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Title: Voltage drop question
Post by: Mark Cadwallader on April 13, 2017, 07:21:58 pm
I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.
Title: Re: Voltage drop question
Post by: Tim McCulloch on April 13, 2017, 07:29:10 pm
I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.

You should use the 120v table as that is the voltage your load(s) operate at.

You are correct that additional connections increase the resistance of the circuit but I'd argue that it's not enough to worry about... However (you knew that was coming, right?) there is a Code limit to how may connections may be utilized to make your feeder run.  The connection at the line end (service) does not count, nor does the connection at your distro, but the limit is 2 connections IIRC.  The example I'm recalling was for a 250 ft run you'd need 2x 100 ft feeders and a 50 ft feeder; using 1x 100ft and 3x 50 ft were a no-go.  I'm not on the computer with the NEC or I’d look it up.

For 200 ft, I'd have them rent a generator.
Title: Re: Voltage drop question
Post by: Stephen Swaffer on April 13, 2017, 10:04:04 pm
I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.

What "table" are you referring to?  More out of curiousity than anything.  Voltage drop is calculated using ohms law, E=IR, the voltage you are running at doesn't really matter-until you try to calculate the "percentage" of voltage drop.  What is important to know is the current flowing through each conductor.  If you are running 120 V equipment from a 14-50R spider box, it  is presumably reasonably well balanced between the 2 phases-the voltage drop you would be concerned with is the more heavily loaded side (and remember-you are really concerned with peak load because that is when voltage drop will be the largest.)
Title: Re: Voltage drop question
Post by: Mark Cadwallader on April 14, 2017, 12:53:27 am
What "table" are you referring to?  More out of curiousity than anything.  Voltage drop is calculated using ohms law, E=IR, the voltage you are running at doesn't really matter-until you try to calculate the "percentage" of voltage drop.  What is important to know is the current flowing through each conductor.  If you are running 120 V equipment from a 14-50R spider box, it  is presumably reasonably well balanced between the 2 phases-the voltage drop you would be concerned with is the more heavily loaded side (and remember-you are really concerned with peak load because that is when voltage drop will be the largest.)

Pocket Ref, 3rd Ed., by Thomas J Glover. Page 149; "Wire Size vs. Voltage Drop"; maximum footage @ 120 volts, 1 phase, 2% max voltage drop. Copper wire, solid, 2-conductor, K=11 (77*-121*F).   I recognize that I'm using stranded wire, but K=11 for stranded as well.  There is a similar table for 240 volts on the same page.

I have thought about being sure to balance the load between the two legs, and it "should be" fairly close.  Thank you, gents.
Title: Re: Voltage drop question
Post by: TJ (Tom) Cornish on April 14, 2017, 07:27:19 am
Pocket Ref, 3rd Ed., by Thomas J Glover. Page 149; "Wire Size vs. Voltage Drop"; maximum footage @ 120 volts, 1 phase, 2% max voltage drop. Copper wire, solid, 2-conductor, K=11 (77*-121*F).   I recognize that I'm using stranded wire, but K=11 for stranded as well.  There is a similar table for 240 volts on the same page.

I have thought about being sure to balance the load between the two legs, and it "should be" fairly close.  Thank you, gents.
Make sure you know if your calculator requires using the whole path length or just the simple distance - I.e. A 100' cord has 200' of circuit length.
Title: Re: Voltage drop question
Post by: Mike Sokol on April 14, 2017, 09:17:19 am
Make sure you know if your calculator requires using the whole path length or just the simple distance - I.e. A 100' cord has 200' of circuit length.

