 Good morning and welcome you all to this session of the course. Now, today we will be solving some problems relating to axial compressors. Last class we discussed in brief the principle of operation and the degree of reaction of an axial compressor. Now, today we will see that how we can solve problems and so that our understanding becomes much better. Now, let us concentrate on one problem. This is the problem number one which we will be solving. The conditions of air, let me read out the problem first. The conditions of air at the entry of an axial flow compressor stage are P 1 is given, the inlet pressure and temperature is given. The air angles are beta 1. So, this nomenclature is known to us. This is the blade angle and inlet and beta 2, the blade angle at the outlet that is the rotor blade angle at the inlet and outlet that is 51 degree. Both are 51 degree and alpha 1 is alpha 2 is 8 degree. This is given alpha 1 at the angle of the absolute velocity at the inlet and alpha 2 is the angle of the absolute velocity at the outlet of the rotor blade. The mean diameter and peripheral speed are 0.5 meter is the mean diameter and this is the peripheral speed respectively. That means the peripheral speed at the mean diameter. The mass flow rate through the stage is 30 kg per second. The work done factor is 0.95. All these things are known to us. Mechanical efficiency is 90 percent. Assuming an isentropic stage efficiency of 85 percent, what are to be found out? We have to find out blade height at entry, stage pressure ratio and the power required to drive the stage. This is given for air, the characteristic gas constant 287 joule per kg k and the ratio of specifically to 1.4. Now, sorry to find out the blade height at entry, first of all we have to understand that it is the geometry blade height, which is basically related to the area at the inlet. So, therefore, if we write the equation of mass flow rate that thing we can we have to select this equation to find this. We can write that rho 1 inlet density and the area at inlet and the velocity of flow which is same throughout. Now, here first of all we have to find out rho 1. Now, within a that is the annulus area the blade height is there. The blade height can be found out from the annulus area a. Now, let us find out rho 1. So, rho 1 at the inlet can be found out from the equation of state as P 1 by R T 1 well. So, P 1 is what P 1 is given as 100 kilo Newton per meter square. So, therefore, 100 into 10 Newton per meter square 100 kilo Newton into 10 to the power 3 p and if you write the value of R 287 and T is 300 we get the value of rho 1 and rho 1 comes out to be 1.16 kg per meter cube alright. We get rho 1 how to get V f the flow velocity is it giving assuming an isentropic state air the flow velocity is not given, but we can find out the flow velocity from the expression. We developed earlier that u by V f is what tan alpha 1 plus because we choose this expression flow velocity is not given, but beta 1 and alpha 1 is given. So, that we can find out the flow velocity and u what is u is given 150 meter per second. So, 150 divided by tan alpha 1 plus tan beta 1 that means tan of 8 degree plus tan of 51 degree and if you solve for it you will get a velocity 109.60 meter per second. So, therefore, you can get the annulus area, but before that you put annulus area in terms of this blade height annulus area is pi into the mean diameter into blade height true where from does it come it comes that annulus area is pi into outer diameter square minus inner diameter. If you consider the inner diameter as the diameter of the drum or disc and the outer diameter is the diameter of the casing then this is the annulus area this can be written as pi into d 0 plus d i sorry by 4 d 0 plus d i by 2 and d 0 minus d i by 2. Now, it is this is the mean diameter d 0 plus d i by 2 and this is the blade height. So, therefore, this can be expressed in terms of the mean diameter and blade height that area. So, therefore, now if I substitute here we can write m dot m dot is given 30 kg per second is equal to rho 1 1.16 into v f first I write v f because I have found out v f 109.60 into pi what is that d m d m is given as 0.5 meter 0.5 meter it is given. So, therefore, pi into 0.5. So, only unknown is your blade height h which gives h equals to rather h 1 the blade height at the inlet. So, I write h 1 blade height at the inlet. So, h 1 is 0 equals to 0.15 meter. So, the first part is over the h 1 blade height at entry how to find out stage pressure ratio stage pressure ratio is simple to find out stage pressure ratio to find out what you have to do you have to find out the static temperature the stagnation temperature ratio or the stagnation temperature difference otherwise you cannot find out the static pressure ratio. So, each one is known now let us find out write this formula that C p into delta t s t that means the stagnation