Our company is installing outdoor lighting from a large manufacturer. They have refered to the lighting as "277volt three phase". We clarified "you mean 480volt three phase?". And they insist on using the terminology of 277volt three phase. The poles came in and have 277volt ballasts. Two ballasts per phase. The question is, "is it appropriate to calculate voltage drop at 277 or 480 volts"? All three phases are terminated on a circuit breaker in a control cabinet in the base of the pole and each leg then is wired to two ballasts each with the neutrals coming back to the common neutral lug.
Voltage Drop
- Thread starter bhattey
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bphgravity
Senior Member
- Location
- Florida
Re: Voltage Drop
What portion of the wiring are you concerned with voltage drop? I assume it to be the conductors that serve the pole and not the individual feeds to each ballast. In that case, the voltage drop would be based on 480V.
What portion of the wiring are you concerned with voltage drop? I assume it to be the conductors that serve the pole and not the individual feeds to each ballast. In that case, the voltage drop would be based on 480V.
Re: Voltage Drop
The concern is the wire from the distribution panel to the poles. The leads from the factory wired polebase cabinet (which is where the ballasts reside) to the luminaires are 60' up the pole and are not a factor provided the voltage drop is not excessive to the base.
The concern is the wire from the distribution panel to the poles. The leads from the factory wired polebase cabinet (which is where the ballasts reside) to the luminaires are 60' up the pole and are not a factor provided the voltage drop is not excessive to the base.
bphgravity
Senior Member
- Location
- Florida
Re: Voltage Drop
I still feel that if you are running a 480V feeder to the light poles and then spliting off from there to individual 277V loads, the voltage drop would be based on the feeder voltage.
I still feel that if you are running a 480V feeder to the light poles and then spliting off from there to individual 277V loads, the voltage drop would be based on the feeder voltage.
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Re: Voltage Drop
I use the voltage drop method on page 41 of Ferraz Shawmut Book of Electrical Information. Their procedure is to calculate the 480V (L-L) 3 phase voltage drop and then convert to the 277V (L-N) result.(in this method multiply the 480V result by .577)
I use the voltage drop method on page 41 of Ferraz Shawmut Book of Electrical Information. Their procedure is to calculate the 480V (L-L) 3 phase voltage drop and then convert to the 277V (L-N) result.(in this method multiply the 480V result by .577)
lady sparks lover
Senior Member
Re: Voltage Drop
I would say, 277 because that would prevent someone from changing their mind to 277/1pole you being suck. Be safe!!
Lady
I would say, 277 because that would prevent someone from changing their mind to 277/1pole you being suck. Be safe!!
Lady
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: Voltage Drop
But first, please realize that the ?voltage drop? has nothing to do with the ?voltage.? You take the current in the wires, and multiply by the resistance of the wires, and you get the voltage that is dropped along the wires.
What you are really looking for here is the percentage of voltage drop, as compared to the supply voltage. If you were to drop 15 volts along the wires, and if you compare that to a 277 volt supply, then you have lost 5.4% of your voltage. That would not sound like a good thing. But if you were to drop the same 15 volts along the wires, and if you compare that to a 480 volt supply, then you have only lost 3.1% of your voltage. That might seem acceptable.
But what are the wires of interest here? Where are the volts being dropped? We are looking at the three phase supply wires. Provided that the system is reasonably balanced, the neutral current will be low. Current will leave the source along one of the three phase conductors, and return to the source along another one of them. That path has a supply voltage of 480. So if you are comparing the calculated voltage drop to the supply voltage, you would use 480.
Bryan is right.
But first, please realize that the ?voltage drop? has nothing to do with the ?voltage.? You take the current in the wires, and multiply by the resistance of the wires, and you get the voltage that is dropped along the wires.
What you are really looking for here is the percentage of voltage drop, as compared to the supply voltage. If you were to drop 15 volts along the wires, and if you compare that to a 277 volt supply, then you have lost 5.4% of your voltage. That would not sound like a good thing. But if you were to drop the same 15 volts along the wires, and if you compare that to a 480 volt supply, then you have only lost 3.1% of your voltage. That might seem acceptable.
But what are the wires of interest here? Where are the volts being dropped? We are looking at the three phase supply wires. Provided that the system is reasonably balanced, the neutral current will be low. Current will leave the source along one of the three phase conductors, and return to the source along another one of them. That path has a supply voltage of 480. So if you are comparing the calculated voltage drop to the supply voltage, you would use 480.
Re: Voltage Drop
Why would the current path be from one supply conductor and return on another supply conductor? The loads are all 277v. The current path would be from one line source, through the load (ballast) and return via the neutral. All voltage drop would occur accross each phase conductor to neutral ( 277volt ).
