MWBC & Voltage Drop

Status
Not open for further replies.

Mike01

Senior Member
Location
MidWest
Have a friendly discussion in the office on this one but trying to work out the math part of it. With a MWBC [typically for lighting] I see a lot of 277V fixtures connected to a 480Y/277V MWBC with the thought that the voltage drop is lower compared to dedicated neutral circuits. If the loads are connected as 277V L-N how is the VD any different? If you calculate it at 480V-3Ph you can definitely extend the circuit or add fixtures to remain within a recommended percentage, just trying to figure how when connecting 277V L-N fixtures on a 480Y/277V ckt. the VD changes? :?
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Have a friendly discussion in the office on this one but trying to work out the math part of it. With a MWBC [typically for lighting] I see a lot of 277V fixtures connected to a 480Y/277V MWBC with the thought that the voltage drop is lower compared to dedicated neutral circuits. If the loads are connected as 277V L-N how is the VD any different? If you calculate it at 480V-3Ph you can definitely extend the circuit or add fixtures to remain within a recommended percentage, just trying to figure how when connecting 277V L-N fixtures on a 480Y/277V ckt. the VD changes? :?

If you have balanced and linear wye connected loads, then the net current in the neutral wire will be zero. No current means no voltage drop. The result is that you calculate the voltage drop based on the current in the phase wire only.
It is the same when you put balanced 120 volt loads on a 120/240 3 wire MWBC. The voltage drop in each phase wire is what it would be for a single 240V load of the same power, but a 120V load sees only the voltage drop in the phase wire on its own side.

If instead of an MWBC you run separate neutrals back to the panel, the cancellation of current will only happen at the panel and each 120V load will see two voltage drops, one in the phase wire and one in the neutral.
 

mbrooke

Batteries Included
Location
United States
Occupation
Technician
If you look at the way it works you will actually realize current prefers the other phases more than the ground or neutral. :lol:

If only one phase has a load to neutral on it that load will experience the same voltage drop as a dedicated neutral. However, if the the other 2 loads on the MWBC begin coming on line the voltage will begin to rise at the first load. When the other 2 are at full load the first load will have a voltage drop nearlly the same as if it were connected phase to phase ie, 480 volt drops instead of 277 volt drops. The current in the neutral would also be zero if all 3 are balanced. If a bolted fault was to say occur, and assuming the source can provide as much power as needed (infinite buss) the voltage will actually rise well above 277 volts on the non faulted loads, but zero on the faulted ones. The longer the neutral length (higher impedance) the more the voltage rise will be on the non faulted load. Same goes should the neutral start to make poor contact at its termination. When this happens lighter loads experience an over voltage heavier ones an under voltage.


As a side note I used to live in an house where the MWBC run from the panel to the second floor was over 200 ft. Black went to the lights red to the oulets. The 30 year old window AC had monster motor inrush. Starting it up would cause all table lamps to dim while the ceiling lights on the black phase would actually briefly brighten. I think it came out to be something like a 16 volt dip for the AC starting while the other leg went up by about 5 or 8 volts.

I tested it with a volt meter an space heater. Without the heater on the black phase with only a few lights was about 116. Put the 12 amp space heater on voltage rose on the black phase to about 118. 116 went to about 111 on the red phase. Approx values but you could clearly see the rise and fall.
 
Status
Not open for further replies.
Top