 good morning and welcome you all to this session of the course. Today we will be discussing the axial flow compressor last class we have more or less completed a brief discussion on the centrifugal compressors now we will be discussing the axial flow compressor. The basic principle is mostly the same as that of a centrifugal compressor what is the basic principle that by absorbing the mechanical energy from outside the working fluid which is usually here gains its static pressure that is the basic purpose of a compressor whether it is centrifugal or axial flow compressor. An axial flow compressor like a centrifugal compressor also consists of a rotor and a stator rotor is the rotating part of the compressor where the mechanical energy is being supplied and it consists of a number of blades which rotates and the mechanical energy is supplied in the form of an external torque under which there is a rotational motion of the rotating blades that the blades which rotate and the air acquires the energy and ultimately what happens its velocity and static pressure is increased and the stator part of the compressor which is known as diffuser the function of which is that here we gain the static pressure from the velocity of the air by allowing the air to have a dislarating flow in this passage this known as diffusion process the diffusion takes place where we get more static pressure by virtue of the or at the expense of kinetic energy and finally at the outlet we get air at high pressure but at low velocity so the basic principle is same but the major difference between the axial and the centrifugal compressor is that the flow takes place in the axial direction whereas in a centrifugal compressor you have seen the flow takes place in the radially outward direction so the inlet and outlet of the flow varies in the radial location and varies in the peripheral speed of the rotor but here what happens the flow takes place in the axial direction so therefore the inlet and outlet of the flow takes place at a particular radial location and all such radial locations the flow inlet is there and flow outlet is there that will be explained when I will explain through the diagram. So, the major difference is the entire flow takes place in an axial direction and there are different number of stages each stage consist of a rotor blade or the rotor and a stator consisting of stator blades and this way a stage is considered that a rotor and a stator and number of stages are there. This is the overall structure of an axial flow compressor that differs from a centrifugal flow compressor which is a radially outward flow type it is an axial flow type through number of stages. At the same time you should know that this axial flow compressor the advantage over centrifugal compressor is that it can handle a large amount of air the flow rate is more as compared to the centrifugal compression it can run more efficiently with high flow. So, therefore, where the high flow is required the axial compressor is more axial compressor is important high flow and also the weight is low. Now, let me explain the overall structure of an axial flow compressor you see that this is an axial flow compressor axial flow compressor axial flow compressor. Now, you see an axial flow compressor how does it look here there are two types of axial flow compressor one is the drum type this is the drum type I tell you drum type what is the difference I tell you this is the disc type. Now, basic structure consist like that there is either a drum rotating drum this is the shaft or a disc on which this rotor blades are being mounted this is the rotor blades and this rotor blades this is the drum type is mounted on the rotating drum and with the rotation of the drum or disc in a disc type this is the rotor blades this is the rotor blades. So, rotor blades rotate so these rotor blades and number of blades are there along the periphery of the drum. So, at any axial location you see there are number of blades if you see this view front view you see one blade there are number of blades in the periphery which are being attached to either drum or disc known as disc type and there is a stationary casing this is the stationary casing I write this is the stationary casing this is stationary casing here also this is stationary casing stationary casing and this stator blades are screwed or attached to this stationary casing these are the stator blades and one rotor and one stator one rotor and one stator comprises one stage. So, here also the stator blades are fixed to the casing and there is a inlet row of guide van upstream of the first rotor this is actually not this is stator this is also a stator static part attached to that and these are known as inlet guide vanes which first direct the air properly in the axial direction. So, this is the flow direction this is the air flow like that flow direction. So, this is the flow direction and this is the direction of the axis axis this is the direction of the axis the flow takes place in the axial