Voltage drop

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I have a question on voltage drop.

Here are the perameters.

A sub panel 240/120 volt located 2000' away from main 240/120 volt panel.
Sub panel only feeds a boat lift at 120 volt and a local 120 volt conv receptacle with no load.
Boat lift located 200' away from sub panel
Sub panel fed with (3) #6 copper, phase, phase and nuetral.

I believe the boat lift horsepower is less then 1/2 HP.

What are the calcs to determine?
 
A quick calc using 9.8A for boat lift motor:

9.8A ? 0.44ohms/1000' ? 2000' = 8.6V

...to subpanel.

ETA: That is one-way. Out and back is double that.
 
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The tables tell me that a 1/2 HP motor operating at 120 volts will draw 9.8 amps. Ignoring the convenience receptacle (which you say has no load, but someday it will), and also ignoring the extra 200 feet to the boat lift (with a wire size that you did not name), my spreadsheet calculator tells me the voltage drop will be just under 20 volts. That is 16% of a 120 volt source. So you already have a problem. If you have #10's going the 200 feet to the boat lift, that will add another 5 volt drop, making the total around 20%. I think you need to rethink the design.

Mike Holt's main web site has a VD calculator that you can download, to experiment with different sizes of conductors and load values.

Welcome to the forum.
 
Smart $, I think your version needs to use 4000 feet for the run. Mine already took the two-way length into account. That would put your answer closer to mine.
 
THE MOTOR PROBABLY WOULD NOT START, JUST HUMMMMMM. KICK THE VOLTAGE UP TO 480 VOLT FOR THE 2000 FT RUN
AND TRANSFORM THE VOLTAGE BACK TO 120/240 VOLT. SEE IF THE MOTOR CAN RUN ON 240 VOLTS. YOU COULD USE # 10 CU CONDUCTOR.
 
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Buy a 208V input - 120/240V output encapsulated transformer and put it at the incoming side of the sub-panel. Use taps to adjust the voltage close to what you need. wire output to sub-panel. A lot cheaper than changing cable and what not.
 
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