17 ms·

Even the opening sentences are wrong: > VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get. That is why electr

by kevhito 10y ago

Even the opening sentences are wrong: > VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get. That is why electrical power lines work at high voltage. No. Losses in transmission lines are proportional to current squared [1]. High voltage is used because it results in lower current (for the same power), so lower losses. [1] http://www.bsharp.org/physics/transmission http://www.bsharp.org/physics/transmission

pietjepuk88 10y ago
I think the analogy holds up quite well, even though they don't go into the details that you mention. Losses in water transmission lines for example are also proportional to velocity squared.
- seanp2k2 10y ago
  Voltage is how hard the nozzle sprays, current is the diameter of the hose. High pressure + small hose can be as much flow as a fire hose just gurgling out. Correlates well to how much current a given wire gauge can handle as well. If you hook the garden hose up to a fire hydrant and open it up, the garden hose might explode since it can't handle all the water. A pressure washer might use a thin hose, but the water sprays so hard that it can cut your skin. Idk, that's how I learned it growing up before I learned about stranded vs solid core wire, skin effect, resistance, etc.
tbrownaw 10y ago
Only if you take the wire gauge as fixed. If instead you use the cheapest (smallest) you can get away with, you'll find yourself using smaller wires at higher voltages, with higher resistance and so less then quadratic loss reduction.
bjt2n3904 10y ago
Perhaps a way to help understand this. Power delivered by the battery is this equation: P = I * V So you can have (1 volt @ 100 amps), or (100 V @ 1 amp) and still get (100 watts). If both are the same, wouldn't the low voltage solution be safer, and more desirable? Massaging the same equation a little bit, we can model our losses along the power lines. Substituting V = IR you get... P = (I^2)R It's difficult to make conductors less resistive without using expensive metals or unconventional techniques (like cryogenically cooled wires). So to cut our losses, we'll go with high voltage, low current.
- nbadg 10y ago
  Since we're getting pedantic, as HN is wont to do, I'm not sure if the parent complaint is about the first sentence... > VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get. ... or about the claim that follows: > That is why electrical power lines work at high voltage. The former sentence (aka the hydraulic analogy, [1]) is a very well-established thing. It's not a perfect analogy, but it's quite reasonable. The latter (high voltage power lines) is demonstrably not correct, as the two of you have pointed out. Extending your power formula to the hydraulic world and simplifying the calculus to steady-state algebra: Power = (Volumetric flow rate) * (pressure) Though, to be fair, the analogy breaks down a little there; the Poiseuille equation [2] (the fluid equivalent of V = IR) involves 6 variables instead of 3. But the point is, you still end up with a final equation of similar form to P = (I^2)R. [1] https://en.wikipedia.org/wiki/Hydraulic_analogy https://en.wikipedia.org/wiki/Hydraulic_analogy [2] https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equat...

by kevhito 10y ago

pietjepuk88 10y ago

I think the analogy holds up quite well, even though they don't go into the details that you mention. Losses in water transmission lines for example are also proportional to velocity squared.

seanp2k2 10y ago
Voltage is how hard the nozzle sprays, current is the diameter of the hose. High pressure + small hose can be as much flow as a fire hose just gurgling out. Correlates well to how much current a given wire gauge can handle as well. If you hook the garden hose up to a fire hydrant and open it up, the garden hose might explode since it can't handle all the water. A pressure washer might use a thin hose, but the water sprays so hard that it can cut your skin. Idk, that's how I learned it growing up before I learned about stranded vs solid core wire, skin effect, resistance, etc.

tbrownaw 10y ago

Only if you take the wire gauge as fixed. If instead you use the cheapest (smallest) you can get away with, you'll find yourself using smaller wires at higher voltages, with higher resistance and so less then quadratic loss reduction.

bjt2n3904 10y ago

Perhaps a way to help understand this. Power delivered by the battery is this equation: P = I * V So you can have (1 volt @ 100 amps), or (100 V @ 1 amp) and still get (100 watts). If both are the same, wouldn't the low voltage solution be safer, and more desirable? Massaging the same equation a little bit, we can model our losses along the power lines. Substituting V = IR you get... P = (I^2)R It's difficult to make conductors less resistive without using expensive metals or unconventional techniques (like cryogenically cooled wires). So to cut our losses, we'll go with high voltage, low current.

nbadg 10y ago
Since we're getting pedantic, as HN is wont to do, I'm not sure if the parent complaint is about the first sentence... > VOLTAGE is to electricity what pressure is to water: the more you have of it the more oomph you get. ... or about the claim that follows: > That is why electrical power lines work at high voltage. The former sentence (aka the hydraulic analogy, [1]) is a very well-established thing. It's not a perfect analogy, but it's quite reasonable. The latter (high voltage power lines) is demonstrably not correct, as the two of you have pointed out. Extending your power formula to the hydraulic world and simplifying the calculus to steady-state algebra: Power = (Volumetric flow rate) * (pressure) Though, to be fair, the analogy breaks down a little there; the Poiseuille equation [2] (the fluid equivalent of V = IR) involves 6 variables instead of 3. But the point is, you still end up with a final equation of similar form to P = (I^2)R. [1] https://en.wikipedia.org/wiki/Hydraulic_analogy https://en.wikipedia.org/wiki/Hydraulic_analogy [2] https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equat...