Fluidic Dynamics Dr. Thara Srinivasan Lecture 17 EE C245 Picture credit: A. Stroock et al., PictureMicrofluidic credit: mixingSandia on National a chip Lab Lecture Outline • Reading • From S. Senturia, Microsystem Design, Chapter 13, “Fluids,” p.317-334. • Today’s Lecture • Basic Fluidic Concepts • Conservation of Mass → Continuity Equation • Newton’s Second Law → Navier-Stokes Equation • Incompressible Laminar Flow in Two Cases • Squeeze-Film Damping in MEMS EE C245 2 U. Srinivasan © 1 Viscosity • Fluids deform continuously in presence of shear forces • For a Newtonian fluid, • Shear stress = Viscosity × Shear strain 2 2 • N/m [=] (N • s/m ) × (m/s)(1/m) • Centipoise = dyne s/cm2 U • No-slip at boundaries τw h U τw x -5 2 ηair = 1.8 × 10 N s/m -4 ηwater = 8.91 × 10 -3 ηlager = 1.45 × 10 ηhoney = 11.5 EE C245 3 U. Srinivasan © Density • Density of fluid depends on pressure and temperature… • For water, bulk modulus = • Thermal coefficient of expansion = • …but we can treat liquids as incompressible • Gases are compressible, as in Ideal Gas Law R PV = nRT P = ρm T MW EE C245 4 U. Srinivasan © 2 Surface Tension • Droplet on a surface • Capillary wetting 2r γ γ ∆P 2γ cosθ h = ρgr EE C245 5 U. Srinivasan © Lecture Outline • Today’s Lecture • Basic Fluidic Concepts • Conservation of Mass → Continuity Equation • Newton’s Second Law → Navier-Stokes Equation • Incompressible Laminar Flow in Two Cases • Squeeze-Film Damping in MEMS EE C245 6 U. Srinivasan © 3 Conservation of Mass • Control volume is region fixed in space through which fluid moves rate of accumulation = rateof inflow − rate of outflow rate of mass efflux • Rate of accumulation dS • Rate of mass efflux EE C245 7 U. Srinivasan © Conservation of Mass ∂ρ ∫∫∫ m dV + ∫∫ ρ U ⋅ndS = 0 V ∂t S m dS EE C245 8 U. Srinivasan © 4 Operators • Gradient and divergence ∂U ∂U ∂U ∇U = x i + y j + z k ∂x ∂y ∂z ∂ ∂ ∂ ∇ = e + e + e ∂x x ∂y y ∂z z ∂U ∂U ∂U ∇ ⋅ U = div U = x + y + z ∂x ∂y ∂z EE C245 9 U. Srinivasan © Continuity Equation • Convert surface integral to volume integral using Divergence Theorem • For differential control volume ∂ρ ∂ρ ∫∫∫ + ∇ ⋅(ρU)dV = 0 + ∇ ⋅(ρU) = 0 V ∂t ∂t EE C245 Continuity Equation 10 U. Srinivasan © 5 Continuity Equation • Material derivative measures time rate of change of a property for observer moving with fluid ∂ρ Dρ ∂ρ + (U ⋅∇)ρ + ρ∇ ⋅ U = 0 = + (U ⋅∇)ρ ∂t Dt ∂t D ∂ ∂ ∂ ∂ ∂ = + U ⋅∇ = +U +U +U Dt ∂t ∂t x ∂x y ∂y z ∂z For incompressible fluid Dρ Dρ + ρ∇ ⋅ U = 0 + ρ∇ ⋅ U = 0 EE C245 Dt Dt 11 U. Srinivasan © Lecture Outline • Today’s Lecture • Basic Fluidic Concepts • Conservation of Mass → Continuity Equation • Newton’s Second Law → Navier-Stokes Equation • Incompressible Laminar Flow in Two Cases • Squeeze-Film Damping in MEMS EE C245 12 U. Srinivasan © 6 Newton’s Second Law for Fluidics • Newton’s 2nd Law: • Time rate of change of momentum of a system equal to net force acting on system Rate of Sum of forces