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# Lecture Slides: Lecture Case Studies

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### Lecture Slides: Lecture Case Studies

1. 1. G-L-6: Review Challenge the future 1
2. 2. Contents • Modelling in design • Case study: Mechanics & biomechanics • Case study: Thermodynamics • Case study: Fluid mechanics • Summary Challenge the future 2
3. 3. Modelling in Design what are you studying? Case brief ‘touch’ your thoughts choices what exactly...? what do you foresee? Build abstract model System choices choices what does your model foresee? Thought simulation Experience Solve/ simulate Compare did you both agree? everything you did/ saw/ felt/ remember to be true... Finish! Acceptable ? choices Revisit choices do you need more insights/ data? Experiment Courtesy of centech.com.pl and http://www.clipsahoy.com/webgraphics4/as5814.htm Challenge the future 3
4. 4. Cause ~ Effect Cause Identify cause-effect relationship Understand cause-effect relationship Quantify cause-effect relationship Effect Cup falls off a table, hits the floor What is special about falling off a table? Long distance, high speed, high acceleration at impact? etc ‘It is not the fall that kills you, but the sudden end’... Acceleration causes large forces... stop at the Accelerations during impact The cup is damaged Challenge the future 4
5. 5. Modelling: The basic loop Learn Identify, choose Cause Evaluate Model Effects Solve/Simulate Challenge the future 5
6. 6. Case study: Free standing punching bar Design a Freestanding Punch Bar Make sure the amplitude of the vibration is less than 0.1 meter in 2 seconds after a big punch. Courtesy of http://www.comparestoreprices.co.uk/keep-fit/fitness-free-standing-punch-bag.asp Challenge the future 6
7. 7. A simplified diagram L2 y Resistances from Spring x θ  Resistances from Damper Challenge the future 7
8. 8. Physics behind – Angular Spring ( t) M M (t )  k (t ) Challenge the future 8
9. 9. Physics behind – Angular damper M (t ) M (t )  c d (t ) dt Challenge the future 9
10. 10. A simplified diagram L2 y Resistances from Spring x θ  Resistances from Damper Challenge the future 10
11. 11. System components Input Mass Spring Damper Earth Courtesy of http://www.comparestoreprices.co.uk/keep-fit/fitness-free-standing-punch-bag.asp Challenge the future 11
12. 12. Cause-Effect Cause Cause Cause Cause Cause The lady hits the freestanding punch bag The freestanding punch bag starts to swing The freestanding punch bag starts to swing. The freestanding punch bag starts to swing. The freestanding punch bag starts to swing The freestanding punch bag starts to swing Effect The inertia moment slows its movement Effect The angular spring prevents its movement Effect The angular damper prevents its motion Effect The gravity of the bag tends make the bag fall down Effect Challenge the future 12
13. 13. Modelling y L2 M  0 Resistances from Spring x θ  Resistances from Damper d 2 (t ) d (t )  m  L12   Finput  L2  k  (t )  c   m  g  L1 sin( (t ))  0 2 dt dt Challenge the future 13
14. 14. We choose Mass m  10 Input 1000 0  t  0.1 Finput   0  0 Spring k  10000 Damper c  20 d (t ) d 2 (t )  m  L1   Finput  L2  k  (t )  c   m  g  L1 sin( (t ))  0 dt 2 dt 2 Courtesy of http://www.comparestoreprices.co.uk/keep-fit/fitness-free-standing-punch-bag.asp Challenge the future 14
15. 15. A simplified diagram- Choices Choices: L2 y We only can change the mass and the damper in the system Resistances from Spring x θ  Resistances from Damper Challenge the future 15
16. 16. Solving D L2 y Resistances from Spring x θ  Resistances from Damper θ D Maximal D around 2 second Challenge the future 16
17. 17. Evaluation D L2 y Resistances from Spring x θ  Maximal D around 2 second Resistances from Damper Damping coefficient Mass Challenge the future 17
