 |
|
| Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
| Reply |
|
|
LinkBack | Thread Tools | Display Modes |
|
#1
|
|||
|
|||
Welder circuit breaker rating?
Hey guys,
Got a question about circuit breakers, I have a HarbourFreight stick welder, that is marked 120/240v draws 41 Amps, Does this mean it draws 41 Amps at 120 or 20.5 Amps at 240V? I'm talking about through each leg, if hooked up as 220V. I have it wired the welder hooked up for 220V operating, and have a dbl pole 40 Amp breakers in my panel, it's been working fine for the past year. Is the dbl pole 40 breaker the correct size for this application? Does this mean that each leg of the break is capable of drawing 40 amps each? Should I reduce the size of this break to a dbl pole 25 Amp breaker? By the way, I installed this dbl pole 40 Amp breaker myself last year. Thanks. Jeff -- A7N8X-Deluxe, AMD XP2500+ (Un-locked) 2x256mb Crucial PC3200 DDR ram Palit-Daytona Ti4200/64M AGP |
|
#2
|
|||
|
|||
|
You should size the breaker for the wire, not for the load.
Size the wire for the load.  |
|
#3
|
|||
|
|||
|
As Gigs said, size the breaker for the wire. A 40 amp breaker should be
attached to 8 gauge wire, as I recall it. If the wire is 10 ga or less, replace it with 8 gauge. 40 amps or 50 amps at 220 volts sounds right for a typical stick welder. Mine is wired to a 50 amp breaker. Richard Gigs wrote: You should size the breaker for the wire, not for the load. Size the wire for the load. |
|
#4
|
|||
|
|||
|
Jeff wrote:
Got a question about circuit breakers, I have a HarbourFreight stick welder, that is marked 120/240v draws 41 Amps, Does this mean it draws 41 Amps at 120 or 20.5 Amps at 240V? I'm talking about through each leg, if hooked up as 220V. I have it wired the welder hooked up for 220V operating, and have a dbl pole 40 Amp breakers in my panel, it's been working fine for the past year. Is the dbl pole 40 breaker the correct size for this application? Does this mean that each leg of the break is capable of drawing 40 amps each? Should I reduce the size of this break to a dbl pole 25 Amp breaker? By the way, I installed this dbl pole 40 Amp breaker myself last year. I think you should resize your breaker to a 30 amp double pole breaker. Current flows out of one leg, through the welder, through the work, back through the welder's ground lead, and back through the other leg. Yes, the current flows in both legs. A little buzzbox stick welder can draw 41 amps at 240VAC. The part you didn't mention is how many amps your welder can weld with. Let's suppose it's 250 amps at 25 volts AC. Power is volts times amps, so that's delivered power of 6250 watts. Let's assume your welder's transformer is 20% lossy, so we'll divide 6250 by 0.8 and get 7812 watts of input power. To calculate the current needed from a 220V plug, divide 7812 by 220 to get 35 amps. So you can see that 41 amps at 220 is not unreasonable. You can almost certainly get by with a 30 amp breaker. If you pulled wire that's too small for a 40 amp breaker (did you use 8 gauge wire or 10 gauge wire?) then it may be prudent to downsize the breaker to 30 amps. By the way, I have a little buzzbox welder, and I run it through a 30 amp breaker and about 15 feet of conduit and a receptacle, and I used no. 10 wire. I never ever turn my welder all the way up, or at least I haven't yet, so this setup is fine for the little buzzbox. Grant |
|
#5
|
|||
|
|||
|
What's in this welder described below, a couple of resistors?
Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB ............... "Grant Erwin" wrote in message ... Current flows out of one leg, through the welder, through the work, back through the welder's ground lead, and back through the other leg. Yes, the current flows in both legs. A little buzzbox stick welder can draw 41 amps at 240VAC. The part you didn't mention is how many amps your welder can weld with. Let's suppose it's 250 amps at 25 volts AC. Power is volts times amps, so that's delivered power of 6250 watts. Let's assume your welder's transformer is 20% lossy, so we'll divide 6250 by 0.8 and get 7812 watts of input power. To calculate the current needed from a 220V plug, divide 7812 by 220 to get 35 amps. So you can see that 41 amps at 220 is not unreasonable. You can almost certainly get by with a 30 amp breaker. If you pulled wire that's too small for a 40 amp breaker (did you use 8 gauge wire or 10 gauge wire?) then it may be prudent to downsize the breaker to 30 amps. By the way, I have a little buzzbox welder, and I run it through a 30 amp breaker and about 15 feet of conduit and a receptacle, and I used no. 10 wire. I never ever turn my welder all the way up, or at least I haven't yet, so this setup is fine for the little buzzbox. Grant ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#6
|
|||
|
|||
|
Wild Bill wrote:
What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. - GWE "Grant Erwin" wrote in message ... Current flows out of one leg, through the welder, through the work, back through the welder's ground lead, and back through the other leg. Yes, the current flows in both legs. A little buzzbox stick welder can draw 41 amps at 240VAC. The part you didn't mention is how many amps your welder can weld with. Let's suppose it's 250 amps at 25 volts AC. Power is volts times amps, so that's delivered power of 6250 watts. Let's assume your welder's transformer is 20% lossy, so we'll divide 6250 by 0.8 and get 7812 watts of input power. To calculate the current needed from a 220V plug, divide 7812 by 220 to get 35 amps. So you can see that 41 amps at 220 is not unreasonable. You can almost certainly get by with a 30 amp breaker. If you pulled wire that's too small for a 40 amp breaker (did you use 8 gauge wire or 10 gauge wire?) then it may be prudent to downsize the breaker to 30 amps. By the way, I have a little buzzbox welder, and I run it through a 30 amp breaker and about 15 feet of conduit and a receptacle, and I used no. 10 wire. I never ever turn my welder all the way up, or at least I haven't yet, so this setup is fine for the little buzzbox. Grant ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#7
|
|||
|
|||
|
Bill
If the utility system is not providing the current to accomplish the welding process, who is? lg no neat sig line "Wild Bill" wrote in message ... What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB .............. "Grant Erwin" wrote in message ... Current flows out of one leg, through the welder, through the work, back through the welder's ground lead, and back through the other leg. Yes, the current flows in both legs. A little buzzbox stick welder can draw 41 amps at 240VAC. The part you didn't mention is how many amps your welder can weld with. Let's suppose it's 250 amps at 25 volts AC. Power is volts times amps, so that's delivered power of 6250 watts. Let's assume your welder's transformer is 20% lossy, so we'll divide 6250 by 0.8 and get 7812 watts of input power. To calculate the current needed from a 220V plug, divide 7812 by 220 to get 35 amps. So you can see that 41 amps at 220 is not unreasonable. You can almost certainly get by with a 30 amp breaker. If you pulled wire that's too small for a 40 amp breaker (did you use 8 gauge wire or 10 gauge wire?) then it may be prudent to downsize the breaker to 30 amps. By the way, I have a little buzzbox welder, and I run it through a 30 amp breaker and about 15 feet of conduit and a receptacle, and I used no. 10 wire. I never ever turn my welder all the way up, or at least I haven't yet, so this setup is fine for the little buzzbox. Grant ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#8
|
|||
|
|||
|
In article ,
Grant Erwin wrote: Wild Bill wrote: What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. Don't laugh. I've done it. The "resistor" was a 100-watt light bulb in series with 110-volt power (AC and DC), the welding "stick" was a carbon rod liberated from a carbon-zinc "D" size battery, and the flux was twenty mule team borax. We were welding chromel-alumel thermocouples (with AC and flux) and 3mm diameter platinum capsules (with DC and no flux aside from the carbon monoxide and hydrogen from the carbon rod and moisture in the air), all in the 1970s. Simple, cheap, and worked well. In those days, the University had its own powerplant, and supplied 110/220 volt DC to the labs. When AC power came, all but the labs converted. In the sub-basement of the lab building there was a four by six foot 1900s-era open-face power panel carrying the 220 volt DC power, controlled by huge bare-metal knife switches and cylinder fuses, all bolted to the front surface of a massive sheet of some kind of black thermoset plastic, perhaps a bakelite composition. A terrifying contraption if there ever was - one stumble or slip, and your goose was *really* cooked. Especially if you were wearing any jewelry. Joe Gwinn |
