NCEES 531

[colo_elec]

Ok, so am I getting this right:

Per Phase

VAR = I^2 * X (X is reactance)

3 Phase

Var = 3*I^2 * X

To me this does not seem intuitive as with KW we always utilize root 3. Anyone have a good article on this or brief explanation?

 

[phowardtx]

Ok, so am I getting this right:
Per Phase

VAR = I^2 * X (X is reactance)

3 Phase

Var = 3*I^2 * X

To me this does not seem intuitive as with KW we always utilize root 3. Anyone have a good article on this or brief explanation?

Root 3 is used in Y or Δ connected loads but in this case we are adding up 3 separate single-phase loads so we multiply by 3.

 

[bacchi]

Ok, so am I getting this right:
Per Phase

VAR = I^2 * X (X is reactance)

3 Phase

Var = 3*I^2 * X

To me this does not seem intuitive as with KW we always utilize root 3. Anyone have a good article on this or brief explanation?

The difference is in Phase Voltage to Line voltage.

Considering only reactive load:-

single phase VAR = V-LN * I

Three phase VAR = Sqrt 3 * (V-LL) * I = Sqrt 3 * (Sqrt 3 * V-LN) * I = 3*V-LN*I

= 3* I*X*I = 3I^2*X

 

[jakesaround]

I am in need of some help understanding this problem.

Can someone comment on bacchi's answer, is this correct/incorrect?

It doesn't seem to make sense with my understanding of how to calculate vars.

Thanks.

 

[DK PE]

The difference is in Phase Voltage to Line voltage.
Considering only reactive load:-

single phase VAR = V-LN * I

Three phase VAR = Sqrt 3 * (V-LL) * I = Sqrt 3 * (Sqrt 3 * V-LN) * I = 3*V-LN*I

= 3* I*X*I = 3I^2*X

I follow the substitutions made above and they're mathematically correct... maybe it would help if you posted a little more of the question statement for those of us without a NCEES problem book.

Three phase power really equals 3 (not sqrt3) times the per-phase power. This is always true although we usually find it handier to use something like sqrt3*VL-L * IL * cos theta = total 3 phase power for either delta or wye. For a wye connection, for example, it is also true that 3-phase power = 3* VLine-neutral * IL * cos theta. But in your case it sounds like you know the per-phase VARs so then in this case the total VARS is 3 times the VARs/phase.

If I haven't helped, please post a bit more of the problem statement and I'll give it another try.

 

[DK PE]

Three phase power really equals 3 (not sqrt3) times the per-phase power. This is always true (should have added the qualifier "for balanced systems" here although we usually find it handier to use something like sqrt3*VL-L * IL * cos theta = total 3 phase power ...

 

[jakesaround]

DK,

Yes I think you have helped but for your benefit and others here is the full problem.

Maybe you can take some time to work through it and help me gain some deeper understanding.

NCEES #531

The figure (use your imaginations shows a 500-kV (line-to-line voltage) transmission line linking two systems. The theoretical maximum power that can be transferred from System A to System B is 5,000 MW, which occurs when the power angle "sigma" = 90 degrees. The line current is then proportional to (Va - Vb). The reactive power losses (MVAR) in the transmission line during this transfer would be most nearly:

A) 1,667

B) 3,333

C) 5,000

D) 10,000

The only other information given by the diagram is;

1) The Line reactance XL = 50 ohms.

2) Va = 500kV at angle sigma

3) Vb = 500kV at angle 0 degrees.

Thanks for your help.

 

[BamaBino]

Eq 18-9 in Fink's EE Handbook 13th ed

for 3-phase line drops with balanced loads

[SIZE=12pt]Vdrop L-L=sqrt(3)(IR cos-theta + IX sin-theta)[/SIZE]

in this problem R=0.

 

[DK PE]

I'm not trying to be clever here but since you didn't post the solution and I don't have the book it is a great opportunity to embarrass myself. Here is my rationale...

First let's consider a simpler problem... say a 5kW resistive load on a 240V single phase system at the end of 200' of # 8 Al wire. The current would be 20.8 Amps towards load in hot line, same returning in neutral line and if I asked you to determine real power dissipated in "transmission line" you would say I2 *R where R is (reasonable guess) 0.2 ohms and get 87 W PER LINE or a total of 174W.

Now we make the problem a little more complex... say a 12kW 3-phase resistive load on a 208 Vline system on same size of conductors (but this time three wire vs. two wire) transmission line. Now P = sqrt(3) * Vline * Iline and cos theta goes away since power factor = unity. The line current = 33.3 A and therefore I2 *R = 221W per line and if I asked you the total real power dissipated in the "overall transmission line" you would multiply X3 since we have three physical conductors and give me 665W.

Now the task at hand seems to have a few distracting factors, the statement on "power angle delta" is true but I don't think is required for the solution. I would just claim that following above, if you have the maximum claimed power of 5000 MW transferred on a 500kVline system, you would apply the same rationale as above P = sqrt(3) * Vline * Iline and get a line current of 5774A. Then applying VARS = I2 * X (analogous to P = I2 *R) you would multiply by the given 50 ohm per line reactance which would give you 1667 MVAR but similar to above, this is per line so you finally multiply by 3 and get 5000 MVAR. Notice if your brain slips only a half a cog that the incorrect 1667 MVAR is listed as first answer :brickwall:

Hope this helps... otherwise post the NCEES solution and I'll try to learn from it !

 

[jakesaround]

Thanks for taking the time to work on this. I did kinda leave you hanging out there without the solution but I know you enjoyed the challenge. And don't worry about any embarrassment we are all here to help and learn, and I appreciate your help.

The correct answer is D and here is how they got it.

