Advice on running electrics to shed

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I'll stand corrected (or sitting at the computer) if wrong but I was taught to never use a coiled cord as it increases the resistance creating heat in the wire coils. Either pull all the cable out of the reel or get a shorter cord instead.

Pete
Sorta right- it increases the APPARENT resistance (inductive reactance actually) as it is a 'loosely coupled coil'

https://en.wikipedia.org/wiki/Electrical_reactance - look down towards the bottom for inductive reactance...

The higher the current, the more the reactance- and the hotter it gets...
And yes- it certainly does create heat...
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Flex doesn't mean something is not part of the fixed installation.
Another way to interpret this is that fixed wiring has no movement, it cannot induce any strain into the conductors especially the CPC. Using a flexable cable makes sense in many applications where you are connecting from something fixed to something that has some movement and in cases like the hand dryers there is no reason to use the flex, it could easily be done in PVC conduit or mini trunking but the OEM decided otherwise. Once a cable is flexable you can have issues where many strands of the CPC might have broken due to the flex and cable gland but it would still bell out as continous, now the PAT test would test by passing a current through it which can cause it to just blow and become open circuit which is what could happen under a fault synario that could leave the appliance in a dangerous state, ie live metalwork.
 
Sorta right- it increases the APPARENT resistance (inductive reactance actually) as it is a 'loosely coupled coil'

https://en.wikipedia.org/wiki/Electrical_reactance - look down towards the bottom for inductive reactance...

The higher the current, the more the reactance- and the hotter it gets...
And yes- it certainly does create heat...
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View attachment 190452

My understanding on this is because it's carrying AC current - it's acts kind of like an electromagnet (although without an iron core), and induces it's own reactive current every time the current reverses - hence "inductive reactance".
 
Yep, coiled can get warm, stretched out, not so much.
I did it once myself without thinking. Just a dumb moment.
Plugged an electric heater into a mostly coiled cable reel and 15 or 20 minutes later realised that I'd done something stupid.
When I unspooled the cable the bulk of the flex from inside the coil was too hot to touch and super floppy.

To do some numbers:
In the UK you can buy a 50m extension cable, good commercial brands, no problem.
That is 100m live and neutral return.
In 2.5mm sq extra heavy flex the wire has a resistance of 0.74 ohms
Carrying its rated load of 13A from a standard plug, the cable drops 9.6V along the wire and dissipates 125W (V=IR, Power = IV = I squared R = V squared/R)
Remember the heat put out by an old style 125W lightbulb. Not so much spread out along 50 metres but imagine all that heat trapped inside the coil of wire. It will get very hot very quickly.

Do the same for a lightweight cable reel, same length but 1.5mm square and 10 amp rated (though it will be on a 13A fused plug so relying entirely on the user not to overload it)
Now we have 1.21 Ohms resistance, 12.1V dropped along the cable and almost the same heating, 121 W even though the current is less. Again the cable will quickly get hot if coiled up.

Electric heaters are the killer.
Most of our tools only run for a short time. Anything that runs continually at high power for a long time poses the greatest risk of heating and fire.
We should teach ourselves to be nervous of electric heaters, immersion heaters, and to a lesser degree powerful pump and fan motors because these are big loads that can be left running for hours at a time.


Oh, and this is why the way a cable is installed must be taken into account when you design household wiring.
These dissipation numbers aren't much different for fixed wires, so you could easily have 125W of heat being generated in a long run of twin and earth. No big deal if that wire is in free air to cool it, but buried under 6 inches of loft insulation a wire can become noticeably warm even if it is stretched out and not concentrated on a reel.
 
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I did it once myself without thinking. Just a dumb moment.
Plugged an electric heater into a mostly coiled cable reel and 15 or 20 minutes later realised that I'd done something stupid.
When I unspooled the cable the bulk of the flex from inside the coil was too hot to touch and super floppy.

To do some numbers:
In the UK you can buy a 50m extension cable, good commercial brands, no problem.
That is 100m live and neutral return.
In 2.5mm sq extra heavy flex the wire has a resistance of 0.74 ohms
Carrying its rated load of 13A from a standard plug, the cable drops 9.6V along the wire and dissipates 125W (V=IR, Power = IV = I squared R = V squared/R)
Remember the heat put out by an old style 125W lightbulb. Not so much spread out along 50 metres but imagine all that heat trapped inside the coil of wire. It will get very hot very quickly.

Do the same for a lightweight cable reel, same length but 1.5mm square and 10 amp rated (though it will be on a 13A fused plug so relying entirely on the user not to overload it)
Now we have 1.21 Ohms resistance, 12.1V dropped along the cable and almost the same heating, 121 W even though the current is less. Again the cable will quickly get hot if coiled up.

