How can one determine the adequate diameter of a Cu. cable for the follow of 1 A current?

How can one determine the adequate diameter of a Cu. cable for the follow of 1 A current?
The ampacity of a cable is determined by the temperature rating of the insulation as well as the crosssectional area of the conductor. The ampacity is derated based on the type of raceway and number of conductors in the raceway. The wiring code in your country or municipality will specify these requirements.
Another consideration is voltage drop. For long cable runs you might need a larger size conductor than what is dictated by the ampacity. For that, you would need to do a voltage drop calculation.
1 amp isn't very much. According to this ampacity table
http://www.allaboutcircuits.com/vol_5/chpt_3/2.html
a 20 AWG wire will easily carry 9 amps.
Thanks,
But, I’ve doubt in your previous quote, pl. clarify
Using the handy google calculator to convert units, I get
1 ((bar liter) per minute) = 1.66666667 watts
I’ve the following data:
Pr. =10 g/mm^2 ( 1 Bar), Area = 10 mm^2, Displacement = 1666.6667 mm ( 1 Lt/ Min.)
Now, how can I convert this into watt with the formulae: power is work divided by time?
It is given = 1 litre per minute and bar
Your first problem is that 10 grams per square millimeter is not in the proper units of pressure. Grams is a unit of mass, not force. You would have to multiply that by the acceleration of gravity 9.8 meters per second per second, to get it in units of pressure.Originally Posted by sak
I used the google calculator because it's easy and leaves less chance of making an error. If you insist upon working it out by hand, you would use these conversion factors.
1 bar = 100 000 pascals
1 pascal = 1 newton per (square meter)
1 watt = 1 (newton meter) per second
1 liter = 0.001 cubic meters
1 minute = 60 seconds
If you work this out, you get
100000*.001/60=1.667
You know this is a really good subject. That most never come across. As a manufacturer I did come across this rather startling reality.
A conductor is unfortunately also an element. That is why it heats while conducting.
You actually need to know the voltage as well as the amperage the wire will conduct.
Most of us know that with electric heaters especially fixed ohm heaters, that mimic a conductor, by maintaining rather constant ohms. As you raise the voltage to a heating element the element gets hotter. So does the lines feeding it.
There are two reasons why a heating element gets hotter. One is that as you add more voltage you get more wattage. But there is another reason, it is that you also get a little more amperage from fixed ohm heaters.
But never the less, we always rate heat output from wire elements in watts. We can determine for any element copper or nichrome. What the watts the element will output based upon the ohms it creates.
If you look at volts, amps, ohms, and watts for a couple scenarios you can see what I am saying.
Volts 120, amps 1, ohms 120, watts 120
Volts 2000, amps 1, ohms 2000, watts 2000
Volts 30,000, amps 1, ohms 30,000, watts 30,000
Volts 70,000, amps 1, ohms 70,000, watts 70,000
You can see that although according to some chart an 18 gauge wire can handle 7 amps. If you feed it 70,000 volts at one amp. And the piece of wire is four feet long. I think you might have a hard time explaining how 70,000 watts is going to squeeze through that four foot long wire. And dissipate 70,000 watts.
Sincerely,
William McCormick
If it's an ordinary 18 gauge wire the insulation will break down before you get to 70,000 volts, but there is no problem for the wire itself to handle the 7 amps whether it is 4 feet long or any other length. The wire is not dissipating 70,000 watts. The load is.Originally Posted by William McCormick
You are absolutely right about the fact that the load is dissipating the 70,000 watts. With no heating and subsequent corrosion, no accidents, the four foot long wire is only showing a voltage difference of 0.02554 volts, at one amp, and only turning 0.02554 watts into heat.Originally Posted by Harold14370
But the wire is carrying 70,000 Watts. In case of an accident, that number eighteen wire is a detonator chord waiting to happen. No way to shut it down in time. It will just blow, straight across that gap.
I am from a heating element kind of understanding. And I know that we have to over size the wire, even on low temperature elements. Or the wire cannot feed a heater, without being destroyed by the heat. The smaller the conductor the faster and greater the effect. The larger the conductor the slower the effect.
I was really saying that although you can supply 70,000 watts to a 70,000 watt heater load, the tiny 18 gauge wire will heat and raise in ohms, corrode and escalate into something else. Where a wire sized to deal with 70,000 watts would not cause a bad effect.
It is true, that heating of the wire is another problem, that could be solved with special heat sinks. But in reality that is not the case. Or you would just use larger wire. But you understand what I am saying. The AWG was supposed to be just a guide. Your application is what counts.
The same is true of induction loads. You cannot just use a number eighteen wire on an induction load that is under 7 amps FLA. Because it could heat very easily if it malfunctioned and caused excessive starting to occur. It would then de rate the wire. By accident.
In a power generator plant you could do these things if you monitor them.
In commercial wiring we were never allowed to use anything less then number twelve wire. Number twelve will take a massive hit that will pop 100 amp breakers in some situations upon dead short.
Where a number fourteen wire might explode with regularity.
But the number eighteen wire, would just flash, with that kind of wattage suddenly applied. Even with a one amp breaker or fuse. If the wire was twenty feet long it might create a twenty foot lightning bolt.
Wattage does mean something to wire and fuses as far as sizing.
But for standard house stuff that AWG wire gauge/guide is good.
One other thing is that even number eighteen wire with the right insulation, and only drawing 1 amp, will in fact get hot, from the magnetic fields created around it.
Sincerely,
William McCormick
One thing you can do to check this out. Is look at the High Voltage transmission lines.
They are monstrous lines. I do not even know how they rate them.
Yet on some systems all three lines together carry 1.5 million watts.
So although they could get away with an 18 gauge wire, practicality and workability make that impossible.
Sincerely,
William McCormick
I don’t know how sincere you are! W = V**2/R, at 120V and 1Ohm Wattage shall be 14400 :wink:Originally Posted by William McCormick
If I caught that correctly, you are saying that Volts times Volts divided by ohms is one formula for watts. That is correct.Originally Posted by sak
But I think you added in an extra zero there in your total. I think it should 1,440 watts.
Just saw this post or I would have said something sooner.
Watts are also amperes times amperes times ohms.
And of course volts times amps.
Sincerely,
William McCormick
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