 Welcome to today's lecture. This is continuation of design analysis of orbit motor part 3. Today we should discuss about how to find out, how to formulate the output torque, stress, deformation and gap at contacts. Now, if we look into the figure this is an interview of orbit motor where we have drawn a line through the contacts which are separating high pressure zone from low pressure zone or in other words the expansion mode to the compression modes. Now, what we have done we have joined we have considered a line here and if we consider the width of this star and ring then we may consider this is an beam where the pressure are uniformly distributed over this length joining two points. Then we can consider the total force acting at the middle of this beam so equivalent force. Now, the torque must be equal to this equivalent force F i into the distance r i which is from the instantaneous centre of rotations. Now, here one may confuse that we are considering the torque about the star why we are considering such a distance. The reason is that what we have considered this is the equivalent system we have considered in that equivalent system what we find that as if a there is a fixed drum in which this drum is rotating. Here of course if we consider the drum that is a friction drive which is not positive but let us consider these are the two gears. So, definitely to calculate the torque what we can find that as if this body is rotating about this pivot point by this force. So, definitely to rotate this body about this point whatever the torque that is in vector product r i into F i this is r i into this is F i this is half of the distance. So, that torque must be the output through this body which is nothing but the star. So, we can write down this is the torque. Now, this is sometimes it is called energy method but whatever may be the name in this way we can find out the theoretical torque output of this machine. This is the distance r i we should always remember. Now, I have drawn another figure here in this figure again I have shown here apparently slightly it has rotated by this angle in this case as if it has not rotated it is at this 0th angle if we remember our earlier access system. So, this is the x axis and this is the y axis this is the reference axis which is coinciding with the ring gear. Now, here we must know because we will refer i th chamber or roller etcetera. In this case we have considered this is roller 1, roller 2, roller 3 etcetera. Therefore, this must be the z th roller. In this case as we see in this figure we have considered 7 roller and as well we find the lobes in the star or 6 number of lobes the star in 6. Now, also we consider the chamber 1 which is in between 1 and 2 roller chamber 2 which is between 2 and 3 etcetera. So, this chamber at the 0th position that must be chamber 7 or z th chamber which is in this case 7 th chamber. Now, obviously the instantaneous point of rotation of equivalent drum is now situated on the x axis. Now, this whole body of the star is revolving around this axis in the anticlockwise directions whereas, this body itself rotating in the clockwise directions. So, output will be clockwise looking from this side. Now, in this case we have considered the high pressure zone and low pressure zone separated by this line and these two contacts we may consider active contacts. First of all these two contacts are active contacts. All contacts which is joining the instantaneous centre of rotations and the crew node which is nothing but the centre of the roller then these are the active contacts and among these active contacts there will be 7 active contacts. Among these active contacts these two are you can say the critical contacts which are separating the chambers from high pressure zone to low pressure zone. Now, instead of considering the axis I mean sorry the beam like this we may also consider the all contacts points joined together and we get in that case we will get 7 such line which obviously not equal in length. We may consider however that if we consider the these are the critical joints which are separating high pressure zone to low pressure zone we may consider in this side there will be the pressure each chamber is having equal pressure which is high pressure ok. Here of course I have considered that apparently this is a neutral conditions we do not know what is the pressure has history. It can be shown that even if we consider this is in high pressure zone or if we consider this is high pressure zone the pressure sorry the torque calculation will show the same value ok. Anyway if we consider this is in high pressure these three are high pressure. So, this is experiencing high pressure and again on these lines pressure distribution will be uniform equal pressure but keep in mind while this is moving these lengths are changing. So, these lengths may not be equal are normally not equal and on the other side this blue lines indicates these are separating I mean the chambers adjacent chambers by these contacts and joining these lines we consider the pressure is the outlet pressure you one may consider this is 0 and this is the differential pressure this will also give the same torque value this would give the same torque value. Now, however the equation for this will be different in that case what we are considering say this is I is varying from 1 to z any chambers in that case this is the length. So, force will be at the middle of that and then that multiplied by this distance say this is a typical one is shown here only one chamber I have shown the pressure is being distributed like this considering this equation will be better in a sense that actually in such machines we will find later that the gaps are generated here. So, case