line loss

[socalelect]

for the life of me i cant remember how to calculate line loss

for example 6 ga CU on a 50 amp breaker , with a varying load up to 50 amps at times , making about a 500 foot run yes i know this is way way way way undersized
how would one calculate the line loss for that ?


also another question

say you have a sub with a 100 amp cb in it then the conductors feed another panel at 100 amps , then from there to another 100 amp cb when a overload condtion exist why would it trip the first 100 amp CB and not the one closest to the fault .

closest thing i could figure is the first one see's the worst of the overload first ?

 

[Hv&Lv]

for the life of me i cant remember how to calculate line loss

for example 6 ga CU on a 50 amp breaker , with a varying load up to 50 amps at times , making about a 500 foot run yes i know this is way way way way undersized
how would one calculate the line loss for that ?


also another question

say you have a sub with a 100 amp cb in it then the conductors feed another panel at 100 amps , then from there to another 100 amp cb when a overload condtion exist why would it trip the first 100 amp CB and not the one closest to the fault .

closest thing i could figure is the first one see's the worst of the overload first ?

There are various voltage drop calculators on the internet. here is just one:http://calculator.net/voltage-drop-calculator.html

Let me get this straight, you have three 100 amp breakers in series? With no other loads in the preceding subpanels? Perhaps any loads in the first panel are added to the loads on the second panel and so on to the third panel. This could explain the first breaker tripping rather than the last.

 

[socalelect]

i wasnt so much after voltage drop as i was line loss , there is a energy loss due to the amount of voltage drop ?

voltage drop increses amperage right ?


i was using the 100 amp breakers in series as a example . we have a set up similar to that i dont remember the amperage thoe

there are no loads in the preceeding subs all of the load is in the last one . and in the buildings with 1st and second sub . they were used in a manufacturing process but arent in use currently and have no loads in the panels all the equipment is gone the panels are still there thoe for future use . i think the panels may be 200 amp

 

[gar]

111213-2354 EST

socalelect:

1000 ft of #6 copper at 20 deg C (68 F) has a resistance of 0.3951 ohms. A 500 ft run
uses 1000 ft of wire in the loop.

Square the current and multiply by the resistance. At 50 A the power loss is 2500 * 0.3951 = 987 watts. Actually more because the wire will warm some from the current. But this is only 1 watt per foot so it won't warm much.

At 25 A 1/4 the power loss, and at 100 A 4 times the power. 4 watts per foot is not a lot of heat to get rid of.

A fusing current for #6 copper is listed 668 A, but actually could be much lower. At 600 A the power per foot would be greater than 144 W because I am using room temperature for wire temperature and resistance.

.

 

[socalelect]

gar,
thanks for the math on that , for some odd reason i couldnt remember it


so basically more voltage drop equates to more energy used correct ?

as the voltage goes down the amperage goes up ?

 

[gar]

111214-0746 EST

socalelect:

It is best to view your original question from the perspective of current. If the resistance is a fixed value, R, then P = I²*R. As current increases power wasted goes up much more rapidly than current.

If you compare different lengths of wire of the same size and material, and the same current, then power wasted is proportional to length, but power per foot is a constant.

so basically more voltage drop equates to more energy used correct ?

If the resistance of the wire is constant, then an increase in voltage drop implies an increase in current, and an increase in power loss.

If the length of wire is fixed, and you change its size while keeping the load current constant, then if the wire is made smaller the voltage drop will increase and the lost power will increase.

as the voltage goes down the amperage goes up ?

I don't know.

If you have a constant power load, and you reduce the applied voltage, then the current will go up.

If the load is a constant resistance, then as the applied voltage goes down, then the current also goes down. Ohm's law.

.

 

[don_resqcapt19]

There is a tolerence band for the breaker trip times. For example the trip curve for one 100 amp breaker shows a trip time of between between 1000 seconds and never for about 110 amps of current.

As far as the line losses, I have always used the votage drop times the current to get the power loss.

 

[bob]

As far as the line losses, I have always used the votage drop times the current to get the power loss.

Thats the same equation VD = I x R. power loss = I² x R = VD x I = IxIxR.

 

[Hv&Lv]

As far as the line losses, I have always used the votage drop times the current to get the power loss.

That is the way I figure it also. Easier to use an online calculator with the known amperages I thought...

 

[socalelect]

thank you everyone for dealing with my stupid questions in a time of tardness on my part

bob and don that was the math i was looking for