 I will make a plan and I will let you know. Welcome to today's lecture on proportional solenoid pilot operated two stage pressure relief valve in shock form which is called P P O P R V, P first P for proportional, second P is for pilot, O is for operative and P R V is pressure relief valve. This term is often used for such a valve. However, what we are discussing here, this is a solenoid operated but it might be other type of drive also there. Now, this lecture essentially the continuation of the lecture 15 which we delivered may be last time two weeks back. Now, first of all again let me explain what is the valve. So, we will again I shall again explain construction and operation features. Now, if I look into the sectional view of this valve this is again it is a schematic not exact section neither it is a photographic view this is may be computed develop. Now, here what we find this portion including this portion you can imagine this is an ordinary pressure relief valve. How it operates? This is connected to the system line. So, one side pump and another side may be the system. Now, the flow is coming in here not here do not make a mistake this is going to tank this portion is going to tank and this is the system line. So, here it is shown and then the flow is coming over here and through an orifice here and a construction here it is going through this passage to this place. Here is another orifice this orifice together orifice and construction together we named as A and then this is B another orifice and then we find the pilot stage. So, this orifice is D this is a variable orifice and this has great role on controlling the pressure and then after passing through this orifice the oil which is going through this this variable orifice it is going to the tank. And in between that which is not shown here there is another you can say valve block which is basically set the maximum pressure. When we call the proportionate solenoid pressure control valve or pressure relief valve in this case then this is the force on pilot stage can be given by the proportional solenoid in such a way that with the input usually voltage it might be also current input with the voltage the output that means in this case pressure will be proportionally related usually this operates in 0 to 10 volt and towards the end we would say 8 to 10 volts and towards the beginning 0 to 2 volts it may not be linear but in between from 2 to 8 volts the pressure relation will be linear. The advantage of this valve is that you can one can put easily in the control online control. So, the in fluid power usually such phenomenon this orifice flow then pressure control all such are non-linear highly non-linear but there will be non-linear inside however this is designed in such a way that input output will be linear and which can be computed directly with due to this linear relations practically no time and due to this reason these valves are used with the modern control units or sophisticated control units although performance may not be very close to the servo valves. Servo valve is having better performance considering a servo valve of equivalent size if we say that is performing with 100 percent the proportional solenoid will perform in between something 95 to 98 percent of the desired output right. Now this I have just explained explained or stated how it is performing but if we look into the performance how it works that through this orifice this passage is connected and let us consider this is closed then what is happening the oil is creating a pressure on this main spool which is supported by spring but this force is not enough to open this one so flow is being used for the system. Now suppose there is increase in pressure say I would start like this the functional feature of the valve under study can be understood from the schematic diagram and the this flow this oil is connected to here this means that as long as there is no flow pressure here and pressure here is constant as well as through this orifice C this orifice has no function with the flow main flow but this has a function with the damping of this main spool it is required when there the transient due to this orifice it dampen somewhat the motion of this and it becomes more stable if you remove this orifice C still this will work but there will be more vibration then the valve with this orifice anyway the everywhere pressure remains same so this side of the main spool total force definitely will be more than this so this will remain closed now when the pressure increases then due to the thrust it is already said we have given a force here we have given a force through the voltage there is a proportional solenoid through this end of the spool we can say here so this is giving a force in this directions so it is being closed and it it it remained closed now with the pressure here when this pressure increases the force increases then what will happen this will be compressed and the this is the main spring here this is the main spring and this spring is an auxiliary spring just to keep this is this puppet in position pilot puppet is position this has practically no function over the force whereas this spring is transferring the force from this to this you may ask why it is not direct it can be shown that if we fix this is direct then it is very difficult to get this proportional characteristics anyway so what happens this opens and once this opens the flow begins from here to here to this drain okay so there is a separate drain connection for this passage it also can be connected to the main passage over here the separate passage is function wise it is more or less same now once the flow begins then definitely there will be pressure drop through this passage now with this drop pressure here but look at this flow is through this passage not to this passage yet so whatever pressure here the same pressure will be there so but that pressure is less than this pressure so at one point this pressure force plus spring force totally will be less than whatever force is generated here so what will happen this will open and the main flow will go to the reservoir now this situation will be maintained unless this pressure is not below the whatever force we have given here whatever for what pressure we have given this input okay now the question