How to Solve in Rectangular and Polar Form

synchro

Senior Member
Location
Chicago, IL
Occupation
EE

In the Steinmetz method, vector math in the complex plane, all of the values have a direct analogue in the observable physical reality.
Thanks for that article. It's an interesting story and very informative.

As you go from falling body, pendulum motion, spring and shock absorber, to inductance and capacitance, the math is the same.
Pendulum motion can be described by sinusoidal analysis like in LC circuits when the peak value of the pendulum's angle θ from the vertical is small so that sin(θ) ≊ θ. For larger motions elliptic functions instead of sinusoidal functions must be used for an accurate analysis.
 

dkarst

Senior Member
Location
Minnesota
Lots of interesting discussion but I'm not clear whether anyone has assisted the OP... that is the first question...

What is the load power factor (number and word)?

I guess we could start with tossing out the transformer and just given a single phase 240 VAC source connected to this 88.8 - j66.6 ohm load what is power factor (number and word)?

By determining the phase difference between voltage and current, you can determine the power factor and whether it is leading or lagging (this is the word).

By converting the load impedance to polar form, the current could be shown to be I V/Z = (240/_ 0) / (111 /_ -36.9 ) so the current is leading the voltage which you would expect in a capacitive load.

If the OP would post back what area is unclear, I'm sure someone will help.
 

__dan

Senior Member
Everything has gone to Youtube (Thank God).

 

__dan

Senior Member
I'm watching Youtube now but it's on autoplay. Buffet (Bershire Hathway Chief) is on giving the 2020 report. He's going on about the Civil War losses 6% of all men 18-60 (I'm sure the point to make is look where we recovered to since then).

Th vibe I'm getting is next he is going to say 'Mars is the problem now, we need to invade Mars'.

OMG he just compared the market to a drinking pot smoking neighbor who quotes the price to buy or sell your farm. Oh man who needs HBO.
 
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dkarst

Senior Member
Location
Minnesota
Technically no one has shown the step by step process. Links are good, but I need some help.
Okay, so let's go back to the simpler example I mentioned earlier.

You have a 240 VAC source with a load of 88.8 - j66.6 (this is an 88 ohm resistor in series with a ~ 40 uF capacitor at 60Hz). If you use ohms law and calculate the current in this circuit as I did in earlier post, you will see the current flow is LEADING the applied voltage by ~ 37 degrees. The power factor (number) then is cos 37 degrees or 0.8 and the word they are looking for is LEADING (as opposed to LAGGING).
 

wwhitney

Senior Member
Location
Berkeley, CA
OK, I'm going to try to work through the problem posted for my own practice/learning and for mbrooke's perusal. I will probably duplicate dkarsts's posts. Those more knowledgeable than me, please point out where I go wrong. : - )

1.1) For a balanced three phase load following Ohm's law, the power factor just depends on the impedance and can be computed as in the single phase case. [Is there a meaningful definition of power factor for unbalanced three phase loads?]

For Z = 88.8 - j66.6 ohms, the magnitude |Z| = 111 ohms. [It's a 3-4-5 triangle, times 22.2]. The power factor is (Re Z) / |Z| : Re Z is the resistive component of the impedance which tells you the component of the current that is flowing in phase with the voltage and actually transferring power. |Z| is the magnitude of the impedance and tells you the magnitude of the total current. So their ratio will be real power / apparent power.

Thus the power factor is 88.8 / 111 = 4 / 5 = 0.8. I can never keep the leading/lagging standard straight, but per dkarst it refers to whether the current peak comes before/after the voltage peak, respectively. I will use ph V to mean the phase angle of V (the angle written in the problem after the little angle symbol). So leading power factor would mean ph V < ph I (higher phase angle peaks first), and lagging the opposite.

Looking at the multiplication V = I * Z in polar form, we see that ph V = ph I + ph Z. For Z = 88.8 - j66.6, ph Z < 0. So ph V < ph I, or leading power factor.

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.2 To find I_ab, we just need to properly use Ohm's Law, V_ab = I_ab * Z_ab [Does the new forum software support subscripts and superscripts like the old one did? [sub][/sub] doesn't work. So I will use X_Y to denote X subscript Y. Likewise ^ means superscript, usually exponentiation.]

We are given Z_ab = Z = 88.8-j66.6. We aren't directly given V_ab, but note from the diagram that V_ab = V_an + V_nb, and that V_nb = - V_bn always. Negating a vector just means reflecting the vector through the origin. I.e. in polar form the magnitude is unchanged, and the angle is increased by 180 degrees (or decreased by 180 degrees, since a 360 degree change is no change at all). For polar form I'll use the notation a \ b to mean magnitude a and angle b. In other words, V = |V| \ ph V.