may be that pressure are not distributed over the whole length uniformly it might be the pressure is non uniform in that case if we can suppose they this pressure distribution is non uniform then if we can consider such mapping then we can accurately calculate the load and this distance the distance from that load to this instantaneous center of rotation and this will give us the more accurate torque that is the purpose of considering the pressure distribution in that way. So, here I have written the due to the inner chamber leakage flow through the capillary passage formed by the lobes adjacent to the active contact and at the flow distributed valve the pressure within the chamber may vary not only this we should remember the flow distributed valve is very close to that. So, due to that there may have the pressure change due to leakage through that. So, once the leakage through that all the paths are considered we may have different pressure history instead of the ideal pressure distribution. Now, let us for the simplicity let us consider there is no variation of this leakage then to find out the length we must know this contact points which can be calculated from the geometry for a particular rotation for a given rotation of the output shaft and then we can find out that length and we may consider the pressure p c n that is that chamber the pressure is p c n and the length is l i n and so this will be the force we can calculate the force in this way. Now, we can see the variation here now look at this variation you can see this lengths are varying again we will see this. So, we have just shown one chamber operation rotation for one chamber again you can look this how this distance are being changed and we have considered that this is the midpoint which can be this is the midpoint. So, this geometrically this we can calculate and easily we can find out the forces we can define this distance and planes in the vector form and we can do the vector multiplication. So, in that way torque can be calculated. Now, what we have calculated so far that is the ideal torque output of such machine. Now, in this machine first of all if we consider the assembly of rotor and stator we may consider that this is just exactly fit this means neither it is interference nor any gap at these contacts when these are assembled. In that case we should consider that there is no interference fit nothing is there. Now, due to the pressure we have considered we are considering now the pressure inside once there is a pressure means load and there will be deformations this deformations means there will be may have gap the contact points will be deformed in that case there are we should consider the several losses. One is that at that contacts when this is moving on that earlier we have examined that this is having both rolling as well as friction so sliding rolling and sliding. So, if we consider the combined friction there definitely there will be one pressure loss and also this what the length we are considering ideally that may also not exactly because the contact has now deformed. So, length is changing due to that there will be variations. Now, let us consider that there is some interference fit to reduce the gaps etcetera in that case due to that interference fits another additional torque will be required which will not be available at the output shaft apart from that there may have some error also considering this one can carefully express the output torque in this form. Now, we will see that what are these parameters L i is the ideal geometric length of a line joining two transition contact swing at any instant this one and then L i a effective length of line joining two transition contact points at any instant what is effective length and what is ideal length ideal length what we are calculating initially from the contact points without considering considering any deformation at the contact points. Now, if we consider the there is a deformations then contact points may shift if that can be calculated then we will consider that this is this will be definite the deviation of the ideal torque. So, that is there. Now, then what is MLH it is called load holding a resistive torque without fluid pressure this can be calculated or experimentally can be found it out if there is some interference even if we consider the ideal fit still there may have some frictional resistance to move this rotor. So, that is why we can find that initially to start that motor you need some pressure just to rotate without any load that you may consider the load holding torque that is that this means that we need a torque to start this motor without any load that is load holding torque on the other hand this load holding torque is again we can name that is a starting torque required. Now, this obviously this increases depending on the fittings if the fittings is too tight in that case this part will increase that can be estimated also this is found out normally by from the experiment. Next we consider the theoretical torque at starting pressure that means when it what we will say that we are giving a pressure, but still this is not moving. So, that pressure into the when pressure is working on this machines without moving that is the load holding part, but once it start moving what will happen due to this forces present there there will be a friction. So, considering this friction coefficient or the correction factor we have given the name to account torque loss due to the starting pressure. So, there will be another loss. So, that this means that actual available torque is this much the ideal torque then there is a variation in the this length. So, that part it will be slightly less than this actual torque then this load