The Development of the Simply Supported Feather Spring Valve

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Purdue University Purdue e-pubs International Compressor Engineering Conference School of Mechanical Engineering 1996 The Development of the Simply Supported Feather Spring Valve X. Qian Northern Jiaotong University Follow this and additional works at: https://docs.lib.purdue.edu/icec Qian, X., "The Development of the Simply Supported Feather Spring Valve" (1996). International Compressor Engineering Conference. Paper 1137. https://docs.lib.purdue.edu/icec/1137 This document has been made available through Purdue e-pubs, a service of the Purdue University Libraries. Please contact epubs@purdue.edu for additional information. Complete proceedings may be acquired in print and on CD-ROM directly from the Ray W. Herrick Laboratories at https://engineering.purdue.edu/ Herrick/Events/orderlit.html

THE DEVELOPMENT OF THE SIMPLY SUPPORTED FEATHER SPRING VALVE Qian Xinghua, Professor Mechanical Engineering Dept. Northern Jiaotong University, Beijing, P.R. China ABSTRACT This paper introduces the simply supported feather spring valve without the valve guard which is developed by the writer. The concepts ofthe linear air lift and the linear moment of inertia which are used for the calculation of the spring valve are brought up. The formulas calculating the linear air lift, the effective flow area and the thickness of the spring are worked out, and the calibrating principles of the intensity, bending, Euler load, frequency of the spring are proposed. Keywords: compressor, valve INTRODUCTION The simply supported feather spring valve in this paper changes the traditional structure of a valve, that is it hasn't the valve guard. The new valve has several characteristics: the little clearance volume, little resistance, low noise, the strong adaptability under different working conditions, the simple producing method and the low cost. THE INTRODUCTION OF THE STRUCTURE OF THE SIMPLY SUPPORTED FEATHER SPRING VALVE WITHOUT THE VALVE GUARD Figure 1 is the structure ofthe simply supported feather spring valve. It consists ofvalve set 1, feather spring 2 and support 3. The support 3 consists of two short arc bottomed keyseats. The bottom surface a of the keyseat supports the feather spring, the side surface b limits the lateral motion of the spring. (See figure 2.) The keyseats can be slotted on the roof of the cylinder in air minicompressor, this leads to the combination of the support and the cylinder to a whole.(figure 3). The span of the two keyseats equals to the length of the feather spring which is L. The Width of the spring is B. In the valve seat, there are several equal width valve channels C, the length and width of the valve channels are I and b which are less than Land B. The characteristic of the valve is that it has no valve guard, the support simply supports the feather spring and does not have the function of the valve guard. While there is no air going through the valve, the spring presses close to the valve seat to block up the valve channel, and while the valve is working, the spring is pushed by air and bent, the Valve channel is opened and the air goes into the cylinder through the valve. Because there is no valve guard to limit the spring lift h, the feather spring lift is a variable. 389