Why would the current path be from one supply conductor and return on another supply conductor? The loads are all 277v. The current path would be from one line source, through the load (ballast) and return via the neutral. All voltage drop would occur accross each phase conductor to neutral ( 277volt ).
- Location
- Massachusetts
Re: Voltage Drop
This is the advantage of running a 3 phase 4 wire circuit and balancing the load between all 3 conductors.
If you could balance the load perfectly among phase A, B and C you could remove the neutral.
In this case as long as you keep the load balanced there will be almost no current on the neutral.
This is the advantage of running a 3 phase 4 wire circuit and balancing the load between all 3 conductors.
If you could balance the load perfectly among phase A, B and C you could remove the neutral.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: Voltage Drop
Let me try and explain it this way: The neutral current returning from a Phase A load joins with the neutral current returning from a Phase B load, and these two currents join with the neutral current returning from a Phase C load. That is, the feeder?s neutral conductor will carry the combined current from the three loads. If the loads are balanced, then the three currents will add up to zero. Even if the loads are not exactly balanced, the current in the feeder?s neutral conductor will be small, in comparison to the current in the feeder?s phase conductors.
That is true of each individual load, and correctly describes the current path from the bottom of a given pole to the light and back to the base of the pole. I was speaking about the feeder that serves all loads.
Let me try and explain it this way: The neutral current returning from a Phase A load joins with the neutral current returning from a Phase B load, and these two currents join with the neutral current returning from a Phase C load. That is, the feeder?s neutral conductor will carry the combined current from the three loads. If the loads are balanced, then the three currents will add up to zero. Even if the loads are not exactly balanced, the current in the feeder?s neutral conductor will be small, in comparison to the current in the feeder?s phase conductors.
Re: Voltage Drop
I do understand the balanced neutral resulting in virtually no load on the neutral itself. But do I understand you to say the appropriate voltage to base the vd calc for the feeder conductors to the pole base to be 480v? And this because the voltage is dropped across the phase conductor and not the neutral? Thank you for your time.
I do understand the balanced neutral resulting in virtually no load on the neutral itself. But do I understand you to say the appropriate voltage to base the vd calc for the feeder conductors to the pole base to be 480v? And this because the voltage is dropped across the phase conductor and not the neutral? Thank you for your time.
Re: Voltage Drop
go to this VD cal. site. http://www.electrician.com/vd_calculator.html
if your going to run ungrounded conductor/s (1 or all 3 phases) with an individual grounded conductor for each ungrounded conductor you use 277/1phase VD cal.
if your going to run all three ungrounded conductors and use one grounded conductor(neutral) you will
use the 277/480 3-phase 4-wire to do your VD cal.
good luck. remember to use pigtails if sharing LOL.
go to this VD cal. site. http://www.electrician.com/vd_calculator.html
if your going to run ungrounded conductor/s (1 or all 3 phases) with an individual grounded conductor for each ungrounded conductor you use 277/1phase VD cal.
if your going to run all three ungrounded conductors and use one grounded conductor(neutral) you will
use the 277/480 3-phase 4-wire to do your VD cal.
good luck. remember to use pigtails if sharing LOL.
Re: Voltage Drop
It’s the same either way…
“What you are really looking for here is the percentage of voltage drop, as compared to the supply voltage. If you were to drop 15 volts along the wires, and if you compare that to a 277 volt supply, then you have lost 5.4% of your voltage. That would not sound like a good thing. But if you were to drop the same 15 volts along the wires, and if you compare that to a 480 volt supply, then you have only lost 3.1% of your voltage. That might seem acceptable.”
277V L-N
15V/277V = 5.4%VD
Note: assumed balanced load.
480V L-L
26V/480V = 5.4%VD
Note: 30V was not used because the two 15V “drops” are 120 deg. out of phase. (15*1.732=26)
What if there were fuses protecting the feeder and we blew two of them?
On the surviving leg (L-N) 15+15=30/277=10.8%VD
What happens if we only blow one fuse?
[ June 23, 2004, 02:17 PM: Message edited by: engy ]
It’s the same either way…
“What you are really looking for here is the percentage of voltage drop, as compared to the supply voltage. If you were to drop 15 volts along the wires, and if you compare that to a 277 volt supply, then you have lost 5.4% of your voltage. That would not sound like a good thing. But if you were to drop the same 15 volts along the wires, and if you compare that to a 480 volt supply, then you have only lost 3.1% of your voltage. That might seem acceptable.”
277V L-N
15V/277V = 5.4%VD
Note: assumed balanced load.
480V L-L
26V/480V = 5.4%VD
Note: 30V was not used because the two 15V “drops” are 120 deg. out of phase. (15*1.732=26)
What if there were fuses protecting the feeder and we blew two of them?
On the surviving leg (L-N) 15+15=30/277=10.8%VD
What happens if we only blow one fuse?
[ June 23, 2004, 02:17 PM: Message edited by: engy ]
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