Transformers

Temperature rise test of power transformers

In order to find the winding temperature rise of three phase
power transfomer we need toget the resistance of the winding at the instant of shut down of power.As we can't get the resistance measurement at the instant of shut down we use extrapolating method according to the IEC standard 60076-2:1993.
Could any one with experience in this subject please help us with better explaination of the above extrapolating method.

 

[electricpete]

I don't have direct experience with this method but I believe I understand the simple theory.

The RISE (dT) of winding temperature above ambient can be expressed as a function of time (t) using a relation of the form:
dT=exp(-k*t).
(let me know if you want a derivation of that part).

Take the natural log of both sides:
ln(dT)=-k*t

So if you plot dT on a log scale vs t on a linear scale the relation will be a straight line. If you plot the points you have at t>0 and draw straight line back to t=0 you should be able to predict dT at t=0.

Forgive me if I'm telling you what you already know.

 

[electricpete]

In case it is not be obious that ln(dT)=-k*t describes a straight line on a log/linear scale.... I should point out the slope is -k.

 

[jbartos]

Suggestion: Since various electrical engineering standards are using about the same principles, the following ANSI/IEEE Std C57.12.00.1987 in 7.4 Calculation of Winding Temperature During a Short Circuit on page 31 could be helpful.
For m<=0.6
Tf=(Tk+Ts) x m x (1 + E + 0.6 x m) + Ts
Where
m=(Ws x t)/(C x (Tk + Ts) up to m=0.6
Ws is short-circuit resistance of the winding at starting temperature, watts per pound of conductor material
Tf is final winding Temp (all temperatures are in deg C)
t is time in seconds (not indicated in the standard)
for m>0.6:
Tf=(Tk + Ts) x {sqrt[exp(2 x m) + E x (exp(2 x m) - 1)] - 1} + Ts
where Tk = 234.5 for Cu, 225 for Al
Ts is starting Temperature
E is per unit Eddy Current Loss
C is the average thermal capacitance per pound of conductor material and its associated turn insulation, (W x s/°C). It shall be determined by iteration from either of the following empirical equations:
C=174 + 0.0225 x (Ts + Tf) + 110 x Ai/Ac, for Cu
C=174 + 0.1 x (Ts + Tf) + 360 x Ai/Ac for Al
Ai is cross-sectional area of turn insulation
Ac is cross-sectional area of conductor.
There are some additional relationships. They are available if interested.