How to amplify DC Voltage to be used like supply to 2km?
You need a boost converter to lift your 24V DC to 240V DC. Because these are not 100% efficient the power in to supply 240V DC at 35 watts is likely to be about 40 watts.

Then you need to convert it back down to (say) 24 volts at the far end and this will probably mean the full input power will be about 45 watts.

With 40 watts at 240VDC "pumped" onto the cable by the booster the current will be 167 mA and now you have to decide on the wire gauge: -



If you choose AWG 24 - it has 84.2 ohms per 1000m. Double this for the return path and multiply by 2 for 2km length and you get a total resistance of 337 ohms - this needs to provide 167mA to the load and will accordingly lose voltage on the way.

Simple ohm's law tells you that the volts lost are 0.167 x 436 ohms = 56 volts. This would need to be added to the 240V DC you generate from the boost converter so immediately you are looking to generate more like 296 volts rather than 240 volts.

That's not a problem for the load (or the cable) but your input power has risen to about 50 watts.

AWG 16 has a 2km loop resistance of only 53 ohms so you may choose this instead. With 167 mA flowing the volt drop will be about 8.8 volts so maybe this is a better choice.

Hope this helps you get your head around the iterative process of choosing the right cable to suit your power budget.