is that we could have make it is a direct direct proportional valve is also available instead of pilot stage it is operating this one what is the advantage of this the advantage of this as I told earlier due to this pressure drops here through the different orifices this remains stable for longer period why we need such stability look at this if once the flow begins from here then here is the tank pressure tank pressure is almost 0 0 pressure or maybe a slight pressure for the filter and other things whereas here pressure is high pressure the system pressures then once the flow begins then definitely there will be pressure drop as well as here will be also pressure drop by using so many orifices what happens that here the orifice opening is very much controlled so that maintaining the pressure almost equal to the system pressure the flow the whole flow is diverted to the tank this is a large orifice as you can this is also a variable orifice but this is a large orifice okay so this is basically the operations now I have shown you that by the equations how the pressure is being dropped what the force everything a part of that today I shall discuss the rest of the portions now here again if you look into this this here there is a closed loop feedback here we and we are having a LVDT that is that controls the armature positions for an input this is not closed loop for the whole system this is closed loop only for the proportional solenoid which is displacement of this solenoid is recorded by LVDT and through this it maintains the position with force to generate a particular force in this directions okay now this hydraulic parts the advantage of proportional solenoid valve is that this other hydraulic parts including this one this is the maximum pressure set again I would say suppose this is working say at 10 voltage let us consider maximum pressure can be control is 10 MPa or say at 8 voltage because 8 to 10 that may not be linear at 8 volt say 10 MPa or let us consider 8 MPa then this valve here the setting will be slightly higher than that so that if pressure increases above that then this portion will not work this directly it the flow will continue like this which including this an ordinary valve this is an ordinary valve so this is additional safety feature however this valve can work without this one also until there is say something wrong with the voltage input or otherwise it would work suppose if the pressure increases up to 12 voltage still this will generate only the force for 10 MPa and this flow will be there but with this there will be no chance of damage at this portions so that is why this portion is additionally introduced there now so orifice A and B are considered as the two resistance placed in series in order to achieve the required optimum pressure drop before the pilot stage if we replace we have done of our own experiments and we have seen if we replace one then performance is no longer linear neither the stability is available there so this as these are designed in considering the linearity as well as proper function now what we did that instead of that block there we put another block just manifold you can say so with the whole passage hole and keeping the passage hole intact but this facilitated us to put the connect this one pressure gauge here to measure the pressure here another pressure gauge here to measure the pressure here another pressure gauge or transducer to measure the pressure here and another to measure the pressure here and another one to measure the pressure inside this pilot stage however we did not have much scope of measuring the pressure at these two points very closely what we consider whatever pressure is there the same pressure is here as well as whatever pressure we are measuring here the same pressure is inside that roughly why roughly it is almost giving the pressure drop within this variable orifice may be slight error is there but it was very close to that so our arrangement was like this two major pressure at different points now the primary objective of the analysis is to predict the set pressure supply pressure so what we wanted to do we wanted to study the steady state characteristics so for the steady state characteristics if I know if we know the proportional soil solenoid feature fully completely then from there if we know the orifice area everything can be calculated and we can develop the design procedure of such a valve however it was very difficult to know the real solenoid characteristics neither it was possible to test it separately because it was it was a integral part so what we did by some trial error error we first found out what might be the force what are the forces generated by the armature solenoid armature at different voltage for that what we could do we measured the different orifice area we could measure the spring stiffness we could measure the dimension of the puppets dimension of the solenoids angle etc. this data is available with the manufacturer but the problem is that they will not provide us the data so we took one valve and we measured all such parameters and then we tried to establish the design procedures particularly to examine the steady state performance and also we examined if we change the orifice what orifice dimensions then what will be the changes in the performance in the valve and for a given input voltage to the proportional solenoid the flow rate through the pilot stage that measurement we did so what we did while the oil in between these passages we put a flow meter like this just to measure the flow through the pilot stage all the intermediate pressures along with the pilot stage was measured by placing the transducer at different places at steady state conditions same amount of hydraulic fluid that enters through orifice aid will pass through orifice so using the orifice equations we can write down the equations because we know the pressure at different positions so we can write down the equations of for the flow and we can easily knowing the flow rate we can easily calculate the how much drop at different places even say for that per pass if we know the dimension of the orifices even we can calculate what is the coefficient of discharge at different points now the flow equations that means we are now considering the flow from here to this to this to