Confused about Zs 5 second disconnection time

[Cookie]

Running some calcs I'm confused about what Zs to use when calculating 5 second disconnection times.

In circuits with 1.5mm2-4mm2 cable a fault doesn't seem to pull the voltage down much at the trafo terminals so from an ohms law perspective the current magnitude reflects simple V/(R1 + R2)=I.

However I'm noticing that on mains/distribution board to distribution board circuits that the larger conductors involved (and in turn lower Z) pull the voltage down on the trafo. This means current will be lower than actually calculated? How do I approach this?

 

[Julie.]

Running some calcs I'm confused about what Zs to use when calculating 5 second disconnection times.

In circuits with 1.5mm2-4mm2 cable a fault doesn't seem to pull the voltage down much at the trafo terminals so from an ohms law perspective the current magnitude reflects simple V/(R1 + R2)=I.

However I'm noticing that on mains/distribution board to distribution board circuits that the larger conductors involved (and in turn lower Z) pull the voltage down on the trafo. This means current will be lower than actually calculated? How do I approach this?

Zs is not r1 + r2
It is this value plus Ze, which is the effective transformer impedance

If you don't include the Ze the actual current will be much lower than any random value calculated by ignoring the transformer

 

[Cookie]

Zs is not r1 + r2
It is this value plus Ze, which is the effective transformer impedance

If you don't include the Ze the actual current will be much lower than any random value calculated by ignoring the transformer

My apologies. I was thinking about Ze.

Ze is small relative to R1+ R2 in 2.5mm2 cable so the voltage stays near nominal at the trafo during a fault at the socket. But a fault on 120mm2 cable pulls the voltage down at the transformer terminals to half or less.

 

[pc1966]

While often Ze is though of in terms of earth impedance, what matters is the earth loop impedance.

More fundamentally it is the fault current that disconnects the OCPD. Usually that is what you get from a measured Zs & supply voltage but some cases like a UPS or small generator might appear "stiff" in terms of good voltage regulation, but when actually shorted out give far lower fault current than the PFC measurement would imply.

We have a couple of makes of 5kVA UPS, one measures 2kA PFC and the other about 80A. When I am hacked off by the crappy software I am tempted to find out what the real fault level is, which I expect is close to the 80A value (which won't meet 0.4s for a 13A plug's fuse!), but so far professional sanity has prevailed...

 

[Cookie]

I understand, but my point is that with larger circuits (lower Z) the voltage at the trafo will dip, so technically 230 volts is not an accurate number or the 80% multiplier.

 

[pc1966]

I understand, but my point is that with larger circuits (lower Z) the voltage at the trafo will dip, so technically 230 volts is not an accurate number or the 80% multiplier.

No, as your Ze value includes the source impedance, so you are computing PFC from the supply voltage (no-load EMF) and loop impedance.

If you have a transformer impedance comparable to your total cable R1+R2 then you would see the supply volts drop to around 1/2, but then your Ze is both transformer impedance and R1+R2, so both are dropping their share.

It might not look as simple given the transformer may be far more inductive than the cable, so you would need to measure voltage and phase to get the numbers to add to the no-load 230, but they should do.

 

[pc1966]

The 80% multiplier is for cable heating before the fault (i.e. assuming it runs at 70C, which is a bit high for most installations). Basically you compute your max Zs and then test your cables & supply when cold (or 20C design values) are no more than 80% of this.

True your transformer impedance might to rise as much (inductive component ought to be fairly stable with temperature), so 80% may be on the conservative side here, but for a general rule it works.

In the UK have a 0.95 factor for Uo to get the minimum voltage and so minimum PFC (not quite our -6% tolerance but close).

 

[Cookie]

No, as your Ze value includes the source impedance, so you are computing PFC from the supply voltage (no-load EMF) and loop impedance.

If you have a transformer impedance comparable to your total cable R1+R2 then you would see the supply volts drop to around 1/2, but then your Ze is both transformer impedance and R1+R2, so both are dropping their share.

It might not look as simple given the transformer may be far more inductive than the cable, so you would need to measure voltage and phase to get the numbers to add to the no-load 230, but they should do.

So basically with Ze, I turn the transformer into an series reactor+resistor (in vector form?) and assume 230 volts infinite on the other side of the trafo?

That makes more sense.

The 0.8 vs 5 second time threw me off a bit... in regards to the physics...

 

[Cookie]

The 80% multiplier is for cable heating before the fault (i.e. assuming it runs at 70C, which is a bit high for most installations). Basically you compute your max Zs and then test your cables & supply when cold (or 20C design values) are no more than 80% of this.

True your transformer impedance might to rise as much (inductive component ought to be fairly stable with temperature), so 80% may be on the conservative side here, but for a general rule it works.

In the UK have a 0.95 factor for Uo to get the minimum voltage and so minimum PFC (not quite our -6% tolerance but close).