direction first it comes to the inlet guide vanes then through rotor and stator then rotor and stator a number of rotor blades number of stator blades and number of rotor blades number of stator blades and a combination of rotor and stator comprises the stage this is an overall view of the axial flow compressor the air enters in this way the air enters in this way. So, therefore, air flows through this annulus area formed by the blade passages. So, this is an overall diagram of an axial flow compressor air goes out. So, axial flow compressors there are other things that I will tell you that while designing this annulus area it is made a convergent type that means the annulus area goes on decreasing in the direction of the flow why this is because you know from continuity mass flow rate is written as mass flow rate can be written as rho times the flow area times the velocity of flow v f and this we consider an average flow velocity over a cross section and this is the annulus area the cross section the area provided to the flow and this is the average density at that section. Now, as the flow takes place here by the action of the rotor and stator the pressure increases this is the increasing direction of the pressure. So, density increases so for a given mass flow rate because at steady state from continuity the mass flow rate will be same throughout the machine. So, if we want to make the axial flow velocity constant we have to decrease the cross sectional area and the main motto is to make the axial flow velocity constant despite the increase of rho for which we have to decrease the annulus flow area same for drum type or a disc type this is one of the major considerations in the geometrical design of the compressors the blade height and other things are will come afterwards, but this is one of the main considerations. So, this gives you a overall picture of an axial flow compressor how does it look like what is the basic direction of flow the flow takes place in the axial direction. Now, if you see the number of vanes now before that we consider now if you see the from the top view then you will see a particular rho is like this including the guide vanes it looks like this you see when you look from the top over the periphery then you will see that this is the direction that is these are the rotor this is the rotor vane this is the diffuser and this is the inlet guide vanes rather I think you can see this way this will be the way actually yes. So, therefore, these are the number of blades that you can see or better you see this way no this way it is better. So, that this goes from this in this direction why they have made a flow direction like that I do not understand this direction will be here I think the flow this is a peripheral direction the number of blades are there. So, this will go like this and will come like this I think this is the direction not this is the direction. So, the flow takes place like that there are inlet guide vanes which directs the flow in such a way ultimately it goes through the rotor series of rotor blades and the diffuser this is the diffuser where the velocity is decelerated and the pressure is increased if you see this diagram you see that pressure in the inlet guide vanes decreases slightly because of the skin friction because the inlet vanes there is no energy interactions this is the static vane. So, pressure lot takes place because of the friction then in the rotor the energy interaction takes place it moves. So, therefore, rotor imparts mechanical energy to the fluid by virtue of which its pressure its velocity and similarly if you seeing the velocity velocity should remain almost constant, but this inlet guide vanes cross sectional area is slightly converging type. So, that the velocity slightly increases. So, this is the velocity part I just do it by rate velocity then what happens by virtue of this mechanical energy given to it the both velocity and pressure increases this is the increase in the velocity this is an increase in the velocity in the rotor because it gains energy there and this is the increase in the pressure and this is the static pressure the increase in static pressure takes place because there is an divergence in the area this flow passage area is made in such a way there is an increase in the static pressure while it flows through that this is being reflected or manifested by a change in the relative velocity that I will tell after what then it comes to a series of stator blades known as diffuser what happens in the diffuser we get a rise in pressure that is known as typical diffusion process in respect to this deceleration of flow that means the flow velocity is reduced. So, this is a typical pressure velocity diagram this is the pressure this is the velocity diagram while flowing through the rotor and diffuser blades and upstream this we consider the first row of rotor blade so that upstream of that there are inlet guide vanes now if you consider you come to again this diagram that flow takes place at all radial