this path and here so first of all first we consider the flow through this path q d we have measured and then we can write down this equations this is root under the pressure difference here this pressure we have measured not here we have measured the pressure here because the dimension of this rest constriction could not be measured however there should a row term this row term is included with this term what we have considered ca is a coefficient not coefficient of discharge this as I told including this row figure this is a coefficient this coefficient is for this orifice and this coefficient is for this orifice these two restriction say like a spring resistance in a series so as you know for that purpose we can write down the equation in this way so we develop these equations but this flow again equal to flow through bridge so we can write down this equation the c b also includes that row value I will show that how it is included then through the orifice d we can this is the variable orifice this equation definitely will be different where y p is the displacement of this puppet ok c p is the again the coefficient of discharge and row part and p c p d is the pressure difference between these two points so we have written in this term but this can be detailed analysis considering the area etcetera I will show that area is also there included now flow through e also we can write like this now here I would like to mention that orifice a and b like the if we say this orifice normally we think this is a very thin plate plate sort of things through which there is a hole now orifice may be a straight hole or orifice usually may be in the convergent towards the downstream what we saw these are these two are of that type this is convergent in the downstream and this diameter is very small 0.5 or less however this orifice was slightly longer this e is a small hole diameter is more or less same a slightly higher than this but this was a long one but as we know that for such a small orifice and such a long orifice knowing the diameter of the hole and length ratio we can use different formula like this however here still this formula can be used if we can properly evaluate that c c e now in all this above equations this four equations I have shown c a c b and c e for the three fixed orifices a b and e respectively these are defined in a general form as c i is equal to c d i a i root over 2 by rho. So, this coefficient includes the coefficient of discharge area of the orifice this is the nominal area I would say nominal area in a say we directly we can write this a i is equal to the pi d square pi 4 where d is the diameter of the hole without considering its q of h. Now, i is a b e etcetera for respective orifices now c d i can be evaluated from the earlier equations 4 16 2 if you remember the earlier equations I think it will be 4 15 not 16 the earlier lecture. So, this is 1 by c d square is is equal to 64 l by d where l is the length of the orifice d is the diameter of the hole and r e the Reynolds number plus k as a constant. So, l by d ratio the orifice varies from 1.5 to 10 normally in this case these are these two orifices was very close to that whereas, this was slightly higher, but not 10 exact dimensions I have not shown here k is a constant which can be evaluated from the above equations knowing the value of c d from experiments if we find out we can find out the c d then this value also can be found out and then for a particular type of orifice perhaps we can develop a and empirical relations. So, that we can calculate k any time the coefficient c con for the constriction here after orifice a is expressed as with this relations we put the constriction area here and 1 by k loss we measure simply the loss there and from which we can find out this coefficient. So, as we know the pressure drop at different point to measure the loss over here what we did we removed this one and also we removed the orifice simply we measure the flow and pressure drop at this two points and then we found out what will be the loss there. Now, from equation 21 the drain pressure p d can be readily evaluated in the written line and pressure p r is also known this pressure also measure. So, this is known. So, with such the known values we can calculate the initially what we did we evaluated all coefficient of discharge whenever a orifice is put we first evaluated the coefficient of discharge of all such points. Now, this p r is not very high it is 0.5 to 08 mega Pascal only that of course depends on how much restriction you have put in the written line if there is one filter only then may be one value if filter with say cooler is there then it will be more this depends on the resistance which easily can be measured. For the relation of the p c p a and p a s from equation 21 the pilot property displacement has to be determined first. Now, you see this first we have measured the pressure and next we have found out the c d and from which we have prepared the set of equations. Now, what can be done with those using those equations again we can p c p a p s all we can p s is known value all we can calculate with respect to this displacement y p displacement. Say for example, this means that next for the next calculation we need to calculate the y p how much will be the displacement ultimately we have to put in terms of the displacement all the equations. So, in order to find y p that is the pilot puppets pull displacement p c and p d is replaced by the pilot stage force balance equations. So, this equation 14 if you remember this 14 equation was used in terms of q d and y p obtained from the equation 21. So, first we calculated c d and etcetera then we from knowing the q d we calculated y p I will show this equations. Now, we should recall here what is the geometry for this puppets pull or the orifice area if you remember in the earlier lecture I described that how the flow is occurring here the flow is coming over here and then this flow is going like this and it is ultimately going into the tank through this orifice e and leakage passage. Now, the we know this angle we know this whole diameter. So, area must be this annular passage that which we can calculate from this I mean it is not like that that we are subtracting this area from this area rather we are considering the perpendicular