# Calculating Phase Currents Based on Measured Line Currents - Delta Heater Load

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#### G._S._Ohm

##### Senior Member
Seems like one load is shorted.

Assuming 58 A line current at zero degrees, 58/0
and 76 A line current at 120 degrees, 76/120
and assuming a fault in only one of three loads [Ra, Rb, Rc]
which means Rb = Rc
then 58/0 = (1/Ra)240/0 - (1/Rb)240/120
and 76/120 = (1/Rb)240/120 -(1/Rb)240/240

I get Rb = Rc = 5.47 ohms and then Ra = 0.18 ohm.

If I made no mistakes in going from polar to rectangular and back again, this should be right.
In any case, with the formulas and assumptions stated above, this can be checked.

I'm out of here. . .

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#### Smart \$

##### Esteemed Member
Are the answers that my spreadsheet comes up with for Iab, Ibc, and Ica correct?

If so, I cannot get the wye-delta transformation at the bottom to work out. The phase currents from the wye-delta transformation do not match what the excel solver came up with for Iab, Ibc, and Ica.

I'm thinking the wye-delta transformation only works with instantaneous data.

#### gar

##### Senior Member
110122-0820 EST

The WYE-DELTA transform that I suggested will not solve this problem. In this transform the center point of the WYE moves away from a theoretical neutral as the individual impedances are unbalanced. Sorry for the misdirection.

.

#### mivey

##### Senior Member
110122-0820 EST

The WYE-DELTA transform that I suggested will not solve this problem. In this transform the center point of the WYE moves away from a theoretical neutral as the individual impedances are unbalanced. Sorry for the misdirection.

.
I apologize for not reading the thread but you are not suggesting that an impedance network must be balanced for the wye-delta impedance transform to work are you?

#### gar

##### Senior Member
110122-0954 EST

Mivey:

The WYE-DELTA transform works, but the mid point of the WYE in an unbalanced condition is not the mid point of the line voltages.

Consider a delta consisting of two 100 ohm resistors and 1 megohm for the third resistor.
Let
Rab = 100
Rbc = 100
Rca = 1,000,000
Sum = Rab + Rbc + Rca

Then the result of the transform is
Ra = Rab * Rca / Sum ~= Rab
Rb = Rab * Rbc / Sum ~= 0
Rc = Rbc * Rca / Sum ~= Rbc

Certainly makes sense. Sort of self-evident. And it shows that the wye midpoint shifted to very close to line B.

.

#### jakeself

##### Member
Seems like one load is shorted.

Assuming 58 A line current at zero degrees, 58/0
and 76 A line current at 120 degrees, 76/120
and assuming a fault in only one of three loads [Ra, Rb, Rc]
which means Rb = Rc
then 58/0 = (1/Ra)240/0 - (1/Rb)240/120
and 76/120 = (1/Rb)240/120 -(1/Rb)240/240

I get Rb = Rc = 5.47 ohms and then Ra = 0.18 ohm.

If I made no mistakes in going from polar to rectangular and back again, this should be right.
In any case, with the formulas and assumptions stated above, this can be checked.

I'm out of here. . .
I do not believe one load is shorted. With that scenario of 58A, 76A, 58A for line currents: My excel solver comes up with Rab = Rbc = 5.47 ohms & Rca = 11.1 ohms.

This was just an example of what one scenario might be for line currents. Let me know if you come up with equations to do it. I would like equations to calculate the phase currents on the fly within a PLC, so I can dynamically calculate what heater wattage is connected at the end of each shift. This would let us notify maintenance if any elements have degraded. The excel solver does seem to do it, I would just like equations to calculate it dynamically.

#### mull982

##### Senior Member
Because all of the angles are locked down, this can be solved for magnitude only by inserting a constant of 1/.866=1.15 to the line currents.

1.15 * IA = Iab + Iac
1.15 * IB = Ibc + Iba
1.15 * IC = Ica + Icb

I am pretty sure this is correct. The phase currents through each element will be slightly higher than using a raw magnitude calculation, so this does appear to make sense.
I'm not seeing where your getting the 1.15 multiplier for the line currents from.

So are your 3 equations solvable since we know the line currents? Would you solve by substitution or matrix math? Wouldn't we have 6 unknown's here since for instance Iab is not the same as Iba?

So it appears as Gar said that we cannot use the delta/wye transform because the unbalanced currents pull the neutral point away from zero? I guess this is becasue there is no neutral on the tie the wye down so it is floating.

#### david luchini

##### Moderator
Staff member
I'm not seeing where your getting the 1.15 multiplier for the line currents from.
I don't understand the 1.15 multiplier either. KCL tells us that Ia=Iab+Iac, then how could 1.15*Ia = Iab+Iac at the same time?

So are your 3 equations solvable since we know the line currents? Would you solve by substitution or matrix math? Wouldn't we have 6 unknown's here since for instance Iab is not the same as Iba?
I'd say we still only have 3 unknowns, because we know that Iab=-Iba, etc.