18. 18. Evaluation D L2 y Resistances from Spring x θ  Resistances from Damper 0.1m Safe zone Damping coefficient Mass Challenge the future 18
19. 19. Further thought Questions The client wants you to adjust the design parameter(s) in order to reduce the frequency of the vibration, e.g., enlarge time t’ in the figure. Which parameter(s) do you suggest to change? Challenge the future 19
20. 20. Case study: Honda® U3-X® Honda® U3-X® The Honda® U3-X® is a self-balancing one-wheeled electric vehicle. Question A: Build a mathematical model to describe the vertical movement of the vehicle when it passes a step. We choose: 1. the mass of the U3-X® is mu; the mass of the rider is mr; 2. the vehicle is rigid except the rubber wheel; There is NO suspension system; 3. the spring constant of the rubber wheel is K; 4. the damping coefficient of the rubber wheel is C; 5. the step can be described as a function of time as y(t) in the vertical direction; 6. to neglect the air friction. y(t) Solid rubber wheel Question B: Which parameter(s) do you suggest to change in order to reduce the amplitudes of the vibrations after it passes the step? And why? Challenge the future 20
21. 21. Case study: Honda® U3-X® m=mu+mr K C Step y(t) x1+ y(t) Solid rubber wheel F  0 d 2 x1 (t ) dx1 (t ) dy (t ) )0 m  k ( x1 (t )  y (t ))  c(  2 dt dt dt Challenge the future 21
22. 22. The natural frequency 0  d 2 x(t ) dx(t ) m c  kx(t )  0 2 dt dt   k m c 2 mk d 2 x(t ) dx(t ) 2  20  0 x(t )  0 2 dt dt Challenge the future 22
23. 23. The damping ratio Challenge the future 23
24. 24. Case study: Honda® U3-X® Honda® U3-X® The Honda® U3-X® is a self-balancing one-wheeled electric vehicle. Question A: Build a mathematical model to describe the vertical movement of the vehicle when it passes a step. We choose: 1. the mass of the U3-X® is mu; the mass of the rider is mr; 2. the vehicle is rigid except the rubber wheel; There is NO suspension system; 3. the spring constant of the rubber wheel is K; 4. the damping coefficient of the rubber wheel is C; 5. the step can be described as a function of time as y(t) in the vertical direction; 6. to neglect the air friction. Question B: Which parameter(s) do you suggest to change in order to reduce the amplitudes of the vibrations after it passes the step? And why? Challenge the future 24
25. 25. The free standing punching bar Questions The client wants you to adjust the design parameter(s) in order to reduce the frequency of the vibration, e.g., enlarge time t’ in the figure. Which parameter(s) do you suggest to change? Challenge the future 25
26. 26. The free standing punching bar Questions The client wants you to adjust the design parameter(s) in order to reduce the frequency of the vibration, e.g., enlarge time t’ in the figure. Which parameter(s) do you suggest to change? Challenge the future 26
27. 27. Case studies: The Softball Pitcher The Softball Pitcher Consider a softball pitcher throws a ball: Question: via a motion capture software, it is identified that when θ = π/4, the angular velocity of her right arm is 2 rad/s, the angular acceleration is 1 rad/s. What is the torque she applied on her right shoulder joint in this moment? We choose: 1. her mass is 65 kg; 2. the length of her upper arm is 0.27 m; 3. the length of her forearm is 0.24 m; 4. the length of her hand is 0.1 m; 5. the mass of the softball is 0.2 kg; 6. the position of her right shoulder joint is fixed in the movement; 7. her right elbow joint and her wrist joint is not moving in the process; 8. to use point mass to approximate the mass moment of inertia. Challenge the future 27
28. 28. Case studies: The Softball Pitcher m3 m2 m1 Challenge the future 28
29. 29. Centre of gravity (c.g.) of body segments Proximal side Distal side Challenge the future 29
