Systems ME 597/ABE 591 - Lecture 7 Dr. Monika Ivantysynova MAHA Professor Fluid Power Systems MAHA Fluid Power Research Center Purdue University
Content of 6th lecture The lubricating gap as a basic design element of displacement machines Gap flow calculation. Gap with a constant gap height Gap with variable gap height (slipper) Gap between piston and cylinder, numerical solution of gap flow equations Non-isothermal gap flow Aim: Derivation of basic equations for gap flow and calculation of Gap flow parameters (pressure and velocity distribution, load ability, gap flow and viscous friction) Basic knowledge about the gap design and simulation models 2 2
Lubricating Gap Basic Design Element of Displacement Machines Design & operating parameter Gap height Gap flow Load ability Leakage Performance & Losses Viscous friction 3 3
Lubricating Gaps -examples Swash plate axial piston machine Q SK Gaps seal the displacement chamber Q SB Q SG Movable Piston p i Rotating cylinder block Displacement chamber Q r Valve plate (fixed) Gaps between: piston and cylinder slipper and valve plate cylinder block and valve plate Swash plate (fixed or with adjustable angle) 4 4
Gap Flow Calculation Aim: High load carrying ability Low friction Low leakage flow Gap height lies in a range of some micrometers, whereas all other dimensions are in a range of millimeters Laminar flow of an incompressible fluid in the gap can be described using Navier-Stokes-Equation: y z h=f(x,y) movable surface Assumptions: p - neglecting mass forces p 2 1 h - assuming steady state flow, - change of fluid velocity only in direction of gap height x - pressure is not a function of gap height h fixed surface Pressure Force + Viscosity Force = 0 5 5
Gap with constant gap height Pressure field Forces applied on a fluid element: with τ = µ v z after double integration velocity yields: v = 1 2 µ p x z2 + c 1 z + c 2 h p boundary conditions: z=0 v=0 z=h v=0 v = 1 2 µ p x z2 h z ( ) 6 6
Gap with constant gap height Gap flow: h Q = b v dz = 1 12µ p x b h 3 0 for pressure: Shear stress on surfaces: τ z= 0 = Δp l with: Viscous friction: h 2 p = p 2 p 1 l x + p 1 z = 0h τ = µ v z τ z= h = Δp l F v = τ b x Power loss due to gap flow: h 2 z = h P SQ = Q Δp = 1 12 µ b h 3 Δp2 l 7 0 h l v = 1 2 µ p x z2 h z ( )
Gap with constant gap height boundary conditions: z=0 v=0 x=0 p=p 1 z=h v=v 0 x=l p=p 2 z p 1 =p 2 v 0 p 1 p 2 h Flow velocity: l x b >> h Shear stress on surfaces: Gap flow: Viscous friction: h Q = b v dz = b v 0 h 0 h z dz = b v 0 h h 2 2 = b v 0 2 h 0 F v = τ l b = µ v 0 h l b Power loss due to viscous friction: 8 8 P Sv = F v v 0 = µ l b h v 2 0 τ = µ v z
Gap with constant gap height Boundary conditions: z=0 v=0 x=0 p=p 1 z=h v=v 0 x=l p=p 2 Flow velocity: v = 1 p 2 µ x z2 h z Gap flow: ( ) + v 0 h z p x = µ 2 v z 2 h Q = b v dz = 1 +v 0 h 12µ p x b h 3 2 b 0 Shear stress on surfaces: Power losses: τ = µ v z = p 2 p 1 l P s = P SQ + P Sv = Q Δp + F v v 0 P s = 1 Δp 2 12 µ l 9 9 p 1 z p 1 + p 2 h b h 3 + µ v 0 2 h b l l p 2 v 0 1 ( 2 2 z h ) + v 0 h µ x b >> h
Radial Gap with constant gap height r h r p 1 p 2 x l d Gap width: b = π d Equations from gap with constant gap height can be applied: Gap flow: Q = b h 0 v dz = π d p 12µ x h 3 10 10
