- Location
- Massachusetts
This makes no sense to me, can you explain further?
Neutral or hot both conductors of the circuit have voltage drop.
Maybe I am misunderstanding you.
250MCM for 21.7 amps? Voltage drop calc must be wrong?
- Thread starter Bri22
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This makes no sense to me, can you explain further?
Neutral or hot both conductors of the circuit have voltage drop.
Maybe I am misunderstanding you.
I totally agree the hot and neutral both have voltage drop. That isn't the issue. It's the way voltage drop is defined. When you do a three phase voltage drop calculation, you only include the distance from the source to the load. The same is true for phase to neutral voltage drop calculations. You only use the distance from the source to the load, not the round trip distance. Look up voltage drop in the IEEE Gray Book.
- Location
- Massachusetts
Sorry still not making sense to me.
Regardless of any definition the goal is to limit voltage drop at the load and I don't understand how that can be done correctly without including both directions. :?
- Location
- Illinois
- Occupation
- retired electrician
The formulas for voltage drop that only use the one way distance have a factor of two for single phase loads and a factor of 1.73 for three phase loads somewhere in the formula.
When you do a three phase voltage drop calculation you only use the distance from the source to the load. When you do a single phase, phase to phase calculation you only use the distance from the source to the load. Which agrees with the definition of voltage drop. When you do a single phase, phase to neutral calculation you also use the distance from the source to the load as per the definition. If you look at how Eaton does the calculation, they do a three phase calculation and then they multiply that result by 0.577 which is 1/root 3. Effectively dividing out the root 3 from the three phase answer. They don't then multiply that result by 2 to include the neutral drop.
What's the voltage to ground at the neutral terminal of the load? Close to zero. If you did a voltage drop calculation of the neutral using the load's neutral to ground as your source voltage you'd get a neutral voltage drop that's negligible.
What's the voltage to ground at the neutral terminal of the load? Close to zero. If you did a voltage drop calculation of the neutral using the load's neutral to ground as your source voltage you'd get a neutral voltage drop that's negligible.
- Location
- Illinois
- Occupation
- retired electrician
On a two wire circuit the voltage at the load is the supply voltage minus the voltage drop on each of the conductors. The voltage drop on the neutral raises the voltage on at the load neutral connection and that reduces the net voltage that the load sees.
- Location
- Massachusetts
No.
The voltage from ground to neutral will be as much as the voltage drop on the neutral.
If it's a long circuit you may see quite a few volts of drop even in a well designed circuit.
ptonsparky
Tom
- Occupation
- EC - retired
The way I see it, if VD is zero, you either have a problem or spent way to much money on wire for the EG.
Kirchhoff's voltage law
the sum of V around a closed loop must = 0
Vs 120 L-N +
line out Vdo -
load Vdl -
line return Vdr -
summed = 0
or
Vdl = Vs - Vdo - Vdr
assume Vdo = Vdr = Vdw (wire drop) since i, length and wire R are the same
or
Vdo + Vdr = 2 x Vdw and Vdw = l x i x R (per unit length)
Vdl = Vs - 2 x Vd
total drop or difference of source to load due to wire = 2(l x i x R)
length must be doubled
the sum of V around a closed loop must = 0
Vs 120 L-N +
line out Vdo -
load Vdl -
line return Vdr -
summed = 0
or
Vdl = Vs - Vdo - Vdr
assume Vdo = Vdr = Vdw (wire drop) since i, length and wire R are the same
or
Vdo + Vdr = 2 x Vdw and Vdw = l x i x R (per unit length)
Vdl = Vs - 2 x Vd
total drop or difference of source to load due to wire = 2(l x i x R)
length must be doubled
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Per the IEEE Gray Book, the formula for voltage drop is V = IR cos theta + IX sin theta where V is the voltage drop in the circuit, line to neutral. Followed by "The voltage drop V obtained from this formula is the voltage drop in one conductor, one way, commonly called the line-to-neutral voltage drop. The line-to-line voltage drop is computed by multiplying the line-to-neutral voltage drop by the following constants: Voltage System - Single Phase, multiply by 2; Voltage System - Three Phase, multiply by 1.732."
There is no factor of 2 in the line to neutral calculation. It's one conductor, one way, Period! The single phase voltage drop discussed with the multiplication factor of 2 is for line-to-line single phase circuits.
