So, welcome to the class. In the last class,
we have seen thatwe started with axial compressor.
And for axial compressor, we have seen how
the velocity triangle is, how to find out
the work interaction in case of axial compressor,
and then how to find out stage temperature
rise and pressure ratio for given conditions
ok.
So, then we had obtained a formula in last
class about the pressure rise as P naught
3 upon P naught 1 is equal to 1 plus stage
efficiency into delta T naught s stage divided
by T naught 1 bracket raised to gamma upon
gamma minus 1. So, this is stage temperature
rise. Rather we had found out that this stage
temperature rise is U upon C a upon C p, where
stage temperature rise is U upon U upon C
p into C a U upon C p into C a into tan of
beta 1 minus tan of beta 2. So, this was our
formula for stage temperature rise.
So, we got a formula for pressure ratio as
P naught 3 upon P naught 1, which is 1 plus
efficiency stage efficiency divided by T naught
1 into U C a divided by C p into tan of beta
1 minus tan of beta 2 bracket raised to gamma
upon gamma minus 1. So, here we can see that
for a pressure rise to happen with required
magnitude, there are different factors which
affect.
First factor which affect is the axial velocity,
which is C a axial velocity. Axial velocity
if it is more, then we can see that we can
get more pressure rise. But, axial velocity
is restricted or there is a limitation on
axial velocity due to relative Mach number.
And relative Mach number is defined as relative
Mach number is defined as v 1 divided by a
1. So, this is the relative Mach number at
the blade, so entry. So, this number should
be subsonic such that we should have minimum
number of losses minimum amount of losses,
so for that C a will be restricted.
So, we basically need to find out, what is
a 1. So, basically we know that a 1 for gas
is equal to gamma R T 1, and we should know
what is T 1. So, T 1 we know formula at T
naught 1 upon T 1 is equal to 1 plus gamma
minus 1 by 2 M 1 square. So, we would know
the Mach number at the entry, we would know
the total temperature, then we can find out
T 1. Putting that T 1, we will get a 1. And
then we will get the value of relative Mach
number, which would have restriction to have
any amount of axial velocity. So, only limited
amount of axial velocity will be allowed.
Then the second factor which can help to raise
the pressure more is blade speed which is
U, but blade blade speed is also restricted
by stresses centrifugal stresses in the blade.
And this stress is directly proportional with
U square. So, there should naught be again
very high amount of blade speed. Third factor
which is affecting is this which is called
as deflection. We had seen in last class that
beta 1 minus beta 2 is called as a deflection.
And if we can have larger deflection, then
we can have larger pressure rise, but larger
deflection would create the problem of flow
separation, so there would be problem.
So, how would be the larger pressure direct
larger pressure gradient created? Larger pressure
ratio created, we can see here by creating
deflection. This was the velocity triangle,
which we had seen in earlier case in our earlier
graph, we had seen that the velocity triangle
is of this kind. So, same velocity triangle,
we are trying to draw here. So, we have C
1, and we have V 1, we have C 2, and we have
V 2. And this angle, we had called as beta
1 minus beta 2, since this is alpha 1, this
is beta 1, then this is alpha 2, and then
this is beta 2.
Now, if we want to have higher deflection,
then we should have our V velocity V 2 should
be in the different place. So, we can have
V 2 to be like this. If V 2 is like this,
then we can see that there will be more deflection.
If there is more deflection, then there will
be more pressure rise. So, obviously we can
also know that we would have more delta C
w in this case. So, there will be more work
absorb, and then there will be more pressure
rise.
So, there are three factors which would affect
the pressure ratio. One is axial velocity,
but that is restricted by relative Mach number.
One is blade speed, but that is restricted
by stress in the blade. And another is deflection,
and large deflection would lead to flow separation.
So, these are the restrictions which will
be we should be knowing which would affect
the pressure rise in a stage.
Now, we would see that there is a term which
is called as work done factor. So, work done
factor 
for the compressor, work done factor is related
with the work supplied and work received.
So, work received is always lesser than the
work supplied. So, what we would have is delta
T naught, delta T naught stage which we have
we had found out as U C a upon C p into tan
of beta 1 minus tan of beta 2. This is stage
temperature rise, but that is ideal stage
temperature rise, so we will get actual stage
temperature rise is equal to lambda into delta
T naught stage ok or which is rather theoretical.
So, this is theoretical.
