Short Circuit Calculation Question?

rob w

Hi,

Doing some study for exams at the moment and come across a question which Im not sure im working out correctly. I have attached a screenshot of the question.

For part (i)

I have added total impedances (Ze + Zc1 + Zc2) in complex form and divided into the supply voltage (230 @ 0 degrees).

I calculated Zc2 by using the volt drop tables as below;


Real Component (Resistance) + Imaginary Component (Reactance)
Zc2 = (1.10 x 10^(-3))/sqrt(3) + (0.145 x 10^(-3))/sqrt(3)

= (0.00063 + j0.000083) x 30 metres (length of phase and cpc)

= 0.0189 + j0.00249


So, Ze + Zc1 + Zc2 = 0.025 + j0.0089

and this converted to polar is 0.026 @ 19.6 degrees

then divided that into 230 @ 0 degrees

Answer = 8846 Amps @ -19.6 degrees




Can anyone here shed any light on this, if its the right method or not, sorry if its a bit all over the place? Its doing my head in and need to get it right before exam time! Any help would be great.

Thanks

2011

OP: I have not done this for a while, so I have not looked at your answer as it might confuse me even if it is correct.

Here is the way I would do this:

The table shown does not give impedance values for the cables, instead it gives mV/A/m values.
This is a littel odd, however I would start by calculating the impedance per meter values for each cable (in complex form).
As the cable lenghts are provided the impedance values per cable (down the phase and back along the CPC) can then easily be calculated.

Next, I would tackle part (i)
First calculate total impedance, this is simply adding the real and imaginary componets together (Ze + Zc1 + Zc2).
Then its Ohm's law:

I = V / Z

(with Z in complex form)

Then part (ii) is the same except for the 2.5 component.

Just do a little sanity check and you should find that the answer for part i > the answer for part ii

Does this make sense??

I see that the socket circuit is supplied in steel conduit. This would provide a parallel path for an earth current, but without more information it is not possible to calculate this.

rob w

Yeah your right, it is bit of an odd question and it's annoying me because the lecturer couldn't really explain it himself!

I had done as you said, adding impedances to get total....using ohms law to work out current.....the fault current is obviously smaller for the final circuit as you said......all that does make sense to me.

It's the bit where I use the mv/a/m values to work out the impedance for the cable. I was told to multiply the value by 10^-3 (as its millivolts, thats fine).....but then told to divide by the square root of 3 to get the value for per metre. Is that necessary, and if so should I be using 400v as the voltage inn my ohms law calculation for the current?

Thanks for taking the time to reply by the way!

2011

I agree with the 10^-3 part asit ismV, but I would would not devide by the square root of 3.

I think you should use 230V as this is the potential above earth.
The voltage between phases is 400, but your short is to earth.

rob w

That's what I was thinking too, can't figure out the reason he uses the root 3, doesn't seem right to me!

2011

rob w wrote: »

That's what I was thinking too, can't figure out the reason he uses the root 3, doesn't seem right to me!

The line voltage (400V) divided by √3 gives the phase voltage. I would guess that this is where the √3 (incorrectly) came from.

However the volt drop (shown in the table) is a function of the current and impedance, not the voltage.

I see that you gave the answer in polar form, there is no harm in doing that but I don't think it was necessary so I do not think that you will get extra marks for it.

BrianDug

Real Component (Resistance) + Imaginary Component (Reactance)
Zc2 = (1.10 x 10^(-3))/sqrt(3) + (0.145 x 10^(-3))/sqrt(3)

= (0.00063 + j0.000083) x 30 metres (length of phase and cpc)

= 0.0189 + j0.00249


So, Ze + Zc1 + Zc2 = 0.025 + j0.0089

and this converted to polar is 0.026 @ 19.6 degrees

then divided that into 230 @ 0 degrees

Answer = 8846 Amps @ -19.6 degrees

This is perfectly correct answer & method.

hat's what I was thinking too, can't figure out the reason he uses the root 3, doesn't seem right to me!

I agree with the 10^-3 part asit ismV, but I would would not devide by the square root of 3.

The impedance values given are for a three phase cable, hence why you are dividing it be sqrt to get a single phase impedance value. Its not presented in an ideal manner but it makes you have to think about the problem.

Are you ok to do question 2?

P.S Be careful of how much you round your figures by as you can end up introducing a large relative error into your answer and it could end up been 50A+ off from the exact solution. Its just something to keep it in mind.

2011

BrianDug wrote: »

The impedance values given are for a three phase cable, hence why you are dividing it be sqrt to get a single phase impedance value.

What do you mean?
The phase and earth conductors are exactly the same. Therefore the impedance between earth and phase is the same as between phases.

frankmul

In the ETCI rules, there are two values given for the mV/A/m value for each size of cable, a single phase and a three phase. They are generally the same for smaller cable with some difference for larger cables.
Is it that here you are given the three phase value and if you need it for single phase ie live to earth you need to work out the single phase value using root 3?

2011

frankmul wrote: »

In the ETCI rules, there are two values given for the mV/A/m value for each size of cable, a single phase and a three phase.

I do not have the rules in front of me so I will take your word for it. I would assume that this is more to do will the thermal properties of the cable, heat dissaption etc. I would expect a 3 phase cable to get hotter due to it containing more current carrying conductors, therefore the resistance would be higher and consequently the volt drop.

