 Welcome to today's lecture on design analysis of orbit motor 2, which is geometric volume displacement. In earlier lecture, the in part 1, I have discussed about the geometric design of the star and envelope that is ring of an orbit motor. In this lecture, I shall show how to estimate the geometric volume of a chamber and then swept volume and how to find out the volume displacement rate with the shaft rotation. So, as I discussed earlier, the epitaphoid generated ropey mass that is rotary piston machines orbit and 0 to units, the envelope ring forms the chambers whereas, the epitaphoid star acts as a piston. If you look into this, this is the chamber. So, one particular chamber is shown here which is formed by you can say that the outer member that is envelope that is ring is the cylinder and this is acting as a piston. Unlike the cylindrical piston machines, in this case the area varies and so variation of volume with the shaft rotation is equal to the variation of area multiplied by the thickness of the star and ring which is constant. It is also discussed the number of chambers is equal to the number of chronodes of the envelope that is equal to z. So, the number of chamber will be equal to the number of roller in this case or it might be sometimes integral part of the envelope whereas, we know the lobe of this star will be 1 less than this. So, here there will be 6 lobes if we take z is equal to 7. Again in earlier discussion what we have seen that the ripple is less in case of odd number of chambers. So, while we shall discuss all the formulations particularly we have to be careful formulations are like that z is always an odd number that means number of chambers are always odd. A complete cycle of piston action comprises of two phases namely volume expansion and volume compressions. Let us look into this. So, this is the star and this is the ring. This is modified epithocate what we have learned and this is basically envelope and this modification is nothing, but the constant difference inward shift of the original epithocate generated. So, for which the envelope actually goes through this centre point envelope and then if you modify this envelope what we find this contact portion becomes circular arc. If you notice this rotation you will find that this contact is not up to the full arc it is some angle the maximum leaning angle you can say in both directions from the centre. So, but the thing is that we can replace this we can either make it integral or we can replace this integral part by roller. Advantage of roller I have already discussed which is that we can replace this roller if this is worn out. Now other part of the envelope is that this circle on which this rollers are kept I mean on the outer body this is slightly less than the circle the which is called bit circle slightly less than that. So, this roller does not come out and it is selected on the basis of at the top most position of this roller. Now this we can call that at this position the piston at its top dead centre top dead centre. Now when it will rotate you will find gradually this space is in being increased the black space is being increased that is the change in area. So, if we can estimate the change in area then if we multiply with the thickness of this star and this envelope which is equal nominally which is equal we can estimate the volume displacement and that is that we will learn today. But one important thing here I would like to discuss that if we consider this is the bottom dead centre then at this position you would not find any other chambers in top dead centre. Top dead centre will be which one will be the top dead centre in next while it will start rotating you will find that this one will become the it will reach at top dead centre this one just a small rotation. Let us see you can see this it is already had bottom dead centre and then again say suppose this is the minimum at this position you will find after a slight rotation this is becoming at top dead centre this angle can easily be calculated from the phases of such or bit motor from the analysis of the phases of such or bit motor. It is important to examine for the flow distributor valve action if we analyze the flow distributor valve action the duration of those phases in terms of shaft rotation for a single cylinder as well as the multi cylinder action. So, what we have to do we need to analyze how long a chambers or what with respect to the shaft rotation at what angle this bottom dead centre and top dead centre is occurring and according to that we have to also look into this and the distributor valve they should act accordingly. So, this phase analysis is not shown it will not be shown in today's lecture, but we will accept that when this is in expansion mode then one the while in channels are connected and when it is in the compression mode the while out channels will be connected. Now, in the analysis the initial position of the output shaft that is the shaft rotational angle zeta is taken as 0 what we are considering with this axis when this crest is at top dead centre that angle is 0. The centre c and o lie on the x, x dot axis this is x dash and here is the x. So, it is lying on this axis both this axis of the star as well as the axis of x axis of the envelope they coincides whereas, y axis is away at the centre distance c o whereas, this one is the this one y axis through the centre of the envelope and that is one of its dead zone we have called it the top dead centre and this is in the threshold of expansion phase that means immediately after that if we rotate in the clockwise directions here the expansion mode will start ok. Now this means