temperature is equal to today last class we have discussed that it is the work done factor into u v f by C p is always there u v f into tan beta 1 minus tan where from it comes that this is the work done per unit mass this is equal to C p delta t s t. So, therefore, this is multiplied with a work done factor. So, here what we have we have u we have v f we have beta 1 we have beta 2. So, we can find out delta t s t we can write this what is the value of lambda here lambda is 0.95 it is given well then the u is 150 meter per second v f we have calculated just now 1 190 I am sorry v f is 109 v f will come out as 109 yes yes v f is 109.6 0. So, this is lambda u into v f v f is 190.6 0 that is the v f we have calculated and tan of into tan of 51 degree minus tan of 50. So, this is the value of 10 degree divided by C p. Now, C p value you can calculate here C p is not given even if it is not given you can take that in this problem gamma and characteristic gas constant is given C p can be found out as gamma by gamma minus 1 that means 1.4 divided by 0.4 into 287 that is the joule per kg k. That means if you find out C p it will be 1005 joule per kg this is the standard value of C p for here. So, C p then you can find out delta t s t and that becomes delta t s t becomes equals to if you calculate it I tell you the value 16.37 degree. Then you can find out the pressure ratio R s as I have told that this is the nomenclature and this is the formula again and again we tell this will be 1 plus the isentropic stage efficiency that is eta s is the isentropic stage efficiency delta this formula I have told inlet temperature divided by gamma by gamma minus. So, this can be written as 1 plus the substituted value substitute if you can substitute the value delta t s t 16.37 divided by 300 and if you put the value of gamma then you get the value of the stage pressure ratio as 1.17. Third one is that what is the third one the power required to drive the stage power required is very simple because this is the C p delta t s t C p delta power required is what power per unit mass into the mass flow rate mass flow rate into C p delta t s t. So, delta t s t we know this already we have found out multiply with the C p that means this part that means actually this part. So, this part we know and mass flow rate already we know 30 kg given in the problem 30 kg per second. So, therefore, if we put the value C p this one delta t s t 16.37 degree Celsius and mass flow rate 30 kg per second you get a value of 548.39 kilowatt if you consistent in the unit finally, you get a value of 548.39 you will get this result this is the power. So, this is one example that we can work with next another problem I will tell you where you have some idea about the degree of reaction. So, this problem is this the preliminary design of an axial flow compressor is to be based upon a simplified consideration of the mean diameter conditions. Suppose that the characteristics of a repeating stage of such a designer as follows that means a particular stage these are repeated for different stages stagnation temperature rise is given. That means indirectly the work done on the fluid is given degree of reaction is given flow efficiency is given 0.5, but this is actually not flow efficiency I think it will be better if you write it as flow number V f by u is actually flow number sometimes they write at flow efficiency. However, flow number blade speed that means it is the ratio of flow velocity to the peripheral speed peripheral speed is also separately given 300 that means we get V f multiply 0.5. Then what is the problem assuming constant axial velocity across the stage and equal absolute velocity that inlet and outlet that means V 1 is equal to V 3 that we have already done determine the blade angles of the rotor for a shock free flow that means there is no incident loss. So, how to find a work out this problem now this problem if you work out you see that the degree of reaction is given. Now first of all we know the stagnation temperature rise. So, first of all we find out the work per unit mass is equal to is known that is what that is C p into delta T s t and what is C p? C p is 1 0 0 5 joule per kg k and that is 30 and finally this becomes equal to your work per unit mass. Now how to find out beta 1 and beta 2 so what we have to find out we have to find out beta 1 we have to find out beta 2 the two angles you remember that beta 1 beta 2. So, I tell you again that this one we have to find out we have to find out beta 1 beta 2. So, what we will do we will take some of this relationship tan beta 1 minus tan beta 2 tan is equal to work done where you will get that formula that V f ok I am coming to it again it was done earlier. If you remember that W by M is V W 2 minus V W 1 into U and that becomes is equal to U into V f into tan beta 2 minus tan beta 1. Now, we know U we know V f V f is 300 into 0.5 U is 300 