I obtained 80% regarding coefficient C from IEC 61200-413:





My understanding is that C is derived from the voltage drop of the transformer while a L-E short circuit is occurring at the far end of a circuit inside a building.

Typically a short circuit on a 16-32 amp branch circuit will on average lead to the voltage (between the X0 & X1 of the transformer's terminals) dipping to about 80%. So at 230 volts, the measured voltage will be 184 volts. Assuming phase and earth are the same size from the trafo to the socket, the indirect contact voltage on the case of the faulty appliance will be 92 volts relative to remote earth. So going by the body graph a 0.4 second disconnection time is mandated.

On circuits over 32 amps R1+R2 is much lower ( as an example, 150mm2 cable vs 1.5mm2 cable) and as such the voltage drop is far more severe across the supply transformer.

During a fault on a 100 amp feeder, the voltage between the X0 and X1 terminals may drip down to 30% of the nominal. As such the indirect contact voltage to remote earth at the fault point will be 35 volts allowing for a 5 second disconnection time.

 

[pc1966]

So basically with Ze, I turn the transformer into an series reactor+resistor (in vector form?) and assume 230 volts infinite on the other side of the trafo?

Basically yes, you can convert it all in to a Thévenin equivalent circuit.

That makes more sense.

The 0.8 vs 5 second time threw me off a bit... in regards to the physics...

The two are different aspects, hopefully clearer now!

 

[Cookie]

Basically yes, you can convert it all in to a Thévenin equivalent circuit.


The two are different aspects, hopefully clearer now!

Neat, cool, and easy to do.

I disagree. Disconnection time is dictated by R1+R2.

 

[pc1966]

I disagree. Disconnection time is dictated by R1+R2.

No, the disconnection time is determined by the OCPS and PFC, in turn from Uo and Zs (Ze + R 1 + R2)

What the disconnection time should be is more complicated...

That leads back to your 80% (which is a different 80% to the 0.8 factor I thought) and the question of shock risk. For that you need more information on high current faults as you can't assume the Ze value is symmetric about "true Earth", nor can you always assume that R2 = R1 as in many cases (e.g. the UK T&E cable) it can be quite asymmetric.

 

[Cookie]

No, the disconnection time is determined by the OCPS and PFC, in turn from Uo and Zs (Ze + R 1 + R2)

What the disconnection time should be is more complicated...

That leads back to your 80% (which is a different 80% to the 0.8 factor I thought) and the question of shock risk. For that you need more information on high current faults as you can't assume the Ze value is symmetric about "true Earth", nor can you always assume that R2 = R1 as in many cases (e.g. the UK T&E cable) it can be quite asymmetric.

OCPS? I'm having trouble grasping how perspective fault current is more accurate in determining disconnection time. I would think instead the ratio of the trafo impedance/alternator compared to the impedance of the conductors leading up to the fault point.

Correct, in many cases R2 is of higher Z, but that ratio is taken into consideration when the equations are run.

 

[pc1966]

OCPS? I'm having trouble grasping how perspective fault current is more accurate in determining disconnection time.

Sorry, should be OCPD - over current protection device (fuse, MCB/MCCB, etc)

For any given OCPD the disconnection time depends on the fault current, for a fuse smoothly do, for breakers only smooth to the point of the "instant" magnetic trip then time changes suddenly.

I would think instead the ratio of the trafo impedance/alternator compared to the impedance of the conductors leading up to the fault point.

Correct, in many cases R2 is of higher Z, but that ratio is taken into consideration when the equations are run.

For a linear system, and the grid will approximate that for an end of line fault - the fault current depends on the equivalent source voltage and loop impedance, i.e. PFC = U / Zs.

The resulting fault voltage (and thus shock risk) depends on that fault current and the 'R2' proportion of the loop impedance seen between the points that a person can touch.

Worst case is touching the faulty appliance and something connected to the outside true Earth, as then it is PFC * (Z2 + R2) where Z2 is the return path impedance to the supply earthing point, and R2 the local circuit's return path to the DB.

 

[Cookie]

Sorry, should be OCPD - over current protection device (fuse, MCB/MCCB, etc)

For any given OCPD the disconnection time depends on the fault current, for a fuse smoothly do, for breakers only smooth to the point of the "instant" magnetic trip then time changes suddenly.

For a linear system, and the grid will approximate that for an end of line fault - the fault current depends on the equivalent source voltage and loop impedance, i.e. PFC = U / Zs.

The resulting fault voltage (and thus shock risk) depends on that fault current and the 'R2' proportion of the loop impedance seen between the points that a person can touch.

Worst case is touching the faulty appliance and something connected to the outside true Earth, as then it is PFC * (Z2 + R2) where Z2 is the return path impedance to the supply earthing point, and R2 the local circuit's return path to the DB.