You could write the 3 equations as:

Ia=Iab-Ica
Ib=Ibc-Iab
Ic=Ica-Ibc

#### jakeself

##### Member
Just to give a little more info - I have some metering class CT's ordered so I can get more accurate line current measurements.

We have four three-pole SCRs feeding banks of 1000W IR heating elements. Two of the SCRs each feed 30 elements and two of them each feed 20 elements, 100 elements total. Originally, the machine was wired where the 30-element bays were balanced 10, 10, & 10 parallel between phases. Just using a hand-held RMS current clamp we took some readings of 52.7A, 71.3A, 52.6A on one of the 30-element bays. I have attached a screenshot of the solver result for this example. It could be inferred that there are 5-6 elements bad between phases C & A.

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#### gar

##### Senior Member
110122-1230 EST

What is a three-pole SCR? What is the schematic? Is this phase shift controlled or is the SCR either on the entire +1/2 cycle or totally off for the +1/2 cycle? Obviously an SCR is off on the negative half cycle.

.

#### Rick Christopherson

##### Senior Member
I'm not seeing where your getting the 1.15 multiplier for the line currents from.
The 1.15 comes from 1/cos(30). This term comes from the angle between the phase current(s) and the line current.

The original equations I posted were vector equations. I later reposted and said to solve it for magnitude only you need to add the 1.15 multiplier.
So are your 3 equations solvable since we know the line currents? Would you solve by substitution or matrix math? Wouldn't we have 6 unknown's here since for instance Iab is not the same as Iba?
How you solve them is a matter of personal preference. I don't use matrix math often enough, so I just use substitution.
So it appears as Gar said that we cannot use the delta/wye transform because the unbalanced currents pull the neutral point away from zero? I guess this is becasue there is no neutral on the tie the wye down so it is floating.
I have no idea what this particular transform is about, so I can't speak to it. However, in a sense, I did the same thing when I inserted the 1.15 factor into my magnitude equations.

I do not believe one load is shorted. With that scenario of 58A, 76A, 58A for line currents: My excel solver comes up with Rab = Rbc = 5.47 ohms & Rca = 11.1 ohms.
Given the original currents, I would not expect one resistance to be higher, but rather, it should be lower.

Given the currents listed above, I got resistances of Rab = Rbc = 5.47 and Rca = 2.16. (Actually, my A, B, C's were switched around, so I wrote them to match yours to avoid confusion.)

By the way, I didn't see your posting with the screen capture in it until now.

#### david luchini

##### Moderator
Staff member
The 1.15 comes from 1/cos(30). This term comes from the angle between the phase current(s) and the line current.

The original equations I posted were vector equations. I later reposted and said to solve it for magnitude only you need to add the 1.15 multiplier.
I think 1.15 would work fine in a balanced case, but not in the unbalanced loads in the OP.

In the vector form of Ia=Iab+Iac, with Iab=10<0 and Iac=10<-60, then Ia is solved to be 17.32<-30. Ia is 30 degrees from Iab and from Iac.

If the load is not balanced, such as Iab=10<0 and Iac=5<-60, then Ia is solved to be 13.23<-19.1. The 30 degree separation is not maintained in the unbalanced condition.

Solving the unbalanced load with the magnitude only equation: Ia=(Iab+Iac)/1.15
yields Ia=13.04. It is close to the actual value of 13.23, but not exact.

#### Rick Christopherson

##### Senior Member
I think 1.15 would work fine in a balanced case, but not in the unbalanced loads in the OP.

In the vector form of Ia=Iab+Iac, with Iab=10<0 and Iac=10<-60, then Ia is solved to be 17.32<-30. Ia is 30 degrees from Iab and from Iac.

If the load is not balanced, such as Iab=10<0 and Iac=5<-60, then Ia is solved to be 13.23<-19.1. The 30 degree separation is not maintained in the unbalanced condition.

Solving the unbalanced load with the magnitude only equation: Ia=(Iab+Iac)/1.15
yields Ia=13.04. It is close to the actual value of 13.23, but not exact.
There is a mistake somewhere in my calculations. For working with the magnitude, I missed a minus sign. Once this was discovered, I couldn't resolve the equations. I started working in vectors, but the complexity is too high for a Saturday, so I need to walk away from this for now.

#### jakeself

##### Member
110122-1230 EST
What is a three-pole SCR? What is the schematic? Is this phase shift controlled or is the SCR either on the entire +1/2 cycle or totally off for the +1/2 cycle? Obviously an SCR is off on the negative half cycle.
.
It is a three phase SCR. The SCR is either ON or OFF. It is driven from a PLC output. It has a 1-second cycle and is ON for a percentage of that 1-second. So if it is set to 60% it will be ON for 0.6 seconds and OFF for 0.4 seconds, repeatedly. We are planning to hold it ON solid, not pulsing, when taking the current measurement.