THE DISTRIBUTION OF THE LINEAR Affi LIFT When the simply supported valve is working, the feather spring will be bent at the effect ofthe air lift. The bending depends on the distribution and the value of the air lift. Because the distribution load was used to study the deformation of plate, to correspond to the up mentioned statement, the concepts ofthe linear air lift q and the linear moment of inertia J is brought up. The air lift per unit length ofthe feather spring (the width of the spring is 1 unit) is called linear air lift (unit is N/M). The moment of inertia of I unit width feather spring is called linear moment of inertia (unit is m3 ). The spring is affected only by the air lift F. The air lift pressed on the line of!1l long at X point is (see figure 4) F(x) = P B L1L L1P(x) p-air lift coefficient, B- the width of the feather spring, &'(x)- the dynamic pressure on the face surface of the feather spring. The linear air lift at X point of the spring is q(x) = F(x)j BM = f3!1p(x) Let B = 1 we can get q(x) = f3l1p(x) (1) If the mean dynamic pressure on the face surface ofthe feather spring is L1p, the mean linear air lift will be q = f3l1p IfEq.(l) is divided by Eq.(2), We can get: q(x) = LJp(x) q LJp Let Q = q(x)jq (3) P=~~v~ ~ Q - relative linear air lift, P - the relative dynamic pressure on the face surface. - - Substituting Eq.(3) and Eq.(4) into Eq.(2), we will get: Q(x) = P(x) (5) From Eq.(5) we can see that the distribution law ofthe spring's relative linear air lift Q(x) is the same as the relative dynamic pressure P(x) on the face surface of the spring. Q and P equal at the same point on the spring. - - - - [1]. Figure(5) is the P-X, P-Z curve ofthe cantilevered feather spring. In figure a, the abscissa is X = xj L, in figure b the abscissa is Z = z/ B, the ordinates of the two figures are P. Curve 1 correspond to the round valve port and curve 2 and curve 3 correspond to the two galley valve ports. The relative length I I L of the three ports respectively equal to 0.16, 0.47, 0. 77. Because I I L of this simply supported valve is about 1, from curve 3 we can conclude that on the whole feather spring, Q is well-distributed, the linear air lift q of the new valve is also well-distributed (see figure 6), and Eq.(2) should be used while calculating q. If the relative pressure loss 3 and the working pressure p is known, q can also be expressed as q = P8 p ( 6) THE CALCULATION OF THE FEATHER SPRING LIFT AND THE EFFECTIVE FLOW AREA (2) 390

The lift and the effective flow area depend on the spring bending functiony = y( x, e). 1. Feather Spring Lift It The feather spring lift It is the maximum departure from the valve seat when the spring is at the maximum bending.(figure 6) h =1Hz+ Yma.-.,.Jh= L(a-sin 8)12 LJ!t- the vertical distance while the spring gliding down. Yma-.- the maximum bending. Because Yma.-.1 L and angle a of the feather spring is little, ijh is ignored, to simplify the problem,presuming that h = Yma'. Under the effect of the well-distributed linear air lift q, the equation of the bending of the simply supported feather spring is (7) E- modules of elasticity, J-linear moment of inertia. If the thickness of the spring is a (unit ism) then J=a 3 /l2 Substituting x = L I 2 into Eq.(7), that comes to the feather spring lift formula It= SqL 2. The Effective Flow Area Av 4 32Ea~ (8), (9) The effective flow area Av is composed of two bow shaped areas after the deformation of the feather spring. Ifthe number ofworking feather spring is i and their length are~, L:., L; respectively, then If all the feather springs as long as L, then Av = ~ ( 11). L5 SEa~ From Eq.(9), Eq.(lO), Eq.(ll), we can see that hand Av will change in different working conditions. They will adjust themselves with the changes ofthe working conditions. THE CALCULATION OF THE LINEAR AIR LIFT The linear air lift q depends on Lip. According to literature An p ncmap. e 2. " L..f-' = -[ (sm +-sm28)l p=pjrt 8 a4v 2 p- gas density, p -gas pressure, T- gas temperature, a- flow coefficient ofvalve, Cmpiston mean velocity,ap -piston area, 2 -crank radius divided by length of connecting rod. (12) 391

Substituting Eq_(l2) into Eq.(2) and considering the relationship ofeq.(ll) and (10), we can work outq: ~ fip 57 CmA. A. l q ===a-{~[ P (sm 8+-sin 2B)fp RT 64aLL; 2 Ifthe effect of A. is ignored (when 2=0.2, the error is less than 4%), the maximum linear air lift qma."- takes place at the position where the angle of the crank is 8=== 90 or 8~ 270. (13) (14) THE CALCULATION OF THE SPRING RATE AND THE THICKNESS OF THE FEATHER SPRING Substitute Eq.(6) into Eq.(l4), the thickness a ofthe spring can be worked out according to the valve parameter 8 which has been given. The definition of the spring rate C of the feather spring is C = q B /.I h and substituting Eq.(9) into it, we can get the formula of the spring rate C::::: 32BE«' j5l 3 (16) THE CALCULATION OF THE PREBENDING OF LARGE SPAN FEATHER SPRING The feather spring will hang down a little by the effect of its gravity and create a bending jjh, so it will not press close to the valve seat, this leads to the loss ofthe actual suction capacity, this is because in the beginning compression stroke, some of the air in the cylinder will flow back to the suction cavity through the gap formed by the bending jjh. Because the bending jjh of the large feather spring is large and can not be ignored, to cancel out it, an opposite pre-bending- LJh must be applied on the spring. Suppose the density of the feather spring is Ps' the pre-elasticity Ah.C should be equal to the spring's gravity ablps. That is jjhc === ablps Substitute Eq. ( 16) into it then comes to the formula of pre-bending: jjh = 5 PsL 4 (17) (15) 32Ea 3 THE DESIGN OF THE ARC BOTTOM OF THE KEYSEAT The bottom surface of the keyseat of the simply supported feather valve should meet the following requirements: 1) It can support the free state feather spring to press close to the valve seat. 2) It should ensure the touch at the both ends of the spring. 392

