Load Calculation for Transformer Secondary (Delta) with unbalanced single phase loads

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dinos

Member
Greetings,

I've attached a sketch of a theoretical problem whose precise solution may be obvious. though a technical detail or two is eluding me.

In the end, I'm interested in calculating the total current flowing in lines L1, L2, L3 and in resistive loads RA, RB, RC (the winding currents WA, WB, and WC are also of interest but I presume they are the same as RA, RB, RC?)

Thanks in advance for any guidance...
 

bob

Senior Member
Location
Alabama
Greetings,

I've attached a sketch of a theoretical problem whose precise solution may be obvious. though a technical detail or two is eluding me.

In the end, I'm interested in calculating the total current flowing in lines L1, L2, L3 and in resistive loads RA, RB, RC (the winding currents WA, WB, and WC are also of interest but I presume they are the same as RA, RB, RC?)

Thanks in advance for any guidance...

The phase currents are 480/RA, 480/RB and 480/RC. The currents are not in phase so you will have to add them vectorially to get the line currents.
 

dinos

Member
The phase currents are 480/RA, 480/RB and 480/RC. The currents are not in phase so you will have to add them vectorially to get the line currents.

This is one instance of where I have a question...

In terms of Winding WB, doesn't it produce 480V/RB=10A through RB in addition to 480V/(100+1)=4.75A through RA and RC which are in parallel with RB?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
081229-2021 EST

dinos:

Redraw your drawing as 3 secondaries not connected together. Connect each load resistor across its secondary. You will get the results described by bob in post #2.

Next connect the L2 end of L1-L2 secondary to the L2 end of L2-L3 secondary. Has this changed any current flows? No.

Now connect the L3 end of L3-L1 secondary to the L3 end of L2-L3 secondary. This leaves no connection between L1 of L3-L1 and L1 of L1-L2. Has this changed any current flows? No.

What is the difference voltage between L1 of L3-L1 and L1 of L1-L2?

Now you should be able to answer your own question.

Do you know how to do the vector addition of the currents that bob mentioned?

.
 

dinos

Member
081229-2021 EST
Redraw your drawing as 3 secondaries not connected together. Connect each load resistor across its secondary. You will get the results described by bob in post #2.
Next connect the L2 end of L1-L2 secondary to the L2 end of L2-L3 secondary. Has this changed any current flows? No.
Now connect the L3 end of L3-L1 secondary to the L3 end of L2-L3 secondary. This leaves no connection between L1 of L3-L1 and L1 of L1-L2. Has this changed any current flows? No.
To this point I'm with you...

What is the difference voltage between L1 of L3-L1 and L1 of L1-L2?
My guess is 480...

Now you should be able to answer your own question.
Unfortunately, I'm still stuck. I redrew the diagram in a fashion that relates to your step-by-step suggestion (attached). When I get to the last connection at L1 it seems to me that a 2nd resistive path (drawn in red) from the '+' side of WB, through L2, through RC, through L3, through RA, and through L1 to the '-' side of WB is then established.

Do you know how to do the vector addition of the currents that bob mentioned?
Yes...I just can't seem to wrap my head around there not being current in that 2nd "red path" from the WB winding.
 

bob

Senior Member
Location
Alabama
To this point I'm with you...

My guess is 480...

Unfortunately, I'm still stuck. I redrew the diagram in a fashion that relates to your step-by-step suggestion (attached). When I get to the last connection at L1 it seems to me that a 2nd resistive path (drawn in red) from the '+' side of WB, through L2, through RC, through L3, through RA, and through L1 to the '-' side of WB is then established.

Yes...I just can't seem to wrap my head around there not being current in that 2nd "red path" from the WB winding.
Look at the way the loads are shown in the 1st dwg at this link. It may clear up your question.
http://www.tpub.com/content/et/14093/css/14093_55.htm
 

dinos

Member
Look at the way the loads are shown in the 1st dwg at this link. It may clear up your question.http://www.tpub.com/content/et/14093/css/14093_55.htm
In looking at that drawing, I still don't see why the current from a given winding wouldn't follow 2 paths through the connected load.

As an example, if I look at winding 'a' in the attached drawing, I know there is 480V across it.

Resistor 'a' has that same 480V applied across its terminals.

Why then would a 480V voltage applied at the terminals of resistor 'a' not cause current to flow through resistors 'b' and 'c'?
 

wirenut1980

Senior Member
Location
Plainfield, IN
In looking at that drawing, I still don't see why the current from a given winding wouldn't follow 2 paths through the connected load.

As an example, if I look at winding 'a' in the attached drawing, I know there is 480V across it.

Resistor 'a' has that same 480V applied across its terminals.

Why then would a 480V voltage applied at the terminals of resistor 'a' not cause current to flow through resistors 'b' and 'c'?


