3 phase voltage drop

[Michael D]

Can anyone refresh me on how to calculate voltage drop with this scenerio: 480 volts 3 phase, 400 amp motor load, 800 feet, 600 mcm ran open air in an underground rock quarry?

 

[lauraj]

Voltage drop = (1.73 x K x I x L) / cma

(1.73 x 12 x 400 x 800)/600,000 = 11.07 volts

Should be good to go!

 

[Michael D]

Thank you very much. You settled a good arguement! (in MY favor!)

 

[barclayd]

That gives you the 'steady state' voltage drop.
There will be considerably more during motor starting.

 

[Smart $]

All things being equal--load, distance of run, size and type of conductors, etc.--what is the difference in voltage drop between the two wiring methods depicted below as realized by each load?

 

[LarryFine]

All things being equal--load, distance of run, size and type of conductors, etc.--what is the difference in voltage drop between the two wiring methods depicted below as realized by each load?

I'd have to say 2:1, with the shared-neutral scheme being the 1 (with all phases balanced, of course).

The same applies to 240/120v single-phase circuits.

 

[don_resqcapt19]

The voltage drop in B is twice that in A.
Don

 

[Smart $]

Larry and Don,

Thanks for the confirmation. I had to check my sanity

However, everyone I've ever seen do a VD calc has used one-way distance for L. Why is that?