that the centres are interchanged from their original initial position during generation. So, in the previous lecture I have explained that how this generation is done in that generation technique what we took initially that the centre of the star was here or the centroid fixed for star was here and the centre of the star sorry the ring here or envelope was here whereas, when we have started operations we have just interchanged the centre. This means that while we are we shall consider the geometry for the envelope we have to take care of this transformations ok. Now the positive z axis is along this z do not confuse this z is not the number of loads, but this is the z axis x axis y axis and z axis is in the upward perpendicular direction of the plane of the paper that is in this it is from the here will be the z axis which is coinciding with the axis of the output shaft. The output shaft is also directed outward so z axis is coinciding with that which I have just told you. Now we have taken some axis on the star and ring what are those we have taken a C S 1 C S 2 C S z be the central axis of the cylinders this means on the envelope we are calling that envelope is acting as a cylinder. So, on the envelope from C the centre of the envelope to the centre of this portion that is you can say the centre of the chambers that we have designated as C S. Now according to the number of chambers we have number the chambers also we have given the S 1 S 2 S 3 etcetera. Now here I have mentioned as a jth chamber any jth chamber, but only with respect to the figure which I have drawn we should say this is chamber 1 this is chamber 1. In the phase analysis I have shown that this is the chamber 1 next chamber is chamber 2 whereas the chamber on the x axis along the x axis when the rotation is 0 in that case this chamber is chamber 7. So, this is you say if you put this number this is h 7 this is S 1 and C is the centre. Similarly we have taken O T 1 O T 2 capital T O T z minus 1 be the central axis through the crest of the convex portion of the epitrocordial lobes. So, this is the convex portion the if you take the most point the crest point then we number is O T 1 O T 2 etcetera, but if you look into this case we have taken as if the O T 1 is here it can be named otherwise this is the T into z minus 1 because there will be 6 lobes 1 lobe less than z. So, that is why number will up to the z minus 1. Similarly we have also taken O small t 1 O small t 2 O t 2 that is the that is through the lowest point of this profile that is the concave portions we have taken this. Now these lines while we are analyzing the phases then which one is coinciding with which one depending on that we are giving the number we are assigning this is easy to understand, but in the present analysis our main concern is to estimate this area at any instant. We will find this area and you may find that these are not much useful here, but while you are writing something presenting something you can say suppose O S 1 has coincided with O T 1 capital T 1 or O T 2 like this that is easy to understand. To estimate full active volume of a chamber the volumes at top dead center and bottom dead center are calculated. Suppose if we would like to find out what will be the volume expansion of a chamber we calculate the volume at top dead center and bottom dead center. Top dead center means when the piston is at the top that means this is the top dead center whatever the volume at that condition that is not the active volume that is not the varying that is a constant volume that will be always there. So, we have to calculate separately this volume as well as we have to calculate when is the maximum volume at bottom dead center and then if we subtract this we will get the volume of a chamber. When the shaft rotational angle is equal to 0 O T 1 coincides with C S z in that case C S 7 if we take this z is equal to 7 and chamber z is at its top dead center. Now, after an angle that is it a 0 it a 0 which is calculated as pi by z into z minus 1 we will arrived and another chamber which is given by z minus z plus 1 by 2 th chamber reaches at its bottom dead center. But I have doubt I think this will be minus this will be true if we consider this is chamber 1, but we have consider this is chamber 1. So, possibly this will be z minus 1. So, that we can examine let us see again say if we let us consider this is 1 no this is 7th. So, 1 2 3. So, third one is reaching there. So, this will be z minus 1 by 2 not the z plus 1 it will be z minus 1, but this angle say this was at its top dead center when the shaft rotational angle was 0. Whereas, this will be at its bottom dead center then the shaft rotation can be calculated as pi divided by number of lobes on the envelope into the number of lobes on the a bit ok that is star the ring number. So, suppose in this case we have taken 7 and this is 6. So, pi divided by 42 it is around 4.3 degree or 4.27 degree for after which this angle will be a this chamber will be at its bottom dead center that means the maximum volume. So, what we can do we put the shaft rotation angle 0 we consider this 7th chamber we calculate the volume and next moment after 4 degree of rotation we consider the chamber 3 and we calculate the volume then this volume minus this volume will give the volume of the one chamber one maximum volume displaced of a chambers. So, this we have to do, but we must calculate this area now. So, this what I have discussed this is told here. Now, next slides what we will say that referring to this figure the area A j A j means area of the jth chamber act an instant when this has rotated by any triangle it is bounded by this