Remember Kirchoff's Current Law: The sum of the currents entering a node must equal the sum of current leaving a node.

It will also help to assign current directions to your diagram for clarity. Let:

Ia = 480V/1 ohm = 480 Amps @ 0 degrees (flowing left to right across Ra)

Ib = 480V/10 ohms = 48 Amps @ 120 degrees (flowing down to up across Rb)

Ic = 480V/100 ohms = 4.8 Amps @ -120 degrees (flowing up to down across Rc)

Let I_L1, I_L2, and I_L3 flow from left to right.

Using KCL:

I_L1 = Ia - Ib

I_L2 = Ib - Ic

I_L3 = Ic - Ia


Clear as mud?:D
 

mull982

Senior Member
In looking at that drawing, I still don't see why the current from a given winding wouldn't follow 2 paths through the connected load.

As an example, if I look at winding 'a' in the attached drawing, I know there is 480V across it.

Resistor 'a' has that same 480V applied across its terminals.

Why then would a 480V voltage applied at the terminals of resistor 'a' not cause current to flow through resistors 'b' and 'c'?

dinos

What you are thinking about the two current paths being in parallel would be true if the delta configuration were fed with a single phase source. Lets say that you only had a single phase source represented by WB. Therefore this source would produce a voltage across RB, and the combination of RA+RC thus having two parallel current paths and thus what you are visualizing.

Because the source is a three phase source this changes things a bit. If you look at the picture that Bob posted you can see that with three legs from the source the current coming in on L1 will split across both RA and RB and thus return on Lines L2 and L3 rather then going across the parallel path you are describing.
 

dinos

Member
I'm with you on all counts except for one concept which isn't sinking into my stubborn brain...
Remember Kirchoff's Current Law:
Ia = 480V/1 ohm = 480 Amps @ 0 degrees (flowing left to right across Ra)
Ib = 480V/10 ohms = 48 Amps @ 120 degrees (flowing down to up across Rb)
Ic = 480V/100 ohms = 4.8 Amps @ -120 degrees (flowing up to down across Rc)

I see the winding currents as:
Ia = 480V/1 ohm + 480V/110 ohms = 484.4 Amps
Ib = 480V/10 ohms + 480V/101 ohms = 52.8 Amps
Ic = 480V/100 ohms + 480V/11 ohms = 48.4 Amps

I've revised my original sketch (attached) to try and show again how I am coming up with 52.8 amps for winding Ib instead of 48 amps. In the sketch I've broken the L3 connection to the load, leaving only L1 and L2 connected. The winding Ib current in this case must be 52.8 amps (if it isn't I'm in real trouble!). So my question is, if I re-establish the L3 connection, why would the current in the Ib winding go from 52.8 amps down to 48 amps?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
081230-1300 EST

dinos:

If I take a theoretically perfect delta source and open one corner, then the voltage difference between the two points formed by opening the corner is zero. If it was not zero what do you think would happen?

Thus, whether that corner is open or closed makes no difference in the currents in the individual loads. You do not consider the delta load resistors connected in parallel when connected to the delta source.

Any time you have a voltage source, AC or DC, it is assumed to have an internal impedance of 0. In the real world all power sources have an internal impedance that is not zero, but the internal impedance is usually small compared to the load, and in an analysis may be assumed to be 0.

Any loads across an ideal voltage source have a current thru each load defined by the voltage source voltage and the impedance of the load.

.
 

wirenut1980

Senior Member
Location
Plainfield, IN
I'm with you on all counts except for one concept which isn't sinking into my stubborn brain...


I see the winding currents as:
Ia = 480V/1 ohm + 480V/110 ohms = 484.4 Amps
Ib = 480V/10 ohms + 480V/101 ohms = 52.8 Amps
Ic = 480V/100 ohms + 480V/11 ohms = 48.4 Amps

I've revised my original sketch (attached) to try and show again how I am coming up with 52.8 amps for winding Ib instead of 48 amps. In the sketch I've broken the L3 connection to the load, leaving only L1 and L2 connected. The winding Ib current in this case must be 52.8 amps (if it isn't I'm in real trouble!). So my question is, if I re-establish the L3 connection, why would the current in the Ib winding go from 52.8 amps down to 48 amps?

A delta connected three phase load is not the same as a single phase voltage source feeding parallel resistances. A delta connected load can be visualised as 3 single phase voltage sources and 3 single phase resistances separated from each other. When connected together to form a three phase delta load, they are not added arithmetcally, but rather vectorally because of the 3 phases are hitting their peak of the sine wave at different times.

I also had a hard time visualising this coming out of college after looking at single phase circuits all my life. Keep at it!:smile:
 

mull982

Senior Member
Assuming the load currents to be the following:

RA= 480A @ 0deg
RB= 48A @ -120deg
RC= 4.8A @ -240deg

I come up with the following line current for L1,L2,L3. These currents are assuming unity power factor:

L1 = 506A @ -26deg
L2 = 50.6A @ -145deg
L3 = 486A @ -210deg
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
081231-0855 EST

All loads are assumed to be resistive.

