Transformer Impedance/Resistance specifications and temperature
Transformer Impedance/Resistance specifications and temperature
(OP)
Hi...
If a transformer test report specifies an impedance at a temperature other than expected, is there a required correction? I know that there are temperature adjustments for copper loss and resistance.... This would imply some affect on impedance as well.
In some cases the impedance is simply specified at an off temperature. In others, when an impedance is specified for an 85 degree rise vice a 75 degree rise, for example, I have seen corrections for MVA but not temperature.
Thx,
- john
If a transformer test report specifies an impedance at a temperature other than expected, is there a required correction? I know that there are temperature adjustments for copper loss and resistance.... This would imply some affect on impedance as well.
In some cases the impedance is simply specified at an off temperature. In others, when an impedance is specified for an 85 degree rise vice a 75 degree rise, for example, I have seen corrections for MVA but not temperature.
Thx,
- john






RE: Transformer Impedance/Resistance specifications and temperature
Where R%=Ohmic part involved copper loss
and X(%)=Inductive part involved leakage inductance (Dimension of the coil and distances beetwen HV and LV winding
Taking as an example From Cutler Hammer Three phase Transformers at 65°C Rise and 75°C Temp reference
KVA Z% %R X%
112.5 5.0 1.71 4.7
150 5.0 1.88 4.63
225 5.0 1.84 4.65
300 5.0 1.35 4.81
500 5.0 1.5 4.77
If we have a XMFR 500KVA at 75°C(Reference) and wish to change %Z al 85°C, you must correct only %R
Express in Watts %R=1.5%*500KVA=7500W=Wtot-cu. But these losses are divided in Wjoul and Wadditional (Get these values from Test report)it means Wtot-cu=Wjoul+Wadditional
Expressing at new temperature
Wjoul at 85°C=Wjoul at 75°C*(235+85)/(235+75)
Wadd a 85°C=Wadd at 75°C*(235+75)/(235+85)
So Wtot-cu=Wjoul+Wadditional at 85°C
Expressing in %R=Wtot-cu at 85°C/(KVA*10) In this case KVA=500
Finally %Z at 85°C= sqrt((R% at 85°C)^2+(X%)^2)=Vcc(%) at 85°C
Where %X is the same value at 75°C
RE: Transformer Impedance/Resistance specifications and temperature
The I^2R loss component increases with temperature and the stray-loss component diminishes with temperature.
RE: Transformer Impedance/Resistance specifications and temperature
regards,
- john