pad mounted transformer impedance 4

[EEuno]

I have a 300 KVA distribution pad mounted transformer that has tested with a 4.42% impedance, slightly above the ANSI limit of 4% +7.5% or 4.3%. Besides fault current let-thru and higher regulation can anyone speak to the effect on efficiency and the energy losses of the transformer? Is the economic consideration sufficient to reject this transformer? The transformer is most likely going to operate at light loading most of the time.

 

[alehman]

Impedance of small pad-mount transformers vary considerably and is not directly related to efficiency. Did your purchase spec include an efficiency requirement? Does the manufacturer claim compliance with NEMA TP-1? Ask the manufacturer for no-load energy loss, which is a standard test.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.

 

[Kiribanda]

EEuno,

I donot know from which ANSI standard you have sid that a 300 kVA pad mounted transformer shall have only 4% impedance.In fact the relevant IEEE std is IEEE C57.12.26 which specifies the impedance can be between 1.2-6.0.Therefore, if your short circuit results are ok with 4.42% then there cannot be a problem.

 

[prc]

Your issue is a transformer with impedance above the guaranteed limit.It will not have any effect on efficiency. The only impact is marginally higher regulation on loading and slight unbalance during paralleling with other transformers. Since your transformer is going to work at light load most of the time, even that is also not a concern.

Of course you have the right to reject as guaranteed technical parameter is outside the limit.But technically absolutely no problem in use.

In IEC world,tolerance for this range of impedance is plus minus 10% instead of 7.5% as per ANSI.

 

[pwrtran]

You will have little less inrush capacity and little higher Var loss.

 

[jack6238]

My lord.

What are the enginners out there doing? Here are the fundamentals for calculating the energy losses, and efficiency of a transformer as a function of load, power factor, and X/R values. Please help me to understand why these question are being asked??

1.Get the transformer test sheet.You cannot calculate the data without this information. You can guess and estimate but the data sheet is the answer. Any good Power System Enginner would ask for this data when he /she purchased a tranformer.

2.Look at at the data sheet and find the load watts loss. this is normaly gives in watts. Also find the percent Z of the transformer and the no load watts loss.

3.Divide the watts loss by 10 times the KVA rating and get the% R of the transformer. Example, 2500 KVA transformer with a % Impedance of 6.47 and a load loss watts of 26495 watss would have a % R of 26495/25000 of 1.0598. the % X would be [6.47^2-1.0598^2]= 6.38 % X. Now you have the parameters of the transformer.

4. Calculate the efficiency of the transformer at different values of load and power factor.

5. For this transformer you will calculate the maximum efficiency is at a 37.@ % load and the efficiency is 99.12 %

6. The transformer regulation is , of course, a function of the per unit loading and power factor of the load. For this transformer at a .75 per unit load and a power factor of the load at 0.8, the regulation is 3.56 % and at 1.0 per unit is 4.77 % The efficeincy at 0.75 and 0.8 PF is 99.02% and 98.81 %.

If you need to know how to calculate this, you will need some help.

Get:
1. J&P Transformer Book
2.Intreperiting the Transformer Data Sheet-Jack Woodham
3.A Practical Guide To Short Citcuit Calculations-Condrad St. Pierre.
4.Symmetrical Componments-Wagner & Evans


Jack

 

[prc]

Jack,What is that "Interpreting trf data sheet" Is it a paper or book? Where will you get it?

 

[wareagle]

jack
Is this a typo
"5.For this transformer you will calculate maximum 'efficiency is at a 37.@ % load and the efficiency is 99.12%