 Welcome to today's lecture, analysis of flap and nozzle valves. This is continuation of our previous lecture, which was on three-way spool valve as well as introduction to flap and nozzle valve. Now, if we recapitulate, then three-way spool valves used with unequal area pistons to provide direction reversal. Now, in most of the cases, we have found that four-way valves are very common used in fluid power. If we control say rotary actuator say any motor, in that case we prefer four-way valve, because there the direction reversal is not very frequent. In comparison to that where the linear pistons are used, then due to the frequent reversal it is found that three-way valve which is cheaper than four-way valve can be used. Also, this has advantage with the linear actuator of unequal areas. Unequal areas means in it is head side, usually there is no rod or even it might be on the either side of piston. There are piston rod, but that piston rods are of unequal area. Now, also we learned that this is, this will be most advantageous if the control pressure is half of the system pressure. Now, why it is like that? We are trying to control something, then apparently we are having pressure on the both the side of the actuator and then we trying to control. The main reason is that this can be explained like this. Say for example, we would like to say suppose this is an indicator, that indicator we would like to put a particular positions. Now, if I try to control with one hand, definitely we can do it, but it is always better if you try to control this by two hands that means creating pressure from both the side and we are moving this. So, for the position control and the as well as with the force control, it is always better that this pressure from both the side and that is why in you will find in the servo control, there is pressure on the both the side. Obviously, there will be loss, definitely there will be flow and pressure loss, but from the control point of view, this is unavailable. Now, in three ways spool valve, what we find that the oil, this is the supply well and this supply well has direct entry to the rodent side of the piston and there is entry to the other side or the piston head side through this control flow. This means that while we are opening this side, if we keep it closed, then supply pressure is going this side and this is being moved up in upward direction. Whereas, if this is opened, then oil is going through this and this orifice is controlled in such a way, part of the oil will go to the control side or in other words head side of the piston to control the motion of this piston and part of the flow will go back to tank. Now, for this pressure ratio which is perhaps the best for the control of such three way valve and this pistons or nickel pistons, the area is also half, I mean area ratio is half or in other words the area of this piston divided by the area of the ring side is equal to 2. Now, these valves are made again critical centre for better response in servo valve. Now, you see this what is critical centre valve? Critical centre valve means this is just critically lapped that means width of this ideally width of this groove and width of this land of this pool is equal ideally dimensionally if you consider the nominal dimensions they are equal. However, to maintain the tolerances because even if this is closed, there will be radial clearance through which there will be leakage usually you will find if you measure the dimensions very accurately, dimensions of this land is slightly more than the groove width. Anyway, this critical centre valve will have very less bandwidth that means at the null positions, there will be less loss as well as response will be very quick. This means that if you move in either directions, the response will begin almost immediately. So, as we need frequent movement I mean reversal motion of this pistons. So, critical centre spool valve is the best. Now, if we recall our earlier development of the formulations, then we can write down the load flow for the displacement is in the positive directions that means in this directions, then this equation is written as the CD the coefficient of friction and this is the total area of the orifice. How? The w is the width and x p is the spool displacement. Now, w is width in this case is the total peripheral length of this spool or so to say inside of this groove that means in this case, directly we can get w is equal to pi into d that is diameter of this spool or d p whatever it might be and then this flow here this load flow we are considering this p 1 and p 2 is the pressure difference. So, that will be the flow pressure here is p 1 here it is p 2. Now, for the other directions then the flow is load flow is given by the p c is the pressure control pressure there then other pressure is the 0. So, p c into 2 by rho root under root and this is the again area of the orifice CD should be determined experimentally, but normally for such a spool where this edges are very sharp normally this CD can be taken the von Mises model which is around 0.62, 0.63 something like that. So, normally we do not need to go for any special experiment for such valves and, but the w in case of the full open valve that means the full groove the w is equal to pi into d as I have told, but sometimes what it is there instead of this groove continuous groove we may have rectangular hole. Then the length of this hole along the spool slave is equal to the this height of the spool land for critical center. However, this groove width we sum up to get w. So, for example, there are 4 groups which perhaps only spreaded over 15 degree or 30 degree then we simply add this w to get the orifice opening area. And if the if there is a some circular drill hole or some other type of hole then these are calculated accordingly. However, in most of the cases we will find that this is the linear port opening that means rectangular port we will get this rectangular port