At each line there are two current vectors differing by 120 deg. Thus the vector sum at a line is
Isum = ( (Ia+Ib/2)^2 + (0.866025*Ib)^2 )^-2
where Ia is current of one load and Ib is current of the other load
Ib/2 comes from sin 30 = 1/2, and 0.866025*Ib is from sin 60 = 0.866025 .

I1 = ( (480+24)^2 + (0.866025*48)^2 )^-2
I1 = ( (504)^2 + (41.5692)^2 )^-2
I1 = ( 254016 + 1728 )^-2
I1 = ( 255744 )^-2
I1 = 505.711

I2 = 50.571 This logically follows from I1 because the resistances are both 10 times larger.

I3 = 482.42

I do not know why mull982 and I have a different answer for I3.

.
 

dinos

Member
081231-0855 EST
I1 = ( (480+24)^2 + (0.866025*48)^2 )^-2
I1 = ( (504)^2 + (41.5692)^2 )^-2
I1 = ( 254016 + 1728 )^-2
I1 = ( 255744 )^-2
I1 = 505.711

I2 = 50.571 This logically follows from I1 because the resistances are both 10 times larger.

I3 = 482.42
I used a formula based on the law of cosines for the calculation and came up with the same values...

I1 = (480^2+48^2 - 2*480*48*Cos(120))^-2

I1 = 505.7
I2 = 50.6
I3 = 482.4
 

mull982

Senior Member
081231-0855 EST

All loads are assumed to be resistive.

At each line there are two current vectors differing by 120 deg. Thus the vector sum at a line is
Isum = ( (Ia+Ib/2)^2 + (0.866025*Ib)^2 )^-2
where Ia is current of one load and Ib is current of the other load
Ib/2 comes from sin 30 = 1/2, and 0.866025*Ib is from sin 60 = 0.866025 .

I1 = ( (480+24)^2 + (0.866025*48)^2 )^-2
I1 = ( (504)^2 + (41.5692)^2 )^-2
I1 = ( 254016 + 1728 )^-2
I1 = ( 255744 )^-2
I1 = 505.711

I2 = 50.571 This logically follows from I1 because the resistances are both 10 times larger.

I3 = 482.42

I do not know why mull982 and I have a different answer for I3.

.

I am rusty with the math so I took the lazy way and just plotted the different phasors in AutoCAD. I may have accidently entered one of the angles wrong.

I have always seen that for a balanced 3-phase delta load the current in the delta windings were always 1.73 times less then the line currents (example of 3-phase motor) Would this multiplier apply to balanced single phase loads as well or just for 3-phase loads, and you always have to do the vector addition for single phase loads?
 

jghrist

Senior Member
I'm with you on all counts except for one concept which isn't sinking into my stubborn brain...


I see the winding currents as:
Ia = 480V/1 ohm + 480V/110 ohms = 484.4 Amps
Ib = 480V/10 ohms + 480V/101 ohms = 52.8 Amps
Ic = 480V/100 ohms + 480V/11 ohms = 48.4 Amps

I've revised my original sketch (attached) to try and show again how I am coming up with 52.8 amps for winding Ib instead of 48 amps. In the sketch I've broken the L3 connection to the load, leaving only L1 and L2 connected. The winding Ib current in this case must be 52.8 amps (if it isn't I'm in real trouble!). So my question is, if I re-establish the L3 connection, why would the current in the Ib winding go from 52.8 amps down to 48 amps?

First of all, forget about the transformer winding currents. They are indeterminant (I'll explain this later).

Assuming a perfect transformer, with balanced 480V, the voltage across each load resistance is 480 volts. The current in each load resistor will be 480V?R, simple application of Ohm's Law. The voltages are 120? out of phase; so are the currents. The line currents can be calculated by vector addition as has been done by others.

The transformer winding currents are indeterminant because it is possible to have a zero-sequence current circulating in the delta windings. This could come about by a grounded wye primary winding with unbalanced loads on the primary system. The transformer would act as a grounding bank and you would get currents circulating in the delta windings. The circulating currents add to zero at each terminal, so they do not contribute to and are not affected by the line currents.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
081231-1410 EST

mull982:

I suspect you had a slight error in the drawing of the vectors in Autocad, or Autocad produced the wrong result.

I just tried it in Fastcad and got 482.41791 .


dinos:

The lines to draw for this calculation are:
One line from 0,0 to 480,0 .
A second line 4.8 long from the 480,0 point at an angle of 120 deg from the first line.
Next get the hypotenuse from point 0,0 to 482.4, 4.1568 (end of the second line).

.
 
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