How phase load balancing affects losses in the distribution lines?
The most efficient way for a three-phase line or even a V-phase system to operate is when the loading on each line is said to be balanced. As what we constantly say, distribution line loss is dependent mainly to the two main variables, the line resistance and the line current.
This known method for minimizing losses in a distribution system is somewhat cheaper compared to other known method. However, the effect of this effort can be very substantial in the part of the electric utility. So, why does an unbalanced loading occur in the first place? Unbalance loading in a three-phase or V-phase system usually happens when proper load monitoring is being neglected. Since a number of a combination of a single-phase and three-phase loads connects to the utility sometimes appropriate phase connection are being overlooked.
Usually, the most significant time that this unbalanced loading can affect the system is during peak loads. During this time, the maximum losses caused by unbalance current are measured. Unbalanced system not only contributes losses to phase conductors but also to its neutral lines (grounded system) since an unbalanced three-phase lines results to a residual current found in the neutral lines.
In phase balancing, the electric utility need not possess a complicated procedure in implementing this action. Basic electrical knowledge and a small arithmetic is more than enough to be use as a tool in load balancing practice. In addition, utility needs only the help of their linemen crew to make the actual reconnections given that proper planning and strategic loading placement is already been formulated.
Example: At peak load, the following phase-ampere readings are taken on a one-mile, three-phase, 2/0 ACSR distribution line: Phase A, 40 amperes; Phase B, 130 amperes; and Phase C, 100 amperes. Assuming that most of the load is beyond this mile of line, what loss savings will result if the load is balanced at 90 amperes on each phase?
Solution: A vector calculation reveals that approximately 80 amperes are returning through the neutral and ground circuit of the unbalanced line. Conductor resistance data are the following; Phase conductor = 0.716 ohms, neutral conductor = 0.234 ohms. The resistance of the ground-return path is approximately equal to the three-phase resistance subtracted from the single-phase resistance. In the present example, using a phase conductor resistance of 0.706 ohms and a ground return resistance of 0.230 ohms, the following loss levels are determined for the unbalanced line;
Phase A: (40^2) x 0.706 = 1.13 kW
Phase B: (130^2) x 0.706 = 11.93 kW
Phase C: (100^2) x 0.706 = 7.06 kW
Neutral/Ground: (80^2) x 0.234 = 1.50 kW
Total Losses = 21.62 kW
If the line is balanced at 90 amperes per phase, the total losses will be three times the losses on each phase conductor, since ground return current will be negligible. Therefore:
Balanced Circuit Losses = 3 x (90^2) x 0.706 = 17.16kW
The savings in losses are:
21.62kW – 17.16kW = 4.46kW
The loss savings could be worth about $600 per year for a typical electric utility distributor.
Similar to the method of single-phase to three-phase conversion, load balancing is related in the sense that the objective of this method is to distribute the current to all of the phases available. The I squared R formula greatly dictates whether the system is having a high or low loss distribution system.
is 0.706 ohm is constant?
ReplyDelete0.706 ohm is dependent upon the conductor being used, in the problem its the resistance of 2/0 ACSR per mile. The neutral conductor usually takes 33% of the resistance of the line.
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