 Good morning and welcome you all to this session of the course. Today, we will discuss the compressor characteristics. Now, compressor characteristics or performance characteristics of compressors are usually expressed in terms of the ratio of the stagnation pressure and temperature. Now, let us first have again a look that how do you make the nomenclature. This is the stagnation pressure at inlet to the compressor and this is the outlet from the compressor. We will use the same nomenclature that is at the outlet of the diffuser and similarly the stagnation or total temperature is T 1 T and T 3 T is the outlet temperature total temperature. So, with this nomenclature the compressor performance characteristics come today we will discuss compressor performance characteristics. So, performance characteristics for a centrifugal compressors are usually expressed in terms of the ratios of the total temperature or stagnation temperature. These are output parameters as a function of input this is output parameters and this is input parameters usually this is done input parameters this is output parameters. As a functions of n the rotational speed sorry here I can write n the rotational speed the size of the compressor d the mass flow rate m. So, this speed dam size of the compressor and the mass flow rate these are the parameters input parameters and the ratios are expressed in terms of this parameter the speed the size and the mass flow rate. So, now you see that this can be expressed in a functional relationship like this that a function of if we think in terms of the functional relationship we can write this n d m separately p 1 t p 3 t is case of this temperature we include r r t 3 t and these are the variables which define the performance of a centrifugal compression. Now, let me explain first these are the input parameters as I have told that this is the rotational speed this is the size that is the overall diameter of the impeller this is the mass flow rate m this is the total pressure at the inlet to the compressor this is the total pressure at the outlet of the compressor outlet from the diffuser and this is the total temperature at the inlet total temperature outlet and they are multiplied with r they are because of the two things it is multiplied with r t has a fundamental dimension temperature, but if you multiply with r r t becomes equal to p by rho and its dimension as a whole can be expressed in terms of m l t because p by rho is v square l square you can very well know that p by rho is l square v square. So, therefore, if you find out the dimension of r t it becomes a v square l square that means p by rho is v square the dimension wise ok dimension wise this dimension is v square l square by t square sorry l square by t square. So, therefore, it is l square by t square so therefore, multiplying with r taking care of the r as a whole to reduce the fundamental dimensions and at the same time taking care of the physical concept that t 1 t and t 3 t are very important parameters describing the centrifugal pump performance. Again another logic is there that you know that work done per unit mass or energy added per unit mass is given by change in c p times this c p times t 3 t or t 2 t whatever you call that means it is the c p times the t and c p is what c p is proportional to r that means c p in case of specific heat at constant pressure it is gamma by gamma minus 1 into r gamma is the specific heat ratio that means it takes care of r that means t a alone has got no function if you take multiplied with c p c p t is the index of the energy c p this t 3 t minus t 1 t is the work or energy input. On the other hand r t if you take together this reduces the fundamental dimension by 1 and things become little simple. So therefore, if you do this way you can now explain that the entire another thing very important that why you have not considered density because it is a compressible flow machines from the beginning I am telling that the density is very important in a compressible flow machines but I am not including density density is implicitly included because p is included r t is included so their ratio is the density. So therefore, density is not included explicitly it is implicit. So therefore, all the variables describing the centrifugal pump performance is are there how many 1 2 3 4 5 6 7 and fundamental dimensions are how many fundamental dimensions are 3 m l is equal to 3 that is m l t since we have considered the product r t. So therefore, by Buckingham's pi theorem number of pi terms will be number of pi terms will be 7 minus 3 4 why you are doing this dimension analysis by applying Buckingham's pi theorem because we want to express this relation in terms of non-dimensional variables rather than dimensional variables which will be reducing the number from 7 to 4. Now by applying the standard procedure of dimensional analysis what we do we take 3 repeating variables here what are the repeating variables we take the repeating variables as d we take the repeating variables as p 1 t and we take the repeating variable as r t 1 t this d p 1 t and r t 1 t as taken as the repeating variables this are taken as re they are taken as you just see that repeating variables they are taken