I need help in per unit to SI conversion for wye-delta transformer in matlab simulink model "100kW grid connected PV model"

11 views (last 30 days)
I was trying to check the pu to SI conversion in the following model.
I used the follwoing formula for finding the pu to actual impedenace of the wye-delta transformer.
For the wye side, I found conversion okay based on base 100kVA and 25KV (L-L)
However, for the delta side, I failed to prove it based on base 100kVA and 260V (L-L).
Please anybody help and share with me by checking this conversion on the delta side.
Is it okay or not? What internal formula have they used to get this conversion?
Please let me know.

Accepted Answer

Brahmadev
Brahmadev on 30 Aug 2023
Hello Ismail,
I understand that you want to understand the conversion from “p.u.” to “SI” for Winding 2 of the Three Phase Transformer block. It is done using the same concept as in Winding 1.
When converting Resistance from p.u. to SI on the D1 side (winding 2) of the transformer
Base power = 100 kVA/3 = 33.3e3 VA
Base voltage = 260 V RMS
Base current = 33.3e3 /260 = 128.2 A RMS
Base resistance = 260/128.2 = 2.028 Ω
Base inductance = 2.03/(2π*60) = 0.00539 H
Quantity in SI units = base value*base value in p.u.
R2(Ω) = R2(p.u.) *Base resistance = 2.028*0.001 = 0.002028 Ω
L2(H) = L2(p.u.)*Base inductance = 0.03*0.00539 = 0.000161 H
You can also refer to Example 1: Three-Phase Transformer from the Documentation link below. It shows the conversion for a 25kV/600V 300kVA Transformer.
I hope this helps!

More Answers (0)

Products


Release

R2018b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!