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International Compressor Engineering Conference School of Mechanical Engineering

1984 Dynamics of the Swash Plate Mechanism J. F. Below

D. A. Miloslavich

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Below, J. F. and Miloslavich, D. A., "Dynamics of the Swash Plate Mechanism" (1984). International Compressor Engineering Conference. Paper 437. https://docs.lib.purdue.edu/icec/437

This document has been made available through Purdue e-Pubs, a service of the Purdue University Libraries. Please contact [email protected] 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 DYNAMICS OF THE SWASH PLATE MECHANISM

John F. Below, Chief Engineer, Rix Industries

David A. Miloslavich, Project Engineer

ABSTRACT speed operation. In a multistage compres­ sor, size and location should be Several mechanisms have been used to con­ determined for an even distribution of vert the motion of a rotating shaft to the load. This reduces wear, lowers horse­ reciprocating motion needed to drive the power input, and reduces weight and cost of a conventional compressor. by eliminating the need for a flywheel.

The most common mechanism is the ­ 3rd Stage shaft in which pistons are driven up anq down in a direction normal to the drive 1st Stage The dynamics of the shaft centerline. 2nd are well developed and readily available in design handbooks. The swash plate, another mechanism, produces recipro­ cating motion in a direction parallel to the centerline of the rotating driveshaft. Compressors of this type are called axial piston compressors. the dynamics of the This paper investigates Bearing swash plate and develops the mathematical relationships necessary for evaluating and for sizing stresses, bearing loads, Motor and locating counterweights for a proper running balance.

INTRODUCTION (See Figure l)

The swash plate can be made up in many different configurations. This paper deals with perhaps the most common construction. That being a bearing where the inner race is fixed to the shaft at an angle and ro­ tates with it. The outer race is re­ First Stage strained from rotating by the thrust rider. Piston As the shaft rotates, the outer race pre­ cesses with a wobble producing a back and Second Stage Third Stage forth motion in an axial direction at ev'ery Piston Piston point on the bearing periphery. A compres­ compres­ sor can be formed by linking the Swash Plate sion pistons to the outer bearing ring.

Controlling the distribution of masses of pistons, thrust riders, and counterweights, provides a means for obtaining a dynamic balance. The primary and secondary unbal­ Figure l anced forces and moments can be reduced to zero by geometry. The effect is a smooth A Three stage swash Plate Compressor running machine with a capability of hiqh

76 THE COMPRESSION PROCESS y axis Modelling the compression process allows for a determination of local loads and stresses. These can be combined based on the orientation of the pistons on the swash plate to obtain the resultant or overall loads and stresses. This is necessary to properly size the bearings, crankshaft, and other load transmitting parts of the compressor.

The first step is to determine piston position, x, as a function of crank angle, 8. Refer to Figure 2 for this analysis. The analysis is performed for one stage, in this case the first stage. The constants are fixed by design. The variables change with crank angle.

S X a ~ arctan ~ = arctan b 2 (r - s 2 ) Figure 2 Swash Plate Geometry therefore: substituting -sin 2 e = cos 2 8-l and taking the square root; X s s b X b r2 . z e)-l:z (r2 - s2)~ (r2 - s2)li X = r COS 8 ( S2 - SLn x2) >, c (r2 - since x varies between the limits defined by s a more appropriate form for this b c cos 9 == (r2 - x2) ~ cos 9 equation would be 2 s • 2 by substitution X = S COS 8 1( - r2 SLn 8 ) ->, 2 2 x == s (r - x )~ cos 9 (r 2 - s 2 )~ note from this expression that piston motion is not rearranging terms and squaring purely sinusoidal, yet sinusoidal motion would be a good first approximation.

For a multistage or multipiston compressor each can be analyzed by substi­ tuting (9 - en) for e in this expression where en equals this - 1 piston position on the swash plate (this also assumes r is equal for all pistons.)