area bounded by g h b d g h b d. Now, it is necessary to express the chamber area with respect to the axis fixed on the envelope we will find out this area considering this axis the fixed axis the fixed to the envelope or the ring. So, the axis x c x dash and y c y dash fixed on the envelope are also axis of the fixed reference frame that means this is fixed. So, this you can say the axis of the fixed reference. So, we will transform all coordinates with respect to this to estimate this area. Assuming the simplest modified form of the envelope what does it mean we have assumed this envelope is nothing, but that there is a circular ring and this is on a ring with the inner circle of this one. So, this is the simplest form which we have considered and the instantaneous area A j of jth chamber is given by A j is equal to A 1 minus A 2 where A 1 is the area bounded by the curve joining the points c d g h b c. So, we first we shall calculate c d g h b and again c this area we shall calculate minus the area bounded by the lines and epitropodal curve joining b c d. So, if we subtract now this area from this area we will get this instantaneous area of this. Now, how it is done? So, geometrical this is pure geometric analysis referring to this figure what we find the this is the area indicating c d g h b c is equal to area c e g h then f then c first we consider this area. So, this is like that we are considering this area. Now, next we subtract c e g d c now we subtract this area and then we add c f b c this area and then finally, we subtract h f b h f b h thus that area we exclude. So, ultimately we are getting this area. So, this you can check yourself this equations and you will find that we are calculating ultimately this area. Now, area in right hand side can be expressed as follows this is purely geometric analysis first we have considered this one this is we have taken pi r s where r s is the radius of this inner circle pi r s square by z that is the whole area minus r s square this is we have taken one seventh of this area because this area will be one seventh and then we have taken r s square arc cos you can you can just understand this how it is this is the cos inverse arc cos mean cos inverse of this angle. So, this is this will give one pocket and then next we shall consider the another pocket we will get this one. So, this you just you have to take this figure and this you have to understand here a 0 square is this is the a 0 this is square then r s square we have taken this this square minus r m square is this one cos inverse of that into r s square will give you this area. Now, plus r m into a 0 r m into a 0 into 1 minus this will give this area. So, we are going to get this area out of that next we consider the area c e g d c c e d g and again c e d g and c that means we are getting this area this is again if you can find out that first this is a triangle we have consider a 0 into this is perhaps not r 0 it will be r m theta r m this will be r m. So, this is 7 this is r m r m that means we are getting this perpendicular a 0 is this one. So, we are getting the area of this triangle this one then plus half r m square into cos inverse of this angle that we are getting perhaps this area this area we are getting this area minus no minus this area we are getting this area and then minus this. So, we are ultimately getting this area here now next we will consider c f b c c f b c this is simply half into a 0 r m by sin phi j. These angles are called leading angle that can be calculated for rotation of this shaft which I have discussed in the last lecture and finally h f b h is nothing but this angle. So, this area this is r m square into cos of that plus half r m into phi g that means we have considered this area first and then we have considered this area. So, now we are we will substitute this value in equations 832 and in non-dimensional now here is the question of non-dimensionalization which I have shown in earlier case what we find that area any distance we have put a bar. The bar means it is the non-dimensional value non-dimensional parameters. So, in that case what is done each and every parameter say if it is a length that it divided by r 0 what is capital R 0 what is capital R 0 capital R 0 is the radius of the centroid the bigger centroid the centroid for the envelope. So, we have divided by R 0 to make non-dimensional of any length a 0 r m etcetera. So, this means that while we are considering area suppose we have calculated this one what will be the actual area then a 1 bar into R 0 square capital R 0 square will be the area that you should remember now in the non-dimensional form this can be expressed as that I am not reading it. So, this when this equation is available you can just sum up and non-dimensionally we will arrived into that. So, now we have calculated only one area a 1 in the next page what we do we find out how they a 2. Now, for this a 2 the coordinates of b and d with respect to x y where this angle this angle is with the reference to the original generation of profile. Epitocids you just look at this angle indicates the shape rotation whereas, this angle with respect to the original development of the profile. Now, what you can find say this angle for the generations we take for the j th chamber psi j is equal to beta, beta is in this case pi by z that means half of this angular spread between two rollers plus the angle of rotation. So, beta plus angle of rotation we will consider for this one similarly, for the other one this angle this angle we will consider beta plus 2 pi by z plus this angle then subrotations. So, in the original formula of the profile we will put