|
#9
|
|||
|
|||
|
|
|
#10
|
|||
|
|||
|
My comment was to point out that the current from the service panel was not
passing through the welder's output leads. A contact between service power and earth ground would be deadly. The welder's transformer in a common AC welder isolates the utility current (while at the same time decreasing the input voltage, and increasing the output current capacity). The output leads are floating and isolated.. although there is a very slight leakage current in almost any transformer insulation, typically measured in microamps. A side note about transformer leakage.. when you see a transformer marked Hi Pot, it's indicating that the transformer has passed a high voltage potential test to insure that the leakage (between input and output) is within safe limits. WB .............. "larry g" wrote in message ... Bill If the utility system is not providing the current to accomplish the welding process, who is? lg no neat sig line "Wild Bill" wrote in message ... What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB .............. "Grant Erwin" wrote in message ... Current flows out of one leg, through the welder, through the work, back through the welder's ground lead, and back through the other leg. Yes, the current flows in both legs. A little buzzbox stick welder can draw 41 amps at 240VAC. The part you didn't mention is how many amps your welder can weld with. Let's suppose it's 250 amps at 25 volts AC. Power is volts times amps, so that's delivered power of 6250 watts. Let's assume your welder's transformer is 20% lossy, so we'll divide 6250 by 0.8 and get 7812 watts of input power. To calculate the current needed from a 220V plug, divide 7812 by 220 to get 35 amps. So you can see that 41 amps at 220 is not unreasonable. You can almost certainly get by with a 30 amp breaker. If you pulled wire that's too small for a 40 amp breaker (did you use 8 gauge wire or 10 gauge wire?) then it may be prudent to downsize the breaker to 30 amps. By the way, I have a little buzzbox welder, and I run it through a 30 amp breaker and about 15 feet of conduit and a receptacle, and I used no. 10 wire. I never ever turn my welder all the way up, or at least I haven't yet, so this setup is fine for the little buzzbox. Grant ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#11
|
|||
|
|||
|
I don't subscribe to the proposal that current flows back out the other
input lead. As I've stated before, I always have to ask Ok, THEN where does it go? The concept that there is a return line is counter-productive to understanding electrical power circuits. If all current travels back into the earth, or some other whimsical theory, nothing would operate as it does. Earth ground connections are for safety only, to insure that a person is never the only path between an electrical potential and earth ground. In the application of a transformer with a 240VAC input, the supply conductors do just that, they supply.. there is no return. Because it's AC, the two conductors take turns being different potentials to each other (easier to just think that they take turns being zero). When the other line's potential is not zero, current flows through the xfmr's primary winding (back 'n forth within the winding, not out to somewhere else). In a hydraulic circuit, the oil is returned to the pump (by way of a reservior usually), and the hydraulic circuit is basically a closed loop. Electricity ain't that way. In an application where the transformer input is 120VAC, the neutral isn't a return line. It's there to establish a potential. No potential, no worky. For either xfmr, the output of the secondary winding floats until circuit potentials are established. The potential between one of the secondary terminals (either one) and earth ground is essentially zero. No potential, no current (aside from the microamp xfmr leakage, assuming that the xfmr is not faulty). Some digital voltmeters are flakey when measuring similar points, and give false voltage readings, but there is absolutely no usable current present. When a circuit is established, the secondary winding terminals are both supply lines, there is no return. Yeah, but my AC amp clamp meter shows current in the neutral line. Does the meter show which direction it's flowing? Didn't think so, 'cause it's alternating. An oscilloscope won't help. There's no point in trying to tag a sinewave peak to see it being used up. You can't count the number of sinewaves that a table lamp uses. It's not helpful to consider sinewaves as quantitiy of consumables. Ever heard of a sinewave storage capacity for an AC capacitor? 'Course not. WB ................ "Grant Erwin" wrote in message ... Wild Bill wrote: What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. - GWE ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#12