as you can see repeating variables and following the dimension and analysis with this 3 as the repeating variables if we combine with n then you get a pi term that is your task you can do n d divided by root over r t 1 t I am now writing this that means this takes the n now d is the repeating variable when you combines with m this 3 repeating variables d p 1 t r 1 t then the second pi term comes as m root over r t 1 t divided by d square into p 1 t then if you take this p 3 t is the one then automatically when obviously when one of the repeating variables as the same dimension it is a thumb rule and if you do it you will get that p 3 t by p 1 t is the pi 3 and similarly pi 4 4 pi term which will be found out with what is that p 3 t then t 3 t then with this you will get since since t 3 t and t 1 t are the same dimension we automatically get t 3 t by t 1 t this is by thumb rule it will always come and if you follow this dimension analysis analysis you will find the same thing now therefore the equation can be written as some functional relationship of the non-dimensional term that means n d by root over r t 1 t m root over r t 1 t divided by d square p 1 t p 3 t by p 1 t t 3 t the ratio of total pressures and the ratio of total temperature is equal to 0 before proceeding further I like to tell you that these two these two are very clear they are the ratios of total pressure and total temperature these two have some physical significance for example this pi 1 n d what is the physical significance r t 1 t what is the physical significance now n d is proportional to the tip speed of the impeller a rotational speed into the impeller diameter and root over r t 1 t I told you that this sound speed is given by root over gamma r t so it is proportional to the sound speed acoustic speed in the medium relative to the flow so therefore this is proportional to u by a and which is known as Mach number based on rotor speed so therefore this pi term signifies physically some sort of Mach number based on rotor speed now this pi 2 if we write here that pi 2 m root over r t 1 t divided by d square p 1 t now this can be written m can be written in terms of flow velocity and the density rho the flow area a and the flow velocity v f root over r t 1 t divided by d square p 1 t now one can write p 1 t by rho is proportional to root over a proportional to r t 1 t p by rho is r t so though this is total pressure and this is this density so therefore it is r t 1 t so therefore this can be written as by cancelling that this is proportional to a v f then this will be cancelling out d square root over r t 1 r t 1 t because this square root and this is this is under root this is not under root so root over so therefore this will be proportional to v f by again a that means this is Mach number based on flow velocity this is known as flow Mach number this is known as rotor speed Mach number that means physically this pi term represents a Mach number based on rotor speed velocity and this pi term represents a Mach number based on flow velocity this is just for your physical implication now one thing that if we express this relationship try to express for a particular machine then d is not necessary to be included d is constant we can drop d and moreover it is for a particular gas for example the air then r also can be omitted here r is there here r is there so therefore for a given machine with a given gas the same relationship which is used non-dimensionally this can be written as some other function for example the function a function of n by root over t 1 t then this will be m root over t 1 t follow it clearly by p 1 t p 3 t by p 1 t t 3 t by t 1 t and for a given machine for a with a given gas the relationship can be expressed like that now you see here these two are not truly dimensionless or truly non-dimensional because we have dropped the term r and d but what happens is that even if they are dimensional term but we take the help of the non-dimensional analysis to reduce the number of terms so the number of variables are reduced where some of the variables which combines other primary variables may not be non-dimensional but that does not matter for a given size and for given fluid we can use this as a functional relationship of the performance so usually what happens is that the performance now is expressed like this that a family of curve is generated as the ratio of the pressure with the I will not tell this is non-dimensional mass flow rate this is normalized mass flow rate the word normalized does not mean non-dimensional different families of curves I am not doing the curve at present different families of curves for different values for different values of n by root over t 1 t in one family another family is the different value the same thing the same thing this m root over t 1 t by p 1 t and here is the ratio p 3 t by p 1 t with the different values of m root over t 1 t by p 1 t that means two families of curves for both the ratio of total pressures and total temperatures for different parametric values for each family of the normalized rotational speed and normalized mass flow rate this is basically the way the performance parameters that performance characteristics of a centrifugal pump is expressed