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Does the length of cable de rate the cable as it creates a greater resistance and as a result heats up the cable reducing its current carrying capacities?
 
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Spoon

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Does the length of cable de rate the cable as it creates a greater resistance and as a result heats up the cable reducing its current carrying capacities?
No.
What is does affect is the voltage. That's why you calculate the volt drop of circuits.
If the volt drop is too high then the cable size will have to increase.
 

Simon47

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No. Think about it - if you keep the current the same but double the length, then you double the resistance and double the heat generated. But, you have also doubled the length of the cable - so the amount of heat generated per length of cable is exactly the same.
 

pc1966

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Not directly, as already said the heat loss per unit length is the same.

But when it comes to coping with the overload under fault conditions the situation is rather different. The let-through fault "energy" (the I2t term) for an OCPD generally increases at small overloads because the 't' term increases much faster than the 'I^2' term decreases.

So for fault conditions a long cable can end up suffering more thermal stress because it took a lot longer for the OCPD to disconnect. For a fuse the curve is fairly simple, dropping down and flattening off as you get to the worst-case PFC actually having the least I2t term!

But a MCB is a lot more complex, and it depend on which region you are in (thermal trip, or "instantaneous" magnetic trip). Here is an example from the Hager commercial catalogue:
Hager-B-curve-MCB.png
You can see that when you hit the magnetic trip region at {3-5 * In} then the fault energy drops massively as the disconnect time drops from 5-25 seconds to tens milliseconds or so.

However, unlike a fuse the MCB does not disconnect that much faster as PFC increases to allow a shorter 't' to offset the greater 'I^2' aspect. So once you get beyond a couple of kA fault currents then a similar rating BS88 fuse lets through less energy.
 
Strange way of wording it Jako but in simple terms yes. I think this is what you are getting at!
If you have for example a 3% VD with 20A at 50m and extend the cable, that 3% volt drop can only be achieved by de-rating the circuit current (effectively de-rating ccc of the cable for a particular circuit). As the circuit current would be less than before, I'm not sure of the significance of any heat generated!
 

pc1966

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I think we are all looking at the OP's statement from differing points of view!

Typically there are several different limits on a cable that have to be taken in to account when designing the circuit:
  1. The thermal limit of CCC. This is a product of both the cable thickness (hence I2R loss per unit length heating it) and the environment it is in (thermally insulated, in duct, open air, etc) that alters how easily the heat escapes, but in itself is independent of the length. (post #3)
  2. Voltage drop limit. Here a long cable may end up much heavier than the CCC limit simply to avoid too much VD (posts #2 & #5).
  3. Fault disconnection time limit. This is depends on both the OCPD characteristics and the Zs value (i.e. PFC) which depends on the cable resistance (R1 + R2) and that is length-dependant. So (much like VD) can result in additional conductor size being needed to meet safe disconnection times.
  4. The fault let-through energy (post #4) depends on the disconnection time and so has a thermal aspect for survival - the adiabatic calculation to see what a fault would do the insulation as an infrequent stress which might require a shorter disconnect time than the shock-protection limits in the wiring regs.
So while a cable's thermal CCC is independent of length (just the current and thermal environment), the VD and fault disconnection/adiabatic limits are length-dependent factors to consider.
 
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My question was if the circuit say was run in 1.5 cable and can take 10 amps if I had a long run would that reduce that 10amps sorry for being unclear. Thanks for your help guys
 

pc1966

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My question was if the circuit say was run in 1.5 cable and can take 10 amps if I had a long run would that reduce that 10amps sorry for being unclear. Thanks for your help guys
Yes, at a point you would find it fails to meet the VD or disconnection limits.

So as the length increases from near-zero you have a constant limit, the thermal CCC, but then you reach one or more points when you need to derate it to maintain either the required VD (say 5% or 3% for lights), or to allow adequate fault clearance.
 
I think we are all looking at the OP's statement from differing points of view!
Agreed pc1966, the wording of the question does'nt help and of course many other things to take into consideration but got the impression trainee asking a simple question looking for a simple answer! We all know though nothing is that simple or straightforward when it comes to circuit design.
 
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Yes, at a point you would find it fails to meet the VD or disconnection limits.

So as the length increases from near-zero you have a constant limit, the thermal CCC, but then you reach one or more points when you need to derate it to maintain either the required VD (say 5% or 3% for lights), or to allow adequate fault clearance.
So say for an example it could take 10 amps at 10 meters if we went up to 100meters would we expect that to drop to day 5 amps for example?
 

pc1966

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Probably in the majority of cases the VD limit will also deal with the disconnection time indirectly. For example, if you need 10A and have a 10A MCB or similar then at 5% VD your fault current is (up to) 20x the MCB rating, so a long cable that has its size increased to maintain VD is keeping the PFC up and clearing time down.

Where that gets in to difficulties is when you have a higher resistance CPC (e.g. T&E with 2.5 L/N and 1.5 E, or SWA with a steel CPC that is a much poorer conductor) or on high current circuits when your cable's R1+R2 is getting down to the supply's Ze value.
 
So say for an example it could take 10 amps at 10 meters if we went up to 100meters would we expect that to drop to day 5 amps for example?
Maybe, you would need to limit the circuit current but you really need to do some proper circuit calcs to be sure. It wont drop automatically to 5A as in your example.
 

pc1966

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So say for an example it could take 10 amps at 10 meters if we went up to 100meters would we expect that to drop to day 5 amps for example?
It depends on the application:
  • Is it 3% or 5% drop you have to meet?
  • Are you on 230V single phase or 400V three-phase?
They will give you different current limits for the same cable!
 