|
|||
|
|||
|
Think whatever you like, Bill. Whatever floats yer boat. Just size both
conductors sufficient to carry the full amps, not half the amps, and all will be good. - GWE Wild Bill wrote: I don't subscribe to the proposal that current flows back out the other input lead. As I've stated before, I always have to ask Ok, THEN where does it go? The concept that there is a return line is counter-productive to understanding electrical power circuits. If all current travels back into the earth, or some other whimsical theory, nothing would operate as it does. Earth ground connections are for safety only, to insure that a person is never the only path between an electrical potential and earth ground. In the application of a transformer with a 240VAC input, the supply conductors do just that, they supply.. there is no return. Because it's AC, the two conductors take turns being different potentials to each other (easier to just think that they take turns being zero). When the other line's potential is not zero, current flows through the xfmr's primary winding (back 'n forth within the winding, not out to somewhere else). In a hydraulic circuit, the oil is returned to the pump (by way of a reservior usually), and the hydraulic circuit is basically a closed loop. Electricity ain't that way. In an application where the transformer input is 120VAC, the neutral isn't a return line. It's there to establish a potential. No potential, no worky. For either xfmr, the output of the secondary winding floats until circuit potentials are established. The potential between one of the secondary terminals (either one) and earth ground is essentially zero. No potential, no current (aside from the microamp xfmr leakage, assuming that the xfmr is not faulty). Some digital voltmeters are flakey when measuring similar points, and give false voltage readings, but there is absolutely no usable current present. When a circuit is established, the secondary winding terminals are both supply lines, there is no return. Yeah, but my AC amp clamp meter shows current in the neutral line. Does the meter show which direction it's flowing? Didn't think so, 'cause it's alternating. An oscilloscope won't help. There's no point in trying to tag a sinewave peak to see it being used up. You can't count the number of sinewaves that a table lamp uses. It's not helpful to consider sinewaves as quantitiy of consumables. Ever heard of a sinewave storage capacity for an AC capacitor? 'Course not. WB ............... "Grant Erwin" wrote in message ... Wild Bill wrote: What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. - GWE ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#13
|
|||
|
|||
|
ROTFLMAO
And the earth is to flat! "Wild Bill" wrote in message ... I don't subscribe to the proposal that current flows back out the other input lead. As I've stated before, I always have to ask Ok, THEN where does it go? The concept that there is a return line is counter-productive to understanding electrical power circuits. If all current travels back into the earth, or some other whimsical theory, nothing would operate as it does. Earth ground connections are for safety only, to insure that a person is never the only path between an electrical potential and earth ground. In the application of a transformer with a 240VAC input, the supply conductors do just that, they supply.. there is no return. Because it's AC, the two conductors take turns being different potentials to each other (easier to just think that they take turns being zero). When the other line's potential is not zero, current flows through the xfmr's primary winding (back 'n forth within the winding, not out to somewhere else). In a hydraulic circuit, the oil is returned to the pump (by way of a reservior usually), and the hydraulic circuit is basically a closed loop. Electricity ain't that way. In an application where the transformer input is 120VAC, the neutral isn't a return line. It's there to establish a potential. No potential, no worky. For either xfmr, the output of the secondary winding floats until circuit potentials are established. The potential between one of the secondary terminals (either one) and earth ground is essentially