now I will show you how does it look in case of a pressure ratio now I will show you that the very important a very important car p 3 t that is the pressure ratio p 1 t versus the non-dimensional mass flow normalized sorry normalized mass flow p 1 t the car looks like this I tell you the car looks like this the car looks like this I will explain let me first draw the label the car now I explain the three points are important point in understanding this now what happens the if you make an experiment and draw the points you will get a car like that initially it increases with a positive slope reaches a maximum then it has a negative slope continuously decreases and probably at high values of mass flow rate it touches the where the pressure ratio is one actually the pressure ratio starts from one now try to understand physically the fact that when the mass flow rate is zero there is a pressure ratio why this is because in a centrifugal pump try to understand when the mass flow rate is zero means that you stop the delivery valve here then what happens the impeller goes on rotating so therefore a centrifugal head or centrifugal the energy is imparted on the fluid in terms of a pressure rise so therefore a pressure rise will take place in the impeller because of the centrifugal action which we called as centrifugal head is impressed on the fluid is imposed on the fluid so fluid may not move in the diffuser there will be a static fluid a static field is there pressure field and that pressure is due to the churning action of the fluid in the impeller which imposes a static pressure rise because of the centrifugal action this we call as centrifugal head that means the centrifugal head because of the rotation of the impeller is imposed on the fluid even if the valve is closed here so a pressure ratio will be developed so that is the pressure ratio by the action of the impeller rotation which is shown here at this point now when we slowly open the valve of the delivery line then what happens the flow commences and when the flow commences again you see that the flow takes place through the diffuser vents so and also the vaneless space now when the diffusion process takes place through vaneless space and the diffuser vents as a hole which is shown in the diffuser then what happens again pressure rise takes place because of the diffusion process so therefore the rise in pressure takes place as we increase the mass flow rate as we increase the mass flow rate the rise in pressure takes place this means that the diffuser contributes its quota diffuser contributes this quota to the pressure rise because of the diffusion process so because of that the pressure rise increases and reaches a point maximum why beyond which if you increase the mass flow rate it will not be manifested in terms of the pressure rise this is because of the frictional losses as I explained earlier in the last class that frictional losses composed of skin friction loss at the same time the losses due to separation ok along with that the incidence losses are there so altogether the losses increases for which an increase in mass flow rate is not manifested with an increase in pressure ratio rather by a decrease in pressure ratio and this point corresponds to the maximum efficiency of the compressor so below which the compressor efficiency drastically falls because of the losses and if we go on increasing the mass flow rate for a given rotational speed for example here we give a rotational speed I tell you this is valid for one rotational speed one normalize rotational speed I show this particular curve for a given rotational speed there may be a point which may or may not be obtained in practice but there may be a point for a given rotational speed if I go on opening the valve wider and wider the mass flow rate may be such that the pressure ratio may be unity that means there is no pressure rise the entire energy given to the compressor to handle is being used to overcome the frictional losses in handling a huge mass flow rate ok so that particular point may not be available for a given speed n but it is theoretically envisaged can be envisaged so physically it is possible for a given n there may be a point which gives a mass flow rate where pressure drop pressure ratio is unity that means the entire energy is utilized to overcome the friction so therefore ABC three points are important and this is a particular curve and this way we can generate a family of curve with different rotational speed and similarly with different normalized mass flow rate so the characteristic curve is like that which has a positive slope maximum point corresponds to maximum efficiency then there is a negative slope here the most important thing now I will discuss is the instability of this part of the characteristic curve now this part of the characteristic curve is having a positive slope and usually this part is unstable and is very difficult to have this part of the curve in practice rather this part which is associated with a negative slope that pressure rise and mass flow rate curve is stable how I explain now let us consider a compressor like this let us come consider a compressor like this this is impeller this is diffuser this is totally the