A heap of high-current, normal-conducting cables lies in a waste bin by the side of the road at CERN. 10 points to any former students of mine who can explain with mathematics why high-current cables must be so thick. There will be no hints.
One Reply to “Photo a day: normal conducting”
Not a former student, but this one’s easy: I^2R is the power (heating) loss, so you want to minimize both I (which is why we use high voltage for long distance power transmission), and R (which is why, to your question, we use big, honkin’ wires.) Since the resistance of a cable is R=(rL)/A (where r is resistivity of the conductor, L is length, and A is cross-sectional area), then the only way you can cut R is by some combination of lowering resistivity (yep, superconductors are good here), minimizing cable run lengths, or increasing the cross sectional area of the cable. I’ve been in the solar industry for several years now, and the wiring losses in many modern solar arrays are staggering, especially at today’s copper prices. This isn’t helped much by “safety” regulations that limit the DC bus voltage to 600V (US) or 1000V (Europe). (Although there are starting to be many more bipolar systems that run +/- 600 or 1000V, keeping the middle rail within a few volts of ground, so you can effectively get a 1200 or 2000V bus…) Staggeringly, most all of the solar industry building unipolar arrays still hasn’t figured out what the telegraph/telephone companies figured out about two centuries ago now: You really want the negative side at ground to avoid corrosion and allow sacrificial anodes! This is one big reason most modern solar arrays have an effective life of less than 20 years – the wires literally turn into hollow straws, until they can no longer carry the current…)
Not a former student, but this one’s easy: I^2R is the power (heating) loss, so you want to minimize both I (which is why we use high voltage for long distance power transmission), and R (which is why, to your question, we use big, honkin’ wires.) Since the resistance of a cable is R=(rL)/A (where r is resistivity of the conductor, L is length, and A is cross-sectional area), then the only way you can cut R is by some combination of lowering resistivity (yep, superconductors are good here), minimizing cable run lengths, or increasing the cross sectional area of the cable. I’ve been in the solar industry for several years now, and the wiring losses in many modern solar arrays are staggering, especially at today’s copper prices. This isn’t helped much by “safety” regulations that limit the DC bus voltage to 600V (US) or 1000V (Europe). (Although there are starting to be many more bipolar systems that run +/- 600 or 1000V, keeping the middle rail within a few volts of ground, so you can effectively get a 1200 or 2000V bus…) Staggeringly, most all of the solar industry building unipolar arrays still hasn’t figured out what the telegraph/telephone companies figured out about two centuries ago now: You really want the negative side at ground to avoid corrosion and allow sacrificial anodes! This is one big reason most modern solar arrays have an effective life of less than 20 years – the wires literally turn into hollow straws, until they can no longer carry the current…)