And then this leads to delta T naught actual
is equal to lambda into U upon C a divided
by C p into tan of beta 1 minus tan of beta
2 ok. And this lambda actually has a range
generally in between 0.98 to 0.85, this number
is less than 1. So, actually we will get lesser
temperature rise than the theoretically which
we would have got, otherwise if there would
be that much amount of work which would be
have got stored in the compressor.
But, this lambda as variation over number
of stages, and this is mean value of lambda,
and this lambda mean decreases over number
of stages. So, as the number of stages increase,
lambda mean decreases. So, lesser and lesser
temperature rise in actual case would be observed
than the comparator compare theoretical value,
if we have higher value of number of stages.
So, this is one more term which we should
be knowing in case of axial compressor.
Now, moving on to the next topic of discussion,
which is called as degree of reaction degree
of reaction. We would define degree of reaction
as a R, suppose R is degree of reaction or
in some places degree of reaction is also
defined as lambda, so that we would term as
capital lambda maybe as in other case. Here
it is also small lambda, and then we will
define this as R or maybe capital lambda.
So, degree of reaction is defined as temperature
rise in the rotor which is static temperature
rise in the rotor, static temperature rise
in rotor divided by total temperature rise
in stage. So, this isthe definition of degree
of reaction ok or static enthalpy rise in
the rotor divided by total enthalpy rise in
the stage. So, this is what, so it is so it
is comparing the pressure rise in the rotor
divided by total complete pressure rise in
the stage. So, this is how degree of reaction
is defined.
Now, what we should be remembering is what
we have in case of axial compressor is like
this, we first have a rotor, then have stator.
So, then this is 1, this is 2, and this is
3 for us ok. So, 1 to 2 is rotor, 2 to 3 is
stage. So, we would say that this is delta
T A, so temperature rise static temperature
rise is delta T A, and this static temperature
rise or rather we can call it as delta T 1
2, and this is delta T 2 3. So, this is what
the temperature rise which would be taken
place in the rotor, and the stator ok.
And then we know that at one, we have T naught
one total temperature, at 2 we have T naught
two total temperature, at 3 we should have
T naught 3, but T naught 3 is equal to T naught.
Since in the stator, there is no work interaction
ok only, there is diffusion of the energy,
and it would rise the pressure. So, we have
this as a variation of total temperature in
a stage, this is the change in static temperature
in the stage. So, we have basically static
temperature rise in the rotor divided by total
temperature rise in the stator as the formula
for degree of reaction.
So, we have delta T 12 divided by delta T
naught 13 delta T naught 13. We should keep
it in mind that there is no total temperature
in the numerator, total temperature used only
in the denominator for the fact that total
temperature rise take place only in the rotor,
otherwise degree of reaction would have been
always one ok.
So, now we will see that we know T naught
1 is equal to T 1 plus C a square divided
by twice C p, axial velocity square divided
by twice C p, this is the temperature, total
temperature in stage 1 ok. We are assuming
that the flow is axial completely at the inlet.
So, we are having similarly T naught 2 is
equal to T 2 plus C a square upon twice C,
which is a rather if we go to 3. If we go
to 3, then we can write T naught 3 is equal
to T 3 plus C a 1 C a 3 square divided by
twice C p. C a 3 is here it is entry to the
next stage, but we know that C a 1 is equal
to C a 3.
So, in that particular case, what we would
have is delta T naught 1 3 is equal to delta
T 1 3. So, total temperature rise in the stage
is equal to static temperature rise in the
stage, so degree of reaction R can again be
written delta T 1 2 divided by delta T 1 3.
So, we will have delta T 1 2 divided by delta
T 1 2 plus delta T 2 3, so this would be the
formula for the degree of reaction.
Now, we should be knowing rather the delta
T naught 1 3 in case of axial compressor.
So, delta T naught 1 3 which is equal to delta
T 1 3 is equal to or rather can be found out
from stage work, which is C p into delta T
naught stage. And then that is for our case,
this is U C a into tan of beta 1 minus tan
of beta 2. So, we knew that delta T naught
stage is equal to U C a divided by C p into
tan of beta 1 minus tan of beta 2. So, we
know how to find out delta T naught stage,
which is delta T naught 13, and which is equal
to delta T 13.
Now, we we got the formula of R like this.
So, we found out the denominator, we need
to find out the numerator. So, numerator can
be found out by considering the energy equation
between the rotor, and the stator such that
we know that this is rotor, this is stator
this is 1, 2, and 3. Let us apply energy equation
between steady flow energy equation between
point station 1 and station 2. We can write
down that h 1 plus C 1 square C 1 square by
2 plus q is equal to h 2 plus C 2 square by
2 plus w, but this is an adiabatic process,
so q is not there. Further this w is received
by the system, so we have h 1 plus C 1 square
by 2 is equal to h 2 plus C 2 square by 2
minus w stage.