zero. No potential, no current (aside from the microamp xfmr leakage, assuming that the xfmr is not faulty). Some digital voltmeters are flakey when measuring similar points, and give false voltage readings, but there is absolutely no usable current present. When a circuit is established, the secondary winding terminals are both supply lines, there is no return. Yeah, but my AC amp clamp meter shows current in the neutral line. Does the meter show which direction it's flowing? Didn't think so, 'cause it's alternating. An oscilloscope won't help. There's no point in trying to tag a sinewave peak to see it being used up. You can't count the number of sinewaves that a table lamp uses. It's not helpful to consider sinewaves as quantitiy of consumables. Ever heard of a sinewave storage capacity for an AC capacitor? 'Course not. WB ............... "Grant Erwin" wrote in message ... Wild Bill wrote: What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. - GWE ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#14
|
|||
|
|||
|
If you think about it, only the electrons on my side of the transformer and those
in the ground that might mix in - flow in my circuits. The power company induce current flow in the transformer. Without circular flow of some kind, only a static or no current flow condition exists. Now - if I can only charge the power company for use of 'my electrons'. :-) Martin Grant Erwin wrote: Think whatever you like, Bill. Whatever floats yer boat. Just size both conductors sufficient to carry the full amps, not half the amps, and all will be good. - GWE Wild Bill wrote: I don't subscribe to the proposal that current flows back out the other input lead. As I've stated before, I always have to ask Ok, THEN where does it go? The concept that there is a return line is counter-productive to understanding electrical power circuits. If all current travels back into the earth, or some other whimsical theory, nothing would operate as it does. Earth ground connections are for safety only, to insure that a person is never the only path between an electrical potential and earth ground. In the application of a transformer with a 240VAC input, the supply conductors do just that, they supply.. there is no return. Because it's AC, the two conductors take turns being different potentials to each other (easier to just think that they take turns being zero). When the other line's potential is not zero, current flows through the xfmr's primary winding (back 'n forth within the winding, not out to somewhere else). In a hydraulic circuit, the oil is returned to the pump (by way of a reservior usually), and the hydraulic circuit is basically a closed loop. Electricity ain't that way. In an application where the transformer input is 120VAC, the neutral isn't a return line. It's there to establish a potential. No potential, no worky. For either xfmr, the output of the secondary winding floats until circuit potentials are established. The potential between one of the secondary terminals (either one) and earth ground is essentially zero. No potential, no current (aside from the microamp xfmr leakage, assuming that the xfmr is not faulty). Some digital voltmeters are flakey when measuring similar points, and give false voltage readings, but there is absolutely no usable current present. When a circuit is established, the secondary winding terminals are both supply lines, there is no return. Yeah, but my AC amp clamp meter shows current in the neutral line. Does the meter show which direction it's flowing? Didn't think so, 'cause it's alternating. An oscilloscope won't help. There's no point in trying to tag a sinewave peak to see it being used up. You can't count the number of sinewaves that a table lamp uses. It's not helpful to consider sinewaves as quantitiy of consumables. Ever heard of a sinewave storage capacity for an AC capacitor? 'Course not. WB ............... "Grant Erwin" wrote in message ... Wild Bill wrote: What's in this welder described below, a couple of resistors? Everyone had better hope that utility system current is not passing through the welder's output leads. Obviously it is not. WB (snicker) Wow what a goober I hacked up that time! I have to laugh at the fact that electrical engineers are capable of the dumbest things sometimes. OK, let's amend it: current flows down one leg, through the primary of the transformer and back through the other leg. Point is, the same current flows in both legs. - GWE ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- -- Martin Eastburn @ home at Lion's Lair with our computer lionslair at consolidated dot net NRA LOH, NRA Life NRA Second Amendment Task Force Charter Founder ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#15
|
|||
|
|||
|
There is no circular flow, this is a misconception.