locations now if we consider the flow velocity v f is uniform over all radial locations and if we consider the flow at any mean height that means at a mean height of the blade at a mean radius from the centre here the flow is basically two dimensional the flow has got an axial component and a tangential component the radial component is relatively less. So, if the radial component is less we can consider the two components of flow mainly in the axial another is tangential which is perpendicular to this plane of the figure and at the same time if you consider the v f at any cross section along the blade height at different radial location if you measure from the centre line is constant that means the uniform velocity distribution which is also constant in the axial direction for which the annulus area is reduced that is different, but radially if we consider it is uniform and two component of velocity then it is almost a two dimensional flow this is a two dimensional representation of a three dimensional flow. So, with this representation now you can have a look of the velocity triangle now if you see that the same thing that we already shown that the a blade passage formed by the rotor and the stator you see this rotor and the stator that means this part if we draw like this that means if we see from the top a this side no oh sorry sorry now it is now it if we see from the top the two rotors earlier things were visible no. So, two rotor blades and two stator blades just a representative figure I show you this is what a representative blade passage found by two rotor blades and two stator blades of a stage now this is a stage this is a stage this is a stage now what happens the stage at the inlet to the stage the air comes with some velocity v 1 with an angle alpha 1 which is nothing but the velocity coming out of the stator of the previous stage with some angle alpha 1 and this angle is made in the axial direction this is the axial direction you understand this is the axial direction. So, this is the well axial direction this is the axial direction axial direction axial direction now what happens the rotor blade shape is like that and if you see the velocity triangle you see the velocity triangle is that here we can write that v r 1 is the relative velocity as you know that is v 1 minus oh not visible that is the problem with this that they have not this is not actually made in this proper ray v r 1 is v 1 minus u. Now, another difference is this from the centrifugal compressor here the if you see this again that the flow takes place axially if we this diagram is drawn at a section taken at the mean radius mean height of the blade. So, therefore, here what happens the inlet and outlet takes place at the same radius we can take it at different radial locations. So, therefore, when we see the picture from the top we see that the inlet and outlet of the air flow takes place at a given radius. So, the peripheral speed of the rotor is same depending upon the radius at that point. So, therefore, u remains the same you see the inlet velocity triangle is like that this is the absolute velocity with which the air strikes the rotor blade with an angle alpha 1 with the axial direction. Then if you make the vector diagram vector triangle diagram that velocity triangle then this is your v r 1 this is the v r 1 I think you can see this is v 1 this is v r 1 and this is the u and this is the v w 1 this part this is the wheeling component or tangential component of velocity v 1 this is your velocity triangle. And this velocity because of smooth flow without incident loss this beta 1 should glide that means should match the angle of the blade at the inlet and this is the beta 1. So, beta 1 is the relative velocity angle which is the angle of the blade at the inlet. Similarly, if you see the outlet diagram the outlet is made like this so that the flow area increases in the direction of flow this is the flow direction. So, this is the flow velocity v f 1. So, at the inlet now at the outlet if you see the velocity diagram this is the relative velocity v r 2 this is v 2 this is the same u and this is v w 2 this is alpha 2 and beta 2. Now, beta 2 is less than beta 1 this beta 2 is less than beta 1 fluid is directed more towards axial direction this happens because of the camber of the blade in a way that the annulus area increases the annulus area that means this area sorry increases sorry decreases no sorry annulus area increases because of this increase in the velocity v r 2. So, what happens here you see that v 1 and v 2 that another important thing is that v r 2 then v 2 v 2 is more than v 1 v 2 has to be more than v 1 because in rotor the fluid gains the energy, but v r 2 v r 2 is less than v r 1 why this is because there is an increasing static pressure if v r 2 has to be less than v r 1. So, this area has to be more. So, make the annulus area more the camber of the blade has to be such that beta 2 is less than beta 1 these are the important considerations of this velocity triangle. Now, when