holding torque and then when it is starting there is a friction due to that there is another loss. So, this is the total loss. Now, this we have considered that there is no deviation in the eccentricity and if we consider further deviation in the eccentricity there will be further loss. So, this amount and this much amount can be can also be considered. So, this is a proposed model to estimate the available torque, but we may this measuring such this error or shift in centre distance is very difficult. So, this we may neglect or in other words even if possible if we can measure or we can estimate that then whatever additional loss that can be attributed to this one. Now, the instantaneous torque efficiency now we can this is the ideal torque and this is the output torque. So, this may be considered as a efficiency of such again for force equilibrium. So, this each chamber force and this efficiency must be equal to this ideal force at the contacts offered by this is the resistive force this is offered by the elastic deformation of the body. So, we can equilibrium in this way. Now, if we consider this is due to the ideal pressure then this factor is 1 and then we will find this force is equal to the pressure force, but actually we will find that whatever force is there this is more than the pressure force. Now, local deformation occurs at contacts initially due to interference speed f n e and then due to fluid pressure. Therefore, it offers the normal force f n which is expressed as follows f n is equal to f h n and f p n f h n a and f p n are the reaction forces due to the contact deformations for interference speed this is for interference speed and then fluid pressure respectively. Say for example, there is no fluid pressure let us consider. So, we do not need or in other words if we consider that there is no torque output torque is required. So, at 0 pressure ideally just flow is given it should be able to rotate, but what we will find that it requires some amount of torque because there is a interference speed. So, this means that there is already some different contacts are different and due to that the force h n is there. Now, what we do we add fluid pressure because the output torque is there. So, therefore, we should consider at each contacts there are two forces acting originally due to the interference and then due to the pressure. If there is no interference we can consider this part is 0 and normal force equal to due to the pressure. Now, with ideal size of rotor stator gaps occur at few contacts which separate chambers in expansion mode that is chambers with high pressure and even at active contacts separating low pressure chamber from high pressure chamber. Now, here I would like to mention that we are trying to find out the forces and each and every contact. Then we have considered that apparently this one we may consider that this line which is separating high pressure, low pressure then this is a beam. Now, this beam is supported by how many supports are there at least this will be the number of chambers divided by 2 number of chambers plus 1 divided by 2 either this or minus 1 divided by 2 it has to be this is for the odd number of chambers we are considering odd number of chambers. So, by that way say for example, in case of this 7 chamber machines this will be either 3 or 4. Now, in a beam there are 4 supports or 3 supports. So, this will be statically indeterminate problem. Now, also if you look into this the support and their direction of force to find out the direction of force to find out the direction of force what we can do or we can join the instantaneous centre of rotation to the centre of the roller or centre of the crew nodes of the envelope. So, we are getting the direction of the force direction of the support, but if we look into the direction of the support these are not parallel or not perpendicular to the beam what we have considered this is at angle. So, the problem to find out this the loads is not an easy task we have to consider all the directions. Now, again looking into the direction of the force we may find that in some supports some supports are not taking the load at all and as well we will also see that a gap is generated. Now, this gap say this is the high pressure side so definitely beam supports are in the low pressure sides and these two are critical and we may find at some instances they are a gap is generated. Once the gap is generated you will find there will be the leakage. So, estimating such leakage is essential to improve the performance of such motor. Also what we the valve is very adjacent to these. So, they are also the possibility of side leakage like this here one is leakage through this contacts another is the side leakage from this valve ports considering all such things we can have a redistributed pressure particularly where this capillary passage is there we can have a distributed pressure over this chamber and that distributed pressure if you put on that we will find a different ports. However, depending on the pre deformation due to the interference speed of star and ring the gap may not occur at all for a reasonably high operating pressures. What we can do we can make this as an interference speed this is basically such machines are called form closed that means even if without any force there will be all contacts all seven contacts in this case without any force and geometrically that for contacts will be maintained. Now, if we put the interference speed interference speeds means actually at some point we will describe that this rollers are over size as well as the star are over size. So, in that case what will happen each and every contact should assembly is difficult, but that is done with some technique heating