The subject began as a discussion of a 240VAC circuit for a welder. Since 240VAC appliances do not require an earth ground (for anything other than safety, not current) to produce an output , there is no return path to earth ground. The earth ground lead of the appliance's power cord isn't connected to the welder's transformer terminals. It's connected to the case of the welder to make it a safe appliance to use. In the event of an internal fault, the case remains at earth ground potential (zero volts). When the current level reaches the circuit breaker rating, the circuit breaker will open. The case is earth grounded so that the operator doesn't die in the event of an internal fault in the welder. In normal operation, there is no return to anywhere, and definitely not to the utility company transformer. The appliance causes the power to be expended/used/dissipated internally. There is no trickery or deceit going on here. The 100W on the top of the incandescent light bulb basically means that the bulb consumes 100W. Ten of 'em left on for an hour is 1 KWH that you'll be charged for. In the winter, you've received the light plus the benefit of the heaters. In the summer, you pay to have the heat removed from the house by an air conditioner. Gotcha. The water heater element wattage rating means the same thing. Anyone can take a stab at an explanation of where the current goes when it goes back thru one of the supply lines by calling it a return, but they have no explanation of where it goes. Just because it's a common belief that it must return, don't make it fact. The fact that some of the utility power distribution points are grounded is because power companies have to deal with lightning strikes, line breaks and transformer failures. They have to pay for insurance too, and they don't want household lines to suddenly jump up into the kilovolts range in the event of a fault/failure or accident. The ground rods driven into the dirt out at the pole or across the street (or other points for underground service) don't create a return path for your household appliances. If I connect one terminal of a AA battery to an earth ground, is the battery charging or draining? What's going on with the hypothetical ground electrons? Pushing? No. Pulling? No. Flowing in either direction? No. A battery is an example of an isolated supply. We know the battery is capable of providing useable current. Is it flowing? No. Most homes are powered from the isolated secondary winding of the utility company's transformer. (I already understand that there is a center tap, and it's the home service panel's neutral bar). That supply is a floating source, just as the secondary of an ordinary power transformer is (explanation of a common power xfmr in earlier post). There is no usable current path from a floating xfmr secondary winding to earth ground (microamp xfmr leakage excluded). There isn't much point in injecting electrons into a dicussion of electrical power, but I'll go along. When the spark occurs at your finger tip when you touch a door knob mounted in a wooden door, where do the electrons go? I don't care either, but I'll read the explanations. WB .................. "lionslair at consolidated dot net" "lionslair at consolidated dot net" wrote in message ... If you think about it, only the electrons on my side of the transformer and those in the ground that might mix in - flow in my circuits. The power company induce current flow in the transformer. Without circular flow of some kind, only a static or no current flow condition exists. Now - if I can only charge the power company for use of 'my electrons'. :-) Martin ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
|
#16
|
|||
|
|||
|
Although the NEC allows conductors supplying a welder to carry more
than their rated capacity, I would never size the breaker for more than the ampacity of the wire. Remember the breaker will also carry more than its rated capacity on a duty cycle of 20%. If you " protect " 14 gauge wire with a 40A breaker, you might have the insulation melt before the breaker trips. Dan Ned Simmons wrote: In article 1119797867.565455.246460 @g14g2000cwa.googlegroups.com, says... You should size the breaker for the wire, not for the load. Size the wire for the load. Yes, *but*, the NEC allows the conductors supplying a welder to carry more than their normal rated ampacity. For example, if we assume the welder in question with a primary rated current of 41 amps has a 20% duty cycle rating, then the supply conductors need only have a rated ampacity of 18.5A. So you could conceivably have an installation with 14ga wire protected by a 40A breaker feeding the welder. |
|
#18
|
|||
|
|||
|
There are several types of circuit breakers. One type is a thermal
