Shaping Processes for Plastics Past Paper PDF

Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13 SHAPING PROCESSES FOR PLASTICS Review Questions 13.1 What are some of the reasons why plastic shaping processes are important? Answer. The reasons include (1) many of t...

Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13 SHAPING PROCESSES FOR PLASTICS Review Questions 13.1 What are some of the reasons why plastic shaping processes are important? Answer. The reasons include (1) many of the processes are net shape processes; (2) in general, less energy is employed than in metalworking processes; (3) lower temperatures are required to process plastics than metals or ceramics; (4) there is great flexibility in geometry; and (5) painting and other finishing processes are generally not required. 13.2 Identify the main categories of plastics shaping processes, as classified by the resulting product geometry. Answer. The categories are (1) extrusion, (2) molding, (3) forming of continuous sheets and films, (4) fibers, (5) foamed products, and (6) discrete formed sheets and films. 13.3 Viscosity is an important property of a polymer melt in plastics shaping processes. Upon what parameters does viscosity depend? Answer. Viscosity of a polymer melt depends on (1) temperature and (2) shear rate. Also, (3) the molecular weight of the polymer affects viscosity. 13.4 How does the viscosity of a polymer melt differ from most fluids that are Newtonian. Answer. A polymer melt exhibits pseudoplasticity, which means that its value decreases with increasing shear rate. 13.5 What does viscoelasticity mean, when applied to a polymer melt? Answer. Viscoelasticity is a combination of viscous and elastic properties which cause the melt to exhibit memory - the tendency to return to its previous shape, as exhibited by die swell in extrusion. 13.6 Define die swell in extrusion. Answer. Die swell is the tendency of the extrudate to expand in cross-sectional dimensions immediately on exiting the die orifice. It results from the viscoelastic properties of the polymer melt. 13.7 Briefly describe the plastic extrusion process. Answer. In plastic extrusion, a polymer melt is compressed to flow through a die orifice and thus the continuous length of the plastic assumes a cross-sectional shape that is approximately the same as that of the orifice. 13.8 The barrel and screw of an extruder are generally divided into three sections; identify the sections. Answer. The sections are (1) the feed section, in which the feed stock is fed from the hopper and heated; (2) the compression section, in which the polymer changes to a viscous fluid; and (3) the metering section, in which pressure is developed to pump the plastic through the die orifice. 13.9 What are the functions of the screen pack and breaker plate at the die end of the extruder barrel? Answer. The functions are to (1) filter dirt and lumps, (2) build pressure, (3) straighten the flow and remove memory of the polymer melt. 13.10 What are the various forms of extruded shapes and corresponding dies? Answer. The shapes are (1) solid profiles, such as rounds and L-shapes; (2) hollow profiles, such as tubes; (3) wire and cable coating; (4) sheet and film; and (5) filaments (continuous fibers). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 1 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13.11 What is the distinction between plastic sheet and film? Answer. The distinction is based on thickness. Sheet stock has a thickness greater than 0.020 in (0.5 mm), while film stock is less than 0.020 in (0.5 mm) thick. 13.12 What is the blown-film process for producing film stock? Answer. The blown-film process is a widely used process for making thin polyethylene film for packaging. It combines extrusion and blowing to produce a tube of thin film. The process begins with the extrusion of a tube that is immediately drawn upward while still molten and simultaneously expanded in size by air inflated into it through the die mandrel. 13.13 Describe the calendering process. Answer. Calendering is a process for producing sheet and film stock out of rubber or rubbery thermoplastics such as plasticized PVC. In the process, the initial feedstock is passed through a series of rolls to work the material and reduce its thickness to the desired gage. 13.14 Polymer fibers and filaments are used in several applications; what is the most important application commercially? Answer. Textiles. 13.15 Technically, what is the difference between a fiber and a filament? Answer. A fiber is a long, thin strand of material whose length is at least 100 times its diameter; a filament is a fiber of continuous length. 13.16 Among the synthetic fiber materials, which are the most important? Answer. Polyester is the most important commercially, followed by nylon, acrylics, and rayon. 