Cable Calc Question

[viperrowe2606]

Hi all, really hope someone can help me as I am doing something wrong when it comes to cable calcs, I am currently working my way through my lvl 3 EAL. Would someone be able to work through this question so i can see where i am potentially going wrong. Thanks in advance.

TN C S = 0.35

Ambient temp = 20

3ph DB (BS EN 60898)

3ph circuit for car lift, wired in pvc singles in trunking with 3 other circuits, circuit length 8M.

3 ph supply to a two post vehicle lift: 400 V 3 kW with a p.f of 0.85,

My calcs so far:

3000/ 1.732x4000.85 = 

Ib = 5.094A

In = 6A (type C) (6A ≥5.094A)

6A/0.65(cg)

Iz = 9.23A

It = Cable Size = 1mm2

38x5.094x8/1000 =

Voltage drop = 1.54V

36.20(R1+R2) x 1.20(multiplier table I3 OSG) x 8 / 1000 = 0.34

0.34+0.35(Ze) = 

Zs = 0.69

230/0.69 = 

Isc = 333A

Adiabatic (0.2s as 3ph?) = 1.29mm (needs to be same or smaller than line?)

I must be doing something wrong, please help.

 

[Fleeting]

Your nominal voltage (Uo) is 230v so your disconnection time for a TN system is 0.4 sec. I haven't done the calculation but consider if your calculated size of cpc is 1.29 and your line conductor is 1.0 what does this suggest. Fault current will use the line and cpc so if your calculated cpc results in 1.29 then this must equate the 1.0 is undersized. Using the adiabatic formula and assuming the line conductor is correctly sized the cpc should not exceed the line conductor csa.

 

[viperrowe2606]

This is my headache, my tutor has said that disconnection time should be taken as 0.2 as it’s 3ph which I questioned as it’ll be the line voltage surely. The problem is every time i increase line size R1+R2 values to say 1.5/1.5 which I thought would be sufficient my cpc size is still coming out high on the adiabatic. I’m obviously being thick somewhere. I started the calc by using 0.4s but still throwing it out

All other calcs I’ve attempted in this way have came out “great success” but this one is throwing me 

 

[Phoenix]

The disconnection time in the adiabatic is not the maximum allowed disconnection time (i.e. 0.4sec) but the actual disconnection time that is achieved with a fault of that magnitude

 

[viperrowe2606]

This is what I put to my tutor who was flabbergasted by the t= equation. Do I need the t= equation as this would give me a time of around 0.1s

 

[Phoenix]

What equation are you using? You might have to re-arrange it to suit what you are trying to do

 

[viperrowe2606]

Full cable calcs for the above circuit I’m lost as I’m being fed cross information. I thought for the adiabatic equation you have to work out 

1) fault current 230/Zs

2) disconnection time t= K*XS*/I*

 

[Fleeting]

Your disconnection time is 0.4sec for the given circuit.

 

[viperrowe2606]

Right so not 0.2s like I was informed even using 0.4s disconnection time my cpc is coming out larger than line conductor in the adiabatic even when stepping up the R1+R2 values given in I3 OSG

Thanks so much for your replies btw!!! This has been causing me grief, making me feel like I should swap over to become a plumber

 

[Fleeting]

Even though you have a line to line voltage off 400 the nominal voltage is 230 so 0.4 sec applies. Are you using the adiabatic equation and have you ascertained the k factor for use in the equation. 

 

[viperrowe2606]

Yeah using adiabatic I’ve done the adiabatic with the Isc for R1+R2 1mm/1mm

1mm/1.5mm

1.5mm/1.5mm

all came out higher than the line csa. All other calcs I’ve done have been ok. 
k=115

 

[Fleeting]

Yes I have just done the calculation and it is larger than the line even at 0.1sec for a considered instantaneous trip. I think you would have to quote the "rule of thumb" table in BS7671 whereby the line conductor up to and equal to 16.0 should have an equally sized cpc. I would also suggest 1.0 is undersized for a power circuit, see BS7671. I think you need to confer with your tutor but show them your workings.

 

[viperrowe2606]

So I just use the rule of thumb rule then? I initially done the calc using 1.5mm/1.5mm but as this was also coming out high I went around in circles with it. I just assumed i was doing something wrong and was afraid I was going to get "dumbed". Like i say ive followed the rest of the calcs the same way and got correct results, it was just this one that i struggled with. Thanks so much again for the replies and help man!!! 

    1 other quicky, Volt drop calc I would use 3ph value for and check scope with 400V for correct? 400V/0.05?

 

[Phoenix]

Have a look at the second paragraph in 434.5.2 and note what it says in relation to instantanous operation

 

[viperrowe2606]

Ive just done the calc again going up in size until 4mm/4mm (R1+R2 = 9.22) This works but i feel as though 4mm is overkill surely?

 

[Fleeting]

I would speak to your tutor with regard to just selecting a cpc which equates to the line conductor and explain your workings. I would assume the three phase circuit is balanced so use the method you use for single phase, an unbalanced circuit is far more complicated and you haven't been given any information to suggest this is the case.

 

[viperrowe2606]

This is exactly what I spoke to my tutor about, so the t=k2 X S2 / I2 calc is the maximum time the line conductor can withstand the fault for? is this correct?

I feel so dumb right now

 

[Phoenix]

This is exactly what I spoke to my tutor about, so the t=k2 X S2 / I2 calc is the maximum time the line conductor can withstand the fault for? is this correct?

I feel so dumb right now

Yes. What do you get if you plug your values for K, S and I into it? (confirm what values you are using for each as well)

 

[viperrowe2606]

k=115

S=1.5mm2

I = 396.5

1152 X 1.52 / 396.52 = 0.189s

 

[Phoenix]

Looks right to me, so your 1.5mm² cable will be able to survive a fault of 396.5A for 0.189s. Now how long will the breaker take to operate with that fault flowing?