Rate of momentum efflux accumulation of acting on control + = from control momentum in volume volume control volume dp d ∑F = = ∫∫∫ Uρ dV + ∫∫ Uρ()U ⋅n dS dt dt V S Net momentum Net momentum accumulation rate efflux rate EE C245 13 U. Srinivasan © Momentum Conservation • Sum of forces acting on fluid ∑ F = pressure and shear gravity force stress forces • Momentum conservation, integral form d ∫∫ ()− Pn + τ dS + ∫∫∫ ρgdV = ∫∫∫ ρUdV + ∫∫ ρU()U ⋅n dS S V dt V S EE C245 14 U. Srinivasan © 7 Navier-Stokes Equation • Convert surface integrals to volume integrals 2 η ∫∫τdS = ∫∫∫η∇ U + ∇()∇ ⋅ U dV S V 3 ∫∫− PndS = ∫∫∫− ∇P dV S V ∫∫ ρU(U ⋅n)dS = ∫∫∫ρU(∇ ⋅ U)dV S V EE C245 15 U. Srinivasan © Navier-Stokes Differential Form 2 η ∫∫∫− ∇P + ρg +η∇ U + ∇()∇ ⋅ U dV = V 3 ∂U ∫∫∫ ρ dV + ∫∫∫ρU()∇ ⋅ U dV V ∂t V DU ∫∫∫ ρ dV V Dt DU η ρ = −∇P + ρg +η∇2U + ∇()∇ ⋅ U EE C245 Dt 3 16 U. Srinivasan © 8 Incompressible Laminar Flow • Incompressible fluid ∇ ⋅ U = 0 DU η ρ = −∇P + ρg +η∇2U + ∇()∇ ⋅ U Dt 3 DU ρ = −∇P + ρg +η∇2U Dt EE C245 17 U. Srinivasan © Cartesian Coordinates DU ρ = −∇P + ρg +η∇2U Dt x direction ∂U x ∂Ux ∂Ux ∂Ux ρ +Ux +U y +Uz = ∂t ∂x ∂y ∂z 2 2 2 ∂Px ∂ Ux ∂ Ux ∂ Ux − + ρg x +η + + ∂x ∂x 2 ∂y 2 ∂z 2 EE C245 18 U. Srinivasan © 9 Dimensional Analysis DU ∇P η∇ 2 U = − + g + Dt ρ ρ • Each term has dimension L/t2 so ratio of any two gives dimensionless group • inertia/viscous = = Reynolds number • pressure /inertia = P/ρU2 = Euler number • flow v/sound v = = Mach number • In geometrically similar systems, if dimensionless numbers are EE C245 equal, systems are dynamically similar 19 U. Srinivasan © Two Laminar Flow Cases U τw τw Low h High h τw P τw P y y Ux Ux EE C245 U Umax 20 U. Srinivasan © 10 Couette Flow • Couette flow is steady viscous flow ∇ ⋅ U = 0 U = U i between parallel plates, where top plate is x x moving parallel to bottom plate U = Ux ( y)i x • No-slip boundary conditions at plates 2 2 DUx 1 ∂P η ∂ Ux ∂ Ux = − + gx + + Dt ρ ∂x ρ ∂x 2 ∂y 2 2 U = 0 at y = 0 ∂ U x x 2 = 0, and ∂y U x = U at y = h U y τw h U U x = τ x EE C245 w U 21 U. Srinivasan © Couette Flow • Shear stress acting on plate due to motion, $, is dissipative • Couette flow is analogous to resistor with power dissipation corresponding to Joule heating ∂U x τ w = −η = ∂y y=h U τw τ A ηA h τ R = w = w Couette U h 2 PCouette = RU EE C245 22 U. Srinivasan © 11 Poiseuille Flow • Poiseuille flow is a pressure-driven flow between stationary parallel plates • No-slip boundary conditions at plates 2 2 DUx 1 ∂P ∂ Ux ∂ Ux = − + gx +η + Dt ρ ∂x ∂x 2 ∂y 2 ∂P ∆P ∂2U ∆P = − x = − , U = 0 at y = 0,h ∂x L ∂y 2 Lη x U = y x τw Low Highh Ux P τw P Umax = EE C245 23 U. Srinivasan © Umax Poiseuille Flow • Volumetric flow rate Q ~ [h3] h • Shear stress on plates, $,is Q = W ∫ U dy = dissipative 0 x • Force balance • Net force on fluid and plates is zero ∂Ux ∆Ph since they are not accelerating τ w = −η = • Fluid pressure force ∆PWh (+x) ∂y