Let's call v = 34500 / (sqrt 3). So V_ab = v \ 0 + v \ 120 = v * (1\0 + 1\120). Drawing that out makes it clear that 1\0 + 1\120 = 1\60, an equilateral triangle. So V_ab = v\60.

Now we want to multiply (rather divide) two vectors (rather complex numbers), one which we have in polar form (V_ab) and one in rectangular form (Z_ab). We could do the computation in either polar or rectangular coordinates, but since the result is requested in polar form, let's use polar form. So we need to convert Z_ab to polar form. In general the vector x + jy has the polar form sqrt(x^2+y^2) \ arctan(y/x) [draw out the right triangle in the complex plane]. We get Z_ab = 111 \ (arctan(-3/4) = -36.9 degrees).

Division requires computing the multiplicative inverse, and to do that to a complex number in polar form, we have to take the reciprocal of the magnitude, and then negate the angle. [As seen in the usual complex polar form R*e^(ja), the reciprocal is R^(-1) * e^(-ja)]. So Z_ab^(-1) = 1/111 \ 36.9

Finally, we have I_ab = V_ab / Z_ab = V_ab * (Z_ab^(-1)) = v\60 * 1/111 \ 36.9. To multiply in polar form, we just multiply the magnitudes and add the angles. That means I_ab = v/111 \ 96.9 amps = 179 \ 96.9 amps.

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.3 Wikipedia tells me that the complex power S is defined as S = V * I', where I' is the complex conjugate of I (often denoted with an overbar, but I can't type that). For a load following Ohm's law we get S = Z * I * I'. Now I * I' = |I|^2 (the magnitude of I, squared), so S = Z * |I|^2.

For the single phase load Z_ab, we have already calculated I_ab. That means S_ab = (179)^2 * (88.8 -j66.6) = 2860 - 2140j kVA. The full 3 phase power S should just be S = 3 * S_ab, or S = 8580 - 6430j kVA.

[As a mathematician, I'm pretty sure that I have the techniques correct so far, but not sure I haven't inadvertently made a sign error somewhere.]

Cheers, Wayne
 
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wwhitney

Senior Member
Location
Berkeley, CA
1.2 To find I_ab, we just need to properly use Ohm's Law, V_ab = I_ab * Z_ab [Does the new forum software support subscripts and superscripts like the old one did? [sub][/sub] doesn't work. So I will use X_Y to denote X subscript Y. Likewise ^ means superscript, usually exponentiation.]

We are given Z_ab = Z = 88.8-j66.6. We aren't directly given V_ab, but note from the diagram that V_ab = V_an + V_nb, and that V_nb = - V_bn always. Negating a vector just means reflecting the vector through the origin. I.e. in polar form the magnitude is unchanged, and the angle is increased by 180 degrees (or decreased by 180 degrees, since a 360 degree change is no change at all). For polar form I'll use the notation a \ b to mean magnitude a and angle b. In other words, V = |V| \ ph V.

Let's call v = 34500 / (sqrt 3). So V_ab = v \ 0 + v \ 120 = v * (1\0 + 1\120).
OK, I got that wrong. V_bn = v \ -120, so V_nb = v \ 60, not v \ 120. I'll redo the previous two posts from that point presently.

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.2 (Take II) To find I_ab, we just need to properly use Ohm's Law, V_ab = I_ab * Z_ab [Does the new forum software support subscripts and superscripts like the old one did? [sub][/sub] doesn't work. So I will use X_Y to denote X subscript Y. Likewise ^ means superscript, usually exponentiation.]

We are given Z_ab = Z = 88.8-j66.6. We aren't directly given V_ab, but note from the diagram that V_ab = V_an + V_nb, and that V_nb = - V_bn always. Negating a vector just means reflecting the vector through the origin. I.e. in polar form the magnitude is unchanged, and the angle is increased by 180 degrees (or decreased by 180 degrees, since a 360 degree change is no change at all). For polar form I'll use the notation a \ b to mean magnitude a and angle b. In other words, V = |V| \ ph V.

