Lecture 24 : Axial Flow Compressor I

[Music] [Music] good morning today i'll discuss on axial flow compression contents of the lecture are two dimensional pitch line design and analysis lockdown factor stage temperature rise stage oceantropic efficiency and stage pressurize i'll be limited here to the discussion of two dimensional line design and analysis of axial flow compressions the basic component of an action flow compressor and the combination of rotor and the stator thus a row of rotor blades followed by a row of stated blade is called a stage the inlet guide vein igv which is a row of stator blade is often provided ahead of the first stage of the compressor here i have indicated the picture of a multi-stage compressor of a real gas turbine engine where you can appreciate the series of rotor blades of aerofoil constructions the schematic of a high pressure multi-stage axial flow compressor has been presented in the figure here where one can appreciate that the router is connected to the half while the stator is connected to the casing furthermore the stage has been indicated here which is the combination of the rotor and stator here a igv is presented ahead of the first stage furthermore i have drawn the central plane by a dotted line which is a circumferential axis symmetric plane between the hub and the casing one should note that there is a contraction of flow annulus from the low pressure to the high pressure into the compressor such that the axial velocity is maintained at a reasonably constant level despite of increase of density of air in an axial flow compressor the flow tends to be high while the pressurized per stage is low and it also maintains high efficiency for an actual compression of high up to tip ratio that is if the blade height is considered small compared to the half radius then the radial variation in velocity and pressure may be neglected and thus the performance of a compressor stage is calculated from the bleeding at the mean radial section which is called the pitch line design now let us consider the flow to the compressor as compressible and the velocity diagram is presented here here the velocity diagram of the compression stage so let us consider the mean circumferential section and if we stretch that circumferential section on a plane of paper will get series of played in this case i have considered first the igv followed by the rotor and then the stator thereby the station 1 represent the exit from the igv and inlet to the rotor and thus the station 2 indicate the exit from the rotor and inlet to the following stator and station 3 indicates the exit from the stator as usual said be the actual direction and theta be the peripheral direction now at station 1 i have drawn the velocity vector as w1 plus u1 is equal to c1 c1 makes an angle alpha 1 with the z-axis similarly w1 makes an angle beta 1 with the z-axis thus alpha 1 is called absolute inlet air angle and beta 1 is the relative air angle at inlet similarly if i draw the velocity triangle at xz to the rotor it would be w2 plus u2 is equal to c2 and alpha 2 is the absolute angle at exit to the rotor and similarly beta 2 is a relative angle exit to the rotor now look beta m is the angle that vectorial mean velocity w m mixed with the z axis and thus d which is the drag will be parallel to factorial mean velocity and lift is the perpendicular to factorial mean velocity w m so thus i have indicated here d and l is the dragon lift on the rotor where u is the tangential velocity or the blade speed similarly for the stator here at station three c3 is the absolute velocity that exit from the stator and alpha three is the angle that c3 makes with the z-axis now i'm sure you have appreciated the velocity triangle of a stage that is particularly the velocity triangle at inlet to the rotor and exit to the rotor now let us apply energy equation for compressible flow through a compression rotor here i have referred the velocity triangle at the inlet and exit of the rotor as i have explained earlier now the stagnation enthalpy rise across the rotor row of the compressor stage can be obtained from the euler equation where the work input per unit must flow all the specific work input can be expressed as w is equal to u into c theta 2 minus c theta 1 that's equal to u into w theta 1 minus w theta 2 where c theta is the tangential component of c and w theta being the tangential component of w and u is the blade speed now from the steady flow energy equation applied between the station 102 which is valid for compressible irreversible flow i can write 0 2 minus h 0 1 that is enthalpy rise across the rotor equal to u into w theta 1 minus w theta 2 thus c p into t 0 2 minus t 0 1 is equal to u into c z within bracket tan beta 1 minus tan beta 2 now c z is the actual component of velocity now this relation i can get from velocity triangle now please note that flow at the exit from the rotor would be deflected in the stator bed however there will be no work input in the stator therefore the enthalpy increase across the stage would be cp into del t 0 where delta 0 is the temperature rise across the stage would be is equal to cp into t03 minus t01 that is equal to cp into t02 minus t01 that is temperature rise across the rotor is equal to u into c is 8 within bracket 10 beta 1 minus 10 meter 2 thus the state loading is defined as i is equal to cp into delta 0 divided by u square where delta 0 be the total temperature rise across the stage that is equal to c z by u into tan beta 1 minus tan beta 2 equation number three now if the absolute velocity of air leaving the stage that is c3 is made equal to c1 while alpha 3 is equal to alpha 1 then the increase in total temperature delta 0 is same as that of the static temperature delta s across the stage such a design is called the repeatating stage thus for a refitating stage delta 0 would be equal to delta s where delta 0 is the total temperature rise across the stage and delta s be the static temperature rise across a stage and again for the repeting stage please remember c c must be equal to c 1 and alpha c is equal to alpha one thus for a refrigerating stage i can write psi is equal to c p into delta zero divided by u square is equal to c p into del t s by u square that is equal to c z by u which is equal to 5 row coefficient into 10 beta 1 minus 10 meter 2 now from the velocity triangle i can write phi is equal to c z by u that is equal to c z divided by w theta 1 plus c theta 1 which yields 1 divided by 10 beta 1 plus 10 alpha 1 referring to the vertices triangle this reduced to 1 by 5 is equal to 10 beta 1 plus 10 alpha 1 similarly i can write phi is equal to c z divided by w theta 2 plus c theta 2 this is 1 by phi is equal to 10 beta 2 plus tan alpha 2 equation 4 now can be written as i is equal to 1 minus 5 into 10 beta 2 plus tan alpha one c z by u is equal to flow coefficient five