# 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

colo_elec said:


> 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

colo_elec said:


> 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

bacchi said:


> 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

DK PE said:


> 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

Kahrlo said:


> 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

Kahrlo said:


> Kahrlo said:
> 
> 
> 
> 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
Click to expand...

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 &amp; B)



electric said:


> Kahrlo said:
> 
> 
> 
> 
> 
> Kahrlo said:
> 
> 
> 
> 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
> 
> 
> 
> 
> 
> Click to expand...
> 
> 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?
Click to expand...


----------



## GabeM

electric said:


> Kahrlo said:
> 
> 
> 
> 
> 
> Kahrlo said:
> 
> 
> 
> 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
> 
> 
> 
> 
> 
> Click to expand...
> 
> 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?
Click to expand...

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!!


----------



## Power12

The Voltage phase-neutral lags the Voltage phase-phase by 30 deg


----------



## Pusta

Power12 said:


> The Voltage phase-neutral lags the Voltage phase-phase by 30 deg


Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0

Am I missing something here?


----------



## cconnawa

Pusta said:


> Power12 said:
> 
> 
> 
> The Voltage phase-neutral lags the Voltage phase-phase by 30 deg
> 
> 
> 
> Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0
> 
> Am I missing something here?
Click to expand...


What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va.	The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.

The way I solved this problem was like this:

*1st. *

Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.

*2nd.*

Calculate the per phase reactive power I^2 * X (per phase) = 3,333

*3rd. *

Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000

*Notes and Comments:*

1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.

2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.

3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..


----------



## Pusta

cconnawa said:


> Pusta said:
> 
> 
> 
> 
> 
> Power12 said:
> 
> 
> 
> The Voltage phase-neutral lags the Voltage phase-phase by 30 deg
> 
> 
> 
> Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0
> 
> Am I missing something here?
> 
> Click to expand...
> 
> 
> What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va.	The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
> 
> The way I solved this problem was like this:
> 
> *1st. *
> 
> Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
> 
> *2nd.*
> 
> Calculate the per phase reactive power I^2 * X (per phase) = 3,333
> 
> *3rd. *
> 
> Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
> 
> *Notes and Comments:*
> 
> 1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
> 
> 2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
> 
> 3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
Click to expand...

I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.


----------



## cconnawa

Pusta said:


> cconnawa said:
> 
> 
> 
> 
> 
> Pusta said:
> 
> 
> 
> 
> 
> Power12 said:
> 
> 
> 
> The Voltage phase-neutral lags the Voltage phase-phase by 30 deg
> 
> 
> 
> Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0
> 
> Am I missing something here?
> 
> Click to expand...
> 
> 
> What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va. The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
> 
> The way I solved this problem was like this:
> 
> *1st. *
> 
> Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
> 
> *2nd.*
> 
> Calculate the per phase reactive power I^2 * X (per phase) = 3,333
> 
> *3rd. *
> 
> Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
> 
> *Notes and Comments:*
> 
> 1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
> 
> 2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
> 
> 3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
> 
> Click to expand...
> 
> I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.
Click to expand...

You can do it that way, and you will still get the same answer. It's just another step you don't need which can potentially be an error while you are performing the calculations. There are certain instances for which the -30 phase shift from line-line to line-phase is really important, for example when the question asks you for the phase angle, and the phase angle is included in the answer choices. But in general, so long as you are consistent in keeping 120 degrees between the phases, the magnitude will be the same. Plug in the -30 on both V_A and V_B voltages and you'll see you get exactly the same answer after you do the math.

Here is another instance where there was the same confusion and I answered it similarly:

/&gt;http://engineerboards.com/index.php?showtopic=19446


----------



## Pusta

cconnawa said:


> Pusta said:
> 
> 
> 
> 
> 
> cconnawa said:
> 
> 
> 
> 
> 
> Pusta said:
> 
> 
> 
> 
> 
> Power12 said:
> 
> 
> 
> The Voltage phase-neutral lags the Voltage phase-phase by 30 deg
> 
> 
> 
> Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0
> 
> Am I missing something here?
> 
> Click to expand...
> 
> 
> What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va.	The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
> 
> The way I solved this problem was like this:
> 
> *1st. *
> 
> Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
> 
> *2nd.*
> 
> Calculate the per phase reactive power I^2 * X (per phase) = 3,333
> 
> *3rd. *
> 
> Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
> 
> *Notes and Comments:*
> 
> 1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
> 
> 2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
> 
> 3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
> 
> Click to expand...
> 
> I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.
> 
> Click to expand...
> 
> You can do it that way, and you will still get the same answer. It's just another step you don't need which can potentially be an error while you are performing the calculations. There are certain instances for which the -30 phase shift from line-line to line-phase is really important, for example when the question asks you for the phase angle, and the phase angle is included in the answer choices. But in general, so long as you are consistent in keeping 120 degrees between the phases, the magnitude will be the same. Plug in the -30 on both V_A and V_B voltages and you'll see you get exactly the same answer after you do the math.
> 
> Here is another instance where there was the same confusion and I answered it similarly:
> 
> http://engineerboard...showtopic=19446
Click to expand...

You’re correct. I get the same magnitude when I take into account the 30 degree shift; however the overall angle is different. I guess in this case its best to leave the shift out.

Thanks for all your help!


----------

