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

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.

Answer 1

[Brahmadev]

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.

https://in.mathworks.com/help/sps/powersys/ref/per-unit-and-international-systems-of-units.html

I hope this helps!

[ismail hossain]

Thank you so much for the detailed calculation.

It became very clear.