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

Status
Not open for further replies.

jakeself

Member
I have an unbalanced IR Heater load setup as a delta load. Voltage is 3-phase 240VAC. There are multiple heaters in parallel between each pair of phases. We are simply looking for a method to tell how many bad heating elements there are.

I can measure the line currents easily, but I believe I need to convert these to phase currents to properly find wattage used between two phases. I can calculate line current magnitudes based on phase currents assuming 120 degrees of separation between phases. But, I cannot figure out how to calculate phase current magnitudes based on line currents. Using a recommendation I read on here from forum member topgone, I created an excel sheet that will use the add-in "solver" to find the phase currents required to match up with our measured RMS line currents.

While the excel solver does seem to work, I cannot figure out the equations to do it mathematically. Can anyone help me with a mathematical method to calculate phase currents from measured RMS line currents with an unknown-unbalanced delta load?
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
I recall a post on this not too long ago. It is impossible to calculate the phase currents of an unbalanced delta load knowing just the line current values.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110120-2104 EST

Go to page 487 of "Basic Electrical Measurements", Melville B. Stout, 1950. Here are equations for WYE-DELTA TRANSFORMATIONS.

Since you have a resistive load you can calculate the impedances for the WYE load from line current and 138.6 V. Then perform the transformation to get your delta impedances.

.
 

jakeself

Member
110120-2104 EST
Go to page 487 of "Basic Electrical Measurements", Melville B. Stout, 1950. Here are equations for WYE-DELTA TRANSFORMATIONS.
Since you have a resistive load you can calculate the impedances for the WYE load from line current and 138.6 V. Then perform the transformation to get your delta impedances.
.

I believe this only works for a balanced load. I can only get it to match if the line currents are equal. Perhaps it is because the line-neutral voltages are not equal for an unbalanced load.
 

jakeself

Member
Here is a link to my spreadsheet
https://docs.google.com/leaf?id=0Bx...IyOWYtMDllODgyNWRmNjAy&hl=en&authkey=CNrTzt0F

If you have the add-in "solver" enabled, you can update the orange cells with the measured line currents and voltages. After running the solver, it will find the necessary phase currents to create those line currents.

I added a section at the bottom where I attempted finding the Wye impedances first. The Wye-Delta transformation only matches when the line currents are balanced.
 

G._S._Ohm

Senior Member
Location
DC area
I have an unbalanced IR Heater load setup as a delta load. Voltage is 3-phase 240VAC. There are multiple heaters in parallel between each pair of phases. We are simply looking for a method to tell how many bad heating elements there are.

I can measure the line currents easily,
I cannot figure out the equations to do it mathematically.
Good problem.

The source is Y?
The phase/line voltage is 240/416 V?

What are your line currents?
What are the maximum heater wattages when they work?
I need actual values to check my work.

I get a boatload of hits on Google for unbalanced 3 phase systems and my text implies the analysis of this is only slightly more difficult than a balanced system. It may involve solving three simultaneous equations.
You should hear back in a day or two.
 
Last edited:

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
xxx. It may involve solving three simultaneous equations. xxx.
I did not give this a lot of thought (disclaimer - I am neither above nor below being screwed up)
Assuming the loads are resistive, aren't there 6 unknowns:
phase angles of the line currents, magnitudes of the single phase delta currents - that is six.

ice
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
xxx The Wye-Delta transformation only matches when the line currents are balanced.
That should not be the case. The theory (model) holds for a gernric unbalanced cases. Try a google search on "Wye Delta transformation". One of the top hits I saw looked pretty generic/unbalanced to me. It was a Wiki - I am not a fan of anything in Wiki being gospel - but it looked pretty through.

ice
 

Rick Christopherson

Senior Member
Maybe I am not understanding how this is connected, but if the load is Delta, then it makes no difference whether the source is wye or delta, the problem remains the same. The connection between source and load is Delta-Delta.

The problem is simple, and is 3-equations with 3-unknowns. This is resistive, so phase angle can be assumed to be constant. Use Kirchoff at each node.

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

mull982

Senior Member
Maybe I am not understanding how this is connected, but if the load is Delta, then it makes no difference whether the source is wye or delta, the problem remains the same. The connection between source and load is Delta-Delta.

The problem is simple, and is 3-equations with 3-unknowns. This is resistive, so phase angle can be assumed to be constant. Use Kirchoff at each node.

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

Your equations make sense but my matrix math is extremely rusty. Can you solve for the unknown phase currents by just having the 3 line current?
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
Maybe I am not understanding how this is connected, but if the load is Delta, then it makes no difference whether the source is wye or delta, the problem remains the same. The connection between source and load is Delta-Delta.

The problem is simple, and is 3-equations with 3-unknowns. This is resistive, so phase angle can be assumed to be constant. Use Kirchoff at each node.

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

It's true the phase angle of Iab = the phase angle of Vab. So what is the phase angle of Ia, Ib, Ic?

Another way to say this is:
You listed 6 variables. Each is a vector (or 'phasor' if you like) with a phase angle and a magnitude.

The magnitudes of Ia, Ib, Ic are known (measured). The phase angles of Iab, Ibc, Ica are known. They are equal to the phase angles of Vab, Vbc, Vca respectively.

We don't know the phase angles of Ia, Ib, Ic. For example, let's look at Ia. If Rab is infinite (all ab heaters open), then Ia = Ica and phase angle Ia = phase angle Vca. If Rca is infinite( all ca heaters open), then phase angle Ia = phase angle Vab. The phase angle of Ia if very much dependent on the unbalance.

There are six unknowns.

ice
(I got to get back to work - later guys and girls)
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
xxx my matrix math is extremely rusty. Can you solve for the unknown phase currents by just having the 3 line current?
For resistive elements, the circuit is determinate. Yes it is complex matrix math - not bad, but somewhat messy.

The wye-delta transformation gar spoke of is a lot easier to see.

ice
(really this one is it for a while):)
 

mull982

Senior Member
For resistive elements, the circuit is determinate. Yes it is complex matrix math - not bad, but somewhat messy.

The wye-delta transformation gar spoke of is a lot easier to see.

ice
(really this one is it for a while):)

So we are saying that we can use the Delta to wye impedance transformation even for unbalanced impedances?

I guess that does make sense considering the equations for the conversions do factor in all the phase impedances.
 

Rick Christopherson

Senior Member
It's true the phase angle of Iab = the phase angle of Vab. So what is the phase angle of Ia, Ib, Ic?
Because it is a resistive system, all of the phase angles are locked into their respective voltage's angles. There are no reactive components available to produce a phase shift.

The problem is slightly more complex than I made it seem, but I believe at a minimum, solving it using just magnitudes might give you a proportional answer, which is what the OP is looking for.

I am looking at it a little closer, but I have this feeling that a squareroot-3 is all that is missing between proportional and absolute numbers. It has been way too many years since I have done this.
 

Rick Christopherson

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.
 

jakeself

Member
Here is a link to my spreadsheet
https://docs.google.com/leaf?id=0Bx...IyOWYtMDllODgyNWRmNjAy&hl=en&authkey=CNrTzt0F

If you have the add-in "solver" enabled, you can update the orange cells with the measured line currents and voltages. After running the solver, it will find the necessary phase currents to create those line currents.

I added a section at the bottom where I attempted finding the Wye impedances first. The Wye-Delta transformation only matches when the line currents are balanced.

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.
 
Last edited:

G._S._Ohm

Senior Member
Location
DC area
Correction: if you can give me the line currents and load voltages I should be able to tell you all the load resistances. It's going to be two or three pages of equations.
 
Last edited:
Status
Not open for further replies.
Top