U is 300 meter per second well and you can see I think you can see very well and V f is 300 into 0.5 that is 150 meter per second tan beta 2 tan beta 1 beta 2 and beta 1 we have to find out W by M we know. So, therefore, if we put this here and this value here we get a relationship tan beta 2 minus tan beta 1 equals to what tan beta 2 minus 0 point you get everything is known. Now, we have to get another relationship relating beta 2 and beta 1 that will come from the relationship of degree of reaction. Now, degree of reaction if you remember is given by U by V f degree of relation is V f by U V f by 2 U if you remember that today we have done it or last class we have done it tan beta 2 tan beta 1 plus beta 2. So, we have done this thing. So, here all we know V f by U V f by U is 0.5 that means 0.6 that means rather I write here I write here it will be better 0.6 is 0.5 V f by U is 0.5 it is given in the problem by 2 into tan beta 1. So, it is a simple problem which gives a relationship tan beta 1 plus tan beta 2 is it is what 1.2 to 2.4 it is very simple. So, this is one relation and this is one relation. So, as simple as this and finally, if you solve for tan beta 1 and tan beta 2 finally, you get beta 1 is 56.92 degree and beta 2 is 40.86 degree these are the answers you can check. So, therefore, with the degree of reaction you can that is only that we have to deal with the equations which equations will be required depending upon the parameters given. So, these are the problems which clear makes your concept more clear. Now, after this I will start the fans and blowers. So, fans and blowers there is nothing much different from that of compressor. The different is like this as I told earlier at the beginning of my fluid machines class that the when the output for a compressible flow for example, that handling air at the outlet of a machine where energy is given and we get the fluid at the outlet of the machine having higher energy acquired stored energy. If that is mostly in the form of high pressure static pressure, but less velocity we call it as compressible whether it is centrifugal it is a pressure, but when the basic purpose is to have fluid with very high velocity where we have to deliver high rate of flow with high velocity of flow. Therefore, they are the machine must deliver the fluid with higher energy by absorbing energy from outside mostly in the form of the kinetic energy with high velocity, but relatively much less pressure static pressure those machines are known as fans and blowers. So, fans and blowers and compressor of same kind in a sense that they take energy from outside and the fluid flowing through it gains its internal energy, but for fans and blowers these energies mostly in the form of kinetic energy rather than pressure static pressure is more of essentially rather than static pressure. Now, amongst fans and blowers fans are those machines where the static pressure at the outlet of the machine is few millimeter of water gauge you can understand that atmospheric pressure is 10 meter of water gauge it is so less few millimeter of water gauge whereas in case of blowers it is something more than 1000 millimeter of water gauge it is very less 10 meter of water gauge is atmospheric pressure. So, this is the pressure gauge pressure that means above the atmosphere at the outlet why this pressure is so low and why the pressure is required first of all pressure is required because these machines deliver air for some purpose that means air has to flow it may be required that to make to supply the air through a duct. So, therefore, we have to overcome the friction of the duct that means this flow has to take place through a duct it has to overcome the frictional resistance not only through a duct through a room. So, therefore, downstream resistance has to be frictional resistance has to be overcome to deliver that flow. So, these machines develop that pressure sufficient to overcome the resistance now in case of blower it handles more amount of air where less amount of air is handled and the circulated or being sent through a duct we employ fans for which the frictional resistance is less. So, therefore, a relatively less static pressure is required at the delivery end of the machine whereas in case of blowers which handles more air and at a high velocity it may have to be conveyed or transported through a long duct the frictional resistance is much more that means it has to overcome more frictional resistance so that the pressure drop is more as a consequence the static pressure at the outlet of the machine has to be higher this is the reason for which blower for blower static pressure is higher because it handles more amount of air. So, this is the difference between blower and a fan. So, fan and blower are basically the compressors, but in this case the static pressure at the end of the machine is much lower as compared to its