Which Residential Voltage & Frequency Arrangement Is Best?

[Don Kelly]

Good point

[Andrew Gabriel]

In UK terminology, a "Fault current" is a current limited mainly by the supply impedance (L-N fault) or the earth fault loop impedance (L-E fault), with the fault itself having a very low impedance.

If the fault itself has any appreciable impedance, then it generates an "overload current", not a "fault current" (at least, in UK terminology).

Arc impedance, if ballasted only by the supply impedance, rapidly collapses towards zero as the current rises to high levels.

[Tony Williams]

The microswitches in our Triton shower are the worst of both worlds..... fast opening and a small gap. :(

One of them switches the 20A element and the other 10A. Doing some ball-park sums shows the sort of problem.

Suppose the 100ft of T&E cabling has an inductance of 100uH, and the stray capacitance is effectively 1000pF. This will form a resonant circuit that will ring when the switch is opened. The energy stored in the charged inductor kicks off the ring, and then the inductor and capacitor exchange energy as they resonate. To get the peak transient voltage we need only consider the first exchange of energy, from the charged L into the C.

Inductor energy= 0.5*L*I^2. Capacitor energy= 0.5*C*V^2.

Suppose the switch opens at the peak of the 50Hz sine, when a full 20*1.414A is flowing through the inductor.

100E-6*28A*28A = 1000E-12*V^2.....Giving V = 8854 volts.

The resonant frequency of 100uH+1000pF is about 503KHz, and the Vpk is reached at the peak of the first quarter sine. ie, The voltage spike goes from roughly zero to 9KV in about 0.5uS..... quite a nasty dV/dT for a switch that has only just started to open.

Ball park numbers only (and many approximations), but it does give a flavour of the problem. The saving grace for Triton is that the switch only opens rarely at the full 28A, and the 50Hz sine means that the arc is only sustained for between 5 to 10mS. Still, it shows why those tiny microswitch contacts have such a short life. :(

[BFoelsch]

Yes, this business of AC loads being more inductive than DC loads is a red herring. In a DC magnet, the current is limited only by the resistance, so typically coils of many, many turns are used to get the resistance and ampere-turns up and the current consumption down. The inductance is wickedly high, but the high series resistance keeps the time constants manageable. A moderately large DC machine may have field coil inductances of hundreds of henrys. The quick-break air switch causes all kinds of problems on DC circuits. The quick-break makes the switch smaller but that is the only benefit. Switches for large motor fields have auxiliary contacts to connect a resistor across the coil as the current is interrupted to prevent damage to the coil. Unless you have worked in an old steel mill that ran on 250 VDC you have no concept of how delightfully simple AC control is.

I was involved in the development of vacuum contactors for MV motor control applications. Our first generation vacuum bottles had pure tungsten contacts. We ruined motors left and right, because upon opening in a vacuum there was no arc and hence extremely high dI/dt. Later versions alloyed a small amount of other metals with the tungsten to provide a small but controlled arc to dissipate some of the stored energy in the bottle. We tried other techniques, like using a small amount of inert gas to provide an arc path but the special contact alloy was the most successful. To put this in perspective, the final design, which is used to this day, interrupts 5kV

360 amp loads with a contact separation of less than 1/2 inch. Vacuum contactors also have a specification called "chop current," which is the current level at which the intentionally developed arc extinguishes, typically in the neighborhood of 15 amps.

[Ben Miller]

It is always determined by the supply. Fault current is the voltage at the generator (the "voltage applied to it") divided by the total impedance from the generator to the fault. This includes all of the transmission and distribution impedances. You MUST know the available short-circuit current from the supply at a defined location, generally the transformer primary or secondary or the service connection, in order to determine the current at the fault. When you do a short-circuit analysis in a facility, you get that information from the utility.

Ben Miller

[BFoelsch]

There isn't one.

In the US at least, the "AC only" switch rating, with its slow break properties, came about in the early 1950's as a side effect of customer demand for an audibly quieter switch. The AC only rating of the "quiet switch" was necessary to prevent their use on the DC systems which were still in use in downtown areas of some large cities. A further improvement was the "silent switch," which was a tilting mercury tube. If you look through issues of the Popular Mechanics type magazines of the era, a frequent article was on putting the new "quiet" switches in the baby's room.

The slow break of the AC only switch exists because it can be tolerated, not because it has any redeeming qualities.

[meow2222]

only true if your fault has zero impedance. IRL they normally dont, thus IRL the current through a fault is usually determined by the fault. Possibly a terminology issue.

NT

[meow2222]

Yes. The issue is that a carbonised plastic fault is much more likely to develop into a high current arc on 240 than 110, which trips the overcurrent protection. On 110, there is more chance of just resistance heating, which is much less likely to trip anythng, and thus more likely to start a fire.

NT

[Don Kelly]

YAHOO! You have hit it right on the head!

From other comments you have made, it appears that you are quite knowledgeable in the area of circuit breakers.

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Don Kelly @shawcross.ca remove the X to answer

[Ben Miller]

It is always true for any fault impedance. If the supply impedance is much lower than the fault impedance, the current will primarily depend on the fault, but you can't make a general statement about this. The type of fault (bolted, arc, etc), location within the electrical system, internal wiring details, and supply system characteristics are all factors.

Perhaps. Can you describe what you are talking about?

Ben Miller