path through which the leakage is going. So, we have taken this component pi d p is the peripheral area d p is the diameter of this hole and pi p is the angle. So, we use this equation to estimate the area. Now, here another point is there what is the z angle here in normal course we can expect z angle will be pi p, but in real case it is slightly higher than that. That means there is a separation of the flow here. So, flow you can say that it might be slightly higher than this. So, actually we should take that angle here, but there is no way we can measure this angle. So, what we do we take this angle later we put a correction factor knowing experimentally what is the flow actually occurring there. Then with further derivation equation 14 is transformed into a third order polynomial equation for y p and is expressed as I am not showing the details of this equations from the note you can or if you just go back to the earlier lecture from there you can calculate and ultimately we can put this equation in this form. This is the third order polynomial equations where a can be put k s p and c p and b is put within these terms these are coefficients these are all coefficients where this is the force driving force and c is c v p then eta q d 3 by 2 2 pi d p by rho sin 2 theta p. So, these derivations again I am not showing here these derivations can be obtained once we exercise these equations then automatically this terms will come and which are put in this form and d is equal to minus a p into q d a p is the area of this I mean this whole not this or if we share a p is the this whole diameter which is a known factor q d is also known. Now as in equation y p equation 24 clearly there will be three roots one of that will be positive and real and which can be considered as pilot puppet displacement. So, if you solve these equations then we will have three values one of that which is positive and real we will consider that value for the puppet displacement. So, combining equation 21 to 24 the different pressure are finally expressed as this p c the pressure here is equal to p b that means p c and p b this zone pressure is derived by q d is the drain flow divided by c p y p square plus p d then c p we have evaluated earlier y p we are calculating from these equations solutions and then clearly we can calculate the pressure there and p a is q d by c b square plus p b and p s is q d square by 1 by c square plus 1 by c con square plus p a this all the equations written from the earlier we will have the equation we wrote the equations and the this just we have written this equation in terms of pressure. So, this easily can be evaluated and after that the variation of all the forces involved in the force baron equations with the common voltage for the pilot stage illustrated here which we can derive this force equations earlier we have shown equation 14 this earlier lecture we have shown now here we have plotted now if you look into this figure now here system flow is this much q s that is that was the system flow and for that f d r f this is f d r these are different forces f d r is the force form for solenoid f s p is the spring force f f p s is the flow force am I clear f d r is the force generated by the armature I mean that solenoid proportional solenoid f s p is the main spring force not this one there is a spring this side this is not the main spring main spring is here. So, that force and then f p s is the flow force here and f p r p is the flow force here and is the force generated by the pressure here. So, these are plotted so this will be the nature of the forces here now individual forces are evaluated analytically from their respective equations this which we have done earlier it is observed that throughout the entire operating range the pressure force f p r p largely follows the driving force f d r this is the driving force the spring force remains nearly constant in between the spring force after some initial increment. So, this means that at the say for example, when the valve is being operated may be the voltage input 4 to 7 or 8 rather here the linear portion is 2 to 8 for this valve. So, for voltage input from 2 to 8 what you will find this spring almost acting as a solid bar and the armature force is being generated by the balanced by the pressure force and flow force again I will repeat that why we need that spring if we give that force directly by this pole there will be nonlinearity this nonlinearity will be reduced if we use a spring in between that the flow force is found to be increasing with the increase in common voltage. Obviously, if the force is increased then what will happen there is a much pressure drop with much pressure drop there is the increase in flow rate. So, there will be increase in the flow force. However, the magnitude of flow force is negligible in comparison with the other forces this is again that although this force is increasing flow force, but still it is less than the other forces unlike the pilot stage in the main stage the flow force has a significant role on the steady state force balance of the puppet. Now, here what we are showing this is the main puppet we are showing if you recall the valve configurations this is the main flow through which the main flow is going to drain in case of relief. So, in this case the flow force has significant role the main reason is that quantity of flow is much much higher here and definitely as I told for the balance for the longer stability this valve is designed this means that this orifice area is very small. So, that there is enough pressure drop to maintain the system pressure close to the system say in case of direct valve what happens once this puppet opens the pressure drops inside then again it close and again it opens. So, there will be chattering sound, but in case of this balanced piston I am pilot operative valve what is called two stage valve this remains balance for longer time. So, that is why there will be the less orifice area much flow and due to that here will be the high flow forces. Now, here the flow force for different input conditions are evaluated using the expression 12 here the equation 12 I am not showing you should compare with the earlier lecture and this is according to the equations 12 whereas the from the equations the other equations there is