Any electricians handy?

[flaboy-]

I am just a farmer plain and simple :lol: but here in Texas, we don't have 208V 2 phase, Its either 240V 2 phase, or 208V 3 phase.

Are you sure you are an electrical engineer??

No, :lol: and you are correct and I stand corrected. I deal mainly with transmission and generation plants and think in plant voltage (3 phase) and not down to the home.

KenB, are you trying telling me length of the wire has nothing to do with ampacity of said wire?

 

[farmerjohn]

Just giving you a hard time.

Take care,

 

[KenB]

I am just a farmer plain and simple :lol: but here in Texas, we don't have 208V 2 phase, Its either 240V 2 phase, or 208V 3 phase.

Are you sure you are an electrical engineer??

No, :lol: and you are correct and I stand corrected. I deal mainly with transmission and generation plants and think in plant voltage (3 phase) and not down to the home.

KenB, are you trying telling me length of the wire has nothing to do with ampacity of said wire?

Yes! I am not saying that the length of the wire will not have a voltage drop. With the voltage at the load being less than at the source. Example--- the 2400 watt heater will draw 20 amps if the voltage is 120, with a 10% voltage drop the voltage at the heater will be 108 volts now the current will be 22.22 amps. The amps or ampacity depends totally on the load connected to it.
So if one needs to have X amount of volts at a given load the wire size will need to be larger the farther the load is from the source.

 

[Bama]

As an election myself I have to agree with KenB. The heater and incondescent lights don't really care if the voltage is a little less. Yes the amperage will go up but the breaker will prevent it from going up to a dangerous level. Biggest problem will be with electric motors. They will run on a lower voltage just not efficient or as long, or if it gets to low they wont run at all. 50 amps is the correct breaker but I would drop down a little 35-40 would be my choice. Next question is what type of #6 wire. If your running 220 and then using 110 at the barn do you have a neutral.? Some states require both a neutral and a ground wire with a ground rod at the barn.

 

[dun]

As an election myself I have to agree with KenB. The heater and incondescent lights don't really care if the voltage is a little less. Yes the amperage will go up but the breaker will prevent it from going up to a dangerous level. Biggest problem will be with electric motors. They will run on a lower voltage just not efficient or as long, or if it gets to low they wont run at all. 50 amps is the correct breaker but I would drop down a little 35-40 would be my choice. Next question is what type of #6 wire. If your running 220 and then using 110 at the barn do you have a neutral.? Some states require both a neutral and a ground wire with a ground rod at the barn.

Yes I'm running a neutral for the 220. I'm one of those people that also uses a ground and aneutral with 110. May not be necesarry, but I then know that it's right.

dun

 

[Bama]

great you'll be fine.

 

[tom4018]

http://www.elec-toolbox.com/calculators/voltdrop.htm

 

[Earl Thigpen]

As you know most power supplies can run at +/- 10% with no problem. I believe the NEC is ok with up to 5% VD.

Speaking of the +/-. When we were running the generator, the flourescent lights didn;t work too well but all of the elctric clocks gained a bout 6 minutes every 5 hours.
Since we've gone back to the grid everything is running right again.

That's because of the frequency of the power from your generator, not the voltage. The generator was running too fast so instead of 60Hz you were running 62 or 63Hz. Electric clocks use synchronous motors and are line frequency dependent. High RPM on the generator would not have caused the Flourescent lights to go crazy (I don't think). I don't believe the ballasts in the light fixture is that critical of frequency (+5Hz).

 

[Earl Thigpen]

I am just a farmer plain and simple :lol: but here in Texas, we don't have 208V 2 phase, Its either 240V 2 phase, or 208V 3 phase.

Are you sure you are an electrical engineer??

No, :lol: and you are correct and I stand corrected. I deal mainly with transmission and generation plants and think in plant voltage (3 phase) and not down to the home.

KenB, are you trying telling me length of the wire has nothing to do with ampacity of said wire?

Yes! I am not saying that the length of the wire will not have a voltage drop. With the voltage at the load being less than at the source. Example--- the 2400 watt heater will draw 20 amps if the voltage is 120, with a 10% voltage drop the voltage at the heater will be 108 volts now the current will be 22.22 amps. The amps or ampacity depends totally on the load connected to it.
So if one needs to have X amount of volts at a given load the wire size will need to be larger the farther the load is from the source.

Folks, this is Ohm's law. If the load (R) remains constant and the voltage (E) goes down then the current (I) goes up. The formula is R=E/I.

 

[dun]

Folks, this is Ohm's law. If the load (R) remains constant and the voltage (E) goes down then the current (I) goes up. The formula is R=E/I.

Reading that formula again after all these years made me think of Marys Fuzzy Coat

dun

 

[flaboy-]

Ok, after recalculating using the correct phase and voltage I come up with 40AMPs being safe. I calculated this using a power factor of .85 which like some said really doesn't matter with the heater and incandesent lamps. I also used and allowable 5% voltage drop and came up with 208 feet at 40 amps using 6 gauge wire.

Now with all that said I know a few of us have loaded a circuit above the "safe recommended capacity".

