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Thread: Voltage Drop - Difference between Three Phase and Single Phase

  1. #1
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    Voltage Drop - Difference between Three Phase and Single Phase

    I've seen this issue raised lots of times in lots of ways. However, none of the instances I can find address the significance (or lack thereof) of the issue. Let me attempt to frame this issue in a specific way:

    Taking into account the following definitions:

    • Vd = Voltage Drop
    • K = Resistance in ohms of one circular mil foot of conductor, or 12.9 for Copper at 75 degrees C, and 21.2 for Aluminum at 75 degrees C
    • L = One way length of circuit (ft)
    • I = Current (Amps)
    • Cm = Cross sectional area of conductor (cmils)

    The equation to find voltage drop in a single phase circuit is:

    Vd = (2K x L x I)/Cm

    The equation to find voltage drop in a three phase circuit is:

    Vd = (1.73K x L x I)/Cm

    As seen in the two equations above, the only difference between the two is the constant, which changes from 1.73 to 2. In a perfectly balanced three phase delta system, the above three phase equation applies. For an unbalanced three phase delta system, this is no longer true. Rather than discuss phasor arithmetic or symmetrical systems analysis, I want to pose this question:

    Would the resulting voltage drop calculation for an unbalanced three phase delta system fall somewhere in between the calculation for a single phase system and a balanced three phase system? Another way of putting it is, if you could boil down the unbalanced circuit to a constant, as the two formulae above contain, would the constant always reside between 1.73 and 2?

    Specifically, this is for lighting, not motors, and assuming normal conditions, not short circuits or open circuits. Thank you for your time.

  2. #2
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    You are asking for too much. If the system is unbalanced, then the currents in the three phases (and the neutral, if there is one) will not be equal to each other. Therefore, the voltage drop along each wire will be different. There will be no single constant that works for all three phases. The only way to know what is happening is to use symmetrical components.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  3. #3
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    The Specifics of the Problem

    Thank you for your quick response. The lighting circuits I'm trying to evaluate are missing information that would be impractical to acquire (no budget for a field trip). I know the wire size, quantity of luminaires, wattage of luminaires, system voltage, and distance between poles. What I don't know are the actual distribution between the phases. The worst case scenario in this case then, is that the circuit is unbalanced.

    I'm trying to work around the missing information problem by establishing general parameters to evaluate the system. What I'm hoping for is to formulate a useful approximation. Since symmetrical components requires full knowledge of the circuit, I would have to construct several scenarios to see under what condition the voltage drop is the worst. The only way I know how to calculate symmetrical components is by hand (on a side note, if you know of any free or cheap application that can perform symmetrical component calculations, I'd love to hear of it; I can't seem to make it work in SPICE, and I've been unsuccessful in building anything in Excel).

    If, per chance, the voltage drop for an unbalanced three phase delta circuit is always contained within the range of a single phase circuit and a balanced three phase circuit, then it becomes easy to evaluate--the most conservative estimate would simply be to treat such a circuit as a single phase circuit. However, if the voltage drop in an unbalanced three phase circuit can stray outside of this range (that is, the voltage drop can be worse than a single phase circuit), then what is the worst case? It can't be infinite, I don't think, which means there is a finite quantity of voltage that a cable can lose per foot, no matter how badly the circuit is balanced. I'm looking for this value.

    If I can't find an answer, I'll just have to figure out how to get out the site and open up those poles and see. I'll make do. But from an intellectual standpoint, the nature of this problem intuitively feels like it has a solution, and that's what I'm hoping to find out.

  4. #4
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    Quote Originally Posted by That Man View Post
    • Vd = Voltage Drop
    • K = Resistance in ohms of one circular mil foot of conductor, or 12.9 for Copper at 75 degrees C, and 21.2 for Aluminum at 75 degrees C
    • L = One way length of circuit (ft)
    • I = Current (Amps)
    • Cm = Cross sectional area of conductor (cmils)

    The equation to find voltage drop in a single phase circuit is:

    Vd = (2K x L x I)/Cm

    The equation to find voltage drop in a three phase circuit is:

    Vd = (1.73K x L x I)/Cm

    As seen in the two equations above, the only difference between the two is the constant, which changes from 1.73 to 2.
    You are missing a major point in your statement. There is no relationship between
    the constants 1.73 and 2. If you look at list you show L = One way length of circuit (ft)
    The 2 is to make up for the extra conductor in the return path. In the 3 phase equation the 1.73 is used because of the phase angle difference between the phase voltages. You also need to remember that you are working with KVA converted to amps. If you have 100 kva at 240 volts single phase I = 417 amps. A 100 kva at 240 volts 3 phase I = 240 amps.

