current phase angles shifted 30deg from voltage angles

[mull982]

I am looking at both voltage and current phase angles on a wye system. I know that if the PT's on a wye system are connected in a L-L fashion and therefore the voltgage inputs are referenced as L-L voltages then the L-N voltages on the system will be shifted by 30deg from these measured L-L voltages.

When measuring currents on a wye system the currents are always referenced as what I refer to as line currents and are essentially the L-N currents on each phase. So if on my given relay the voltage was being measured by a L-L input then I would expect my currents to be shifted 30deg (assuming unity PF) from the voltage angle simply becuase my current would be in phase with the L-N voltage readings which are shifted by 30deg from the L-L voltages.

What if however the voltage inputs to the relay were referencing a L-N voltage input and the 3 PT's were connected in a L-N configuration. Would I still expect to see this 30deg shift when looking at the line currents from CT's? I would expect no shift (again assuming unit PF) since the measured line currents would be in phase with the L-N voltage phasor.

Someone was trying to tell me that even if measuring L-N voltages there would still be a 30deg phase shift on a wye system. I just could not see this, so I'm looking for a second opinion.

 

[Smart $]

Sorry, but I don't have a second opinion

My first and only opinion is in agreement with yours, TTBOMK and based on the information you provided.

 

[jghrist]

I also agree with your assessment. The only caveat being that the currents will be shifted if the power factor is not unity. Reactive loads will have the line current lagging the line-neutral voltage by an angle equal to the arcosine of the power factor (in per unit).

 

[steve66]

Wouldn't it depend on if your loads are line to line, or line to neutral??

For a line-line load, the line current would be in phase with the L-L voltage.

For a L-N load, the line (or neutral) current would be in phase with the L-N voltage.

 

[mull982]

Wouldn't it depend on if your loads are line to line, or line to neutral??

For a line-line load, the line current would be in phase with the L-L voltage.

For a L-N load, the line (or neutral) current would be in phase with the L-N voltage.

I guess I should mention that all of the loads for the system I was referring to in the OP were Delta connected loads. Esentially the primay of a Delta-Wye transformer.

This may change things as you mention?

 

[steve66]

I guess I should mention that all of the loads for the system I was referring to in the OP were Delta connected loads. Esentially the primay of a Delta-Wye transformer.

This may change things as you mention?

That makes it a little more tricky. The line current would be the combination of two different phase currents (load currents). For example, the A phase current would be the sum of the current flowing from A to B and the current flowing from A to C. Without actually doing the vector math (and without looking it up), I think the line current would still be 30 degrees out of phase with one of the L-L voltages.

 

[LarryFine]

I think every wye system I've seen had the voltage taps or PT's (if there were any) and the CT's wye connected. I'd think that, as long as the PT's and CT's are arranged the same way, they'd measure accurately regardless of the load arrangement.

I imagine that Delta loading on a wye system is merely seen as a balanced load. There's no reason for that to be measured inaccurately. Delta supplies, on the other hand, have no neutral to reference, and the voltage tapping must be wired Delta.

Added: Are there Delta-only meters?

 

[skeshesh]

I think every wye system I've seen had the voltage taps or PT's (if there were any) and the CT's wye connected. I'd think that, as long as the PT's and CT's are arranged the same way, they'd measure accurately regardless of the load arrangement.

I imagine that Delta loading on a wye system is merely seen as a balanced load. There's no reason for that to be measured inaccurately. Delta supplies, on the other hand, have no neutral to reference, and the voltage tapping must be wired Delta.

Added: Are there Delta-only meters?

Some PTs can be connected in open delta, I was looking at some configurations from a SEL relay manual and you can probably find it online. The only difference in device settings I saw was the transform ratio. It's a project I'm doing on the side at home so I don't have the info at the office, but I can get you model # etc. when I get home. If I remember correctly though, they're not "delta-only" but the PT can be connected as wye or open-delta.