Right, but you're leaving out the source impedance regarding touch potential. The greater the voltage dip at the spades of the transformer during a fault the longer the disconnection time is permitted in theory and in practice.

 

[pc1966]

Right, but you're leaving out the source impedance regarding touch potential.

No, the source impedance is part of what determines the PFC, and that in turn with "R2" determins what the touch voltage is.

So for a fixed R2 and the supply Z increases, so does the PFC drop, and then PFC*R2 drops as well.

 

[Julie.]

Touch potential can be quite different to what you may think.

Let's say you have a transformer and supply cable to a property, say the neutral conductor is 0.25 ohm to the star point and the line plus winding is 0.75 ohm, then within the property the circuit r1+r2 is 1.3 ohm.

In the event of a solid fault this would be 2.3 ohm and at 230v would be 100A fault current.

Take the case of a TT arrangement, the local earth obtains its connection via the ground/earth itself from the star of the transformer.

So, the neutral point at the property would rise to 100A x 0.25 ohm = 25V (to earth)
The line would drop to 100A x 1.3 ohm = 130V ABOVE the neutral so 155V (to earth)

Now take the case of a TN-C-S, in this case the combined neutral/earth conductor is usually tied to earth locally via connections to gas/water pipes, multiple earth rods within the supplier network etc.

So even though there is still 25V dropped in the neutral supply cable, the actual local earth is very close electrically to the supply cable at the property NOT the end at the transformer's star point.

So the neutral point at the property would remain close to zero volts measured to the local earth
The line would again go to 130V above neutral, but this time that's 0V so we only see 130V to local earth.

In practice the figures are obviously smaller - I just picked nice figures to illustrate it.

The multiple connections between local earth and the combined neutral/earth connection (PEN) result in lower resistances, and the holding of local earth close to that of the PEN local to the properties.

In the US most properties are similar to the TT system here - even if it has a utility supplied earth, it's often an additional conductor ( TN-S) so very much like TT in terms of this volt drop calculation.

It's the main reason we use the term earth and not ground, ground would be the ground at the substation star point/ground rods - in the US this would be a similar point in terms of volt drop at the property when there is a fault; but in the uk and much of Europe the local ground would be tied close to the point where the PEN is separated; so not the same point as the transformer star. Hence the differential between ground (substation) and earth.

 

[Cookie]

No, the source impedance is part of what determines the PFC, and that in turn with "R2" determins what the touch voltage is.

So for a fixed R2 and the supply Z increases, so does the PFC drop, and then PFC*R2 drops as well.

Right, supply impedance determines PFC. Same applies to touch voltage.





The lower the transformer Z the higher voltage at its output for a short circuit inside the building.

 

[Cookie]

Touch potential can be quite different to what you may think.

Let's say you have a transformer and supply cable to a property, say the neutral conductor is 0.25 ohm to the star point and the line plus winding is 0.75 ohm, then within the property the circuit r1+r2 is 1.3 ohm.

In the event of a solid fault this would be 2.3 ohm and at 230v would be 100A fault current.

Take the case of a TT arrangement, the local earth obtains its connection via the ground/earth itself from the star of the transformer.

So, the neutral point at the property would rise to 100A x 0.25 ohm = 25V (to earth)
The line would drop to 100A x 1.3 ohm = 130V ABOVE the neutral so 155V (to earth)

Now take the case of a TN-C-S, in this case the combined neutral/earth conductor is usually tied to earth locally via connections to gas/water pipes, multiple earth rods within the supplier network etc.

So even though there is still 25V dropped in the neutral supply cable, the actual local earth is very close electrically to the supply cable at the property NOT the end at the transformer's star point.

So the neutral point at the property would remain close to zero volts measured to the local earth
The line would again go to 130V above neutral, but this time that's 0V so we only see 130V to local earth.

In practice the figures are obviously smaller - I just picked nice figures to illustrate it.

The multiple connections between local earth and the combined neutral/earth connection (PEN) result in lower resistances, and the holding of local earth close to that of the PEN local to the properties.

In the US most properties are similar to the TT system here - even if it has a utility supplied earth, it's often an additional conductor ( TN-S) so very much like TT in terms of this volt drop calculation.

It's the main reason we use the term earth and not ground, ground would be the ground at the substation star point/ground rods - in the US this would be a similar point in terms of volt drop at the property when there is a fault; but in the uk and much of Europe the local ground would be tied close to the point where the PEN is separated; so not the same point as the transformer star. Hence the differential between ground (substation) and earth.

I'd have to disagree. If the person is outside in the back lot, on wet grass, the person will see those 25 volts drop across the neutral in addition to the internal R2 volt drop at the outside socket. A local earth rod does not change that unless you are standing on the earth rod itself.


In terms of being inside the building the earth rod again makes no difference when everything is bonbded to the MET. The 25 volts are only eliminated via bonding of pipes and rebar to the MET, not earth rod connected to the MET.