Distribution Feeder I2R Loss and Loss Factor

ScopeLoss is quadratic in current while the current varies all day, so the average of the square is not the square of the average, and the ratio between them is the LOSS FACTOR. It is bounded at both ends by quantities anyone can name: a perfectly flat load loses at its load factor, a load that is either at peak or entirely off loses at the square of it, and every real load sits between. The long-standing utility approximation blends the two as 0.3 times load factor plus 0.7 times its square, and that blend is what turns a peak loss into an annual energy. Take the peak loss for all 8,760 hours and the answer comes out several times too high; use the load factor alone and it is still tens of percent high. The practical consequence cuts in a direction people do not expect. A feeder with a poor load factor loses much LESS energy than its peak loss suggests, which means the savings from reconductoring, from moving a capacitor bank, or from balancing phases are smaller than a peak-based estimate promises. The economic case for any of them has to be built on the loss factor, not on the peak, and a payback computed the other way will not arrive. Conductor loss on one balanced three-phase feeder with the load treated as CONCENTRATED AT THE FAR END. A real feeder has load distributed along it, and for a uniformly distributed load the effective loss is about a third of the concentrated value -- a correction large enough to matter and one this does not apply, so read the answer as an upper bound unless the load genuinely is at the end. It does not include transformer core and copper losses, which on a distribution system are usually the larger share of total losses, nor neutral, secondary, or service losses, nor unbalance, nor the temperature dependence of the conductor's own resistance. The 0.3 and 0.7 coefficients are a widely used approximation rather than a measurement, and utilities carry their own. The utility's loss study and its metered load data govern.

Formula or tablepeak loss = 3 x peak current squared x total resistance; loss factor = k x load factor + (1 - k) x load factor squared, k entered (0.3 common; RUS 1724D-107 and 1724E-301 use 0.2); annual loss energy = peak loss x loss factor x 8,760 hours; energy delivered = peak demand x load factor x 8,760 hours.

Source and editionThe three-phase I2R loss relation and the standard distribution loss-factor approximation, by name. The 0.3 and 0.7 coefficients are a widely used approximation rather than a measurement, and utilities carry their own. Load treated as concentrated at the far end, so the answer is an upper bound. The utility's loss study and its metered load data govern.

Free accessA square, a blend of two terms, and a multiplication by the hours in a year; no utility loss study is reproduced.

What governsEstimate. AHJ and licensed professional govern.

API field namespeak_current_a, resistance_ohm_per_mile, length_miles, load_factor, energy_cost_per_kwh, peak_demand_kw, total_resistance_ohm, peak_loss_kw, loss_factor, annual_loss_kwh, annual_cost, peak_all_year_kwh, loss_percent_of_delivered

• Load concentrated at the far end a uniformly distributed load loses about a third as muchdistribution loss practice
• Conductor loss only transformer core and copper losses are usually the larger sharethe utility's loss study
• The 0.3 / 0.7 coefficients are an approximation utilities carry their own from metered datathe utility's loss study