compressor try to understand this compressor okay and what we do this is the delivery and we control a daily a valve delivery valve here and this is further downstream of the compressor where the compressor is discharging here this is the downstream the two thing you have to understand downstream of compressor compressor downstream of compressor and this is the compressor delivery this is the delivery end delivery end now consider a case that the compressor is running with a given speed this valve is open part partially at steady state some flow is there and compressor is discharging steadily and let the operational point is on the positive part of the curve let this point is D now what happens by any chance if there is a reduction in flow in the compression by any disturbance or any closure of the valve then what happens a decrease in flow rate here if you see is accompanied by a decrease in pressure ratio because this is in the positive slope the figure tells like that so therefore the delivery pressure will fall immediately now what happens you see again that this part of the curve is such initially it is very steep then finally it becomes flat as it happens for a curve which has a maximum then it reaches a maximum because gradient has to be zero here so if this point is little bit on the steeper side of the curve then this pressure falls rapidly so the delivery end pressure falls rapidly while due to the reduced mass flow rate the downstream side where this compressor is delivering air does not fall that rapidly so therefore what happens as a result this pressure becomes higher than the delivery pressure that means a pressure gradient for flow is generated in the reverse direction this is a high pressure and this is a low pressure so therefore the flow starts from the downstream end of the compressor to its delivery side that means to the compressor you understand because of this if this does not fall rapidly whether this falls what happens in this part there may be a point and usually it happens so that this is reduced more rapidly than that at downstream when the flow takes place like this then what happens the net flow through the compressor delivered by the compressor is reduced by the opposing flow so therefore the flow rate is still reduced and the pressure is still reduced in turn it affects in reduction of delivery pressure again the reverse flow is increased and this way what happens this makes the flow in the compressor totally zero there is no flow that means compressor cannot deliver air anymore but still the delivery side there is a pressure pressure ratio that I explained because of the impeller action and by that time what happens is the mass flow delivery is totally shut down is reduced gradually gradually to zero then the delivery side pressure is reduced when the delivery side pressure is reduced at this condition a then what happens this pressure becomes high and it takes up again repeats the flow in the positive direction and therefore it starts repeating the cycle that means it starts flowing in this direction again an instability in reducing the flow causes the flow reversal so therefore what happens a small disturbance in reducing the flow in this zone makes a repeating cycle that means the flow reversal takes place again flow the flow comes in again flow goes in this direction comes this direction goes this direction so this type of flow reversal takes place when the operating point is on the steeper side of the positive slope part of the characteristic curve and this is known as surging of the compressor clear this is very important thing now you see this instability type of instability known as surging is not there in the negative slope part of the curve because here what happens if there is a decrease in mass flow rate this is associated with increase in pressure decrease in mass flow rate increase in pressure so no way flow reversal that means from the downstream to the compressor side can take place so therefore this part is unstable and another thing I told you since the slope is steeper initially and then flat so there is not necessarily that the point has to be immediately that down upstream of this V left side of the V that means there may be a point here even there may be a part of this positive slope where the surging will not occur that means surging may not start when the operating point falls just left of V the maximum efficiency curve there may be some point away some distance away from the maximum point where from the surging can start the onset of surging is there so this can be well understood by this particular figure well this can be well understood by this particular figure now if I draw a show you this figure you see that as I told earlier that this is the curve the characteristic curves now this point that means if I find out the such onset of such point D for all curves of the family for different values of the parameter and if they are joined this is the search line is the locus of the starting of the search point that means this part of the curve for a given value 0.6 of this n by root over t 1 t is the stable part is the stable part is the stable part is the stable part and this cross point at the maximum efficiency that means this