13.17 Briefly describe the injection molding process. Answer. Injection molding is a process in which a polymer is heated to a highly plastic state and forced to flow under high pressure into a mold cavity, where it solidifies. The molding is then removed from the cavity. 13.18 An injection-molding machine is divided into two principal components. Name them. Answer. The components of an injection-molding machine are (1) the injection unit and (2) the clamping unit. 13.19 What are the two basic types of clamping units? Answer. The clamping units are: (1) mechanical toggle clamp and (2) hydraulic. In addition, there are hydromechanical units which combine hydraulic and mechanical actuations. 13.20 What is the function of gates in injection molds? Answer. The function of gates in an injection mold is to constrict the flow of molten plastic into the cavity, which increases the shear rate and reduces the viscosity of the polymer melt. 13.21 What are the advantages of a three-plate mold over a two-plate mold in injection molding? Answer. As the mold opens, the three-plate mold automatically separates the molded part(s) from the runner system. 13.22 Discuss some of the defects that can occur in plastic injection molding. Answer. The defects include (1) short shots, in which the polymer melt solidifies before filling the cavity; (2) flashing, in which the polymer melt is squeezed into the parting surfaces between the mold halves and around ejection pins; (3) sink marks, in which the surface is drawn into the molding Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 by contraction of internal material; and (4) weld lines where the melt has flowed around a core or other convex detail in the mold cavity and met from opposite directions, thus resulting in mechanical properties that are inferior to those in the rest of the part. 13.23 Describe structural-foam molding. Answer. Structural-foam molding is an injection molding process in which a gas or gas-producing ingredient is mixed with the polymer melt prior to injection into the mold cavity; this results in the part having a tough outer skin surrounded by a foam core. 13.24 What are the significant differences in the equipment and operating procedures between injection molding of thermoplastics and injection molding of thermosets? Answer. The differences in injection molding of thermosets are (1) shorter barrel length, (2) lower temperatures in the barrel, these first two reasons to prevent premature curing; and (3) use of a heated mold to cause cross-linking of the TS polymer. 13.25 What is reaction injection molding? Answer. Reaction injection molding involves the mixing of two highly reactive liquid ingredients and immediately injecting the mixture into a mold cavity where chemical reactions leading to solidification occur. The two ingredients form the components used in catalyst-activated or mixing-activated thermoset systems. 13.26 What kinds of products are produced by blow molding? Answer. Blow molding is used to produce hollow, seamless containers, such as bottles. 13.27 What is the form of the starting material in thermoforming? Answer. Thermoforming starts with a thermoplastic sheet or film. 13.28 What is the difference between a positive mold and a negative mold in thermoforming? Answer. A positive mold has a convex shape; a negative mold has a concave cavity. 13.29 Why are the molds generally more costly in mechanical thermoforming than in pressure or vacuum thermoforming? Answer. In mechanical thermoforming, matching mold halves are required; while in other thermoforming processes, only one mold form is required. 13.30 What are the processes by which polymer foams are produced? Answer. There are several foaming processes: (1) mechanical agitation - mixing a liquid resin with air, then hardening the polymer by means of heat or chemical reaction; (2) mixing a physical blowing agent with the polymer - a gas such as nitrogen (N2) or pentane (C5H12) which can be dissolved in the polymer melt under pressure, so that the gas comes out of solution and expands when the pressure is subsequently reduced; and (3) mixing the polymer with chemical compounds, called chemical blowing agents, that decompose at elevated temperatures to liberate gases such as CO2 or N2 within the melt. 13.31 What are some of the general considerations that product designers must keep in mind when designing components out of plastics? Answer. Some of the general considerations are the following: (1) Plastics are not as strong or stiff as metals and should not be used in applications where high stresses will be encountered. (2) Impact resistance of plastics is general good, better than many ceramics. (3) Service temperatures of plastics are limited relative to engineering metals and ceramics. (4) Thermal expansion is greater for plastics than metals; so dimensional changes due to temperature variations are much more significant than for metals. (5) Many types of plastics degrade from sunlight and certain other forms of radiation. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 Also, some plastics degrade in oxygen and ozone atmospheres. Finally, plastics are soluble in many common solvents. 13.32 (Video) According to the injection molding videos, what are the four primary elements that influence the injection molding process? Answer: The four primary elements that influence the injection molding process are the molder, material, injection machine, and the mold. 13.33 (Video) According to the injection molding video, name the four types of mold design most common in industry. Answer: The four types of mold design most common in industry are (1) cold runner two-plate mold, (2) cold runner three-plate mold, (3) hot runner mold, also known as runnerless mold, and (4) insulated runner mold. 13.34 (Video) According to the injection molding video, what is the most common type of injection molding machine used in industry? Answer: The most common injection-molding machine used in industry is a hydraulic, three- platen system. 13.35 (Video) According to the blow molding video, what materials are used in blow molding? Name three. Answer: Materials are used in blow molding are high density polyethylene (HDPE), medium and low density polyethylene (MDPE, LDPE), polyethylene terephthalate (PET), polypropylene (PP), polyvinyl chloride (PVC), thermoplastic elastomers (TPE), polystyrene (PS), polycarbonate (PC), fluoropolymers (PTFE), and polyimide/nylon. 13.36 (Video) List the four most common blow-molding processes according to the video on blow molding. Answer: The four most common blow-molding processes are (1) extrusion blow molding, (2) injection blow molding, (3) biaxial stretch blow molding, and (4) co-extrusion blow molding. 13.37 (Video) List the stages of extrusion blow molding according to the video. Answer: The stages of extrusion blow molding are (1) plasticizing the resin, (2) parison production (or preform production in biaxial stretch blow molding), (3) parison or preform inflation and cooling, (4) ejection from the blow mold, and (5) finishing and trimming. 13.38 (Video) Name the four types of finishing operations performed on plastics, according to the plastics finishing video. Answer: The four types of finishing operations performed on plastics are degating, deflashing, cleaning, and decorating. 13.39 (Video) What are the different processes that can be used to apply decorations to plastic parts according to the plastics finishing video? Answer: The processes that can be used to apply decorations to plastic parts include painting, plating, vacuum metallization, pad printing, hot stamping, silk screening, and fill and wipe. Multiple Choice Quiz There are 29 correct answers in the following multiple choice questions (some questions have multiple answers that are correct). To attain a perfect score on the quiz, all correct answers must be given. Each correct answer is worth 1 point. Each omitted answer or wrong answer reduces the score by 1 point, and Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 each additional answer beyond the correct number of answers reduces the score by 1 point. Percentage score on the quiz is based on the total number of correct answers. 13.1 The forward movement of polymer melt in an extruder barrel is resisted by drag flow, which is caused by the resistance to flow through the die orifice: (a) true or (b) false? Answer. (b). Drag flow is the forward motion of the melt caused by the Archimedian screw principle in the barrel. The resistance to forward flow is called back pressure flow. 13.2 Which of the following are sections of a conventional extruder barrel for thermoplastics (three best answers): (a) compression section, (b) die section, (c) feed section, (d) heating section, (e) metering section, and (f) shaping section? Answer. (a), (c), and (e). 13.3 Which of the following processes are associated with the production of plastic sheet and film (three correct answers): (a) blown-film extrusion process, (b) calendering, (c) chill-roll extrusion, (d) doctor blade method, (e) spinning, (f) thermoforming, and (g) transfer molding? Answer. (a), (b), and (c). 13.4 The principal components of an injection molding machine are which two of the following: (a) clamping unit, (b) hopper, (c) injection unit, (d) mold, and (e) part ejection unit? Answer. (a) and (c). 