slalom skier
Member
- Location
- Royal Palm Beach FL. USA
To figure voltage drop on feeders coming off of a multi meter to 120/240 volt panels which voltage would I use along with load calculations?
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Voltage drop calc.
- Thread starter slalom skier
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ramsy
NoFixNoPay Electric
- Location
- LA basin, CA
- Occupation
- Service Electrician 2020 NEC
Same voltage divider as using for load calc Watts.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Seems to me this is a bit tricky with mixed 120V and 240V loads. Say your 120V loads are evenly split between legs, so you have X amps of 120V loads on each leg, and Y amps of 240V loads.
For the 240V loads, the worst case is simple enough: check the voltage drop for X+Y amps @ 240V. That's the case that everything is operating at full load.
For the 120V loads, the worst case is actually all the 120V loads on one leg at full load, the loads on the other leg off, and all the 240V loads at full load. Then the voltage drop would be based on 120V and would be the sum of the one-way voltage drop for X+Y amps on the ungrounded leg, plus the one-way voltage drop for X amps on the grounded conductor.
The 120V case controls if we assume the grounded conductor is not larger than the ungrounded conductors. If they are equal in size, the 120V case voltage drop would be the same as (2X + Y)/2 amps (average current on the two conductors) @ 120V. Which is the same as (2X + Y) amps @ 240V, and 2X+Y > X+Y (the 240V case). And if the grounded conductor is smaller than the ungrounded conductors, then the voltage drop for the 120V case gets larger, while the 240V voltage drop is unchanged, so the 120V case still controls.
Cheers, Wayne
For the 240V loads, the worst case is simple enough: check the voltage drop for X+Y amps @ 240V. That's the case that everything is operating at full load.
For the 120V loads, the worst case is actually all the 120V loads on one leg at full load, the loads on the other leg off, and all the 240V loads at full load. Then the voltage drop would be based on 120V and would be the sum of the one-way voltage drop for X+Y amps on the ungrounded leg, plus the one-way voltage drop for X amps on the grounded conductor.
The 120V case controls if we assume the grounded conductor is not larger than the ungrounded conductors. If they are equal in size, the 120V case voltage drop would be the same as (2X + Y)/2 amps (average current on the two conductors) @ 120V. Which is the same as (2X + Y) amps @ 240V, and 2X+Y > X+Y (the 240V case). And if the grounded conductor is smaller than the ungrounded conductors, then the voltage drop for the 120V case gets larger, while the 240V voltage drop is unchanged, so the 120V case still controls.
Cheers, Wayne
I have always converted everything to VA or watts and assume i'll do a balanced panel or close enough.Seems to me this is a bit tricky with mixed 120V and 240V loads.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Yes, but I believe the worst case voltage drop (percentage wise) occurs when all the 240V loads are at full load, and half of the 120V loads are at full load, just the ones on a particular leg. So you need to track them separately and calculate for that case, with respect to 120V base voltage.I have always converted everything to VA or watts and assume i'll do a balanced panel or close enough.
Cheers, Wayne
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Simple example: 120/240V supply, 24 kVA total load, of which 12 kVA is 240V loads (50A) and 12 kVA is 120V loads (50A on each leg). Ungrounded supply conductors have a resistance of 25 milliohms one way, grounded conductor a resistance of 50 milliohms.
All loads on. 100A * 25 milliohms = 2.5V VD on each of L1 and L2. The -120V / 0V / 120V supply becomes a -117.5V / 0V / 117.5V supply. 2.1% VD for both 120V and 240V loads.
240V loads on, along with just the 120V loads on L2. Current L1 / N / L2 = 50A / 50A / 100A. Voltages become -118.8V / 2.5 V / 117.5V. 1.5% VD for the 240V loads, but 4.2% for the 120V loads. [Would be 3.1% VD for the 120V loads if the feeder didn't have a reduced size grounded conductor.]
Cheers, Wayne
All loads on. 100A * 25 milliohms = 2.5V VD on each of L1 and L2. The -120V / 0V / 120V supply becomes a -117.5V / 0V / 117.5V supply. 2.1% VD for both 120V and 240V loads.
240V loads on, along with just the 120V loads on L2. Current L1 / N / L2 = 50A / 50A / 100A. Voltages become -118.8V / 2.5 V / 117.5V. 1.5% VD for the 240V loads, but 4.2% for the 120V loads. [Would be 3.1% VD for the 120V loads if the feeder didn't have a reduced size grounded conductor.]
Cheers, Wayne
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