which is here the hole is rectangular or the full groove. Now, flow gain is same, but pressure sensitivity is half that way 4 way critical center valve. This is if we recall the 4 way valve analysis and compare the flow gain that is the 3 different coefficients valve coefficients then flow gain coefficients shows that here this is pressure sensitivity is half whereas flow gain will be same as that of the 4 way valve. Due to this reason error to overcome loads will be more in case of 3 way valve this will be more. Then why the why the 3 way valve first of all it is very less expensive and it is better or quick control can be achieved by this arrangement 3 way valve arrangement. However, this is for the control we have to look into this error. It also can be shown that dynamic load errors are almost double than in comparison to 4 way central valve. That limits the application of 3 way servo valves this means that we need to control the error there is no way that we will allow the error that means that will be drifted, but the question is that to eliminate such error how much time we can spend. If there is we need to control it in very short time very quick response in that case better to choose 4 way valve otherwise where we can have we can allow more time for such control for such eliminating errors in that case we can go for 3 way valve. Now what is flapper nozzle valve we find that in flapper nozzle valve that instead of in case of the 3 way spool valve we have seen that spool is controlling the flow. In that case that means basically it is allowing the oil to go to the ring side ring end side of the piston and while we are trying to control that motion of this piston we are allowing the oil on the other hand side, but that flow we are controlling by 3 way spool valve. In that case what we find say if we compare with that previous one then oil is allowed to go into road in road this sorry this piston head side, but that is again controlled through an orifice and part of the oil is allowed to go back to the tank through an orifice the opening of which is controlled by a cantilever beam that is basically a plate very thin plate is mounted pivoted on a pin this pivot and then this is moved by some mechanism in that case we call the torque motor is there. So, this means that like the other one like the spool valve one this supply oil can freely go to the rod end side if it would like to come in the head side in that case first of all it is going through a fixed orifice the here orifice is fixed in case of spool valve this orifice is also varying this is also varying, but in this case this is a fixed orifice and fixed orifice and this is being controlled. Another thing is there in such control that the instead of this opening which is circular obviously this orifice area the important area is that cut end area cut end means if we consider the diameter which is d n pi d n is the periphery, periphery into that height is the cut end area and that area is important in this case not the directly this orifice area. Then flapper valves are usually used in low pressure applications where the pressure is relatively low. Now, it allows higher leakage losses and such valves are of low cost and less sensitive to duct what it is in case of the most crucial problem in hydraulic is the duct particles. In case of servo valve where this components are almost match part this means that sleeve diameter and this spool diameter is made such that we get a less sensitive minimum leakage loss and very high sensitivity in any flow. In that case if duct comes in between then problem becomes either this will damage the spool or the whole operation will be stopped due to this duct particles and thus controlling this duct or if the duct comes in breaking that duct into smaller particles or removing allowing this duct to go in this other side all such mechanism is done to design a very good valve. On the other hand there is also research is that if we allow this duct to go with the flow and easily it can go through this leakage path then it is seen that in instead of spool valve if we use this flapper nozzle valve where this opening is relatively more the particle can go out directly and it does not stop the machine at least stop the function at least although there will be some disturbance. So, in that way flapper nozzle valve is better than spool valve. So, this is the biggest advantage using the flapper nozzle valve. Now, flapper driven by torque motor as I told there will be a torque motor. Torque motor means it is not rotating fully and its output of course output of motor is torque in that case this torque motor means it is just giving a small amount of torque actuation is small in one direction it can rotate in the opposite direction say if the current sense is reverse and these are used to move such flapper. Now, again in that case what we look into this? In that case what we look into this? This is comparable with 3-way spool valve and we have that single flapper jet application of the 3-way spool valve mechanism. This is acting as a 3-way spool valve, but in most of the cases the application of this flapper valve is at this pilot stage of 4-way valve. What it is? I will show later that if we use this jet in also in the other directions then we can imagine here a spool 4-way spool valve. Now that 4-way spool valve again is being used to actuate another spool valve which is also 4-way which is actually the main stage of the valve because in many cases the load is so high that if you use a single stage spool then it becomes difficult to control that spool. The spool force is very high and with high force control becomes difficult then to move that spool 4-way spool we had another stage where spool is very small. We need to actuate that spool valve with very less amount of force which is controlled by this flapper knowledge and nozzle and this nozzles are at both the sides and then that main stage controls that pilot stage control the main stage