Synchronous Machines 2, Machines, Electrical Engineering, GATE Video Lecture

hi guys in this lecture we will continue
from where we left off and we will look
at some more type of problems that can
be asked based on synchronous machines
so last time we looked at parallel
operation so this is one more type of
problem that can be asked on parallel
operation usually when alternators are
connected in parallel the purpose behind
that is to supply higher value of load
so the alternators will generate a total
power which is equal to the load power
but here we are neglecting any losses in
the transmission which means apparent
power so or some of apparent power of
generators should be equal to some of
apparent demand okay and as you can see
I have placed the phaser sign above
which means they shall be added as
phasers or you can say their real and
imaginary parts will be individually
added like this summation of real power
generated is equal to real power demand
and summation of reactive power
generated is equal to reactive power
demand while taking reactive power keep
in mind the concept of lagging and
leading usually for lagging reactive
power is positive for leading it is
negative okay but in real power there is
no problem of sign generated power is
always positive and demand is negative
so let us see the problem that we have
been given to 550 kV alternators operate
in parallel to supply the following
loads so these two are the loads so this
basically is demand one demand to one
machine is supplying this is the
supplied power so PG one we need to find
the power factor of other machine now
one thing I know is there are two
generators so total generated real power
should be equal to total demand real
power total generated real power is how
much 200 + PG - total real demand power
250 + and rate which means PG - is 150
kilowatt now in order to determine the
power vector we must also have the
knowledge about the reactive power so
first we should derive the reactive
power for first load that is QD 1 as you
can see it is lagging power factors
reactive power shall be positive we know
reactive power is given like this P into
tan of Phi and power factor is cos of
Phi so I can determine Phi by taking cos
inverse of power factor so QD 1 I am
directly writing 250 into 10 and Phi
will be cos inverse power factor is 0.6
okay
similarly QD 2 it is leading so it shall
be taken as negative - 110 cos inverse
0.5 similarly qg 1 it is lagging so it
shall be positive 210 off cos inverse
0.9 now we know q g2 will be equal to QD
1 + QD 2 minus Q g1 if you do this you
will figure out that Q g2 will come out
to be 63 point 2 6 kV AR now we have
real power of generator - as well as the
reactive power so we can figure out the
power factor angle Phi it is qg 2 by p
g2 it is actually tan of Phi 2 similarly
if you want to find out cos of Phi 2 it
will be cos of
an inverse qg2 by p g2 q g2 is 63 0.26
P g2 is 150 so from here it comes out to
be 0.92
but now whether it is lagging or leading
as you can see reactive power is
positive so it should be lagging so that
is how we can derive the power sector of
any machine when we talk about the
parallel operation is it fine
moving forward the next type of problem
that can be asked is based on efficiency
and loss is inside the synchronous
machine so before finding efficiency of
any machine what you need to remember is
power flow how does the power flow
inside a machine so first if we talk
about generator in generator we have
shaft input and we have input at the
field the shaft input causes rotational
loss and then it is converted to develop
the power now what is developed power EF
I a soft sigh what is this I it is the
internal power sector angle or the angle
between induced EMF and current it is
not equal to power factor angle because
power factor angle is the angle between
terminal voltage and armature current it
is between induced EMF and armature
current then we have copper loss and
then we have output power now what is
this rotational loss it is actually
equal to friction and windage loss plus
core loss so remember this field power
so whatever is the real power in the
field is actually taken as a loss why is
it taken as a loss because the function
of field winding is to generate the flux
which should require only reactive power
for whatever is the real power consumed
in the field winding
that makes no use and hence it is taken
as losses similarly for the motor
how shall we draw the power flow diagram
for motor we will have electrical input
we will have field input after
electrical input we will have copper
loss which will lead to the develop the
power again it is same efi a cause of
