Calculate Phase Shift From Voltage Readings

Status
Not open for further replies.

AHarb

Member
Location
Atlanta, GA
Hi,

We had something occur yesterday that I couldn't figure out. We were tying two distribution circuits together and had an issue when "phasing" the normal open point. We use phasing sticks that measure the voltage from one open point to another to verify that the phases being tied are identical. In this scenario, we were tying the circuits together through a large (5000 kVA) transformer. The measured voltages across the open points for phases A and B were 0 V indicating that there were no phase shifts and they could be tied. The measurement taken across the phase C open points showed 4 kV. The line/ground measurements were 15 kV at all 6 points so it seems like there was a small phase shift at the phase C normal open point. We think this was due to the large 5000 kVA (14.4 kV/7.2 kV wye/wye) transformer. Can we calculate the phase shift with the given information? It seems like it should be possible, but I can't remember how to do it.

Thanks in advance.

Adam
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160115-1440 EST

Draw vector diagrams.

Draw a circle with a 15 kV radius. All of your wye vectors should radiate from the center of the circle to points on the circle based on all line to neutral (ground) measurements being equal at 15 kV.

Assume the radial vectors for phases A and A', and B and B' coincide as your measurements imply, then create two C vectors C and C' with their tips spaced by 4 kV. This is a large angle difference.

Get all the line to line measurements and see those results can correlate with the vector diagram. On the surface your measurements look strange.

.
 

Phil Corso

Senior Member
AHarb...

If synchronized the two distribution-circuits are connected to a single source! What, then, is the distance from each distribution-circuit to that source?

Regards, Phil Corso
 

mivey

Senior Member
AHarb...

If synchronized the two distribution-circuits are connected to a single source! What, then, is the distance from each distribution-circuit to that source?

Regards, Phil Corso
Not that far if you are thinking what I think you are thinking!
 

AHarb

Member
Location
Atlanta, GA
Not that far if you are thinking what I think you are thinking!

Our distribution circuits are fairly short. I can't get the exact values because our modeling software server is down. I've used Good to give estimates of the circuit lengths. The first circuit is about 3.4 miles long and the other is about 3.6 miles long. I'm not sure I follow your logic about the circuit lengths.
 

AHarb

Member
Location
Atlanta, GA
Did you check for blown fuses on capacitor banks?

We didn't look for that, but that could be a problem. All of our "switched" capacitor banks should be offline during this time period and all of our "fixed" banks should have been taken out of service. It wouldn't be the first time that we've found one incorrectly in or out of service. I'll have to check the capacitor banks and see if that may be causing our issue.
 

AHarb

Member
Location
Atlanta, GA
160115-1440 EST

Draw vector diagrams.

Draw a circle with a 15 kV radius. All of your wye vectors should radiate from the center of the circle to points on the circle based on all line to neutral (ground) measurements being equal at 15 kV.

Assume the radial vectors for phases A and A', and B and B' coincide as your measurements imply, then create two C vectors C and C' with their tips spaced by 4 kV. This is a large angle difference.

Get all the line to line measurements and see those results can correlate with the vector diagram. On the surface your measurements look strange.

.

I think I understand how to draw the vectors, but I'm not sure how to solve for the angle between the two. It seems like I remember learning how to solve this in college, but it's been a while. I may have to start digging through my old text books.

I cannot remember if they took the line to line measurements or not while we were out there. It's too late to go back and get the measurements since the open point has been tied and this will now be a "normal "closed" switch for a few years. Assuming we did take line to line measurements, it should've been around 25 kV.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
If you know the length of two vectors with a common endpoint but you do not know the angles you can "solve" graphically by drawing two circles of the appropriate radius and finding the point(s) where they intersect.
There is a trigonometric solution too, but it is more complicated.
In other words I am not going to provide it. :)
 

mivey

Senior Member
Our distribution circuits are fairly short. I can't get the exact values because our modeling software server is down. I've used Good to give estimates of the circuit lengths. The first circuit is about 3.4 miles long and the other is about 3.6 miles long. I'm not sure I follow your logic about the circuit lengths.
If by a switching difference or some weird means we have a difference in length from the same source of about 130-140 miles you would have a difference of about 4 kV due to the wave on the long leg being delayed enough for the phase to differ by a little over 16 degrees (using a 3,030 mile long wavelength).

Interesting to see and can be a factor when tying an uneven loop but doesn't apply in your case.
.
 

mivey

Senior Member
We didn't look for that, but that could be a problem. All of our "switched" capacitor banks should be offline during this time period and all of our "fixed" banks should have been taken out of service. It wouldn't be the first time that we've found one incorrectly in or out of service. I'll have to check the capacitor banks and see if that may be causing our issue.
I'm just trying to find something other than phase shift due to length since you said volt drop was not an issue. Not even sure an open cap would do it. I would suspect the measurement but we can keep thinking of other possibilities
 

mivey

Senior Member
I think I understand how to draw the vectors, but I'm not sure how to solve for the angle between the two. It seems like I remember learning how to solve this in college, but it's been a while. I may have to start digging through my old text books.

I cannot remember if they took the line to line measurements or not while we were out there. It's too late to go back and get the measurements since the open point has been tied and this will now be a "normal "closed" switch for a few years. Assuming we did take line to line measurements, it should've been around 25 kV.
arcsin(4000/14400) gave me 16.13 degrees. I should have used arcsin(4000/24942) to get 9.23 degrees. That would correct my previous post to about 80 miles so we still need something other than a length difference.
 

mivey

Senior Member
arcsin(4000/14400) gave me 16.13 degrees. I should have used arcsin(4000/24942) to get 9.23 degrees. That would correct my previous post to about 80 miles so we still need something other than a length difference.
...and for a 3,030 mile wavelength that calc is 9.229d / 360d x 3030 miles = 77.67 miles
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
...and for a 3,030 mile wavelength that calc is 9.229d / 360d x 3030 miles = 77.67 miles
Just for my edification, that 3030 mile wavelength is 186,000 miles/sec divided by 60 cycles/sec, derated for the slightly lower than c speed of electricity in copper, right?
 

mivey

Senior Member
Just for my edification, that 3030 mile wavelength is 186,000 miles/sec divided by 60 cycles/sec, derated for the slightly lower than c speed of electricity in copper, right?
Yes. It was a figure I remembered for propagation on a particular transmission line but I figured it was close enough for the distribution line. I could have calcated a standard distribution line configuration but I did not feel like it since it was not a practical answer to the problem at hand.
 
Status
Not open for further replies.
Top