Transformer ratings: input or throughput? 5

[bacon4life]

Are transformer nameplate ratings the input or the throughput ratings. For example on a 115 to 13.8 kv 42/56 MVA substation transformer the nameplate ratings are:

HV:
115000 kV * 281 A * sqrt 3 = 55.97 MVA

LV:
13800 kV * 2343A * sqrt 3 = 56.00 MVA

Using an impedance of 11% with a 56 MVA load at 0.90 pf, the input to the transformer would be 60.3 MVA due to approximately 8 MVAR used in series reactance of the transformer.

Would this be overloading the transformer by 7% or would it be the design loading?

Std C57.12.00-2000 states that:
"5.4.1 General
The rated kVA of a transformer shall be the output that can be delivered for the time specified at rated secondary
voltage and rated frequency without exceeding the specified temperature-rise..."

The standard leads me to think the output should be 56MVA, but I don't know how to reconcile the 56MVA limit on the input.

Thanks,

 

[davidbeach]

I don't know the answer to your question, but that strikes me as one of those "if you have to ask..." type of questions. If you are so close to the limit that you are concerned with such nuances you have too much load for your transformer.

 

[RRaghunath]

bacon4life,

I think the MVA rating indicated on the name plate is output rating. As long as the output is limited to the rated value. The input doesn't matter, irrespective of the load power factor.
In ANSI, as I understand the voltage ratio on the transformer name plate doesn't reflect the transformer turns ratio, which happens to be the case in IEC and is relevenat when you are making such calculations. This is because the secondary voltage indicated on the name plate is load-voltage in case of ANSI and not the no-load voltage.
For your example, the no-load voltage would not be 13.8kV and thus the same transformer if we rate as per IEC would have higher MVA rating.

 

[prc]

The answer depend to which standard your transformer was made.

In case your are in US and product as per ANSI , transformer rating is the output.Input naturally will be more than output.But if you are in IEC world, rating is the input to transformer.This change in IEC came around by end 60's. Till then rating was output.I dont know why IEC experts decided like that at that tim.But you will see even textbooks, blindly follow efficiency= rating /rating+losses.(out put/input as in case of all other electrical equipments)

 

[RRaghunath]

Thanks prc. I didn't know that the change in IEC stipulation came in 60s. In ANSI, the no-load voltage of the transformer secondary winding (for transformer design purposes) is calculated correponding to a load power factor 0.8pf, I guess.

 

[Marmite]

PRC is right. I certainly didn't appreciate there was a difference.

This is the explanatory text on ratings from IEC 60076-1.

The interpretation of rated power according to this subclause implies that it is a value of apparent power input to the
transformer — including its own absorption of active and reactive power. The apparent power that the transformer delivers to the
circuit connected to the terminals of the secondary winding under rated loading differs from the rated power. The voltage across the
secondary terminals differs from rated voltage by the voltage drop (or rise) in the transformer. Allowance for voltage drop, with regard
to load power factor, is made in the specification of the rated voltage and the tapping range (see clause 2 of IEC 606).
This is different from the method used in transformer standards based on US tradition (ANSl/IEEE C57.12.00), where “rated kVA”
is “the output that can be delivered at ... rated secondary voltage ...”. According to that method, allowance for voltage drop has to be
made in the design so that the necessary primary voltage can be applied to the transformer. In addition, ANSI/IEEE specifies, under
“Usual service conditions”: “load power factor is 80 % or higher” (quotation from 1987 edition).

Regards
Marmite

 

[bacon4life]

Thanks all for the help.

Raghun, could you please expand a bit ANSI nameplate voltage not being the turns ratio?

 

[RRaghunath]

bacon4life,

In ANSI, the transformer name plate indicates load voltage for secondary winding. The load voltage means the voltage that is available when the transformer is carrying rated load at 0.8pf.
Hence, the secondary voltage at no-load will be more than the name plate voltage, to compensate for the load voltage drop in the transformer and provide rated voltage at the terminals.
In your case, the 13.8kV is load voltage of the secondary and the no-load voltage could be 14.5kV or so.
Thus, the transformer turns ratio is given by (115/14.5) and not (115/13.8). Hope I am clear.

 

[stevenal]

From IEEE C57.12.00 2006:

5.5.2 Voltage ratings
The voltage ratings shall be at no load and shall be based on the turn ratio.

 

[ppaya]

I don't think that P.F.=0.8 is important and this is first time I hear someone limiting Tx load on nameplate by P.F.
(please correct me with reference paper if you find one). Load current on the nameplate is apparent current! What p.f. may be is not limitation. You may have 0.7 p.f. or 0.9 p.f. or any other p.f. for that matter and apparent current magnitude will not change. This means ratio active/reactive current component may be changing - apparent power/current stay the same.