y=h 2L balanced by shear force at the walls of 2$WL (-x) • Wall exerts shear force (-x), so fluid Q must exert equal and opposite force U = on walls, provided by external force Wh 2 h ∇P 2 τw U = = 3 Umax Ux 12ηL EE C245 τw W 24 U. Srinivasan © 12 Poiseuille Flow • Flow in channels of circular cross section 4 2 2 (r − r )∆P πro ∆P U = o Q = x 4ηL 32ηL • Flow in channels of arbitrary cross section × 2 L 4 Area ∆ = 1 ρ f Re = dimensionless Dh = P f D ()2 U D D Perimeter Dh constant • Lumped element model for Poiseuille flow ∆P 12ηL Rpois = = 3 EE C245 Q Wh 25 U. Srinivasan © Velocity Profiles • Velocity profiles for a • Stokes, or creeping, flow combination of • If Re << 1 , inertial term may pressure-driven be neglected compared to viscous term (Poiseuille) and plate motion (Couette) flow • Development length • Distance before flow assume steady-state profile EE C245 26 U. Srinivasan © 13 Lecture Outline • Today’s Lecture • Basic Fluidic Concepts • Conservation of Mass → Continuity Equation • Newton’s Second Law → Navier-Stokes Equation • Incompressible Laminar Flow in Two Cases • Squeeze-Film Damping in MEMS EE C245 27 U. Srinivasan © Squeezed Film Damping • Squeezed film damping in parallel plates • Gap h depends on x, y, and t F • When upper plate moves moveable downward, Pair increases and air is squeezed out • When upper plate moves h upward, Pair decreases and air is sucked back in fixed • Viscous drag of air during flow opposes mechanical motion EE C245 28 U. Srinivasan © 14 Squeezed Film Damping • Assumptions • Gap h << width of plates • Motion slow enough so gas moves under Stokes flow • No ∆P in normal direction • Lateral flow has Poiseuille like velocity profile • Gas obeys Ideal Gas Law • No change in T ∂(Ph) 12η = ∇ ⋅[](1+ 6K )h 3P∇P • Reynold’s equation ∂t n • Navier-Stokes + continuity + Ideal Gas Law • Knudsen number Kn is ratio of mean free path of gas molecules λ to gap h • Kn < 0.01, continuum flow h EE C245 • Kn > 0.1, slip flow becomes important • 1 µm gap with room air, λ = 29 U. Srinivasan © Squeezed Film Damping • Approach ∂(Ph) h 3 • Small amplitude rigid = []∇2P 2 motion of upper plate, h(t) ∂t 24η • Begin with non-linear partial differential equation • Linearize about operating h = h0 + ∂h and p = P0 + ∂P point, average gap h0 and average pressure P 0 ∂p h 2P ∂2 p 1 dh • Boundary conditions = 0 0 − • At t = 0, plate suddenly ∂t 12ηW 2 ∂ξ 2 h dt displaced vertically amount 0 z (velocity impulse) 0 ∂P y • At t > 0+, v = 0 p = and ξ = • Pressure changes at edges P0 W of plate are zero: dP/dh = 0 EE C245 at y = 0, y = W 30 U. Srinivasan © 15 General Solution • Laplace transform of response to general time-dependent source z(s) 96ηLW 3 1 = F(s ) 4 3 ∑ 4 sz(s ) n odd S π h0 n ()1+ αn plate velocity b 96ηLW 3 π 2h 2P F(s ) = sz(s ), b = , ω = 0 0 S 4 3 c 2 1+ ωc π h0 12ηW damping cutoff 1st term for small constant frequency EE C245 amplitude oscillation 31 U.