Let's call v = 34500 / (sqrt 3). So V_ab = v \ 0 + v \ 60 = v * (1\0 + 1\60). You can do the vector addition either in rectangular form [1\0 = 1 + j0, 1\60 = 0.5 + j sqrt(3)/2] or geometrially [tip to tail, the two vectors form an isoceles triangle with an obtuse 120 degree angle], but either way 1\0 + 1\60 = sqrt(3) \ 30. Thus V_ab = 34500 \ 30

Now we want to multiply (rather divide) two vectors (rather complex numbers), one which we have in polar form (V_ab) and one in rectangular form (Z_ab). We could do the computation in either polar or rectangular coordinates, but since the result is requested in polar form, let's use polar form. So we need to convert Z_ab to polar form. In general the vector x + jy has the polar form sqrt(x^2+y^2) \ arctan(y/x) [draw out the right triangle in the complex plane]. We get Z_ab = 111 \ (arctan(-3/4) = -36.9 degrees).

Division requires computing the multiplicative inverse, and to do that to a complex number in polar form, we have to take the reciprocal of the magnitude, and then negate the angle. [As seen in the usual complex polar form R*e^(ja), the reciprocal is R^(-1) * e^(-ja)]. So Z_ab^(-1) = 1/111 \ 36.9

Finally, we have I_ab = V_ab / Z_ab = V_ab * (Z_ab^(-1)) = 34500 \ 30 * 1/111 \ 36.9. To multiply in polar form, we just multiply the magnitudes and add the angles. That means I_ab = 34500/111 \ 66.9 = 311 \ 66.9 (amps).

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.3 (Take II) Wikipedia tells me that the complex power S is defined as S = V * I', where I' is the complex conjugate of I (often denoted with an overbar, but I can't type that). For a load following Ohm's law we get S = Z * I * I'. Now I * I' = |I|^2 (the magnitude of I, squared), so S = Z * |I|^2.

For the single phase load Z_ab, we have already calculated I_ab. That means S_ab = (311)^2 * (88.8 -j66.6) = 8580 - j 6430 (kVA). The full 3 phase power S should just be S = 3 * S_ab, or S = 25700 - j 19300 kVA.

Cheers, Wayne
 

dkarst

Senior Member
Location
Minnesota
1.2 (Take II) To find I_ab, we just need to properly use Ohm's Law, V_ab = I_ab * Z_ab [Does the new forum software support subscripts and superscripts like the old one did? [sub][/sub] doesn't work. So I will use X_Y to denote X subscript Y. Likewise ^ means superscript, usually exponentiation.]

We are given Z_ab = Z = 88.8-j66.6. We aren't directly given V_ab, but note from the diagram that V_ab = V_an + V_nb, and that V_nb = - V_bn always. Negating a vector just means reflecting the vector through the origin. I.e. in polar form the magnitude is unchanged, and the angle is increased by 180 degrees (or decreased by 180 degrees, since a 360 degree change is no change at all). For polar form I'll use the notation a \ b to mean magnitude a and angle b. In other words, V = |V| \ ph V.

Let's call v = 34500 / (sqrt 3). So V_ab = v \ 0 + v \ 60 = v * (1\0 + 1\60). You can do the vector addition either in rectangular form [1\0 = 1 + j0, 1\60 = 0.5 + j sqrt(3)/2] or geometrially [tip to tail, the two vectors form an isoceles triangle with an obtuse 120 degree angle], but either way 1\0 + 1\60 = sqrt(3) \ 30. Thus V_ab = 34500 \ 30

Now we want to multiply (rather divide) two vectors (rather complex numbers), one which we have in polar form (V_ab) and one in rectangular form (Z_ab). We could do the computation in either polar or rectangular coordinates, but since the result is requested in polar form, let's use polar form. So we need to convert Z_ab to polar form. In general the vector x + jy has the polar form sqrt(x^2+y^2) \ arctan(y/x) [draw out the right triangle in the complex plane]. We get Z_ab = 111 \ (arctan(-3/4) = -36.9 degrees).

Division requires computing the multiplicative inverse, and to do that to a complex number in polar form, we have to take the reciprocal of the magnitude, and then negate the angle. [As seen in the usual complex polar form R*e^(ja), the reciprocal is R^(-1) * e^(-ja)]. So Z_ab^(-1) = 1/111 \ 36.9

Finally, we have I_ab = V_ab / Z_ab = V_ab * (Z_ab^(-1)) = 34500 \ 30 * 1/111 \ 36.9. To multiply in polar form, we just multiply the magnitudes and add the angles. That means I_ab = 34500/111 \ 66.9 = 311 \ 66.9 (amps).

Cheers, Wayne
Thanks for helping MBrooke out, I had done some of the calcs but hadn't typed anything in.