velocity. So, you see a typical centrifugal fan or blower here I will show you simultaneously I do not know it may be little difficult I will try it yes this is the thing. So, let me show you this is the thing let me just a minute let me make these things let me show you like this let this is a typical centrifugal fan or blower now centrifugal fan or blower consist of an inlet this is the impeller the same principle impeller which imparts the energy to the fluid then after the impeller there is a spiral casing known as volute chamber spiral or scroll casing this is the volute where this gains the energy in terms of both velocity and pressure rise at the end of the impeller high velocity and high pressure some of the velocities are then converted in the volute chamber to static pressure depending upon the requirement at the outlet of the machine. So, this is the picture this can be shown if we take a section like this then it is this is the impeller you see it. So, this is the inlet the flow at the inlet takes place through a nozzle that means there is a little acceleration of the flow before it enters into the impeller. Now, impeller it goes radially in the same way that it happens in a centrifugal compressor that means it is shock or it is induced the axial direction then the flow change in the radial direction. So, this is the radially outward flow then it comes out and this is the volute casing and this is the outlet. So, this is the main component of a centrifugal fan and compress a centrifugal fan or blower. So, after this I will tell you about the typical type of blades in the in a blower or fan. Now, blower or fan both are the same machines as I have told only the static pressure rise is different for blower and fan and depending upon the flow rate required the fan blade shape is different. Now, three types of blade shapes from the fluid mechanical principles are used I tell you one is known as forward swept blade you see this left one what is that if you consider the motion in this direction that means this is the direction of rotation that means the peripheral speed is in this direction the tangential. So, curvature of the blade is in the direction of the motion this is known as forward swept blade another type of blade is this is known as radial blade this is known as this is known as this is forward swept blade forward swept blade forward swept blade forward swept blade. So, this type of blade is known as radial blade radial blade what is the radial blade radial blade they are outlet that means at the outlet the blade is radial, but at the inlet there is a curvature and this curvature is forward swept why because this is the direction of the motion the curvature in the direction of the motion. So, radial blades are radial outward flat, but it is curve at the inlet which is forward swept another is the backward swept that means is curvature this is the direction of the motion its curvature is in the opposite direction of the motion. So, velocity triangles will immediately change that I will tell you you can understand because the direction of velocity and the curvature of the blades are in the opposite direction or relative directions are changed. Now, before that I tell you the forward swept blades are used where large flow rates are required relatively large flow rates and higher pressure rises required as compared to backward swept blade whereas radial blades are preferred where the fluid that is air used contains more impurities and dust this is because of the fact that these are less prone to the blockage and they work more efficiently with the dust laden gas. So, therefore this radial blades are preferred for that now if you see the velocity triangle. So, velocity triangle at the inlet for example in the radial you see this is same for all the blades what is that there is the inlet velocity is axial absolute velocity direction this is the axial direction axial this has got no component in the tangential direction that is v w 1 component is 0 v w 1 is 0 everything that is the problem here v w 1 is 0 v w 1 is 0 ok v w 1 is 0 you see that can be seen v w 1 is 0 can be seen it seen v w 1 is 0. So, radial blades so entry for all blades are same that is v w 1 is axial this is the relative velocity which matches the angle of the blade this is the peripheral velocity this makes the vector diagram that is the velocity triangle. Now, at the outlet also you see this is the peripheral speed this is the direction of the flow a direction of the rotation speed this velocity that is the relative velocity matches the blade angle that means it is radially outward and this is the absolute velocity alpha 2 is the absolute velocity angle v w 1 is the angle of the blade at the inlet and this is alpha this is 90 degree this is the beta 2 which is radial this is alpha 2 is the angle of the absolute velocity. Now, let us see the velocity triangle for the forward swept