one element and cooling the other element that assembly is possible, but once it is assembled then there is whatever may be the small deformations will be there. Now, with the increase in pressure that deformation will gradually decrease and at one point with the higher pressure definitely there will be gap, but we can design such machines that up to the working pressure there will be no gap at the contacts. Only disadvantage of such machines will be there at starting you need a starting torque higher starting torque, but once it reaches at its working zone you are not require no leakage as their efficiency will be higher although some initial torque were required, but you may find that overall efficiency is more or overall performance at least better than the ordinary I mean without interference speed machines. As the material of star and ring are usually still it may be assumed that they have same Poisson ratio and modulus of elasticity and then we can now we can consider the deformation at a contact. This is along the normal directions that means in this directions this will be the deformation so this expression will give us this deformation if the force are known here we do not know how much is the deformation how much is the force we will we are trying to find out the force at each and every contacts. Now, a n is the bandwidth half bandwidth which is given by this formula this is from Hurgian contacts stress theory so you can this is perhaps known to you we have only written in this form here this is the contact radius of the star and this is the contact radius of the ring, but you should remember this plus sign means both are concave. Now, if one of that but if you look into the star this is having convex as well as concave when this is a concave then this there will be a minus sign minus plus sign so you have to take care of that part. Now, what is this one h a h a is a depth after which this deformation occurs this is a complex things, but this we have to assume for such machines what might be this depth h is average compliance depth in this case we have consider this depth is up to the centre of the roller so this is equal to r m maximum but this is an arbitrary however if you consider this is half of that or double of that you will find this overall contribution to this will be negligibly small. So, it is a reasonably good to assume this h a is equal to r m so in all calculations we consider this is equal to r m this radius already I have explained to you radius of curvature. Now, for the dimensionless analysis this formula is further simplified which is given by this developed by Cornell and this can be written in this form. So, this I have given this article as a reference one can read to how this was derived once we reach here then we can non dimensionalize it is like that. Now, earlier when we are non dimensionalizing a dimension we are simply dividing that one by the radius of the base circle or the describing circle of the ring here that is r 0 we are considering this radius r 0 of the biggest circle here it is a green circle whereas when we are trying to non dimensionaling this force what we do we consider this force normal force divided by b that means this is coming per width and then divided by del p i which is coming as a per unit pressure and then r 0 is this radius radius of the biggest circle this look into this unit this is area and this is pressure so this must be also force. So, this becomes a non dimensionalize that means once we calculate this deformations or force for a machines considering this approach then for actual pressure and actual width we can find out what will be the actual deformation or actual force and beauty of using such non dimensional analysis is that you can analyze the machines for a particular geometry and then you can design you can consider the pressure actual width flow rate everything later to give the physical size of this machines keeping the geometric relation same. Now, this is the deformations in the same way we can express in the non dimensional form. Now, the contact deformation on star at the active contracts nth loaf of ring can be expressed this is actually this is in the vector form. So, we can know not it not in the vector form we would say that this is this what we are doing the actual deformations total deformation from there we are trying to find out what is the deformation in the normal directions we have considered any deformations at any directions then by this multiplication we are finding out what is the magnitude of deformation along the normal line. So, for that we need this operations and then also this actual area of the contact can be find out because we have calculated the half bandwidth. So, total bandwidth will be this much and b is the width of this star or ring. So, this we get actual area. Now, then finally we can express this formula the f n non dimensional is equal to 0.44 into delta n look at this this is not for the both star and ring material as a steel we have considered that. So, Poisson ratio is around 0.3 with that we get this relations this is very simple relations any force in the normal direction is equal to 0.44 times of the deformations. So, this analysis become easy if the calculation of contact deformation forces etcetera caused due to the interference feet or oversize star and ring it is assumed that excess size is different in the direction of the normal to active contacts that is in the direction of line joining the instantaneous centre of rotation to the centre of low or roller in ring now here I would like to mention you see if you consider a simple cylinder and a hole and cylindrical body perfectly round and they are in interference feet you may consider deformations over