it comes to the stator blades then what happens this is the absolute velocity v 2 with it comes makes an angle alpha 2 in the axial direction. So, stator blade does not move. So, there is no velocity triangle simply the velocity direction is changed as far as the camber of the stator blade or the curvature of the stator blade. Here also the stator blade velocity is change in such a way that alpha 3 is reduced from alpha 2 it is directed more towards axial direction and the design is made such a way that alpha 3 becomes again equals to alpha 1. So, that it can smoothly glide or enter to the rotor blade of the next stage. So, we make this alpha 3 is equal to alpha 1 these are the important considerations. So, this way you can draw the velocity triangle at inlet and the velocity triangle at outlet of the rotor blades and this is the velocity how does it change in the stator blade. Now, we can write certain formula from this, but problem is that this cannot be shown here at the same diagram. Let me adjust it if you can see that thing this is the state. Now, if you can say if you can see now oh this blades you may not you may discard. So, that you can go little here. So, now if you see this triangle what we can write from the now v f 1 as I have told in the design is made v f 2 and v f that means this axial flow velocity. Now, from the geometry of this tan alpha 1 is what this base divided by v f 1 tan beta 1 is this base divided by v f 1. So, if you add this two things then we get tan alpha 1 plus tan beta 1 is equal to this plus this is the u u by v f because v f 1 is v f 2 is v f. So, this is I am writing v f similarly from the outlet velocity triangle tan alpha 2 if you make the tan alpha 2 from the simple geometry tan beta 2 is equal to u by v f u by v f. Now, if you make this two equal then we can write tan alpha 1 plus tan beta 1 is equal to tan alpha 2 plus tan alpha 2 plus tan alpha 2 plus tan alpha 2 plus tan alpha 2 tan beta 2 that we can write or we can write tan alpha 1 minus tan alpha 2 is equal to tan beta 2 minus tan beta 1 this is one very important relationship this two this relationship and this is a geometrical relationship this two relationship we can get from the two velocity triangles. Now, if we come to the work here itself I can write now work if you write here now work per unit mass w by m work per unit mass is nothing, but what is work per unit mass we know from the Euler's equation that v w 2 u 2 that is v w 1 u 1 all right here u 2 is u 1 and that is equal to u. So, therefore, it becomes equal to v w 2 minus v w 1 into u why because u 1 is equal to u 2 is equal to u now we see that if this is true then we can write w is equal to what is v w 1 minus v w 2 let me see again now v w 1 minus v w 2 is what is v w 1 minus v w 2 from this diagram now v w 1 can be written as this one v w 1 v w 2 v w f 1 tan alpha 1 similarly v w 2 is v f 2 tan alpha 2 v f 1 v f 2 is v f. So, therefore, this equals to v f tan alpha 1 v w 1 and v w 2 v f tan alpha 2. So, if you follow this velocity triangle then you can write this is equal to u v f into v w 2 that is tan alpha 2 will come first minus tan alpha 1 again from the equality as I have done already that tan alpha 2 minus tan alpha 1 is tan beta 1 minus beta 2. So, we can write u v f tan of beta 1 minus tan of beta 2 where from because we had this thing are earlier shown that that earlier we shown this thing that actually this is a problem here this is tan alpha 2 tan alpha 1 minus tan alpha 2 is tan beta 2 minus tan beta 1. So, therefore, tan beta 1 minus tan beta 2 tan although this is the work done per unit mass sorry or energy added per unit mass whatever way you can write it is your concept energy added. Now, if you consider that the change in the stagnation temperature which we called is the total temperature I give the nomenclature delta t s that means change in the stagnation temperature of the compressor total change in that c p is the total energy added to the air on the working fluid per unit mass that will equate this one v f tan alpha 2 minus tan alpha 1 I can write tan beta 1 minus beta 2 tan r does not matter either of this two I can write that two are equal. So, therefore, the increase in this stagnation temperature is given by this formula which is very important that increase in the tan alpha 2 minus tan alpha 1 divided by c p. Now, what happens is that the actual stagnation temperature rise is less than this and this is taken care of by a coefficient lambda by a coefficient lambda first of all let me write that which is less than lambda this is lambda less than 1 lambda less than 1 by this coefficient lambda which is known as the work done factor this is the lambda which is known as the work done factor lambda is work done factor work done factor now this is mathematical what is the physical meaning of that and this lambda equation that means that the theoretically the energy per unit mass or work done per unit mass which is calculated is not actually imparted to the liquid sorry to the fluid