So, we would have h 2 minus h 1 is equal to
rather, we would have w stage h 2 minus h
1 is equal to C 1 square by 2 minus C 2 square
by 2 plus w s ok. So, we have C p into delta
T 1 2 is equal to w stage plus or we can also
write it as minus C 2 square half here, and
then we can write it as C 2 square minus C
1 square. So, this is the formula, till now
what we have got. So, here we got the numerator,
here we got the denominator.
So, now R is equal to delta T 1 2 which is
equal to delta T naught 1 3 is equal to w
s minus half. So, we can multiply both the
sides by C p, so we will get Cp into delta
T naught 1 2 divided by C p into delta T naught
1 3, so this is here. So, we can write it
as w s minus half C 2 square minus C 1 square,
but this factor is w s.
So, we will have w s. So, what we get is one
minus half 1 minus 2 w s into C 2 square minus
C 1 square, but we can know that w s is equal
to this. So, we can use that and say that
1 minus C 2 square minus C 1 square divided
by U C a into 2 into tan of beta 1 minus tan
of beta 2 ok. So, this is the formula till
time for R.
Now, we have to find out what is C a, what
is C 1. We will go back to the velocity triangle,
what we had drawn. So, in this velocity triangle
if we see what is C 1, and what is C 2. So,
C 1 is C a sec alpha 1, and C 2 is equal to
C a sec alpha 2. So, from the velocity triangle,
we know that C 1 is equal to C 1 is equal
to R is equal to 1 minus C a square sec square
alpha 2 minus C a square sec square alpha
1 divided by 2 into U into C a divided by
tan beta 1 minus tan beta 2.
So, we have R is equal to 1 minus, then this
formula can be furtherreplaced sec square
alpha in tan square alpha, and then we can
get it again replaced using the relation of
sec and tan, we can get it as C a upon twice
U into tan square alpha 2 minus tan square
alpha 1 divided by tan beta 1 minus tan beta
2.
But, in earlier sections, what we had seen
from the velocity triangle that there is a
relation which says that tan alpha 1 plus
tan beta 1 is equal to tan alpha 2 plus tan
beta 2. So, we can write down that relation
which says that tan alpha 1 plus tan beta
1 is equal to tan alpha 2 plus tan beta 2.
So, we get tan of beta 1 minus tan of beta
2 is equal to tan of alpha 2 minus tan of
alpha 1.
So, we can write down it over here, and we
can say that R is equal to 1 minus C a upon
twice U into tan square alpha 2 minus tan
square alpha 1 divided by tan of alpha 2 minus
tan of alpha 1. So, we get 1 minus C a upon
twice U into tan of alpha 2 plus tan of alpha
1. So, this is our formula for R, further
this formula can also be reduced in terms
of betas. So, this is how we can find out
R for the for the compressor, which is the
degree of reaction for a compressor.
This formula can further be utilized to understand,
how the R is going to vary with respect to
the blade height. So, here we know that our
formula for degree of reaction is this. And
here we have C a tan alpha 2 and C a tan alpha
1 ok, we will do it. After one more derivation
which is called as the free vortex theory
ok. So, till time what we were trying to do
was for the two dimensional case, where we
had seen that we were working with the velocity
triangle on a plane which is parallel to the
axis of the compressor. So, we were having
this analysis as two dimensional.
But, we can do further a glimpse of three
dimensional analysis, where we will use a
theory which is called as free vertex condition
or free vortex theory ok. So, we will say
that this is free vortex condition. And we
will derive this condition, as per this condition.
Now, this condition is applied for the compressor
axial compressor as well as for the axial
turbine. So, this part of discussion is common
between axial compressor and axial turbine.
So, what do we mean by free vortex condition,
we using this condition we are trying to find
out the relation between the triangle velocity
triangles, whatever we would have drawn in
a plane at a condition. So, this is the blade
as we know, this is a blade, and this is a
stator, this is rotor, this is stator. We
know that we are working at this height of
the blade. So, if we are working at this height
of the blade, then we are drawing the velocity
triangle here and here, then knowing the velocity
triangle at these locations how to find out
the velocity triangle at other locations or
the of the rotor. So, this is what our objective
through this free vortex condition is.
