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Coiled Mains Lead - what's the problem?


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Tony Jones - 2017-08-01 7:00 PM

 

Interesting thread this. It does make me wonder about the LIKELIHOOD of any damage or danger (what professionals would call "risk assessment.")

Having travelled in France for years, and now living here, I've seen scores of mains cables thinner than mine, only partly uncoiled (or conversely uncoiled at great distances, across several pitches, paths or roadways!). And never yet witnessed a fire.

Have I (and my "aire-neighbours") just been lucky, or is the actual (mathematical) risk lower than we Brits think it is?

SorryTony, but luck doesn't really come into it! The risk is science based, and a consequence of choice, not luck.

 

The choices are:

1) the cable size used for the EHU, and

2) the load in Amps to which you subject the cable.

 

I have a 20 metre, 2.5mm, "13A" extension cable for use around the house. It is clearly labelled:

Max current:

Fully wound: 5 Amp (1150W).

Fully unwound: 13Amp (2990W).

 

Anyone who is unsure of the EHU cable size, and/or of their connected load, would be best to fully unwind the cable - whether or not they've seen one catch fire. :-)

 

I think the main difference between continental motorhomers and us in the UK, is that we tend to make greater use of EHU because they are mainly rated at 13A, which induces folk to take a "I'm paying for it, so I'll get the maximum from it" attitude, meaning 3kW kettles, hairdryers, electric hobs, even underfloor heating, whereas many continental bollards are only rated at 5 - 6A, and trip under those loads. So, the continentals have been conditioned into using less electricity, and consequently can get away with 1.5mm EHU cables, which, only being subjected to light loads, remain perfectly adequate.

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Dave 'R' - 2017-08-01 6:46 PM

 

Gram, It's my job.

 

I'm not doubting you, Dave, but this has intrigued me for some time (particularly when the subject of overheating mains cables comes up).

 

Is there a significant difference between the resistance and the impedance of, say, a 1 kW wound heater element?

 

G

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hallii - 2017-08-02 11:36 AM

 

Another factor is voltage, lower voltage increases the current, some of the voltages on continental sites are surprisingly low, 190v on one!

 

H

 

This is a misconception. If you lower the voltage the power consumed (wattage) will be lower... imagine if you connected a 100w 230v light bulb to a leisure battery, do you think the 230v bulb would still consume 100w at 12v? If so 8.3 amps would flow. No In reality the bulb would be a much lower wattage at 12v.

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Dave 'R' - 2017-08-02 9:55 PM

 

resistance is simple with a DC circuit, but when it comes to AC circuits power factors, impedances, inductances and cable capacitances come into play as well as the voltage drop due to the resistance of cables depending upon circuit configuration.

 

I have no problem with that, Dave.

 

But, is there a significant difference between the resistance and the impedance of a 1 kW wound heater element?

 

G

 

 

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Charles - 2017-08-03 5:04 AM

 

A 100w 230v light bulb would consume 67w at 190v... phew, I hate maths.

 

Assuming a purely resistive light bulb, and the element is the same resistance at both 230 V and 190 V; I make it 82.6 W at 190 V.

 

I hate maths too!

 

G

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Gram - 2017-08-0

 

I'm not doubting you, Dave, but this has intrigued me for some time (particularly when the subject of overheating mains cables comes up).

 

Is there a significant difference between the resistance and the impedance of, say, a 1 kW wound heater element?

 

G

 

The simple answer is NO. A wound heater element will only have minimal inductance, which would only become significant at radio frequencies.

 

Alan

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Gram - 2017-08-04 9:22 PM

 

Charles - 2017-08-03 5:04 AM

 

A 100w 230v light bulb would consume 67w at 190v... phew, I hate maths.

 

Assuming a purely resistive light bulb, and the element is the same resistance at both 230 V and 190 V; I make it 82.6 W at 190 V.

 

I hate maths too!

 

G

 

We are discussing a "purely resistive light bulb". I am sorry but you are overlooking the fact that the resistance of the filament is proportional to the absolute temperature of the filament. A 230V tungsten light bulb operating at at 190V will not reach normal operating temperature. Hence the resistance will be lower, resulting in a higher current and energy consumption than a straight application of Ohm's Law would suggest.

 

If the device was a water heater, the element of which is not subject to such a large temperate change, the your calculations would more likely be correct.

 

Alan

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Andy&Lou - 2017-08-01 2:15 PM

 

I always work under the assumption that If I plan to use the heater or electric hob then I must unwind the cable.

If I am just charging the battery and running the fridge then there is no need to unwind it.

 

These discussions seem to suggest that I have made a reasonable assumption.

 

Andy

 

I agree but do not forget the electric kettle. Even a "low energy" one will take more power than the battery charger and fridge combined. However provided that the kettle is not used repeatedly, it would probably not be a problem.

