# Equivalent admittance between points in a power network

Hi,
I am looking for a closed-form equation for the equivalent admittance between any two points in a n-bus power system with a given Ybus
matrix.
Any ideas?
Thanks! Emily
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"emily" wrote in message
Hi,
I am looking for a closed-form equation for the equivalent admittance between any two points in a n-bus power system with a given Ybus matrix.
Any ideas?
Thanks! Emily
Sorry The y method that I gave before DOES'T APPLY in general.
you will have to use the Zbus but once you have this, it is simple.
a)note that you can throw away all of the Zbus except the desired submatrix (do you see why) | Zii Zij | | Zji Zjj | this is a simple rake and the equivalent Z is easy to find Alternatively consider a unit current IN at i and OUT at j (Ii=1, Ij=-1) and solve V=ZI for to get Vi-Vj which is the Thevenin impedance between i and j Note that the inverse will give a pi as in the previous example.
Don Kelly cross out to reply
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emily wrote:

Just a thought. Invert the Ybus matrix to an impedance (rake equivalent) matrix:
[I] = [Y][E]
to
[E] = [Z][I]
and then solve for [E] with [I] = [0,..0,Ia,0,...,0,Ib,0,...]
--
Paul Hovnanian snipped-for-privacy@hovnanian.com
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"emily" wrote in message
Hi,
I am looking for a closed-form equation for the equivalent admittance between any two points in a n-bus power system with a given Ybus matrix.
Any ideas?
Thanks! Emily
If you are simply looking at the transfer impedance , you have it in the corresponding element of the Ybus. Since this is so simple, I assume that you are looking for something else. Are you looking at the network with all sources set to 0 - essentially seeking the Norton equivalent between terminals i,j? If so, I suggest that you take 1/((1/(Yii-Yii) + 1/(Yjj-Yij)) +Yij (1) This will agree with the Zbus method as given by Paul (expressed as a rake with the reference node open. Here is a simple example: 1 2 o-----------j0.01------o --------j0.1---------o----j0.1-------o | | | | V1 j0.5 j0.5 V2 | | | | o------------------------o----------------------o-----------------o reference
Ybus = -j (22 -10) Zbus =j(0.05729 0.02604) (-10 22) (0.02604 0.05729) Zbus may be represented by a T 1 2 o------j0.03125--o------j0.03125-------o | 0.02604 o--------------------o----------------------reference
From the Ybus Y11-Y12 =Y22-Y21  leading to Yin  Note that the Zbus T model with the reference floating gives Y12 input  In general it would take more effort to find Zbus than to simply use the expression (1) above Without having explored a more complex case, I would suggest that using either Zbus or expression (1) above would result in the same value but it looks as if (1) applies in all cases. Life would be easier with better diagrams but consider node i connected directly to node j and also to a blob representing a network which also lies between i and j through any and possibly all nodes other than i, j-- all essentially related to Norton/Thevenin
Don Kelly cross out to reply
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"Don Kelly" wrote in message
Hi,
I am looking for a closed-form equation for the equivalent admittance between any two points in a n-bus power system with a given Ybus matrix.
Any ideas?
Thanks! Emily -------------------------------- If you are simply looking at the transfer impedance , you have it in the corresponding element of the Ybus. Since this is so simple, I assume that you are looking for something else. Are you looking at the network with all sources set to 0 - essentially seeking the Norton equivalent between terminals i,j? If so, I suggest that you take 1/((1/(Yii-Yii) + 1/(Yjj-Yij)) +Yij (1) ********************* Correction :Change sign of Yij from the Ybus before applying (1) or use 1/((1/Yii+Yij) +(Yjj+Yij)) -Yij Following has been corrected ****************************
Don Kelly cross out to reply
This will agree with the Zbus method as given by Paul (expressed as a rake with the reference node open. Here is a simple example: 1 2 o-----------j0.01------o --------j0.1---------o----j0.1-------o | | | | V1 j0.5 j0.5 V2 | | | | o------------------------o----------------------o-----------------o reference
Ybus = -j (22 -10) Zbus =j(0.05729 0.02604) (-10 22) (0.02604 0.05729) Zbus may be represented by a T 1 2 o------j0.03125--o------j0.03125-------o | 0.02604 o--------------------o----------------------reference
From the Ybus Y11-Y12 =Y22-Y21 =-j12 leading to Yin =-j16 Note that the Zbus T model with the reference floating gives Y12 input =-j16 In general it would take more effort to find Zbus than to simply use the expression (1) above Without having explored a more complex case, I would suggest that using either Zbus or expression (1) above would result in the same value but it looks as if (1) applies in all cases. Life would be easier with better diagrams but consider node i connected directly to node j and also to a blob representing a network which also lies between i and j through any and possibly all nodes other than i, j-- all essentially related to Norton/Thevenin
Don Kelly cross out to reply