So, for that we practically are considering
this radial variation. And since this is an
axial compressor or more intention was that
it has only axial velocity. So, it is not
having any radial velocity, but with this
non-radial velocity being accounted, we will
try to derive a constraint. So, for that we
will first consider, there exist a radial
equilibrium and radial equilibrium leads to
1 upon rho dp by dr dp by dr is equal to C
w square by r. So, centrifugal force is balanced
by the pressure gradient in the r direction
for the compressor at a location ok. So, this
is our assumption we could have derived that,
but this can be taken to start with as a [vocalized-]
assumption for radial equilibrium to derive
the free vortex condition.
Now, let us consider h naught is the total
enthalpy at a station which is equal to h
plus C square by 2, but we know that C can
be decomposed into two components which is
C a square plus C w square, further we can
differentiate this h naught, and then we get
dh naught by dr is equal to dh by dr plus
C a into d C a by d r plus C w into d C w
by d r plus C w into dcw by dr, this is equation
number 1.
Now, we know our thermodynamic relation which
is combined first law and second law as per
that Tds is equal to dh minus vdp, so that
can be written as dh minus dp by rho, then
we can write here as dh is equal to Tds plus
dp by rho. Now, here we can see that there
is a term dh by dr. So, we can differentiate
this equation, then we can get dh by dr, so
T into ds by dr plus ds into dT by dr plus
1 upon rho into dp by dr minus, we would have
1 upon rho square into dp into d rho by dr,
this would be the term.
But, then this term is a second order, further
this term is also second order will for the
sake of convenience, we will neglect them
being their magnitudes to be very small. So,
we get dh by dr is equal to T into ds by dr
plus 1 upon rho dp by dr, but we know that
1 upon rho dp by dr is equal to C w square
by r.
So, what we get after this putting equation
2 in equation 1, we get dh naught by d r is
equal to T into ds by dr plus C w square by
r plus Ca into d C a by d r plus C w into
d C w by d r. Now, we know that when we are
considering centrifugal compressor our C a
is taken as constant with respect to r. So,
this term is negligible in magnitude or 0
for us.
Further we are considering subsonic compressors.
So, there is no entropy variation in r direction.
So, these two terms would get cancelled. Further
we are also considering that the dh naught
by dr is negligible which is saying that we
are having equal amount of work to betaken
by the compressor at different heights. So,
considering these terms to be negligible in
the respect of other two terms, we get C w
square upon r is equal to minus C w into dCw
by dr, so which says that minus dr by r is
equal to d C w by C w.
After integration, this expression leads to
the fact that C w into r is equal to constant.
So, if we know C w at a height, then we can
find out C w at some other height. So, this
is how this formula helps us to work out for
finding out the velocity triangle known at
one point to be finding out velocity triangle
to be evaluated or estimated at the other
height.
This formula can be having thought to having
relation with the formula for what we have
derived for the degree of reaction. So, degree
of reaction formula, what recently we derived
has 1 minus C a upon 2 U into tan alpha 2
plus tan alpha one. So, r is equal to 1 minus
U into 2 C a into tan of alpha 2 plus tan
of alpha 1, so it is R is equal to R is equal
to C a by U C a by U. So, we have 1 minus
twice U into C a tan alpha 1 plus tan alpha
2 into C a.
If we see velocity triangle, so C a tan alpha
1, C a tan alpha 1 would be C w 1. So, R is
equal to 1 minus twice U, and this is C w
1 plus C w 2. And now we will multiply both
the numerator and denominator by R. So, we
will get 1 minus R upon twice U into C w 1
plus C w 2, which would lead to 1 minus r
C w 1 plus r C w 2 divided by r into U. Then
we know that r 1, and we can write it as r
2, since r 1 and r 2 are same, but this is
a constant. Since this is a constant.
We get R is equal to 1 minus 1 upon r into
U c that this constant is K 1, but we know
that U is equal to pi D N by 60. So, U is
equal to pi r N into 2 by 60. So, it turns
out that R is 1 minus K 2 upon r square. So,
this is the relation of degree of reaction
with respect to R. So, if R is increased,
degree of reaction will increase. If R is
decreased, degree of reaction will decrease.
So, this is how we when do our 2D analysis
for the velocity using the velocity triangle
at mid bed height, free vortex theory helps
us to translate that analysis with the constraint
that we are having negligible change in axial
velocity, we are having negligible change
in in the total enthalpy in the radial direction.
So, we can take that C w into R is equal to
constant. So, these are the topics which we
will be needing for understanding the axial
compression. And here we end the topic on
axial compressor. We will meet for the next
topic in the next class.
Thank you .