 

[gar]

091123-1503 EST

Consider this:

A Y source of 100 V L to N, a resistor of 100 ohms connected from L to N. The current in the line is 1 A and in phase with the L to N voltage. The power dissipated in the resistor 100 W.

Now connect the same resistor L to L. the L to L voltage is 173.2 V and the line current is 1.732 A. Because the load is resistive the current in the line has to be in phase with the L to L voltage and 30 deg shifted from the L to N voltage. The power dissipated in the resistor is 173.2 * 1.732 = 300 W.

Next get the component of the line current in phase with the L to N voltage on the assumption there is a 30 deg phase shift. That component is 1.732*1.732/2 = 1.5 A. Note: 1.732 is the sq-root of 3. Also 0.866 is 1.732/2 and the cos of 30 deg, or the sin of 60 deg. And by contrast the sin of 30 deg is 0.5 and the cos of 60 is 0.5.

1.5 A and 100 V is 150 W. But we have two lines supplying the resistors and thus 2*150 = 300 which corresponds to the power calculated from 173.2 V and 1.732 A.

Next observe that an equivalent circuit from a power perspective of the 100 ohms L to L is a 150 ohm resistor from L1 to N and another 150 resistor from L2 to N. However the phase angles of the currents in the lines relative to some fixed reference, such as the L1 to N, are different depending upon on where the resistors are placed, one resistor L to L dissipating 300 W, or two resistors L1 to N and L2 to N.

A different set of conditions. Three 100 ohm resistors connecting each line of a 3 phase system to neutral. Total power dissipated is 300 W, and 1 A line current from each line and in phase with that line to neutral voltage. Alternatively put 3 resistors of 300 ohms in a delta across this source. What is the power dissipated in each resistor? What is the current in each resistor? What is the current in each line? What is the phase angle of the line current relative to its L to N voltage? Now all the line currents are in phase with the L to N voltages whether Y or delta connected.

A more precise value for the sq-root of 3 is 1.73205081 .

.

 

[chris kennedy]

Thats a great post Gordon. I never saw the relationship of the numbers before.

 

[mull982]

091123-1503 EST



1.5 A and 100 V is 150 W. But we have two lines supplying the resistors and thus 2*150 = 300 which corresponds to the power calculated from 173.2 V and 1.732 A.

Next observe that an equivalent circuit from a power perspective of the 100 ohms L to L is a 150 ohm resistor from L1 to N and another 150 resistor from L2 to N. However the phase angles of the currents in the lines relative to some fixed reference, such as the L1 to N, are different depending upon on where the resistors are placed, one resistor L to L dissipating 300 W, or two resistors L1 to N and L2 to N.

.

Why did you go from one resistor to two resistors in this part of the explanation? Were you trying to show that two resistors connected L-N from different phases would be the same as the combination of these two resistors connected L-L? If so, then why would 100ohms line to line be the same as (2) 150ohm resistors connected L-N. Is this because a wye and delta connected resistances convert by a factor of 3?

A different set of conditions. Three 100 ohm resistors connecting each line of a 3 phase system to neutral. Total power dissipated is 300 W, and 1 A line current from each line and in phase with that line to neutral voltage. Alternatively put 3 resistors of 300 ohms in a delta across this source. What is the power dissipated in each resistor? What is the current in each resistor? What is the current in each line? What is the phase angle of the line current relative to its L to N voltage? Now all the line currents are in phase with the L to N voltages whether Y or delta connected.

.

Power in each resistor = V^2/R = (173.2V)^2 / 300 = 100W across each
resistor (Total of 300W)

Current in each resistor = V/R = 173.2 / 300 = .57A

Current in each line = Iphase * 1.73 = .57A * 1.73 = 1A

It looks like the phase angle of the line current will be the same as if the loads were connected L-N so this proves that even with a delta connected load, line currents are always in phase with L-N voltages on a wye system, and line currents are always 30deg out of phase with the L-L voltage readings.

On the contrary L-L currents in a wye system will always be 30deg out of phase with the L-N voltages, and in phase with the L-L voltages.

