Mains supply : what is its reactance?

[Roger44]

Hello
Just supposing we could short circuit the mains supply with a perfect switch going from infinite resistance to zero ohms in picoseconds, and that we live 80 km from an oil-powered generating station with sub-stations somewhere along the line. Does anybody have a ball-park figure for the reactance (R+Xinductance) in the mains supply that would slow down the current rise?

Thanks for any ideas on this one.
Roger

 

[Tony Stewart]

0.168+j 0.789 per mile on 138kV line

Household breakers are rated for 10kA and will switch quickly and vaporize a thick blade screwdriver tip shorting out by prying off a staple on wire leaving vaporized copper specs melted onto plastic glasses's lens.

Just ask me
Rec Room experience learning in early 80's

 

[BobW]

I believe some of the older residential plug-in breakers are rated less than 10kA, maybe as low as 6kA. However, bolt-on breakers should all be minimum 10kA.

The limiting factor for short circuit current is generally the nearest transformer that feeds your building. Just to be confusing, they typically measure transformer impedance in percent. In simple terms, this means that a transformer with 5% impedance will pass 95% of the power, and 5% will be lost. Percent impedance values can be converted back into resistance values.

If you're in a residential building, then use 10,000 amps as the maximum available short circuit current. It's unlikely to be any higher than that. At one time, a pole mounted transformer would typically have an impedance of 5%, but the electrical utility in my area is now using transformers having an impedance in the 1.5% range, which is a cause for concern, since this can allow a short circuit current that exceeds the interrupting capacity of the breakers in the house.

If you're in an industrial location, the available short circuit current could be in the 50kA range when transformed down to 120/240 volts. However the building's power distribution system, if designed properly, will have various means to limit this to the 10kA range. In other words, if you're working with voltages in the 120/240 range, 10kA is a good design value for short circuit current.

Caveat: This is typical North American standard, but I suspect it won't vary much in other locations.

 

[tcmtech]

In the vast majority of real world conditions your local electrical wirings natural resistances are going to be the limiting factor in the peak amps you can get out of the system at any single point.

Theoretically on a 240 VAC line to line short right off your main distribution panel assuming your main power transformer was less than 10 feet away from the house and connected to your panel with 4/0 copper wire you could get a peak of ~25,000 amps.

Realistically I have doubts that in a typical residential wiring system you could even hit 5,000 amps.

 

[BobW]

Realistically I have doubts that in a typical residential wiring system you could even hit 5,000 amps.

I agree. My suggestion to use 10kA as a design value, was being exceedingly cautious.

 

[Grossel]

The limiting factor for short circuit current is generally the nearest transformer that feeds your building. Just to be confusing, they typically measure transformer impedance in percent. In simple terms, this means that a transformer with 5% impedance will pass 95% of the power, and 5% will be lost. Percent impedance values can be converted back into resistance values.

In europe, and what I've learn and also what most transmission transformers actualy are stamped, the percent values does not relates to power loss. Rather, the percent values stand for primary voltage (in percent to transformer primary rated voltage) when the transformers secondary windings are shorted and the primary windings is feeded with a current equal to the primary windings rated current.
However I'm open to beleive that the transformers stamps may folow other rules in your country.

Also, the percentage of loss vary as load changes.

 

[Roger44]

Hi,
If I were to take 7500A as the max current, that gives 0,03 ohms as supply impedance for 230V RMS. Would anyone dare to make a least worst guess as to the ohmic and inductive content?

 

[tcmtech]

As with anything inductive there is a inherent raise time so without knowing that its impossible to say for sure.

Treating it as pure resistance equation .03 ohms across a 230 VAC line would theoretically give you , (230 x 1.414) /.03 = ~10841 amps.

 

[Roger44]

How to measure the rise time? I could connect 100 meters of 2.5mm cross section cable (= known resistance) across a 32 A circuit breaker, and visualise how the 230V mains signal rises when I trip the circuit breaker. I've got an old analogic oscillo' and also a maybe not fast enough ADC.

But would I be wasting my time? First of all, the current rise depends on the moment in the cycle, so I would have to repeat the measurements. Then would the time delay for the rise of current up to nominally 32A be clearly visible/measureable with my limited equipement? And if it were measurable, would it be long enough to calculate R and L in the mains supply? Would my 32A circuit breaker resist many repetitions without altering its value? (they don't cost much) Would the Elecricity Boards 500mA circuit breaker suffer from the repetition, as I'm sure it would trip too each time.

Any thoughts on this idea?

