IEC transformer damage curve calculation 2

[healyx]

Hi All,

Does anyone have a source for a numerical example of how to calculate the damage curve of a transformer using IEC 60076-5.

I have found heaps of examples for IEEE/ANSI curves, but no luck with the IEC calc.

Specifically I'm not understanding where the J term (current density) in the equation comes from and if there are typical values and also what to use as the initial temperature. The standard uses ambient plus allowable temperature rise which is a bit confusing....

 

[7anoter4]

J it is "current density" the current flowing in a conductor [A] divided by cross sectional area of the conductor [mm^2].
I did not ever use this formula but you can deduce it theoretically.
The heat produced in a conductor by current passing through it is R*I^2*t for a definite time t or R*I^2*dt for an infinite small time interval.
We'll neglect SKIN and PROXIMITY effect and the conductor resistance will increase only with the temperature: Rt=Rr*[1+alpha*(T-Tr)]
Rr=ror*length[m]/Amm2 [ ror units has to be =ohm.mm^2/m=ohm.cm/100] at reference temperature Tr [usually 20dgr.C].
Since ror it is in ohm.cm Rr=ror/100*length[m]/Amm2
Let's say alpha will get the same value from Tr up to Tm.
R*I^2*dt- heat produced in the conductor produces the conductor heat energy rising by:
TCAP*Vol*dT- conductor infinite small temperature difference rising.
TCAP-See IEEE-80/2000- thermal capacity[J/cm^3/dgr.C.]
Vol=conductor volume=Amm2*length[m]=[cm^3]
In an adiabatic heating no energy dissipation will be considered.
I^2*ror/100*length[m]/Amm2*(1+alpha*T-alpha*Tr)*dt=TCAP*Amm2*length*dT
And integrating both parts we'll get:
I^2*ts=TCAP*(Amm2)^2/ror/alpha*10^2*LN[(1-alpha*Tr+alpha*Tm)/(1-alpha*Tr+alpha*Ta)]
If we'll put Ko=(1-alpha*Tr)/alpha and J=I/Amm2 the above formula will be:
J^2*ts=TCAP/ror/alpha*10^2*LN((Ko+Tm)/(Ko+Ta))
If (Ko+Tm)/(Ko+Ta)<1.42 the error is less than 1% for the following approximation:
LN((Ko+Tm)/(Ko+Ta))=[approx.]2*[(Tm+Ko)-(Ta+Ko)]/[(Tm+Ko)+(Ta+Ko)] then:
J^2*ts=TCAP/ror/alpha*10^2*2*[(Tm+Ko)-(Ta+Ko)]/[(Tm+Ko)+(Ta+Ko)]
Let's say K1= 2*TCAP/ror/alpha*10^2
J^2*ts=K1*[(Tm+Ko)-(Ta+Ko)]/[(Tm+Ko)+(Ta+Ko)]
From here by algebraic transformations we''ll get:
K1*(Tm-Ta)=J^2*ts*(Tm+Ta+2*Ko) further:
K1*(Tm-Ta)=J^2*ts*(2Ta+2*Ko+Tm-Ta)
K1*(Tm-Ta)/(2Ta+2*Ko)=J^2*ts*[1+(Tm-Ta)/(2Ta+2*Ko)]
and at the end:
Tm-Ta=J^2*ts*(2Ta+2*Ko)/(K1-J^2*ts)
Tm=Ta+ 2*J^2*ts*(Ta+Ko)/(K1-J^2*ts)
Tm=1 Ta=o
For copper:
Ko=(1-0.00393*20)/0.00393=234.45
K1=2*3.42/1.724/0.00393*100=100955

 

[healyx]

Thanks 7anoter4,

Do you have a concrete example? What are typical values for J?

 

[prc]

IEC 60076-5 ed3.0 2006 specifies two types of over current capacity.One dynamic short circuit withstand capacity and other is thermal withstand capacity.For oil immersed transformers with copper windings,the maxium average winding temperature shall not exceed 250C during a symmetrical over current(short circuit current) for two seconds.The end temperature is calculated assuming that the entire heat generated in winding is stored in copper with out any dissipation( an assumption accurate for 2 seconds)

As per clause 4.1.5 of standard,the final winding temperature theta1= theta 0 + {2(theta0+235 )/ 106000/ JE2 x t} - 1 where thetao=initial winding temperature,J =current density during over current (A/mm2) t= current duration (sec)

In reality during short circuit current flow ,winding failure is due to dynamic mechanical forces and not due to thermal damage.

 

[healyx]

So what are typical values for J?

 

[prc]

Difficult to say. It depends on the impedance of transformer and the working J.Usually the working J wil be 1.5 to 4 A/mm2. The J to be used in the above formula will be 1/(per unit transformer impedance +system impedance) of the working J.

 

[healyx]

Thanks Prc,

What is the typical initial temperature (theta0)? I only have the ed2.0 version of the standard which only does the mechanical damage section. What is required for the thermal section?

 

[7anoter4]

I have not the latest version of this standard-as prc has. However, in the 2000 version in section 4.1.4 "Maximum permissible value of the average temperature of each winding" it is stated:
"If the measured winding temperature rise is not available, then the initial winding temperature thetao shall correspond to the sum of the maximum permissible ambient temperature and the temperature rise allowed for the winding insulation system".
From Table 3 – Maximum permissible values of the average temperature of each winding after short circuit
theta0 could be "Insulation system temperature" and theta1 "Maximum value of temperature".
I think prc will agree with me.

 

[prc]

Theta0 is the average temperature of the winding at the moment of short circuit. It will be ambient temperature at the time + winding temperature rise. For calculation purposes, you can take max ambient temp + average winding temperature rise at full load.

 

[healyx]

Thanks guys