  • Thread Starter Thread Starter
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It depends on the application:
  • Is it 3% or 5% drop you have to meet?
  • Are you on 230V single phase or 400V three-phase?
They will give you different current limits for the same cable!
Okay thank you. I understand that you need to up the cable size to satisfy voltage drop and disconnection times. Just when you look in the regs it says a cable can carry a given amount of current depending on its installation method and external influences but the main question I was trying to clear up was that when the length of the conductor increases so does the resistance meaning the conductor gets hotter and effecting the current carrying capabilities. Thanks for all your in-depth answers
 

Simon47

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No, the conductors don't get hotter.
As already said, for a given current, the increase in heat generated will be matched by the increase in length. The heat per length of the cable remains the same. So 1m or 100m - for a given current being carried, the cable will run at the same temperature.
But as described above - at some point as the length increases you will have to use a larger cable to meet volt drop or fault carrying requirements, but not for temperature rise reasons while carrying the normal load current.
 
Resistance will cause heat like a high resistive joint etc.

But it's a bit chicken egg - Resistance increases with length of conductor - Resistance causes heat - Heat causes more resistance.

but under normal operation the cable is ok but under fault the cable could fry.

Edit: unless your CSA is too small in the first place I.e bellow your protective device then fry.
 
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davesparks

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the main question I was trying to clear up was that when the length of the conductor increases so does the resistance meaning the conductor gets hotter and effecting the current carrying capabilities.
As has been said already this is incorrect, the conductor does not get hotter as it gets longer.
The conductor will reach the same temperature.

The heat is evenly distributed around the surface area of the conductor, for every increase in length there will be a proportional increase in surface area.
 
Surely an excessive volt drop from a very long run would cause the conductors to heat up.
 

pc1966

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Surely an excessive volt drop from a very long run would cause the conductors to heat up.
Heat per unit length is exactly the same. Total wasted power is higher, of course!

But under fault conditions a higher overall resistance and resulting lower fault current means the cable can rise to a much higher temperature due the longer disconnection time of the fuse/MCB.
 

littlespark

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If your really long length of cable is up on a drum, then yes.... it would be hotter.
Actually would generate the same amount of heat, but wouldn’t be able to disappate.

The reason extension leads may state “10A when fully unwound“
 

Simon47

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OK, lets consider an artificial example with numbers made up to keep the maths simple.
Lets say you have a cable that has a resistance of 1 ohm/metre (lets just assume that's the loop resistance, i.e. the sum of 0.5R down one wire and 0.5R up it's partner), and you pass a steady 1 amp through it. It will therefore have a volt drop of 1 volt/metre, and generate 1 watt/metre of heat.
So if the cable is just 1m long, you'll have 1W of heat to dissipate. Make that 10m long but still pass the same 1A, you now generate 10W of heat, but have 10m of cable to dissipate that heat - i.e. still the same 1W/m. Make that 100m and the same holds - now 100W of dissipated heat, but over 100m of cable, and so still 1W/m.
Same cable, same 1W/m of heat to dissipate = same temperature rise in the conductors.

As I said, a very artificial example with "quite lossy" cable. You can see that if your supply is 240V, then at the end of a 10m length of cable, you are down to 230V - a loss of 4.1%. But if it were 100m of cable, you'd be down to just 140V left.

It also illustrates what's also been said about the effect on cable protection. Lets say you had a hard fault at the end of a 10m cable, you now have 240V dropped across 10R, and will pass 24A. If you had a 6A MCB protecting the cable, then it's possibly going to take quite a while to trip - and until it does, then the cable will be dissipating 5760W (I^2R=24*24*10), or 576W/m. Asking the same cable to dissipate 576W/m instead of 1W/m absolutely will make it heat up more :eek:
Were the cable only (say) 1m long, then under fault conditions the current would be more like 240A (assuming zero source impedance) - so the cable would generate over 57kW of heat (I^2R = 240 * 240 * 1). But the 6A MCB would trip "almost instantaneously" with such an overload.

Now, as already hinted at, there is the complication that the load drawn may well vary with voltage. A resistive load (such as a heater or incandescent light bulb that's lit) will decrease power and current as the voltage decreases. A switch mode PSU will typically increase current to maintain constant output power as input voltage reduces. In extreme (such as the 100m long cable above), an incandescent light bulb will initially take less power and current, but at some point the resistance of the filament will decrease and current draw may go up - typically you can expect an inrush of around 10x the "lit" current for incandescent light bulbs when switching on from cold.

Some railway modellers use that characteristic of light bulbs to protect sections of track - with DCC*, you could power (in theory) the whole layout from one supply/controller, but without any protection/sectioning, a fault anywhere on the layout would kill the whole thing. By using a low voltage (i.e. 12V) bulb in series with a track section, the normal train current won't light the bulb and it's cold resistance will be low enough not to cause problems. But should there be a short, that will light the bulb, it's resistance will increase and limit the current (while allowing other sections of track to continue working), and as a bonus the light will indicate both the fault and it's location.
* Digital Command and Control. Power is applied to the rails continuously, with a control signal superimposed upon it. Multiple items can run non one section, all fed from one power supply but controlled independently by the control signal.
 
No. Think about it - if you keep the current the same but double the length, then you double the resistance and double the heat generated. But, you have also doubled the length of the cable - so the amount of heat generated per length of cable is exactly the same.
Yes but is the cable run vertically? Then the hotter air will rise and heat your cable more.
Actually the voltage drop is a good way to indicate power lost along the length. Power loss is the heat generated. But also voltage is important for say running a motor. The motor will overheat a bit for a 5% voltage drop below nameplate voltage; if the cable is undersized for the length then the increased current due to the voltage drop may cause more voltage drop as the cable heats, causing more current until either there just isn't enough power left or the cable burns.
 
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