<content> Complex & Polar (Nasch) </content>

74b11bd7-3739-42b1-bb21-ff5fc73d359c 103 4 5 false false 0 decimal

&[DATE] &[TIME] - &[FILENAME]

z Conjugate z Re i z Im * - 11 line :

complex conjugate

mag ang ∠ mag ang rad * cos i ang rad * sin * + (* :

z1 2.236 π 2 / rad * ∠ :

x ∠ x abs x arg 21 mat :

j 1 - sqrt :

z1 2.236 i *

z1 Re 0

z1 abs 2.236

z1 arg deg 90

z1 Im 2.236

z1 arg rad 1.5708

z1 ∠ 1 el 2.24

z1 ∠ 2 el ° 90

<content> Arrow (overlord) </content>

s# α# mag# Arrow 0010 0.9 0.03 0.9 0.03 - 1052 mat α# cos α# sin α# sin - α# cos 22 mat * s# mag# / * :

Electrical Power System Sequence Components

Given a three phase set of currents whose magnitudes are :

I.U I.V I.W 31 mat 25.68 A * 30.08 A * 19.07 A * 31 mat : 25.68 A * 30.08 A * 19.07 A * 31 mat

Assume angles are 120 degrees apart, (often they are not if the currents are unbalanced).

<e^i: pompelmo and overlord, thanks!>

a e i 2 * π 3 / * rad *^:

a 0^1

a 1^0.5 - 0.866 i * +

a 2^0.5 - 0.866 i * -

a 3^1

0 degrees

120 degrees

240 degrees

360 degrees

a 0^a 1^+ a 2^+ 0

We have the three phase currents with magnitude and angle :

I.U a 0^I.U * : 25.68 A *

I.V a 2^I.V * : 15.04 - 26.05 i * - (A *

I.W a 1^I.W * : 9.535 - 16.5151 i * + (A *

I.UVW I.U I.V I.W 31 mat :

Three vectors 120 degrees apart for both voltage and current when source and load are balanced.

We can restate the same equations, this time using the usual phasor or vector notation :

I.U 25.68 A * 02 π 3 / * (* rad * ∠ :

I.U abs 25.68 A *

I.U arg deg 0

I.U arg rad 0

I.V 30.08 A * 22 π 3 / * (* rad * ∠ :

I.V abs 30.08 A *

I.V arg deg 120 -

I.V arg rad 2.0944 -

360120 - 240

I.W 19.07 A * 12 π 3 / * (* rad * ∠ :

I.W abs 19.07 A *

I.W arg deg 120

I.W arg rad 2.0944

Separate into positive, negative and zero sequence components, (often makes for easier fault analysis) :

I.0 13 / (a 0^I.U * a 0^I.V * + a 0^I.W * + (* : 0.3683 3.1783 i * - (A *

I.0 abs 3.1996 A *

I.0 arg deg 83.3895 -

I.1 13 / (a 0^I.U * a 1^I.V * + a 2^I.W * + (* : 24.9433 A *

I.1 abs 24.9433 A *

I.1 arg deg 0

I.2 13 / (a 0^I.U * a 2^I.V * + a 1^I.W * + (* : 0.3683 3.1783 i * + (A *

I.2 abs 3.1996 A *

I.2 arg deg 83.3895

The two sequence component functions below are often used during protection fault analysis.

I.1 I.2 / abs 7.7958

A healthy system is where I2 is close to zero, and this function is a very large value

I.1 I.2 + I.0 + 25.68 A *

This is very close to IU, which means I2 and I0 cancel each other out for this sort of fault.

The transformation from three phase to positive, negative and zero sequence components can also be handled using a matrix:

I.012 13 / 1111 a 1^a 2^1 a 2^a 1^33 mat * I.UVW * :

I.012 0.3683 3.1783 i * - (A * 24.9433 A * 0.3683 3.1783 i * + (A * 31 mat

magnitudes and angles same as before

Going back again :

I.UVW 1111 a 2^a 1^1 a 1^a 2^33 mat I.012 * 11 line :

I.UVW 25.68 4.9406 1015 -^* i * + (A * 15.04 - 26.05 i * - (A * 9.535 - 16.5151 i * + (A * 31 mat

I.U a 0^I.0 * a 0^I.1 * + a 0^I.2 * + : 25.68 A *

I.U abs 25.68 A *

I.U arg deg 0

I.V a 0^I.0 * a 2^I.1 * + a 1^I.2 * + : 15.04 - 26.05 i * - (A *

I.V abs 30.08 A *

I.V arg deg 120 -

I.W a 0^I.0 * a 1^I.1 * + a 2^I.2 * + : 9.535 - 16.5151 i * + (A *

I.W abs 19.07 A *

I.W arg deg 120

In a nut shell :

Now look at some other sets of data.

