Hi friends in this video we are going to
solve a problem based on efficiency of a
single phase transformer so the
statement of problem is like this 40 kVA
transformer have iron loss 450 VAT and
full load copper loss of 850 VAT if the
power factor of the load of 0.8 lagging
calculate full load efficiency second
the load at which maximum efficiency
occurs and third the maximum efficiency
so it is a problem based on formula of
efficiency but before solving that we
will write the formula first so see the
procedure to solve this type of problem
what I will do first I will write a
formula of efficiency so efficiency
equal to X into kV a rating of a
transformer x power factor upon same X
into kV a rating of a transformer x
power factor plus iron loss plus X
square copper loss at full load so what
we will do we have to list out all the
things that those are required for
solving this equation so first kV a
rating of a transformer is given
next iron loss is also given full load
copper loss is also given
so WC you FL is eight fifty back and
part of the load is also given zero
hundred lagging so all the things are
given except one and that is X so what
is this X X is nothing but fraction of
full load what is the meaning of this
sometimes load applied is half load or
75% of load or 60% of load or sometimes
it may be full load so if it is a full
load that means X is equal to one if it
is a half load X equal to 0.5 if it is
quarter load X is equal to 0.25 like
that you have to find out the values of
X and it is given indirectly so in our
problem
if you check calculate full load
efficiency so in the first part they
have said they want a full load
efficiency that means value of x equal
to 1 so here for first case I can say X
equal to 1 so all the things we have
listed out so let's substitute the full
load efficiency can be calculated as X
is 1 kV a rating is 40,000 power factor
is zero point 8 divided by 1 into 40,000
multiplied by 0.8 plus iron losses 450
plus X is 1 through so 1 square and
copper losses full load is 850
so if you solve this you will get
efficiency equal to 0.96 one normally we
talk in terms of percentage so percent
efficiency equal to ninety six point one
we simply multiply this value by 100 so
full load efficiency we got ninety six
point one let's go to the second part
then second part we are supposed to
calculate the fraction of load at which
efficiency is maximum so for maximum
efficiency fraction of load X is given
away root of iron loss upon copper loss
at full load so this is equal to root of
iron losses for 50 full load copper loss
is 850 if you solve this you will get X
equal to zero point seven two seven six
so this would be the fraction of full
load in order to get maximum efficiency
so KVA
for maximum efficiency equal to X
multiplied by K be a rating of a
transformer so if I substitute the value
X is zero point seven two seven six and
transformer kV is forty so if we solve
you will get load KVA for maximum
efficiency equal to twenty nine point
one zero four three so this much load
you should apply in order to get maximum
efficiency let us calculate what is that
maximum efficiency
phurba lies once again the same maximum
efficiency is nothing but X for maximum
efficiency x kv aerating x power factor
divided by x for maximum efficiency x kv
aerating x power factor plus iron loss
plus x square and obviously this x where
efficiency is maximum x WC you FL so the
only difference is that whenever we
calculate a maximum efficiency we are
suppose to calculate X it is not given
so that X you have to calculate and same
formula you have to use so we got X for
a maximum efficiency 0.727 6kv editing
is same 40000 power factor is 0.8 only
divided by zero point seven two seven
six multiplied by forty thousand
multiplied by 0.8 iron loss is four
fifty plus X square is zero point seven
two seven six square multiplied by eight
five zero if you solve you will get
maximum efficiency 0.96 to a in terms of
percentage it is ninety six point two
eight percent
this is the required answer thank you
