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So we have a 5kW generator for our film set. We need to get it as far away from the set as possible. As well as leave the option for 25m consumer extension cord for lights and haze machines.

So what I need is to work out the max length I can get with a little voltage drop as possible.

We would not be using all 5kW on set, probably only around 4kW but I would like to leave some headroom.

Am I right in saying that 6 mm\$^2\$ pair with earth will be adequate for this? I have Google'd this for a few days but not found a definite answer to this.

Thanks guys.

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You can use a resource like this page to find the resistance (ohms) per meter for wire of any particular diameter: http://chemandy.com/calculators/round-wire-resistance-calculator.htm

By knowing the current (whatever you are powering?) and the total resistance of the wire (in both directions) then you can use Ohm's Law to calculate the voltage-drop you can expect. I have a handy Ohm's Law calculator online here: http://www.rcrowley.com/eirp.htm

How much voltage-drop your loads can tollerate depends on what they are, how you are using them, and for how long. Remember that if you overload the extension cords, they will overheat and become dangerous.

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6mm copper wire (slightly smaller than AWG 2) has a resistance of about 0.594 mΩ/m, or 1.188 mΩ/m for the two wires in an extension cable.

5 kW would be about 21 Arms @ 240 V. Let's say we can tolerate a 5% drop, or 12 V. This means we can have at most a total of 12 V / 21 A = 571 mΩ of resistance.

Therefore, the maximum distance with this cable would be 571 mΩ / 1.188 mΩ/m = 480 m.

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  • \$\begingroup\$ 480 meters at 240v? \$\endgroup\$ – Jack Ogara Apr 3 '16 at 1:17
  • \$\begingroup\$ This answer seems to assume that the OP means wire with 6mm diameter. In the UK at least when one talks about "6mm" electrical wire they actually mean wire with a CSA of 6mm^2, \$\endgroup\$ – Peter Green Apr 3 '16 at 1:18
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    \$\begingroup\$ Those wacky UKers! Why don't they say what they mean? Anyway, in that case, it would be closer to AWG 10, with a resistance of 5.6 mohm/m, giving a total allowed distance of about 100 m for a 5% drop at 21 A. \$\endgroup\$ – Dave Tweed Apr 3 '16 at 1:22
  • \$\begingroup\$ Did you already take into account the drop in the return path (neutral) also? Or should you still divide by two? \$\endgroup\$ – PkP Apr 3 '16 at 2:10
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    \$\begingroup\$ @PkP: Was it not clear in my first paragraph that I was accounting for the doubled drop across the two wires in the cable? \$\endgroup\$ – Dave Tweed Apr 3 '16 at 2:12

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