to the air here a less amount is being imparted because of this lambda factor which is less than 1 which is known as work done factor this is because of the fact that we assumed at the beginning this axial flow is uniform over the entire blade height this is not so and that creates a problem now let us see that thing that if we consider the this way if we consider the axial velocity axial velocity axial velocity axial velocity that is v f and if we consider this is the radial direction this is the radial direction r and if we consider these as the blade height rotor blade height this is the this is the blade height this is the height of rotor height of rotor blade. Let us consider the first stage. Now, according to our two dimensional assumptions, we have considered that this is the dotted line I am showing. This is the axial velocity distribution. This is the axial velocity distribution over the blade height. This is the blade height. Here, I cannot write because the space is not there. So, here it could have written the ordinate is height of the, that is the radial direction and this is the height of the rotor blade. I just make a hatch like that for your understanding. So, uniform distribution, but what happens in practice because of the three dimensional effect, the velocity distribution is not uniform. The flow velocity distribution, this is the axial flow velocity Vf. So, distribution becomes like this. Some peak key at this centre and then again it reduces. So, this becomes the distribution of the axial velocity. Actual distribution becomes like that instead of this uniform one. If you go to further stage, for example, a fourth stage, number of stages are there or fifth stage. Some blade height also will change. As you go on that, there is a decrease in the area. However, I show you with the same height fourth stage for your understanding. So, if this is the Vf, this is the blade height, the same diagram, but it looks like that this is the uniform one, the axial velocity. This becomes more peaky little this side and after certain stage, this velocity distribution which is skewed with a peak becomes almost same like the fully developed flow type. But because of this distribution not being uniform, what happens? The work done per unit mass is changed which is taken care of by the lambda. Now, after this I will show you that you can have some idea that if I express the work done per unit mass W by Vf. Now, W per unit mass is now we can write W per unit mass as U Vf into tan beta 1 minus as I have done earlier tan beta 2. This I can write. Now, we know that tan alpha 1 plus tan alpha 2 is tan beta 1 plus tan beta 2 that we know or we can replace tan beta 1. Another result I know that what is this? That tan alpha 1 again we know that tan alpha 1 plus tan beta 1 is U by Vf that I know. Now, I can replace tan beta 1 in terms of tan alpha 1 and can get this expression U into now Vf. Now, Vf tan beta 1 I can write tan U by Vf minus tan alpha 1 minus tan beta 2. This is the way I can write. Why you will understand now? If you take U here then you can write U minus this Vf cancels Vf into tan. Now, this tan alpha 1 and tan beta 2 this alpha 1 and beta 2 this is fixed. This is the outlet angle of the rotor blades and this is the alpha 1 that means this is the outlet angle of the stator blades or the angle of the absolute velocity at the inlet of the that is the inlet of the rotor blade. So, if these are fixed then it is seen that for a given peripheral speed at any height the work done per unit mass depends upon Vf. So, when there is a reduction in the flow velocity work done per unit mass is increased when there is an increase in flow velocity it is decreased. So, therefore, with this if you compare this you will see that it changes with the radial locations. Here in this radial locations here the work done per unit mass is decreased while the work done per unit mass is increased because of a reduction in Vf near the tip and the root. But the overall effect of this reduction of the work done per unit mass at all radial locations where it is more than the axial velocity because we have made the calculation based on the uniform axial velocity is counter ways the gain at the root and the tip because of the reduction in flow velocity Vf by this formula. So, what happens finally the work done per unit mass based on the uniform flow velocity is being reduced by a factor known as work done factor clear. Now, this work done factor ultimately decreases with increasing number of stages this work done factor a typical graph is like that usually this value lies from 1.8 to 1 and with number of stages number of stages let I start with stages 2, 4, a 6, 8, 10, 12 like this the work done factor goes like that this for your idea this is the work done factor lambda. So, more number of stages lambda is less and less number of stages lambda is high. Now, we come to the pressure rise pressure rise is very simple again in this page itself I will do now if we consider the two pressures just like