that we have seen that earlier in the servo valve section. So, this is there at the basic application of such flapper nozzle valves. However, this is also used for 3-way valve used as a 3-way valve for actuator control of unequal area. Now the pressure flow curves have better linearity although there are losses more loss but we get better linearity. So, control becomes easier also performance of these devices are quite predictable and dependable that means where we do not need very quick response we can allow such losses then for the low cost applications this is better than spool valve 4-way spool valve. Now single jet flapper valve which we have learned in earlier lecture is also called as 3-way flapper valve. We should basically call it flapper valve but once a single jet valve is there then we call it 3-way flapper valve. Now for that load flow equation where which we have derived earlier what we find that load flow divided by this orifice area that is the fixed orifice A0 into the this is the coefficient of discharge there. This area is nothing but pi D0 square by 4 and C D0 D0 is the coefficient of discharge here. Then we get PC by PS that is the control pressure by the system pressure this will be definitely less than 1. So, this term is always real term and then XF divided by XF0 where XF is the flapper motion and XF0 is the initial gap there initial gap. Now this PC by PS is usually found 0.6 is good for such flapper nozzle valve and in case of spool valve 0.5 and in this case 0.6 is better. Orifice ratio at the null point is derived as this C D0 is the C F is here the coefficient of discharge into A F is the this curtain area and divided by the fixed orifice area into the coefficient of discharge there. And if this ratio is maintained 1 at null position then the performance of such flapper nozzle single jet flapper nozzle valve will be the best. Now again in practice we have this that fixed gap divided by diameter of this orifice is usually 1 by 16 that means you can imagine this this diameter if this is say 1.6 millimeter in that case this will be how much this will be 0.1 millimeter this gap will be 0.1 millimeter. And usually this flapper nozzle valve you will find that of that range I will we will see this later. And F1 F1 is the force due to this flow here remains also PC and AN AN is the area of this this hole not this curtain area just compare with this AN AN is the force and AN AN is this area. Now what we find also that if we differentiate this force with the flapper nozzle movement then this becomes 4 pi into C D F square P S into X F0. And this is equivalent to the spring coefficient of fluid spring. If we compare this value then we find as if this is a fluid spring but it is in the negatives it is a negative spring what does it mean in case of positive spring this force increases with the when this is complex say this is or the tensile phase it is being tension but it is other way in that case if we reduce the length then the force will increase. And due to that control becomes a problem. So, how it is controlled in case of single flapper jet we usually use a spring here in the opposite direction. But if we use the double jet in that case opposite side also there is a nozzle. So, we can control the flow from both the sides and this balancing problem will not be there it will become very easy to control the motion of this flapper. Therefore, double jet is very common where we use the flapper jet valve. And as it is mentioned here this is usually used to control the pilot stage of a main stage servo valve or so. Now if we look into the what is the forces on the flapper this is one there will be static pressure force. Now here if we look into this flapper nozzle valve with double jet then you can see that how this it is working. Suppose if we would like to move this flapper in this directions we need to have more force than that this means that F 2 should be greater than F 1. You see if we would like to move you can make this is 0. But if we would like to move this one with a controlled motions then this should have a force and this should have also a force. Now this is done in valve operation. There is also dynamic pressure or the force due to the fluid velocity. What is static pressure? Simply static pressure is will be equal to P 1 into this area not the curtain area. Curtain area that orifice area is used for the load flow and other things. But normally when we are calculated the force then we consider this area which is equal for both the nozzles. However the pressure will be different whereas dynamic pressure is that if while we are considering the dynamic pressure in that case we have to consider the velocity of the fluid which is impinging on this flapper. Now referring to this figure then and considering the Bernoulli's equations F 1 that is the force can be derived as F 1 is equal to P 1. We are calculating this force F 1 then P 1 then rho u 1 square into a n a n is area of this orifice whole diameter. Now what is here in the above equation right hand side right hand one first one is the static and second one is the dynamic part. So this is the static part that means P 1 a n is the static force and half a n rho u 1 square is the dynamic force u 1 being the velocity at the plane of nozzle diameter that means this velocity while how this oil is going it is going like that it is coming like this and then going in this direction. Now we should measure the velocity along this perpendicular direction and at the vicinity of the surface of the flapper that velocity is called u 1. Now it is expressed as big equation that is Q s by a n and then C d f the Q s is the flow through this nozzle which may be Q 2 and Q 4 here pi d n then no here this is the while we are calculating this flow as I told that we will consider this cut end area. So this is the cut end area into this 2 by rho into P 1 under root that other pressure is 0. So we consider the P 1 whereas a n is the area of this hole. So this