size then we will have rotational loss
and then we will have shaft power okay
so entire power flow diagram is just
reversed now let us see the problem that
is given to us it is saying of 150 kv m
so that should be our output power or
output rating because in kv a only the
apparent power can be given and
mechanical power does not have any
imaginary term so it is electrical power
and for an alternator electrical power
is output power and also remember do not
directly use rating in any equation
unless otherwise you are given full load
half load quarter load because these
loads are in comparison or in reference
to the full load or the rated power but
if you are given certain load power you
have to use that as a load then the
rating will not be used in the problem
now if you see we have friction and
windage laws and the core loss these two
will combine Li be called as rotational
loss so that is 850 watt we have field
winding resistance we have field winding
voltage now we know in the field winding
we have DC supply so what is the field
loss field winding voltage square by
field winding resistance so this is 375
what we need to find the alternator
efficiency eight point eight power
vector and half of the full load now let
me calculate full
current which is actually rated current
150 kb/s of 115 to 10 to the power 3 now
remember this is a parent power so there
is no need to use the power factor I can
directly use s is equal to root 3 VL IL
and since it is star connected IL will
be same as the phase current okay but if
this power was real then we will use P
is equal to root 3 v Li L cos Phi so
this is a very common mistake we do not
consider that either the power is s or P
we directly use it is it fine root 3
into 500 so this current shall come out
to be it is 6 point 6 ampere now this
current shall be taken as half so so
this will come out to be half eighty six
point six
I am taking half because it is a half
load okay
now this current will cause copper
losses so what will be copper losses
three ia square ra remember this factor
of three because it is a three phase
machine three into it is 6.6 square into
RA is how much 0.03 so this comes out to
be how much 675 this is actually three
hundred twelve point five now we have
all the losses let us calculate output
output is given as half of the full load
which means 75 kV and at point eight
power factors of P is equal to s cos Phi
so how much will this come out to be it
will come out to be sixty kilo watt we
have output we have losses we can
calculate efficiency how output by
output plus loss
losses 850 is rotational loss three
hundred twelve point five is core loss
or a field loss and then 675 is the
copper loss so this will come out to be
point nine seven or you can say ninety
seven percent so that will be the
percentage efficiency of the alternator
is it fine how you have to consider
different losses let us look at another
problem based on losses it is saying a
415 volt 3-phase delta connected
synchronous motor runs at rated voltage
with an excitation of 500 volt now
remember this is Delta connected which
means line voltage is equal to phase
voltage which means the phase voltage is
now 415 volt EMF is 500 volt synchronous
impedance is 0.5 plus j2 friction
windage and iron losses are thousand
watt so zs is 0.5 plus j2 ohms this
comes out to be two point zero six one
at an angle of seventy five point nine
six three degree ohm now we have to find
the maximum shaft power output now sense
of friction losses are constant maximum
output power can also be corresponding
to the maximum developed power because
as we saw in synchronous motor power
flow diagram shaft power will be
developed power minus rotational losses
so maximum shaft power is corresponding
to maximum developed powers
if rotational losses are constant so
what is the expression for developed
power in a synchronous motor three EF VT
by mod Z s cos of Delta minus theta s
- 3ef square by mod zs cos of theta s
okay that is the formula now to maximize
this develop the power what you should
take Delta must be equal to theta s
theta s is basically angle of impedance
and Delta is the power angle so what is
P max 3 let us take out common EF can be
also taken common zs which is two point
zero six one can be taken common so
first one is VT 415 minus EF into cos of
theta L theta as we just now derived
seventy five point nine six three so
this power comes out to be two hundred
thirteen point seven five kilowatt okay
now we know the friction and the iron
losses are one kilowatt which means
shaft power will be two hundred twelve
point seven five kilo watt I have just
subtracted one kilo watt corresponding
to the losses which gives me the total