 

[tcmtech]

Why?

 

[BobW]

Have you checked the specs of your 32A breaker to see what its rated interrupting capacity is?

 

[Tony Stewart]

How to measure the rise time? I could connect 100 meters of 2.5mm cross section cable (= known resistance) across a 32 A circuit breaker, and visualise how the 230V mains signal rises when I trip the circuit breaker. I've got an old analogic oscillo' and also a maybe not fast enough ADC.

But would I be wasting my time?

yes, you haven't learnt how to do high speed , high current measurements yet with coax and 50 OHm terminators.

First of all, the current rise depends on the moment in the cycle,

NO, but the current will depend on V(t)/Z(f) where the Z(f) will have a characteristic impedance ( I think a few hundred ohms ) rise time of I is dependant on distributed inductance, V=Ldi/dt

so I would have to repeat the measurements. Then would the time delay for the rise of current up to nominally 32A be clearly visible/measureable with my limited equipement?

maybe it will be in the sub-microsecond range

And if it were measurable, would it be long enough to calculate R and L in the mains supply?

that's not necessary to short the line to measure this. Even turning on the stove and measure current and voltage drop would measure Rsource up to the breaker panel.

Would my 32A circuit breaker resist many repetitions without altering its value?

there are thermal and electromagnetic types latter, which are common and very fast. Your short circuit current will be in the thousands of Amps.

(they don't cost much) Would the Electricity Boards 500mA circuit breaker suffer from the repetition, as I'm sure it would trip too each time.
Any thoughts on this idea?

Yes if you short it out too many times at >1Ka, perhaps reduce life from 1k trips to 100 trips before contact resistance changes and breaker gets warm, but trip level NO.

 

[tomizett]

As a side note, you can get a piece of equipment to directly measure the short-circuit current of your mains (Prospective Fault Current I think they call it). A friend of mine picked one up at a car boot sale or something - looks like an AVO style meter with massive terminals on the top and a scale in kA. He did try it out in his house, but I forget the result...

The point of interest here though seems to be measuring the inductance of the line, rather than the combined R+L. Rather than trying to measure the rise time directly, could you not make a tuned circuit by connecting a cap across the line? If you had a phase reference for the un-loaded line, you could just measure the phase shift, and... I'd imagine that the phase shift would be small though, and so hard to measure. And how would you know your reference phase? Would it be possible to sync a PLL from the line, then apply your load, and measure lag from PLL the line? And I suppose you'd also need to measure the resistance in order to do the the calculations. I'm sure there has to be a less hair-raising way to measure this than actually shorting the mains.
I'm sure I read a thread somewhere to do with measuring and distributing mains phase information - apparently it's something utility companies go in for in order to monitor monitor the grid.

In the book Power Electronics (Mohan, Robbins, Undeland), they keep making mention of "a worst case line inductance of 5%" or somesuch, but I've no idea %5 of what! There must be rules of thumb that are used by power supply designers though, at it's related to the design of snubbers etc.

 

[Tony Stewart]

5% refers to the drop in voltage due to impedance of line at rated current.
Generally a 10% drop or "sag down to 90%" is permitted split between feeders and distribution to end customer's drop point.
At the any point in the network Vsag = Zf/(Zf+Zs) for impedance of source Zs and fault Zf.
It can be measured as the minimum RMS for one or two consecutive cycles.

Since the Zf is critically dependant on the distance to the nearest transformer, Vsag can be re-written in terms of distance between fault and dist. transformer. The critical distance to fault to reach 10% drop gets drastically shorter as line voltage is transformed down, where it may be >50km on a 400kV line ~ 4km on a 33kV line and ~35m on a 400V line.
Feeder impedances vary between 200~450 mΩ/km. This is called the critical distance.

If your fault is 30m from transformer @ 250 mΩ/km ( or 25o μΩ/m)
then your line impedance is Zf= 30* 250μΩ= 7.5 mΩ
If line voltage is 120V, then fault current, If = 120/7.5m = 16kA

ref. Chapter 38 by Math Bollen
The Electric Power Engineering Handbook
Third Edition
Edited by
Leonard L. Grigsby

One can measure sag vs load current before trying 0 Ohm fault to compute source impedance. Then try a 130V gas tube on 120V line, wait for transient and enclose safely and wait for the fireworks. I know a senior competent designer who used a gas tube for line protection on an exterior RF repeater. You can image the excitement when the unit was returned with a 4 " crater on the PCB with everything vaporized. He didn't know gas tubes were negative resistance devices like SCR's and it clamped the line voltage.
Solution: use a PTC current limiter & choke before the gas tube.
Tony Stewart
I was Operations Mgr at the time and put on my Engineering hat to resolve the issue quickly.