There is no factor of 2 in the line to neutral calculation. It's one conductor, one way, Period! The single phase voltage drop discussed with the multiplication factor of 2 is for line-to-line single phase circuits.
- Location
- Massachusetts
No, there is something wrong with that explanation even if you put 'period' on the end of it.
line to neut IS single phase
mult x 2
equal current flow in both line and neut
Both MUST have a drop
since both have i and R
note they say the line-neut drop is ONE conductor ONE way
must account for return drop
Or no current
thought problem
gnd/neut are common/bonded ref negative - at 0 V
source is + line and neut - 120 V
load is opposite sign of source 10 ohm purely R
line and g/n wire are 1 ohm each
total loop 10 + 1 + 1 = 12 ohm
so i = 120/12 = 10 A
starting at g/n = 0 V
Work towards load neg along wire: assume Vdrop is 0 from g/n to load - since we don't mult x 2 and therefor not a factor in the total loop Vdrop
make KVL work
- Location
- Illinois
- Occupation
- retired electrician
That is saying that the voltage drop on the neutral does not change the voltage at the load and we all not that is not physically possible.
There is no difference on how you calculate the voltage drop on a two wire circuit...you have to account for the drop on both conductors and this does not change just because one of the conductors is a grounded conductor.
kwired
Electron manager
- Location
- NE Nebraska
I think he is right but not making his point easy to see. I assume he is trying to get to the fact that with a multiwire circuit voltage drop isn't as simple to calculate as it is for a two wire circuit or a balanced three phase three wire circuit.
If that is not what he is after then I am throwing that into the mix.
I don't think he's looking at it that way
120 V source
2 legs at 1 ohm each for a total of 2 ohm
load at 10 ohm
loop 12 ohm
i = 120/12 = 10 A
he's saying the V across the load is 120 - 1 x (1 x 10) = 110
not 120 - 2 x (1 x 10) = 100
you double the distance to account for the drop of each leg
KVL
- Location
- Placerville, CA, USA
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- Retired PV System Designer
<Begin lecture. There may be a quiz.>
1. One of the problems this thread is circling around is that most available calculators allow you to input a selection such as "single phase", "three phase" and "DC" and then they use that input to make an assumption as to what calculation will give you the answer that you really need. And nowhere do they tell you explicitly what they are doing, but they always ask for the physical distance of the circuit rather than one or two way wire length.
If the calculator in question is a spreadsheet you can look at its formulas and logic to see what it does. With an online calculator that is usually not an option.
If you really need to know what it is doing you need to run a test calculation for each voltage system specification and compare it to a hand calculation where you know what you are inputting for the wire length.
2. Another problem is what to do in general with a line to neutral load.
If you have a two wire connection from load to supply there will be voltage drop on both wires. Period. And the drop in voltage seen by the load and the power lost to wire resistance will be based on the loop wire length.
If you have a line to neutral connection where the neutral is part of a "full boat" MWBC, whether single or three phase, AND the total of all of your single phase loads is roughly balanced across the MWBC, then there will be very little voltage drop on the neutral since there will be very little current. The result is that the voltage drop will correspond closely to the one way wire length.
3. When dealing with percentage voltage drop rather than absolute voltage drop you have to be careful to use the denominator voltage which corresponds to the calculated numeric voltage drop (the numerator).
For example, if you have a single isolated 120V load on a 120/240 single phase system and the resistance is such that a 10A load drops 1.2V on one wire, then you will have a 2.4V total drop out of 120V for a percent drop of 2%.
If you have a 240V 10A load the voltage drop on each wire will be 1.2V, for a total voltage drop of 2.4V and a percentage drop of 1% with exactly the same wire gauge. But you will be delivering twice the power to the load.
If you have two balanced 120V loads in place of the 240V load you will have a voltage drop of 1.2V for each, and a %VD of 1.2/120 or 1%.
The figures are consistent.
4. When you introduce three phase, you need to be even more careful about whether you are talking about load current on each line-to-line phase or talking about the current flowing in the conductor.
If you put a single resistive load phase to phase on a three phase supply you have the unfortunate complication that the two line currents are equal to the load current and they are out of phase in opposite directions and the total voltage drop, the vector sum, will have a magnitude which is exactly twice the magnitude of the "VD" on each phase line individually. (Not to be confused with the VD on the line conductor when you have a balanced triad of line to line loads.) A properly configured calculator should NOT introduce a factor of 1.73 into this calculation but will instead give you the same answer as the two line single phase result.