And the interesting thing about voltage drop and overheated extension cords is that they really don't care about the actual voltage applied to them. It's the current that actaully creates the voltage drop and heating. Here's a video I made for the RV industry using a 3-volt Glo-Melt transformer to apply 30 amperes of current though a piece of 16 gauge extension cord. My transformer is only drawing around 100 watts (less than 1 amp) on the 120-volt side, but making 30 amps at 3 volts on the secondary side. The cable doesn't care if the one end of the wire has 120 volts and the other end has 117 volts, or the one end has 3 volts and the other end has 0 volts. It's all about the differential voltage which in this case is 3 volts. Of course in a actual hookup that would be 2 times 3 volts for both the hot and neutral run in the same extension cord, which totals 6 volts. And my short length of cable (maybe 10 feet) would have to multiplied to the length of the actual extension cord run. So I'm guessing that if you dead shorted the far end of a 100 ft, 16 gauge extension cord it might draw 30 amps of current. What's dangerous in the RV industry is that you can buy a 30-amp to 15-amp dogbone adapter that will actually let you load a skinny extension cord with 30 amperes. As you can see, in just 5 minutes or so it will approach the boiling point of water. https://www.youtube.com/watch?v=WZznobYGF_c
Title: Re: Voltage drop question
Post by: Mark Cadwallader on April 14, 2017, 09:44:57 am
The book provides the Ohm's Law formula for voltage drop on the prior page, which shows the double length of the run, and walks through the calculation.  The tables are built for showing what is the max length of different gauges of line that will not exceed a 2% drop.

I'm now wondering what the other providers have done for other events with the stage in the usual place. I think I'll suggest the stage be placed in a different location, nearer the 14-50R. 
Title: Re: Voltage drop question
Post by: Jay Barracato on April 14, 2017, 10:47:16 am
And the interesting thing about voltage drop and overheated extension cords is that they really don't care about the actual voltage applied to them. It's the current that actaully creates the voltage drop and heating. Here's a video I made for the RV industry using a 3-volt Glo-Melt transformer to apply 30 amperes of current though a piece of 16 gauge extension cord. My transformer is only drawing around 100 watts (less than 1 amp) on the 120-volt side, but making 30 amps at 3 volts on the secondary side. The cable doesn't care if the one end of the wire has 120 volts and the other end has 117 volts, or the one end has 3 volts and the other end has 0 volts. It's all about the differential voltage which in this case is 3 volts. Of course in a actual hookup that would be 2 times 3 volts for both the hot and neutral run in the same extension cord, which totals 6 volts. And my short length of cable (maybe 10 feet) would have to multiplied to the length of the actual extension cord run. So I'm guessing that if you dead shorted the far end of a 100 ft, 16 gauge extension cord it might draw 30 amps of current. What's dangerous in the RV industry is that you can buy a 30-amp to 15-amp dogbone adapter that will actually let you load a skinny extension cord with 30 amperes. As you can see, in just 5 minutes or so it will approach the boiling point of water. https://www.youtube.com/watch?v=WZznobYGF_c
We are currently doing simple resistive circuits in physics. I like to over emphasize that while we analyze in voltage (kirchhoff's loop law) the fundamental principle is the conservation of charge, I.e. current.

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Title: Re: Voltage drop question
Post by: Stephen Swaffer on April 14, 2017, 12:50:36 pm
In Mike's case, a 3 volt drop on a 3 volt supply = 100% voltage drop, on the 120 volt supply a 3 volt drop is 2.5%-usually considered acceptable-the chart is using a conservative 2%-but that would be the only difference between the charts is on the percentage calculation.

On the 3 volt circuit, you have 0% effeciency in your transmission of power, on the 120 volt 97.5%.  If you used the same current on the same wire at 240 volts you would still have 3 volts dropped, but now the voltage drop as a percentage is 1.25% resulting in an effeciency of 98.75 %.

Normally, the load will remain constant, not the current-so if you double the voltage, you halve the current resulting in half the voltage drop.  Now, in the example above you only have .625% voltage drop.  Which shows the advantage of running at a higher voltage.
Title: Re: Voltage drop question
Post by: Mike Sokol on April 14, 2017, 07:28:34 pm
In Mike's case, a 3 volt drop on a 3 volt supply = 100% voltage drop, on the 120 volt supply a 3 volt drop is 2.5%-usually considered acceptable-the chart is using a conservative 2%-but that would be the only difference between the charts is on the percentage calculation.