this value and this value to get the coordinates of these two contact points. So, therefore, this coordinates can simply be written in this form if you use this one you will find this coordinates. So, this is being transformed from the original coordinates what we have calculated with respect to the reference frame. First of all we have to calculate these two and then we will transform in this coordinate system. Next here I have mentioned this beta may be pi by z or any other angle only we have to take care of the proper geometry. Now, next we will express the A 2 in the dimensional non-dimensional form. Now, this is from this epitropoidal geometry it can be shown that this integration will give us this area A 2. If you would like to find out this area then knowing this coordinates of these two we can differentiate like this. Now, here I have presented this formula this is from the earlier lecture. So, this is the formula for expressing the coordinates of the envelope at any this psi angle. Now, in that case we have calculated this angle separately and we are adding to that. Remember we have to calculate the leaning angle while we are trying to calculate this coordinates of any contact point that is you have to also take the formula in the earlier lecture to calculate this properly. Now, if we evaluate this integral we will arrived into this formula, but still one integration part will be there for which this can be expressed in this form. Now, this integration is not presented in the any closed form closed from solution is not given here. So, we have to go for numerical integration to evaluate this, but may be a mathematician can he can find out something, but otherwise we have to go for this numerical integration. This is that you can find out this also this is not difficult to find out this is the leaning angle that means this angle. So, finally, before going into calculation this chambers here one interesting note is given that now we have calculated in the as a to this area. Now, at any instant if we add all such areas together that must give the area of the rotor which is important to find out the envelope I mean total area under the profile of the star which can be simply if we add all the area calculated like this at any instant we can simply get this area of the star. Now, the second term in is nothing, but the area bounded by d o b this equation can be modified to the form this one reference you will find that one reference he developed the formula in other way it was to me it was more cumbersome, but I have simplified this formula and to one we arrived into in this form which we will see in the next slide. So, what I have done I have substituted all the findings all the formulation in non-dimensional form and ultimately we get this will be the area of a chamber at any instant that means when the shaft is rotated by this angle we will get this angle or so to say I would consider if we consider any j and whether it is rotation is 0, 1 degree whatever it might be if we substitute this values we will get this area of a particular chamber. If we consider the next chamber we have to take all such values accordingly. The rate of change of this area with respect to the output shaft rotation is expressed in general form for both inward and outward modifications as follows. Now, what we have done we have first calculated this area. Now, here we have derived this as a rate of change of this area with respect to shaft rotation which now becomes in this form. If we look into this this is basically you will find that this motion of this epitropodal area this has no connection of the other geometric relations it is obvious the fixed part is eliminated in this case it clearly indicates that whatever shape we take for this envelope keeping this active envelope portion intact that means instead of taking this portion and this fixed portion as a this circular even if we take a big hole or even if we make a curve like this this equation will remain same. So, because this is only the variation of the volume which is working volume. Now, if we get this one then as a here I would like to mention one thing that this modification may be inward and outward this profile can be modified outward also. See in case of wonkel engine this is modified in the outward directions like that is small modification of course. So, to accommodate this whatever may be the modifications in that case of obviously it is inward directions what we have done we have used a equation force phi 0 in fact this phi 0 we can put along this axis. So, when this angle is pi that means this becomes minus 1 this is for outer modification that means in this formula for the orbit motor what we have considered we will put minus 1. So, this will become plus whereas, if we modify in the outward directions we will put this is equal to 1 that means in that case this angle is 0 and in inward modification this phi 0 angle is pi. So, it is minus 1 that we have to remember. Now, again so this variation this is obviously will valid for whether in compression mode and whether it is expansion mode, but we have to identify very carefully as long as this is in say expansion mode we will add with the other chambers which is in expansion mode because this machines in orbit machines is also a DC machines not alternating machines all the flow is being mixed. Suppose at an instant we find the chamber 1 chamber 2 and chamber 3 this is in the positive value. So, we will put this DAP that is in expansion mode and whenever we will get say for the chambers we will get this is minus we will put that separately and we will that make that this is in the compression mode, but the