB instead of subtracted it so thanks for pointing that out! I find that a much more obvious way to solve this than solving to a pseudo neutral point then converting back to line values but it never hurts to have an extra trick in the bag. Good luck Friday to all!


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## soma (Oct 22, 2012)

As Katag did, loop analysis can also be done but since this is a balanced system we can skip an extra piece of calculation of drop in line bB.


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## Silkworm (Oct 22, 2012)

Soma,

The load is delta connected so when you convert to Y (line to ground) the voltage will lag by 30 deg since the reference voltage is 0 deg. You will see this if you draw the phasor diagram. NCEES method is correct.

Dean


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## soma (Oct 22, 2012)

Silkworm said:


> Soma,
> 
> The load is delta connected so when you convert to Y (line to ground) the voltage will lag by 30 deg since the reference voltage is 0 deg. You will see this if you draw the phasor diagram. NCEES method is correct.
> 
> Dean


Dean,

NCEES method would have been correct if the load was 3-phase, 4 wire, Y-connected. In delta connection phase and line voltages do not have 30 deg. shift. I agree with you that in y connection there is a phase shift of 30 deg between line and phase voltages.

Soma

Soma


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## Silkworm (Oct 22, 2012)

Soma,

In a 3 phase system, line to neutral voltages are always phase shifted by 30 degrees behind line to line voltages. The only reason they convert to Y configuration is because it's convienant. You could work with the phase to phase voltages but then you need to apply Kirchhoff's voltage rule around the Ia and Ib loop..more chance of making a mistake that way. Both result in 12.95kV.


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## ruffryder (Oct 22, 2012)

I think we are getting confused with the definition of phase voltage vs line to neutral voltage. I had this discussion with my boss, who used to be a professor, and he railed on me for using the term line voltage and phase voltage as it is too ambiguious as to what the voltage actually is being discussed. I think I like his opinion that phase voltage should never be used as it can be confusing. We should always use either line to line or line to neutral in discussing voltages.

What is everyone's thoughts on this? Soma, I hate to pick on you, but it seems this is where some confusion is?


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## Insaf (Oct 22, 2012)

In practice, when power is supplied to a load, connection arrangement may not known. It can be treated as you are supplying power to a black box, so questions of wye or delta have no importance anymore. In that case any connection arrangement can be solved as Y connection considering a virtual neutral point (n point of NCEES solution). Then multiplication of phase to neutral value by 1.732 will give line to line value.

So NCEES solution is correct (and less time consumimg)


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## thammond (Oct 23, 2012)

Thanks for all the help. Good luck to everyone on Friday.


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## Lielec11 (Oct 24, 2012)

soma said:


> It's a delta connected load, so voltages do not shift nor do they change magnitude. Just use Vab = VAB + IaA * Z line. Though the answer is correct, the method used in the solution is wrong.


I thought I was going crazy looking at the solution because I thought the same as you.. I think my mind is just burnt out, need to relax before Friday,,,


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## AAGR2013 (Oct 12, 2016)

ruffryder said:


> I think we are getting confused with the definition of phase voltage vs line to neutral voltage. I had this discussion with my boss, who used to be a professor, and he railed on me for using the term line voltage and phase voltage as it is too ambiguious as to what the voltage actually is being discussed. I think I like his opinion that phase voltage should never be used as it can be confusing. We should always use either line to line or line to neutral in discussing voltages.
> 
> What is everyone's thoughts on this? Soma, I hate to pick on you, but it seems this is where some confusion is?


totally agree! in this delta connection, its phase voltage is also 12.5kv, so no √3 need to be divided by, mistake! We have to use line-neutral, line-line term.


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## knight1fox3 (Oct 12, 2016)

AAGR2013 said:


> totally agree! in this delta connection, its phase voltage is also 12.5kv, so no √3 need to be divided by, mistake! We have to use line-neutral, line-line term.


Hey way to revive a 4+ year old thread!!! :thumbs:


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## snickerd3 (Oct 12, 2016)

at they are looking through the old stuff instead of asking the same thing again


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