 

[Smart $]

It looks like the phase angle of the line current will be the same as if the loads were connected L-N so this proves that even with a delta connected load, line currents are always in phase with L-N voltages on a wye system, and line currents are always 30deg out of phase with the L-L voltage readings.

On the contrary L-L currents in a wye system will always be 30deg out of phase with the L-N voltages, and in phase with the L-L voltages.

Whether a system is wye or delta does not matter. The line currents on a balanced system with unity power factor are in phase with wye voltages.

Obviously the line currents' phasing changes for unbalanced systems and/or ones with other than unity power factor.

 

[gar]

091124-0854 EST

mull982:

In the first part of the explanation it was just arbitrary for the progression that I followed.

I started with some simple numbers that could be calculated without paper or calculator. Something needed to be constant so I first let the resistor be constant. But it is also interesting that in moving the one fixed resistor from L to N to L to L that the power dissipation increases by a factor of 1 to 3. Something that might not not be intuitively obvious.

Also in the process of this particular resistor position move that one needs to keep in mind that the current thru a resistive load will always be in phase with the voltage across the resistance. This in turn means the phase angle of the line current changes dependent upon on the location of the resistance, if phase angle is referenced to a fixed time base, like the L1 to N voltage.

Next I wanted to show that if I changed the single 100 ohm load resistor connected L to L to two resistors in series, but with the center point of the resistors connected to neutral, and such that the same total power was dissipated, that I could move the phase angle in L1 back to in phase with the L1 to N voltage. This also provided an exercise in calculating the value of the resistors to make this equivalent power circuit. It further points out that for the same total power dissipated for the two circuits that the L1 line phase angle changed.

To answer your question of why two 150 ohm resistors in series dissipate the same power as one 100 ohm resistor you simply look at the voltage applied across each resistor. Rather than pull some number out of the air you need to look at the circuit and see why the relationship exists. Basically it is a right triangle calculation problem.

Draw three equal length 120 V vectors from a point, and spaced at 120 degrees. Next draw a vector between the end points of two of the 120 vectors. The angle between the new vector and one of the others is 30 deg. How do we know it is 30 deg. Draw a line perpendicular to this fourth vector and passing thru the original intersection point of the first three vectors. This is obviously at an angle of 60 deg from two of the first three vectors. Thus, since the sum of the angles of a triangle is 180 we have 180 - 90 - 60 = 30 deg.

From two of the original vectors, the fourth vector, and the perpendicular line we have two identical right triangles. The hypotenuse has a length of 120. One side is 120 sin 30, and the other side is 120 cos 30. The sin of 30 = exactly 0.5, and the cos 30 = exactly 0.5*sq-root 3 = approximately 0.866025403785. The L to L voltage is 120 * 2 * 0.5 * sq-root 3. This is how one gets the factor 1.732 for the relationship of L to N to L to L.

If this does not answer your first question, then fire back with a variation on the question. A back and forth discussion becomes a learning experience.


The second part of your post looks good. Of course this only applies to a balanced resistive load.

.

 

[mull982]

091124-0854 EST



Next I wanted to show that if I changed the single 100 ohm load resistor connected L to L to two resistors in series, but with the center point of the resistors connected to neutral, and such that the same total power was dissipated, that I could move the phase angle in L1 back to in phase with the L1 to N voltage. This also provided an exercise in calculating the value of the resistors to make this equivalent power circuit. It further points out that for the same total power dissipated for the two circuits that the L1 line phase angle changed.

To answer your question of why two 150 ohm resistors in series dissipate the same power as one 100 ohm resistor you simply look at the voltage applied across each resistor. Rather than pull some number out of the air you need to look at the circuit and see why the relationship exists. Basically it is a right triangle calculation problem.

I understand everything but these two paragraphs. In the 1st paragraph are you saying that instead of having a 100ohm resistor connected L-L dissipating 300W but with the current being 30deg out of phase with L-N voltage, that instead you could have two 150ohm resistors connected L-N and still dissipate 300W however now the currents through these resistors would be in phase with the L-N voltages? Is this what you are saying?

