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Ivybridge_I7
Cyber-Crime Researcher OpSec
Premium Member
join:2004-06-09
Daytona Beach, FL

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Ivybridge_I7 to AKA06

Premium Member

to AKA06

Re: Extension cord for chest freezer (electrical question)?

said by AKA06:

I've got a 7 cu. foot chest freezer that I'd like to install in a utility room. The best available spot is located far enough away from an outlet that I will need some kind of extension. I know extension cords are not ideal for appliances, so if I use one I want to be sure that it exceeds what I will need so that it is safe.

Call a qualified master electrician and run a dedicated outlet to that freezer. If the compressor in that freezer shorts out and decides to draw a huge amount of current. the heat made across a long extension cord could start a potential fire.

Think ohms law, voltage drop across the resistance of the copper in the extension cord (from end to end, three wires. The bigger the gauge of wire, the less voltage drop) means a increase in current, therefore heat

Volts/ resistance =current
TheMG
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Canada
MikroTik RB450G
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TheMG

Premium Member

said by Ivybridge_I7:

Think ohms law, voltage drop across the resistance of the copper in the extension cord (from end to end, three wires. The bigger the gauge of wire, the less voltage drop) means a increase in current, therefore heat

Volts/ resistance =current

While that may be partially true from a purely theoretical standpoint, it is most certainly not applicable here.

The extension cord is of low enough resistance that it does not act like a current limiting resistance in the circuit, since the effective resistance and impedance of the load connected to it is significantly higher than the resistance of the extension cord.

A decrease of resistance of the extension cord will cause a small decrease in voltage drop across the cord, but a very minimal difference in the amount of current that the appliance will draw. In fact with motor/compressor loads, the current can actually go DOWN as the voltage at the motor is increased. A motor that is being overworked, such as a motor which is powered with insufficient voltage, can draw large amounts of current. Actually this also goes true for most electronic devices since they have regulated power supplies which will continue to deliver a constant power to the device regardless of input voltage fluctuations therefore the higher the voltage at the load, the lower the current.

Since less voltage is dropped across the cord and the current remains relatively the same due to the connected load, the amount of heat dissipation in the extension cord is LESS with a thicker (lower AWG number) extension cord.

whizkid3
MVM
join:2002-02-21
Queens, NY

whizkid3 to Ivybridge_I7

MVM

to Ivybridge_I7
said by Ivybridge_I7:

voltage drop across the resistance of the copper in the extension cord (from end to end, three wires. The bigger the gauge of wire, the less voltage drop) means a increase in current, therefore heat

Unfortunately, you've either got that backwards, or have worded it very confusingly. As TheMG See Profile said,

The larger the wire, the less heat dissipated from the wire. The right formula is:

P = I^2 x R

Heat is measured in watts, as it is a measurement of power. Lower the resistance with larger wire, and you will lower the power dissipated from the wire as heat. The load regulates the current draw. Any change in the total current draw of the circuit is negligible, as the resistance in the wire is minuscule compared to the impedance of the load. Think of the current in the above equation as a constant. Your wording would be correct, if one was talking about the heat dissipation from a toaster coil, which is not in series with any significant load.

That being said, an extension cord is not the solution to the problem of not having a receptacle where you need one. Extension cords are for temporary use. Leaving an extension cord in place permanently powering a load on the floor in one's basement is asking for trouble.

Ivybridge_I7
Cyber-Crime Researcher OpSec
Premium Member
join:2004-06-09
Daytona Beach, FL

Ivybridge_I7 to TheMG

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to TheMG
said by TheMG:

While that may be partially true from a purely theoretical standpoint, it is most certainly not applicable here.

The extension cord is of low enough resistance that it does not act like a current limiting resistance in the circuit, since the effective resistance and impedance of the load connected to it is significantly higher than the resistance of the extension cord.

