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skippyv

dangers of long line voltage drops

skippyV
9 years ago

Hello,
We are designing a tool that requires a small DC voltage but needs it at the end of a long umbilical. And using batteries aren't an option.

So I'm asking if there are concerns with designing a system that is very inefficient with regards to line loss. Besides the obvious one of wasting energy.

For example, 1000 ft of 14 AWG wire at 5 amps creates a voltage drop of 12.87v with a voltage source of 24v.
Which means about 54% energy loss.

But if the wire's ampere rating is sufficient for that load - are there any other factors that need to be considered?

The example values I used are what we are considering using for this project.

Thanks for any input.

Comments (7)

  • bus_driver
    9 years ago

    Be aware that an electrical "circuit" means just that- the current must return to the source. So 1000 feet of wire will create a circuit that is not more than 500 feet between the source and the device that uses the power.
    Voltage is the "pressure" that pushes the amperage (volume) through the conductor. Lower voltage means that lower amperage will be available for use at the device. The lost power in the wire is converted to heat. The results of that vary depending on the type of device using the power. Motors will try to draw higher amperage if the voltage is low. Overheating of motors is typically the result.

  • Ron Natalie
    9 years ago

    Voltage drop on a conductor isn't dependent on the voltage at the source. It's a function of the current and the resistance (the size of the wiring and it's material and the length).

    Ampacity rating is only slightly related to voltage drop. Ampacity is determined by how much heat the conductor can tolerate (given the installation). Heat is a function of current for a given wire (and essentially heat is where the energy lost in the voltage drop goes).

  • paulusgnome
    9 years ago

    If the load current will vary with operation of the tool, so too will the voltage to the tool, this being a product of the cable resistance. Disconnect the tool, and you will see 24V at the end of the cable. Depending on the nature of the tool, this may or may not be a problem.
    It is possible to use some tricky electronics to correct for the voltage drop, but most people will find it simpler and easier to stump up for some fatter conductors.

  • bus_driver
    9 years ago

    My solution would include bringing 120 volts AC to near the point of use and have the transformer and rectifier at the point of use. If the DC amps is relatively high, a 240 volt supply might be better.
    Lots of details not included in the original post.

  • skippyV
    Original Author
    9 years ago

    Thank you all for your inputs. They were most helpful.

    @bus_driver, I didn't even think of the return path! Thanks for that. I'll have to adjust my calculations. (slaps forehead - DOH!)

    @ronnatalie, I mentioned current ratings only because it was an issue to consider. We wouldn't want our conductor to melt ;).
    But the question's purpose was to determine if there were other factors to consider.

    This circuit plan has unique requirements that unfortunately I don't have permission to share many details with.

    The load is not a motor - just 5v (or lower) electronics. And we can predict initial startup "spikes" and of course typical current loads. We don't expect much (if any) variation in the load.

    AC transformer solutions we are well aware of. However they are not an option due to reasons I can't go into.

    So ultimately - based on these responses - they aren't any potential hazards that we haven't considered.
    Which is good to hear!

    Thanks again to everyone.

  • SaltiDawg
    9 years ago

    "For example, 1000 ft of 14 AWG wire at 5 amps creates a voltage drop of 12.87v with a voltage source of 24v.
    Which means about 54% energy loss."

    Actual the loss in power to the load will vary as the square of the voltage, (For a given load, halving the voltage at the load due to line loss will reduce the power by 75%.)

    This post was edited by saltidawg on Sun, Aug 17, 14 at 20:20

  • suburbanmd
    9 years ago

    You want something like Power-Over-Ethernet (PoE), but with different parameters. PoE sends 48V over a 100 meter run of very thin wire (24 or 26AWG) and converts it to 5V or 12V at the far end.