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I have a football field to wire up. It measures 900m x 450m. I need to connect multiple light weight devices that use a 100-220vAc adapter that convert the voltage to 24v for the daughter device.

What i was thinking was to use a Single Phase 220v to 480v Step up transformer at the starting point and to have step down treansformers at the nodal points. A 10AWG copper cable laid on the border of the field to connect the main transformer to all the following ones in a parallel setup. I have around 15 of those adapters to plug. The devices use 24v 0.5A load.

Am i planning something wrong here? Is there a better approach for this solution?

I dont want to plug in 15 UPSs at the nodal points, it would be best to have one on/off to power all of them 15devices on/off. Picture explaining setup

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    \$\begingroup\$ Are you sure you need the step up to 480V? Are there appropriate building regulations about this? \$\endgroup\$ – pjc50 Mar 4 '15 at 9:58
  • \$\begingroup\$ 2 things: obtain the minimum mains voltage allowed by your wall adapters (generally those are transformers+rectifiers+regulator with wide input), and calculate the maximum loss in your 10AWG wires (there are calculator online depending on length). Pretty sure the drop is not that much, and the corresponding lost power not that important. But you can also put wires in parallel. \$\endgroup\$ – Mister Mystère Mar 4 '15 at 10:08
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    \$\begingroup\$ @mister two wires in parallel is generally not allowed. If one wire breaks, the other will become a dangerous heater - not too hot to burn out fast, not enough current to trip anything. If you need more current, just go up a gauge. \$\endgroup\$ – tomnexus Mar 4 '15 at 10:13
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    \$\begingroup\$ Is this to electrocute players that fall over the touchlines? \$\endgroup\$ – Andy aka Mar 4 '15 at 10:24
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    \$\begingroup\$ Are you sure you have the dimensions right? 900m is nearly a kilometer (more than 1/2 mile.) You could fit dozens of football fields into an area of the dimensions you mention. If it is just one football field it could be done by a competent amateur. But if it is dozens of football fields you should consult a professional. \$\endgroup\$ – Level River St Mar 4 '15 at 14:20
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It may be simpler than you think.

The wall adapters you're using are 24V@0.5A, which is 12W. Assuming 0.8 efficiency, that's 15W that mains has to provide. Under 220V, that's I=68.2mA*.

You have 15 adapters distributed along 2.5km, that's 166.7m of wires between each adapter. AWG10 wires have a resistance of 3.277Ohms/km, so on each line there will be R=546.3mOhms between each adapter. As shown on the equivalent circuit below (where I've modelled 3 adapters out of 15), the voltage drop experienced by each adapter will be twice the voltage drop in the line resistances (single and return).

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the resistance values for the adaptors are for illustration purposes only (they're likely to change in reality), worst case current 'I' calculated above will be used instead. This is simplified this way:

schematic

simulate this circuit

In your case, there will be 15 resistors. Let i be the index of the adapter considered, in your case between 1 and N=15. The total voltage drop at adapter i is therefore:

$$\Delta V(i)=\sum\limits_{k=0}^{i-1} 2R(N-k)I$$ $$i \in 1,2,...,15$$

The 15th adapter will have the biggest drop, but it is only: enter image description here

Or 4% of the input voltage. Since your adapters work from 100 to 220V, you're fine. No need to do anything other than wiring them around the football field.

Just to check, the maximum power dissipated (~in terms of density) is in the first segment before the first adapter: $$P_{dissipated}=\frac{\Delta V(1)^2}{2*R}=1W$$ Or 0.6mW every 10cm - not a heater, not a problem.

You should probably check the efficiency of the wall adapters and redo the calculations but it shouldn't be so bad that all this is thrown out the window.

*: Only if the voltage drop is neglectable compared to 220V. Otherwise, it's an iterative process: a second iteration with I=Power_device/(220-total voltage drop)=71.1mA will give you a more accurate voltage drop value, then a third one an even more accurate etc. In this case I got 1) 8.94V 2) 9.32V 3) 9.335V; I think it converges to about 9.34V. This will actually be an overestimate since it is assumed all taps draw the same current (the biggest, after the total voltage drop) when in fact they draw less and less as you get closer to the supply.

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  • \$\begingroup\$ If someone is to reproduce this analysis for a larger voltage drop (like by putting more poverful devices instead of 15W), the 15th device will draw current not off 220V, but off, say, 150V. In this case a 15-watt device would draw 100mA. Now I see you did a "worst case of best case" (max power, low voltage drop), but you should at least mention the "worst case of worst case" too (max power, high voltage drop). \$\endgroup\$ – Eugene Ryabtsev Mar 4 '15 at 16:02
  • \$\begingroup\$ Absolutely true. I've updated my answer above. \$\endgroup\$ – Mister Mystère Mar 4 '15 at 17:21
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A better option would be to use two phases and go halfway around the perimeter with one phase and cover the other half with a wire in the other direction. This way you can use two fuses, provide more current and do not need a transformer (depending on your power output needs).

To power all of the devices off and on at once you might want to consider a three phase switch on your distribution point.

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  • \$\begingroup\$ So a 2phase line, 1Phase powering left side 2nd phase powering otherside without stepping up anything? \$\endgroup\$ – user2967920 Mar 4 '15 at 10:04
  • \$\begingroup\$ Exactly! I'm not sure about the US but in Europe the usual setup consists of three phases (L1, L2 and L3), a neutral point (N) and a protective conductor (PE). You should have a similar setup from which you take two phases, the neutral and the protective conductor. \$\endgroup\$ – Jan Krüger Mar 4 '15 at 10:08
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    \$\begingroup\$ No particular need for two phases, but what's important is to go halfway around, each way, to minimise voltage drop. \$\endgroup\$ – tomnexus Mar 4 '15 at 10:14
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    \$\begingroup\$ In the UK, for domestic use, you normally join the two sides up to create a 'ring main' so two loads on one side don't cause an excess drop as some current can flow the long way round; probably not as useful if you know the loads are even. \$\endgroup\$ – Pete Kirkham Mar 4 '15 at 12:37

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