ap with variable gap height z Hydrodynamic force p 1 =p 2 =p 0 = 0 Boundary conditions: z=0 v=v 0 x=0 p=p 1 z=h v=0 x=l p=p 2 Flow velocity: with p x = µ 2 v z 2 v = 1 p 2 µ x z2 h z ( ) τ = µ v z v 0 h z + v 0 Gap flow: h Q = b v dz = 1 +v 0 h 12µ p x b h 3 2 b 11 11 0
Gap with variable gap height Pressure distribution in x-direction: (1) p F from Q we obtain: = 1 +v 0 h 12µ p x b h 3 2 b (2) x p x = 12µ Q + 6µ v 0 b h 3 h 2 (3) v 0 p( x) = 6µ v 0 h 2 12µ Q b h 3 dx (4) with: (5) 12 12
Gap with variable gap height p( x) = 6µ v 0 h 2 12µ Q b h 3 dx (4) with: (5) using: (6) in our case: (7) where n=-2 for the first term in Eq. (4) and n= -3 for the second term in Eq. (4) and after integration: 6µ v 0 h 2 h 1 ( 1) l 1 12µ Q 1 h 2 h 1 x + h 1 b h 2 h 1 ( 2 ) h 2 h 1 + c 2 l l x + h 1 l h h Boundary conditions: x=0 p=p 0 p( x) = 13 13 (8)
Gap with variable gap height 6µ v 0 h 2 h 1 ( 1) l c = p 0 + 6µ v l 0 h 2 h 1 p( x) = p x ( ) h 1 ( ) = 6µ l ( ) v 0 h 2 h 1 1 h 2 h 1 x + h 1 l 6µ Q l b h 2 h 1 h + ( ) h 1 2 Q b h 2 + v 0 h 1 Q b h 1 2 b h h 2 1 h l (9) 12µ Q + p 0 (10) ( 2) h 2 h 1 l 1 + c 2 x + h 1 h 2 h 1 = ( h h 1 ) l x (8) (11) (12) (13) finally we get: Boundary conditions: x=0 p=p 0 ( ) = 6µ x ( ) v h + h 1 0 p x h h 1 h h 1 Q b + p 0 (14) 14 14
Gap with variable gap height ( ) = 6µ x ( ) v h + h 1 0 p x h h 1 h h 1 Q + p 0 b Using boundary conditions for x=l is p=p 0 and h=h 2 from Eq. (14) follows : (14) h 1 p F x p 1 =p 2 =p 0 =0 Q = v 0 b h 2 h 1 h 2 + h 1 (15) p 1 v 0 h 2 p 2 x h Substituting Eq. (15) into Eq. (14) the pressure p(x) yields: p( x) = 6 µ x v 0 h h 2 + p h 2 0 h 2 + h 1 (16) when h=h 1 =h 2 p(x)=p 0 No load ability! 15 15
Gap with variable gap height Load capacity due to hydrodynamic pressure field generated in the gap: l 0 ( ) F = b p p 0 l 0 F = b 6 µ v 0 F = 6 µ b l2 h 1 h 2 dx ( ) ln h 1 2 (17) h h 2 ( ) x dx h 2 h 2 + h 1 2 h 1 h 2 v 0 h 2 h 1 + h 2 Maximum pressure force for h 1 /h 2 =2.2 yields: p x ( ) = 6 µ x v 0 h 2 h h 2 h 2 + h 1 + p 0 (18) (19) h 1 p 1 p h v 0 F h 2 x p 2 (16) p 1 =p 2 =p 0 =0 x F max = 6 µ l2 1.2 2 2 h 2 2.4 ln2.2 3.2 v 0 = 0.16 µ l2 h 2 2 v 0 (20) 16 16
Example A gap between two circular plain and parallel surfaces with a central inflow as shown in Fig.1 is given. Both surfaces are fixed. The inner radius is R 1 and the outer radius is R 2. Calculate the flow Q through the gap for a given gap height h, when the pressure at the inner radius is p 1 and the pressure at the outer radius is p 2. And determine the pressure on the radius r= 15 mm. The following parameters are given: p 1 =20 MPa R 1 =12 mm p 2 =0.5 MPa R 2 =25 mm h = 20 µm Dynamic viscosity of the fluid: µ =0.0261 Pa s Boundary conditions: p 1 p 2 p 2 r z=0 v=0 r=r 1 p=p 1 z=h v=0 r=r 2 p=p 2 17 17 r Fig.1
Example From force balance on a fluid element follows: p r = τ z with τ = µ v z After double integration the velocity yields: v = 1 p 2 µ r z2 h z Gap flow: Therefore: and p 2 h ( ) Q = 2 π r v dz = π r h 3 0 p r = 6 µ Q π r h 3 (1) (4) dp = 6 µ Q dr π r h 3 p 1 R 2 R 1 6 µ (5) p r Boundary conditions: z=0 v=0 r=r 1 p=p 1 z=h v=0 r=r 2 p=p 2 (3) p 1 p 2 p 2 r (6) r 18 18