we did earlier that p 3 t and p 1 t if we consider three at our this thing then what happens that this is our sorry this is dotted this should be this is our isentropic one this is our actual one that means if this is t 1 t this is t 3 t dash and this is t 3 t. So, our delta t s t is t 3 t minus t 1 t, but our pressure rise p 3 t by p 1 t which usually we write here is the r s here s suffix is given per stage per stage the pressure rise is the pressure ratio is p 3 t not rise is p 1 t that is related to this t 3 t dash by t 1 t to the power gamma by gamma minus 1 and again we know that again we know that eta c we know that the isentropic efficiency again it is stage efficiency which is defined as t 3 t dash minus same thing I am repeating again and again t 3 t minus t 1 t which gives that t 3 t dash by t 1 t from here is 1 plus eta s into t 3 t minus t 1 t divided by t 1 t. So, therefore, we have to place it here so that we get r s is equal to same expression 1 plus eta s and this is defined as delta t s t so that it is expressed in term of this in terms of this thing t 1 t to the power gamma by gamma minus 1. Now, with this I tell you the overall principle of action of a stage of an axial flow compression what happens in the rotor and the stator now we come to a thing which is known as degree of reaction which is very important degree of reaction what is mean by degree of reaction now try to understand one thing again I will repeat the thing which I told you earlier in the fluid machines class when I discuss the hydraulic machines the machines using water now one has to understand that any fluid machines has a rotor and stator whether it is a turbine or it is a compressor or pump. So, the basic purpose of the stator or the diffuser in a pump on compressor is to change the velocity to static pressure, but the question comes whether in a rotor the static pressure will change or not first of all try to understand in terms of pressure static pressure will change or not. So, static pressure will change in the rotor depending upon the rotor construction if it is a radial flow machines machine type rather I will tell machine type there is a radial flow machine automatically the pressure changes because of the centrifugal action of the centrifugal head that is the centrifugal force when the radial location changes the peripheral speed is changing. So, therefore this is manifested in terms of the increase in a static pressure I explained so many times that a radial pressure grade is imposed when the fluid has a tangential velocity and it changes its radial location, but in an axial flow machines when there is no change in the tangential flow from inlet to outlet not necessarily there will be a change in the static pressure the change in the static pressure will depend upon the rotor design and construction. That means if we have to change the flow area if you change the velocity relative to the rotor inflow course of flow to the rotor passage then only there will be a change in the static pressure and that is the measure of reaction in a reaction turbine that whether there is a change in the static pressure in the rotor itself or not. For example, in impulse turbine if you remember a water turbine that pelton turbine the water jet is striking the rotor that is the pelton wheel at atmospheric pressure throughout pressure is same there is no change in the pressure this is known as impulse turbine whereas in Francis turbine when it goes through the runner blade there is a change in the pressure similar in the centrifugal pump always there will be a change in pressure in the impeller this is because of the radial flow in a rotating or tangential flow field similar is the case of centrifugal compressor, but in axial compressor the question comes whether the flow passage changes in course of flow through the rotor so that the relative velocity changes or not. I told you that if the relative velocity changes in a sense that if vr2 is less than vr1 then we can consider that there is we think that there is a change in the static pressure and when there is a change in the static pressure there will be a change in the static temperature also. So, therefore whether there is a change in the static pressure or not that will call the machine is reaction type or not usually there is a change in static pressure and the static temperature while flowing through the rotor vr2 is made less than vr1 and therefore the question of reaction comes and the degree of reaction in this context is defined as this way the in terms of enthalpy it is defined the change in enthalpy let us consider per unit mass in the rotor divided by the change in the stage delta h rotor plus delta h stator this is known as delta h stage. Since the enthalpy change for an ideal gas is equal to the cp into change in the temperature and since cp is constant here we are considering the ideal gas it is independent of the temperature we can cancel it out and reality also the cp does not vary much with the range of temperature this can