Electric heaters are the killer.
Most of our tools only run for a short time. Anything that runs continually at high power for a long time poses the greatest risk of heating and fire.
We should teach ourselves to be nervous of electric heaters, immersion heaters, and to a lesser degree powerful pump and fan motors because these are big loads that can be left running for hours at a time.


Oh, and this is why the way a cable is installed must be taken into account when you design household wiring.
These dissipation numbers aren't much different for fixed wires, so you could easily have 125W of heat being generated in a long run of twin and earth. No big deal if that wire is in free air to cool it, but buried under 6 inches of loft insulation a wire can become noticeably warm even if it is stretched out and not concentrated on a reel.
In fact, the Au/NZ electrical standards calculator accounts for the various possible ways a cable could be installed and figure that into the wire sizing needed for a run...
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https://www.jcalc.net/cable-sizing-calculator-as3008

Much easier than when I first started out- we had a 'book of tables'- that you went through looking for the right value to select....
 
Sorta right- it increases the APPARENT resistance (inductive reactance actually) as it is a 'loosely coupled coil'

https://en.wikipedia.org/wiki/Electrical_reactance - look down towards the bottom for inductive reactance...

The higher the current, the more the reactance- and the hotter it gets...
And yes- it certainly does create heat...
View attachment 190450
View attachment 190451
View attachment 190452
I would've thought another factor is just that when the lead is uncoiled, it is likely exposed to the air, and the heat is easily dispersed, whereas if you coil it up the heat is concentrated.
 
I would've thought another factor is just that when the lead is uncoiled, it is likely exposed to the air, and the heat is easily dispersed, whereas if you coil it up the heat is concentrated.
Correct, that also makes a contributing factor- but the 'apparent resistance' from the inductance means it is more than a simple resistance measurement would indicate... (so say you measured the resistance and it calculated at say 100w dissipated in the cable, in fact you would get more than that 100w actually being dissipated in the rolled up lead... add in the heat retention, and thats how you get them going into 'meltdown' like the pics from before...)
Heat retention is an issue however- which is why electricians have to 'derate' cables depending on their surrounding (running in conduit, underground or surrounded by insulation all means a cable can carry less current safely- there's tables for that as well)

If you go to that link I provided earlier, you can play with the settings for how a cable is surrounded and actually see what different materials being around the cable does to the current limit it can safely carry at various current levels...
 
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Re: coiled extension leads - I don’t think inductive reactance plays much of a part. Reactance is lossless, for one thing, and the currents within the cables are opposing, and would therefore cancel. Try using a coiled extension lead as an electromagnet.

As above, the cable can’t dissipate heat properly.
 
Re: coiled extension leads - I don’t think inductive reactance plays much of a part. Reactance is lossless, for one thing, and the currents within the cables are opposing, and would therefore cancel. Try using a coiled extension lead as an electromagnet.

As above, the cable can’t dissipate heat properly.
Although this is talking about powerlines, inductive reactance can indeed cause conductors to heat up (powerlines have the issue of it causing them to 'droop' or sag)
In electric power systems, inductive reactance (and capacitive reactance, however inductive reactance is more common) can limit the power capacity of an AC transmission line, because power is not completely transferred when voltage and current are out-of-phase (detailed above). That is, current will flow for an out-of-phase system, however real power at certain times will not be transferred, because there will be points during which instantaneous current is positive while instantaneous voltage is negative, or vice versa, implying negative power transfer. Hence, real work is not performed when power transfer is "negative". However, current still flows even when a system is out-of-phase, which causes transmission lines to heat up due to current flow.
https://en.wikipedia.org/wiki/Electrical_reactance
A coil is an inductor and a rolled up cable is indeed an inductive coil (loosely coupled and low efficiency, true, but still an inductor) and as such still has a phase shift in the voltage/current curves...
 
“In electrical circuits, reactance is the opposition presented to alternating current by inductance and capacitance.[1] Along with resistance, it is one of two elements of impedance; however, while both elements involve transfer of electrical energy, no dissipation of electrical energy as heat occurs in reactance; instead, the reactance stores energy until a quarter-cycle later when the energy is returned to the circuit. Greater reactance gives smaller current for the same applied voltage.”

See the middle sentence in bold, from your wiki link. Resistance dissipates heat, not reactance.

Humour me: power up a coiled extension lead, and touch something thin and metal around it. If there’s any significant net inductance going on, it will vibrate. I think you’ll find it won’t.
 
“In electrical circuits, reactance is the opposition presented to alternating current by inductance and capacitance.[1] Along with resistance, it is one of two elements of impedance; however, while both elements involve transfer of electrical energy, no dissipation of electrical energy as heat occurs in reactance; instead, the reactance stores energy until a quarter-cycle later when the energy is returned to the circuit. Greater reactance gives smaller current for the same applied voltage.”

See the middle sentence in bold, from your wiki link. Resistance dissipates heat, not reactance.

