I've been asked to carry out alternative hand fault calculation to align with Amtech. I have the upstream fault current 10kA, pf 0.95, length 95m and the impedance of the conductors 4 core 95mm copper 90 C.

My issue is how to apply the upstream fault current without knowing the R and X.
Please help

Have you already calculated this on Amtech? Why not extract the data and formulas from the programme, so you can analyse the figures.

I always double check their figures, as they have been wrong numerous times.

All the detailed formulas can be found within the manual

Regards

Jobbo

Hi Jobbo, I don't even have the Amtech Software as the cable calc report was submitted to me to review. I've searched for the manual online but can't find one. Do you know what formula is used for the max fault current? I'm getting figs using IEE guidance note which are 20% lower than Amtech and I can invalidate Amtech till I know precisely why we have the variance.

Perhaps you could detail the Amtech report figures and your own, so we can help you establish the variant

What is the source of supply used on the report?


Regards

Jobbo

The source of supply is an upstream board (400 Line V) with 10000A fault current. The cable to the load end is a 4 core 90 degrees C thermosetting armoured Cu, 95m in length. Amtech gives 6154A for the max fault current but my calculations yields 5473. We used 95mm2 cable with BICC data giving It = 263A for a design current of 225A, with V = 240V, which I assume Amtech is using.

The source of supply is an upstream board (400 Line V) with 10000A fault current. The cable to the load end is a 4 core 90 degrees C thermosetting armoured Cu, 95m in length. Amtech gives 6154A for the max fault current but my calculations yields 5473. We used 95mm2 cable with BICC data giving It = 263A for a design current of 225A, with V = 240V, which I assume Amtech is using.

I have just checked with my Amtech and get a 6054 A fault current using a 400V supply at 10kA at the origin with a Ze of 0.0231.

I did a manual calculation and got a fault current of 5355A at the far end of the cable for 230/(Ze + Z1).

I will try and find the calculation that Amtech uses in their manual.

-------------------------
John Peckham

http://www.astutetechnicalservices.co.uk/

Where is our design duo when we need them? Come on OMS and Zs point out where I am going wrong?

-------------------------
John Peckham

http://www.astutetechnicalservices.co.uk/

Edited: 21 July 2012 at 06:27 PM by John Peckham

I have just had a look in the Amtech manual and they use Uo/(Ze + Z1) also.

I have just used their calculated values of Ze (0.02309) and Z1 (0.02464) and get 4818As which is different to their displayed figure of 6054A.

-------------------------
John Peckham

http://www.astutetechnicalservices.co.uk/

Cheers John.

John,
The Z1 value 0.02464 may be for 90 deg C which would have to be corrected to 20 deg C. Also Z1 I suspect may be the total impedance for both the phase and neutral conductors. For max fault, the neutral Z will not be utilised.
Not sure how Amtech does it though!

First the OP has given a power factor of 0.95. This must be at the load end as the source power factor on short circuit will be much lower.

If you put 10kA into Amtech it assumes a source power factor of 0.25. It gives Ze as 0.023 ohms but this is nearly all reactive.

Check out
pf = R/Ze

So, Re = Ze * 0.25 = 0.0058 ohms

and, Xe = Ze * sin(arcCos(0.25)) = 0.023 * 0.97 = 0.0223 ohms

For a three phase fault

If = U0/Z where Z is the complex impedance of the supply and one line conductor. (check GN6)

So we have.

Z = the phasor sum of Ze and Zcable. Ze is nearly all reactive and Zcable is nearly all resistive

Z = SR[(Zcable*Zcable) + (Ze*Ze)]

Zcable is approximately R1 which Amtech give as 0.03 ohms and Ze is approximately 0.0223 ohms

this give Z = 0.038 ohms

and , If = Uo/0.038 = 6084A

Come on chaps get with the programme

Edit - Some corrections made

Regards

Geoff Blackwell

Edited: 21 July 2012 at 07:26 PM by GeoffBlackwell

Uti, I thought about this while I was driving this evening. My first question was to query which end you were calculating at but I think you've got that covered. When you say to align with amtech, do you mean that you are expected to come up with the same answer? Unless it is one of your own amtechs I think it is ok to come up with a different answer. I'm assuming this is a cable analysis report from Pro and not from Single Cable. For your peace of mind, have a look at some of the other reports....you might see different numbers on them because Amtech has got a few bugs.

In order to come up with the same answer, you would need to know exactly how the calc was entered and which, if any functions were set to automatic. It is on the automatics that things will change because each one causes a minute daisy chain of adjustments. I don't use automatic functions at all for inputting. Amtech makes assumptions which you might not have.

How the load was entered is also a factor, one box will change several others, was this entered onto the motor section etc.

You'd need to see a screen shot of the entered data and then push the calculation up and down to see how it changes things. Ask for one?

I don't have amtech here and having just found the troubled mate on her doorstep smoking fags and drinking stella with the local drunks I'm not very logical. But I get 6052 with a triangle drawn on the back of an evelope and R1 stolen from a previous calc which may have been rounded up or down at some stage.

Give or take my somewhat laisse faire attitude right now..Tell them you've come up with a different number. You can't change physics but you can juggle amtech to give you exactly the numbers you want to see. There's a reason why you are being asked to validate and it's probably got something to do with the client trusting you more than a bit of binary.

And oh, Try 230V not 240V.

Zs

Edit: See you in about a week Chaps. Taking a few days off from things electrical to carry out an art commission.

Edited: 22 July 2012 at 01:12 AM by Zs

It would appear that one this occasion, and most others, Amtech is correct. . As Geoff has pointed out you cannot do a simple addition. Amtech does use a PF for the Tx. of 0.25 as default setting. So the current will be out of phase with the current through the mainly resistive cable so the fault current will be the phasor sum of the 2.


You get the same result on Amtech Single cable and Amtech Pro.

-------------------------
John Peckham

http://www.astutetechnicalservices.co.uk/

Like to rephrase that?

Regards

Thanks all for your contributions. Using a source short circuit pf of 0.25, I get 6.053 kA and Amtech yields 6.154. However, assuming a pf of 0.190 gives exactly 6.154. Tried these in other Amtech reports and the pf seem to lie between 0.185 and 0.25 to arrive exactly at Amtech's figures. As I don't have Amtech myself, I assume it uses the ratio of the Transformer R/X and thus ignores the cable from the Transformer's LV board to the sub board upstream of the circuit. Could have been the user's error though.

Well, I've been rehoming a few chickens, John - but Geoff has clarified load and source end power factor and the impact of predominantly source reactance and line resistance on downstram fault level.

I get the same numbers (within a few Amps)

Regards

OMS

-------------------------
Failure is always an option

Out of interest, why have you been asked to do this?

Just to learn how cable calcs are done as part of my development.