interestingly we will find that summation of these positive and negative will always constant. That means flow in flow out we will remain constant and another interesting information I would like to give you here although it is not shown, but if you take 1 chamber to the corresponding chambers you may find that the volume in and volume out is not same or in other words when say 1 chamber is in the expansion mode. Let us consider in this case or if I consider the fixed axis in that case this angle will be nothing, but these 2 pi by z that is the angle for any phase is expansion on compressions. Now, this phase what we will find 1, 2, 3, 4, 5, 6, 7 degree say such we will make it the interestingly we will find that for a particular angle the compression and expansion of a chamber is not equal whereas, in case of ordinary piston machines this is equal. So, whatever may be the rotations whether it is suction or compressions you will find a single degree of rotations this volume is constant. In this case these volumes are neither constant nor it is matching with the expansion phase and compression phase whereas, if we consider the overall that is matching and it has to match otherwise this machine will become impossible it will not rotate smoothly. Anyway our purpose is to find out the swift volume. So, what we it is to be noted that this rate of change of area is independent of the shape of the inactive envelope. This is inactive envelope may be say suppose this is the contact point up to this. So, this portion is active and this circular arc and this circular arc all inactive. So, it really this formula is independent of this area whatever it might be sorry whatever the area in equation 9 which is different for other approximated curves is only useful to calculate the area and volume at an instant. Only thing suppose if you would like to find out the trapped volume unused trapped volume then you have to consider the geometry. The expression of for flow rate speed ripples etcetera for such a rotary piston machines with different kinematics that is zero tour orbit units with star output ring output etcetera can is can easily be derived with the help of equivalent system concept. What it is? Now orbit motor what we have seen the ring remains fixed and output is taken through the star. Star is having two rotations one rotation is the revolving actions around the central axis that is the axis of the outer ring and it has also rotation about its own axis which is the output in case of orbit motor and orbit this version is only used as a motor because other pump version is not efficient or beneficial from the transmission point of view volume displacement point of view it needs if we make a pump with orbit principle it will need large torque for slow speed which is not available from the engine. So, it is not beneficial on the other hand when we make it fixed axis then which is called zero tour also a special name is assigned g roller when we use this type of roller when it is integral usually call it is zero tour or we should call in general case it is zero tour unit that is fixed axis unit. What is fixed axis unit? In that case both rotates about their own axis, but even they rotates about their own axis you will find this compression expansion that variation of this area will occur. So, that can be used as a pump as well as that can be used as a motor also. Now, what we have done we are analyzing all such area and everything with respect to the orbit motor, but it is possible that if we know the kinematics. So, same formula also can be used for the other kinematics that means zero tour units out let us see. Now, what we do the general flow we can express that what we have derived that this is 2 pi nr nr is the output rotations the b is the width. So, d A p and d gamma this gamma is general angle of rotations we have given. So, this simply if you add them you will find out what is the rotations and in case of orbit motor this will put eta and this is the output rotation of the shaft. Now, this angle is nothing, but transmission ratio into this angle this transmission ratio we have derived separately which is not given here. So, directly if you would like to use this formula you had to know this transmission ratio in terms of the gearing in between this as star and ring which is tabulated form available one of the references. Now, in case of obviously in case of orbit motor we will put this as a 1 orbit motor this as a 1. So, we can easily calculate this one or simply this will be replaced by this one in case of orbit motor. Now, here I have written is in case of 1 for orbit unit. So, without following the table you can still calculate this if you follow this one for 0 to width unit width ring rotation when the ring is rotating simply you put 1 by 1 by z minus 1. So, we will get this value and then this flow rate will be for that 0 to unit width ring rotations for the star rotations this will be 1 by z and we will find this same expression. In case of orbit motor this is 1 however, the generalized dimensions less shift volume V s can be expressed for unit width width is given by B of star ring as follows. What we do in case of so this first we calculate this I t is 1 for say orbit motor and then we multiply with this. This is flow rate volume in and we integrate up to this angle and then we simply find out. So, this is I would say rather it is a complicated form to find out the swept volume, but automatically from the flow rate we should be able to get the swept volume. So, this is shown here, but I will show you the simplest way of calculating the swept volume. Now, this is again the same thing I have described