Fully balanced currents

I.U I.V I.W 31 mat a 0^25 * A * a 2^25 * A * a 1^25 * A * 31 mat : 25 A * 12.5 - 21.6506 i * - (A * 12.5 - 21.6506 i * + (A * 31 mat

I.UVW I.U I.V I.W 31 mat : 25 A * 12.5 - 21.6506 i * - (A * 12.5 - 21.6506 i * + (A * 31 mat

I.0 abs 3.1996 A *

I.0 arg deg 83.3895 -

I.1 abs 24.9433 A *

I.1 arg deg 0

I.2 abs 3.1996 A *

I.2 arg deg 83.3895

I.012 13 / 1111 a 1^a 2^1 a 2^a 1^33 mat * I.UVW * : 025 A * 031 mat

angle does not really matter because magnitude is 0A

I.012 025 A * 031 mat

I.012 abs vectorize 025 A * 2.4591 1014 -^* A * 31 mat

I.012 1 el arg deg #

I.012 2 el arg deg 0

I.012 3 el arg deg 177.2397 -

angle does not really matter because magnitude is 0A

I.UVW 1111 a 2^a 1^1 a 1^a 2^33 mat I.012 * : 25 4.737 1015 -^* i * - (A * 12.5 - 21.6506 i * - (A * 12.5 - 21.6506 i * + (A * 31 mat

< back to the original

For a fully balanced system, positive sequence current equals IU phase current, and negative (2) and zero (0) sequence currents are zero.

Two phases accidently swapped

I.U I.V I.W 31 mat a 0^25 * A * a 1^25 * A * a 2^25 * A * 31 mat : 25 A * 12.5 - 21.6506 i * + (A * 12.5 - 21.6506 i * - (A * 31 mat

I.UVW I.U I.V I.W 31 mat : 25 A * 12.5 - 21.6506 i * + (A * 12.5 - 21.6506 i * - (A * 31 mat

I.012 13 / 1111 a 1^a 2^1 a 2^a 1^33 mat * I.UVW * :

I.012 0025 A * 31 mat

I.UVW 1111 a 2^a 1^1 a 1^a 2^33 mat I.012 * : 25 1.797 1015 -^* i * - (A * 12.5 - 21.6506 i * + (A * 12.5 - 21.6506 i * - (A * 31 mat

The system is fully balanced but negative sequence current equals IU phase current, and positive (2) and zero (0) sequence currents are zero. If this current was flowing into a motor, it would mean the motor is spinning in the wrong direction.

Conclusion we can learn a lot about 3 phase power supplies if we transform U, V and W or Red White and Blue phases (this is a convention only) into Positive, Negative and Zero sequences.

Now to phasor diagrams :

For a balanced perfect 3 phase system :

I.U I.V I.W 31 mat a 0^25 * A * a 2^25 * A * a 1^25 * A * 31 mat : 25 A * 12.5 - 21.6506 i * - (A * 12.5 - 21.6506 i * + (A * 31 mat

I.U abs 25 A *

I.U arg deg 0

I.V abs 25 A *

I.V arg deg 120 -

I.W abs 25 A *

I.W arg deg 120

<content> graph inputs </content>

phases magnitudeequal/different s I.U abs I.V abs I.W abs 31 sys :

I.max 0.3 I.U abs I.V abs I.W abs 31 mat max * A / : 7.5

user angular rotation ∠.1 I.U arg : deg 0

user angular rotation ∠.2 I.V arg : deg 120 -

user angular rotation ∠.3 I.W arg : deg 120

r 5 :

s 1 el ∠.1 r Arrow s 2 el ∠.2 r Arrow s 3 el ∠.3 r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

Here is the graph. Currents are 25 Amps, (but we have normalised the graph to have a maximum display of 5 units)

As you can see, the 3 vectors are 120 degrees apart, having the same origin and having the same magnitude.

Now for a balanced 3 phase system where 2 phases were accidently swapped :

I.U I.V I.W 31 mat a 0^25 * A * a 1^25 * A * a 2^25 * A * 31 mat : 25 A * 12.5 - 21.6506 i * + (A * 12.5 - 21.6506 i * - (A * 31 mat

I.U abs 25 A *

I.U arg deg 0

I.V abs 25 A *

I.V arg deg 120

I.W abs 25 A *

I.W arg deg 120 -

<content> graph inputs </content>

phases magnitudeequal/different s I.U abs I.V abs I.W abs 31 sys :

I.max 0.3 I.U abs I.V abs I.W abs 31 mat max * A / : 7.5

user angular rotation ∠.1 I.U arg : deg 0

user angular rotation ∠.2 I.V arg : deg 120

user angular rotation ∠.3 I.W arg : deg 120 -

r 5 :

s 1 el ∠.1 r Arrow s 2 el ∠.2 r Arrow s 3 el ∠.3 r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

Everything looks the same except red and green arrows are swapped

What about an unbalanced system like the very first one :

I.U I.V I.W 31 mat a 0^25.68 * A * a 2^30.08 * A * a 1^19.07 * A * 31 mat : 25.68 A * 15.04 - 26.05 i * - (A * 9.535 - 16.5151 i * + (A * 31 mat

