The Short Circuit MVA level of a 132/33kv substation is 20.31pu on 100MVA base. 2 x 60MVA Transformers were installed in the substation and the two packed up within a very short time. What factors do we consider in choosing a transformer for this substation please.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Short Circuit Rating 6
- Thread starter haisa
- Start date
-
2
- #21
The above mentioned by prc IEC 60076-5 chapter 4.2.3 presents the formula for calculate the asymmetrical test current peak. At first, the amplitude î of the first peak of the asymmetrical test current is calculated as follows:
î = I × k ×√2
where I=U/(Zs+Zt) the rms value as prc said, U[kV]; Zs =system impedance [ohm]=U^2/Sscsystem Zt=Ur^2/Srt*zt%/100 [ohm]
Ur[kV] it is the rated voltage of the referred winding and Srt=transformer rated power[MVA].
k=(1+(e^–(φ+π/2)R/X)sinφ) where:
e is the base of natural logarithm;
φ is the phase angle which is equal to arctan X/R, in radians.
That means for X/R=∞ k=2 and this is the theoretical limit.
Usually for II category [2.5-100 MVA] k=1.8 and for III category[above 100 MVA]=1.9
î = I × k ×√2
where I=U/(Zs+Zt) the rms value as prc said, U[kV]; Zs =system impedance [ohm]=U^2/Sscsystem Zt=Ur^2/Srt*zt%/100 [ohm]
Ur[kV] it is the rated voltage of the referred winding and Srt=transformer rated power[MVA].
k=(1+(e^–(φ+π/2)R/X)sinφ) where:
e is the base of natural logarithm;
φ is the phase angle which is equal to arctan X/R, in radians.
That means for X/R=∞ k=2 and this is the theoretical limit.
Usually for II category [2.5-100 MVA] k=1.8 and for III category[above 100 MVA]=1.9
- Status
Similar threads
- Question