Exactly. However, because the heating effect is uniform across the length of the extension cord (resistor), my little 10 ft piece of test cable should exhibit the same localized temperature rise as a 100 ft cable. So if you take a 100 ft of 16 gauge extension cord and make pass 30 amperes of current, it should heat up similarly to my short short cable with 3 volts passing 30 amperes of current. I've simply created a small sample from the middle of a larger sample.
Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 14, 2017, 08:23:59 pm
Exactly. However, because the heating effect is uniform across the length of the extension cord (resistor), my little 10 ft piece of test cable should exhibit the same localized temperature rise as a 100 ft cable. So if you take a 100 ft of 16 gauge extension cord and make pass 30 amperes of current, it should heat up similarly to my short short cable with 3 volts passing 30 amperes of current. I've simply created a small sample from the middle of a larger sample.

And if that 100' extension cord is coiled, the heat won't dissipate so it can get hot enough to melt the insulation.
Title: Re: Voltage drop question
Post by: Ed Hall on April 14, 2017, 11:20:10 pm
And if that 100' extension cord is coiled, the heat won't dissipate so it can get hot enough to melt the insulation.

Is that why large feeder, 4/0 etc., isn't coiled but excess is laid in a figure 8 or is that done for inductive reasons?

I'm thinking inductive reasons because if 4/0 would get hot it should be larger or multiple runs.
Title: Re: Voltage drop question
Post by: Kevin Graf on April 15, 2017, 08:48:45 am
In a cable with a closely spaces pair of wires, the inductance in the send conductor cancels the inductance in the return conductor.
Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 15, 2017, 12:56:35 pm
Is that why large feeder, 4/0 etc., isn't coiled but excess is laid in a figure 8 or is that done for inductive reasons?

I'm thinking inductive reasons because if 4/0 would get hot it should be larger or multiple runs.

I always figured it was because laying heavy cables (and hoses) in a figure 8 is a lot easier than the over/under technique.
Title: Re: Voltage drop question
Post by: Mike Sokol on April 15, 2017, 10:06:17 pm
Is that why large feeder, 4/0 etc., isn't coiled but excess is laid in a figure 8 or is that done for inductive reasons?

I've run the numbers before, and there's just not enough induction with any possible feeder coil to create heat at 60 Hz. However, as noted above, a bunch of wire colled together closely, especially in a roadcase, will get hot simply due to the trapped heat and lack of circulating air. That's why we always pull all the feeder cables out of the roadcase if it's going to be loaded to near capacity.
Title: Re: Voltage drop question
Post by: Ivan Beaver on April 16, 2017, 08:53:58 am
Is that why large feeder, 4/0 etc., isn't coiled but excess is laid in a figure 8 or is that done for inductive reasons?

I'm thinking inductive reasons because if 4/0 would get hot it should be larger or multiple runs.
Yes a coil of cable will help keep the heat in a smaller area, but the inductance is the main reason.

When you coil it up, you form a nice inductor that can produce lots of heat-both to itself and internal objects.

Here is a fun little test.

Take a standard passive loudspeaker apart to where you can get to the xover, it doesn't matter what the model is.  A stronger one is more fun however. :)

Now grab a screw driver and run music through the cabinet.

Stick the screw driver in the middle of one of the coils.  The effect will vary depending on where the coil is in the circuit and what it is used for.

YES, the screwdriver will change the value of the coil it is inserted in, but it should be fine for this little test.

Hopefully you will be able to "feel" the energy inside the coil, because the screwdriver will be bouncing around with the music.

Leave it there a little bit, could be a few seconds or a minute or so.

Now remove the tip and touch it.  BE CAREFUL, it could be REALLY hot.

That should give you an idea of inductive heating.

Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 16, 2017, 10:49:47 am
Yes a coil of cable will help keep the heat in a smaller area, but the inductance is the main reason.

And when you lay it in a figure 8, now you have two inductors. Sure, it's less inductance (per coil) and it may not be as even as a round coil (I don't know the effect of a teardrop shaped "coil"), but you haven't solved a whole lot as far as induction. I mean, what's the difference between a 50' feeder round-coiled and a 100' feeder in a figure 8?