 

[Roger44]

Hello, and very many thanks for all these ideas.
I quite agree with "I'm sure there has to be a less hair-raising way to measure this than actually shorting the mains."
Measuring sag versus load current by switching on all my purely ohmic radiateurs would give a measureable RMS voltage drop, maybe more than a volt, and therefore allow me to calcalate line source impedance. As Tomisett says, there remains the problem of knowing what (small) part is inductive in order to predict rise time in event of a short circuit.

Knowing that L will be small, how about simply a battery powered HF generator +variable air cap between phase and neutral to find a series resonant frequency? It might be too flat to be detectable though. And would L at HF be representative of L at 50Hz, thinking about it that seems to me to be the biggest stumbling block.

 

[Tony Stewart]

It's about 80% inductive so that should be easy to calculate.
**broken link removed**

Look up SIL Surge Impedance Loadimg

 

[Roger44]

To some extent I have found the answer on the Net :

here : http://www.acoustica.org.uk/other/mains_Z.html . " IEC725:1981 models the European domestic mains supply as having an impedance of (0.4+j0.25)ohms. Surprisingly, measurement shows that the UK agrees with or somewhat betters the model on the whole, at something like (0.25+j0.23) ohms."

here : http://www.compliance-club.com/archive/old_archive/000610.html "Impedance of the Low Voltage Network

This is defined in IEC725:19815 as
Z=(O.4+jO.25)ohms = O.4ohms + O.8mH

This represents the entire impedance of the generating system including the LV, MV & HV networks as seen by the consumer. For major countries, using the 230V supply system, at the extremes the UK is the best at less than (0.25+j0.23)ohms and Ireland is the worst at (1.03+j0.55)ohms both at 90% of locations. In the UK only 2% of consumers have supply impedances above the standard value whereas in Ireland 40% have. These are 1981 figures but there is no reason to think that they have changed substantially over the last 19 years".

There also several videos on Youtube where you can see measurements with a 800$ Fluke meter showing plug Z of rarely better than 0,4 ohm, done by people like you or me. The meter also displays the calculated short circuit current, several hundred A, not in the thousand A range.

Plugging R=0,4 and Xl=0.25 ohms into my computer software gives the following 3 curves for a sudden short-circuit between phase and neutral, at 2, 5 and 8 mS into the cycle.

**broken link removed**

Obviously the circuit breaker will intervene early but I don't know when. In these 3 cases, the current reached 50A at 213, 127 and 234 uS after the short-circuit. Maybe somebody knows about circuit breaker speed to see if these 50A ramp up times check out. Note that short-circuit occuring at voltage peak would give the slowest current ramp up.

 

[Tony Stewart]

I would have expected much lower, Ω, as our grid generates much higher current in residential outlets.

Having conducted ( no pun intended a short circuit test once myself in my younger days, it vaporized a heavy duty screw driver tip in milliseconds with a bang and tripped the breaker. I shorted out the wires prying off a staple near the breaker panel.

You can't do that with 50A or even 100A. thru a 1cm wide blade as I have done before starting a car by using the same screwdriver to bypass the solenoid.

The vaporized screwdriver tip left copper spray embedded onto the surface of my plastic glasses like a fine sputtered spray paint.

 

[profbuxton]

One can measure the "fault loop impedance" using a test device which puts a short circuit on the active -neutral for a VERY short time and displays the result to give a reading. Not fully cognizant how it is done in the device. This is one of the required tests when energizing a new installation here.
One can use such a tester or do a calculation of loop impedance by measuring wiring resistance of the installation. Purpose is to ensure that under short circuit conditions protective devices will operate corrcctly.

 

[Roger44]

Tony, what you described with your screwdriver is exactly what I would expect too. For the mains bang, all depends on how long it takes for the circuit breaker to open the circuit. If we take the short-circuit occuring 2mS after the start of the cycle case above, the current will have shot up to 600A if it takes 3mS for the breaker to open. I quote 3mS as an example, I've no idea how fast they react.

"Not fully cognizant how it is done in the device" From the videos, the Fluke device puts several 100 volts into the line. At that voltage it can't be more than a short burst of waves or pulses. Pity it does't show the R and Xl content seperately. I suppose they use a voltage similar to the mains voltage in order to have the same currents up where iron core saturation occurs.