blade the velocity triangle for the forward swept blade is this one is this one this is the it is visible it is visible velocity diagram let me draw the velocity diagram here I think for you it will be forward swept blade I write. So, this is the inlet diagram it is already same as that that means there is no tangential component it is axial and this is the now here forward swept this is the u the direction of u is this. So, therefore, the direction of u this is v r 1 and this is the u and this is the v r 1 this is the beta 1 v r 1 is called to v 1 u 1 minus u 1 very good because they differ it is a radial flow machines radial outward. Now, at this end what will happen this will be my radial flow sorry relative velocity directions and this will be the diagram will look like this. So, this will be obviously, from the v r 2 this will be u 2 which will be higher than u 1 and this v 2. So, this is the simple diagram because the velocity is in this direction. So, it has to be like that in this case this is defined as the this is defined as the beta 2 that means in the positive sense u 2 is this direction with this is an obtuse angle this is this angle beta 2 with the tangent and this angle is alpha 2. So, this is the velocity triangle which is shown here like this I think you can see, but this diagram I think is very difficult to see the things are not properly shown not legible however. Now, for a backward swept backward swept this can be visible now backward swept 1 I again draw it here for the backward swept what will be that. So, this curvature is in the opposite direction that means this is like this this is in the opposite direction, but the direction of peripheral velocity is like this. So, here also. So, this is the same thing that this is the inlet velocity triangle this is the inlet velocity triangle this is v r 1 this is v 1 let this is this is u 1 sorry u 1 and this angle is beta 1. So, the outlet angle for example, here or here I can draw the velocity this relative velocity is this then what is this velocity relative velocity is this direction that is v r 2 then this velocity is in this direction this velocity is in this direction. So, therefore, the absolute velocity will be. So, this looks very odd. So, little more the relative this is u 2 this will be v 2. So, this will be now if you draw it in scales for a given value of u 2 at the outlet and for a fixed value of v f you will see that for forward swept blade the component that v 2 is more and the component that v w 2 that this one v w 2 is much more compared to this this is v 2 this v w 2. So, therefore, you see v w 1 is 0 therefore, work done per unit mass is again v w 2 u 2 since v w 1 u 1 v w 1 is 0. So, therefore, in this case this is understandable. So, in this case v w 2 is more than this and that is the reason that forward swept blade is used where we require more work and more flow will be available and more static pressure rise will be there. So, more work will be imparted to the can be imparted to the fluid flowing through it. So, three types of possible blade configuration forward swept radial blade and backward swept blades are possible. So, with this now other treatments I tell you in the axial flow fans and compressors are almost identical to that of the centrifugal compressors which I have told in case of centrifugal blowers and centrifugal fans. Now, with this I will go on solving a problem let us see this problem is centrifugal fan a centrifugal fan running at 1500 that is 1500 rpm has inner and outer diameter of the impeller as 0.2 meter and 0.24 meter that is the inner diameter of the impeller outer diameter of the impeller the absolute and relative velocities of air at entry are 20. So, absolute and relative velocities of air at entry are this respectively and those that means exit entry and exit absolute and relative velocities are given. The flow rate is 0.6 kg per second the motor efficiency is 80 percent determine the stage pressure rise degree of reaction and the power required to drive the fan assuming the flow to be incompressible with density of air as 1.2 kg per meter. Now, this particular problem I tell you will make your total understanding of the fan or blow are clear. So, let us must see that what this problem tells now this problem tells that velocity absolute and relative is given. Now, if you recall this type of problem what you will do first you have to find out the stage pressure rise. So, how to find out the stage pressure rise? So, you have to find out. So, first of all you find out the stage pressure rise by which expression you have to find out the stage pressure rise stage pressure rise means what the total pressure rise in the stage. So, total pressure rise in the stage depends upon the work done total work done if you first find out the work done per unit mass what is the work done per unit mass. Now, if you remember the work done per unit mass is