the contraction of the cylinder is uniform over the periphery expansion of the cylinder is also deformed uniformly over the periphery. In this case we we had to depend I mean such interference is taken care by few contacts only. So, most likely it is not uniformly distributed and if you think if you try to visualize there is a possibility of that unstable conditions that means for such a for interference feet such interference feet ultimately you will find this star will try to take a stable position what is that stable position you will find one of the chamber will be either in the outer dead centre or what of dead centre here you can say this is at the outer dead centre. So, this is one stable conditions next another next stable condition will be this one of these two chamber will be inner dead centre. So, which is precisely can be found out by the pi divided by z into z minus 1. So, every of these two chamber will be after that angle this will be in the stable conditions or in other words you will find this machine if you leave this machine it will try to go to that position for its stability. Now, in other conditions suppose if you hold that shaft at the other conditions what will be the situation in the machines this machine due to the pressure it is giving a stable torque you will find apart from that there is some unbalance amount within this machines that means to hold the shaft at that particular position you need some unbalance torque although the magnitude is may not be very high, but this unbalance torque is there. So, in the force analysis definitely such contact forces giving this unbalance torque you will find out. So, while you are calculating automatically this should be should also be taken care of. So, here I have shown that the equation using equation 9 and 10 one can try to find out the forces, but this we have to follow a particular technique that how we can find out this forces. Let F h and b the reaction force due to the contact deformations for interference at nth low then we can have that summation of all such forces due to the interference at no pressure must be equal to 0. So, we must satisfy that condition to find out this force and then we can consider this load hold holding torque must be equal to that that force into the coefficient of friction into this R n in that case to calculate this R n we have to consider this contacts point to this distance, because here this is the normal force. So, this force is in this transverse directions and that multiplied into R n will give the load holding torque. You see the coefficient of combined rolling and sliding friction at contacts. This means that to rotate the shaft even there is no fluid pressure the required torque is m L h. Specularly for such geometric form closed machine this summation of forces may not become 0 in all the angular position of shaft rotation and the rotor is virtually floating as the rotor is virtually floating. The summation will become minimum floating rotor will assume a comfortable position by deviating from its ideal geometric positions which I have explained. It will actually we will try to take a position for which one of the chamber is either at the top dead center or the inner dead center or outer dead center or the inner dead one is outer one is inner one is a top dead center one is bottom dead center whatever name you can give it to this. It is better to give this is outer dead center and this is inner dead center. We have also verified this in numerically and that can be shown. Now, in calculating this forces f p n part that is the force due to pressure. First of all an arbitrary small deflection delta d n is considered in the dimension numerical competitions. We will consider any arbitrary value small deformation in opposite directions of the resultant applied force. That means suppose this is the direction of the force. We have considered these two points say shaft has rotated by a small angle for which these two are the critical contact points. So, we know these lines we know this direction we consider the transverse directions of this. Then we consider deformations are taking place in this directions. That means whatever deformation is there deformation is taking in this directions which is perpendicular to this first we consider. Let us give a very small because this is dimensionless quantity. Let us take 0.01 is the deformations numerical value. Now, what is there due to these deformations we can find out the component along the normal directions. What will be the deformation at these points? Amount of this deformation then from there we can calculate what might be the contact forces there. So, we can find out that resultant forces there. Now, we consider that resultant force. So, we have considered only in this directions. Now, let us consider that these deformations we are rotating. Rotating means may be only pi angle about this center. We are considering these deformations we have rotated. For all such positions we will find these forces. We consider which is the minimum force for which conditions for the same not one deformations for which direction of the deformations the summation of force is minimum. Actually that is that would be the directions of this I mean deformations because there is no meaning that it will be the higher side. It will be always in the lower resistive path lower sides. So, we find out first that positions. Once we find out that positions we then that definitely not in the directions of the load this is in the resultant in the some other directions. Now, the component of that force that summation of this resistive force due to the deformations is contributing towards the torque. But the other component which is in the transfer bed directions