shaft power output so remember this
thing whenever you are dealing with a
synchronous machine question please
carefully look at whether the machine is
star connected or Delta connected that
will immensely help you in finding the
solution next type of problems are based
on voltage regulation in voltage
regulation do not use the methods that
we use for transformer which is I into R
cos PI plus X sine Phi this method
should not be used in case of
synchronous machine why because it is an
approximate method and this approximate
will approximation will deviate by a
large amount as compared to actual
result for the case of synchronous
machine so what should we do to find the
voltage regulation
find out magnitude of EMF minus mod V by
mod V this I told you in the previous
video that what you can do for Maudie
you can use direct equation what was
that V cos Phi plus I a are a whole
square plus V sine Phi plus minus I a
excess square under root okay plus was
for lagging and minus was for leading is
it fine so that is how you can calculate
magnitude of EMF and then we have to
remember the condition for maximum
voltage regulation for maximum voltage
regulation theta must be equal to Phi
and for zero voltage regulation cos of
theta plus Phi is equal to minus I is
EA's by 2 V phase okay these two
condition you should always remember for
the case of voltage regulation now let
us see the problem determine the voltage
regulation of a 2,000 volts single phase
when you see the single phase it means
we do not have to use any factor of root
three giving a current of 100 amperes
per point eight power factor leading and
from the test result field current of
100 ampere is produced by full load
current hundred ampere is produced on
short circuit by a field excitation of 2
point 5 ampere which means short circuit
current is 100 ampere EMF of 500 voltage
produced by on open circuit by same
excitation EEOC is 500 from here I can
calculate that is 500 by hundred there
will be no factor of root 3 at a single
phase alternative 5 ohm
now armature resistance is pointed
we can calculate access under root five
square minus point eight square which
will come out to be how much four point
nine three five six now the next step is
to calculate the EMF in order to
determine the voltage regulation so if
you want to calculate by phasor it shall
be two thousand current is 100 ampere at
0.8 power factor leading point eight
corresponds when angle of thirty six
point eight six into zs which is point
eight plus J so this will be ohm four
point nine three five six okay this
comes out to be one eight two two point
five seven at an angle of fourteen point
zero six degree volt now we have to find
the voltage regulation it will be one
eight two two point five seven - mm by
two thousand in two hundred percent so
this comes out to be minus eight point
eight seven percent okay so as you can
see for a leading load usually the
voltage regulation comes out to be
negative in the next type of problem it
is based on group characteristics what
is meant by troop characteristic is when
you increase the load on any alternator
its the speed and the frequency of
output will decrease usually we assume
that the frequency decreases linearly
with power like this so this is our no
load frequency this is our full load
frequency corresponding to full load
power output now here you have to
remember the power output of any machine
is equal to no load frequency - actual
frequency
no load frequency - full load frequency
into full load power output this is the
straight line gov equation governing the
troop characteristics and what is droop
actually it is no load frequency - full
load by full load in 200% similarly in
the denominator there can also be no
load frequency which you have to judge
based on the question so let us see the
problem two generators rated 500 and 600
megawatt are operating in parallel the
troop characteristics of their governor
are 4% and 5% from no load to full load
assuming generators are operating at
sixty Hertz at no load which means no
load frequency is 60 Hertz so what is
full load frequency of first one it has
a group of five percent sorry four
percent 0.96 into sixty 57.6 Hertz what
about second one it has a group of five
percent so it will be 0.9 5 into 60
which is 57 Hertz now I can form the
equation for power no load - f upon no
load - full load which is two point four
into P full load P full load of first
alternator is five hundred similarly P -
no load minus F by no load - full load
into P 2 or P full load - is 600 now it
is saying a load of 1100 megawatt is
shared by the alternators which means P
1 plus P 2 is 1100 so now what you have
