on state resistance for isolator/disconnector

i have designed a 2000Amp ACB recently, and got 6 μΩ across the whole product per phase, including 5x2 contact tips (AgW), top and bottom conductors, moving contacts (5 per phase) and the usual maximum contributor, the 5 copper braids per phase. In an isolator you will have less components in series, and maybe the contacts will be knife type (the ACB contacts are butt type). You may extrapolate linearly (inverse of course) as you are looking for much higher current ratings. i agree linear is oversimplification, but you just want a ballpark figure you say.

The Holm conference papers may give you the theoretical calculations if you need that. If you can lay your hands on Ragnar Holm, even better

Thanks Mr. Sridhar.

Can u give me some idea about switching over voltage ,when breaking dc current by DC circuit breakers. Actually in my application, we will break high dc current and the load is purely inductive. the current on interruption will be commutated to a resistor connected parallel to dc circuit breaker. So, I just wanted to have some idea about wht overvoltages may appear across the breaker. I hope u will provide me some useful guidance.

You are welcome. Glad to have been of some help.

The overvoltages that the breaker sees will depend on the inductance of the circuit and the speed with which the breaker quenches the DC arc .. to wit, V= L*di/dt. If you are in IEC country, i can help with more details of normal and transient recovery voltages that may appear across the open circuit breaker. Please let me know.

You may also check whether the breaker has (a) special arc-blowout means like blowout coils or permanent magnets, and (b) whether any means are there to mitigate the severity of the arc...

Thanks again, ur guidance is really helpful.... As our application is with DC, Will the transient recovery voltage come into picture. i am not having much knowledge on this,so want to confirm from u. Generally, these are present in ac power systems. The over voltage L*dI/dt will depend on the opening speed of the breaker, say if my breaker is opening in 60-80 ms, and current 30kA , with load inductance=300 mH, then the voltage appearing across the breaker will be very high; is this correct? I will check for the special arc-blowout means or whether any means for reducing the severity of arc.

You are quite correct.

Coming to TRV, here is what IEC says:

Prospective transient recovery voltage (of a circuit) (441-17-29)
The transient recovery voltage following the breaking of the prospective symmetrical
current by an ideal switching device.
Note.- The definition assumes that the switching device or the fuse. for which the prospective transient recovery
voltage is sought. is replaced by an ideal switching device, i.e. having instantaneous transition from zero to
infinite impedance at the very instant of zero current. i.e. at the "natural" zero. For circuits where the
current can follow se~ral different paths. e.g. a polyphase circuit. the definition further assumes that tne

breaking of the current by the ideal switching device takes place only in the pole considered.

Rather long, sorry

Anyhow, you get the drift. i expect even DC will have some sort of transient, which is the very voltage you are worried about, isn't it?

One thing. Though the total time is 60-80ms, the arcing time is usually far less. i haven;t studied high power DC devices, but in good AC circuit breakers, it is as low as 10ms sometimes (and there are current-limiting CBs, with times of even 1.6ms!)

300ms is a very high inductance, are you sure ? i have seen these in DC lifting magnets.

Please feel free to ask if i can be of some more help. Cheers

the ARTICLE FOR TRV was really long:) bt yes u r right that there will be some sort of transients in the circuit which may create problem. The load inductance is surely 300mH. The power supply delivering this high current will have ripple of high frequency, does this have any relation with the surge generated in current interruption? I m really messed up, becoz many a times i thought about these questions bt i couldnt solve them.

Well, join the club

i have been in switchgear design for 43 years and i still don't know how really DC can be broken. Though i have done some blowout designs using coils and magnets, these were in the order of a few hundred amps and time constants of 15-100ms.So i am no expert.

Anyhow, my feeling about the high-frquency ripple is .. (a) with a high inductance is the system, will they not get smoothened a bit? and (b) there may be a reactance effect due to the changing value of current.

Can you give me some details of the application and the circuit breaker details if they are not confidential? Maybe some things will become clearer....

Some knowledge on DC circuit breakers if you are interested....

AREVA HVDC

May not be relevant to you, still ....useful.

Again thanks for the links, but in these links, the current interruption is by means of creating artificial zeros and then interrupting them . This is quiet complex, rather what is preferred is to use dc circuit breakers, air circuit breaker can directly interrupt dc current by arc elongation through arc chutes....Scheron is having current rating of 6kA only. higher number will be needed to paralleled. Well can u suggest me that for ratings of such dc breaker how do we calculate the short time rating of the breakers.... there is a high current power supply connected to series with a dc circuit breaker(a resistor is also connected in parallel) ...which is feeding to highly inductive load.

Yes of course, i did not mean that you need something as complex as that Areva HVDC marvel of technology. Just some peripheral knowledge.

Now, if the current rating you require is far higher than 6kA, the temptation to parallel breakers is plain risky. The last breaker to open, even by a millisecond, will take the brunt of the arc. Not a good idea.

Your requirement is definitely in a special category. There is resistor in parallel to the breaker, i am perplexed. Please elaborate ?

ya there is a resistor in parallel to the DC CB but it is not the closing resistor. the function is similar as discharge resistor , it will be extracting the inductive coil's inductive energy AND DISSIPATING it into the resistor body in the form of heat energy. This event is seen in case of fault in the operation... Sir, were u also associated with GE, USA.

Ok, i still must say that i have not really a good idea of your aookication, but you are coy about revealing it in an open forum, no issue, don't worry

Yes, i was associated with GE Industrial Systems, Plainville, CT, in the last four years of my career, since we were designing many products for them.

Have you checked the Ferraz Shawmut links? They have to be what you need....

Sorry about the typo .. should have been "Application" .. looks hilarious by shifting one key

Teaches me not to submit the comment without a careful look, and using my reading glasses...

Yes Sir, I have already read about the Ferraz products. They are having the disconnectors that we require. I feel privileged to talk with you sir. Can I have your contact number or email-id.

Since CR4 frown upon bandying personal details in open fora. i will mail you my details. And thank you, the privilege was equally mine, since (a) i learnt something i did not know before and (b) your promptness in reading the posts and responding have been so good as to be an example for a "Good CR4 thread" ... i hope CR4 admin is monitoring. Cheers

thanks sir for the compliments,I m also new to this world of DC SYSTEM. These discussions were interesting becoz of ur answers and my questions. Looking forward for your mail.

Sir, still i am waiting for ur email...

Maybe this will meet your requirement ?

ferrazshawmut_Very_High_Power_Disconnectors.pdf

and the main website of the company

ferrazshawmut

We have a common paractice to consider each contact having resistance of 20μΩ. Now you have to check that from where you can measure the contact resistance of that isolator and how many contacts are in between your selected points for testing. Simply apply the formula- n X 20µΩ, where n= no. of contacts.

Thanks friend...well is similar method is applicable for breakers also? (multiplying the number of contacts with 20micro)

Yes, it is. I wanted to mean each joint or contact can be considered as 20μΩ.