here while you are using this formula. The average geometric volume displacement Q A average you see this if we estimate this one with respect to this there we will find the flow fluctuation. So, what we should do I mean once we calculate this we simply multiply with the width factor and because this is for the unit volume unit width. So, we have to multiply here with B and then the rotational speed. Now, the alternative method of calculating swept volume this is just to calculate the swept volume what we can do? We can calculate the BDC and TDC at area of the chambers at BDC and TDC and which I have earlier explained that if you for the geometry we have considered if you consider this angle is equal to 0 then you can first calculate the this area at chamber 7 which is the chamber at top dead center and then after a rotation of pi by z into z minus 1 angle you can take the other one that is z minus 1 by 2 that is in this case this will be 3. Third chamber after this after that rotation of pi by z into z minus 1 and then you will get the bottom dead center area. So, and then what you would do the working volume is nothing but the area difference of these two areas into the width. Now, the swept volume is calculated then as follows for orbit motor what we do simply this volume of one chamber into this action in one revolution that means how many such chamber actions will be there in case of orbit motor z into z minus 1 time. So, you simply you multiply this you will get the swept volume orbit motor interestingly in orbit motor this is one respect to the other that means either star rotating with respect to the ring or ring rotating with respect to the star. So, in that case there will be single formula to find out this swept volume whether the star is rotating or ring is rotating whether whereas in case of zero tour unit this will be either v c into z or v c into z minus 1 depending on ring and star rotation respectively because in that case suppose your star is output both are rotating star is output. In that case you will find that when the star one full rotation is done actually 6 chambers are displacement is there not the 7 chambers whereas if the ring is rotating then 7 I have mentioned 7 or 6 in that case z minus 1 and z. So, carefully if we use we can calculate the swept volume of the ring I mean zero tour depending on which one is the output. Let this star ring of an orbit motor have the following data we will take a numerical example. Now in that case we have taken z is equal to 7 e 0 bar is 1.625 r m bar is equal to 0.405 beta b is approximately 0.7 it is slightly not 0.7 some value is there and r s means this one is 1.55 whereas r 0 is 19.6805 approximately 20. So, you can imagine what is the actual dimensions then this is r 0 is the radius of the inner centroid centroid for the epitroquate which is 16.869, but this is not required because we have non-dimensional lies with respect to this one. Interestingly say we use a 0 r m bar, but we never use this C 0 bar because C 0 bar is nothing but 1 by z because this you will find this and relation r 0 by capital R 0 by small r 0 is nothing but z by z minus 1. So, as their difference will be 1 z. So, this means that C 0 bar is 1 by z. So, in the formula you will find that not C 0 bar is used only 1 by z is used. So, with this data if we calculate this area what we have done for the swift volume a z we have taken first a 7 we have calculated and then this is again I have made a mistake this will be minus 1 by 2 the third chambers and this is a 3 and we will get this now this area these two areas are a TDC and a BDC respectively. Therefore, we will now use equation 14 to calculate the working full volume of a chamber which becomes this and you see this I have not shown the calculation, but to calculate this area we have used this formula and ultimately we have found out that this full volume of a chamber working volume will be around 2 cc very close to that. So, therefore, the swift volume of an orbit motor of with this data it will be 84 cc per revolution the flow required is 84 cc. So, you can imagine that in case of the fixed axis it will be either 7 times less or 6 times less than that for one revelations this is one. Second thing look at the size, size is say a 0 is 1.625 if you multiply with this 20 it will become 20 plus 12 30 or may be 35 millimeter here and this might be another say 20. So, it is 55 to 60 millimeter at the most. So, diameter will be 120 millimeter which is less than 5 inch 120 millimeter will be something like this. So, this is the size of the motor whereas and thickness is only 14 millimeter if you multiply with this it is close to 14 millimeter, but you see this how much volume is required for one rotation. Obviously, this is due to the gearing action in inside. So, in this way you can calculate the swift volume also you can calculate you have to obviously you have to make the numerical integration to find out the volume displacement of such machines. So, these are the references I would say that this paper you will not be able to follow because this is in this is old as well as this is in German language I think not in not even German language it is something it is not exactly German language I think these two people we are from Austrian people. They this if you read this paper you will have some idea about this fixed axis machine not orbit motor also Colburn he derived such displacement formula for fixed axis and Thoma he is I have used this how to phase is changed another this you can read this book also, but if you would like to know that phase differences and volume change etcetera. So, you can read this paper. Thank you.