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.
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- Thread starter haisa
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When you loose one, or one gets taken out of service, what does the load happen to be on the other? Typically 132/33kv substations have 3 transformers each loaded to 2/3 of their capacity, ie a 100MVA load requires 3 50MVA transformers. Two other options are load shedding or forced load transfer to bring the remaining unit back into its rating.
Also new units must be of the same impedance when paralleled. MVA rating can be higher or smaller, but impedance must be the same in parallel and the taps set similar. Preferably identical X/Rs to the old units, but not as critical. I assume the units you have in their right now are 12.5% on the base (air natural oil natural) MVA?
In terms of short circuit you need to know the rating of the 33kv breakers, their X/R ratio interrupting rating, buss bar (short circuit force withstand) and equipment down line of the 33kv as it may see an increase in current during faults. Second you need to know the source strength of the 132kv supply if not using an infinite assumption in modeling 33kv short circuit.
When simulating faults, always do L-L-L; L-L; L-L-G; and L-G.
A but more info is needed, but there are a number of ways to go about selecting transformers or another transformer.
In terms of short circuit you need to know the rating of the 33kv breakers, their X/R ratio interrupting rating, buss bar (short circuit force withstand) and equipment down line of the 33kv as it may see an increase in current during faults. Second you need to know the source strength of the 132kv supply if not using an infinite assumption in modeling 33kv short circuit.
When simulating faults, always do L-L-L; L-L; L-L-G; and L-G.
A but more info is needed, but there are a number of ways to go about selecting transformers or another transformer.
- Thread starter
- #4
What I want to know is whether the fault level of a substation has any effect on the choice of transformer to be installed at that substation. In this case the fault level of this substation according to my calculation is about 35KA while the short circuit withstand current rating of the transformers is 3.2KA for 2 seconds. Can a Manufacturer produce a transformer of 60MVA, 132/33KV rating that can withstand 35KA short circuit current.
Hi,
Transformers are built in order to be able to withstand a short circuit limited only by it's own impedance for 2 seconds (IEC and ANSI)
Notice the difference betweeen total shortcircuit current on the faulted bus and shortcircuit current flowing through the transformer
Transformers are built in order to be able to withstand a short circuit limited only by it's own impedance for 2 seconds (IEC and ANSI)
Notice the difference betweeen total shortcircuit current on the faulted bus and shortcircuit current flowing through the transformer
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- #6
35ka of fault current would indicate an internal bolted fault within the transformer (primary) itself, and nothing else. In such a case a transformer should withstand 35ka in terms of not failing catastrophically provided the relaying clears as it should.
Now in terms of secondary short circuits outside the transformer itself the current is limited well below 35ka due to the transformer's own impedance. Most power transformers can easily survive multiple short circuits for a few seconds through out their life time. In fact it is usually the mechanical stress that concerns the manufactures well more than the thermal aspect- they do take that into account very well.
Now in terms of secondary short circuits outside the transformer itself the current is limited well below 35ka due to the transformer's own impedance. Most power transformers can easily survive multiple short circuits for a few seconds through out their life time. In fact it is usually the mechanical stress that concerns the manufactures well more than the thermal aspect- they do take that into account very well.
bacon4life
Electrical
While the transformer can limit damage to itself thanks to the self impedance, it is also important to consider the fault rating of other equipment connected to the bus. Having multiple low impedance transformers in parallel can require the use of more robust switchgear, elbows, etc or the addition of current limiting fuses to the use of current limiting reactors.
The effect of increased system short circuit level to the transformer itself is as under:
a) Which side of the transformer (primary or secondary side) the short circuit current is going to be increased? The terminal box and the connected cables (or bus duct) of that particular side should be capable of withstanding the new short circuit current.
b) The other side would witness a marginal increase in the short circuit current.
c) The transformer through fault current withstand capability and its characteristics, remain almost un-affected. Even if the thru fault current increases marginally, (due to the increased SC level at one side), the transformer design value takes care of this. Hence no issue.
a) Which side of the transformer (primary or secondary side) the short circuit current is going to be increased? The terminal box and the connected cables (or bus duct) of that particular side should be capable of withstanding the new short circuit current.
b) The other side would witness a marginal increase in the short circuit current.
c) The transformer through fault current withstand capability and its characteristics, remain almost un-affected. Even if the thru fault current increases marginally, (due to the increased SC level at one side), the transformer design value takes care of this. Hence no issue.
Power Transformers will not and need not withstand the system fault current (fault MVA/ line to line voltage x1.732). Transformer windings need to withstand only fault current corresponding to combined impedance of (system impedance + transformer impedance) when a 3L fault occurs at secondary terminals. If system fault current is less, then system impedance will be high and hence current passing through the windings will be marginally less.
Let us consider a case of 100 MVA load to be catered by a station at say 132/33 kV. With an impedance of 10% , the secondary fault level in 100 MVA transformer will be 100/0.1 =1000 MVA, assuming infinite fault level at 132 kV grid. If the secondary breaker cannot handle this ,then transformer impedance has to be increased or transformer MVA has to be reduced. Number of transformers is decided as Mbrooke explained-(n-1) principle. Even with the failure of one unit, the entire load has to be handled by balance transformers. So you can have 2x 100 MVA units or 3X50 MVA units. If the 132 kV fault level is 10,000MVA (43.7 kA)then the fault MVA seen by transformer will come down to 100/(10% +1%) =909 MVA. But when the secondarys of two 100 MVA units are paralleled with common secondary breaker, then fault level will be 2000 MVA. So either impedance should be increased to 20 % or use 3x50 MVA transformers with 15 % impedance to limit fault level.
