MIT2_72s09_lec14

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 2.72 Elements of Mechanical Design Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Reading and plans Shigley-Mischke sections None Today: Actuators: Hydraulic and Electromagnetic Energy transfer and scale Hydraulic / fluidic DC Permanent magnet motors Perhaps…. wrap up of MEMS © Martin Culpepper, All rights reserved 1 Friction-based machines Purpose: Do work at a given rate, Energy - Power Physics: Energy and mass conservation/balances Characteristics of import Load Speed Bandwidth Cost © Martin Culpepper, All rights reserved 2 And there can be other issues of import… Image removed due to copyright restrictions. Please see http://www.onefunsite.com/images/donkey.jpg © Martin Culpepper, All rights reserved 3 Consequences Please see trigirl. “Crane Drops Steamroller on Car!” May 8, 2007. LiveVideo. Accessed November 25, 2009. http://www.livevideo.com/video/16A18C6512B945C29547A8658E890AF1/crane-drops-steamroller-on-car.aspx © Martin Culpepper, All rights reserved 4 An unpleasant (I hope) example © Martin Culpepper, All rights reserved 5 Common actuators for mechanical systems Biological Pneumatic/Hydraulic Electromagnetic Electrostatic Piezo Thermal © Martin Culpepper, All rights reserved 6 Biological People powered machines Energy Power Load Speed Bandwidth Why is it important to understand what humans can do? © Martin Culpepper, All rights reserved 8 Hydraulics Basic principles Sub-system design Pump Motor Output shaft Τ2=? Tp = ? Dp = ½ in3/rev Dm = 2 in3/rev ωp = 100 rpm Apiston = 1 in2 ω2 = 100 rpm Nm = 40 vp = ? N2 = ? p1 Pump p2 p3 Motor p4 Flow direction Reservoir © Martin Culpepper, All rights reserved 10 Examples: Real but practical ;) ? Image removed due to copyright restrictions. Please see http://darkdiamond.net/wp-content/uploads/2006/08/115638952415_hugegundam1.JPG http://gizmodo.com/ © Martin Culpepper, All rights reserved 11 Other less than practical examples Please see HydraulicGuitar. “Hydraulic Guitar.” September 10, 2006. YouTube. Accessed November 25, 2009. http://www.youtube.com/watch?v=Elt1XriaQXU © Martin Culpepper, All rights reserved 12 Other less than practical examples Please see any video of a hydraulic low rider assembly. © Martin Culpepper, All rights reserved 13 Other less than practical examples Please see arefadib. “The Flying Steamroller.” October 17, 2006. YouTube. Accessed November 25, 2009. http://www.youtube.com/watch?v=sKGRRIiR5xA © Martin Culpepper, All rights reserved 14 Hydraulic systems in machines Advantage: High force/torque and routing of power Disadvantage: Leaking and wear due to contaminants Liquids & gases in fluid-based machinery Hydraulics: Pneumatics: © Martin Culpepper, All rights reserved Fluid is a liquid Fluid is a gas 15 Example: Piston pump doing work Hydraulic machines can be used to do work Load on the system extracts energy from the liquid Pressure increases between the input and output components Pressure is used to do work on internal