circuit breaker. You would not want to use a 90a thermal circuit breaker with 10 gauge wire. Copied from Electronic Design article on circuit breakers. "Appropriate for equipment where high-surge currents accompany the start of motors, thermal circuit breakers use bimetal technology to discriminate between safe switch-on currents/transients and prolonged overloads. A typical thermal breaker must trip within one hour at 140% of its rating. A latching mechanism makes these devices highly tolerant of shock and vibration." lets see 140 percent of 90 amps is 126 amps. So you think it might be alright to run 126 amps through 10 gauge wire for an hour? Granted most circuit breakers are not thermal breakers, but without knowing the breaker type, I would not recommend sizing the breaker for more current than the wire is rated for. I have no problem recommending using a welder on a circuit with wire that can not handle the input current of a welder on a 100 % duty cycle. Dan Nate Weber wrote: That's the entire point of the duty cycle, You can hit the wire with a high current but before it overheats, either you stop welding or the welder trips on thermal and allows the wire to cool down. Alot like pulse welding where ya set your peak amperage to 150 - 200% normal amperage and can still make a good weld. Don't forget also that you can size you breaker up to 200% the rated input current of the welder, on a miller thunderbolt, one could be running 10ga supply conductors on a 90A breaker and still be within code. Everything depends on the particular welder you are using, The welder will not draw more than the rated input amps unless there is a dead short and the welder will not draw high currents for any length of time due to the duty cycle. Change the welder and you are back to square one. Nate -- Http://www.Weber-Automation.net:8000 |
|
#19
|
|||
|
|||
|
I think with the typical residential installation in the U.S.A. you are
going to be relying on the thermal trip characteristics of the breaker. But with good equipment, it should not be as bad as it might sound. My house has a Square D QO load center, which is generally considered a good brand and line, and is used in commercial buildings as well as residences. The breakers have both thermal and magnetic trips. The higher the overcurrent condition, the faster the thermal trip. Moderate overcurrent conditions will indeed take a while to trip, in the case of a 50A QO breaker, something on the order of 40 seconds at 150%. The magnetic trip does not come into play until something close to 20 times the rated current is going through the breaker, when it trips nearly instantaneously. The magnetic trip is there only to deal with catastrophic problems like dead shorts in the equipment or line. That said, there are a lot of other brands and lines out there, and some may have very different trip characteristics. FWIW, I would use the largest wire within reason that will fit the terminals of the breaker and receptacle if for no other reason than to minimize voltage drop. On a short run, it won't make much of a difference in cost. On a long run, you may need the bigger size to maintain a reasonable input voltage or have the required ampacity after derating for ambient temperature. |
|
#20
|
|||
|
|||
|
lets see 140 percent of 90 amps is 126 amps. So you think it might be alright to run 126 amps through 10 gauge wire for an hour? Granted most circuit breakers are not thermal breakers, but without knowing the breaker type, I would not recommend sizing the breaker for more current than the wire is rated for. I have no problem recommending using a welder on a circuit with wire that can not handle the input current of a welder on a 100 % duty cycle. Dan If the welder has a max rated input current of 48.5 Amps and a Duty cycle of 20% how is anyone going to pull 126 amps for an hour? I could dead short the welder leads at max output and draw 48.5 amps for about 2 or so minutes. After that it would cool down for 8 minutes. -- Http://www.Weber-Automation.net:8000 |
|
#21
|
|||
|
|||
|
|
|
#22
|
|||
|
|||
|
Bill,
It is preposterous to say that the electrons flow out and go somewhere mysterious and never return during normal operation. There is definitely a return on a 240V circuit going to a welder. Electrons flow down one leg, thought the transformer winding in the welder, then back up the other leg. Re-examining the situation 16.6mS later, of course the current is flowing the opposite direction. But whatever electrons flow out on one leg of the circuit come back up the other side. Thats why its called a "circuit". I'm not sure what point you are shooting for here, and you are correct in many cases, i.e. saying that there is no current flow to earth ground. But you are mistaken in saying there is no return. A conductor can provide a return path without being at earth ground potential. Similarly on 120V circuits, the return path is the neutral line, which is also not necessarily at earth ground potential (indeed, because of the return current flowing across its resistance). -Holly |
|
#23
|
|||
|
|||
|
Yeah, folks keep insisting that half of the supply leads are returns, but no