Thanks to all for the refresher on A/C theory. Just goes to show ya, if you don't use it you loose it. :shock:

 

[dun]

Now with all that said I know a few of us have loaded a circuit above the "safe recommended capacity".

You would have loved the wiring in the milk parlor on this place. They used doubled up 14 ga for 220. In the shop all gounds were removed and there was mixture of 14 and 16 gauge wire in one continuous circuit strung around the building with 30 some odd receptacles.
And maybe the best part, in the attic of the parlor, no junction boxes or wire nuts. All connections were twisted togehter and layed on pieces of 2X4. Not having building or electrical codes makes somethings a lot esier but boy can it get scary.

dun

 

[flaboy-]

Now with all that said I know a few of us have loaded a circuit above the "safe recommended capacity".

You would have loved the wiring in the milk parlor on this place. They used doubled up 14 ga for 220. In the shop all gounds were removed and there was mixture of 14 and 16 gauge wire in one continuous circuit strung around the building with 30 some odd receptacles.
And maybe the best part, in the attic of the parlor, no junction boxes or wire nuts. All connections were twisted togehter and layed on pieces of 2X4. Not having building or electrical codes makes somethings a lot esier but boy can it get scary.

dun

Ha, I remember my fathers barn. Wire nuts everywhere. Switches and receptacles just hanging out of the boxes by the wiring. Someone gave him some 2-wire cable and he did some re-wiring with that. :shock:

 

[Earl Thigpen]

Folks, this is Ohm's law. If the load (R) remains constant and the voltage (E) goes down then the current (I) goes up. The formula is R=E/I.

Reading that formula again after all these years made me think of Marys Fuzzy Coat

dun

Doesn't it? :lol: Dun, where is your switch for this apparatus? It needs to be pretty close to your load. Breaker box at the source and on-off switch close to the equipment. I know this means you're going to have to go out in the nasty weather to turn it on or off but that's what needs to be done. And that's another formula not nearly as simple as Ohm's law.

Sounds like you've got everything figured out.

 

[preston39]

Some old timers have done some scary things. dun reminds me of one of our old barns with wires hanging everywhere with many lights and duplex's.

We were locating the starting duplex of each circuit to install GFI's and found that we had a 3.0 copper svc cable from the meter into the service panel. Each wire in the casing is as big as the right index finger. Got to looking and found it has two/200 amp main fuses. Couldn't believe we had a 200 amp service. The service panel is an old ITE and is small with only 5 circuits and no expansion slots.

 

[dcara]

Please bare with me as I don't get to do this to often. My appologies in advance.

Earl wrote:

Folks, this is Ohm's law. If the load (R) remains constant and the voltage (E) goes down then the current (I) goes up. The formula is R=E/I.

Your equation is correct (although in the wrong format) but your statement is wrong. If you rearrange your equation to solve for I (current) instead of R then you get.

I=E/R

Which says that for a given R, as E (voltage) drops then so does I (current).

KenB wrote:

Example--- the 2400 watt heater will draw 20 amps if the voltage is 120, with a 10% voltage drop the voltage at the heater will be 108 volts now the current will be 22.22 amps.

The reason this can happen (and seem to defy Ohms law) is that the resistance of the heater is actually decreasing with voltage in order to maintian a constant power consumption. However, there is a limit to the heaters ability to perform this way. Below a certain voltage the heater will cease to operate.

For reference, the corollary to ohms law for power is Power=volts x current (P=ExI). Other equations for power can be derived from combining Ohms law also. Such as P= voltage squared divided by the resistance, or, the current squared times the resistance.

A good example is an incadecent light bulb. The off/cold resistance of a 100W bulb measures about 10 ohms. At 120v this would be 12 amps (1440 watts)!! For just an instant at turn on the bulb does indeed draw 12 amps which is what heats up the filament. As the filament heats up it increases in resistance until it stabilizes at about 144 ohms which results in a 100 watt continous power usage while it is on.

Motors also have large initial current draws but reduce their current needs as they spin up due to something called a "Back EMF". I won't go into that now as I'm sure I have already put most of you to sleep.

As far as to what melts wires, and/or kills people, it is the power. High current alone does not melt wires. There has to be a voltage drop across a resistance, or, enough current through a resistance (i.e power in either case) that results in heat generation.

Now back to your regular scheduled programing.

 

[KenB]

Please bare with me as I don't get to do this to often. My appologies in advance.

Earl wrote:

Folks, this is Ohm's law. If the load (R) remains constant and the voltage (E) goes down then the current (I) goes up. The formula is R=E/I.

Your equation is correct (although in the wrong format) but your statement is wrong. If you rearrange your equation to solve for I (current) instead of R then you get.

I=E/R

Which says that for a given R, as E (voltage) drops then so does I (current).

KenB wrote:

Example--- the 2400 watt heater will draw 20 amps if the voltage is 120, with a 10% voltage drop the voltage at the heater will be 108 volts now the current will be 22.22 amps.

The reason this can happen (and seem to defy Ohms law) is that the resistance of the heater is actually decreasing with voltage in order to maintian a constant power consumption. However, there is a limit to the heaters ability to perform this way. Below a certain voltage the heater will cease to operate.