But it is likely that the calculator will assume that you are talking about a balanced three phase load, in which case the factor of 1/1.73 will come into play and you will have a reduced power loss, just as you would have a lower loss for a balanced line to neutral load in a single phase 120/240 situation.
5. If you consider line to neutral loads in a balanced three phase wye configuration, the voltage drops will all be 10A time a single wire length, just as for balanced line to neutral loads on a 120/240 MWBC. But the voltage you divide by to get %VD must be the line to neutral voltage not the line to line voltage. (Which voltage did you enter into the calculator so that it could give you %VD???)
<End of lecture, resume discussion>
1. One of the problems this thread is circling around is that most available calculators allow you to input a selection such as "single phase", "three phase" and "DC" and then they use that input to make an assumption as to what calculation will give you the answer that you really need. And nowhere do they tell you explicitly what they are doing, but they always ask for the physical distance of the circuit rather than one or two way wire length.
If the calculator in question is a spreadsheet you can look at its formulas and logic to see what it does. With an online calculator that is usually not an option.
If you really need to know what it is doing you need to run a test calculation for each voltage system specification and compare it to a hand calculation where you know what you are inputting for the wire length.
2. Another problem is what to do in general with a line to neutral load.
If you have a two wire connection from load to supply there will be voltage drop on both wires. Period. And the drop in voltage seen by the load and the power lost to wire resistance will be based on the loop wire length.
If you have a line to neutral connection where the neutral is part of a "full boat" MWBC, whether single or three phase, AND the total of all of your single phase loads is roughly balanced across the MWBC, then there will be very little voltage drop on the neutral since there will be very little current. The result is that the voltage drop will correspond closely to the one way wire length.
3. When dealing with percentage voltage drop rather than absolute voltage drop you have to be careful to use the denominator voltage which corresponds to the calculated numeric voltage drop (the numerator).
For example, if you have a single isolated 120V load on a 120/240 single phase system and the resistance is such that a 10A load drops 1.2V on one wire, then you will have a 2.4V total drop out of 120V for a percent drop of 2%.
If you have a 240V 10A load the voltage drop on each wire will be 1.2V, for a total voltage drop of 2.4V and a percentage drop of 1% with exactly the same wire gauge. But you will be delivering twice the power to the load.
If you have two balanced 120V loads in place of the 240V load you will have a voltage drop of 1.2V for each, and a %VD of 1.2/120 or 1%.
The figures are consistent.
4. When you introduce three phase, you need to be even more careful about whether you are talking about load current on each line-to-line phase or talking about the current flowing in the conductor.
If you put a single resistive load phase to phase on a three phase supply you have the unfortunate complication that the two line currents are equal to the load current and they are out of phase in opposite directions and the total voltage drop, the vector sum, will have a magnitude which is exactly twice the magnitude of the "VD" on each phase line individually. (Not to be confused with the VD on the line conductor when you have a balanced triad of line to line loads.) A properly configured calculator should NOT introduce a factor of 1.73 into this calculation but will instead give you the same answer as the two line single phase result.
But it is likely that the calculator will assume that you are talking about a balanced three phase load, in which case the factor of 1/1.73 will come into play and you will have a reduced power loss, just as you would have a lower loss for a balanced line to neutral load in a single phase 120/240 situation.
5. If you consider line to neutral loads in a balanced three phase wye configuration, the voltage drops will all be 10A time a single wire length, just as for balanced line to neutral loads on a 120/240 MWBC. But the voltage you divide by to get %VD must be the line to neutral voltage not the line to line voltage. (Which voltage did you enter into the calculator so that it could give you %VD???)
<End of lecture, resume discussion>
Although on a balanced 3 ph system the 3 line i's mathematically cancel at the neutral and no i flows to the X0 (zero seq reactance) that does not mean a loads i evaporates
the sum is 0 but each current has value, a composite of other ph loads
if balanced ia + ib = -ic and so on
it returns on a different phase so a V drop results
total roundtrip length must be consider
with ph considered sqrt3 vs 2
I still use 2 lol
the sum is 0 but each current has value, a composite of other ph loads
if balanced ia + ib = -ic and so on
it returns on a different phase so a V drop results
total roundtrip length must be consider
with ph considered sqrt3 vs 2
I still use 2 lol
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