V w 2 u 2 minus V w 1 u 1 without going for a change in temperature stagnation temperature. So, far we did this is because this problem is told that assuming the flow to be incompressible. So, for an incompressible flow better we do this type of analysis that this is the this is true for any fluid incompressible or compressible if you remember that this work done per V w 1 u 1 that is work done that is w by m can be written as different components if you remember that is V 2 square that is work done on the fluid minus V 1 square by 2 plus u 2 square minus u 1 square by 2 plus V r 1 square minus V r 2 square by 2 this I explained in details in one earlier class of our fluid machines that this can be a split like this from the geometry of the velocity triangles why it is required because this gives a very clear understanding this is the gain in kinetic energy where the absolute velocity here in terms of compressor in terms of pump in terms of fans blowers where the fluid gains energy V 2 is always greater than V 1. So, this is the gain in the kinetic energy this is the gain in the kinetic energy or dynamic we sometimes tell as dynamic head it is per unit mass usually energy per unit mass or unit weight we call it as head and this two is the gain in the static head why it is called static head or static energy which is not very much used that means this is manifested in terms of the increase in the pressure of the fluid this is because of the change in this is the change in the peripheral velocity at outer radius and inner radius and the fluid therefore when it reaches at outer radius from inner radius gains in static pressure and that change in the pressure energy is given by u 2 square minus u 1 square by 2. Similarly, the change in static pressure because of the change in the relative velocity this is again another part of the diffusion that means change in the relative velocity takes place where V r 1 is more than V r 2 that means V r 2 is less than V r 1. So, passage is made in such a way that V r 2 is less is a diverging passage for which the pressure is gained. So, this pressure is gained out of the momentum change or the change in the kinetic energy relative to the blade passage. So, this two this is due to the centrifugal action the radial pressure gradient is imposed I told many times and this is because of the diffusion from the change in the kinetic energy V r 1 is higher V r 2 is lower. So, therefore pressure is lower and pressure is higher at the outlet. So, the combination of this two is the static head static energy and in this connection I like to tell you again this is a very important concept for all of you that in a radial flow machine. So, that it is outward or inward even if the passage is uniform there is no change in the relative velocity still there is a change in the static pressure rate because of the change in the u the peripheral velocity in case of a turbine there is a loss in pressure in case of a pump or compressor there is a rise in pressure. So, therefore any radial flow machines outward or inward has to have some pressure change in the rotor at least by this one, but along with that if you make a diffusing passage in case of compressor there will be an additional pressure rise. So, this two sum up give the static head or static energy this is very important I want to repeat it again. So, here what happens I know V 2 I know V 1 I know V r 1 I know V r 2 because all the relative velocities and inlet and outlet is given and again I know u 2 u 1 because the impeller diameters are given and this speed are given I am not solving this problem. So, I just write u 2 is pi d 2 n and u 1 is pi d 1 n n is given rotational speed 1 500 1500 r p m. So, that we can 1500 r p m so that we can do this thing better I write here. So, that u 1 u 2 I find out u 1 this way u 2 this way we can find out and then what we do we can find out the work done. Now, if we know this work done for an incompressible situation we can write that delta p stage is nothing, but rho times this work done per unit mass work done per unit mass. That means, you just multiply with rho this with rho you get the delta p per stage total in the stage you find out the delta p fine. Now, static pressure rise is this one so delta p static in the rotor because this is in the rotor delta p static in rotor is equal to rho into let this part I denote at x rho into x. So, this part is denoting at x rho into x. So, if you find out this by putting this value you get delta p total stage that means impeller and the diffuser total delta p stage because the work is done only in the impeller delta p total is 221.11 Newton per meter square and delta p in rotor or static pressure rise static ok. So, that is static pressure across the stages 110.71 Newton per meter square. Now, here this delta p static divided by total one is known as this is the static pressure rise in the rotor and this