parts of hydraulic devices Power is input/extracted via shaft (motor) or rod (cylinder) Power out 0 Motor © Martin Culpepper, All rights reserved Cylinder Pressure gauge 16 Volume flow rate and displacement Displacement (D) Displacement = volume of fluid moved / cycle Cycle = rotation (drill pump) or stroke (cylinder) Q = volume moved per unit time D Volume Cycle(s) * f Cycles * second = = Q Volume second F is the frequency of a machine’s cycle o For hydraulic pumps, f = speed of the shaft = ω/(2π) o For hydraulic motors, f = speed of the shaft = ω/(2π) o For cylinders, f = strokes/second © Martin Culpepper, All rights reserved ω [rad/s] ω [rad/s] f [Hertz] 17 Displacement: Physical example D = volume pumped per cycle 1 Cycle = expansion + contraction rinitial r final Vinitial Vinitial 4 = ⋅ π ⋅ (rinitial )3 3 Vfinal = Vfinal Displacement = D = Vinitial − Vfinal = © Martin Culpepper, All rights reserved [ ( 4 ⋅ π ⋅ r final 3 ( )3 4 ⋅ π ⋅ (rinitial )3 − r final 3 )3 ] 18 Incompressibility Incompressible fluid: Compressible fluid: © Martin Culpepper, All rights reserved 19 Why is incompressibility important? Mass balances The mass density (ρm) of fluids changes with pressure (Δp) Compressible fluids: exhibit large (Δρm) for small (Δp) If (Δρm) is large, it is possible to store significant mass in a machine This complicates our analysis Σmin = Σmout © Martin Culpepper, All rights reserved d + mstored dt 20 Why is incompressibility important? Energy balances All fluids store energy when compressed (similar to a spring) Compressible fluids store A LOT of energy (think balloons!!) Stored energy complicates analysis (calculating can be difficult) ΣEin = [ΣEout ] + ΣEstored © Martin Culpepper, All rights reserved 21 Example: “Locked” piston positions Incompressible fluid: Compressible fluid: © Martin Culpepper, All rights reserved 22 Incompressibility Bulk modulus: Measures of resistance to Δvolume β =− dp ⎛ dV ⎞ ⎜V ⎟ initial ⎠ ⎝ For small (incremental) changes in volume :β ≈ − Δp ⎛ ΔV ⎞ ⎜ Vinitial ⎟ ⎝ ⎠ Example: Fluid in tube exposed to pressure increase ΔV V final Vinitial © Martin Culpepper, All rights reserved 23 Incompressibility Hydraulics, pneumatics and incompressibility Low β, usually compressible High β, usually incompressible Pneumatics = gas: Hydraulics = liquid: What makes a good assumption? Depends on the error you are willing to live with Example: Incompressibility of water (e.g. H2O) β H 2O = 2.2 ×109 N m2 = 3.2 ×105 (ΔV V ) ≈ − Δβp lbf in 2 Example for H 2O where Δp = 2500 psi, ΔV V = −0.006 = −0.6% Is this OK? © Martin Culpepper, All rights reserved 24 Volume flow rate, Q Link between mass flow rate & volume flow rate: Q = time rate of volume flow through a hydraulic system From mass conservation From 8.01 Qi = mi ρ mi Σmin ρ m in = = ρ mi ⋅ Ai ⋅ v i ρ mi Σmout ρ mout d (Vstored ) ~ 0 dt d mstored d + dt → Qin = Qout + (Vstored ) dt ρ mstored Mass densities are equal and cancel out of equation if fluid is incompressible For incompressible flow in a pipe : Ai n ⋅ vi n = Qi n = Qout = Aout ⋅ v out © Martin Culpepper, All rights reserved 25 Vane pumps Series of vanes extending radially from rotating core Vanes can slide in/out or deform depending upon design How a sliding vane pump works: Step 1: Fluid enters when volume between vanes is increasing Step 2: Fluid travels when volume between vanes does not change Image removed due to copyright restrictions. Please see http://pumpschool.com/images/vnsteps.gif © Martin Culpepper, All rights reserved credit: pumpschool.com Step 3: Fluid exits at when volume between vanes is decreasing 26 Pump types: Piston How it works: Step 1: Piston forces fluid out during initial stroke Step 2: Valves change fluid path (only allows flow into pump) Step 3: Piston recharged with fluid, cycle starts again What is the displacement? © Martin Culpepper, All rights reserved 27 Pump types: Piston How it works: Step 1: Piston forces fluid out during initial stroke Step 2: Valves change fluid path (only allows flow into pump) Step 3: Piston recharged with fluid, cycle starts again Images removed due to copyright restrictions. Please see http://www.animatedsoftware.com/pics/pumps/wobble.gif http://www.flexicad.com/bilder/Rhino_Galerie/Kolpenpumpe.jpg www.animatedsoftware.com/pumpglos/wobble.htm © Martin Culpepper, All rights reserved http://gallery.mcneel.com/fullsize/11155.jpg 28 Pump types: External gear pump Only one gear is driven, the other spins free Which way does the flow go? Step 1: Fluid comes in at ? Step 2: Fluid travels through ? Step 3: Fluid exits at ? What is the displacement? © Martin Culpepper, All rights reserved 29 Pump types: External gear pump Bushing Bearings O-Ring a) Pump Body © Martin Culpepper, All rights reserved a) Gear and Shaft 30 Hydraulics Exercise Competition: Pump Form group In 10 minutes, make best estimate of gear pump displacement Hand in answer/analysis at end of exercise (with all names) Sketches, calculations, etc… must be handed in before bell sounds 91 R10 10 H1 R5 H2 10 50 10 R10 13 5 32 H7 13 5 H5 R14 70 R14 H3 H4 26 10 H6 12 4 10 Gear Pump Cavity Plate with Dimensions All Dimensions in mm © Martin Culpepper, All rights reserved 2 10 50 28 14 R10 0 10 6 8 Minutes 32 Hydraulics power Power example: Pump at steady state Tshaft Ain pin Aout Pump xin pout xout Ai n ⋅ vi n = Qi n = Qout = Aout ⋅ v out © Martin Culpepper, All rights reserved 34 Example: Pump at steady state Work done on fluid via shaft input Pressure force does work on fluid entering pump Tshaft Ain pin Aout Pump vin © Martin Culpepper, All rights reserved Pump does work on Fluid exiting pump pout v out 35 Example: Pump at steady state ΣPin = [ΣPout ] + d (Estored ) → Pinlet + Pshaft = [Poutlet + Ploss ] + d (Estored ) dt dt Pinlet = [Fin ]⋅ vin = [ pi n ⋅ Ai n ]⋅ vin Tshaft Pshaft = Tshaft ⋅ ωshaft Poutlet = [Fout ]⋅ v out = [ pout ⋅ Aout ]⋅ v out Ain pin Aout Pump pout If can assume A&B, Ploss & d(E)/dt can be neglected A. B. d (Estored ) <<< Pin − Pout dt vin v out Ploss <<< Pin − Pout Substituting