to do is you have to substitute the
values of P 1 and P 2 into this equation
now both values are derived in terms of
F so we can calculate F
it will come out to be fifty seven point
three Hertz from there we can calculate
P 1 pi 61 point two megawatt and P to
five thirty eight point eight so that is
how the load will be shared among the
two generators so for this you only have
to remember the equation for power in
terms of frequency the method is always
same Express power in terms of frequency
add both powers equal to the load power
that will give you frequency from
frequency we can determine the power of
each generator the next category of
problem is based on synchronous motor
which is cylindrical rotor till now we
were discussing about only alternator
okay nine synchronous motor what does
change in synchronous motor the
equivalent circuit essentially remains
same like this this is EF VT but here I
is going inwards so here EF is VT minus
ie into Z okay so similarly if you want
to write the magnitude of EMF it will be
V cos Phi minus Iara square plus V sine
Phi minus plus IX is cos square okay -
for lagging and plus for leading so that
is how you can determine the magnitude
of EMF in case of a synchronous motor is
it fine
- here corresponds to lagging plus for
leading otherwise if you remember the
formula for alternator what to do is
replace IA by minus of IA and then you
can get the same expression
because everywhere there was plus it
will become - that we also you can
derive this expression okay the power
expression will remain same so let us
see and the other thing that you have to
remember is here the power angle Delta
is negative okay it is saying a thousand
kilo watt three phase star connected
three point three kV which means phase
voltage is 3300 by root three 24 volt 50
Hertz synchronous motor has synchronous
reactance of three point two four ohm
per phase the resistance is zero the
motor is fed from infinite bus bars at
3.3 kV which means per phase voltage is
3300 by root three its field excitation
is adjusted for unity power factor
operation which means pi is zero
calculate the maximum power that can
delete that the motor can deliver with
excitation remaining constant so for
excitation to be constant first was what
we must know is what is the excitation
of what is the excitation EMF if you
have terminal voltage to be determine
EMF you need armature current okay so
let us find out armature current at
rated value because it is saying rated
load so what is the armature current
under rated condition thousand kilo watt
by root 3 into VL into cos Phi cos Phi
is 1 because Phi is 0 or unity power
factor so here I am using P is equal to
root 3 VL il cos of Phi as you can see
the power given is in terms of real
power so this current comes out to be
one seventy four point nine five ampere
once I have this I can find out EMF
3,300
by root 3 minus 174 0.95 it has an angle
of 0 into J 3 point 2 4 okay that will
be my EMF so you have to calculate its
magnitude which will be under root 3,300
by root 3 square plus one seventy four
point nine five into three point two
four square that way we can calculate
the magnitude now I have already told
you how to derive the maximum power
output of a synchronous motor P max is
how much it will be e V by XS because
here R is zero and if you are using both
the line voltages you need not have a
factor of three but if you are using
phase voltages so it will be three e
phase we face upon axis if we have phase
voltages okay so once you have a phase
we already know V faces 3300 by root
three we can calculate the maximum power
output by substituting the values is it
fine so that is how we calculate the
maximum power or power in a synchronous
motor the next kind of problem based on
synchronous motor is for power factor
correction now for power factor
correction what happens we have a load
originally which will have a certain
real power and reactive power then we
use synchronous motor it will have
certain synchronous real power and
reactive power usually we take this as
negative they call we have to reduce the
real
if power in order to improve the power
factor so now the new apparent power is
PL plus p sm + j ql - q sm so this we
call as p2 this we call as p1 so
basically the objective is to find Q SM
it will be P 110 or Phi 1 minus P - 10
of 5 - this formula you always have to
remember and this p sm is the real power
of synchronous motor now assume that
this p sm was 0 that is synchronous
motor does not consume any real power
then what happens is this P 1 will be
same as the P 2 because PS m will be 0
so it will become P times tan Phi 1
minus tan of Phi 2 that is case if the