is pi d n square that is the diameter of this nozzle opening divided by 4 and if we wait further this area will arrive into this expression. Now combining this above 2 equations and simplifying what we get f 1 is equal to P 1 into 1 plus 16 C d f square x f 0 minus x f whole square divided by d n square into a n. Similarly if we equate the f 2 in the same way we will get this f 2 will be expressed in this form. So this is not difficult but look into this here we get minus sign here plus sign why it is minus signs in that case we have considered that this is the positive motions. That means x f is positive in this directions. So while this flapper is moving in this directions with theta positive also then this length is being decreased. So physically this initial distance minus this full movement will be used to find out the cut end area. In this case due to this motion this height is being increased so this will be plus sign here. Now if x f itself is negative then automatically this will be corrected. So this equation we can use this is the general form of the equations. Now then this controlling force as I told that force from both the directions. So this must be f 1 minus f 2 this might be negative also because f 2 might be more than f 1 but anyway this can be expressed in this form. Simply we have subtracted this from this one and this is the final form of the force. Now combining the again two equations and sorry this is the perhaps the same perhaps by mistake this has been copied. So we will go to the next slide. Now a good flapper valve has usually x f 0 by d n is 1 by 16 that which we have learnt in earlier lecture. So this gives this is a very good value for designing such valve and also the flapper normally works near the null point both p l and x f are significantly small. That means normally this is for as we are trying to control the pilot stage in that case load flow load pressure is small as well as this motion is also very small because in case of because this is basically the high pressure process the main valve. In that case what we have found this valve opening is very small this x f this motion of the valve is very small but that will generate a very large flow due to the pressure difference. Anyway this if these are too small therefore the right hand side of the equation 6 earlier 6 second and third terms are much smaller in comparison to the first term and then finally we you have to see if you look into that equations and then we can neglect two terms and we can have this is the usable equations to estimate the controlling force. Then we calculate the torque on flapper that means how much torque we need to operate this one. The equation of motion of flapper is in dynamic terms can be written as this T d is equal to this is the inertia and then this square theta by dt square this is the torsional spring stiffness into theta and this is the force into r is this arm. Now here I would say that this flapper may also deflect while we are calculating this torque we do not care about that but while we are really calculating with the motion of theta and we are trying to relate x f 0 we have to careful about that where T d is the torque required to drive the flapper. Now as x f is much less than r f there is some problem with this compatibility this is actually x f is much less than r that means this length and x f is the spool moment and for that we can consider tan theta actually we have to consider the tan theta which is x f by r is equal to almost equal to theta and therefore this we can combine this 7, 8, 9 these three equations and we will get the torque equation in this form. You can have a look into this equation and this equation is used for estimating the torque. Flow force due to the fluid impingement on flapper give negative spring action which already we have learnt the negative spring rate is usually 5 kilo Newton per meter and at 7 megapascal with x f 0 is not 0.075 millimetre. You see this how small it is even less than 0.1 millimetre and then actually this is this value is just a realistic value this is to understand this what are the parameters parametric values of such flapper nozzle valve. And k by r square k is the torsional spring thickness must be greater than the spring rate to have effective controllability that you can understand if this is not greater then we can not be able to control it. So, we according to this value once we decide this value first of all we estimate this one and from there we have to estimate this one and accordingly we can design or select a torque motor for a flapper nozzle valve. Now, after determining the flow gain requirement for the main system main stage valve for direct applications main stage valve means that means this flapper valve we are using as a pilot stage or if we are using this directly for controlling the actuator. What we do the nozzle orifice diameter can be selected using the flow gain equations we use directly this flow gain equations and from there we find out what will be the diameter of the nozzle. Say what how much flow gain we need from there we can have this value as well as have this value and CDF we can have from the experience we have to take because we have not yet designed the this nozzle. However, we know this what oil is being used what will be the system pressure and from there we estimate the diameter of the nozzle and rearranging this equation give is in this form. Now, x of 0 should be as small as possible to have better pressure sensitivity and minimum leakage on the other hand it should be large enough to give a passage to the duct because basically flapper valve is used to allow the duct particle size of about 120 micron in general purpose application and 10 micron in precision servo valve applications. You see normally in ordinary valve general purpose valve we can allow the particle of 120 micron in that case usually you will find after the pump we can allow this flow directly to the valve and before the