30. 30. Mass of body segment Following Biomechanics and Motor Control of Human Movement Segment Hand Forearm Upper arm F'arm+hand Upper limb Foot Shank Thigh Foot + shank Lower Limb Head, neck, trunk Head, neck, arms, trunk Head and neck Segment Centre of Mas Centre of M Total Body WeigSegment lengtSegment le 0.006 0.016 0.028 0.022 0.05 0.0145 0.0465 0.1 0.061 0.161 Proximal 0.506 0.43 0.436 0.682 0.53 0.5 0.433 0.433 0.606 0.447 Distal 0.494 0.57 0.564 0.318 0.47 0.5 0.567 0.567 0.394 0.553 0.578 0.66 0.34 0.678 0.626 0.374 0.081 m3 m2 m1 m1  70*0.028  1.96 m2  70*0.016  1.12 m3  70*0.006  0.2  0.62 L1  0.436*0.27  0.11772 L 2  0.27 L3  L 2  0.43*0.24  0.3732 L 4  L 2  0.24  0.51 L5  L 4  0.1*0.506  0.5606 Ref. http://books.google.nl/books?id=_bFHL08IWfwC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=snippet&q=mass%20segment&f=false Challenge the future 30
31. 31. Case studies: The Softball Pitcher M  0 m3 m2 m1 d 2 (t ) (m1L1  m2 L3  m3L5 )  (m1L1  m2 L3  m3L5) g cos  (t )  M  0 dt 2 2 2 2 Challenge the future 31
32. 32. Case studies: The coke can Design brief Take a can of Coca-Cola® (0.355Liter) out from the refrigerator and put it on the table: Question: Make a quick estimation of the time span during which the Coca-Cola® stays cool (≤ 8°C, which is nice to drink). Courtesy of http://www.coca-cola.com/ Challenge the future 32
33. 33. Components Coke Coke can The System Air Challenge the future 33
34. 34. Cause-effect Heat Heat Heat Challenge the future 34
35. 35. Physic behind - Conduction q kA T L Challenge the future 35
36. 36. Physic behind - Convection q  hAT h Fluid properties The velocity of Fluid The typical values of h Air: 10~100 Water: 500 to 10,000
37. 37. A simple sketch
38. 38. Understand the problem R5  1 hairTop Atop R6  R7  R8  R1  1 hairSide Aside R2  R3  Lcan / 2 K can Aside R4  1 hwater Aside 1 Lcan / 2 K can Atop hwater Atop
39. 39. Our choices
40. 40. Model de R5  1 hairTop Atop R6  R7  R8  R2  R3  Lcan / 2 K can Aside R4  Lcan / 2 K can Atop 1 hwater Atop Cause 1 hwater Aside Air temperature is higher than can The coke can absorbs heat The can(side) temperature is higher than Coke Effect Heat transferred from air to the can: q1 Tair  TcanSide (t ) m Temperature rises: q2 C Heat transferred to coke: q3 dTcanSide (t ) TcanSide (t )  TCoke (t )  0
41. 41. Model de R5  1 hairTop Atop R6  R7  R8  L /2 R2  R3  can K can Aside 1 R4  hwater Aside Lcan / 2 K can Atop 1 hwater Atop Cause Air temperature is higher than can top The coke can top absorbs heat The can top temperature is higher than Coke Effect Heat transferred from air to the can: q4 Tair  TcanTop (t ) Temperature rises: q5 Heat transferred to coke: q6 dTcanTop (t ) TcanTop (t )  TCoke (t )
42. 42. Model R6 CanTop TcanTop(t) R7 q6 CanSide TcanSide(t) R8 R5  1 hairTop Atop R6  R7  R8  Lcan / 2 K can Atop 1 hwater Atop Coke  Tcoke(t) q3 q7 e R5 R2 R3 Cause R4 L /2 R2  R3  can K can Aside 1 R4  hwater Aside The can(side) temperature is higher than Coke The can(top) temperature is higher than Coke The temperature of coke arises Effect Heat transferred to coke: q3 Heat transferred to coke: q6 TcanSide (t )  TCoke (t ) TcanTop (t )  TCoke (t ) Heat absorbed by coke: q7 dTCoke (t )
43. 43. Review R6 CanTop TcanTop(t) R7 q6 CanSide TcanSide(t) R8 R5  1 hairTop Atop R6  R7  R8  Lcan / 2 K can Atop 1 hwater Atop Coke  Tcoke(t) q3 q7 e R5 R2 R3 R2  R3  R4 Lcan / 2 K can Aside R4  1 hwater Aside Tair  TcanSide (t ) dT (t ) T (t )  TCoke (t )  mcanSideCcanSide canSide  canSide 0 R1  R2 dt R3  R4 Tair  TcanTop (t ) R5  R6  mcanTop CcanTop dTcanTop (t ) TcanTop (t )  TCoke (t )  0 dt R7  R8 TcanSide (t )  TCoke (t ) TcanTop (t )  TCoke (t ) dTCoke (t )
44. 44. Solving Can Top temperature (Green) Can Side temperature (blue) 8 degrees Coke Temperature (Red) 1575 seconds
45. 45. Can we find a quick solution? Can I make a step advantage?