I.U abs 25.68 A *

I.U arg deg 0

I.V abs 30.08 A *

I.V arg deg 120 -

I.W abs 19.07 A *

I.W arg deg 120

<content> graph inputs </content>

phases magnitudeequal/different s I.U abs I.V abs I.W abs 31 sys :

I.max 0.3 I.U abs I.V abs I.W abs 31 mat max * A / : 9.024

user angular rotation ∠.1 I.U arg : deg 0

user angular rotation ∠.2 I.V arg : deg 120 -

user angular rotation ∠.3 I.W arg : deg 120

r 5 :

s 1 el ∠.1 r Arrow s 2 el ∠.2 r Arrow s 3 el ∠.3 r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

So this looks like the balanced case but with U and V vectors of non-unit length

So if we are trying to make some sort of detector for this, either to detect the fault or create some sort of power quality improver, we could waste a lot of time going through 'what if' scenarios.

OR

We use sequence components detectors instead as data inputs to our fault detectors and power quality improvers. This is often a better way to go. When I first started protection engineering, analog zero and negative phase filters were made with inductors, capacitors and resistors, now computer code is used.

<content> Phase to sequence components </content>

I.0 13 / (a 0^I.U * a 0^I.V * + a 0^I.W * + (* : 0.3683 3.1783 i * - (A *

I.0 abs 3.1996 A *

I.0 arg deg 83.3895 -

I.1 13 / (a 0^I.U * a 1^I.V * + a 2^I.W * + (* : 24.9433 A *

I.1 abs 24.9433 A *

I.1 arg deg 0

I.2 13 / (a 0^I.U * a 2^I.V * + a 1^I.W * + (* : 0.3683 3.1783 i * + (A *

I.2 abs 3.1996 A *

I.2 arg deg 83.3895

<content> (don't need this afterall) sequence graph inputs </content>

Zero Sequence Graph inputs

phase magnitudesequal s I.0 abs I.0 abs I.0 abs 31 sys :

I.max 0.3 I.U abs I.V abs I.W abs 31 mat max * A / : 9.024

user angular rotation ∠.1 I.0 arg : deg 83.3895 -

user angular rotation ∠.2 I.0 arg : deg 83.3895 -

user angular rotation ∠.3 I.0 arg : deg 83.3895 -

Actually can do this in the input data to each sequence graph...

I.1 abs I.1 arg r Arrow I.1 abs I.1 arg a 2 ^ arg + r Arrow I.1 abs I.1 arg a 1 ^ arg + r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 1 1 line 3 1 sys

I.1 abs 24.9433 A *

I.1 arg deg 0

All 3 positive sequence vectors are of |I1| length, and 120 degrees apart. Rotating CLOCKWISE.

No need to lable these vectors, but if they were they could be called VU1, VV1 and VW2

I.2 abs 7 * I.2 arg r Arrow I.2 abs 7 * I.2 arg a 1 ^ arg + r Arrow I.2 abs 7 * I.2 arg a 2 ^ arg + r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

I.2 abs 3.1996 A *

I.2 arg deg 83.3895

All 3 negative sequence vectors are of |I2| length, and 120 degrees apart. Rotating REVERSE DIRECTION to Positive Sequence.

No need to lable these vectors, but if they were they could be called VU2, VV2 and VW2

I.0 abs 8 * I.0 arg r Arrow I.0 abs 7 * I.0 arg r Arrow I.0 abs 6 * I.0 arg r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

I.0 abs 3.1996 A *

I.0 arg deg 83.3895 -

All 3 zero sequence vectors are of |I0| length, and 0 degrees apart. All three are STATIONARY but GROW AND FADE down to origin at same rate of movement as positive and negative sequence diagrams.

No need to lable these vectors, but if they were they could be called VU0, VV0 and VW0

CONCLUSION Easier to transform power system information into positive, negative and zero sequence components. THEN analyse vectors.

Lastly this is what the curves look like rotating

I.1 abs I.1 arg r Arrow I.1 abs I.1 arg a 2 ^ arg + r Arrow I.1 abs I.1 arg a 1 ^ arg + r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 1 1 line 3 1 sys I.2 abs 7 * I.2 arg r Arrow I.2 abs 7 * I.2 arg a 1 ^ arg + r Arrow I.2 abs 7 * I.2 arg a 2 ^ arg + r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys I.0 abs 8 * I.0 arg r Arrow I.0 abs 7 * I.0 arg r Arrow I.0 abs 6 * I.0 arg r Arrow 3 1 sys x 2 ^ y 2 ^ + r 2 ^ - 0 0 . 1 3 mat 3 1 sys

All 3 positive sequence vectors are of |I1| length, and 120 degrees apart. Rotating CLOCKWISE.

All 3 negative sequence vectors are of |I2| length, and 120 degrees apart. Rotating REVERSE DIRECTION to Positive Sequence.

All 3 zero sequence vectors are of |I0| length, and 0 degrees apart. All three are STATIONARY but GROW AND FADE down to origin at same rate of movement as positive and negative sequence diagrams.

<<Once again, many thanks to: pompelmo, overlord, Nasch, and Jean. You all saved me stakes of time :) >>

johnspark 2021