A decrease of resistance of the extension cord will cause a small decrease in voltage drop across the cord, but a very minimal difference in the amount of current that the appliance will draw. In fact with motor/compressor loads, the current can actually go DOWN as the voltage at the motor is increased. A motor that is being overworked, such as a motor which is powered with insufficient voltage, can draw large amounts of current. Actually this also goes true for most electronic devices since they have regulated power supplies which will continue to deliver a constant power to the device regardless of input voltage fluctuations therefore the higher the voltage at the load, the lower the current.

Since less voltage is dropped across the cord and the current remains relatively the same due to the connected load, the amount of heat dissipation in the extension cord is LESS with a thicker (lower AWG number) extension cord.

But you are talking about inductive and capacitive (start up capacitor) loads running on 120 AC R.M.S.(170 peak) You have no way of knowing the total resistance across the extension cord two wires (hot-neutral) that are in series from the circuit breaker to the load. The inrush current just from the inductive load of the motor starting, could be over 15 amps for a few seconds. Those two wires of the extension cord are the weak point in the circuit chain. You need to measure the total series resistance across both the hot-neutral wires and the longer the cord, the more voltage drop and in theory more heat being dissipated.

I still thing that spending the money and doing it right with a dedicated (15 amp) outlet using a romex or BX from the circuit breaker is the best way to go.
Ivybridge_I7

Ivybridge_I7 to whizkid3

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to whizkid3
said by whizkid3:

The larger the wire, the less heat dissipated from the wire. The right formula is:

P = I^2 x R

Heat is measured in watts, as it is a measurement of power. Lower the resistance with larger wire, and you will lower the power dissipated from the wire as heat. The load regulates the current draw. Any change in the total current draw of the circuit is negligible, as the resistance in the wire is minuscule compared to the impedance of the load.

Okay you would be right if you where talking about D.C. and the voltage drop is fairly constant across the extension cord.

We are not talking about direct current here

We are talking about A.C. where the load sees a voltage that goes from 0 to 170 volts peak (120 R.M.S. times 1.414) on the first half of the sine wave and then does the opposite going from 0 to -170 Peak on the other side of the sine wave. That's 340 volts peak to peak voltage. Now the voltage doesn't stay the same and varies during the 60 cycles. When the load is light on the outside transformer, the voltage could go as high as 130 voltage AC R.M.S (183 Volts peak or 366 peak-2-peak) Because A.C loads uses the average 120 AC (root means square) of the peak, the voltage drop across the extension cord is not 100 percent constant and will change as the load and line voltage varies over time.

Don't mix up Alternating current with Direct current when it comes to how they relate to loads. This is is why Phasor Algebra is used in the calculation of Alternating current induction and capacitive loads.

Ohms law can only be used as a base to do simple load and voltage drop calculation, but in reality it doesn't show the big picture of what is real happening in the A.C. circuit.

whizkid3
MVM
join:2002-02-21
Queens, NY

4 edits

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whizkid3

MVM

said by Ivybridge_I7:

Okay you would be right if you where talking about D.C. and the voltage drop is fairly constant across the extension cord.
etc....

No, I am right and know that I am; because I do voltage drop calculations for AC circuits every week and get paid to do them and take full personal liability for mistakes . There is absolutely no need to start talking about peak-to-peak, vs, RMS, phasors, etc. Heat production is steady in AC wiring and averages are just fine for what we are doing here. Yes, loads fluctuate, and hence so does the voltage drop and the heat output. But when you design a circuit or component thereof (like a wire size), you size it for the maximum load. You do it once. Although the instantaneous energy transfer differs, the heat exits the wire and its insulation in a much slower, steady fashion.

Phasors? I don't need them. Phasor diagrams and algebra were developed for people who don't have the education and steady practical experience in advanced engineering mathematics. When I need something I can just derive it using calculus or Euler's equations. Phasors are just a way of explaining complex math for people that can't do it. Phasors are a practical tool. But they are not needed for understanding such a simple subject as voltage drop. AC or DC. Any electrician worth his or her salt, can easily do design voltage drop calculations simply by using some tables in Ugly's. They don't need phasors, And they know that increasing the size of the wire, decreases the voltage drop and heat disipation. Its basic. Sorry for being blunt, but you seem to have just enough knowledge to be dangerous, but absolutely no real experience in sizing wiring for voltage drop. (Or you would have never said something so backwards in the first place, as "The bigger the gauge of wire, the less voltage drop". Its simply wrong.

