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I'm not sure exactly which of your posts you are referring to, but if it's the one I've pasted at the end of this message, I'll add a few comments for what's it's worth. This seems to be turning into some sort of thought-experiment rather than a general question. I don't believe you've included enough specific information for an astute and absolute answer. For example, you state that 110 volts and a 15 amp breaker with 14 gauge wire will only operate eight "plugs." In this context, what does "only" mean? Unacceptable voltage drop, a device that won't operate, or a tripped breaker? When caclulating wire runs and sizes, there is a relationship between length of run, gauge, material and estimated temperature of wire, and the subsequent voltgage drop. Depending on what device is to be operated, sometimes a certain drop can be tolerated, and sometimes not. If a motor is involved, ampacity is another factor. As far as how the flow of water relates to all this, It's used as a simple analogy - and not intended to connote a mirror-image of electron theory. And, perhaps, I don't understand your question? Below is the posting I am referring to. "Hello Everyone Again, I agree that the comparsion to water works best. On the practical side. Is the following correct?
Using a 110 volt (pressure of water) 15 Amp(volume of water) breaker. The wire size of 14 guage (diameter of pipe means only a fixed volume) will only operate eight plugs (sprinklers).
However using a 110 volt(pressure of water) 20 Amp (volume of water)) breaker. The wire size of 12 guage ( diameter of the pipe is greater meaning more volume) will operate 11 plugs (sprinklers)
Am I correct in my understanding and description?
How does watterage inter into the entire scheme?
Thanks, Benjamin J. "Joe" Browning"
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