I still dont totally see how the two 150ohm resistors give you 300W. With 100V across each resistor power across each resistor is 100^2 / 150 = 66.6W giving a total wattage of 133.3W. This does not equal 300W. If I use half of 173.2W across each resistor I do not arrise at this answer either.

 

[gar]

091124-1351 EST

mull982:

Neither do I. That is why a discussion is good. I was not working on paper and did the wrong things in my head.

Yes we want 150 W with 100 V applied. So it is 10,000 / 150 = 66.67 ohms for each resistor.

Actually my mistake is your benefit because it made you question my results.

.

 

[mull982]

091124-1351 EST

mull982:

Neither do I. That is why a discussion is good. I was not working on paper and did the wrong things in my head.

Yes we want 150 W with 100 V applied. So it is 10,000 / 150 = 66.67 ohms for each resistor.

Actually my mistake is your benefit because it made you question my results.

.

O.K. I'm with you now. So we will not have 300W total between the two resistors, and now our current will be in phase with the L-N voltages.

Good Example!

Out of curiosity would there ever be a benefit for getting the current in phase with the L-N voltages by rearrnaing the circuit as such?

 

[skeshesh]

Okay. I was starting to think that if I suddenly can't do basic circuit analysis maybe I should give up electrical engineering and see how fast I can handwash a car! It was seriously bothering me, I had to stop working and take out pen+paper and go at it... I really thought I was finally losing it! In any case, we could make things more hairy and try talking about adding reactive elements, if you guys want. I havent looked over laplace transforms in a little while, but should be interesting to look over that stuff again.

 

[mull982]

Okay. I was starting to think that if I suddenly can't do basic circuit analysis maybe I should give up electrical engineering and see how fast I can handwash a car! It was seriously bothering me, I had to stop working and take out pen+paper and go at it... I really thought I was finally losing it! In any case, we could make things more hairy and try talking about adding reactive elements, if you guys want. I havent looked over laplace transforms in a little while, but should be interesting to look over that stuff again.

I'll give the reactor a try.

I would think that if we replaced the 100ohm L-L resistor with a reactor then the line current would be 90deg out of phase with the L-L voltage. That would mean with the 30deg phase shift from L-N voltage, the line current would then be aprox 120deg out of phase with the L-N voltage.

If we had placed a single resistor or any combination of resistors between L-N then the line current would be 90deg out of phase with the L-N voltage and 120deg out of phase with the L-L voltage angles.

Am I correct with this?

 

[gar]

091124-1451 EST

mull982:

I don't know that there is any real advantage. It was just an easy way to move the current phase angle around.

If each of the two hot lines and the neutral all have the same line resistance, then the two 66.7 ohm resistors produce less line loss than the one 100 ohm resistor. If my calculation is correct the ratio of line power dissipation is 5.0625/6.0000 . Again if my calculation is correct if all the copper in the neutral is equally distributed into L1 and L2 and the 100 ohms is between L1 and L2, then this is favored from a power loss point of view. See what values you get.

.

 

[skeshesh]

I'll give the reactor a try.

I would think that if we replaced the 100ohm L-L resistor with a reactor then the line current would be 90deg out of phase with the L-L voltage. That would mean with the 30deg phase shift from L-N voltage, the line current would then be aprox 120deg out of phase with the L-N voltage.

If we had placed a single resistor or any combination of resistors between L-N then the line current would be 90deg out of phase with the L-N voltage and 120deg out of phase with the L-L voltage angles.

Am I correct with this?

Yes, the only detail I'd add is (and I say this just as a matter of electrical theory) we could also talk about whether reactive element is inductive (imp=jwL) or capacitive (imp=jw/C): the current(L-L) will be 90deg out of phase with the voltage(L-L), but the current will lag behind the voltage waveform for the inductive load while with a capacitive load the current will lead the voltage. Now here's the question: is the L-L current leading or lagging the L-N voltage, since that would make the difference between 90+30=120 or 90-30=60deg?