Let us consider a case of 100 MVA load to be catered by a station at say 132/33 kV. With an impedance of 10% , the secondary fault level in 100 MVA transformer will be 100/0.1 =1000 MVA, assuming infinite fault level at 132 kV grid. If the secondary breaker cannot handle this ,then transformer impedance has to be increased or transformer MVA has to be reduced. Number of transformers is decided as Mbrooke explained-(n-1) principle. Even with the failure of one unit, the entire load has to be handled by balance transformers. So you can have 2x 100 MVA units or 3X50 MVA units. If the 132 kV fault level is 10,000MVA (43.7 kA)then the fault MVA seen by transformer will come down to 100/(10% +1%) =909 MVA. But when the secondarys of two 100 MVA units are paralleled with common secondary breaker, then fault level will be 2000 MVA. So either impedance should be increased to 20 % or use 3x50 MVA transformers with 15 % impedance to limit fault level.
In my opinion, 35 kA it is short-circuit current when the transformer secondary is short-circuited but the current it is measured on 33 kV side when 3.2 kA it is measured at 132 kV side. You don’t need 35 kA but only 3.2 kA there.
You have to take only one transformer into consideration and to use the actual short-circuit
impedance[about 12.5%].
You have to take only one transformer into consideration and to use the actual short-circuit
impedance[about 12.5%].
Generally in transformer specifications, only the primary fault level is given for considering in the transformer fault calculations. In Transformers, the forces from fault current depends on ampere-turns and hence the same whether HV or LV winding. Normally standby transformer is put in parallel for better overall efficiency and keeping it warm.
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- #14
Mrooke said:
Hi Mbrooke, i've done some research:
IEC: 60076-5 (Power transformers - Ability to withstand short circuits)
ANSI: IEEE Std C57.12.00 (IEEE Standard for Standard General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers)
IEEE Std. defines four categories of transformers based on their size:
ANSI Category I: from 5 to 500 kVA single phase or from 15 to 500 kVA three phase
ANSI Category II: from 501 to 1667 kVA single phase or from 501 to 5000 kVA three phase.
ANSI Category III: from 1688 to 10000 kVA single phase or from 5001 to 30000 kVA three phase
ANSI Category IV: above 10000 kVA single phase or above 30000 kVA three phase
For Category I t=1250/I^2
"For Category II, III, and IV units, the duration of the short-circuit current as defined in 7.1.4 is limited to 2 s, unless otherwise specified by the user. "
7.1.4.1 Category I: "The symmetrical short-circuit current shall be calculated using transformer impedance only except that the maximum symmetrical current magnitudes shall not exceed the values listed in Table 16. "
7.1.4.2 Category II: "The symmetrical short-circuit current shall be calculated using transformer impedance only"
7.1.4.3 Categories III and IV: "The symmetrical short-circuit current shall be calculated using transformer impedance plus system
impedance, as specified by the transformer user."
PS: It's not as simple as I stated in my previous comment
Hope this clarifies
Regards,
Juan
Juan BC, A few decades back, this fault current withstand capability was 5 sec, then reduced to 3 sec and now at 2 sec. But in reality, power transformers fail to with stand fault currents not due to thermal incapability, but from dynamic mechanical forces during the first one or two cycles when asymmetry of nearly 2.55 enhances forces seen by windings by a factor of 6.5, compared to symmetrical fault current.This is the reason duration of short circuit test on power transformers at high power labs is only 0.25 sec and not 2 sec. A transformer withstanding assymetrical fault current for first 2-3 cycles will withstand over current for several seconds.
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- #18
The peak value of a sine wave is √2 (1.414) of the RMS value. A fully offset asymmetric peak will be 2 x √2 = 2.828.
As the X:R ratio increases the asymmetric current will approach 2 x 2.82 of the RMS value.
The magnetic forces increase as the square of the current.
(√2 x 2)[sup]2[/sup] = 8 times the magnetic forces of the forces developed by the RMS current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
As the X:R ratio increases the asymmetric current will approach 2 x 2.82 of the RMS value.
The magnetic forces increase as the square of the current.
(√2 x 2)[sup]2[/sup] = 8 times the magnetic forces of the forces developed by the RMS current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
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- #19
Maximum radial and axial forces in windings vary as the square of the current flow. So for an asymmetry of 2.55, the forces will increase by 2.55 square =6.25 times. But I made a mistake. The asymmetry factor 2.55 is the peak of asymmetry compared to RMS fault current. But during RMS current flow, the maximum forces correspond to peak of RMS value. So the force multiplier will be (2.55/1.414) square= 3.25 times only. Hope waross will agree.
The asymmetry factor varies according to X/r ratio of Transformer. It will vary 1.76 (X/r=2) 2.46(X/r=10) 2.55 (X/r=14) As per IEC 60076-5 Power Transformer- Ability to withstand Short circuit, Clause 4.2.3, if X/r >14, asymmetry factor to be taken as 2.55 up to 100 MVA 3 phase rating and 2.69 for higher ratings.
The asymmetry factor varies according to X/r ratio of Transformer. It will vary 1.76 (X/r=2) 2.46(X/r=10) 2.55 (X/r=14) As per IEC 60076-5 Power Transformer- Ability to withstand Short circuit, Clause 4.2.3, if X/r >14, asymmetry factor to be taken as 2.55 up to 100 MVA 3 phase rating and 2.69 for higher ratings.
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