i like to state here that the exit flow angles of blade remains almost constant here beta 2 being the exit flow angle of the rotor and alpha 1 be that of the stator thus the summation of 10 beta 2 and tunnel alpha 1 will be invariant hence following the earlier equation psi would be approximately equal to 1 minus some constant into phi that is ziphi relation would be linear for a wide range of mass flow rate further note that the static pressure rise may be evaluated from the static temperature rise of the stage if the stage efficiency is known or assessed now the total temperature rise of the stage or the static temperature rise can be evaluated from the specific work input from equation 4 as delta 0 is equal to delta s is equal to c z into u divided by c p into tan beta 1 minus 10 beta 2 where delta 0 be the total temperature and del p is with the static temperature across the stage however the state temperature rise in the actual case will be less than that of the calculated value from equation five into the three dimensional effect in the compressor annulus this fracture has been taken care by introducing lambra which is called walk down factor this work done factor is a measure of ratio of actual work absorbing capacity of the stage to its ideal value as calculated by equation five here i like to pause stating that the equation 5 has been evaluated following the assumption that the radial distribution of axial velocity is constant within the however that relation is not truly valid that's why i've stated here that the radial distribution of axial velocity is not truly constant across the annulus but becomes increasingly peaky as the flow proceeds settling down to a fixed profile at about four stage of the compressor this happens because of growth of boundary layer on the hub and the kc thus in nutshell a factor lambda will be introduced in equation 5 to take care the effect of three dimensionality within the annulus the actual distribution of velocity has been illustrated here please note that the boundary layer will grow on the half and the casing and therefore the velocity in the annulus would not be constant so at this first stage i have indicated a velocity profile from hub to t while at stage fourth the velocity profile would be settled at a value where the central region the value of velocity would be higher than the hub and the casing this i have illustrated in the rds slide now the work capacity can be stated as or the specific work input can be expressed as w is equal to u into c theta 2 minus c theta 1 which can be written as u into u minus w theta 2 minus e theta 1 following the velocity triangle and thus i combined w b equal to u within the bracket u minus c z within bracket 10 beta 2 plus tan alpha one where tan beta two would be the exit four angle from the rotor and tan alpha one will be the exit flow angle from the stator i have also indicated that exit flow angle from a blade which remains almost constant that is summation of tan beta 2 plus 10 alpha 1 remains invariant for a wide range of flow thus following this equation an increase of cz would result in a decrease of work input and vice versa thus the effect of an increase of cz in the central region of the annulus would reduce the work capacity this reduction should however be compensated by increase of work capacity in the region of deep and hub because of reduction of season unfortunately the influence about an error on the hub and the casing and the tip board that produced because of chip clearance have an at first effect on this compensation process and the next effect is a loss in the total work capacity his lambda into w is the measure of actual work input to this stage thus the actual stage temperature rise is given by delta 0 is equal to del ts is equal to lambda into c z into u divided by c p within bracket tan beta 1 minus tan beta 2 that's equation number six where delta 0 is the total temperature and del ts be the static temperature across the stage and lambra is the walk down factor furthermore the variation of work done factor with number of stages has been indicated in the figure below which illustrates that as number of stages increases the walk down factor decreases and after a certain number of stages it should almost saturate that is here after 16 the value of work done factor almost means constant thus the equation 6 will give us the total temperature or static temperature rise of a stage and after evaluating this stage temperature rise we can assess the pressure ratio of the stage if efficiency is known i have derived here the expression of stage pressure ratio now the stage pressure rise can be evaluated from the known temperature rise if the isentropic efficiency of the stage it is is assessed let us consider now the stage isentropic efficiency beta is and then i can write it as is equal to isentropic temperature rise divided by actual temperature rise now referring the h's diagram of a stage in a compressor then is can be written as t 0 3 s minus t 0 1 divided by t 0 3 minus t 0 1 t 0 3 s is the isentropic temperature rise as indicated here so it is can be written as let me take common t 0 1 from numerator so it will be equal to t 0 1 into t 0 3 s divided by t 0 1 minus 1 now ratio of t 0 3 s by t 0 1 can be represented as pressure ratio v 0 3 by p 0 1 thus eta s can be written as t 0 1 within bracket p 0 3 by p 0 1 to the power gamma minus 1 by gamma minus 1 divided by delta 0 delta 0 is the temperature rise of the stage equation 6 gives the value of delta 0 thus from here i can write p03 by p01 which is the pressure ratio of the stage thus p03 by p01 to the power gamma minus 1 by gamma is equal to 1 plus eta is delta 0 divided by t01 t01 is the inlet temporal temperature at the stage hence the stage pressure ratio can be expressed as p03 by p01 is equal to within bracket 1 plus eta s delta 0 by t 0 1 to the power gamma minus 1 by gamma where delta 0 is the temperature rise across a stage and d01 is the total temperature of the fluid at nt to a stage furthermore the stage pressure ratio can be determined only if isentropic efficiency of the stage eta is is known to us now if all stages are designed for the same work input then delta 0 is equal to psi into u square divided by cp would be the same for different stages however the pressure ratio would be different considering the fact that the state pressure ratio depends on t 01 that's going to conclude here that the distribution of temperature across different stages determine the compressor performance further please note here that even if a compressor is designed with identical stages where delta 0 being the same then pressure rise would be different for different stages thus the questions remains unanswered how to evaluate the overall pressurize of a compressor this will be discussed in the next section thank you for listening [Music] [Music] you