Thanks for explanation, however, of the 'big picture' of what 120V AC does and what a sine wave is. I work with AC voltages from 120V to 138kV. The voltage drop math and results are the same. Sorry, but when it comes to electric power, I think you may be mistaking me for someone who is easily baffled by BS. (By the way, I don't use simple Ohm's law for voltage drop. My calculations include the reactive effects of the wire and its raceway if its in one. Most of the time, however, Ohm's law, is perfectly sufficient, unless its a long run in steel conduit.)

Ivybridge_I7
Cyber-Crime Researcher OpSec
Premium Member
join:2004-06-09
Daytona Beach, FL

Ivybridge_I7

Premium Member

said by whizkid3:

Thanks for explanation, however, of the 'big picture' of what 120V AC does. I work with AC voltages from 120V to 138kV. The voltage drop math and results are the same. Sorry, but when it comes to electric power, I think you may be mistaking me for someone who is easily baffled by a bunch of BS.

No B.S. my good friend , I just know A.C. theory when you apply it to parallel and series circuits. Induction and Capacitive loads are a very interesting subject, more on the engineering level. I am sure you would at least take in consideration the inrush current when the compressor motor starts. You know the electromagnetic flux that actual starts the motor when the current level is at it's peak.

whizkid3
MVM
join:2002-02-21
Queens, NY

4 edits

whizkid3

MVM

said by Ivybridge_I7:

I am sure you would at least take in consideration the inrush current when the compressor motor starts. You know the electromagnetic flux that actual starts the motor when the current level is at it's peak.

No, actually inductive inrushes of motors or transformers are never taken into account for heat dissipation from wiring. They are extremely short term effects, that happen infrequently and contribute nothing to be concerned about when one is concerned about heat.

Inrush is, however, taken into account for voltage drop and wire sizing calculations to ensure there is enough voltage available to start a motor. They are also taken into account for designing protective relaying, and fuse & breaker sizing to ensure that the inrush will not trip the circuit's protection. But we are talking about bigger stuff here. If it were a grave issue to be concerned about here, the manufacturer wouldn't put a NEMA 5-15 plug on the end of his little appliance. Bro, we are talking about sizing a short extension cord. After a while, experience is the best judge of the need to increase the wire size to accommodate inrushes. The only other realistic choice is to go to automated software, because its quicker than hand-calculating the time-varying inrush effects on voltage drop and heat dissipation using calculus. That is simply not necessary for a small circuit like this. If you have a long run of wire; you go one size up. No big deal.

I do agree. Inductive and capacitive effects are very interesting from an engineering standpoint. Once one learns the 'relatively easy' differential equations behind the physics of L & C, however, they easily see that the math behind this is uniform and the same equations govern many things in the physical world. I am more fascinated by non-linear effects in AC circuits, the frequency domain, Maxwells equations, large scale conceptual design and solving actual (large) construction problems. Voltage drop calculations, IMHO, are rote work that I more often pass off to others. After a while, sizing wire gets as boring as pulling wire.
whizkid3

whizkid3 to Ivybridge_I7

MVM

to Ivybridge_I7
said by Ivybridge_I7:

You have no way of knowing the total resistance across the extension cord two wires (hot-neutral) that are in series from the circuit breaker to the load. ... You need to measure the total series resistance across both the hot-neutral wires and the longer the cord, the more voltage drop and in theory more heat being dissipated.