into energy balance (top equation on sheet) [ pin ⋅ Ain ]⋅ vin + Tshaft ⋅ ω shaft = [ [ pout ⋅ Aout ]⋅ vout + ~ 0]+ ~ 0 © Martin Culpepper, All rights reserved 36 Power example: Pump at steady state [ pin ⋅ Ain ]⋅ vin + Tshaft ⋅ ω shaft = [ [ pout ⋅ Aout ]⋅ vout + ~ 0]+ ~ 0 Ai n ⋅ vi n = Qi n = Qout = Aout ⋅ v out Qi n = Qout = Q Tshaft [ pout − pin ]⋅ Q = Tshaft ⋅ ω shaft Ain pin Pump vin © Martin Culpepper, All rights reserved Aout pout v out 37 Hydraulics System example Power example: Pump at steady state Dm = 0.5 in3/rev 3 Dp = 0.5 in /rev Tm = ? Tp = 10 in-lbf ωm = ? ωp = 1000 rpm 2π p1 p2 Pump p1 = 10 psi p3 Motor p2 = ? psi p3 = 10 psi Density = ρ Reservoir © Martin Culpepper, All rights reserved 39 DC permanent magnet motors DC Permanent magnet motor T vs. ω for Black & Decker Screw Driver 4.0 T [N-m] = - 0.012 [N-m/rpm] * ω + 3.678 [N-m] Torque [N-m] 3.0 2.0 1.0 0.0 0 50 100 150 200 250 300 350 ω [ rpm ] © Martin Culpepper, All rights reserved ⎡ ω⎤ T(ω ) = TStall ⎢1 − ω NL ⎥ ⎣ ⎦ ω rpm Data 310 210 190 167 148 121 67 33 0 T Nm Data 0.00 0.90 1.08 1.47 2.04 2.30 2.92 3.32 3.60 T Nm Fitted -0.09 1.11 1.34 1.63 1.85 2.17 2.83 3.23 3.63 % Error N/A 23 24 10 -9 -6 -3 -3 1 41 Understanding the model y Simple 1 loop model Goal: understand trends x z 2 B T, ω r FE FB 1 r V1 V4 Low loss in wires Steady state Single loop No ferrous cores Snap shot with loop plan in the y-z plane 3 FB i Assumptions FE L 4 © Martin Culpepper, All rights reserved 42 Point we will study Points we are studying D +TMAX A T=0 D A B C C D -TMAX B Torque curve of simple loop © Martin Culpepper, All rights reserved Side view of simple loop 43 Forces y Force on wire x z 2 B T, ω r L points in direction of current flow L FE MAGNETIC FLUX DENSITY, B FB 1 r V1 CONDUCTOR / WIRE CURRENT, iwire 3 FB i V4 FB = i ⋅ (L × B ) FE 4 © Martin Culpepper, All rights reserved MAGNETIC FORCE Lorentz Force L FE = q ⋅ E + q ⋅ v × B 44 Torque inducing forces on wire y Force on wire x z 2 B T, ω r FB 1 r V1 V4 © Martin Culpepper, All rights reserved T = 2 ⋅ (r × F ) B 90o in y-z plane T = 2 ⋅ r ⋅ (i ⋅ L ⋅ B ⋅) ⋅ sin (θ r − F ) T = 2⋅r ⋅ L 4 Torque at ω = 0 3 FB i FB = i ⋅ (L × B ) (V1 − V4 ) Battery (V1 − V4 ) Battery R ⋅L⋅B T ⋅R = 2⋅r ⋅ L⋅ B 45 dV E=− dx Lorentz force y Force due to E & B FE = q ⋅ E + q ⋅ (v × B ) T, ω E = v ⋅ B ⋅ sin (θ v − B ) x z 2 B r FE 1 r V1 3 i V4 FE 4 © Martin Culpepper, All rights reserved Wire 1-2 (V2 − V1 ) ω L = (r ⋅ ω ) ⋅ B Wire 2-3 v × B not along r V2 ω = V3 ω Wire 3-4 L (V4 − V3 ) ω L = (r ⋅ ω ) ⋅ B 46 Induced voltage due to rotation y ΔV due to rotation, ω x z 2 B T, ω (V2 − V1 ) ω L r FE 1 r (V4 − V3 ) ω L = (r ⋅ ω ) ⋅ B V2 ω = V3 ω = (r ⋅ ω ) ⋅ B V1 3 i V4 FE 4 © Martin Culpepper, All rights reserved (V4 − V1 ) ω L L = 2 ⋅ (r ⋅ ω ) ⋅ B (V1 − V4 ) ω = −2 ⋅ (r ⋅ ω ) ⋅ B ⋅ L 47 Total voltage y ΔV due to rotation, ω x z 2 B T, ω r FE FB 1 r V1 3 FB i V4 (V1 − V4 ) ω = −2 ⋅ (r ⋅ ω ) ⋅ B ⋅ L ΔV due to battery Total potential diff. (V1 − V4 ) Battery T ⋅R = 2⋅r ⋅ L⋅ B ΔV = (V1 − V4 ) Battery + (V1 − V4 ) ω FE 4 © Martin Culpepper, All rights reserved L T ⋅R ΔV = − 2 ⋅ (r ⋅ ω ) ⋅ B ⋅ L 2⋅r ⋅ L⋅ B 48 Torque – ω relationship y Total potential diff. x z 2 B T, ω r FE r V1 3 FB i V4 FE 4 © Martin Culpepper, All rights reserved Ohm’s law ΔV = i ⋅ R FB 1 T ⋅R ΔV = − 2 ⋅ (r ⋅ ω ) ⋅ B ⋅ L 2⋅r ⋅ L⋅ B Total potential diff. i⋅R = T ⋅R − 2 ⋅ (r ⋅ ω ) ⋅ B ⋅ L 2⋅r ⋅ L⋅ B T- ω relationship L 4 ⋅ r 2 ⋅ L2 ⋅ B 2 T = 2⋅i ⋅ L ⋅ r ⋅ B − ω R 49 Torque – ω relationship cont. y T- ω relationship x z 2 B T, r FE r V1 V4 T- ω relationship 3 FB i Stall torque TStall = 2 ⋅ i ⋅ L ⋅ r ⋅ B FB 1 4 ⋅ r 2 ⋅ L2 ⋅ B 2 T = 2⋅i ⋅ L ⋅ r ⋅ B − ω R ω T = TStall FE 4 ⋅ r 2 ⋅ L2 ⋅ B 2 − ω R L 4 © Martin Culpepper, All rights reserved 50 Torque – ω relationship cont. y T- ω relationship x z 2 B T, ω r FE FB 1 Motor constant T vs. ω for Black & Decker Screw Driver 4.0 3 i FE 4 © Martin Culpepper, All rights reserved Tstall L T [N-m] = - 0.012 [N-m/rpm] * ω + 3.678 [N-m] T = TStall − 3.0 Torque [N-m] FB Loop constant T- ω curve r V1 V4 T = TStall 4 ⋅ r 2 ⋅ L2 ⋅ B 2 − ω R TStall ω NL ω 2.0 1.0 ωNL 0.0 0 50 100 150 200 ω [ rpm ] 250 300 350 51 Scaling Follow up on micro-actuator lecture Electrostatics How does electrostatic physics scale? UE = ε o ⋅ L ⋅ L ⋅V 2 2⋅ z How does ratio of FElectric scale to FBody? FElectric 1 ~ FBody L What does this mean for MuSS interaction? What happens if you downsize each by factor of 10? © Martin Culpepper, All rights reserved 53 Electrostatics U Electric − z = ε o ⋅ L ⋅ L ⋅V 2 2⋅ z FElectric − z dU =− dz Fbody = ρ ⋅ V FElectric − z = 3 ε o ⋅ L2 ⋅V 2 2⋅ z2 FElectric 1 ~ FBody L Table removed due to copyright restrictions. Please see http://www.sizes.com/built/clean_rooms.htm © Martin Culpepper, All rights reserved 54 Semi-intuitive example Cooling… Heating… © Martin Culpepper, All rights reserved 55 Thermal behavior How does thermal physics scale (small Bi #)? e ⎡ ⎛ h⋅ A ⎞ ⎤ ⎢ −⎜ ⎜ ρ ⋅V ⋅c ⎟⋅t ⎥ ⎟ ⎠⎦ ⎣⎝ θ T − Tinf = = θ inf Tinitial − Tinf © Martin Culpepper, All rights reserved h ⋅ L Convection Bi = ~ k Conduction 56 Thermal behavior How does thermal physics scale? dT − h ⋅ A ⋅ (T − Tinf ) = ρ ⋅ c ⋅ V ⋅ dt e ⎡ ⎛ h⋅ A ⎞ ⎤ ⎢ −⎜ ⎜ ρ ⋅V ⋅c ⎟⋅t ⎥ ⎟ ⎠⎦ ⎣⎝ τ~ h⋅L Bi = k θ T − Tinf = = θ inf Tinitial − Tinf ρ ⋅V ⋅ c h⋅ A →L Is this a good or a bad thing for MEMS actuators? For the STM? © Martin Culpepper, All rights reserved 57 Fluidics How do fluid-based physical phenomena scale? π ⋅ r 4 ⋅ Δp Q= 8⋅ μ ⋅ L Q = U ⋅π ⋅ r 2 D L 8 ⋅ μ ⋅U Δp = − ⋅L 2 r High pressure change over narrow flow paths… © Martin Culpepper, All rights reserved 58 Fluidics Reynolds number D L Re = ρ ⋅U ⋅ D μ D = 50 μm Ratio of inertial forces to viscous forces U = 500 μm/s L = 1000 μm ReAir and ReH2O << 1 What does this mean: Heavily damped Limits response time (ms vs. μs) © Martin Culpepper, All rights reserved 59 ...
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.72 taught by Professor Martinculpepper during the Spring '09 term at MIT.

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