For reference, the corollary to ohms law for power is Power=volts x current (P=ExI). Other equations for power can be derived from combining Ohms law also. Such as P= voltage squared divided by the resistance, or, the current squared times the resistance.

A good example is an incadecent light bulb. The off/cold resistance of a 100W bulb measures about 10 ohms. At 120v this would be 12 amps (1440 watts)!! For just an instant at turn on the bulb does indeed draw 12 amps which is what heats up the filament. As the filament heats up it increases in resistance until it stabilizes at about 144 ohms which results in a 100 watt continous power usage while it is on.

Motors also have large initial current draws but reduce their current needs as they spin up due to something called a "Back EMF". I won't go into that now as I'm sure I have already put most of you to sleep.

As far as to what melts wires, and/or kills people, it is the power. High current alone does not melt wires. There has to be a voltage drop across a resistance, or, enough current through a resistance (i.e power in either case) that results in heat generation.

I have to disagree about this statement.

It is the current that melts wires, it takes voltage and resistance to have current.
Example you will have the same amount of power with 12volts and 100 amps as you would have with 120 volts and 10 amps.
You could safely use #12 wire with the 120 V. 10 A.
Try it with 12 V. 100 A. and you will have melted wires

Now back to your regular scheduled programing.

 

[dcara]

You are correct about the wire melting in the 100 amp case but misunderstand why. The wire melts in the 100A case because there is a large amount of power dissipated in the wire due to the resistance of the wire. 12ga is aabout 1.6 ohms per 1000 ft.. The power distribution for the 2 scenarios you mentioned are as follows considering a 10ft pc of wire:

Case 1
Source power = 120V x 10A=1200W
Power dissipated in wire = current squared x resistance = 10 x 10 x .016 = 1.6 watts
Power delivered to load = 1200-1.6=1199.4 watts

Case 2
Source power = 12V x 100A=1200W
Power dissipated in wire = current squared x resistance = 100 x 100 x .016 = 160 watts!
Power delivered to load = 1200-160=1040 watts

You might be tempted to say that it is the high current producing the high power given that the resistance is the same in both cases. But it actually the combination of current and resistance, which for our purposes, IS the definition of power. Also note that if we cool copper to just a few degrees kelvin it becomes a super conductor ( i.e the resistance becomes extremely small). With such low resistance the wire can then easily handle 100A without melting.

 

[KenB]

You are correct about the wire melting in the 100 amp case but misunderstand why. The wire melts in the 100A case because there is a large amount of power dissipated in the wire due to the resistance of the wire. 12ga is aabout 1.6 ohms per 1000 ft.. The power distribution for the 2 scenarios you mentioned are as follows considering a 10ft pc of wire:

Case 1
Source power = 120V x 10A=1200W
Power dissipated in wire = current squared x resistance = 10 x 10 x .016 = 1.6 watts
Power delivered to load = 1200-1.6=1199.4 watts

Case 2
Source power = 12V x 100A=1200W
Power dissipated in wire = current squared x resistance = 100 x 100 x .016 = 160 watts!
Power delivered to load = 1200-160=1040 watts

You might be tempted to say that it is the high current producing the high power given that the resistance is the same in both cases. But it actually the combination of current and resistance, which for our purposes, IS the definition of power. Also note that if we cool copper to just a few degrees kelvin it becomes a super conductor ( i.e the resistance becomes extremely small). With such low resistance the wire can then easily handle 100A without melting.

In your first post you said P=I x V , basic ohms law, agreed.
It is true that the resistance or the wire may change, but it is still basicly the current part of the equation (basic ohms law ) that will cause the heat to melt the insulation.

You just wrote "You are correct about the wire melting in the 100 amp case but misunderstand why"

I do understand why. It may be some what of a combination of things, but the basic one being the current in the conductor. You can change the conductor by cooling it, using a larger conductor, etc. But it is still the current that would melt the insulation of the 10 foot piece of #12 AWG with THHN insulation in ambient air that I was refering to in my post.

 

[dcara]

This is a simple matter of exact mathematical definitions, not "somewhat of a combination". It is the dissipated power in the wire that creates the heat; and, power IS defined as the product of both current squared AND resistance. To say that "It is the current that melts wires.." Is akin to saying that it is the ham (or the eggs) that makes a ham and egg breakfast. Your either having ham AND eggs, or your not.

We have been discussing 12AWG wire. But suppose we were discussing a different material that had 100 times the resistance we could still achieve the same power dissipation by increasing the voltage but at a much lower current.

Example:
12AWG at 10ft and 1.6 ohms/1000ft = 0.016 ohms
Voltage required to produce 100A through 0.016ohms = I times R = 1.6 volts
P=I^2 x R = 100 x 100 x .016 = 160 watts

R x 100 = 0.016 x 100 = 1.6 ohms
V required to produce 160 watts through 1.6 ohms = square root of (P x R) = sqrt(160 x 1.6)=16 volts
I = V divided by R = 16/1.6=10 amps

So we have produced the same 160 watts power dissipation with 10 amps as we did with 100 amps by just varying the resistance and voltage.