Unbalanced resistive loading of three phase system 1

[david707]

Can anyone prescribe the method of calculating the Line currents in a three phase Mesh connected system with equal resistive loads connected between Line 1,2 and line 2,3 with Line 2 as the common connection. Wish to be able to calculate all line currents. I suspect Line 1 and Line 3 currents equal V/R but cannot fathom Line 2 by calculation or vector method.
Many thanks
david707

 

[peterb]

The line 2 current is the vector sum of the line 1 and line 3 currents. Draw the circuit and the vectors and you will clearly see the conditions. In other words, line 2 current magnitude is the same as the other two lines.

 

[BJC]

Try this one - it got me through a lot of homework.

http://search.barnesandnoble.com/bo...?endeca=1&userid=2T4O41Y0LB&ean=9780070113558

 

[jghrist]

The way I see it, line 2 current is the vector difference between line 1 and line 3 currents, with a magnitude of sqrt(3) times V/R. I could be wrong; I haven't checked my work.

 

[david707]

Many thanks to Pereb but this is the solution I came to but it doesn't agree with actual figures taken from a practical situation. Actual situation as follows:-
Line volts = 208v
Resistive load = 20 ohms
Loads connected between Line 1,2 and 2,3
Mesh/Delta source configuration
Readings obtained, Line 1 current = 10.4 amps = Line 3 current
Line 2 current measured at c18 amps.

 

[peterb]

Oops! Vector SUM[\b] of the two is in fact sqrt(3)*I1 - I managed to confuse myself and subtracted the vectors. This corresponds with the readings that you took (10.4*1.73 = 18A).

 

[jghrist]

I got messed up with my current directions. What I actually did was subtract the current from 2 to 3 (which is the negative of line 3 current I3) from the current from 1 to 2 (which is line 1 current I1). So actually line 2 current I2 is -(I1+I2) which is sqrt(3)*V/R. At least I got the right answer even though the explanation was wrong.

 

[david707]

Many thanks chaps. I have been this route but when I actually draw the vector diagram the root 3 x I1 or I3 line current solution does not map out. I can contrive it but the vector solution does not make sense. I agree (almost) withjghrist (So actually line 2 current I2 is -(I1+I2)) where it should be -(I1 + I3). At the risk of becoming a bore by banging on about this, I'd be most grateful if someone could demonstrate the vector solution.
Many thanks again chaps,
David707

 

[peterb]

OK, here goes -
-Draw the equilateral voltage triangle 1-2-3
-As the load is resistive, the current vectors I12 and I32 are in phase with the respective voltages V12 and V32
-Add the vectors I12 and I32; the resultant is a vector with magnitude {sqrt(3)*I12}, at an angle leading V12 by 30 degrees (or lagging V32 by 30 degrees)

Hope this helps.

 

[david707]

Many thanks again, peterb. I agree with your method but the sticking point hinges around the direction of the current vectors for I1-2 and I3-2. This is best exemplified by a slightly different visual vector presentation. as follows: Draw the 3 Line Voltage vectors, V1, V2 and V3 at an angle of 120 degrees to each other and, then as you agree, insert the I1 and I3 Line current vectors in phase with their respective voltage vectors. Mapping the resultant then of the two current vectors does not produce an I2 vector of magnitude root3*I1 The magnitude of the resultant is seen to be the same as the I1, I3 vector. . A (root3*I1) resultant can be obtained by mapping the resultant of I1 and -I3 or indeed -I1 and I3.
The only justification for changing the sign of the I3 vector is by being strict about the directions of the three line currents as seen on the circuit diagram. This has been at thye centre of my problem throughout notwithstanding the fact that the measured value of I2 is c18 amps.
Cheers
David707

 

[peterb]

David707 -
The crux of the visualization problem is that you are drawing the wrong vectors. For your scenario, you need to draw the currents in phase with V12 and V32 respectively (60 degrees apart) - the way the loads are connected, NOT V1 and V3 (120 degrees apart). Having done that, I think that all will become clear.

 

[Shortstub]

A simple approach is to consider two current loops supplied by three voltage terminals, Va, Vb and Vc. The upper loop current is Iab in a cw direction, while the lower loop current is Ibc in a cw direction. For sequence A-B-C set Vab as V@120 deg, and Vbc as V@0 deg. Then Iab=(V/R)@120deg, and Ibc=(V/R)@0deg.

Then the phase currents are,

Ia=Iab.
Ib=-Iab+Ibc.
Ic=-Ibc.

The corresponding values are:

Ia=(V/R)@120deg.
Ib=sqrt(3)(V/R)@-30deg.
Ic=(V/R)@180deg.

 

[david707]

Shortstub - Many thanks indeed for your contribution. It confirms my thoughts described in my last response that the whole thing hinges on maintaining a convention on current direction. I'm now very happy to acceopt the (measured) fact that the common current Ib is sqrt3(3) (V/R) lagging 30 deg on Ibc. I note there are some minor (but confounding) differences between (what I presume to be US) and UK conventions for Delta Circuit layout and resulting vector convention. Still, all quiet on the Western front now and I can now sleep again without drawing vectors in my dreams..Thanks again all
David707