synchronous motor does not consume any
real power that is the only formula that
you need to remember for the case of
power factor improvement or the power
factor correction so see it is saying an
industrial plant takes a load of 5
megawatt so load power is 5 megawatt and
power factor is 0.8 lagging from this I
can form SL it will be 5 plus J into
3.75 okay based on this cos Phi I have
calculated the reactive power now it is
decided to replace one 500 kilo watt
induction motor by a synchronous motor
of same rating okay means this plant is
taking a load of 5 megawatt in that we
have certain induction motors out of
those induction motors you are replacing
one induction motor by a synchronous
motor so here
power of induction motor and
Kronus motor is same which is 500
kilowatt now both have an efficiency of
90% noise efficiency is given it is
clear to me this is the output power
which means the input power can be
derived by dividing it by efficiency so
it will be five fifty five point five
five kilowatt that is the power input of
the induction in the synchronous motor
now induction motor operates at a power
factor of 0.9 lagging but synchronous
motor at 0.8 leading which means the
reactive power of induction motor will
be five fifty five point five five into
ten off cos inverse zero point nine and
it should be positive why because it is
lagging power factor so it is 269 point
zero six seven k VAR what about reactive
power of synchronous motor by fifty five
point five five into ten or cos inverse
zero point eight and it will be negative
because it is operating at leading power
factor minus 416 point six seven kv AR
now what happens to the total load in
the total load we are removing an
induction motor but we are adding a
synchronous motor both have the same
power which means it will remain same as
pls which is five megawatt what about
reactive power we are removing induction
motor but we are adding synchronous
motor so it was three point seven five
minus two sixty nine or if you keep it
in MVA are 0.269 and you are adding a
synchronous motor my
Oh point four one six - because there is
a minus sign so it becomes three point
zero six five now we have the new real
and the reactive power we can calculate
the new power factor so you can
calculate first five which is tan
inverse QL - by pl - then cos of Phi
will come out to be zero point eight
five two six leg by leg because this
reactive power is positive okay
so that is how we can improve the power
factor of any plant from point A to
point eight five by the use of a
synchronous motor is it fine moving on
to the next type of problem till now we
have been discussing about cylindrical
rotor machine but similarly we have
another construction in synchronous
machine that is salient pole so what is
the difference in salient pole the
salient pole motors or generators do not
have a uniform air gap across the rotor
they have one direct axis and they have
one quadrature axis that direct axis
reactance is always greater than the
quadrature axis reactance due to lesser
air gap along the direct axis so here
what you need to remember is first of
all the formula for ten of size size the
internal power factor angle V sine Phi
plus minus ie X cube by V cos Phi plus
ia RA
plus is for lag - is for lead okay this
is for generator for motor what you will
do you will replace IA by - IA so what
will happen this will become minus plus
this will become minus okay that is for
motor 9 generator and motor
what you have to remember this sy is
equal to Delta plus Phi okay now in
generator lagging this one will be
positive this one will be positive this
one will be posted leading this one will
be positive positive negative okay
similarly motor lagging negative
negative positive motor leading negative
negative negative okay so remember Phi
and Delta will always have the same sign
and for generator it will be positive
for motor it will be negative five will
be positive four lagging five shall be
negative four leading is it fine so
based on that we can use direct formula
force I in terms of Delta and Phi now
let us see this problem it is a salient
pole synchronous generator with
negligible resistance and has the
following per unit parameters here we
need to calculate load angle and
excitation EMF whenever RA is zero you
have to remember excitation EMF is V cos
Delta plus ID XD now what is this ID
direct axis current is sine Delta plus
Phi or you can say ie into sine of sigh
okay so now first let us calculate load
angle that is Delta 4 Delta I need to
find out
sy so tan of Phi is V sine Phi rated
voltage 1 PU sine of Phi power factor is
pointed so sine will be 0.6 plus ia ia
is how much one PU they call