equation 18 we get a different value one is one the equations we developed from the general equations another with respect to the valve in terms of force and other things. So, why this difference is there definitely as I told we could not consider all the parameter inside. So, there will be some differences now these differences are observed in all cases who have analyzed such valve and therefore, we need to rehabilitate the equations considering a factor for these corrections which we called correction factors. So, the flow forces obtained from two approaches are clearly illustrated in the both the figures earlier figures and this figures also and then the inequality essentially demands for a correction factor which should be introduced in the expression of flow force. Now, we will look for that one now the modified equations of flow force is then can be written as with a factor we have these equations again we have rewritten and here we have earlier we developed these equations without this one. If you remember this equation we developed earlier where we did not have this one. Now, we have to calculate this one this kappa this factor this is the correction factor. Now, where k the correction factor k or kappa whatever we call it I think we used kappa the correction factor is the ratio of flow force obtained from equation 18 to the ideal flow force given by equation 12. 12 gives the ideal flow and 18 gives the obtained flow from these geometries and other things that means, 18 equation without this kappa was there. Now, in some previous investigations this in the reference earlier reference we have given today I also give the references this say Davies it is 1994 this is 1998 this is even 2002. They faced similar problem and we also we carried out such analysis in the year of 2004 yeah 2004 and this time we when we had to face this problem we found they also faced similar problem, but their valve was different. So, we had to evaluate this value of our own as shown in figure 8 the factor k is not constant, but is found to be varying with the supply pressure this is not the figure 8 the earlier figure 8 this kappa value is not constant in found to be found to be varying with supply pressure. So, measuring the supply pressure we can find out this values. Now, to find out this what first of all we should know what is the angle this poppet angle here. So, knowing this poppet angle due to the typical poppet configuration the flow angle may not be same as the half poppet angle for all the input conditions I mean here again this we wanted to mean that z angle is not always matching with the phi s. So, due to that this correction factor is required the actual flow angle seems to be larger than this geometric angle of the poppet large flow angle enhance the flow force the assumption of uniform pressure distribution on the external poppet surface may not be applicable here the characteristics of the flow force correction factor may be used as a reference for the sensitivity analysis of the valve. So, this not only this correction factor from the correction factor again we can examine the sensitivity of the valve what is sensitivity that changing the parameter like a spring constant and other value what will be the change in the flow or pressure etcetera. An attempt has been made to calculate the armature displacement. So, if we think in terms of design of such valve. So, now we need to consider what is the armature displacement without the details of the constructional and functional features of the solenoid the displacement is predicted from the geometric details of the pile of threads and the developed driving force model you see as I told that it was very difficult to measure the displacement of course if we can calibrate if you could calibrate the LVDT and if you could the take the output from LVDT then it could be possible to measure the displacement, but whatever the information we already have we have the geometric relation inside we have the spring stiffness we have the flow and from there we can find out what will be the displacement and from that equations definitely it is possible to find out the armature displacement and what we established and with that armature displacement we again reexamine for different pressure setting and we found that result was correct. So, we consider that is the armature displacement and ultimately we developed all the equation in terms of armature displacement also. Now, here this first of all to do that what we measured that we measured this length so this is a fixed geometric length and then say this is the say this point is touching over here when it is 0 positions. So, what we measured the LP we measured from the top to this positions that means if we measure this distance that is LP was when LP was 0. Now, this is you can say this is the opening from the orifice point and then this is the position of this seat I mean this step and here also this step these two are some fixed dimensions and then this is the dimension for the compressed spring which we the free length F LP is the free length of the spring minus this displacement and here again we had to measure this this cap dimensions because the spring was sitting from here to here only. So, this dimension we had to include and then by equating we were calculating the displacement of the armature. So, here the critical values are these because what is the compressed length of the spring and to find out this armature length and we had to consider this length also. So, these are all fixed length geometric so this will not vary, but once with the force this will vary this will vary and this will vary. Now, next the equation the armature displacement can be written in this form I am not going into the details knowing these dimensions you can easily find out say this is the L I O this is the LP. Now, it is not exactly matching this I mean because this is written in may be one format this is other format, but these are corresponds to this. So, L C is here these dimensions L L is this one L B is this one. So, you will be able to understand why this figure is enlarge I think you cannot read it, but there is slightly mismatch anyway this is this dimension which can be estimated this is the fixed this is also fixed dimensions this L C is also fixed dimensions this is also we have