Maybe people THINK they are solving an inductance problem, but that just shows a lack of understanding. I'll admit to a lack of understanding myself, but I'm not afraid to challenge "conventional wisdom" when it doesn't make sense to me.
Title: Re: Voltage drop question
Post by: Kevin Graf on April 16, 2017, 11:47:10 am
Oh dear, oh dear.
A cord or cable is not an inductor. The field around one conductor is almost completely canceled by the opposite polarity field around the other conductor. 
Title: Re: Voltage drop question
Post by: Jean-Pierre Coetzee on April 16, 2017, 11:58:52 am
Oh dear, oh dear.
A cord or cable is not an inductor. The field around one conductor is almost completely canceled by the opposite polarity field around the other conductor.

This, put a clamp meter around a twisted pair cable and you will see what he's talking about. Equal current flows in each conductor just in opposite polarity, if the cable isn't twisted there will be some current due to one conductor being closer than the other but honestly it is negligible.

The real problem with coiling a cable is the lack of air around the cable causing heat to build up. If you coil power cable just derate it by x(you will need to ask the smart people by how much) and you should be just fine, otherwise lay it out straight.

The figure 8 coil has more surface area to dissipate heat from, this is the entire reason.

There is a lot of research done on the position of the ground conductor and inductive leakage between the current carrying conductors and the ground conductor depending on where it is in the cable, this is a bigger concern for sound guys IMHO.

EDIT: Same applies for 3 phase, the 3 phases cancel each other out. You might actually notice some current from triplen harmonics generated by dimmers/switch mode power supplies and such but should still be pretty low.
Title: Re: Voltage drop question
Post by: Matthew Knischewsky on April 16, 2017, 01:49:51 pm
Yes a coil of cable will help keep the heat in a smaller area, but the inductance is the main reason.

When you coil it up, you form a nice inductor that can produce lots of heat-both to itself and internal objects.

Here is a fun little test.

Take a standard passive loudspeaker apart to where you can get to the xover, it doesn't matter what the model is.  A stronger one is more fun however. :)

Now grab a screw driver and run music through the cabinet.

Stick the screw driver in the middle of one of the coils.  The effect will vary depending on where the coil is in the circuit and what it is used for.

YES, the screwdriver will change the value of the coil it is inserted in, but it should be fine for this little test.

Hopefully you will be able to "feel" the energy inside the coil, because the screwdriver will be bouncing around with the music.

Leave it there a little bit, could be a few seconds or a minute or so.

Now remove the tip and touch it.  BE CAREFUL, it could be REALLY hot.

That should give you an idea of inductive heating.
Title: Re: Voltage drop question
Post by: Ed Hall on April 16, 2017, 02:49:26 pm
Oh dear, oh dear.
A cord or cable is not an inductor. The field around one conductor is almost completely canceled by the opposite polarity field around the other conductor.

An extension from a distro, yes. The question I had was about the feeder leading from the service entrance or generator to the distro. Think camlocks. They are single conductor relatively high amperage.

Thanks to all who replied. I have learned a lot here and that why I ask. To learn.
Title: Re: Voltage drop question
Post by: TJ (Tom) Cornish on April 16, 2017, 05:17:24 pm
An extension from a distro, yes. The question I had was about the feeder leading from the service entrance or generator to the distro. Think camlocks. They are single conductor relatively high amperage.

Thanks to all who replied. I have learned a lot here and that why I ask. To learn.
The feeder, just like extension cords, has multiple conductors. Unless you make a separate coil for each conductor, the same cancellation occurs.
Title: Re: Voltage drop question
Post by: Mike Sokol on April 16, 2017, 08:23:10 pm
I don't have time to look for it right now, but I did do the math on this once, and IIRC unless you coil a single feeder wire tightly around an iron core of some sort, the amount of inductance is way too low for any heating effect at 60 Hz. And anything with a neutral and hot in the same sheath would certainly cancel out most external magnetic fields. It would be pretty simple to set up an experiment to prove this one way or another, but I'm heading to Holland and Paris for a few weeks and there's lots of packing to do tomorrow. I'll pick this thread up once I get back.
Title: Re: Voltage drop question
Post by: Stephen Swaffer on April 16, 2017, 10:04:06 pm
Is that why large feeder, 4/0 etc., isn't coiled but excess is laid in a figure 8 or is that done for inductive reasons?

