Per-Unit System - How to explain this...?

Why Rm = 100 p.u. and Lm = 100 p.u.? Pls advise.

Please refer to this website: **broken link removed**

Or, as below...

Example 1: Three-Phase Transformer

Consider, for example, a three-phase two-winding transformer. The following typical parameters could be provided by the manufacturer:

Nominal power = 300 kVA total for three phases
Nominal frequency = 60 Hz

Winding 1: connected in wye, nominal voltage = 25 kV RMS line-to-line
resistance 0.01 p.u., leakage reactance = 0.02 p.u.

Winding 2: connected in delta, nominal voltage = 600 V RMS line-to-line
resistance 0.01 p.u., leakage reactance = 0.02 p.u.

Magnetizing losses at nominal voltage in % of nominal current:
Resistive 1%, Inductive 1%


The base values for each single-phase transformer are first calculated:

For winding 1:

Base power
300 kVA/3 = 100e3 VA/phase

Base voltage
25 kV/sqrt(3) = 14434 V RMS

Base current
100e3/14434 = 6.928 A RMS

Base impedance
14434/6.928 = 2083 ohm

Base resistance
14434/6.928 = 2083 ohm

Base inductance
2083/(2*60)= 5.525 H


For winding 2:

Base power
300 kVA/3 = 100e3 VA

Base voltage
600 V RMS

Base current
100e3/600 = 166.7 A RMS

Base impedance
600/166.7 = 3.60

Base resistance
600/166.7 = 3.60

Base inductance
3.60/(2*60) = 0.009549 H



The values of the winding resistances and leakage inductances expressed in SI units are therefore

For winding 1: R1= 0.01 * 2083 = 20.83 ohm; L1= 0.02*5.525 = 0.1105 H
For winding 2: R2= 0.01 * 3.60 = 0.0360 ohm; L2= 0.02*0.009549 = 0.191 mH


For the magnetizing branch, magnetizing losses of 1% resistive and 1% inductive mean a magnetizing resistance Rm of 100 p.u. and a magnetizing inductance Lm of 100 p.u. Therefore, the values expressed in SI units referred to winding 1 are

Rm = 100*2083 = 208.3 kohm
Lm = 100*5.525 = 552.5 H

Answer to my own question:

Nominal voltage = 1.0 pu
1% of nominal current = 0.01 pu

Rm = (Nominal voltage)/(1% of nominal current) = 1.0/0.01 = 100 pu

Please advise if my answer is wrong. Thanks.

here are good example 4 u

smxx said:

here are good example 4 u

Click to expand...

None of them explains my question. Anyway, thank you so much for sharing...

powersys said:

Answer to my own question:

Nominal voltage = 1.0 pu
1% of nominal current = 0.01 pu

Rm = (Nominal voltage)/(1% of nominal current) = 1.0/0.01 = 100 pu

Please advise if my answer is wrong. Thanks.

Click to expand...

I think is correct...
When you read this kind of problems, the aim is only to develop a mathematical modeling, but if you remember the magnetizing branch gives the ability to the transformer (according the nominal KVA) about the inrush current...

In fact I agree with your reasoning...

Hi madeza,

Thanks for your feedback. The info in my first post was extracted from Matlab's help file of transformer model. For Rm the explanation should be OK. But do you think our explanation can be applied for Lm as well?

powersys...
I was wondering if every element (i mean: resistance, inductance) is treated as impendance...
By the way I think it coul be applied as Rm...
I think thoses values (Rm=100pu; Lm=100pu) are impendances bases related magnetizing branch...
Do you agree with me...?

Good luck....

Hi madeza...

To be frank, I would like to agree with you, but I think it would be better if we could confirm the explanation is correct or not.

It's the first time I see the Rm and Lm are given as the following form:
Magnetizing losses at nominal voltage in % of nominal current:
Resistive 1%, Inductive 1%
Therefore, I don't understand it... Thanks.

hi powersys
It is correct SL=Sr=.01Sb Xm=Rm=U²/SL=U²/(.01Sb)=100 Zb
Rm and Xm Align in parallel with source and take .01Ib current.
it is better for a transformer to have large leakage impedance.