be written in terms of delta t the static temperature change in the rotor divided by static temperature change in the stage that means in rotor plus delta stator now this is the degree of reaction when there is no change in static temperature or static pressure degree of reaction is 0. So, degree of reaction is 0 for a reaction machine there will be always a degree of reaction since delta t rotor has got a value more than 0. So, this is the measure of the extent by which the total the fraction by which the total change that is total change in the static temperature of the stage is taking place to the rotor now with this definition let us now see the how we can find out an expression for this now let us consider now again the definition since we give this s for stage and a rotor and stator this things we will not write. So, simply we tell that a stands for stator for our convenience and b stands for rotor and we write this definition as delta t this is a for example you write sorry this you write rotor this will be better and this you write stator then you write delta a delta t a plus delta t b now you see that work done per unit mass is nothing, but c p into delta t stagnation. Now, if we make v 1 is v 3 now you see this diagram we told that alpha 3 is alpha 1, but in our design it v 3 is made v 1 now see that it approaches with some v 1 from the earlier stage then while it passes through the rotor it gains energy and is v is increased v 2 is greater than v 1 then this v 2 is again by the diffusion process that means this is decelerated to get a rise in static pressure is that means v 3 is less than v 2 is less than that and it is also displaced more towards axial direction this deceleration is made in such a way v 3 comes back again to the original v 1 if we make a design like that that the absolute velocity at the discharge of the or at the outlet of one stage becomes equal to that of its inlet velocity at this stage you understand. So, that we can write v 3 is v 1 in that case we can write delta that means what is t total t stagnation temperature at 3 for example t 3 t now I write the total temperature t here you see then t 1 t t 1 t is t 1 static plus v 1 square by 2 c p similarly here you see t 3 t the total temperature is t 3 plus v 3 square by 2 c p. So, if v 1 is v 3 that means the dynamic equivalent temperature that means the velocity equivalent temperature are same that means the difference in the total or stagnation temperature is difference in their static temperature. So, that delta t s t that is the stagnation temperature is delta t static and here s is written that is per stage that means delta t stage that means this is delta t stage that is static temperature per stage is equal to the stagnation temperature that means this can be written as delta t s and this is nothing but the work done formula that means if you remember that this what is this formula u v f into tan alpha 2 minus tan alpha 1. So, therefore c p delta t s is given by this correct bracket is there there will be a bracket now this is correct now how to find out this for delta t a so therefore now I find out the value for delta t a now rotor the energy is actually given in the rotor. So, w by m is written in terms of the total one that means u v f again I am writing tan alpha 2 minus for your convenience tan alpha 1 and this becomes is equal to c p delta t s static temperature rise per stage because the static temperature rise equal to the stagnation temperature now if I write this w by m that is u v f into tan alpha 2 minus tan alpha 1 in terms of the static temperature change of the stator I am not permitted because stator static temperature so entire energy use not to increase static temperature but at the same time to increase its velocity v 2 square by v 1 square. So, an energy balance in the rotor gives that rotor energy takes and the air static temperature is increased this is the energy change due to the increase in static temperature that is the static enthalpy rise plus the kinetic energy change. So, two summation of this two from a steady flow energy equation is the energy input per unit mass basis so this is that so therefore, we can write c p delta t a is equal to u v f tan alpha 2 minus tan alpha 1 minus half v 2 square minus v 1 square now you see v 2 v 1 from the velocity triangle now from the velocity triangle if you recall that v 1 is what if you see this triangle so this is alpha 1 in terms of alpha 1 the cosine of alpha 1 is which one v f by v 1. So, v 1 is v f by cosine alpha 1 and v 2 is v f v f is same as I have told earlier by cosine alpha 2 that means this is v f by cosine alpha 1 this is v f by cosine alpha 2. So, with this thing I can write this equal to u v f tan alpha 2 minus tan alpha 1 minus half this is v f square I take v f common v f square into 1 by cosine that means sec sec square alpha 2 minus sec square alpha 1 I think I am correct sec square alpha 2 now you know again from the trigonometric relationship that sec square alpha minus tan square alpha equal