I'm thinking inductive reasons because if 4/0 would get hot it should be larger or multiple runs.

If you look up "ampacity" in the NEC, you will see a chart with 60 deg C, 75 deg C, and 90 deg C ratings.  Any wire, from 20 awg or smaller to 4/0, or 500 MCM or larger has resistance and will create heat depending on the current flowing through it.  Those ampacities are based on how hot the wire will get under the conditions listed-typically 3 current carrying conductors in a raceway.  If you follow code correctly, you will have to derate the ampacity by the number of conductors in a raceway-by the time you get to 10, the derate is a full 50%.  This is essentially what you are doing when you coil a wire-putting a bunch more wires inthe same space-while a coil is not in a raceway, it still is limited in the amount of heat it can dissipate.
Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 16, 2017, 10:58:18 pm
Oh dear, oh dear.
A cord or cable is not an inductor. The field around one conductor is almost completely canceled by the opposite polarity field around the other conductor.

Of course a bundled cable won't exhibit significant inductance.

But what about feeder cables that aren't bundled?
Title: Re: Voltage drop question
Post by: David Sturzenbecher on April 16, 2017, 11:18:38 pm
Of course a bundled cable won't exhibit significant inductance.

But what about feeder cables that aren't bundled?

Feeders are still in relatively close proximity to one another.


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Title: Re: Voltage drop question
Post by: Tim McCulloch on April 17, 2017, 03:53:14 am
Of course a bundled cable won't exhibit significant inductance.

But what about feeder cables that aren't bundled?

And Code requires that feeders of single conductor cables NOT be bundled, presumably because of heat. 
Title: Re: Voltage drop question
Post by: David Sturzenbecher on April 17, 2017, 06:20:29 am
And Code requires that feeders of single conductor cables NOT be bundled, presumably because of heat.

What defines a "bundle" here? Every feeder that I have seen over 5ft has been held together with tape every 10ft or so.


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Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 17, 2017, 10:47:57 am
If you look up "ampacity" in the NEC, you will see a chart with 60 deg C, 75 deg C, and 90 deg C ratings.  Any wire, from 20 awg or smaller to 4/0, or 500 MCM or larger has resistance and will create heat depending on the current flowing through it.  Those ampacities are based on how hot the wire will get under the conditions listed-typically 3 current carrying conductors in a raceway.  If you follow code correctly, you will have to derate the ampacity by the number of conductors in a raceway-by the time you get to 10, the derate is a full 50%.  This is essentially what you are doing when you coil a wire-putting a bunch more wires inthe same space-while a coil is not in a raceway, it still is limited in the amount of heat it can dissipate.
I recently had to macgyver a fuse for the rear channel of my surround system (kids don't try this at home)  :-[. The smallest wire I could find was 30ga (actually one strand from some 22ga stranded) that I soldered to the outside of the open fuse. Doing some WWW research I discovered that my DIY fuse was rated for >10A...  Only slightly better than a penny in the fuse box...

I swapped out a good fuse from one of the other identical amps in case something was actually wrong with that one amp, and put the DIY fuse in one of the other amps that didn't blow a fuse... My new fuses arrived a few days later and my house didn't burn down in the meantime.  ;D

Of course the fuse current for wire (when the metal melts) is probably higher than the rated ampacity which will be derated based on insulation temperature capability, heat dissipation, and other safety factors.

JR
Title: Re: Voltage drop question
Post by: Mark Wilkinson on April 17, 2017, 12:37:21 pm
I recently had to macgyver a fuse for the rear channel of my surround system (kids don't try this at home)  :-[. The smallest wire I could find was 30ga (actually one strand from some 22ga stranded) that I soldered to the outside of the open fuse. Doing some WWW research I discovered that my DIY fuse was rated for >10A...  Only slightly better than a penny in the fuse box...

