How can low voltage, high current (kA) power be dangerous?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 Spehro 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.