NCEES Solution:

 

[BamaBino]

I think this is a poor problem.

After reading it, you have to assume that System PF angle is 90 degrees.

 

[DK PE]

Well, after digging out my copy of Stevenson, I understand their solution at least although it isn't the best question I've seen. You could use Ia = (Eg /_ sigma - Vt ) / JXg and you apply it to the system rather than generator as power angle is usually referred to. The trick is the voltage is the line-neutral using this equation. I guess the good thing is back to your orginal question, once you correctly solve for the line current, the total MVAR is 3X the per line... maybe I at least helped that a bit. :blush:

 

[DK PE]

In an effort to partially redeem myself after blowing the line current calculation while trying to help the other day, I've attached below the way I would do the problem. Using Wildi Chap 25 Sec 23 as a reference... again, remember once you obtain the line currents to multiply by 3, not sqrt(3).

 

[Kahrlo]

sqrt of 3 is only used when using line to line voltages and line currents..

Apparent power= sqrt(3) x Vl-l x I line

3 is used for a single phase voltage and line to line current for delta or line to neutral for wye connection..

Apparent power= 3 x Vl-n x I l-n (wye)

Apparent power= 3 x Vl-n x I line (delta)

Since the above problem asks for the reactive power lost, and since the reactance is represented as a single phase, I^2 * X, the total 3 phase reactive power lost is 3 times of that

 

[Kahrlo]

sqrt of 3 is only used when calculating 3 phase power using line to line voltages and line currents..Apparent power= sqrt(3) x Vl-l x I line

3 is used for a single phase voltage and line to line current for delta or line to neutral for wye connection..

Apparent power= 3 x Vl-n x I l-n (wye)

Apparent power= 3 x Vl-n x I line-line (delta)

Since the above problem asks for the reactive power lost, and since the reactance is represented as a single phase, I^2 * X, the total 3 phase reactive power lost is 3 times of that

 

[electric]

sqrt of 3 is only used when calculating 3 phase power using line to line voltages and line currents..Apparent power= sqrt(3) x Vl-l x I line

3 is used for a single phase voltage and line to line current for delta or line to neutral for wye connection..

Apparent power= 3 x Vl-n x I l-n (wye)

Apparent power= 3 x Vl-n x I line-line (delta)

Since the above problem asks for the reactive power lost, and since the reactance is represented as a single phase, I^2 * X, the total 3 phase reactive power lost is 3 times of that

I am still confused about this problem based upon the explanation and formula mentioned in Stevenson (page 108) which is;

Q=Vt/Xd (Ei*cos delta - Vt)

For this problem since delta =90, Q=Vt^2/Xd. With line voltage Q=square of 500kV/50=5000.

What am i doing wrong here?

 

[Nucky]

My 2 cents on this..

The problem is a power flow problem. It's asking for the power flow when the power angle = 90 deg (This is actually the max limit, any more than this and the system goes unstable).

Anyway, if you solve the power flow equations at this power angle, you would find that:

Real Power transferred from A to B = 5000 MW

Real Power transferred from B to A = -5000 MW

Reactive Power transferred from A to B = 5000 MVAR

Reactive Power transferred from B to A = 5000 MVAR

In other words, 5000 MW of real power is being sent across the line and 10000 MVAR of reactive power is lost in the line (5000 MVAR is sent from each source A & B)

sqrt of 3 is only used when calculating 3 phase power using line to line voltages and line currents..Apparent power= sqrt(3) x Vl-l x I line

3 is used for a single phase voltage and line to line current for delta or line to neutral for wye connection..

Apparent power= 3 x Vl-n x I l-n (wye)

Apparent power= 3 x Vl-n x I line-line (delta)

Since the above problem asks for the reactive power lost, and since the reactance is represented as a single phase, I^2 * X, the total 3 phase reactive power lost is 3 times of that

I am still confused about this problem based upon the explanation and formula mentioned in Stevenson (page 108) which is;

Q=Vt/Xd (Ei*cos delta - Vt)

For this problem since delta =90, Q=Vt^2/Xd. With line voltage Q=square of 500kV/50=5000.

What am i doing wrong here?

 

[GabeM]

sqrt of 3 is only used when calculating 3 phase power using line to line voltages and line currents..Apparent power= sqrt(3) x Vl-l x I line

3 is used for a single phase voltage and line to line current for delta or line to neutral for wye connection..

Apparent power= 3 x Vl-n x I l-n (wye)

Apparent power= 3 x Vl-n x I line-line (delta)

Since the above problem asks for the reactive power lost, and since the reactance is represented as a single phase, I^2 * X, the total 3 phase reactive power lost is 3 times of that

I am still confused about this problem based upon the explanation and formula mentioned in Stevenson (page 108) which is;

Q=Vt/Xd (Ei*cos delta - Vt)

For this problem since delta =90, Q=Vt^2/Xd. With line voltage Q=square of 500kV/50=5000.

What am i doing wrong here?

500 kV is the line voltage; you need to use the voltage drop across the reactance, which is phasor Va minus phasor Vb.

A couple people here mentioned that above 90 degrees between Va and Vb, the system goes unstable. How does the system go unstable?

 

[mull982]

I simply used V^2 / X to find the total system reactance, where V^2 is equal to the phasor VA-VB. Is this a correct way of doing it as well, or is this just a conincidence?

Also when using VA-VB/X to solve for current and then using I^2 *x to solver fore reactance works when using the L-L voltages of VA and VB there was not need to multipy answer by 3 to get correct answer.

 

[Pusta]

This may be a stupid question and maybe I am missing something, but why do we not take into account the 30 degree phase shift when converting from phase to phase voltage to line to neutral voltage.

Thanks!!