 

Alan

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Brian Kirby - 2017-08-02 12:33 PM

 

I have a 20 metre, 2.5mm, "13A" extension cable for use around the house. It is clearly labelled:

Max current:

Fully wound: 5 Amp (1150W).

Fully unwound: 13Amp (2990W).

 

.

 

Brian,

 

The cable size that you quote seems a bit OTT. In my professional opinion 1.5mm^2 cable rated at 15A would be adequate in this application.

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G

 

....... you are overlooking the fact that the resistance of the filament is proportional to the absolute temperature of the filament. A 230V tungsten light bulb operating at at 190V will not reach normal operating temperature. Hence the resistance will be lower, resulting in a higher current and energy consumption than a straight application of Ohm's Law would suggest......

 

Alan

 

Absolutely.

 

I covered that in my assumptions.

 

G

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Alanb - 2017-08-05 12:58 PM

 

Gram - 2017-08-0

 

I'm not doubting you, Dave, but this has intrigued me for some time (particularly when the subject of overheating mains cables comes up).

 

Is there a significant difference between the resistance and the impedance of, say, a 1 kW wound heater element?

 

G

 

The simple answer is NO. A wound heater element will only have minimal inductance, which would only become significant at radio frequencies.

 

Alan

 

Do you agree, Dave?

 

G

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Dave 'R' - 2017-08-02 9:55 PM

 

resistance is simple with a DC circuit, but when it comes to AC circuits power factors, impedances, inductances and cable capacitances come into play as well as the voltage drop due to the resistance of cables depending upon circuit configuration.

 

The above is generally true, but introduces too many technical terms that are not relevant to EHU cables.

 

The major loads are resistive in nature. e.g. electric space and water heating, electric kettle, perhaps an electric toaster, and the fridge 240V element. Modern switch mode battery chargers may constitute a partly capacitive load.

 

In general therefore we only need to consider resistance, and that has been amply dealt with previously in this thread.

 

Alan

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In general therefore we only need to consider resistance, and that has been amply dealt with previously in this thread.

 

Alan

 

I'm not sure that Dave'R' is with you on that one, Alan.

 

Now that I have got the hang of editing the quotes!

 

Nice pun, though!

 

G

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Gram - 2017-08-05 2:07 PM

 

Alanb - 2017-08-05 12:58 PM

 

Gram - 2017-08-0

 

I'm not doubting you, Dave, but this has intrigued me for some time (particularly when the subject of overheating mains cables comes up).

 

Is there a significant difference between the resistance and the impedance of, say, a 1 kW wound heater element?

 

G

 

The simple answer is NO. A wound heater element will only have minimal inductance, which would only become significant at radio frequencies.

 

Alan

 

Do you agree, Dave?

 

G

 

Gram,

 

You do not seem to be happy with my brief answer to your question, so here are some figures..

 

Due to work in progress I currently have easy testing access to the water heating electrical element of the Truma Combi in my motorcaravan.

 

The water heater is rated at 450W 230V. By calculation this indicates a resistance of 117.6 Ohms. A recent measurement gave 118 Ohms, but that included the test leads.

 

I have a meter that can measure inductance, but the lowest range is 4 mH (milli Henrys). With some rounding up, the highest inductance reading for the heater element was 0.3mH.

 

Inductive reactance Xl, measured in Ohms is given by using the formula Xl= 2 x Pi x f x L, where L is inductance in Henrys, and f is the frequency in Hertz (Hz).

 

The resulting reactance must be added in quadrature with the resistance to obtain the combined impedance, Z.

 

This comes down to the Square Root of ((R squared)+(XL squared)). (Pythagoras's Theorem)

 

Using the values of 117.6 Ohms and 0.3mH the result is Z=117.600038. No significant difference!

 

To be generous, and assume that my inductance measurement is too low by a factor of 10, the result becomes Z=117.6038. Again no significant difference between impedance and resistance, with the water heater element.

 

Alan

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trevor166uk - 2017-08-06 9:21 PM

 

 

Can I just safely assume it is best to fully unwind the mains lead? It seems sensible to me.

 

 

I think it's best to assume that it's best to unwind it - then it won't matter if it's wrong.

 

I've heard of coiled cables overheating - but I've never heard of an uncoiled cable overheating.

 

;-)

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trevor166uk - 2017-08-06 9:21 PM

 

Having read through this thread it all seems to be getting very technical, too much so for me.

 

Can I just safely assume it is best to fully unwind the mains lead? It seems sensible to me.

 

 

Trevor,

 

Please may I apologise for my technical input. I did try to avoid it, by giving the simple answer, and also considered a PM.

 

Your question has been answered in a previous post.

 

May I make one further technical point, that I do not think has been emphasised previously.

 

Simply that heating of any cable is proportional to the square of the current carried, so doubling the current will multiply the heating effect by FOUR!

 

 

Alan

 

 

 

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