A Better Understanding Of Load And Loss Factors

A Better Understanding of Load and Loss
Factors
(Summary)
The primary focus of this paper is to find a simplified way to accurately determine
the I R losses in a distribution system. These lossescan be accurately computed by
determining the average load on each component during each hour and doing a
separate loss calculation for the same time period. The problem of th is approach iscomputational complexity. A simpler approach consists in the use of loss and load
factors to estimate these losses.
2

The Load Factor (FLD) is determined using a load profile (typical daily load overa
period of time, usually 24 hours) on the desired component or feeder. The load factor is
determined by taking the ratio of the average load (kW) and peak load (kW) over the
period of question.Similarly, the Loss Factor (FLS) can be determined by taking the ratio
of the average and peak losses. If the feeder resistance is know and constant, both FLD
and FLS can be calculated (assumingconstant power factor and voltages). The FLD is
calculated by taking the ratio between average and peak current and, F LS by ratio of
average and peak currents square.
Since the FLS depends on theactual load, the equation (3) of the paper is
suggested to calculate the loss factor using only data contained in the load profile. The
parameter “a” introduced in this equation can be determined byequation (4) or
equation (7) in terms of 1) average and peak currents, 2) current square average and
peak current squared. The possible values “a” are between 0 and 1. W hen FLS = (FLD)2
or average ofthe current square is equal to the average current the “a” has a value of
zero. Also, when FLS = FLD the “a” parameters has a value of 1. Both cases occur the
load is constant. The paperdemonstrated that FLS ≥ (FLD)2 and the following
characteristics have been determined:
1. If “a” = 0
a. FLS = (FLD)2
b. FLS = 1 and FLD = 1
c. Load constant
2. As “a” approaches 0, the load becomes more...