VD CaLculation

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augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Not sure how you get "phaseS" on 277, but using Mike's chart, I come up with (3) 600s.
Of course, that depends on how much voltage drop you can tolerate (the 600s were based on 3%)
 

kwired

Electron manager
Location
NE Nebraska
If you have all three phases present, then you need to figure voltage drop based on line to line loading of the balanced portion of the load. The neutral will only be carrying unbalanced current and any voltage drop on it will depend on just how much current the neutral is carrying. If you want to know exactly what the drop is for each conductor you will have a more complicated calculation as you will basically need to figure each conductor separately based on the load of each conductor.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
In my opinion VD does not depends so much on conductor cross section area-for 500 mcm or 750 mcm-the VD has to be the same: the resistance drops from 0.0268 to 0.01891 but in the same time the reactance rises from 0.03019 to 0.03354 [ohm/600 ft].
So for 500/3=167 A[3 parallel cables/phase] the drop will be 2.4%for 500 mcm and 2.3% for 750 mcm[ if 277 V is the supply voltage phase-to-neutral-480/277 V].
For 500/2=250 A and 2*750 mcm per phase it is only 3.42% drop.[0.8 p.f, 75dgr.C conductor temperature, flat formation].
The voltage drop depends- slightly- on p.f. , on the conductor temperature and on distance between cables in conduit[cradle or bunch layout].
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
Correction:
Instead of "but in the same time the reactance rises from 0.03019 to 0.03354 [ohm/600 ft.]."
has to be "and in the same time the reactance drops from 0.03224 to 0.03125 [ohm/600 ft.]."
and instead of "So for 500/3=167 A the drop will be 2.4%for 500 mcm and 2.3% for 750 mcm[ if 277 V is the supply voltage phase-to-neutral-480/277 V]."
has to be "So for 500/3=167 A the drop will be 2.5%for 500 mcm and 2.2% for 750 mcm[ if 277 V is the supply voltage phase-to-neutral-480/277 V]. "
Instead of "the VD has to be the same" has to be "the VD has to be close"
The differences resulted from wrong average overall cable diameter appreciation.
:ashamed1:
 
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