pump we have only a strainer which is in the range of at the most 150 micron or something like that 120 micron this value if we use this value then probably that is 120 micron you can understand it is 120 micron particle size means the filter we have used it is a mesh of where the diagonal length probably is of that size 120 micron or so. But in case of servo valve where the servo applications we are doing this gap is very small it can if the particle size more than 10 micron then it becomes difficult to move that valve in that case usually a high pressure filter is used after the pump and before this valve. So, we need a high pressure valve again using a high pressure valve definitely a loss to the system, but this we cannot compensate because such a control is definitely expensive, but we need to control say for example, machine tools even missile control etcetera keeping the curtain area about one fourth of nozzle orifice area is a good design. So, this is just we are considering the valve design. So, this value it is from the experience if we maintain such relations then the valve design will be better. Now, this means that pi dn xf0 is less than one fourth into pi dn square by 4 or specific results and the simplification results in xf0 is less than equal to dn by 16 which we have already which we are using the yes this relation already we have used. Anyway the fixed upstream orifice is usually short tube orifice with length to diameter ratio is 2 to 4. Again another important factor is that say this is the orifice opening and this is the length of this nozzle what should be the length we can know this is not this orifice. We are talking about this other orifice here that has to be there to control this flow both the side this orifice will be fixed orifice will be along with this tube. Now, in that case usually there a short tube orifice with length to diameter 2 is to 4 is used that means orifice are of different type orifice are used in sub valve control. In that case we can imagine that that orifice is that within the tube there is a plate. In that plate usually there is a hole simply drill hole and the ratio of the length to diameter is 2 to 4. If diameter is 4 unit then length is only 2 unit say 2 millimeter thickness mean at the most 4 millimeter diameter is the hole, but usually you find very thin plate and very small hole is used and usually a straight hole in some cases also hole with an inclined entry or may be this side it is used normally straight hole is the base. Now, from this diameter and the length ratio we can find out the coefficient there. There are different models, but we will see some realistic data the upstream orifice diameter d 0 is derived using this equations. We would like to maintain this ratio is equal to 1 and from there this is a f is the that is the curtain area and a 0 is this upstream origin sorry orifice area. Now, this again can be written in this form and then simplifying the above equations we get d 0 is equal to that is the fixed orifice diameter is equal to 2 root over c d f by c d 0 into d n x f 0 d n is the diameter of this nozzle. Now, other parameters what we find is you can see that c d 0 that is the coefficient of discharge at the this orifice. You see this is in the range of 0.8, 0.9 where normally we find that 0.6 is the good value for the orifice that means this is actually capillary short tube where it is written itself because in comparison to in case of these orifices the length is very small very thin plate and then there is a hole whereas, in this cases certain length is there, but it is found that using some such short type short tube orifice is better for the from the control point of view and as we find whereas, this is 0.6 to 0.85 that is if we consider the coefficient of discharge here this is 0.6 to 0.85 and usually c d f by c d 0 is preliminary design that means when there is no movement of the x b is equal to x f is equal to 0 in that case this ratio is a good design. Now, if we consider the realistic value say x f 0 is 25 micron you can imagine how small it is then d n is 0.4 millimeter only whereas, d 0.18 millimeter and the flow there is 0.45 into 10 to the power minus 6 meter cube that means it is no it is some 4.5 shall we call this meter cube means in relation to the c c 10 to the power 6 so 4.5 c c. So, if it is 50 micron then this is 0.8 and this is 0.36 and 17.25 c c and if it is 75 micron then 1.2 0.54 and this is 39 this is to have an idea that what usually the flow say you can consider this is perhaps for the direct valve and this is may be for the pilot stage and this might be for the pilot stage of very big valve. Now, again so this is sorry this is not properly much if we consider that some realistic data more realistic data on this what we find this is d n of course it is given in inch 0.893 inch 0.896 means here it is a quite large one and then there the Reynolds number are say 1000, 2000 and 3000 and for that what we get the coefficient of discharge is varying from 0.6 to almost 1.9598 something like that and it is more at large Reynolds number and that means the high velocity flow and it is less with the less Reynolds numbers. Now, and this is the ratio of L into x f 0 what is L? L is the we can see that this is the nozzle actually and this is the curtain height and this nozzle means if we say suppose if we decrease this L that means it is almost a tapered hole in that case that this say L is less means this is we are in this side so coefficient discharge will be less whereas, if you go on increasing this coefficient of discharge will be more and this alpha angle this value is not given here I have no idea also what is the this alpha angle is there. So, this is some realistic value to understand how this nozzle is designed and this is mainly we have followed this merits book this is hydraulic control system and also some idea is taken from the blackburn or a fourth and say errors fluid power control books. Thank you for listening.