Page 1 of 9 GEORGIA INSTITUTE OF TECHNOLOGY George W. Woodruff School of Mechanical Engineering ME 4210 MANUFACTURING PROCESSES AND ENGINEERING Fall 2008 Final Exam: December 10 Directions: • Books, notes, and devices that communicate are prohibited. • You may use any type of calculator. • There are five (5) problems. • The exam is 2:50 long. • List all assumptions • Write legibly in the “Blue Book” or on the paper provided.Page 2 of 9 Problem #1 (25 points) Consider the fabrication of a solid plastic rod by extrusion. The rod has a diameter of 5 mm. After extrusion, the rod is pulled through a water bath that is maintained at 30°C. The rod is extruded at 12 cm/second. a) Determine the screw rpm. b) Determine the water bath’s length, if the center of the rod must be at 45°C when it is removed from the bath. c) Discuss the effect of the polymer flow on the final part’s properties during use. Polymer properties: Thermal conductivity (k) = 0.123 W/m-K Density (ρ) = 1040 kg/m3 Specific heat (c) = 2100 J/kg-K Viscosity (μ) = 200 Pa-s Temperature of extrudate = 235°C Barrel diameter = 35 mm Screw flight depth (H) = 5 mm Screw channel normal width (w) = 25 mm Screw helix angle (θ) = 17 degrees Screw axial length = 1.1 m Die land length = 28 mm Screw flow equation (drag and pressure flow) (Newtonian) Round channel flow equation (pressure flow) (Newtonian) ⎥⎥⎦⎤⎢⎢⎣⎡−=dzdpHHvwQzμ1223LpRQΔ=μπ84Page 3 of 9 Rectangular channel flow equation (pressure flow) (Newtonian) Q = flow rate (m3/s) w = width of flight or channel (m) Η = height of flight or channel (m) R = radius of channel (m) μ = viscosity (N-s/m2) L or dz = channel length (m) Δp or dp = pressure drop (Pa) vz = velocity along flight (helix) z = direction along flight (helix) length along a helix = axial length/sinθ velocity along helix = velocity of barrel (or screw) x cosθ Cooling time for a cylindrical part ⎟⎟⎠⎞⎜⎜⎝⎛−−=WEWMcTTTTRt 7.1ln7.122απ Cooling for a rectangular part ⎟⎟⎠⎞⎜⎜⎝⎛−−=WEWMcTTTTltπαπ4ln422 α = thermal diffusivity = k/ρc k = thermal conductivity ρ = density c = specific heat R = radius (round part) 2l = thickness (rectangular part) TM = injection temperature TE = ejection temperature TW = mold temperature LpwHQΔ=μ123Page 4 of 9 Problem #2 (20 points) Consider the machining of the outside of a cylindrical mirror with a single point tool on a lathe. The mirror is a cylinder with an average diameter of 350 mm. The outside surface of the mirror needs to have an arithmetic average (AA) roughness of 150 nm. Your cutting tool has a radius of 50 μm. The cutting tool cannot be used above 1320°C. Room temperature is 21°C. The specific cutting energy of the mirror metal is 3.5 W-s/mm3. Determine the following: a) The appropriate feed rate b) The appropriate lathe rpm Material Properties Density = 4510 kg/m3 Specific heat = 519 j/kg-K Thermal conductivity = 17 W/m-K Machining Power = u x MRR u = specific cutting energy MRR = material removal rate (volume/time) for turning = speed x feed x depth of cut for drilling = feed rate x area of cut for milling = area of cut x workpiece velocity Temperature rise 3/104.0⎟⎠⎞⎜⎝⎛=ΔαρVtcuT α = thermal diffusivity = k/ρc V = cutting velocity t0 = feed for turning on a lathe; t0 = depth of cut for orthogonal cutting good for b/t0 > 5 b = depth of cut for turning on a lathe; b = width of cut for orthogonal cutting Arithmetic average (AA) roughness Peak to valley roughness f = feed r = tool radius Velocity (lathe), V = π d N d = diameter N = revolutions/unit time Taylor’s tool life equation: VTn = C V = velocity T = life Machining time (lathe), tm = L/fN L = axial length of cut f = feed rate (length/revolution) N = speed (revolutions/unit time) Power = force x speed Power = torque x angular speed rfRoughness3182≈rfRoughness82≈Page 5 of 9 Problem #3 (25 points) Consider the sand casting of an iron fire hydrant. It can be modeled as a cylinder with one closed end. In this case, the closed end is down, so in cross section, the mold appears to be a “U”. The fire hydrant has the following dimensions: The outer diameter is 30 cm; the wall thickness is 2 cm; the height is 65 cm (from the bottom to the outside of the top). Room temperature is 30°C. The metal is poured at 50°C above its melting point. Do the following: a) Estimate the time required from the maximum temperature to the completion of solidification. b) Estimate the gate diameter required. Discuss your answer and its implications. Casting Information and Equations Data for solid materials (room temperature) Material Specific heat (C) (J/kg-oC) Density (ρ) (kg/m3) Thermal conductivity (k) (W/m-oC) Thermal diffusivity (α) (m2/s) Sand 1160 1500 0.60 3.45 x 10-7 Aluminum 900 2700 202 8.31 x 10-5 Nickel 440 8910 92 2.35 x 10-5 Magnesium 1070 1700 156 8.58 x 10-5 Copper 390 8970 385 1.1 x 10-5 Iron 441 7125 42.7 1.36 x 10-5 Data for liquid materials Material Melting point (oC) Latent heat of solidification (fusion) (Hf) (J/kg) Specific heat (C) (J/kg-oC) Viscosity (μ) (Pa-s) Aluminum 660 396,000 1050 1.3 x 10-3 Nickel 1453 297,000 730 4.01 x 10-3 Magnesium 650 384,000 1380 1.04 x 10-3 Copper 1083 220,000 520 2.1 x 10-3 Iron 1251 211,000 340 5.25 x 10-3 You may assume that the liquid and solid metals have the same densities and thermal conductivities.Page 6 of 9 Bernoulli's equation Reynold's number (keep below 20,000) Conservation of mass Aovo = A1v1 Cooling time for small Biot numbers (lumped parameter) (hl/k <<1) Solidification time (t) for an insulating mold 22,int_14⎟⎟⎠⎞⎜⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−Δ=castingcastingmoldmoldmoldinitialmoldpomeltingcastingcastingAVCkTTHtρρπ Solidification time (t) for a conducting mold kcasting ~ kmold αcasting ~ αmold AVkhcastingerface<<int ()AVTThHtetemperaturmoldetemperaturmeltingfcasting__interface−Δ=ρ μρVd=RefghvPghvPooo+++=++1211222ρρρρ⎟⎟⎠⎞⎜⎜⎝⎛−−=∞∞21lnTTTTAhcVtρPage 7 of 9 Chvorinov’s rule for solidification time Insulating mold: t = K (V/A)2 Conducting mold: t = K (V/A) t = cooling time K = a constant V = volume A = area Thermal diffusivity: α = k/ρC Filling rate time constant: τ ~ velocity / length Heat transfer