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Topic Title: Two-Phase Supply-
Topic Summary: Wondering how to calculate KVA on a Two Phase Supply
Created On: 28 March 2014 08:02 AM
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 28 March 2014 08:02 AM
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Sanft

Posts: 37
Joined: 22 August 2011

Good Morning Gents- Have quite a scenario here that I haven't come across before- I'm a quite young in the industry to be fair =]

We have had a report from a school that their main head was 'smoking' and the fire brigade had come out to isolate the electrics, overloading springs to mind-

The School has 2x 100A Fuses and 1x Blank in the main head, there are only two phases coming into the building (L2, L3 and N), we have left a Metrel power analysing meter just to measure the current that is being drawn over the course of a week as the Supplier didn't want to know anything about this without physical proof- wonderful bunch of chaps they are,

Turns out, approximately, that L2 looked to range between 80A - 140A and L3 looked to be 90A - 140A, the worrying thing being that on the offchance we clamped the Neutral conductor as we had a spare clamp as L1 was missing, and the Neutral Conductor seems to range from around 75A - 105A which worries me a little.

My theory is a little hazy on this matter, but is the Neutral carrying so much current because L1 is absent? and there for the 'imbalance' between L1+2+3 is quite off as there is effectively 0A on L1?

This is the main part though- the client has asked me to calculate KW/KVA, but being that there are only two phases; is this done any differently? And if so could you point me in the right direction please?

Many Thanks in Advance,

Kind Regards,

Andrew
 28 March 2014 12:20 PM
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statter

Posts: 126
Joined: 06 February 2013

The neutral current is indeed high because there is no load on L1.
You can calculate the kVA a phase at a time if you use the max values above its 230x140 + 230x 105 = 57kVA.

You could take this on a three phase 100A supply if you can balance the loads but depending on the nature of the loads and how well you can balance them it might be a bit tight. This would also reduce the neutral current.
 28 March 2014 02:58 PM
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Sanft

Posts: 37
Joined: 22 August 2011

I think that site is looking to upgrade the supply regardless, the cable in the road from the Energy Supplier is apparently only a Two-Phase Cable (Road was previously just a domestic dwelling road, the school was built some years after and has been expanding ever since)

I was under the impression that the Sq. Root of 3 would have to be used in some instance to calculate KVA, Its been a while since I've tampered around with Three-Phase theory- Using your Mathematics above, would it be simply as this...

L2 = 140A Max, 140A x 230v = 32200 VA (32.2KVA)
L3 = 140A Max, 140A x 230v = 32200 VA (32.2KVA)
Neutral = 105A Max, 105 x 230v = 24150VA (24.15KVA)

32.2KVA + 32.2KVA + 24.15KVA = 88.55KVA?

It just feels too basic that this is worked out like this, maybe I over-complicate these things, I was expecting Sq. Root of 2 to be thrown into this somewhere haha-

Thanks for the advice =]
 28 March 2014 04:03 PM
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AdrianWint

Posts: 262
Joined: 25 May 2006

The flaw here is your neutral calculation. The voltage between neutral & neutral (yes, I do mean neutral) isn't 230V ... its 0V!

The power is only drawn from the phases, so in your case the power is 32.2 + 32.2 = ~65kVA

If it where a single phase system you wouldn't do, say, 20A * 230V = 4600VA & then add another 4600VA for the neutral...........

The sqrt(3) relationship only comes into play if you use the line-line voltage eg. If the line current is 20A & the line-line voltage is 400V then the power is:

sqrt(3) * 20 * 400 = 13.8 kVA

which is the same as:

(20*230) + (20*230) + (20*230) = 13.8 kVA

Edited: 28 March 2014 at 04:09 PM by AdrianWint
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