Ok sorry :), but IME no way  30ga carries 10A . 
One of my past lives was elevator adjustor / troubleshooter at large...
Sometimes shorts would appear intermittently, based on momentary circuit logic, that couldn't be metered statically. 
It was easy to chew up a box of buss cartridge fuses trying to hunt the short down.
# of strands of 18ga to temporarily use as substitute during troubleshooting only was pretty common knowledge/practice. 
I don't remember it all so well anymore, but 1 strand carried less than 6A
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 17, 2017, 01:57:07 pm
Ok sorry :), but IME no way  30ga carries 10A . 
I remember it like it was just last week, because it was...

30ga copper wire was something like 10+A...Aluminum or steel with higher resistance was less current.  I was surprised.

I remember from my old technician days when I would open up TO-3 transistor to analyze failure modes. If the 1 mil base wire melted that was more than one amp (rule of thumb from 50 years ago so not trustworthy). 
Quote
One of my past lives was elevator adjustor / troubleshooter at large...
Sometimes shorts would appear intermittently, based on momentary circuit logic, that couldn't be metered statically. 
It was easy to chew up a box of buss cartridge fuses trying to hunt the short down.
# of strands of 18ga to temporarily use as substitute during troubleshooting only was pretty common knowledge/practice. 
I don't remember it all so well anymore, but 1 strand carried less than 6A
OK you made me look it up again http://www.powerstream.com/wire-fusing-currents.htm (http://www.powerstream.com/wire-fusing-currents.htm)  30ga copper 10.2A, aluminum 7.6A, iron 3.15A

18ga comes in different flavors, there is a 16 strand 30ga @ 10A each strand, 7 strand 26ga @ 20.5A each (copper).  There is also a 41/34 so 41 strands of 34ga @ 5.1A each...  So if you had 41 strand 18ga you are in your <6A ballpark.

There is also 18ga test lead wire with 192/41,   65/36,  413/44, etc...

I was surprised by the fuse current of 30ga wire, wish I had some 36ga or 44ga handy to make fuses with, but I didn't.

JR
Title: Re: Voltage drop question
Post by: Mark Wilkinson on April 17, 2017, 02:23:25 pm
I remember it like it was just last week, because it was...

30ga copper wire was something like 10+A...Aluminum or steel with higher resistance was less current.  I was surprised.

I remember from my old technician days when I would open up TO-3 transistor to analyze failure modes. If the 1 mil base wire melted that was more than one amp (rule of thumb from 50 years ago so not trustworthy).  OK you made me look it up again http://www.powerstream.com/wire-fusing-currents.htm (http://www.powerstream.com/wire-fusing-currents.htm)  30ga copper 10.2A, aluminum 7.6A, iron 3.15A

18ga comes in different flavors, there is a 16 strand 30ga @ 10A each strand, 7 strand 26ga @ 20.5A each (copper).  There is also a 41/34 so 41 strands of 34ga @ 5.1A each...  So if you had 41 strand 18ga you are in your <6A ballpark.

There is also 18ga test lead wire with 192/41,   65/36,  413/44, etc...

I was surprised by the fuse current of 30ga wire, wish I had some 36ga or 44ga handy to make fuses with, but I didn't.

JR

Nice research, thx.
That's a good example of you are the engineer, and I'm just the 'learned it on the job' dude !

I figure my remembrance is based on finer stranded 18ga. 
We did try to use pretty flexible wire, to manage all the long pulls through conduit and junction boxes.
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 17, 2017, 02:48:09 pm
Nice research, thx.
That's a good example of you are the engineer, and I'm just the 'learned it on the job' dude !

I figure my remembrance is based on finer stranded 18ga. 
We did try to use pretty flexible wire, to manage all the long pulls through conduit and junction boxes.
Except that I'm an engineer (some might dispute that?) who mostly learned on the job...

I love the modern internet, there is a world of information at our fingertips if we just ask the right questions.

It wasn't this easy back in the day.

JR
Title: Re: Voltage drop question
Post by: Mark Wilkinson on April 17, 2017, 03:00:32 pm
Except that I'm an engineer (some might dispute that?) who mostly learned on the job...

I love the modern internet, there is a world of information at our fingertips if we just ask the right questions.

It wasn't this easy back in the day.

