Z12 = [1.5609 + 0.67155i 0.5203 + 0.22385i 0.5203 + 0.22385i
0.5203 + 0.22385i 1.5609 + 0.67155i 0.5203+ 0.22385i
0.5203+ 0.22385i 0.5203 + 0.22385i 1.5609 + 0.67155i] ohms/km of 3 km length.
Need to know each step to convert Z12 to Y12 for this three-phase line.
Can you specify what is Z12?
If Z12 is Zbus, then Y bus will be its inverse.
You can use Octave or MATLAB software to find the inverse.
Md Abdul Raheman it is 3-phase impedance matrix of the line connected to bus 1 and bus 2.
To derive the Y-bus matrix for a 3-phase line of an unbalanced system, we can follow these steps:
Step 1: Convert the given line impedance matrix Z12 to an admittance matrix Y12 using the formula:
Y12 = inv(Z12)
where inv() is the matrix inverse.
Step 2: Calculate the geometric mean distance (GMD) of the line using the formula:
GMD = ((d1d2d3)^1/3)
where d1, d2, and d3 are the distances of the three phases.
Step 3: Calculate the positive, negative, and zero sequence impedances using the formula:
Z1 = Z2 = Z0 = Z12 * GMD
Step 4: Calculate the positive, negative, and zero sequence admittances using the formula:
Y1 = inv(Z1)
Y2 = inv(Z2)
Y0 = inv(Z0)
Step 5: Derive the Y-bus matrix using the following equations:
Y12 = Y1 + Y2 + Y0
Y21 = Y12
Y22 = Y1 + Y2 + Y0
Ybus = [Y11 Y12 Y13; Y21 Y22 Y23; Y31 Y32 Y33]
where Y11, Y12, Y13, Y21, Y22, Y23, Y31, Y32, and Y33 are the elements of the Y-bus matrix.
Therefore, applying the above steps for the given line impedance matrix Z12 = [1.5609 + 0.67155i 0.5203 + 0.22385i 0.5203 + 0.22385i 0.5203 + 0.22385i 1.5609 + 0.67155i 0.5203+ 0.22385i 0.5203+ 0.22385i 0.5203 + 0.22385i 1.5609 + 0.67155i] ohms/km of 3 km length, we can obtain the Y-bus matrix for the 3-phase line of an unbalanced system.
Salim H. Jassam Sir, hope your well. Thank you very much sir for your detailed explanation. I'll follow the steps and verify if the desired result is coming or not, and will get back to you if it is not coming.
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