of that force that definitely generating an unbalanced torque of this motor and you will in such calculation you will find when one chamber in bottom dead center or at the top dead center this unbalanced force is 0. So, after calculating such this force now if we would like to quantify suppose we have a particular pressure operating at a particular pressure we know other dimensions then by multiplying this unit deformations for the given deflections we can find out what will be the actual deformation what will be the actual force what will be the actual torque what will be the actual unbalanced torque. So, this is the same what I have described it is written here and for dimensional analysis the following waiver formula same approach is followed when a particular model of known dimension in analyze. Now, what I have done in a particular case I have taken later for the numerical calculations one can find that non dimensional analyze and the dimensional analyze using such formula and if you calculate that there will be definitely some deviations. Now, we have calculated few such machines now in this in such machines if you look says for example, set one that is one machine dimensions in that case you can see this is the dimension. Now, this is you can find that plus some value what does it mean actually this is a coral thickness to thickness this if you use a simple caliper you will find that these two there will be some portion of the curvature which will be the tangent of which will be parallel. So, we measure this distance now for an ideal case whatever the distance that we consider the ideal dimensions now if we increase this dimensions by this much only this is a practical case we have taken this much then this certainly this is an interference bit and what we find for such machine we get this is the amount of torque we can find out we can have out of this machine this is again Newton meter per 10 millimeter of width per mega Pascal this will be the output torque. So, if you suppose you have 15 millimeter. So, simply you multiply with 1.5 and pressure say another 15. So, 1.5 into 15 into this this will be the output torque of such machine ideal output torque if you consider the second machine in that case this is slightly loose bit actually I would say if you consider a machine without any interference it should have some clearance. So, this machine is without an interference whereas, if you come to this machines this is giving this is the torque this can be calculated the formula we have proposed. Now here is just to show the dimensions different dimensions and then you can see these are the contact forces directions are shown then in model calculations the deflection at active contact without fluid pressures may occur due to the various reasons. For an example in a set star and ring may be over cut the roller or low radius in star may be less than the ideal geometric size and the star is also cut with an over size cutter but the roller placed at the same pitch circle with lower radius this means that when we call this is an interference machine we can create such interference by several means one is that keep all the rollers dimension everything ideal simply you increase the size of this. So, this is one way of interference another is that keep this is intact just you reduce the roller radius keep the roller size also same only the pitch circle on which the roller is placed just to reduce that one. But you have to remember the way you have created this interference in your geometric calculation you have to consider that properly to find out the deformation due to this interference. So, we should know that how it is the interference is created. So, this is described here the projection of distance between two points on the normal at the ideal contact point give the deflections or gap at the contact points. The forces are calculated and the result resultant torque required to rotate the star about the shaft axis is calculated using equation 14 which we have described earlier. Now, the resultant deviation of star center from its ideal geometric position is assumed arbitrarily then considering an infinitely small circular zone around the ideal point the actual center is shifted to verify the possible minimum unbalanced load conditions which I have described. And in that way calculation is done and then we can find out at no pressure if there is any torque is required due to the interference pitch. We can have then there is one constant torque load holding torque average and another unbalanced torque in this calculation geometric and kinematics of the rotor and star are to be considered we have to consider as I told this load holding torque would help in breaking when ill-made flow motor is stopped with outlet is blocked. So, in some cases this is required in other cases it is not required. So, for the particular requirement knowing the particular requirement if we want to design a machines then we have to calculate or we have to follow such calculations. So, I have described here definitely the starting pressure would be more in such motor with interference pitch star and ring. Now, I have shown the result also in that result what we find we have calculated in both way and we have considering this geometry and in that case what we find this is the load holding torque you can see this is a newton meter very small amount of torque, but there is as well unbalanced torque. This means here you can see this is the z pi divided by z into z minus at that point you can find that no unbalanced torque is there whereas, if you want to keep machine at that positions you will find that some unbalanced torque is of this much will be required and again this is varying in negative and