Efficiency
Transformers
which are connected to the power supplies and loads and are in operation are
required to handle load current and power as per the requirements of the load.
An unloaded transformer draws only the magnetization current on the primary
side, the secondary current being zero. As the load is increased the primary
and secondary currents increase as per the load requirements. The volt amperes
and wattage handled by the transformer also increases. Due to the presence of
no load losses and I2R losses in the windings certain amount of electrical
energy gets dissipated as heat inside the transformer.
This
gives rise to the concept of efficiency. Efficiency of a power equipment is
defined at any load as the ratio of the power output to the power input.
Putting in the form of an expression, while the efficiency tells us the
fraction of the input power delivered to the load, the deficiency focuses our
attention on losses taking place inside transformer. As a matter of fact the
losses heat up machine. The temperature rise decides the rating of the
equipment. The temperature rise of the machine is a function of heat generated
the structural configuration, method of cooling and type of loading (or duty
cycle of load). The peak temperature attained directly affects the life of the
insulations of the machine for any class of insulation.
These
aspects are briefly mentioned under section load test. The losses that take
place inside the machine expressed as a fraction of the input is sometimes
termed as deficiency. Except in the case of an ideal machine, a certain
fraction of the input power gets lost inside the machine while handling the
power. Thus the value for the efficiency is always less than one. In the case
of a.c. machines the rating is expressed in terms of apparent power. It is
nothing but the product of the applied voltage and the current drawn. The
actual power delivered is a function of the power factor at which this current
is drawn.
As the
reactive power shuttles between the source and the load and has a zero average
value over a cycle of the supply wave it does not have any direct effect on the
efficiency. The reactive power however increases the current handled by the
machine and the losses resulting from it. Therefore the losses that take place
inside a transformer at any given load play a vital role in determining the
efficiency. The losses taking place inside a transformer can be enumerated as
below:
1.
Primary copper loss
2.
Secondary copper loss
3.
Iron loss
4.
Dielectric loss
5.
Stray load loss
These are
explained in sequence below.
Primary
and secondary copper losses take place in the respective winding resistances due
to the flow of the current in them. The primary and secondary resistances
differ from their d.c. values due to skin effect and the temperature rise of
the windings. While the average temperature rise can be approximately used, the
skin effect is harder to get analytically. The short circuit test gives the
value of Re taking into account the skin effect. The iron losses contain two
components Hysteresis loss and Eddy current loss. The Hysteresis loss is a
function of the material used for the core. Ph = KhB1.6f For constant voltage
and constant frequency operation this can be taken to be constant. The eddy
current loss in the core arises because of the induced emf in the steel
lamination sheets and the eddies of current formed due to it. This again
produces a power loss Pe in the lamination. Where t is the thickness of the
steel lamination used. As the lamination thickness is much smaller than the
depth of penetration of the field, the eddy current loss can be reduced by reducing
the thickness of the lamination. Present day laminations are of 0.25 mm
thickness and are capable of operation at 2 Tesla.
These
reduce the eddy current losses in the core.This loss also remains constant due
to constant voltage and frequency of operation. The sum of hysteresis and eddy
current losses can be obtained by the open circuit test.The dielectric losses
take place in the insulation of the transformer due to the large electric
stress. In the case of low voltage transformers this can be neglected. For
constant voltage operation this can be assumed to be a constant. The stray load
losses arise out of the leakage fluxes of the transformer. These leakage fluxes
link the metallic structural parts, tank etc. and produce eddy current losses
in them. Thus they take place ’all round’ the transformer instead of a definite
place, hence the name ’stray’. Also the leakage flux is directly proportional
to the load current unlike the mutual flux which is proportional to the applied
voltage.
Hence
this loss is called ’stray load’ loss.This can also be estimated
experimentally. It can be modeled by another resistance in the series branch in
the equivalent circuit. The stray load losses are very low in air-cored
transformers due to the absence of the metallic tank. Thus, the different
losses fall in to two categories Constant losses (mainly voltage dependant) and
Variable losses (current dependant). The expression for the efficiency of the
transformer operating at a fractional load x of its rating, at a load power
factor of 2 can be written as losses and Pvar the variable losses at full load.