46. 46. Revisit the model Heat absorbed Heat absorbed 1.26% temperature + 1 degree mCoke  CCoke  TCoke (t ) mcan  Ccan  Tcan (t )  0.33  4181  1400.6  0.0195  900  17.6
47. 47. The purpose of models he purpose of models is not to fit the ata but to sharpen the questions. Samuel Karlin
48. 48. Revisit the model Heat absorbed Heat absorbed 1.26% temperature + 1 degree mCoke  CCoke  TCoke (t ) mcan  Ccan  Tcan (t ) Neglect heat absorbed by the can for a quick estimation
49. 49. Cause-effect Heat Heat Heat
50. 50. Neglect the heat absorbed by the can R5  1 hairTop Atop R6  R7  R8  R1  1 hairSide Aside R2  R3  Lcan / 2 K can Aside R4  1 hwater Aside 1 Lcan / 2 K can Atop hwater Atop
51. 51. The new model: Case 2 e R5  1 hairTop Atop R6  R7  R8  1 Lcan / 2 K can Atop hwater Atop Cause R2  R3  Lcan / 2 K can Aside R4  1 hwater Aside The air(side) temperature is higher than Coke The air(top) temperature is higher than Coke The temperature of coke arises Effect Heat transferred to coke: q1 Tair  TCoke (t ) Heat transferred to coke: q4 Tair  TCoke (t ) Heat absorbed by coke: q7 dTCoke (t )
52. 52. Further: R5  hairTop Atop R6  R7  R8  1 de 1 Aside R2  R3  Lcan / 2 K can Aside R4  1 Lcan / 2 K can Atop hwater Atop 1 hwater Aside 0.088 2.6*10-5 3.68 31.5 1.87*10-4 0.63
53. 53. Further: 1 e Aside R5  1 hairTop Atop R6  R7  R8  R2  R3  Lcan / 2 K can Aside R4  1 Lcan / 2 K can Atop hwater Atop 1 hwater Aside 0.088 2.6*10-5 3.68 31.5 1.87*10-4 0.63 Neglect thermal resistance of the can
54. 54. Neglect the thermal resistances of the can R5  R8  R1  1 hairSide Aside R4  1 hwater Aside 1 hairTop Atop 1 hwater Atop
55. 55. The new model: Case 3 Aside R5  1 hairTop Atop R8  1 hwater Atop Cause 1 R4  hwater Aside The air(side) temperature is higher than Coke The air(top) temperature is higher than Coke The temperature of coke arises Effect Heat transferred to coke: q1 Heat transferred to coke: q4 Heat absorbed by coke: q7 Tair  TCoke (t ) Tair  TCoke (t ) dT (t )   mCokeCCoke Coke  0
56. 56. The comparison 97 seconds 98.7% 1359.516 seconds 98.7% both heat ed by the n and the sistances of the can Neglect heat absorbed by the can The original model 1377 seconds
57. 57. A question TCoke (t ) Tair  TCoke (t ) dT (t )   mCokeCCoke Coke  0 R4 R5  R8 dt R5  R8  1 1 1 hairTop Atop 1 hwater Atop
58. 58. We can solve it by hand Model Tair  TCoke (t ) Tair  TCoke (t ) dT (t )   mCokeCCoke Coke  0 R1  R4 R5  R8 dt C (Tair  TCoke (t ))  Simplify Cdt  Separation dTCoke (t ) Tair  TCoke (t )  Cdt   ntegration The result Simplify dTCoke (t ) dt dTCoke (t ) Tair  TCoke (t ) Ct  C1   ln(Tair  TCoke (t )) e  (Ct C1)  Tair  TCoke (t )
59. 59. The Heineken beerkeg The Heineken beerkeg Consider a 5 Liters Heineken® draft beer keg: Question: Quickly estimate the time needed to cool the draft beer to 5ºC when it is put in a refrigerator. Courtesy of www.heineken.com We know: 1. the keg is made of steel, the specific heat capacity of the steel is 460 J/(kg·K), the thermal conductivity of the steel is 43 W/(m·K), the thickness of the keg is 0.1 mm and the mass of the keg is 130.5 g. The pressure inside the keg is 2 bar; 2. the keg has a cylindrical shape and is filled with beer. The diameter of the keg is 16 cm and the height is 25 cm; 3. most of the bottom part is in contact with the air. 4. the density of the beer is 1060 kg/m3, the specific heat of the beer is 4181 J/(kg·K), the initial temperature is 20°C, 5. the temperature inside the refrigerator is 4°C; 6. the heat transfer coefficient between the beer and the keg is 500 W/(m2·K); 7. the heat transfer coefficient between the air and the keg is
60. 60. Case study: The bath tub Design the drain of a bathtub Specify the radius of 6 holes in the drain to make sure that the bathtub can be emptied within 500 seconds.