to measure. Now, where the poppet displacement y p and the pilot main spring compression y h s p this one y p is evaluated from the analytical model we have already developed. Now, the driving force is transmitted to the poppet through the pilot main spring therefore, y h s p is found from the equation of the spring written as y h s p f d r this is the driving force of the armature divided by k h p is a we can say this spring constant of the main spring. Now, this is again you can see are evaluated all three displacement involved this armature this and this in the pilot stage are plotted against the current. So, if we have now plotted these three characteristics now this one is the y h s p that is the spring compressed spring length. So, this displacement of the spring and y p is the displacement of the main spool and y armature is the displacement of the armature. Now, if we look into study this curve this is as you can see this linear and this is also we are getting linear whereas, this y p is not linear with the displacement and current. Obviously, when current is less then displacement will be much higher this is obvious you see this current is less means the load is less and that load definitely this less pressure drop. So, there will be more displacement it can be seen that the armature displacement becomes linear after some initial non-linear region at smaller current range suggesting presence of inherent non-linearity within this valve. You can say this this is in terms of current we have written because ultimately in the armature what is fed that is the current our input is voltage that converted into current and that current relations we find non-linearities here as well as there will be some non-linearities here also. So, from this to this you can say linear and this is we normally keep the operating zone from this to this that means about when the current is 0.7 to 1.6 it corresponds to roughly 2 voltage to 8 volts. So, now as you see this from this point of view we need to modify this our equations for in the usual form and what we find considering that the armature position for different input currents for different driving force that is for the different driving force is independent of pilot means spring stiffness. So, definitely we can examine if we change the spring stress spring stiffness what will happen of definitely the range of voltage might be it will change, but performance may not change very much as I told after an initial compression the spring almost act as a solid var. Consequently the pilot for a proper displacement may be expressed in terms of driving force pilot the main spring stiffness and the armature displacement. So, modifying equation 27 now the expression we can write in this form here directly we have the spring stiffness which we have measured and the force also we can estimate from our input voltage and others and we can calculate. So, with this we have this part we have replaced this one we have replaced by this force. So, now y p can be calculated and in the analysis the pilot flow rate is no longer an input it is not an input our input is the voltage we can say it is an unknown quantity and is to be determined from the analysis say. In fact, y p now we have to estimate for the estimate the performance the pressure drop across the pilot stage is to be evaluated before the evolution of the pilot flow rate from the force balance equations for the pilot stage. The pressure drop may be evaluated as now we can write down these equations this a p is this or if this area a f s is this ring area and we can calculate p c minus p d knowing these two force. So, now we have to finish very quickly. So, I would like to say the from the pilot stage force characteristics the pilot stage flow force is found to be negligibly small in comparison to the other forces. Instead of using equation 3 for the flow force it is quite reasonable to express it in terms of pilot puppet displacement and pressure drop assuming constant coefficient c d p and c v p the final form of the above equation now written in this form whereas, the c d p and c v p is the two coefficient of discharge at this points. So, these are separately evaluated and we can now write down these equations. Once the pilot puppet displacement and the pressure drop are evaluated pilot flow rate can be found directly from equation 14. The individual pressure are then evaluated using the equation 21 to 25 and here we have seen the three different plot. The analytical results what we have done actually we have estimated what will be the sensitivity the analytical results is plus minus 5 percent change in the pilot spring stiffness. This is purely theoretical analysis what we have developed with that why we have shown here you can see the change in pressure and this common voltage. This is 100 percent means the original spring and if we increase this stiffness by 5 percent then it is this one and if we reduce the stiffness by this one we will get this is the curve. This means that by simply changing this spring we will have different pressure range for different voltage, but the nature of curve is more or less same. Let us examine the other also that what will be the changes in q d what we find here for the our original spring this is more or less constant linear whereas, due to these two changes what we find it is becoming highly non-linear. So, and also if we look into the flow force correction factor we find that these two are also different from the linearity and then obviously if we want to use say for example this is 5 percent less we have to very careful probably this design will not be feasible whereas, by increasing the spring stiffness the 5 percent we can still use this. So, this is with this I would say that I wanted to show that how a proportional valve can be analyzed for its steady state performance. We have not yet shown what is the transient performance. Now I think you can understand this from the self study and there is of course the conclusion part which I am not going in details because you have to read it, but I have discussed throughout this presentations what might be the conclusions and finally these are the we these are the main source from where we have developed I have developed this lectures including our own research and thank you for listening.