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Aluminum refineries use electricity to separate aluminum from minerals it naturally occurs in. This electricity typically takes the form of low voltage DC ("low" meaning 4 to 6 volts), at very high current (on the order of tens of kiloamps). This much power poses an electrocution danger, but I don't understand how. If the entire electrical system runs at, say, 5 volts, and the human body acts like a resistor, then how can enough current actually make it through a human body to be dangerous? Similarly, how can an electrical arc through air happen, if it takes hundreds of volts to arc over a very short distance?

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    \$\begingroup\$ The magnetic fields are very strong .The fault currents are very high . \$\endgroup\$ – Autistic Mar 11 '17 at 5:03
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    \$\begingroup\$ A break in the insulation isn't a big issue, but a momentary short will melt metal. Or, evaporate it. \$\endgroup\$ – Whit3rd Mar 11 '17 at 6:15
  • \$\begingroup\$ I like how one of the top answers for this question is a theoretical one (based on inductance), and one is a practical one (based on how we implement that process)! \$\endgroup\$ – Cort Ammon Mar 11 '17 at 16:07
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The voltage for the Hall–Héroult process is inconveniently low (and the current too high) for efficient parallel operation so they use a whole bunch of cells in series.

From this source ("Studies on the Hall-Heroult Aluminum Electrowinning Process"):

The optimum current density is around 1 A cm-2 with a total cell current of 150-300 kA and a cell voltage -4.0 to -4.5 V. A typical cell house will contain about 200 cells arranged in series on two lines.

So the voltage at any given cell with respect to earth can be quite high, and the voltage across a cell if it opens up will be almost 1kV. Currents like that will easily vaporize metal so they can sustain a very long arc if it opens up relatively slowly and does not have a blow-out mechanism (DC is worse than AC).

To understand the efficiency issue- consider a simple full wave rectifier made with 6 silicon rectifiers. It will have a drop of (say) 2V at full current so the loss will be the output current x 2V. At 150kA that's 300kW lost. If you run 200 cells in parallel you would be wasting 60MW. Even at the cheap electricity prices that smelters pay that will add up- of the order perhaps 25-50 million dollars a year. In series, the loss is 'only' 300kW. The capital cost is also much less to make 150kA at 800V vs. 30MA at 4.5V because far more rectifiers and heat sinking would be required.

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  • \$\begingroup\$ Wow +1 . I did not realise that they had so many cells in series . \$\endgroup\$ – Autistic Mar 11 '17 at 9:08
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    \$\begingroup\$ (+1) Impressing data! P.S.: Why do you say "full wave rectifier with 6 silicon rectifiers"? Are you referring to 3-phase full wave rectifiers perhaps? Or is there a typo (6 in place of 4)? \$\endgroup\$ – Lorenzo Donati supports Monica Mar 12 '17 at 11:15
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    \$\begingroup\$ @LorenzoDonati yes at 150kA (!) and 800V you would definitely be using a 3 phase rectifier! \$\endgroup\$ – Level River St Mar 12 '17 at 11:34
  • \$\begingroup\$ The remaining AC voltage of the rectifier is much lower for a three phase full wave rectifier than for a single phase. At such power levels, a symmetric loading of all three phases is essential. The electricity bill would be higher for asymmetric loads. \$\endgroup\$ – Uwe Mar 12 '17 at 19:18
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    \$\begingroup\$ The liquid aluminium of all the cells in series is not potential-free, if this is disregarded at the withdrawal of molten aluminium, large dangerous arcs may occur. \$\endgroup\$ – Uwe Mar 12 '17 at 19:55
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The loop of conductor that carries the 10 kA current has non-zero inductance. That means a large amount of energy is stored in that loop as \$\frac{1}{2}LI^2\$.

If there's a break in the circuit, the inductance will raise the voltage at the break in order to keep the current flowing, while there is still stored energy is available to drive it. This will be enough to sustain an arc, and should the arc become long enough so need a high voltage to sustain it, enough to electrocute a person.

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A more everyday example of a hazardous low-voltage, high-current source is a humble car battery. Why? Even though the voltage (12V give or take) isn't enough to electrocute or even significantly shock you under normal circumstances, the fault currents possible are high enough to cause significant heating of any metal object involved in the fault, leading to serious burns.

As Sphero points out -- 10kA is too inconvenient to handle (imagine the size of the busbars you'd need!) so practical Hall-Heroult apps connect a bunch of cells in series. This means that hazardous voltages are present across the cell-string as a whole (and to ground!) even if each cell only operates at a few volts. Think of this like the difference between a RC LiPo and the Li-Ion pack in a Tesla -- both can put out hazardous fault currents, but the latter can also shock you.

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    \$\begingroup\$ "10kA is too inconvenient to handle (imagine the size of the busbars you'd need!)" - I thought they did use 10kA and more? Imagine the size of a big aluminium smelter and imagine how small the busbars are compared to everything else... \$\endgroup\$ – user253751 Mar 13 '17 at 0:04
  • \$\begingroup\$ @immibis -- I suspect there are plants that are large enough to run 10kA through a bunch of cells in series, yes... \$\endgroup\$ – ThreePhaseEel Mar 13 '17 at 0:07
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    \$\begingroup\$ FWIW -- aluminum plants usually use busbars made out of aluminum. They usually have a convenient source nearby. \$\endgroup\$ – Bryan Boettcher Jul 5 '17 at 17:47

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