Abstract:
Among the three main components of the power system: generation, transmission and
distribution, most pressing problem in the power sector has been with the distribution system.
It is well conceived that almost all the revenue stream comes from distribution system. If this
section is characterized with high system loss resulting poor revenue collection, then the toal
power system tends to collapse. So, for financial viability of the sector, it is vital that the
distribution sector should be taken care of. So, it is vital that serious efforts should be taken to
bring the system loss to a standard level. It is acknowledged that in quantifying the standard
loss of distribution system is not as easy as it has many complex calculations due to wide
range of system variables.
Considering the urgency to find the loss quantity against the various contrast, a model tool
has been devised following heuristic process in this thesis. It is anticipated that in finding
distribution loss an expensive software with expertise are required. The software needs huge
system data for identification and quantification of losses. This is also time consuming.
To get the required job done, a device is required to monitor the system in finding out the
accurate quantity of loss. The thesis is aimed to develop how to find losses of both feeder
and transformer by one item only that is current of low voltage side of the transformer. The
fundamental concept which is considered is that maximum system load flow is 1/3 of the
feeder length. Such consideration of load flow might deviate the actual expected result. This
deviation needs a tool to be used to rationalize the calculated result. The tool which has been
devised in this thesis is termed as “K” factor. “K” factor is the main part of this thesis.
The value of “K” depends on load ratio of feeder when each feeder is divided in two halves.
It is known that loss varies in the feeder with same value of load if the load ratio varies
between 1st half and 2nd half. As per load ratio, “K” is higher, if load in 2nd half is greater
than load in 1st half. It is equal, if load in 2nd half is equal to load in 1st half and lower, if load
in 2nd half is less than load in 1st half.