line is the locus of points of maximum efficiency and this line is the search line so this part of the curve is characteristic curve is stable this is n root over t 1 t and this is with respect to mass now you understand you have understood this thing now next what I like to tell you that another important thing is there on this side of the curve that is in the negative slope there is another interesting point E where we may stop what is that now you consider when the flow rate is increased the pressure drop decreases pressure decreases for example if you make the valve wide open so what happens the flow rate is increased delivery pressure is decreased and a decrease in delivery pressure decreases the now velocity of flow is proportional to mass flow rate divided by area into row so a decrease in the mass flow rate sorry increase in the mass flow rate and a decrease in the density because of decrease in the pressure because you see increase in mass flow rate is associated with decrease in pressure with the negative part of negative slope part this part makes a huge increase in flow velocity and it may so happen that is also not always possible depending upon the value of n that a point may come when the sonic velocity may be attained at some part of the compressor so when the sonic velocity is attained we cannot increase the flow anymore by any change in the downstream this will be explained again in detail in your compressible flow class on that is known as choking that is known as choking so the maximum flow condition choking of flow when the flow at any part becomes sonic that means the compressor will run there is absolutely no problem but no further increase in mass flow rate possible that means there is a point here which will on the characteristic curve which will indicate the limit of the maximum flow rate so therefore I will show you here also along with the surge along with the constant along with the locus of the surge point onset of surge the maximum efficiency point there may be another line this is the joining of the e point there which is the choking or the maximum flow limit choking line or maximum flow line so therefore the stable part of the characteristic line is bounded by left extreme by the locus of maximum a locus of the surge line onset of surge surge line in the extreme right is the choking line maximum flow line and in between is the maximum efficiency line it is clear so this is as a whole is the this is as a whole is the your characteristic curve I think it is all right now we will stop the discussion and we will try to solve some problem here it is now let us see a problem now here also I did a mistake this will be not this will be a repetition of this this will be n sorry this will be m we are writing n by root over t 1 t and this will be for t 3 t by t 1 t this I did a mistake earlier this will be like this the two curves now let us solve a problem so we have discussed the principle of the characteristic curves the concept of surge the surge line the maximum efficiency line and the choking line now let us consider this problem a centrifugal compressor has an impeller tip speed 360 meter per second this is the impeller tip speed determine absolute Mach number of flow leaving the radial vanes of the impeller and the mass flow rate the following data are given so data are given so let us see that Mach number of flow leaving the radial vanes absolute Mach number means based on absolute velocity if we recall the vane now let me recall the vane like this sorry this is the vane it is not oh then this is the vane okay this is the vane now if this is the vane then what is the diagram that the velocity diagram let me say better show these things which I earlier so this is the thing I think this will be better to show like this we can make things like this can you see okay please why I am writing that why I am not doing the radial one here one thing is important that slip factor is given 0.9 that means here due to this slip what happens we have a this is the rotating in this direction u2 so vr2 is not radial because there is a slip so vw2 this is vw2 vw2 and vw2 is less than u2 what is u2 u2 is this one okay this is this is u2 so this is the absolute velocity v2 and this velocity is vf2 this is vf2 so this is the diagram because there is a slip so therefore this is the outlet triangle now what I will do I will write what has to be found out that it has to be found out that Mach number based on absolute velocity at the outlet of the impeller so therefore I have to find out Mach number v2 by root over gamma rt2 okay now how to find out v2 v2 is the absolute velocity now v2 is root over this is vf2 vf2 square plus this is vw2 vw2 square now vw2 is not u2 okay this is u2 this is u2 so vw2 is sigma u2 okay now u2 is 360 meter per second vf2 is given flow area power input factor impeller tip speed flow area Mach flow rate so vf2 is not given impeller tip speed is given radial component of flow velocity is given vf2 is 30 meter per second you see that vf2 is 30 meter per second so therefore you get v2 is equal to 30 square plus 0.9 into 360 square I will not do everything calculations square and this way you will get a value of v2 equals to what is the value of let me