Trfr Short Time Ratings
Trfr Short Time Ratings
(OP)
Hi
I have a 45MVA, 66/11kV trfr where a 3-phase fault on the LV side is 15.0kA and the peak is 33kA. The manufacturer has given the peak witstand rating as 4.5kA. I presume the withstand ratings supplied by the manufacturer is for maximum throughfault currents expressed at the HV.
I simply took the 33kV peak and divided by 6 to get 5.5kA > 4.5kA so is the trfr in trouble? Relevant standard is IEC 60076.5.
Thanks in advance.
I have a 45MVA, 66/11kV trfr where a 3-phase fault on the LV side is 15.0kA and the peak is 33kA. The manufacturer has given the peak witstand rating as 4.5kA. I presume the withstand ratings supplied by the manufacturer is for maximum throughfault currents expressed at the HV.
I simply took the 33kV peak and divided by 6 to get 5.5kA > 4.5kA so is the trfr in trouble? Relevant standard is IEC 60076.5.
Thanks in advance.






RE: Trfr Short Time Ratings
IEC 60076-5 Table 4 – Values for factor k*sqrt(2)].
The peak short-circuit current at 66 kV side could be 4.5 kA in a case like following one:
I"k [66kV]=11/66*I"k[11 kV]=11/66*15=2.5 kA [as you said]
If we shall take the minimum short-circuit apparent power [Table 2] of 3000 MVA we'll get Zs=66^2/3000=1.452 ohm.
Zt has to be 13.75 ohm [14.2% short-circuit impedance Zt=66^2/45*0.142=13.746 ohm.
If copper losses in the transformer will be 5.75%[usually has to be less than 4%]
pk=3*I^2*Rcu=5.75%*45=2.5875 MW
Rcu=pk/3/I^2=2.5875/3/0.3936^2=5.566 ohm/phase
Xt=sqrt(Zt^2-Rcu^2)=12.568 ohm
Xt/Rcu=2.258 fi=ATAN(X/R)=1.1539 rad.
k ×sqrt( 2) = (1+(e–(φ+π/2)R/X)sinφ) × sqrt(2) or [in Excel language]:
k ×sqrt( 2) =(1+SIN(fi)*EXP(-(fi+PI()/2)/(X/R))*SQRT(2)
k ×sqrt( 2) = 1.801
Ipeak=1.801*2.5=4.503 kA
RE: Trfr Short Time Ratings
Rcu=0.4356 ohm and X/R=31.5 [and it corresponds to ANSI C37.010 Tab.8B].
Since transformer S=45 MVA[ see: 3.2.2 Transformers with two separate windings]
it is Category II and if X/R>14 then Ipeak=2.55*I"k=2.55*2.5=6.375 kA.
I think someone forgot to multiply by sqrt(2) and the result was 1.8*2.5=4.5 kA.
If the transformer Ipeak rated is indeed 4.5 kA in a short-circuit case will be severely damaged by dynamical efforts.
RE: Trfr Short Time Ratings
RE: Trfr Short Time Ratings
Thanks for the feedback. Interesting indeed. Few questions:
1. Where does the 5.75% for the trfr come from?
2. Where does the I = 0.3936 come from in the Rcu calc?
The formula for k*sqrt(2) I see is given at the bottom of Table 4. Puzzles me a bit as I have two formulae for calculating Ip.
From AS 3851, Section 8.5.2.1 get k = 1.02+0.98*e^(-3*R/X), here k = x.
From first principles get k = 1 + e^(-pi*R/X)….equation 2
None of these correspond (directly at least) to the one in AS 60076.5.
Yet for X/R = 14, and using equation 2, get k = 1.8!
I agree that it should have been Ip = 2.55*2.5kA = 6.375kA. BUT at the same time it must be ascertained that 6.375kA is greater than the calculated peak, viz 33kA/6 = 5.5kA. If Ip calc = 7kA say, then Irms = 7/2.55 = 2.75kA. The trfr should then have a min Irms rating of 2.75kA and Ip rating of 7kA.
RE: Trfr Short Time Ratings
2) If I"k=2.5 kA and Ip=4.5 kA then sqrt(2)*k=4.5/2.5=1.8 [î = I × k ×sqrt(2)]
Using the formula given in Table no.4 for sqrt(2)*k=1.8 we get X/R=2.25 then Rcu=X/2.25 =12.47/2.25=5.54 ohm
pcu=3*Rcu*Irated^2=3*5.54*0.393648^2=2.575 MW
pcu%=2.575/45*100=5.72%.
High losses are actually high expenses[more kwh to pay for], but on the other hand that means less copper conductor so lower equipment initial price. There is a point where the losses and the transformer price will be minimum.
However, no power transformer copper losses could be so high [usually it is less then 0.4%- it was wrongly said 4%].
4)In IEC 60909-1 ch.2.4.1 the k formula is indeed:
1,02 + 0,98 e−3*(R / X) for 50-60 Hz [or k=1.02*0.98*exp(-3/(X/R)) ]
Then for X/R=2.25 k=1.02+0.98*exp(-3/2.25)=1.278 and sqrt(2)*k=1.808.
It seems to me it is the same result.