is the total pressure rise this is the total energy given. So, therefore the degree of reaction in this case we define as delta p stage sorry delta p static this already I did earlier delta p stage. Now, one thing just 221.11 divided by 110.70 oh sorry it is just reverse 110.71 by 221.11 this will be roughly 0.5. Now, here this is very simple this is numerical, but thing is that what is the concept earlier also if you recall the earlier discussions that the degree of reaction was defined in terms of the total pressure change in the machine that is change in the total pressure in the machine divided by the change sorry change in the static head or the static pressure in the machine divided by the total head or the total energy given in the machine. And this definition holds true for every machine in case of centrifugal compressor and axial flow compressor this was manifested in terms of the enthalpy change where you consider along with the temperature change. But here what happens when you consider the flow to be incompressible with a constant density of air and there is no change in temperature. So, therefore this degree of reaction is better to be estimated by that definition which we discussed in case of hydraulic machines that delta p static divided by delta p stage. So, this is the same. So, therefore we find this now next is the power required what is the next one the and the the stage pressure rise degree of reaction and the power required to drive the fan how to find out the power required power required is very simple power required we can write here power required is the mass flow rate into work per unit mass that means work per unit mass is this one. And you can find out the mass flow rate what is the mass flow rate the flow rate is 0.6 kg per second. So, therefore 0.6 kg per second you know mass flow rate W by M that divided by the mechanical efficiency probably there is a motor efficiency of 80 percent. That means you write the motor efficiency that means you write the motor efficiency in the denominator that means you did the mass flow rate is given as in this problem 0.6 kg per second. Now, W by M which is calculated I have only calculated here I have told you the value of delta p stage actually W by M here W by M calculated will be 184.26 joule per kg. So, therefore this will be 0.6 into 184.26 divided by 0.8 which is ultimately you know 138.19 watt. So, this is the thing that means this problem is looked from a different angle that work done per unit mass is true that is the Euler's equation this can be splitted in this term for an incompressible flow with a constant density it is better to looked upon this way that this is the static head rise and this is the kinetic energy. Some of all these three is the energy input to the machine and that is the total head that is raised in the fluid. So, this is the denominator there per stage and this is the numerator. So, therefore the degree of reaction is this divided by this it was discussed earlier when I discussed the hydraulic machines in general this is the degree of reaction. So, this is worked out to be this numerically delta p stage which is found out rho times the work done per unit mass that is the total pressure change in the machine and the static pressure change in the machine is rho times this part this is x again I repeat this part the static head that means this is the energy per unit mass this into rho is the delta p static. So, delta p static by delta p stage is this one is 0.5 clear now after this I will tell you something else that what is fan loss because sometimes you will see that people tell about fan loss what is fan loss this is a terminology actually. Now, in general for any fluid machines when I discuss the similarity principles one of the earlier lectures in fluid mechanics from all the variables involved in a fluid machines in the physics of fluid machines rather I will tell we derived by application of Buckingham's pi theorem the different pi terms as the criteria of similarity parameters. If you recall that see my earlier lecture note then you will find these terms are like this q by one term is like this n d q another term is g h by there are five terms n square d square another term is rho n d square by mu another term is p by rho n q d phi another term is e by rho divided by n square d square. Now, these are all pi terms this is let phi one this is pi two this was told at length in one of the earlier class there are five four and five. Now, when we derived the pi terms for centrifugal compressors we had four pi term this is because this term we neglected the effect of viscosity we did not take. In fact, the viscosity has less effect in fluid machines why I will tell you now earlier also I told, but this four terms we got, but not in this fashion in a different fashion I told you earlier in last classes this is an essence or corollary of pi theorem that depends