to 1 that relationship and the school level you know. So, therefore, if you use that this becomes u v f tan alpha 2 minus tan alpha 1 minus half v f square then this is also tan square alpha 2 minus tan square alpha 1 now I use this expression for c p delta t s that is stage and this expression delta c p delta t a for the rotor this stage means delta t a plus delta t b then I can write the omega which is equal to delta t a by delta t s as c p c p cancels. So, therefore, this becomes equal to u v f tan of alpha 2 minus tan of alpha 1 minus half v f square tan square alpha 2 minus tan square alpha 1 divided by the total stage which we derived earlier u v f tan of alpha 2 minus tan of alpha 1 this can be written as 1 minus then v f v f will cancel that is v f by 2 u and this will be tan of alpha 2 plus tan of alpha 1 now this is the definition now this tan of alpha 2 plus tan of alpha 1. So, this is the definition of final definition of the 1, but now if we change it to tan beta 1 then what we call that we already have found out this expression if we see this expression at the beginning which we found out that tan alpha 1 tan alpha 2 minus tan alpha 1 is equal to u v f that what we did just a minute u by v f by u is what is that one the very beginning which we did if you see that thing at the beginning u by v f u v f is tan alpha 2 minus tan alpha 1 and tan alpha 2 minus tan beta 1. So, if tan alpha 2 plus tan alpha 1 if you remember this u by v f again I am writing tan alpha 1 plus tan beta 1 from the geometry that is from this one we initially write this one if you add this 2 you twice u by v f is tan alpha 1 plus tan alpha 2 plus tan beta 1 plus tan beta 2. So, again I am going to write that for your benefit tan alpha 1 tan beta 2. So, therefore we write twice u by v f from here is tan of alpha 1 plus tan of alpha 2 plus tan of beta 1 plus tan of beta 2. Now, if tan alpha 1 plus tan alpha 2 is substituted or is just eliminated in terms of tan beta 1 tan beta 2 from this that means tan alpha 1 plus tan alpha 2 is written twice u by v f minus this if you write here then finally you get an expression for the degree of reaction you get finally an expression if you do it is very simple that is nothing difficult you get an expression is equal to v f by 2 u into tan of that is nothing great beta 1 plus tan beta 2 that means simply you make tan alpha 1 plus tan alpha 2 u by v f minus tan beta 1 minus tan beta 2 and then make it clear 1 minus 1 will cancel. So, you will see a result like that this is the final expression for the degree of reaction now let us make an case study that where we get a 50 percent degree of reaction that means 50 percent or half of the total enthalpy or static temperature rise of the stage takes place in rotor in that case we get a very good result that u by v f is equal to tan beta 1 plus tan beta 2 now this is one very very important result now if you compare this result with this results u by v f is tan now you compare this with the result earlier that means this one u by v f is tan alpha 1 u by v f is tan alpha 1 plus tan beta 1 and that just now I wrote earlier also I wrote this is simply from the geometry so with this three relationship you get very beautiful result when you equate this with this that means we get alpha 1 is beta 2 and if we equate this with this then we get alpha 2 is equal to beta 1 and as we told earlier that alpha 1 is alpha 3 that alpha 1 is alpha 3 then that becomes equal to alpha 3 so this is one very important result what does it give this important result gives if I write here that this results if I write here alpha 1 is beta 2 is alpha 3 and alpha 2 is beta 1 then you see what we get alpha 1 is beta 2 beta 2 is the outlet angle of the blade rotating blade rotor blade and that becomes equal to alpha 1 what is alpha 1 alpha 1 is this one which is alpha 3 that means outlet angle of the stator blade that means the outlet angle of the rotor blade equals to the outlet angle of the stator blade again beta 1 beta 1 is the inlet angle of the stator blade which becomes is equal to alpha 2 alpha 2 is the absolute velocity angle at the outlet of the rotor which exactly equals to the inlet angle of the stator blade. That means, the inlet angle of the rotor blade is equal to the inlet angle of the stator blade. So, therefore, inlet angle of this rotor blade is equal to the inlet angle of the stator blade, outlet angle of the rotor blade is equal to the outlet angle of the stator blade. So, they are cambered in the opposite direction, but their inlet outlet angles are same. The inlet angle of one blade is equal to that of the other and outlet angle of rotor blade is equal to that of the stator blade. This type of design of blade is known as symmetrical bedding and it is easy for construction. Therefore, we see for a 50 percent degree of reaction 0.5 we get a symmetrical bedding that means the inlet angle of the rotor and stator blades are same similarly the outlet angle of the stator and rotor blades are same thank you today after this.