and no-load. For simplicity of modeling, both load and no-load real power losses in a distribution system can be modeled as resistors RL and RNL in a shunt circuit as shown in Figure 1.

Load losses represented by RL vary quadratically with system load. They are also referred to as resistive, conductor, copper, or I²R losses. Constant no-load losses, represented by RNL, are driven by the presence or application of AC voltage. The amount of load has no appreciable effect on no-load losses. The vast majority of these losses are from distribution transformer excitation. Underground cable dielectric losses are also a small component of no-load losses.

In a well-run, highly capitalized distribution system, losses are approximately 3 percent to 5 percent. In neglected systems with poor planning criteria and

Figure 1

Distribution System Loss Model

Voltage Source

No-load Losses

Load ^RL Losses

^RNL

Variable
Distribution
System Load

design standards, losses may increase to 5 percent to 7 percent or higher. Peak losses occur during peak load when generation pricing is at its highest and distribution equipment thermal capacity is often stressed.

Main Causes of Distribution Losses

Physicists may study phenomena such as dielectric and induction losses, but the most consequential distribution losses are due to resistive copper losses and transformer excitation.

Conductor power losses result from electron flow (current) through resistance in primary, secondary, and transformer conductors regardless of conductor material (e.g. copper, aluminum, aluminum-alloy, etc.). Feeder voltage, conductor length and size, power factor, load factor, loading, and phase current balance all determine copper losses.

The conductor current contribution to technical power losses is higher than that for conductor resistance as shown in the resistive power loss equation Ploss = I² x R, where: Ploss is power loss in watts; I is current measured in amperes, and R is resistance measured in ohms.

Conductor heating is proportional to the square of the current; therefore, conductor power losses will double if the current increases by only 41 percent as represented in Figure 2, next page.

 

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