In the real world, there is no need to know the resistance of the load or the total resistance of the circuit. We know what the circuit is designed to support in terms of maximum load. Voltage drop is calculated using the current draw of the load, not the resistance. The formulas are very easy and even when they are not, there are already published tables of voltage drop based on load size, wire size, what type of raceway the wire is in, and the power factor of the load. No need to reinvent the wheel here, and start deriving voltage drop equations. Pro's that do voltage drop calculations and wire sizing - electricians & engineers - don't waste their expensive time thinking about it. They get it done, using tried and true methods, and its done accurately. After a while, the need to increase wire size dependent on the size of the circuit and the length becomes gut instinct. Experience is the reason I can tell you off the top of my head, the maximum distance anyone would want to run 480V and not step up to 4160 or higher. Or what is the maximum distance to run a 120V, 15A or 20A circuit, before the wire size has to be increased. Only in extremely unusual and rare situations does anyone have to worry what 'may be' plugged into a 120V, 15A receptacle when sizing the wire for the circuit. The same goes for extensions cords. Most of the pro's here don't need calculations to tell you the size that should be used in a given situation. Yeah, to some its fascinating stuff. When you do it often enough; its very easy and becomes boring quickly.
said by Ivybridge_I7:

I still thing that spending the money and doing it right with a dedicated (15 amp) outlet using a romex or BX from the circuit breaker is the best way to go.

Of course. Agreed.

Jack_in_VA
Premium Member
join:2007-11-26
North, VA

Jack_in_VA

Premium Member

said by whizkid3:

Experience is the reason I can tell you off the top of my head, the maximum distance anyone would want to run 480V and not step up to 4160 or higher.

Really? What about the 2300 volt motor option? 1000 HP 2300 volt motors are very common in industrial plants. How about 13.8 Kv motors usually found on large chillers? Sometimes it's more cost efficient to run high voltage to the site of the 480 volt motors and set a transformer and MCC there. A good design company takes all that into account.

All this OP wanted to do was plug in his freezer. A quality proper sized extension cord would be fine for this. He said it was temporary.

One could make the argument that the POS cords that connect most equipment are extension cords themselves.

Ivybridge_I7
Cyber-Crime Researcher OpSec
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join:2004-06-09
Daytona Beach, FL

Ivybridge_I7 to whizkid3

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to whizkid3
said by whizkid3:

In the real world, there is no need to know the resistance of the load or the total resistance of the circuit. We know what the circuit is designed to support in terms of maximum load. Voltage drop is calculated using the current draw of the load, not the resistance. The formulas are very easy and even when they are not, there are already published tables of voltage drop based on load size, wire size, what type of raceway the wire is in, and the power factor of the load.
I still thing that spending the money and doing it right with a dedicated (15 amp) outlet using a romex or BX from the circuit breaker is the best way to go.
Of course. Agreed.

I never said that you should sit there and do the actual calculation to figure out the inductive or capacitive loads of the motor in the freezer, only that a good understanding of what is going on when the electrons go through the circuit is important.

Once you have a good understanding of Alternating Current and how it relates to circuit theory you will see them differently. Power Factor (ratio) is also good to know because that tell you how efficient the circuit will be with regards to the efficiency of the power applied.

besafe
@verizon.net

besafe to Ivybridge_I7

Anon

to Ivybridge_I7
As antiphising says:
Call a qualified master electrician and run a dedicated outlet to that freezer. If the compressor in that freezer shorts out and decides to draw a huge amount of current. the heat made across a long extension cord could start a potential fire.

Here is proof it happens-
»www.dailyrecord.com/arti ··· RONTPAGE

whizkid3
MVM
join:2002-02-21
Queens, NY

whizkid3 to Jack_in_VA

MVM

to Jack_in_VA
said by Jack_in_VA:

said by whizkid3:

Experience is the reason I can tell you off the top of my head, the maximum distance anyone would want to run 480V and not step up to 4160 or higher.

Really? What about the 2300 volt motor option? 1000 HP 2300 volt motors are very common in industrial plants. How about 13.8 Kv motors usually found on large chillers? Sometimes it's more cost efficient to run high voltage to the site of the 480 volt motors and set a transformer and MCC there. A good design company takes all that into account.

Agreed. But I didn't mention motors. Talking about distribution, diesel generation, etc. While I realize you do, I don't have to deal with motors larger than 600V. Using other voltages would certainly be a better and cheaper option than running additional 4" conduits hundreds of feet.
said by besafe :

Here is proof it happens-
»www.dailyrecord.com/arti ··· RONTPAGE

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