JR

Funny...the engineer part that is..:)

Yeah, information was so much harder to come by 'back in the day'.
Information held alot of power back then, at least relative to today, I think....
Title: Re: Voltage drop question
Post by: Tim McCulloch on April 17, 2017, 03:22:34 pm
Funny...the engineer part that is..:)

Yeah, information was so much harder to come by 'back in the day'.
Information held alot of power back then, at least relative to today, I think....

Topic swerve alert.  You have been warned. ;)

Commensurate with the greatly more accessible information is more accessible dis-information.  Lots of silly stuff on youtube that perpetuates "DIW"  Doing It Wrong.  There are musician/prosumer oriented forums (fora?) that contain dubious advice that gets repeated like gossip in a hair salon/barber shop.

I suppose the good outweighs the bad, at least I hope so.

In the Ye Olde Badde Dayz­® we relied on dead trees & ink.  I'd like to think the s/n was better back then. 8)

/topic swerve alert
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 17, 2017, 03:31:51 pm
Topic swerve alert.  You have been warned. ;)

Commensurate with the greatly more accessible information is more accessible dis-information.  Lots of silly stuff on youtube that perpetuates "DIW"  Doing It Wrong.  There are musician/prosumer oriented forums (fora?) that contain dubious advice that gets repeated like gossip in a hair salon/barber shop.

I suppose the good outweighs the bad, at least I hope so.

In the Ye Olde Badde Dayz­® we relied on dead trees & ink.  I'd like to think the s/n was better back then. 8)

/topic swerve alert
In fact some of the "Audio Myths" I wrote about in the '80s, that were printed on those dead trees are still turning up like bad pennies today.

JR
Title: Re: Voltage drop question
Post by: Kevin Graf on April 17, 2017, 04:18:39 pm
An old chart, shows 26AWG with a fusing current of 20.5 Amps. That would be the same as a pair of 29AWG wires. So one 29AWG wire would be 10.25 Amps.
Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 17, 2017, 04:36:29 pm
An old chart, shows 26AWG with a fusing current of 20.5 Amps. That would be the same as a pair of 29AWG wires. So one 29AWG wire would be 10.25 Amps.

Maybe we need Mike to run some tests with his glo-melt transformer.
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 17, 2017, 04:49:51 pm
An old chart, shows 26AWG with a fusing current of 20.5 Amps. That would be the same as a pair of 29AWG wires. So one 29AWG wire would be 10.25 Amps.
The link I posted earlier shows 26ga copper as 20.5A, 30ga copper as 10.2A 

JR
Title: Re: Voltage drop question
Post by: Kevin Graf on April 17, 2017, 05:28:46 pm
The link I posted earlier shows 26ga copper as 20.5A, 30ga copper as 10.2A 
JR
Oops sorry.
I was going from the same link. But I had only printed the top part of the page (some years ago).
I had looked at the chart some hours ago. But when i got back to the computer, I missed your link.
Interesting the two smaller conductors will carry more current than one larger conductor of the same cross-section area.
I'll go back and print the entire page.
Title: Re: Voltage drop question
Post by: Stephen Swaffer on April 17, 2017, 11:08:35 pm
So, 2 smaller wires of the same cross section will have a greater surface area to dissipate the heat that is generated.
 
The parallel DC resistance of the 2 wires should equal the resistance of the larger wire-so for a given voltage/current the amount of heat generated should be the same-but the greater surface area would allow the 2 smaller wires to run cooler.

Just a basic, practical example of physics in action.
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 18, 2017, 09:56:28 am
So, 2 smaller wires of the same cross section will have a greater surface area to dissipate the heat that is generated.
 
The parallel DC resistance of the 2 wires should equal the resistance of the larger wire-so for a given voltage/current the amount of heat generated should be the same-but the greater surface area would allow the 2 smaller wires to run cooler.

Just a basic, practical example of physics in action.
Physics..."It's the law"... Since resistance is linear with cross sectional area, the two smaller wires adding up to the same cross sectional area as one larger wire will have 1.4x the surface area.  Cross sectional area increases with the square of the radius, while surface area increases linearly with diameter.