Humour me: power up a coiled extension lead, and touch something thin and metal around it. If there’s any significant net inductance going on, it will vibrate. I think you’ll find it won’t.
It however stores that energy and then releases it later in the phase angle...
 
Reactance is lossless
The best way to look at this is that a resistance can produce usable energy, both capacitive and inductive reactances will consume energy so it will cost you but they do not produce any usable energy and is why they are often referred to as parasitic. This is a good reason why our national grid should become HVDC because you then reduce the loses in the transmision lines.
 
It can be dissipated as heat however (nothings 100% efficient) plus the phase angle change can make the apparent current appear higher if the voltage is lower at that point (total output wattage remains the same as demanded by the load)- meaning more heat generated because of the increased current through the resistance of the winding... (ohms law)
This is well known by those working with large high current inductors (like the ones found in offgrid mains inverters) and has to be accounted for in their design...
 
It can be dissipated as heat however (nothings 100% efficient) plus the phase angle change can make the apparent current appear higher if the voltage is lower at that point (total output wattage remains the same as demanded by the load)- meaning more heat generated because of the increased current through the resistance of the winding...
This is well known by those working with large high current inductors (like the ones found in offgrid mains inverters) and has to be accounted for in their design...
Yes, I think you’re talking about power factor, where an inductive load (eg) causes a higher current than would be expected for a given power dissipation. Higher current means more I^2R losses. But note it’s R in the equation, not X or Z.

When you wind an extension lead, you’re making two opposing air-core inductors, which should cancel any displacement/phase shift.

So, the question is, what is the power factor of a coiled extension lead? When I get time, I’ll (briefly) run a heater (PF=1) with one, and measure the overall PF with a power analyser. My guess is very close to 1.
 
Hopefully this might show it a bit more clearly- a purely resistive load (top, labeled RO) has no phase shift, current and voltage remain in phase, where an inductive load (bottom diagram RL) has the current and voltage out of phase- with the same wattage being consumed with the load at both points, you have times in the cycle where the output current is higher than when the stored energy came in- same wattage, but higher current coupled with lower voltage, with higher IR losses in the winding...
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re your last post (added while I was doing this one lol), note that the two 'windings' are physically separated in space (by the insulation), meaning that you won't have a 'perfect' cancellation of the inductance's of the two 'windings' ie the Active and the Neutral conductors inside the cord are 'close' but not occupying the same space (plus they will be randomly twisted so physically be closer or further away than their 'counterpart' opposing 'winding'...
 
OK, seeing as pulling out old OSB in the new workshop is boring, and physics is interesting, I did the test.

2kW filament heater alone:
236.2 V
8.75 A
2058 W
PF: 1.00

2kW heater + 30m coiled extension lead (ten seconds only!):
236.4 V
8.53 A
2071 W
PF: 1.00

As you see, the coiled lead isn’t sucking (or pumping) any VARs. What you say about displacement PF is correct, but doesn’t apply in this case.
 
The way we remembered the phase shift was using CIVIL, with a capacitance the current leads by 90° and with an Inductance the current lags by 90°.
 
OK, seeing as pulling out old OSB in the new workshop is boring, and physics is interesting, I did the test.

2kW filament heater alone:
236.2 V
8.75 A
2058 W
PF: 1.00

2kW heater + 30m coiled extension lead (ten seconds only!):
236.4 V
8.53 A
2071 W
PF: 1.00

As you see, the coiled lead isn’t sucking (or pumping) any VARs. What you say about displacement PF is correct, but doesn’t apply in this case.
Well it is something that designing high current inductors for inverters and such specifically warn against- how accurate (ie significant digits) is your tester???

(I am also wondering why the extra 13w in the total draw appeared??) with added resistance in the line causing a voltage drop at the load/heater, it should have been less current (same resistance, lower voltage = should be a lower wattage drawn... the current has dropped as expected, which means the loads wattage has dropped (W=VxA), yet the total wattage has risen....)
Its going somewhere....
:unsure:
The displayed wattages also don't match the calculated wattages...
:dunno:
Displayed is 2058W, but 236.2V×8.75A=2066.75W
on the cord, displayed is 2071W, yet 236.4V×8.53A=2016.4922W
(this displays the expected drop in total wattages if the load is purely resistive, where the devices displayed wattages do not- the wound up cord is obviously an added inductance, but the PF cant be 1 if the displayed values are correct...

Something doesn't quite add up there...
🧐
 
There will be some rounding errors in the display. Eg any PF of 0.995 or above would display as 1.00.

Also, the meter doesn’t display all the results on one screen, so as the heater (and cable) were heating up, the various values were changing slightly as I recorded them. Still, just a few percent, and not enough to convince me there’s significant reactance in the ext lead. This is most likely because the currents in the cable run counter to each other. Real inductors, like the inverter ones you mention are wound in a single direction.
 
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