rated kv or rated power into X Q it is
0.5 upon V cause 5.8 Iara RA is 0 so it
can be neglected so it will come out to
be how much one point three seven five
okay now Y is equal to Delta plus Phi
where Phi will be thirty six point eight
six and it will be taken as positive
because it is lagging so from here if
you calculate Delta it will be seventeen
point one degree now in order to
calculate EMF first we need ID ID is one
into sine of Delta plus Phi any 17.1
plus thirty six point eight six okay so
when you calculate EMF it will be V into
cos of Delta V into cos of 17.1 plus ID
means sine of fifty three point nine six
into XD which is zero point eight so
this comes out to be one point six zero
to P so these equations you have to
remember one for the internal power
factor angle and other for the induced
EMF the next problem is again based on
salient pole machine but now it is
salient pole synchronous motor okay and
it is asking us the power angle so let
me again apply the same formula Tensai
but it is motor so I will reverse ie it
will be V sine Phi minus ia X Q by V cos
Phi minus I into R okay so what
pens now the the armature current is not
given to us so we have to find armature
current at full load so full load is 50
into 10 to the power 6 by root 3 into
thirteen point eight so this armature
current comes out to be 2 0 9 1 point 8
5 ampere now remember in this formula
whenever you put the voltage it must be
phase voltage which is thirteen thousand
eight hundred by root 3 into sine Phi
will be 0.6 minus 2 0 9 1 point 8 5 xq
will be how much 1 point 3 1 by 13,800
by root 3 into 0.8 armature resistance
is neglected so second term in
denominator is 0 so this comes out to be
zero point 3 2 now for motor we know psy
is equal to Delta plus Phi but we all
know Delta is negative for motors Phi is
negative and Phi will be positive
because it is lagging so Delta will be
how much it will be Phi plus sign okay
one minute so I will be taken as
negative so size coming out to be
positive Delta is negative for a motor
so it will be Delta minus 4 Phi minus I
Phi is how much thirty six point eight
six and Phi is how much tan inverse 0.32
which is 17.7 so it will be nineteen
points one but we know for a motor Delta
should always be negative so when you
write down in examination you will write
minus nineteen point one degree they
cause for motor Delta must always be
negative is it fine in the next one it
is again based on salient pole machine
but now it is talking about power
so for power you should always remember
the formula if armature resistance is
neglected it is e V by X D sine of Delta
plus VT square by 2 1 by X cube minus 1
by X D sine of 2 days okay now in this
case again if you use line voltages you
will get three-phase power but for phase
voltage you will get a single-phase
power okay and the next thing is there
this is called as electromagnetic power
and this power is called as reluctance
power somehow if you wish to derive the
power maximum power or condition for
maximum power do not remember any direct
formula you have to differentiate this
power with respect to Delta and equate
it to 0 that will give you the condition
for maximum power the next thing that
you have to remember is if you add any
reactants or any impedance in series
with the Machine it will equally be
added in direct X's and synchronous X's
sorry quadrature axis C in this case the
generator is connected to an infinite
bus through a reactance of point 2 P you
this is my external reactance excitation
voltage is 1.3 PU infinite bus voltage
is 1 PU directx is reactance is 0.95 but
due to external reactance it will be
1.15
similarly quadrature is 0.7 but due to
external it will be 0.9 what is the
active power supply to the
we just need to plug in the values into
this expression 1.3 infinite bus voltage
1 pu XD is 1.15 sine of Delta that is
sine of 25 minus VT square by 2 1 by X
cube minus 1 by X T sine 2 Delta is sine
of 50 degree so this will come out to be
zero point five seven PU okay so that is
how you calculate the active power
delivered by a salient pole machine
moving to the last problem based on
salient pole machine it is based on slip
test in which we determine the direct
and the quadrature axis reactance for
slip test you should always remember X D
is equal to V Max upon I mean and XQ is
remain upon I max so XD is how much
maximum voltage is 108 minimum current
is 10 to 10.8 ohm XQ is minimum voltage
96 maximum current well that is a 2 so
that is how we can calculate to
reactances of a synchronous machine that
is direct axis and quadrature axis okay
so these are the type of problems that
can be asked overall in case of a
synchronous machine
you