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From: Darryl on 3 Apr 2010 23:12
Hi,
I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.
The equations are Ra+Rb=250
Rb+Rc=500
Rc+Ra=300
But the second one has terms which are to the power of negative 1, and I don't know how or indeed if they can be setup in a matrix to be solved.
The equations are (Z1^-1+(Z2+Z3)^-1)^-1=300
(Z2^-1+(Z1+Z3)^-1)^-1=250
(Z3^-1+(Z1+Z2)^-1)^-1=500
The equations are used in 3 phase power engineering, the one above is for a delta model, the one above that is for a star model
The one I'm not having a problem with is for a star model, and its values can be used to find the values of the above equations by inserting them into another set of equations Z1=Ra+Rc+((Ra*Rc)/Rb)
Z2=Ra+Rb+((Ra*Rb)/Rc)
Z3=Rb+Rc+((Rb*Rc)/Ra)
I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.

So my question is can the second set of equations be solved with matrices, and if so how??
Cheers Darryl
From: Roger Stafford on 4 Apr 2010 03:27
"Darryl " <alfalfaNOT.THIS(a)value.net.nz> wrote in message <hp9026$cnj$1(a)fred.mathworks.com>...
> Hi,
> I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.
> The equations are Ra+Rb=250
> Rb+Rc=500
> Rc+Ra=300
> But the second one has terms which are to the power of negative 1, and I don't know how or indeed if they can be setup in a matrix to be solved.
> The equations are (Z1^-1+(Z2+Z3)^-1)^-1=300
> (Z2^-1+(Z1+Z3)^-1)^-1=250
> (Z3^-1+(Z1+Z2)^-1)^-1=500
> The equations are used in 3 phase power engineering, the one above is for a delta model, the one above that is for a star model
> The one I'm not having a problem with is for a star model, and its values can be used to find the values of the above equations by inserting them into another set of equations Z1=Ra+Rc+((Ra*Rc)/Rb)
> Z2=Ra+Rb+((Ra*Rb)/Rc)
> Z3=Rb+Rc+((Rb*Rc)/Ra)
> I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.
>
> So my question is can the second set of equations be solved with matrices, and if so how??
> Cheers Darryl
-------------------
No, I don't think matrices will help you at all in solving your "delta model" problem, Darryl. Without the help of that clever substitution of the R expressions for the Z's above - I wouldn't have thought of doing it that way - you face an uphill battle in solving for the Z's directly by hand, but it can actually be done and there is a unique solution. Doing so involves manipulating your equations and combining them so that eventually two of the unknowns are eliminated and you have only one remaining equation in one unknown, and then substituting its solution back into previous equations to find the other two unknowns. As you probably realize by now, they are all rational numbers. However, if you do it this way, you should stock up on a large quantity of scratch paper!

You can also solve it the easy way by letting matlab's symbolic toolbox do all the hard work, but perhaps that wouldn't be a suitable way to "show my working for exams"?

Roger Stafford
From: Darryl on 4 Apr 2010 20:18
Gidday Roger,
thanks for taking the time to reply. Its as I feared, ethier way I have a lot of work to do, and the scratch pad is where i make all my mistakes as I'm slightly dislexsec.
But I will go and have a play with the symbolic tool box, its always good to have a backup to check your answers.
Cheers Darryl
From: Roger Stafford on 5 Apr 2010 00:19
"Darryl " <alfalfaNOT.THIS(a)value.net.nz> wrote in message <hpba7q$aoq$1(a)fred.mathworks.com>...
> Gidday Roger,
> thanks for taking the time to reply. Its as I feared, ethier way I have a lot of work to do, and the scratch pad is where i make all my mistakes as I'm slightly dislexsec.
> But I will go and have a play with the symbolic tool box, its always good to have a backup to check your answers.
> Cheers Darryl
----------------
I am very sympathetic with you on that, Darryl. At 84 I also make numerous mistakes in my algebraic manipulations on paper, probably more than you make. I have learned to compensate for this by using the Symbolic Toolbox to check frequently that supposedly equivalent expressions are actually mathematically identical by getting a symbolic zero for their simplified difference. I did that with your problem - it would have taken me forever to arrive at an error-free result otherwise.

I should mention that when I boiled things down to a single equation in Z1, it was actually a quartic equation possessing multiple roots. It looked for a moment as though there would be multiple solutions to the problem. However two of these were zeros for Z1 and a third resulted in an infinite value for Z2, none of which would have been valid in the original equations, so only the one solution remained.

Roger Stafford
From: Derek O'Connor on 5 Apr 2010 04:04
"Darryl " <alfalfaNOT.THIS(a)value.net.nz> wrote in message <hp9026$cnj$1(a)fred.mathworks.com>...
> Hi,
> I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.
> The equations are Ra+Rb=250
> Rb+Rc=500
> Rc+Ra=300
> But the second one has terms which are to the power of negative 1, and I don't know how or indeed if they can be setup in a matrix to be solved.
> The equations are (Z1^-1+(Z2+Z3)^-1)^-1=300
> (Z2^-1+(Z1+Z3)^-1)^-1=250
> (Z3^-1+(Z1+Z2)^-1)^-1=500
> The equations are used in 3 phase power engineering, the one above is for a delta model, the one above that is for a star model
> The one I'm not having a problem with is for a star model, and its values can be used to find the values of the above equations by inserting them into another set of equations Z1=Ra+Rc+((Ra*Rc)/Rb)
> Z2=Ra+Rb+((Ra*Rb)/Rc)
> Z3=Rb+Rc+((Rb*Rc)/Ra)
> I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.
>
> So my question is can the second set of equations be solved with matrices, and if so how??
> Cheers Darryl

---------------------------------------------------


Darryl,

It's 40 years since I was an electrical (power) engineer in Westinghouse, Pittsburgh, and I'm a bit 'rusty' on three-phase networks.

Your problem is called the Delta-Y or Y-Delta transformation for three-terminal networks (Y = star, in US terminology). I can't draw labelled diagrams here so I'll assume you will understand the following (using your labels):

Ra, Rb, and Rc are the impedances of the Y-connected network, and Z1, Z2, and Z3, are the impedances of the equivalent Delta-connected network. The problem is to find Z1, Z2, Z3, in terms of Ra, Rb, Rc, or vice versa.

I can't remember how to do this (simple algebraic manipulation), but I'm looking at one of the few remaining books I have on the subject, "Circuit Analysis of A-C Power Systems, Vol. 1", by Edith Clarke, Wiley, 1943. She gives the transformation formulas on pages 33 and 34. Here are the equations for (your) Y-Delta transformation:

Let

(i) S = RaRb + RaRc + RbRc. Then

(ii) Z1 = S/Rc, Z2 = S/Rb, and Z3 = S/Rc

So, your problem is solved by

(1) Calculate Ra, Rb, Rc from your first set of of equations.

(2) Calculate S, Z1, Z2, and Z3, using (i) and (ii).

Note that if Ra=Rb=Rc=R (balanced circuit) then S = 3R^2 and Z1=Z2=Z3 = 3R, for the Y-Delta transformation, and if Z1=Z2=Z3 = Z, then Ra=Rb=Rc = Z/3, for Delta-Y transformation.

Do check these formulas in a 'modern' book.

Derek O'Connor.
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