tell you the value of v2 here the value of v2 is 325 you can check 325.38 meter per second now to find out the Mach number you require this static temperature here in the formula it is gamma rt2 t2 is the static temperature at the outlet of the vn how to find out t2 now our main objective to find out t2 how to find out t2 now let us must find out the total temperature t2t we know that t2t here t2t is equal to t3t that means outlet total temperature from the compressor that the outlet of the diffuser and we know that cp into t2t minus t1t that is the work done that is is equal to psi sigma that power input factor u2 square divided this is okay so now here you see t1t is given where because the inlet stagnation temperature 300 k is given psi is given power input factor is given power input factor is given 1 now slip is given 0.9 sigma is given 0.9 psi 1 u2 is given 360 meter per second t1t is given the inlet stagnation temperature is 300 k so everything is given except t2t from which we can find out t2t equals to what t2t becomes equal to ultimately if you calculate t2t will be 416 k I am not writing every step because everything I know power input factor is given in the problem slip factor is given in the problem and t1t is given in the problem u2 is given in the problem again I show you the problem that impeller tip speed is 360 meter per second radial component of flow velocity 30 meters slip factor is 0.9 flow area impeller exit is 0.1 square it is not now required power input factor is given isentropic efficiency given inlet stagnation temperature inlet stagnation pressure then r and gamma so here what is required power input factor which is given as 1 the slip factor which is given as 0.9 u2 is 360 meter per second and t1t is 300 k we get t2t now what happened t2 how to find out t2 now t2 has to be found out from the concept of the stagnation temperature what is that t2 plus the temperature dynamic velocity equivalence temperature equivalence of the dynamic head that is v2 square by 2 cp that is the temperature equivalent of the kinetic energy v2 square by 2 that is this plus this is the total temperature from which we can find out t2 is t2t is now found out 416 now v2 already we have found out 325.38 meter per second the value of cp has to be found out from r and gamma which you have read at school level that the specific heat at constant pressure is gamma by gamma minus 1 into r so you know the cp so therefore from here you can find out the static temperature as 363.33 k everything is known so we know the static temperature when we substitute the static temperature here we get the Mach number equals to the Mach number 2 for example the 2 suffix I am using equals to 0.85 all right now the next is the to find out the mass flow rate how to find out the mass flow rate you see the mass flow rate let me keep it here so that you can see it. Now, mass flow rate to find out mass flow rate let us write the mass flow rate mass flow rate is same throughout the machine. Let us write the mass flow rate based on the condition at the outlet of the diffuser, rho 2, the a 2 and the flow velocity. Now, here a 2 is the flow area at the outlet of the impeller, which is given you see here, flow area is given radial component of flow velocity, the mass flow rate impeller tip speed flow area impeller exit that means, a 2 is already given 0.1 meter square. So, what is not given V f 2 is given the radial flow velocity component that impeller exit is given V f 2 is given V f 2 is what 30 meter per second it is all right 30 meter per second what is not given rho 2. So, how to calculate rho 2 now rho 2 is P 2 by R T 2 now T 2 I know the static temperature already T 2 already is calculated here. So, I do not know P 2 how to calculate P 2 now before calculating P 2 you have to calculate the stagnation pressure then if you calculate the stagnation pressure then you can calculate the static pressure. So, how to calculate the stagnation pressure. So, stagnation pressure you can calculate P 2 by P 1 T from your this earlier formula 1 plus eta c into you can calculate like that T 2 T minus T 1 T divided by T 1 T to the power this has been told earlier gamma by gamma minus 1. So, you remember this one that is the pressure this comes from where this comes from the isentropic relationship and then using the isentropic efficiency of the compression well. So, using this relationship we can find out this relationship if you remember this was derived in the class that means I find out this way that if these are the two pressure lines then what happens this P 2 T P 1 T then this is the thing this is the T 2 T T 1 T and this is the T 2 T dash. So, P 2 T by P 1 T is T 2 T dash by T 1 T to the power gamma by gamma minus 1. Now, this T 2 T dash T 1 T is found out by expressing this eta c is T 2 T dash minus T 1 T I repeat again this was done earlier minus T 1 T. So, therefore, T 2 T dash that means P 2 T by P 1 T is T 2 T dash by T 1 T to the power this is the isentropic relationship P T relationship. So, this thing is taken from here T 2 T dash by T 1 T is eta c into this plus 1. So, eta c into this by T 1 T plus 1. So, therefore, we can write this is the thing done earlier now T 2 T minus T 1 T you know already you know T 2 