  5. #5
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    I don't seem to be attracting the sort of responses willing to explore the issue. I was hoping that someone would see the problem I'm facing, and explain how they would solve it in my position. As Bob points out, the relationship between 2 and 1.73 are coincidental. We'll come back to that in a moment. First, please bear with me.

    Let's say we have a single 400W (464W with ballast) luminaire on a phase AB of a three phase lighting circuit. To calculate the current draw, would you use the single phase current calculation or the three phase calculation? I'll wait for a response before I continue.

  6. #6
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    IMO, the current draw on the single phase circuit at 208V is 464/208 no matter if the panel is 3 phase. At least that would be my bet. :grin:

  7. #7
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    Does suemarkp's post help answer?

  8. #8
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    Quote Originally Posted by That Man View Post
    Thank you for your quick response. The lighting circuits I'm trying to evaluate are missing information that would be impractical to acquire (no budget for a field trip). I know the wire size, quantity of luminaires, wattage of luminaires, system voltage, and distance between poles. What I don't know are the actual distribution between the phases. The worst case scenario in this case then, is that the circuit is unbalanced.

    I'm trying to work around the missing information problem by establishing general parameters to evaluate the system. What I'm hoping for is to formulate a useful approximation. Since symmetrical components requires full knowledge of the circuit, I would have to construct several scenarios to see under what condition the voltage drop is the worst. The only way I know how to calculate symmetrical components is by hand (on a side note, if you know of any free or cheap application that can perform symmetrical component calculations, I'd love to hear of it; I can't seem to make it work in SPICE, and I've been unsuccessful in building anything in Excel).

    If, per chance, the voltage drop for an unbalanced three phase delta circuit is always contained within the range of a single phase circuit and a balanced three phase circuit, then it becomes easy to evaluate--the most conservative estimate would simply be to treat such a circuit as a single phase circuit. However, if the voltage drop in an unbalanced three phase circuit can stray outside of this range (that is, the voltage drop can be worse than a single phase circuit), then what is the worst case? It can't be infinite, I don't think, which means there is a finite quantity of voltage that a cable can lose per foot, no matter how badly the circuit is balanced. I'm looking for this value.

    If I can't find an answer, I'll just have to figure out how to get out the site and open up those poles and see. I'll make do. But from an intellectual standpoint, the nature of this problem intuitively feels like it has a solution, and that's what I'm hoping to find out.
    The simple answer to your question is as you surmise... the voltage drop will always be between the factoring of 1.732 and 2 of 3Ø and 1Ø circuits respectively. But Bob is correct in that for a given load, the current used in the formula also changes.

    With that said, I assume such a 3Ø circuit would never be imbalanced more than all the load across two phases. As an example, and expanding on Bob's 100kVA scenario, you have 417 1Ø amperes and a balanced 241 3Ø amperes. If the load were only on two of three phases, the current of half the total load would be on two of the lines (50kVA ÷ 240V = 208A) while the largest current is 1.732 times that (361A) on the third line, which IMO would be the worst case scenario.

    However, if we assume each line conductor size to be the same, we will have a different voltage drop on the higher amperage line than the two lower amperage lines. Since the voltage drop is actually across two lines, you would calculate each line's voltage drop (i.e. one way only)... then sum the results for total voltage drop across the respective lines. You will get a value that is somewhere between calculating the load's voltage drop as all single phase or balanced three phase.
    I'll never get there. No matter where I go, I'm always here.

  9. #9
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    Continuing with the Problem

    Smart $, thank you for taking the time to respond. That is exactly what I was looking for--some sort of confirmation that my intuition's hypothesis was correct or not. Is Dennis's supposition correct? That the current through a single luminaire is simply P/E (assuming Pf=1)?

  10. #10
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    Can someone please verify the following:

    Let's say we have a single 400W (464W with ballast) luminaire on a phase AB of a 480V three phase lighting circuit. What is the current? In this scenario, Phase BC and CA have no load, so are open circuits.

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