Recent content by ShooterBoter

Well, if you (or your client) don't care about the measuring being wrong then I don't understand your issue. Assuming of course you don't have any directional protection (pure over current IOC and TOC; these protection functions don't care about the voltage/angle) But not quite sure what ratio...

What is the transformer vector group? What is your aim with the relay, i.e. at what fault scenario should the relay pick up?

A diagram says more than 1000 words...please draw one. If the disconnect terminals should be open or closed depends on where in the circuit they are located.

"In addition, the phase of the power is not correct in your calculations. Use the difference of the phases of voltage and current, not their sum." This is why you have to use the conjugate of the current, but you always add the angles. (22A @ 20 degrees)*=22A @-20 degrees

Ok, I gave it a try, anyone else?

You wrote that VLL = √3 x VLN. Well, there's you fundamental error. The correct way of putting it should be: |VLL| = √3 x |VLN|. See what I'm aming at? Example: VLN= 230V∠0º VLL= 400V∠30º You stated that "VLL = √3 x VLN": "400V∠30º" = √3 x 230V∠0º, as you can see this is not correct...

Hmm, sorry for my previous reply, thick fingers on a small keyboard is what caused that... Any how; S=VLN x IL ∠∅ Where VLN=Line to neutral voltage IL=Line current ∅=Angle between line to neutral voltage and line current The 3-phase apparent power is; S3P=3 x VLN x IL ∠∅ As you stated we can...

Ok, here it goes: 1) S=Vp x I ¼∅

Not that I advocate the Schneider switchgear, but it does exceed current rating, it's up to 5000A. Link

Schneider-Electrics PIX-H switchgear is one. Don't know if they have any references to offshore, you'll have to ask Schneider.

Well, he/she asked for kVA...so the equation still holds but with wrong term; "P" should be changed to "S".

@Skogsgurra "If it is a wye winding - the ampere-turns do not agree so well..." Why wouldn't the amp-turns match in this case?

Take a look at Elmaster, Link SB

Oops, I totally forgot about this thread... I haven't checked the connection via ethernet Cable, but I know that a Dsub9F-Dsub9F serial cable will work. I can't remember if it's a strait or a null modem Cable I use - need to check this. Kind regards, SB

Usually the battery manufacturer will define a "end voltage". Discharging the battery below this point will probably cause damage to the battery. If you discharge the battery way below this pint you'll not be able to charge it to full capacity again.