T you already know T 2 T T 2 is find out T 2 T is 4 16 k you know T 1 T T 1 T is 300 k given eta c is given in the problem eta c is what eta c is 0.9. So, therefore, T 1 T is known everything is known you find out the P 2 T. Now, after knowing the P 2 T you have to know the P 2 because how because rho 2 is P 2 by R T 2 I wrote it earlier R T 2. So, you have to know the static pressure now how to know the static pressure by the concept of stagnation pressure P 2 by P 2 T is T 2 by T 2 T to the power gamma by where from it comes that means the static pressure is changed to stagnation pressure that means when this is brought rest isentropically that means this process of changing from P 2 to P 2 T is obtained by bringing the fluid isentropically to rest that means therefore the gain in temperature T 2 T from T 2 is made isentropically. So, that the pressure ratio will be related to the isentropic relationship with the temperature ratio this is the relationship between total or stagnation pressure to the static pressure. So, therefore, T 2 T 2 T is known therefore we get P 2. So, we can find out P 2 you understand. So, from here we find out P 2 T P 2 T is known here we find out P 2 T P 2 T is found out here and here we find out P 2 because P 2 T is known. So, finally P 2 is find out first we find out the ratio of the total pressure in terms of this you know everything P 1 T you know you know P 2 T when you know P 2 T you know P 2 from this equation. So, P 2 is find out. So, when P 2 is found out then you can find out rho 2 from P 2 by R T 2 and you can find out the mass flow rate. So, I am not doing things by putting the numerical value, but I tell you the way. So, this value is you check 5.09 kg because this will take more time to solve this by putting this value. So, that is why I am not doing. So, if you put these values you will get the result I think there will be absolutely no problem now only substitute the numerical value and get the results and check the results with this. Now, next another problem I will discuss before I leave you. So, another problem is this one. Let us consider a problem like this the following data are suggested as a basis for the design of a single sided centrifugal compression single sided. Power input factor is 1.04, slip factor 0.9 almost the similar problem, which we discussed earlier rotational speed is 290 revolution per second overall impeller diameter 0.5 meter I t I root diameter air mass flow rate is given inlet stagnation temperature is given inlet stagnation pressure isentropic efficiency. Now, what is to be found out determine pressure ratio of the compression power requirement inlet angles of impeller veins at root and tip radius of the eye. Now, let us find out the most simple thing the pressure ratio how to find out the pressure ratio. Now, pressure ratio to find out what we have to do let us again write the pressure ratio formula. If you write the pressure ratio now I write p 3 t by p 1 t is equal to in terms of this stagnation temperature rise I think this you know again and again I am writing just now I have discussed this just now I have discussed this that this stagnation or total pressure ratio here also the pressure ratio of the compressor total pressure ratio p 3 t by p 1 t has to be found out. Now, therefore what we require we require t 3 t minus t 1 t how to find out t 3 t minus t 1 t again the same formula we know the work done to the fluid per unit mass or the energy added to the fluid per unit mass it is psi per input vector sigma u square u 2 square. Now, this thing can be found out provided psi sigma u 2 square is given what is given in the problem let me see this problem gives rotational speed and overall impeller diameter 290 revolution per second and overall impeller diameter 0.5. So, u 2 is equal to pi into the overall diameter 0.5 into 290 is equal to the rotational speed that means the tangential speed is 455 meter per second that means u 2 is known sigma is given in the problem if you see this slip factor is 0.9 sigma is power input factor is 1.04. So, if you put everything you get the value of t 3 t minus t 1 t which becomes equal to t 3 t minus t 1 t is 193 k here the value of c p is not given in the problem if the problem the value of c p is not given in any case you can use that for air the value of c p is 1.005 kilo joule is the specific heat per k g k. So, therefore, you can use the value of c p that and guess t 3 minus t 1 t eta c is probably given in the where is eta c overall there is this isentropic efficiency 0.78 when you get the isentropic efficiency 0.78 this is this t 1 t is given t 1 t is the inlet stagnation temperature inlet stagnation or total temperature whatever you call 295 k. So, everything is known and this p 3 t equals to what is the value equals to 4.23 this you can check. So, you can find out. So, pressure ratio of the compressor is found a power requirement power requirement is mass flow rate into c p this work done per unit mass t 1 either this or this both the things are same. So, anyway you find out the mass flow rate is given probably otherwise power cannot be found out air mass flow rate 9 k g per second. So, work per unit mass into mass flow rate is the power you know everything. So, power