3) The higher edge of the bottom should make the spring glide down a little under the effect of q, and it should ensure the axial force not beyond the Euler Load under the effect of qma.v.. The lower edge of the bottom should ensure the spring not gliding off from the keyseats and falling down into the cylinder under the effect of qma.v.. CALffiRATING PRINCIPLES OF THE FEATHER SPRING I. The Calibrating Principles of Bending Because Eq.(7) can be used at the condition of Ym._..j L < 0.1, we can get the formula of bending calibration: 2. Intensity Calibration (18) Th. f h.r. e maximum Stress a:m" h.. M. BLz q./ 4 3Lzq. ( 19) 0 t e 1eat er spnng IS a: = ~ = ma.v. :::: max ~- ma.v. W a 1 B/6 2a 1 If the allowable stress is [a], the intensity calibration formula is 3 L 2 q~..., :::; [a:] (20) 3. Euler Calibration ~ 2a- Under the effect of qma.v., the axial force pressed on the both ends of the feather spring by the support should not beyond the Euler Load, so the axial force pressed on the spring by the arc swface should be smaller than Euler Load ff- EJ / L 2. The maximum axial force which is created at the highest position A ofthe arc surface is BLqma.x tgy/2 ( r is the central angle ofthe arc AC, see figure 8 ), it should satisfy the formula: _1( 27i-EJ) r<tg --3- BLqma., 4. The Natural Vibration Frequency Calibration (21) The formula ofthe natural vibration frequency is f = KnL 2 ~PsBa -::f: o. 75-1.25 (22) REFERENCES fo 30Jr EJ 1. Lezhi Yang, Zhe Li, " An Experiment Study On Thrust Distribution of Reed Valve Plate In Reciprocating Compressor "ICECESE, Whuhan, 1990, P765 393

L/2 t x F(x) ~ i v j\ a ::-rg.! The indicator diagram of 'he s1mpi;' sllpporleci feothe: sprlng 'Jalve t"ig.4 The air!!f~ or. the -:eather spri.-:s "ig.9 ~he oxicj! fo~ce equds k "he :::uler Loori :=-ig.s The dist!'~b;.~tlor: of :he!!nee~ ai:!lft c; :md the spring!!ft r ~-- the vert1cai dis~cmce while ~he. ~prlng g~iding ~ow~.,.. The mox1murr. oonams!!-- "he slope of the eicstic cu:-ve rig, 3 The indicai or diagrarr. o7 ihe simply supported feather spring valve of minicompressor fig.8 Determine the arc surface ABC ~ y-z y o ' 3 2 (Q) ~7 : I ~ / ~~==~==~~+-~~x \ I \ I \.../ 3 j5 (0) --------«'---+---"'-----! Fig.5 The distribution of the relative dynamic pressure of the cantilever feather spring E-- relative dynamic pressure 0-- relative linear air lift X-- the relative position of the valve port on X orientation Z-- the relative P,OSition of the valve port on Z orientation X = x/l Z = z/b -1 Fig. 7 The air f!ow back into the suction cavity under effect of the bending h which is created by the gravity of the simply supported feather spring 394