positive directions for one directions you will find that need to torque and other positions is automatically this is coming back to its positions. Now, another interesting point is that this is the dimension analysis and this is non dimension analysis you will find there is a little difference this amount is small in that I would say little otherwise comparison to this small you will find the unbalanced force is coming more here. You see it is also interesting to note that if we if we vary this dimensions it has been seen for some values this unbalanced torque is also minimum as well as this is almost without much peak values. So, those are very interesting. Actually better machines. So, this already we have described that is you will try to assume always in such angles. Now, another interesting point which I have mentioned that this non dimensional approach is more conservative. This means that if we calculate in this way it will show always more stress value etcetera. So, non dimensional approach in that way we will be always safe side in the design not only this helps us because the beauty of using non dimensional analysis is that we can for optimization we can optimize the machine then we can finalize the actual dimensions. But in that process it will be over designed, but it no matter this will always give us a safe design. Now also if we consider the deformations at different contacts because we have to why we are estimating such torque and etcetera we have already calculated the deformations and as we find that there is also gap. These contacts this is a contact one that is a roller one, this is roller two, this is roller four, this is roller three and for such angle of rotations and we find that sometimes this is from deformation to gap for different rollers. And if we look into this, this is the dimensional approach, this is the non dimensional approach, but mind it this non dimensional approach after that we have multiplied this factors to give to find out the actual dimensions. Now this is for the other contacts, this is there are differences but it is more or less similar. Now this is as you see this is the unbalanced force, this is sometimes it is quite high of course this is with respect to some pressure we have considered this is not mentioned here, but you as you can see this both the approach giving more or less same value, but what we find this with the waivers approach this is non dimensional approach this forces as more, but actual force may be less. Now this is I would say that if experimentally experimentally is carried out then what we have tried to find out that whether this approach is correct or not. Now as you can see this is the predicted typical cycle, this is the torque cycle it is giving the test result as you see this is obviously it is very difficult to measure the actual test it is at random it is coming like this. So test result typical sample this is the sample and the actual test record is this one, this only one sample we have matching with this and here actual is like this. Obviously the theoretical curve will not match with that, but it is giving more or less similar trend and it can be fairly accurately we can estimate what will be the torque etcetera of such machines. Now this is using the FEM technique and this is for this is the value and this pressure as you can see this 15 mega Pascal and this is the physical size this will give as the physical size and at 15 mega Pascal we may find that this are the gaps are generated and this as you see this is very critical at that moment this gap is being generated. This gap of course there is no harm because both are in high pressure zone. So there will be no leakage, but this amount of gap is generated there only dangerous part is there, but it is again not always momentarily it will be there. Now this also can be realized from this machines. Now this is obviously not an orbit motor, but this is an zero torque machines. In that zero torque machines this is recently I visited technical university of Roslov. They are called as FEM. They are conducting such experiment using the steel rotors and stator, but this is a very thick some sort of glass they are using for the cover and there with the photography they have found this. Of course in this figure you can see only the machines I will show you another one where you can realize the you can see suddenly you are looking at this bubble what is happening if you look into this when this is changing is phase from high pressure to low pressure zone. This bubbles is definitely where the leakage is occurring that means the gap is occurring. So theoretically what I have tried to prove practically also this is happening. This is a good example for interchangeable leakage. You can see this once again this looking into bubbles means this at the critical contacts there is gap. Critical contacts mean when it is separating high pressure zone to low pressure zone and this is momentarily theoretically it is matching with my theoretical calculations whereas it is the those who have conducted such experiments they have not yet calculated such things. They are only experimentally observing. So these are the reference if you can go through such papers you can find out say for example this paper where I will find that Weber approach and the Hurzian approach how it is simplified and non dimensionalized here this prediction of the starting torque etcetera is shown and this is the you can say this is the first paper who calculated stresses for the zero torque pump mainly but he proposed this formulation part initially. However I personally also calculated such stresses in more rigorous way and in dimensionless form which I have described. So thank you very much for listening.