61. 61. Physics behind - CoM   m2  m1  m1  m 2  0  A1v1t   A2 v2t  0 A1v1  A2 v2  0
62. 62. Conservation of mass   mbathtub  morifice
63. 63. The area of water in the bathtub dh(t )    water  Awater   mbathtub dt R-h(t) Awater  Lengthbathtub  w w  2 R 2  ( R  h ( t )) 2 h(t) w/2 5 Water 2 2 dh(t ) 
64. 64. Physics behind - CoE ow ork Kinetic Energy pa   gh a 2 va pb    gh 2  b v b2   Constant 2 Potential Energy
65. 65. Physics behind – Torricelli's law 2 2 va pb vb  gha    ghb   2  2 pa pa va db ha pa  pb hb  0 pb if d a  db , va can be neglected in the energy equation vb  2 gha
66. 66. Mass flow rate at the orifice  m orifice  norifices  Cd   water  Aorifices  2  g  h(t ) Discharge coefficient
67. 67. Solving    water  Lengthbathtub  2  R 2  ( R  h(t )) 2  dh(t ) dt  norifices  Cd   water  Aorifices  2  g  h(t )  mbathtub  morifice Conservation of Mass  Length  2  R 2  ( R  h(t )) 2  dh(t )  n C   A  2  g  h(t )
68. 68. Solving 2 2 water  Lengthbathtub  2  R  ( R  h(t ))  t of water dh(t )  norifices  Cd   water  Aorifices  2  g  h(t ) dt We choose the radius of each hole is 3mm and discharge coefficient is 0.8 Time
69. 69. Evaluation  Lengthbathtub  2  R 2  ( R  h(t )) 2  water dh(t )  norifices  Cd   water  Aorifices  2  g  h(t ) dt ht of water It is not fast enough! @ 500th second
70. 70. Evaluation 2 2 water  Lengthbathtub  2  R  ( R  h(t ))  me for empty the bathtub Discharge coefficient dh(t )  norifices  Cd   water  Aorifices  2  g  h(t ) dt Which parameter is more effective ? Diameter of the hole
71. 71. Evaluation 2 2 er  Lengthbathtub  2  R  ( R  h(t ))  dh(t )   norifices  Cd   water  Aorifices  2  g  h(t ) dt 500 seconds 700 seconds 900 seconds ischarge oefficient Safe zone Diameter of the
72. 72. Case study: The V-Shaped wash basin e The V-Shaped wash basin Open the stopper located at the orifice in a V-shaped wash basin and let the water go out: 140° Questions: Considering the information below: Predict the remaining height of water 5 seconds after you open the stopper; We know: 1. the shape of the basin is a V shape as the figure and the angle is 140°; 2. the depth D of the of wash basin is 0.5m and the width W is 0.85m, respectively. 3. the initial height of water in the basin is 0.1 m; 4. the area of the orifice is 0.0015 m2; 5. the discharge coefficient is 0.6 for the orifice;
73. 73. Case study: The V-Shaped wash basin 140° The orifice   m washba sin  m orifice   water  D   2  h(t )  tan    dh(t )  Cd   water  Aorifice  2  g  h(t )
74. 74. Case study: Wash basin Design the drain of a wash basin Remove the stopper of the drain in a corner wash basin to let the water go out of the basin. R1 60mm Question A: Predict the remaining height of water after 5 seconds draining; 90° R2 230mm Drain at the bottom Question B: Manufacturing errors lead to systematic variation of the diameter of holes in the drain. This will affect the flow rate. Evaluate the sensitivity of the height of the water at 5th second with respect to the varying diameter of holes in the orifice.