upon the choice of the repeating variable you arrive number of terms are same you arrive some pi terms which may not match exactly the pi terms you want or the way you express the result then a combination of the pi terms is also a pi term. So, different combinations can be made by making the numbers same so that you arrive at different pi term. So, therefore, the pi terms of centrifugal compressors which we derived can be recoupled or rearranged to get the similar type of thing. So, this thing where earlier explained I am not going into detailed of it, but this term I tell you as you know this term represent a sort of Reynolds number because rho n d is the velocity u d by mu that means this u is the peripheral speed that means it is a Reynolds number based on peripheral speed. So, it can be changed to Reynolds number based on other flow velocities also physically it represents a sort of Reynolds number. Now, in fluid machines the flow is highly turbulent so in those turbulent flow the Reynolds number has got very less influence on the flow parameter mainly in the pressure loss or any other parameter the Reynolds number does not have much influence. So, therefore, in centrifugal compressors also we have not included the property mu to find out the dimensionless term. So, therefore, in the fluid machines earlier I also told this term is not of that relevance. So, only these terms are relevance. Now, if you take this term now you see it is not only for fan for any fluid machines that if you now this term is a non dimensional power rho in cube d phi this term you know that e by rho is the square of the acoustic speed or speed of sound in the fluid medium relative to the fluid and this is the square of the peripheral velocity. So, therefore, this is some sort of square of the Mach number Mach number square type of thing. So, therefore, you see that now if you see all this pi terms now you see that for any fluid machines if we make the fluid machines same machines of same size machines of same size machines of same size machines of same size then we can tell that q is proportional to n because d is same from this pi term first pi term which gives q 1 by n 1 is q 2 by n 2 this is school level things from the second pi term we see that head because flow and the head is very important what is head this is g h that means energy per unit mass head means energy per unit weight that means this is the energy either gain by the fluid or developed by the fluid or given away by the fluid depending upon whether is a compressor pump or turbine h is proportional to n square that means h 1 by n 1 square is h 2 by n 2 square similarly now if we take this term then we can write power is proportional to n q which means that p by p 1 by n that means this is the scaling law p 2 by that means we are interested in three quantities q h and p and multiplication of this two is this one. So, if it is proportional to n this is n square it has to be n q that means power is proportional to n q head is proportional to n square q is proportional to n for same side and machine of different sizes, but at same speed different sizes sizes, but at same speed one can derive from this formula that q is proportional to d q h is proportional to d square obviously if you know these two you do not have to see the pi term otherwise pi term is wrong that means it has to be d 5 that means q 1 by d 1 q is q 2 by d 2 q h 1 by d 1 5 is h 2 d 1 square is h 2 by d 2 square and similarly p 1 by d 1 5 is p 2 this is so simple d 2 5 that means the scaling laws with speed for the same size and with size which is represented by the impeller diameter as a representative characteristic size for the same speed is known as this is the scaling based on the similarity parameters of fluid machines and valid for all fluid machines, but usually because this grew like this that initially the fans and blowers when they were designed to make the prototype they use this type of scale. So, till date this scaling laws are known as fan laws in designing the fans. So, therefore while studying the fans we must know what is fan law, fan law is nothing but the scaling laws for the volumetric flow rate the head that means the energy per unit weight or energy per unit mass you can take g h does not matter g is constant and the power with the speed for the same size and with the size for the same speed and known as fan laws. So, I think today we will stop here and we have I like to close the lecture on these fans blowers and last class we discussed the axial flow compressor before that we discussed the centrifugal compressors and including all today I will stop or I will close my lecture series on fluid machines initially hydraulic machines basic principles of fluid machines hydraulic machines and then centrifugal compressors axial flow compressors and the fans and blowers. So, next class we will start the compressible flow. Thank you.