requirement is now 1746 kilo watt all right. Now, the second part is the inlet angles of impeller vents at root and tip. Now, at root and tip if you want to find out the impeller angles then what will happen let us find out and root or tip anywhere at any representative section it may be root it may be tip that means it may be root or it may be tip that either it is this is the root and this is the tip. So, root and tip axial flow velocity is constant the normal flow velocity is constant it is given there we will assume that. However, now what happens is that if we know the e 1 the root or speed and v 1 and v r 1 if this is this alpha just I draw the diagram. So, tan alpha is u 1 by v 1. So, I have to know the flow velocity or the relative velocity tan alpha is o v 1 by sorry tan alpha is v 1 by u 1. So, this is if I know the at root velocity then it will be root angle at the root at the tip it is angle at the tip. Now, root and tip q 1 will be found out based on the root and tip diameter that I know because I know the rotational speed I know the rotational speed I know the tip diameter I know the i root diameter, but here I know I know neither v 1 nor v r 1. So, how can I find out? So, this type of problem is based on a trial what trial how to find out v 1. So, v 1 is not known now if you see from the mass flow rate basis mass flow rate is given rho a f v 1. Now, a f is given how a f is given a f is given because i tip diameter and i root diameter is there. So, one can find out a f as pi into 0.3 square minus 0.15 square by 4 and this become is equal to 0.053 meters. So, I know f. So, I have to this two I know. So, what is that I have to make a trial guess for rho and find out v 1 how to guess for rho this rho for example trial 1 this is a trial method trial 1 I guess rho from the total pressure p 1 t r t 1 t at impeller i I know t 1 t I know p 1 t based on this I find out and this value is given 1.1 bar into 100 by 0.287 into 295 this is the value given here yes 1.1 bar. So, this in terms of kilo joule kilo this Newton per meter square I am converting this unit is there that is why I have written 0.287 which is this value it should be 10 to the power 5 that another 10 to the power 3 then it will be 287 the characteristic gas constant you see that consistent unit it is written and it becomes 1.30 kg per meter cube. Now, if you know this rho 1 first trial this rho 1 rho 1 trial 1 you put that thing rho 1 and if you know the mass flow rate is known mass flow rate is already given in the problem what is the mass flow rate mass flow rate is 9 kg per second you get a value of v 1. Now, when you get the value of v 1 how to iterate there should be a base on which you will iterate when you get the value of v 1 you can calculate the corresponding temperature dynamic temperature by v 1 c p for example, 1 this density you get from this density you find out the v 1. So, from with this density if you find out the v 1 then v 1 for 1 trial will be 131 meter per second. So, when you know the v 1 then what you do with this v 1 you calculate the for 131 second v 1 square by 2 into 1.005 into 10 to the power 3 the corresponding temperature. So, if you know this temperature you can calculate the static temperature t 0 1 minus this 8.5 k t 0 1 we know. So, therefore, we can find out the static temperature and at the same time we can find out this is a little laborious calculation I know, but this is usually done in the design the static pressure as I told you the formula earlier the static pressure and static a total pressure is related to the static temperature and total temperature to the isentropic relation. So, I get p 1 when I get p 1 and t 1 I can find out rho 1 as p 1 by r t 1 that means by guessing a rho based on this stagnation condition I find out a v when v is found out I find out the dynamic equivalent temperature and with that temperature I can find out the static temperature and the static pressure by isentropic relation when these two things are known static pressure and temperature then rho 1 is p 1 by r t that means rho is getting corrected with this rho I find out again the corrected velocity. So, this way both rho and v rho 1 and v 1 1 v 1 1 and rho 1 1 rho 1 is getting corrected so that we get a converged value when we have a converged value then we get v 1 when we get the v 1 which is constant the axial velocity of flow throughout the impeller passage is same that means it is same as the at the root and tip then what we will do when we will use the peripheral speed at the root then we will get the angle of the vane at the root and at the tip then I will get the angle at the tip and if you do so then you will get I am giving you this value the v 1 converged value of v 1 comes out to be 140 meter per second and u tip you find out pi into as the value is given you check the rotational speed into tip diameter 273 meter per second and u root is equal to just half the diameter is half meter per second and alpha root that by tan alpha that formula v 1 by u perpendicular by base v 1 by u it becomes 46 degree and alpha tip becomes 27 degree it is 20 minute in more from where I have taken this is the value so there is an iterative process by which you have to do thank you today up to this