Mike Dolt Enterprises
Mike Dolt Enterprises
User: That Man, Location: Lockport, IL
Asked on: Wednesday, March 30, 2011 at 4:58 pm EDT
Answered on: Tuesday, May 14, 2024 at 3:30 am EDT
Question: 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 DropK = 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 CL = 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)/CmThe equation to find voltage drop in a three phase circuit is:Vd = (1.73K x L x I)/CmAs 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.
Yes: The resulting voltage drop calculation for an unbalanced three-phase delta system will fall somewhere between the calculation for a single-phase system and a balanced three-phase system.
Explanation:In a perfectly balanced three-phase delta system, the phase currents are equal and there is no neutral current. This results in a voltage drop of:Vd = (1.73K x L x I)/CmIn an unbalanced three-phase delta system, the phase currents are not equal and there is a neutral current. This results in additional voltage drop due to the unbalanced current distribution.The constant in the voltage drop equation for an unbalanced system will vary depending on the degree of unbalance. For a slightly unbalanced system, the constant may be closer to 1.73 (balanced three-phase value). For a heavily unbalanced system, the constant may be closer to 2 (single-phase value).However, in general, the constant for an unbalanced three-phase delta system will always reside between 1.73 and 2. This is because the unbalanced system is neither perfectly balanced (1.73) nor completely single-phase (2).
Therefore: For normal lighting loads, the voltage drop calculation for an unbalanced three-phase delta system will be typically between that of a single-phase system and a balanced three-phase system. Flag for review