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Sun, 04 Apr 2010 03:12:06 +0000
solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#732800
Darryl
Hi,<br>
I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.<br>
The equations are Ra+Rb=250<br>
Rb+Rc=500<br>
Rc+Ra=300<br>
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.<br>
The equations are (Z1^1+(Z2+Z3)^1)^1=300<br>
(Z2^1+(Z1+Z3)^1)^1=250<br>
(Z3^1+(Z1+Z2)^1)^1=500<br>
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<br>
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)<br>
Z2=Ra+Rb+((Ra*Rb)/Rc)<br>
Z3=Rb+Rc+((Rb*Rc)/Ra)<br>
I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.<br>
<br>
So my question is can the second set of equations be solved with matrices, and if so how??<br>
Cheers Darryl

Sun, 04 Apr 2010 07:27:05 +0000
Re: solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#732837
Roger Stafford
"Darryl " <alfalfaNOT.THIS@value.net.nz> wrote in message <hp9026$cnj$1@fred.mathworks.com>...<br>
> Hi,<br>
> I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.<br>
> The equations are Ra+Rb=250<br>
> Rb+Rc=500<br>
> Rc+Ra=300<br>
> 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.<br>
> The equations are (Z1^1+(Z2+Z3)^1)^1=300<br>
> (Z2^1+(Z1+Z3)^1)^1=250<br>
> (Z3^1+(Z1+Z2)^1)^1=500<br>
> 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<br>
> 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)<br>
> Z2=Ra+Rb+((Ra*Rb)/Rc)<br>
> Z3=Rb+Rc+((Rb*Rc)/Ra)<br>
> I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.<br>
> <br>
> So my question is can the second set of equations be solved with matrices, and if so how??<br>
> Cheers Darryl<br>
<br>
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!<br>
<br>
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"?<br>
<br>
Roger Stafford

Mon, 05 Apr 2010 00:18:02 +0000
Re: solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#732980
Darryl
Gidday Roger,<br>
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.<br>
But I will go and have a play with the symbolic tool box, its always good to have a backup to check your answers.<br>
Cheers Darryl

Mon, 05 Apr 2010 04:19:03 +0000
Re: solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#733038
Roger Stafford
"Darryl " <alfalfaNOT.THIS@value.net.nz> wrote in message <hpba7q$aoq$1@fred.mathworks.com>...<br>
> Gidday Roger,<br>
> 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.<br>
> But I will go and have a play with the symbolic tool box, its always good to have a backup to check your answers.<br>
> Cheers Darryl<br>
<br>
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 errorfree result otherwise.<br>
<br>
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.<br>
<br>
Roger Stafford

Mon, 05 Apr 2010 08:04:03 +0000
Re: solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#733060
Derek O'Connor
"Darryl " <alfalfaNOT.THIS@value.net.nz> wrote in message <hp9026$cnj$1@fred.mathworks.com>...<br>
> Hi,<br>
> I am trying to solve two 3 term equations with matrices the first one is not a problem as it is linear.<br>
> The equations are Ra+Rb=250<br>
> Rb+Rc=500<br>
> Rc+Ra=300<br>
> 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.<br>
> The equations are (Z1^1+(Z2+Z3)^1)^1=300<br>
> (Z2^1+(Z1+Z3)^1)^1=250<br>
> (Z3^1+(Z1+Z2)^1)^1=500<br>
> 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<br>
> 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)<br>
> Z2=Ra+Rb+((Ra*Rb)/Rc)<br>
> Z3=Rb+Rc+((Rb*Rc)/Ra)<br>
> I can solve the delta model directly with a calculator but I would perfer to be able to show my working for exams.<br>
> <br>
> So my question is can the second set of equations be solved with matrices, and if so how??<br>
> Cheers Darryl<br>
<br>
<br>
<br>
<br>
Darryl,<br>
<br>
It's 40 years since I was an electrical (power) engineer in Westinghouse, Pittsburgh, and I'm a bit 'rusty' on threephase networks. <br>
<br>
Your problem is called the DeltaY or YDelta transformation for threeterminal 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):<br>
<br>
Ra, Rb, and Rc are the impedances of the Yconnected network, and Z1, Z2, and Z3, are the impedances of the equivalent Deltaconnected network. The problem is to find Z1, Z2, Z3, in terms of Ra, Rb, Rc, or vice versa.<br>
<br>
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 AC 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) YDelta transformation:<br>
<br>
Let <br>
<br>
(i) S = RaRb + RaRc + RbRc. Then<br>
<br>
(ii) Z1 = S/Rc, Z2 = S/Rb, and Z3 = S/Rc<br>
<br>
So, your problem is solved by <br>
<br>
(1) Calculate Ra, Rb, Rc from your first set of of equations.<br>
<br>
(2) Calculate S, Z1, Z2, and Z3, using (i) and (ii).<br>
<br>
Note that if Ra=Rb=Rc=R (balanced circuit) then S = 3R^2 and Z1=Z2=Z3 = 3R, for the YDelta transformation, and if Z1=Z2=Z3 = Z, then Ra=Rb=Rc = Z/3, for DeltaY transformation.<br>
<br>
Do check these formulas in a 'modern' book.<br>
<br>
Derek O'Connor.

Mon, 05 Apr 2010 08:30:24 +0000
Re: solving equations with a matrix
http://www.mathworks.com/matlabcentral/newsreader/view_thread/278421#733062
Derek O'Connor
"Derek O'Connor" <derekroconnor@eircom.net> wrote in message <hpc5hj$es4$1@fred.mathworks.com>...<br>
> "Darryl " <alfalfaNOT.THIS@value.net.nz> wrote in message <hp9026$cnj$1@fred.mathworks.com>...<br>
> > Hi,<br>
 snip <br>
> Your problem is called the DeltaY or YDelta transformation for threeterminal 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):<br>
> <br>
> Ra, Rb, and Rc are the impedances of the Yconnected network, and Z1, Z2, and Z3, are the impedances of the equivalent Deltaconnected network. The problem is to find Z1, Z2, Z3, in terms of Ra, Rb, Rc, or vice versa.<br>
> <br>
> 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 AC 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) YDelta transformation:<br>
> <br>
> Let <br>
> <br>
> (i) S = RaRb + RaRc + RbRc. Then<br>
> <br>
> (ii) Z1 = S/Rc, Z2 = S/Rb, and Z3 = S/Rc<br>
> <br>
> So, your problem is solved by <br>
> <br>
> (1) Calculate Ra, Rb, Rc from your first set of of equations.<br>
> <br>
> (2) Calculate S, Z1, Z2, and Z3, using (i) and (ii).<br>
> <br>
> Note that if Ra=Rb=Rc=R (balanced circuit) then S = 3R^2 and Z1=Z2=Z3 = 3R, for the YDelta transformation, and if Z1=Z2=Z3 = Z, then Ra=Rb=Rc = Z/3, for DeltaY transformation.<br>
> <br>
> Do check these formulas in a 'modern' book.<br>
> <br>
> Derek O'Connor.<br>
<br>
<br>
I just realised that without diagrams there is crucial information missing: I assume that you have three terminals labelled 'a','b','c', in your Yconnected circuit, and that Ra = impedance between terminal 'a' and neutral (middle of the Y), etc., and that Z1 = impedance between terminals 'a' and 'b' in the equivalent Deltaconnected circuit, etc.<br>
<br>
Again, check your textbook to make sure that I haven't permuted the a,b,c.<br>
<br>
Derek O'Connor