Example (6) From Eq. (6) follows for the gap flow: (7) For the pressure distribution in radial direction we obtain integrating Eq. (4): (7) And using boundary conditions for c follows: (4) (8) (9) 19 19
Example The pressure distribution in radial direction yields: (9) Substituting the volume flow equation Eq. (7) into Eq. (9) we obtain: (7) p 1 p 2 p 2 r r 20 20
Gap between piston & cylinder Axial and radial piston motion cylinder v p R Z p 2 p 1 >p 2 p 1 ω R z R K Unrolled gap in Cartesian co-ordinates system Gap height due to inclined piston position Gap height p 1 Gap circumference p 2 Gap length 21 21
Gap height function cylinder 22 22
Gap between piston & cylinder After double integration flow velocity yields: Boundary conditions: z=0 v x =0 v y =0 z=h v x =ω R z v y =v p v x = 1 2 µ p x (z2 h z) + ω R z z h 23 23
Gap between piston & cylinder Considering the continuity equation for incompressible fluid: After substituting the mean velocities the Reynolds-equation can be derived: and taking into account time dependent change of gap height the Reynolds equation becomes: This partial differential equation has to be solved numerically 24 24
Numerical solution of gap flow Applying the method of finite differences the Reynolds equation can be written: i j (1) i, j (2) Δx Δy (3) (4) Different methods available: -finite differences -finite volumes -finite elements Substituting Eq. (1), (2), (3) and (4) into the Reynolds- equation we obtain: (5) Iterative solution of Eq. (5) to calculate the pressure p i,j 25 25
Numerical solution of gap flow The gap grid Gap height Gap height Gap circumference Gap length 26 26
Gap Parameter Pressure field between piston and cylinder Load capacity of the gap p [MPa] 0 Gap length Gap circumference with: 27 27
Contact between piston and swash plate Support by slippers Increase of volumetric losses Hydrostatically balanced Hydrodynamically balanced Additional hydrodynamic effect usable Requires oil filled case Special surface geometry can support hydrodynamic effect Lower loading of piston-cylinder Suitable for high pressure and high speed 28 8
Slipper Design Gap flow assuming constant gap height B G slipper balance (0.95 0.99) (1) F GF Flow from displacement chamber to slipper pocket Through a small orifice or laminar throttle (3) (2) 29 9
Slipper Design Load ability F GF due to pressure field under the slipper (1) D G /d G = 1.2 1.3 (2) F GF D G /d K = 1.2 1.3 Viscous friction force: Losses due to friction: Losses due to gap flow: Power loss: 30 10
Slipper Motion Gap flow simulation requires the determination of gap height Pressure field HP HP HP LP LP LP Operating parameter: 31 11
Slipper hold down device Slipper hold down using springs Inside the piston Outside the piston Slipper hold down by cylinder block spring Slipper hold down device Cylinder block spring Slipper hold down spring 32 11
Slipper hold down device Slipper hold down using springs Slipper hold down device Slipper hold down spring Slipper hold down spring 33 13
Slipper hold down device Fixed slipper hold down Slipper hold down spring Spacer 34 14