JR

PS: For a wild card think about skin effect where current doesn't flow in the middle of the wires (crowded out by magnetic field), tiny wires have more skin all else equal.
Title: Re: Voltage drop question
Post by: Jonathan Johnson on April 18, 2017, 11:30:38 pm

PS: For a wild card think about skin effect where current doesn't flow in the middle of the wires (crowded out by magnetic field), tiny wires have more skin all else equal.

An issue at higher frequencies, not so much at 60Hz. And no issue at all with DC.

At extremely high AC voltages (even 60Hz) you also have a corona effect where the magnetic (electric?) field ionizes the air surrounding the conductor, so some current is carried by the air.
Title: Re: Voltage drop question
Post by: John Roberts {JR} on April 19, 2017, 08:58:17 am
An issue at higher frequencies, not so much at 60Hz. And no issue at all with DC.
True, it's an AC phenomenon so not a consideration at DC. For 60Hz the effective skin depth where conduction occurs is around 0.33" deep, so moot for wires less than 0.66" across... IIRC I think some big dog AC wires use a steel center with (lower resistance) aluminum conductors wrapped around the steel core for good strength and good "effective" conduction in the outer 0.33" at 60Hz.
Quote
At extremely high AC voltages (even 60Hz) you also have a corona effect where the magnetic (electric?) field ionizes the air surrounding the conductor, so some current is carried by the air.
I just realized I am guilty of my signature rant... doing an information dump instead of answering the OP's question... ::)

JR
Title: Re: Voltage drop question
Post by: Kevin Graf on April 19, 2017, 03:17:59 pm
For any situation that we are likely to be in, loop inductance will overwhelm skin effect.
Title: Re: Voltage drop question
Post by: Mark Cadwallader on April 19, 2017, 04:58:21 pm

 I just realized I am guilty of my signature rant... doing an information dump instead of answering the OP's question... ::)

JR

I appreciate the additional education I am getting on the general topic. My original question was promptly answered (thank, guys), and the rest is all good.
Title: Re: Voltage drop question
Post by: Scott Holtzman on April 20, 2017, 03:15:18 am
I am sure y'all know the higher the frequency the more skin effect is relevant.  7/8" and larger RF transmission cable uses hollow center conductors to lower weight/cost.  Also seen copper low density foam for center conductor.  It works even at VHF frequencies.



Title: Re: Voltage drop question
Post by: Kevin Graf on April 20, 2017, 08:13:58 am
And even at HF frequencies. That's 3 to 30 MHz.
Title: Re: Voltage drop question
Post by: Guy Holt on May 03, 2017, 05:40:39 pm
You might actually notice some current from triplen harmonics generated by dimmers/switch mode power supplies and such but should still be pretty low.

Where that was once true, it is no longer the case. I am discovering that a lot of LED fixtures use switch mode power supplies that are not power factor corrected (pfc.) In recent testing I did of the fixtures in the inventories of Boston area rental and lighting sales companies, over half were not power factor corrected. With power factors ranging from .45 to .63, these fixtures generated considerable harmonic distortion (THD ranged from 75-85%.) If you don’t take into account the extra current they will draw and the harmonic currents they will generate, you may find unexpected voltage drop, breakers tripping, portable generators running erratically, and the neutrals in three phase systems getting hot. Even those that were pfc generated harmonic currents when dimmed. For instance, the pfc of the new Litepanel Astra 1x1 dropped from .99 to .54 when dimmed 50% (THD increased to 83.2%.)

For those interested in the results of my survey of fixtures, I have an article coming out in the spring issue of Protocol on the harmonic profile of a number of common LED fixtures and how it effects portable generators. If you are not familiar with Protocol, it is the quarterly publication of PLASA – an international organization working to raise standards, improve skills and strengthen the events, entertainment and installation industries. If you can’t find the print edition of the magazine, there are links to it and other articles in my “Production Power on a Budget” series in Protocol at http://www.screenlightandgrip.com/html/hd_plug-n-play_pkg.html (http://www.screenlightandgrip.com/html/hd_plug-n-play_pkg.html).

Guy Holt, Gaffer
ScreenLight & Grip
www.screenlightandgrip.com