The table on the question provides mV/A/m values that has been provided to detrmine inpedance values. At a fixed frquency the impedance is not a function of the applied voltage, therefore I do not see why we should devide the impedance by √3.
Stating that the ETCI do this will not earn marks in this type of exam, whereas a technical explanation of why the problem is tackled in a particular way will.

If the CPC was different that the phase conductors then I could understand why the impedance value would have to be modified. However in this case the question clearly states "Assume the size of each CPC to be equivalent to ites respective phase conductor".

I find some of question a odd for a number of reasons, not just the volt drop table. For example the reference to steel conuit being used. Perhaps this is to prompt the a comment that the prospective short circuit current will be higher than the calculated value due to a further reduction in the earth fault loop impedance.

BrianDug

What do you mean?
The phase and earth conductors are exactly the same. Therefore the impedance between earth and phase is the same as between phases

In the ETCI rules, there are two values given for the mV/A/m value for each size of cable, a single phase and a three phase. They are generally the same for smaller cable with some difference for larger cables.
Is it that here you are given the three phase value and if you need it for single phase ie live to earth you need to work out the single phase value using root 3?

Sorry 2011 I did not phrase my response very well, what I was trying to say is what frankmul has explained.

The questions are not very clear and should be rephrased. The part of question with the conduit will leave you asking more questions than coming to a an actual solution.

frankmul

2011 wrote: »

The table on the question provides mV/A/m values that has been provided to detrmine inpedance values. At a fixed frquency the impedance is not a function of the applied voltage, therefore I do not see why we should devide the impedance by √3.

2011
Your right that the impedance is not a function of the applied voltage. The √3 is not due to the voltage but is due to the relationship of the currents in a three phase system. The voltages are out of phase but so are the currents.
I'm not sure if that clears things up but for an exam, the √3 should be included.

2011

frankmul wrote: »

The √3 is not due to the voltage but is due to the relationship of the currents in a three phase system.

But the question is in relation to a single phase to earth. It does not ask about a short between different phases.

frankmul

The figures in the table are for calculation based on three phase current. The figures take into account the fact that the current are out of phase. These figures do not hold true for a single phase current ie phase to earth. The √3 is the conversion factor used to allow the figures in the table be used for a single phase supply.
Does that make any sense?

2011

frankmul wrote: »

The figures in the table are for calculation based on three phase current. The figures take into account the fact that the current are out of phase. These figures do not hold true for a single phase current ie phase to earth. The √3 is the conversion factor used to allow the figures in the table be used for a single phase supply.
Does that make any sense?

In my view the table provides a volt drop that is proportional only to the current and the length of the cable. For a fixed current and cable length the volt drop will remain constant regardless of the phase angle or number of phases. In other words for a given the volt drop will only vary if the current or impedance changes regardless of the number of phases, or angle between phases.
After all volt drop is simply calculated using Ohm's law, Vd = I x Z

If you look at part (ii) of the question it does not even tell you if the "final circuit outlet" is 3 phase or single phase. Why is this? I would argue this is the case because the calculation is the same.

frankmul

2011 wrote: »

Vd = I x Z

Yes, I agree with that but I wonder is Z value taken from volt drop tables equal to the impedance of a cable?

2011

frankmul wrote: »

Yes, I agree with that but I wonder is Z value taken from volt drop tables equal to the impedance of a cable?

There is nowhere else to get it from.

frankmul

The mV/A/m value is given as 29 for single phase 1.5mm2, can anyone work what the resistance of 100 meters of the cable is?

rob w

frankmul wrote: »

The mV/A/m value is given as 29 for single phase 1.5mm2, can anyone work what the resistance of 100 meters of the cable is?

Haven't got my regs with me, but I think the max ccc for a 1.5mm cable is around 22amps, (could be wrong, just off the top of my head), then using that as the load;

Volt drop over 100m would be : .029 x 22 x 100 = 63.8v (this being worst case)

So using ohms law, z=v/i = 63.8/22 = 2.9ohms


This could be miles off though, correct me if I'm wrong!!

frankmul

Rob
what you did looks right but I was wondering, what is the value if you measured it with a meter or better, calculated it from its resisitivity (forget about the voltdrop table for a minute). I think the values are in the back of a reci guide book?.

Bruthal

frankmul wrote: »

Rob
what you did looks right but I was wondering, what is the value if you measured it with a meter or better, calculated it from its resisitivity (forget about the voltdrop table for a minute). I think the values are in the back of a reci guide book?.

I measured it before on 100 meters of 1.5 pvc/pvc. I think it was 1.5 ohms.

The 2.9 ohms given above is probably for supply and return inclusive.

Bruthal

rob w wrote: »

Haven't got my regs with me, but I think the max ccc for a 1.5mm cable is around 22amps, (could be wrong, just off the top of my head), then using that as the load;

Volt drop over 100m would be : .029 x 22 x 100 = 63.8v (this being worst case)

So using ohms law, z=v/i = 63.8/22 = 2.9ohms


This could be miles off though, correct me if I'm wrong!!

Because its mv per single amp in the table, the mv will = milli ohms. So simply getting 0.029 mv and multiply by 100 = 2.9 mv per amp per 100 meters, the 2.9mv will = 2.9 ohms. So your answer is right.

frankmul

I tried to calculate it using pL/A
Rho, I took as 1.7 x 10^-8
L as 200m phase and neutral
And A as 1.5 x 10^-6

I got 2.26 ohm
Near enough is it?