For a given power factor an expression for _ in terms of the variable x is thus
obtained. By differentiating _ with respect to x and equating the same to zero,
the condition for maximum efficiency is obtained. The maximum efficiency it can
be easily deduced that this Maximum value increases with increase in power
factor and is zero at zero power factor of the load. It may be considered a
good practice to select the operating load point to be at the maximum
efficiency point.
Thus if a transformer is on full load, for most
part of the time then the max can be made to occur at full load by proper
selection of constant and variablelosses.However, in the modern transformers
the iron losses are so low that it is practically impossible to reduce the full
load copper losses to that value. Such a design wastes lot of copper. This
point is illustrated with the help of an example below. Two 100 kVA
transformers A and B are taken. Both transformers have total full load losses
to be 2 kW. The breakup of this loss is chosen to be different for the two
transformers. Transformer A: iron loss 1 kW, and copper loss is 1 kW. The
maximum efficiency of 98.04%occurs at full load at unity power factor.
Transformer B: Iron loss =0.3 kW and full load copper loss =1.7 kW. This also
has a full load of 98.04%. Its maximum occurs at a fractional load of q0.31.7 =
0.42. The maximum efficiency at unity power factor being at the corresponding
point the transformer A has an efficiency of Transformer A uses iron of more
loss per kg at a given flux density, but transformer B uses lesser quantity of
copper and works at higher current density.
Power output
= Power input - Total Losses
Power
input = Power output + Total Lossess
= Power
Output + Pi + Pcu
h = Power
Output / Power Input
= Power Output / [Power Output + Pi
+ Pcu]
Now Power
Output = V2I2cosϕ
Cosϕ = Load power factor
h = [ (VA
rating) x cosϕ ] / [ (VA rating) x cosϕ + Pi + I22
R2e ] x 100
All day efficiency
Large
capacity transformers used in power systems are classified broadly into Power
transformers and Distribution transformers. The former variety is seen in
generating stations and large substations. Distribution transformers are seen
at the distribution substations. The basic difference between the two types
arises from the fact that the power transformers are switched in or out of the
circuit depending upon the load to be handled by them. Thus at 50% load on the
station only 50% of the transformers need to be connected in the circuit. On
the other hand a distribution transformer is never switched off. It has to
remain in the circuit irrespective of the load connected. In such cases the
constant loss of the transformer continues to be dissipated. Hence the concept
of energy based efficiency is defined for such transformers. It is called ’all
day’ efficiency. The all day efficiency is thus the ratio of the energy output
of the transformer over a day to the corresponding energy input. One day is
taken as duration of time over which the load pattern repeats itself. This
assumption, however, is far from being true. The power output varies from zero
to full load depending on the requirement of the user and the load losses vary
as the square of the fractional loads. The no-load losses or constant losses occur
throughout the 24 hours. Thus, the comparison of loads on different days
becomes difficult. Even the load factor, which is given by the ratio of the
average load to rated load, does not give satisfactory results. The calculation
of the all day efficiency is illustrated below with an example. The graph of
load on the transformer, expressed as a fraction of the full load is plotted
against time. In an actual situation the load on the transformer continuously
changes. This has been presented by a stepped curve for convenience. For the
same load factor different average loss can be there depending upon the values
of xi and ti. Hence a better option would be to keep the constant losses very
low to keep the all day efficiency high. Variable losses are related to load
and are associated with revenue earned. The constant loss on the other hand has
to be incurred to make the service available. The concept of all day efficiency
may therefore be more useful for comparing two transformers subjected to the
same load cycle. The concept of minimizing the lost energy comes into effect
right from the time of procurement of the transformer. The constant losses and
variable losses are capitalized and added to the material cost of the
transformer in order to select the most competitive one, which gives minimum
cost taking initial cost and running cost put together. Obviously the iron
losses are capitalized more in the process to give an effect to the
maximization of energy efficiency. If the load cycle is known at this stage, it
can also be incorporated in computation of the best transformer.
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.