For the purposes of this analysis we will 1 assume ideal isentropic compression :X2 (ignoring losses ~ue to valves, dead volume, or heat transfer within the cylinder.)

for an isentropic process n k extracting r 2 ~~ = ( ~~) 2 2 2 x = r cos 9/(~~- 1) + cos 2 s/- 1 cylinder or volume == V = A(s - x) where A is piston area and s - x is the distance from 2 2 2 piston to the head. For a given cylinder r cos 8 + (cos 2 1i2 e - 1) 1- l undergoing compression from position 1 to position 2

77 1/k Ar (s - xr) f~) y Az(s- Xz) ( then 1/k

consider position 1 to be the start of the compression process where P1 ~ inlet pres­ sure and x1 ~ -s, then for any position 2 X (prior to when the discharge valve opens) the following relation holds:

2s p 1/k s - x 2 (p~) or B 2s }k P1 (s-=-x 2 Figure 3 it is useful to know at what point the discharge valve opens (i.e. Pz equals the Mass & Load Distribution discharge pressure) solving for x ~ s - 2s ( ~~) l/k incident with one of the pistons g1ves a less than ideal load distribution. in th1s· case PzP1 represents the compression ratio of that stage. To minimize weight and product of inertia With the relation for pressure, P, as a effects, the locations of four masses function of piston position, x, and piston favors a 90° symmetrical distribution position as a function of crank angle, e, (¢ = 0°). Since both configurations are the thrust load on the bearings can be used, four masses and three loading points, determined by summing the forces P x A, a compromise must be reached. where A is the piston area, for all stages through one revolution of the crankshaft. The kinetics and kinematics of the system lead to relative size requirements of the pistons and the thrust rider. Starting ARRANGEMENT OF PISTON MASSES FOR DYNAMIC with an adequately designed thrust rider, the piston masses and counterweights can BALANCE be sized. A good dynamic balance can be achieved by proper mass and load distribution coupled It is important to note how the compromise with adequate counterweighting. was made to arrive at cp = 15°. Particu­ larly, the size of the thrust rider bear­ Consider a three stage compressor with a ing in comparison to the optimal first single thrust rider. Positions on the stage size (as determined by load swash plate are as shown in Figure 3. capacity), coupled with the most even spacing between A, c, and D, resulted in A 1st Stage Piston ¢ = 15 as the best trade-off (i.e. the B Thrust Rider lighter the thrust rider the better the c = 2nd Stage Piston load distribution. D 3rd Stage Piston Having established the relative position­ cp 15° ing, a moment balance and moment of inertia r radius balance is considered. See Figure 4.

For balance: EM 0 X (Statically) An equal load distribution occurs when EM 0 cp = 30° leaving 1200 between A, C, and D. y However, such a configuration would require the thrust rider B to have zero mass for a and I ~ I XX yy good mass balance. Thus the existence of a thrust rider at a position other .than co-

78 I y YY I yy (4)

Setting I :ex = I yy

2 2 m r + m r + 2m r 2 sin 2 ¢ 2m r 2 2 A B e "' c cos ¢

2 2 rnA + mB + 2me sin ¢ = 2me COS ¢

From the moment balance

Figure 4 (2)

Inertia & Moment Balance Substituting for rnA from eq. (2)

mB + 2me sin ¢ + mB + 2mc sin2 ¢ Assume A, B, e, and D are at an equal radius r from the axis of 2 the machine. = 2mc cos ¢ Let rnA, m , me, and m 8 0 equal the masses at rearranging each position, respectively. 2 2 2mB = 2me (cos ¢ - sin ¢ sin ¢) g = gravitational acceleration. -

2 Thus: mB me (cos ¢ - sin2 ¢ - sin ¢)

EM 0 megr cos ¢ - m gr cos or y 0 ¢ 0 mB m = me c mn (1) 2 (5) (cos ¢ - - sin 2 ¢ sin ¢) l:M 0 :c = mAgr - mBgr - megr sin ¢ Experience has shown that mB, the mass of - nugr sin ¢ 0 = the anti-rotation device (thrust rider), is the most convenient starting point when initiating a new design. Therefore, the piston masses are exp:t·essed in terms of mB. 0 (6) Solving for rnA m = (7) rnA = D m8 - 2me s~n ¢ (2) Since rnA (2) For a moment of inertia balance:

I I (8) x:c YY or setting This produces a situation where the masses mB = l (9) are evenly distribu~ed about the center of rotation. This is necessary for obtaining rnA L852 (10) balance by counterweighting alone. 1. Thus: 647 (11)

2 2 I m r + m r ¢) 2 = 1.647 (12) XX A B + mc(r sin mD

m (r sin ¢) 2 for dynamic balance. + 0 I m r 2 + m r 2 + 2m r 2 XX A B c sin ¢ (3)

79 COUNTERWEIGHTING

For a disc shaped body fixed at an angle ~ to the axis of a rotating shaft a gyratory moment results. This moment rotates with the shaft at angular velocity w. By physically weighing and dimensioning parts, accurate values of II and I can be 0 determined.

Since the direction of moment MG rotat~s For a swash plate mechanism where the inner race rotates and the outer race wobbles with the shaft and its magnitude is without rotating (restricted by the thrust constant it is possible to reach a smooth rider) two moments are of primary concern. running balance by counterweighting.

The radial moment of inertia of the inner Consider two equal masses eccentrically race which rotates with the shaft tends to mounted to the shaft, 180° opposed at decrease the angle ~ because of the gyrato­ separation b and radius a. ry restoring moment produced (MI) . The a = constant moment of inertia of a wobbling outer race b "' constant tends to increase the angle ~ because of mE = constant the inertial moment (M ) generated by the 0 change of direction (i.e. axial reciproca­ tion) . The difference between these two See Figure 6. moments is an unbalanced moment MG which results in rough running (i.e. MG = M0-MI). (See Figure 5). II From the dynamics of a rigid body rotating about an axis we have .. K E I w2 sin ~ cos ~ (1) I

(2) where II = radial moment of inertia of the rotating inner race. and I = radial moment of inertia of the 0 reciprocating (non-rotating) outer race (includes piston and thrust rider masses). I I

Figure 6

Counterweights

As these masses rotate about the shaft a centrifugal force KE is generated for each. I I (4) A couple ME 2KEb/2 results

Figure 5 (5)

Gyratory Moments

80 For a running balance the moments must be CONCLUSIONS equated. l) It is possible to analyze (6) the dynamic behavior of the swash plate mechanism to optimize piston loading and running Thus: balance.

2 2 2) Swash (I - I l w sin t/J cos tj! = MEw (7) plate piston motion is not 0 1 ab sinusoidal, purely yet it is close enough to be a good first approximation. (8) 3) When the number of masses on the swash since II, I can be calculated plate does not match the 0 and tj! is number of loading points (as occurs when known mE, a and b can be proportioned a to separate thrust bearing is used) compromise a best suit the machine design. must be reached when deter­ mining their distribution. Here optimization an technique may be used. 1. Note that counterweighting is indepen­ dent of running speed. BIBLIOGRAPHY 2. Note that the greater the distance b, the smaller the counterweights need to Dynamics, 2nd Edition, J. c. Meriam, be. John Wiley 3. Note and Sons, 1971. also that the center of MG need SKF Engineering not coincide with the center ME Data, SKF Industries, to Inc., 1972. obtain balance. Where they are not coincident a bending An moment is induced Introduction to Mechanical Vibrations in the shaft. If the shaft is Robert with designed F. Steidel, Jr., John Wiley ' an adequate stiffness (giving and Sons, deflections 1971. of approximately .001 inch or less due to the bending moment) any Introduction to Mechanics of Solids, increase in vibration should be Egor negligible. P. Popov, Prentice~Hall, 1968. NOTATIONS (See Figure 2)

The constants are fixed by design. The variables change with crank angle. 8 = crank angle e = 0 at T.D.C. s (variable) = ~ piston length (constant) X piston position, X = S at 8 = 0 and X = 0 at 8 = 90 (variable) r distance from center of crankshaft to piston a (constant) = angle of the swash plate to the crankshaft (constant) c distance from piston to center of wobble plate normal to 2 2 x-axis = (r - x )~ (variable)

81