#### iceworm

##### Curmudgeon still using printed IEEE Color Books
110122-0820 EST

The WYE-DELTA transform that I suggested will not solve this problem. In this transform the center point of the WYE moves away from a theoretical neutral as the individual impedances are unbalanced. Sorry for the misdirection.

.
I apologize for not reading the thread but you are not suggesting that an impedance network must be balanced for the wye-delta impedance transform to work are you?
110122-0954 EST

Mivey:

The WYE-DELTA transform works, but the mid point of the WYE in an unbalanced condition is not the mid point of the line voltages.

Consider a delta consisting of two 100 ohm resistors and 1 megohm for the third resistor.
Let
Rab = 100
Rbc = 100
Rca = 1,000,000
Sum = Rab + Rbc + Rca

Then the result of the transform is
Ra = Rab * Rca / Sum ~= Rab
Rb = Rab * Rbc / Sum ~= 0
Rc = Rbc * Rca / Sum ~= Rbc

Certainly makes sense. Sort of self-evident. And it shows that the wye midpoint shifted to very close to line B.

.
gar -
Help me out. For a Y-D transformation, I'm looking at this as resistor networks.

I'll start assumming the impedances are all resistive. I don't know the Line to Line voltage, so I'll just call it VLL. And the Voltage (call it 'V') from each node to the Y-point is VLL/(√3)

See the attached sheet for the algebra.

And, since I am just dealing with the current and voltage magnitudes to come up with the Y resistances, this is not a vector math problem. And there should not be an issue with the imaginary Y-point being displaced.

Except this does fall apart if I1, I2, or I3 are zero. So except for that case what am I missing?

This is nothing I do for work, it is just an interesting algebra problem using the Y-D transformation.

What say ye (all are invited)?

ice

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#### iceworm

##### Curmudgeon still using printed IEEE Color Books
See the attached sheet for the algebra.
uhhhh..... can't seem to attach the sheet. Though I had, but it didn't. Trying again.

ice

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#### skeshesh

##### Senior Member
Looks good to me ICE, why do you suspect that you're missing something? When the current is 0 there are methods of solving the problem such as letting the limit of the value of the variable (current) go to 0 and evaluating the function using L'Hopital's rule or by graphing the function to see it's behavior while approaching 0. The problem is much more difficult when you have other variables such as than resistive loads,varying phase angles, etc. Take a look at the link Smart \$ posted on page 1, that was a fun thread that was going on for a while...

#### iceworm

##### Curmudgeon still using printed IEEE Color Books
Looks good to me ICE, why do you suspect that you're missing something? When the current is 0 there are methods of solving the problem such as letting the limit of the value of the variable (current) go to 0 and evaluating the function using L'Hopital's rule or by graphing the function to see it's behavior while approaching 0. xxx..
Well - cause gar is pretty good and usually has well thought out responses. And actually when I1, or I2, or I3 go to zero, the problem gets mathematically trivial. Two legs of the Delta are open. So that one really doesn't matter.

gar's issue (I think) is if one leg of the Delta is open.

But I had not thought about "LesHospital" (Differentiate the two functions then divide them - I didn't look it up, is that close to right?)

I have to give this up and go to work for a couple of weeks. I'll check back in and see how you did with it. I think you are on a good track.

ice

#### Smart \$

##### Esteemed Member
...

This was just an example of what one scenario might be for line currents. Let me know if you come up with equations to do it. I would like equations to calculate the phase currents on the fly within a PLC, so I can dynamically calculate what heater wattage is connected at the end of each shift. This would let us notify maintenance if any elements have degraded. The excel solver does seem to do it, I would just like equations to calculate it dynamically.
The geometric solution is the First Fernat Point I mentioned in the other thread I linked to in post #4.

I started to derive the equations but I currently don't have the time to complete them. Diagram I was working from is below (and measures of XA, XB, and XC match the solver answers in Excel). I am under the impression the basis for the equations is several steps of using the law of cosines to determine angles and leg lengths. I do not understand the notation used on the mathworld page, but perhaps someone can make sense of it.

#### skeshesh

##### Senior Member
Yea L'Hopital is basically saying if a function approaches a value as the variable goes to 0 then the derivative of the function will also approach the same value as it goes to zero. It's used to evaluate indefinite fractions, take a look when you get a chance it's basic calculus. I agree with you btw, it is always a pleasure reading posts by Gar as well as many other contributors to this forum who are better educated and experienced that I dare wish becoming.

Smart - Why dont you use equilateral triangles to solve the problem? Of course we're talking about unbalanced phases but if we solve it using an equilateral triangle but in terms of variables the result will be the same (I think).

I'm not sure I'm upto it at this point (between work and etc. there's little room for recreational math time!) but maybe I'll attempt it at some point. Probably won't take all that long if you use matlab or scilab to solve for the variables and test the results by plugging in some known values...

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