75. 75. Case study: Wash basin R1 60mm 90° R2 230mm 0.06  0.23  m washba sin  m orifice Drain at the bottom 1 2 2 dh(t ) 2
76. 76. Case study: Wash basin h(t )  dt Can you continue?  Cd 12  R 2  2  g dh(t ) dt  1 2 2 h(t )    R1  R 2  4 dh(t ) Cd 12  R 2  2  g  dt 1 h(t )   R 22  R12   4 Cd 12  R 2  2  g  h(t ) 1    R 22  R12  4 1 3 2 2 2 dh(t ) 2 4
77. 77. Case study: Wash basin  0.004( R  D / 2), t  5)  ( Cd  24  R 2  2  g  R1 2  R2 D  0.004*1.01( R  D / 2), t  5)  (  2  t  0.25) 2  0.05024 Cd  24  R 2  2  g Sensitivity   R1 2  R2 2  t  0.25) 2  0.05 h( D  0.004*1.01, t  5)  h( D  0.004, t  5)  6 0.004*0.01
78. 78. Sensitivity analysis in simulation In reality, we need at least 3 points 1 step Errors 3 steps 6 steps eps Steps in Euler
79. 79. Case study: The Splash challenger The Splash challenger With a big water tank, the splash challenger is able to spray water through several orifices to amuse children in the playground. The splash challenger Questions: Considering a two-orifice splash challenger as shown in the sketch: A. Predict the remaining height of water 60 seconds after the start of the play; B. Manufacturing errors lead to systematic variation of the diameters of orifices. This will affect the flow rate. Evaluate the sensitivity of the height of the water at 60th second with respect to the varying diameter of orifice 1. H1 D2 H0 D D1 Orifice 2 Orifice 1   We know: 1. the shape of the tank is cylindrical and its diameter (D) is 1 meter; 2. the height difference (H1) between the two orifices is 30 cm; 3. the initial height of water (H0) in the tank is 1 m regarding orifice 1; 4. the shapes of both orifices are circular; the diameter of orifice 1 (D1) is 3 cm and the diameter of orifice 2 (D2) is 2 cm; the discharge
80. 80. This afternoon The Unicycle A unicycle is a human-powered, single-track vehicle with one wheel. Unicycles resemble bicycles, but are less complex. In a market investigation, it was found that two groups of children often use unicycles. They are: Group A: Age 6~7 and Group B: Age 7~8; Observation research indicates that among those children, 75% are Group A children and 25% are Group B children. In designing a unicycle, defining the damping ratio is crucial regarding the comfort of the rider. The damping ratio of a unicycle can be specified as: c ζ= 2 mk where m is the sum of the mass of the unicycle and the mass of the rider, k is the spring constant of the unicycle and c is the damping coefficient of the unicycle .
81. 81. This afternoon The Unicycle Question: Find the optimal spring constant k and the optimal damping coefficient c of the unicycle to satisfy as many children as possible based on the following wish: Professor Jørgen Winkel, a senior ergonomics scientist, concluded that: “For both groups of children, the optimal damping ratio of a unicycle is about 0.3”. In the optimization, we choose: 1. the mass of the unicycle to be 5 kg; 2. the average mass of Group A children to be 20 kg; 3. the average mass of Group B children to be 31 kg; 4. to use gradient descent method in the optimization; 5. to use the values of the gradient directly in the gradient descent method (NOT the unit vector of the gradient); 6. the initial guess of k to be 14463 (N/m) and c to be 1627 (N·s/m); 5
82. 82. We think… curiosity is fun knowledge is power science is easy computers experience can be harnessed are tools hard work is the way to success
83. 83. My problem is complicated?
84. 84. hank You! The Modelling Team