You are correct on the power factor.

I agree with the phase current as well as I calculated 310.8 /_ 67º Amps

I agree with your calc on complex power as well.

One thing interesting is to calc the current which is just V/Z you said "Division requires computing the multiplicative inverse, and to do that to a complex number in polar form, we have to take the reciprocal of the magnitude, and then negate the angle"

Engineers when they have both numerator and denominator in polar form just divide the magnitude and the angle is numerator - denominator. I know in the end both techniques give the same answer... just found it interesting.

Thanks again for helping MBrooke out.
 

wwhitney

Senior Member
Location
Berkeley, CA
1.4 Conservation of charge (Kirchhoff's first law) applied to the lower left node of the delta connected loads tells us that I_a = I_ab + I_ac, or I_a = I_ab - I_ca. I_ca will of course depend on V_ca just as I_ab depended on V_ab. And we have V_ca = V_cn - V_an, just like V_ab = V_an - V_bn.

Note that the inputs to our (as yet unsolved) equation V_ca = V_cn - V_an are almost the same as the inputs to our (solved) equation V_ab = V_an - V_bn. The only difference is that V_cn is rotated 120 degrees CCW from V_an (the corresponding input), and V_an is rotated 120 degrees CCW from V_bn ( the corresponding input). So if we went through the trouble to solve for V_ca and then I_ca, we'd just find the results to be 120 degrees rotated CCW from our previous results.

This shows that I_ca = 311 \ 186.9 (amps). Then I_a = 311 \ 66.9 - 311 \ 186.9 = 311 \ 66.9 * (1 \ 0 - 1 \ 120). Now 1 \ 0 - 1 \ 120 = 1 \ 0 + 1 \ -60, and we previously saw that 1 \ 0 + 1 \ 60 = sqrt(3) \ 30. That means 1 \ 0 + 1 \ -60 = sqrt(3) \ -30.

Thus I_a = 311 * sqrt(3) \ (66.9 - 30) = 539 \ 36.9

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.5 So this depends on knowing what the "Delta-Wye Standard -30 degree phase shift" Is. I'm not 100% sure, but I think it means that the currents and voltages on the secondary (wye) side have a -30 degree phase shift from the corresponding current and voltages on the primary (delta) side.

If so, then if the transformer were 1:1, we'd have I_A = I_a * 1 \ 30 = 539 \ 66.9. As the transformer is instead 4:1, the current I_A will be 1/4 as much, or
135 \ 66.9.

[Or is this off by -1, because I_A is drawn with the arrow pointing towards the primary, and I_a is drawn with the arrow pointing away from the secondary?]

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
1.6 I again invoke conservation of charge, this time at the lower right transformer primary. That tells us I_A = I_AB + I_AC = I_AB - I_CA as usual.

In the same way that I_ca = I_ab * 1\120, it follows that I_CA = I_AB * 1\120. So we get I_A = I_AB * (1\0 - 1\120) = I_AB * (sqrt(3) \ -30) as we prevously saw.

Thus I_AB = 135 \ 66.9 / sqrt(3) \ -30 = 77.8 \ 96.9 (amps). We could also have gotten this answer starting with I_ab, dividing the magnitude by 4, and then phase shifting 30 degrees CCW.

[And again, perhaps I'm off by -1 for the same reason as in 1.5.]

Cheers, Wayne
 

Hv&Lv

Senior Member
Location
-
Occupation
Engineer/Technician
Thank you- I'm digesting this now.

Can anyone tell me where this would be used in the real world?
I didn’t go back and read the posts, quite frankly just Don’t want to..

but your question specifically.
Some relays use rectangular, some use polar.
SEL relays use polar notation, the old form 6 use rectangular I believe.

how do you like to add vectors?
personally, I like rectangular, but that’s the method hammered home in school.

now my spreadsheet does it for me.


just to expand a bit more..
your setting up a Distance relay. You have your wire impedance, distance, and the nominal spacing.
wha values are you going to input?
your resistance and reactance has to be calculated by length and separation to be entered so your relay calculates fault distance correctly.
 

mbrooke

Batteries Not Included
Location
United States
Occupation
*
Correct and correct. However, I'm assure whether to use cold conductor temps or loaded temps. As you know heat and sag effects what the relays sees during a fault. STE + fault is a concern for me (after all, thats when sag shorts in the right of way 😉) Mutual coupling causes zone 1 ground to under reach and zone 2 ground to over reach (if I have it right) 120% Z typical value won't work- 150% I've figured.
 
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