Short circuit on primary side 1

[AusLee]

Hello,

I have 2x1,000kVA transformers in a building fed from the same 15kV ring. Each TR feeds a separate MDB so the secondary sides are not in parallel but a tie breaker is provided so they May be put in parrallel sometime for a given reason.

The local authority wants to read only 1 watt meter so i have the following MV cells arrangement:

Ring In-Ring Out-Metering-Out TR1-Out TR2

The breakers (no fuses) in "Out TR1" and "Out TR2" are 25kA because the transformers are 1000kVA 15kV/400V and network Pcc=500MVA. My question is: what would be the short circuit rating of the breaker in the Metering cell? 25 or 50kA and why?

In case it should be 50kA, can you please point out a product on the market? i know Merlin Gerin SM6 but they go up to 25kA only for equipment under 24kV.

Thanks.

 

[RalphChristie]

AusLee

I am not sure if I understand your arrangments very clearly, but:

A short circuit on the primary side are determined by the source. The source will supply a certain ammount of current during a HV-fault (let say a bolted three phase fault) if the two transformers are in parallel, not in parallel, or even both not energized. That current (fault-current) is 19.28kA. (500MVA - I suppose the local supplier gave you the data) Again, this fault-current is for practical reasons independant on the transformers - but rather dependant on the source.

On the secondary side it is a different story. If the two transformers are in parallel, it means you have actually two sources. Now, if there is a fault between the LV-bus and a transformer, each transformer will supply the fault, but from different sides. If the fault are on a feeder from the bus, then both fault-currents will sum-up in the feeder to a value double that of one transformer.

In my opinion, for the extra HV-cell you can specify a 25kA breaker. (Must be abble to break 19.3kA max)

Regards
Ralph

 

[AusLee]

Hi,

Thank you for the clarification. Ok let me add this: you are right, they did say 500MVA. But i think they copied this from the US where in Reality the network is meshed and you might actually be burning fuel in generation plants to produce gigawatts and be touched by those.

But here the Reality is different: the netowrk is made of only two generation stations, one is 130MVA and the other 2x75MVA on of them is always under repair and the rest never run at full load because the MV distribution network outhere is not in condition.

So my question is: can i consider for personal and technical and for my own belief that there is only some 250MVA and if a 16kA breaker is installed it is not the end of the world? or is the 500MVA something Theoretical that does not have to corresponf to something physical in nature, in particular: the amount of MVAs produced by all the generation plants connected to the substation in concer?

Thanks again!

 

[jghrist]

The 500 MVA is not the full load rating of the source. It is the fault rating. A 130 MVA generator with a subtransient reactance of 10% would have a 1300 MVA fault available at the terminals. To get the fault current available at your substation, you need to consider the impedance of the lines as well. You need to have a fault study done on the system using the most generation that would be on the system at any one time, considering future expansion.

 

[RalphChristie]

It seems you are confused between the load-current(MVA) and the fault-current(MVA). The load current is the current drawn from the supply during normal conditions. This is the sum of the currents supplied to motors, transformers etc. when they are operating normally. The fault-current however, is an abnormal condition when there is a short-circuit or earth-fault on the network. It is only the resistance of the wires that limits this current, and because wire-resistance is normally very low, the current will be very high. (V=IxR) The further you move from the supply, the lower your fault-current will become, because you have more wire (more resistance) between you and your source. The fault-current is much higher than the load-current.

The utility supply customers with the max. fault-level. This current is the highest (max) current you will see during a fault at that specific location. You must rate your equipment (breakers) such that if there is a fault-current of that magnitude, it will safely clear the fault without damage to the breakers. I don't think you will be able to run an fault-study on the utility's side (they have to do it themselves) but you can ask them to do it again and to confirm it.

Normally you choose a breaker size bigger than the max. fault-level.

Regards
Ralph

 

[RalphChristie]

It is very important to use the max. fault-level

Depending on how the utility run their generators, your fault-level can vary. Your fault-level will be probably the lowest when they are running just one 75MVA generator, and the highest if the run all of them in parallel. Maybe all three will never run at the same time, but two will. It all depend on the utility's operating philosophy. You must be sure that your breakers will be able to carry the fault-current (during a fault), when the fault-level is on its maximum.

Also, like jghrist said, consider future expansion - not only in your side, but also from the utility's side.

 

[bonesa]

I think you can apply to company Power Systems Canada and they could help you to solve your problem.This company is mnfs of compact substation up to 34KV and 3000kVA.Mr Ferris
my personal recomandations Daniel