one is saying where the power or current is returning to/going. You present a contradiction, I think, by stating (agreeing) that there is no current path to earth ground in 120V circuits, but then stating that the neutral provides a current return path, although the neutral is actually bonded to the structure's earth ground rod in most instances. Neutral isn't anything other than a reference point. That reference point is essentially zero since it's bonded to earth ground. That essentially zero point is present at one side of a 120V lamp or appliance cord. It stays at the reference point of essentially zero. One exception to this would be at the precise instant of a very close lightning strike, when the reference changes, and often results in damage. In a 120V lamp circuit, the neutral is lower or higher in potential than the home panel's bus bar (either one), depending upon the amplitude of the zero-crossing sinewave. The 120V lamp doesn't care if the voltage potential is lower or higher, it just lights, dims and lights at 60 cycles per second. So, for those that insist that the neutral is a return path, they're still not correct in thinking that the neutral is returning current half the time, as a result from being a lower or higher potential (to the bus bar), because it is still just an essentially zero reference point. The only time the lamp lights is when the line (bus bar) potential is not zero. Where the current goes in a 120V lamp, is into the filament resistance. The neutral is still just an essentially zero reference point. The light and heat are the product of the 120v being dissipated by the filament's resitance, and there is no left over power or current to return to anywhere else. As I said before, I don't see the point in discussing electrons when considering electrical power applications. The duration of a sinewave is not it's rate of travel, it's the frequency at which the AC alternates. The frequency of the cycles has nothing to do with how long a cycle spends in the load, or how fast it passes out to somewhere else. In a 240VAC circuit, what about the other phase, then? Is current from L1 returning "out" on L2, then L2's current returning "out" on L1? Fascinating, if you could show that happening, don't you think so? If everyone would limit the "return" current, they would save energy? What I have said before, is that the two 240VAC input terminals at the welder transformer are basically taking turns being the higher potential, while the difference in potential is 240VAC. The changes in potential (alternating voltage changes) create current in the resistance of the winding, and are dissipated in the transformer's primary winding. AC can be distributed for long distances with only minimal losses. It doesn't flow in a circular pattern, out one conductor and through a load and back out to the other. It's either present, or it's not. Any of this real science can be tested/proven by experimenting with a common power transformer. Since many homes are supplied power from a nearly identical source, it would be simple enough to make an equivalent house service panel setup with an isolated center-tapped 24VAC transformer secondary winding (or secondary voltage of choice). -Holly "But you are mistaken in saying there is no return". "A conductor can provide a return path without being at earth ground potential". For AC, it's not a return path, it is only a different potential. My question remains.. return to where? WB .................. wrote in message oups.com... Bill, It is preposterous to say that the electrons flow out and go somewhere mysterious and never return during normal operation. There is definitely a return on a 240V circuit going to a welder. Electrons flow down one leg, thought the transformer winding in the welder, then back up the other leg. Re-examining the situation 16.6mS later, of course the current is flowing the opposite direction. But whatever electrons flow out on one leg of the circuit come back up the other side. Thats why its called a "circuit". I'm not sure what point you are shooting for here, and you are correct in many cases, i.e. saying that there is no current flow to earth ground. But you are mistaken in saying there is no return. A conductor can provide a return path without being at earth ground potential. Similarly on 120V circuits, the return path is the neutral line, which is also not necessarily at earth ground potential (indeed, because of the return current flowing across its resistance). -Holly ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
| Reply |
| Thread Tools | Search this Thread |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| Popping circuit breaker w/ TS | Woodworking | |||
| circuit breaker popping once every 24 hrs | Home Repair | |||
| circuit is off, circuit breaker is on | Home Repair | |||
| circuit breaker | Home Repair | |||
| Is it a radial or ring circuit? | UK diy | |||
Linear Mode
