There have been 60 deaths and nearly 50 serious shocks reported over the past 13 years involving electrical hazards in and around swimming pools.60 deaths in 13 years. I would think most pools would just trip the breaker if such a thing happened and the deaths come from pools made without breakers, bad ground, electrocuted by exterior lights, shocks from pumps or maintenece equipment, etc. The syriana accident can happen but is very rare.
Once it gets so small that the distance between the 120V source and the ground connection is much shorter than your body length (radio falls in the tub near the grounded end) the situation changes a lot again.But that is, more or less, falling for the "path of least resistance" line of thinking, which is wrong. The current will take every possible path from high voltage to low, the amount of current being limited by the resistance along that path. From one end of the tub to the other, and then all the way back again, is a valid path.
pure water: r = 2.5 x 10^5 ohm metersWhich is a pretty big range.. JackFlash's 100 micro-siemens/cm is 100 ohm·m, right in the middle.
saturated NaCl in water = 0.044 ohm meters
The actual resistance of the body varies depending upon the points of contact and the skin condition (moist or dry). Between the ears, for example, the internal resistance (less the skin resistance) is only 100 ohms, while from hand to foot is closer to 500 ohms.Which means the swimmer is subject to about 50mA of current.. From the same article again:
For currents above 10 milliamps, muscular contractions are so strong that the victim cannot let go of the wire that is shocking him. At values as low as 20 milliamps, breathing becomes labored, finally ceasing completely even at values below 75 milliamps.The swimmer is in a bad way, but it will be death by drowning rather than ventricular fibrillation, I guess.
Assume that current flows from the fault into the water in every direction evenly. The fault is on the wall, and it is approximated a hemisphere of radius rfault. The cross sectional area of the current path will be hemispherical shells of progressively increasing diameter, from a radius of rfault until it fills the depth of the body of water, for the pool that is a radius of 1m. That still leaves the contact resistance at the actual interface between fault and water..Back to the larger problem, we can think of a few scenarios.. Along with the point fault, the ground connection could be either distributed or a point. Also, the point fault (or ground) could be small, like a test probe, or large, like the entire casing of a light fixture.
The area of a hemispherical shell is 4πr2, so the resistance contribution of a shell is (100Ω·m / 4πr2)·dr, integrating from rfault to 1m we get:
16 x (1/rfault - 1) Ω
posted by Floydd at 10:16 AM on December 21, 2006