13.5 The parting line in injection molding is which one of the following: (a) the lines formed where polymer melt meets after flowing around a core in the mold, (b) the narrow gate sections where the parts are separated from the runner, (c) where the clamping unit is joined to the injection unit in the molding machine, or (d) where the two mold halves come together? Answer. (d). 13.6 The function of the ejection system is which one of the following: (a) move polymer melt into the mold cavity, (b) open the mold halves after the cavity is filled, (c) remove the molded parts from the runner system after molding, or (d) separate the part from the cavity after molding? Answer. (d). 13.7 A three-plate mold offers which of the following advantages when compared to a two-plate mold (two best answers): (a) automatic separation of parts from runners, (b) gating is usually at the base of the part to reduce weld lines, (c) sprue does not solidify, and (d) stronger molded parts? Answer. (a) and (b). 13.8 Which of the following defects or problems is associated with injection molding (three correct answers): (a) bambooing, (b) die swell, (c) drag flow, (d) flash, (e) melt fracture, (f) short shots, or (g) sink marks? Answer. (d), (f), and (g). 13.9 In rotational molding, centrifugal force is used to force the polymer melt against the surfaces of the mold cavity where solidification occurs: (a) true or (b) false? Answer. (b). It is the force of gravity in the doubly rotating mold that forces the polymer against the mold surfaces. 13.10 Use of a parison is associated with which one of the following plastic shaping processes: (a) bi-injection molding, (b) blow molding, (c) compression molding, (d) pressure thermoforming, or (e) sandwich molding? Answer. (b). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 5 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13.11 A thermoforming mold with a convex form is called which one of the following: (a) a die, (b) a negative mold, (c) a positive mold, or (d) a three-plate mold? Answer. (c). 13.12 The term encapsulation refers to which one of the following plastics shaping processes: (a) casting, (b) compression molding, (c) extrusion of hollow forms, (d) injection molding in which a metal insert is encased in the molded part, or (e) vacuum thermoforming using a positive mold? Answer. (a). 13.13 The two most common polymer foams are which of the following: (a) polyacetal, (b) polyethylene, (c) polystyrene, (d) polyurethane, and (e) polyvinylchloride? Answer. (c) and (d). 13.14 In which of the following properties do plastic parts often compare favorably with metals (two best answers): (a) impact resistance, (b) resistance to ultraviolet radiation, (c) stiffness, (d) strength, (e) strength-to-weight ratio, and (f) temperature resistance? Answer. (a) and (e). 13.15 Which of the following processes are generally limited to thermoplastic polymers (two best answers): (a) blow molding, (b) compression molding, (c) reaction injection molding, (d) thermoforming, (e) transfer molding, and (f) wire coating? Answer. (a) and (d). 13.16 Which of the following processes would be applicable to produce hulls for small boats (three best answers): (a) blow molding, (b) compression molding, (c) injection molding, (d) rotational molding, and (e) vacuum thermoforming? Answer. (a), (d), and (e). Problems Extrusion 13.1 The diameter of an extruder barrel is 65 mm and its length = 1.75 m. The screw rotates at 55 rev/min. The screw channel depth = 5.0 mm, and the flight angle = 18°. The head pressure at the die end of the barrel is 5.0 x 106 Pa. The viscosity of the polymer melt is given as 100 Pa-s. Find the volume flow rate of the plastic in the barrel. Solution: Qd = 0.5π2(65x10-3)2(55/60)(5x10-3)sin 18 cos 18 = 95,560x10-9(0.3090)(0.9510) = 28.081 x 10-6 m3/s p = 5 MPa = 5x106 n/m2 Qb = π(5x106)(65x10-3)(5x10-3)3(sin 18)2/12(100)(1.75) = 5.804(10-6) m3/s Qx = 28.081 - 5.804 = 22.277 x 10-6 m3/s. 13.2 An extruder has a diameter of 5.0 in and a length to diameter ratio of 26. The barrel heats the polypropylene melt to 450°F, which provides a melt viscosity of 0.0025 lb-s/in2. The pitch of the screw is 4.2 in and the channel depth is 0.15 in. In operation the screw rotates at 50 rev/min and a head pressure of 450 lb/in2 is generated. What is the volume flow rate of polypropylene from the die at the end of the barrel? Solution: A = tan-1(p/(πD)) = tan-1(4.2/(5π)) = 15° Qd = 0.5π2 D2 N dc sinA cosA = 0.5π2 (5.02)(50/60) 0.15 sin 15 cos 15 = 3.9 in3/sec Qb = pπDdc3sin2A/(12ηL) = 450π(5.0)(0.153)sin215/(12(0.0025)(5.0)(26)) = 0.41 in3/sec Qx = Qd – Qb = 3.9 – 0.41 = 3.5 in3/sec Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 6 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13.3 An extruder barrel has a diameter of 110 mm and a length of 3.0 m. The screw channel depth = 7.0 mm, and its pitch = 95 mm. The viscosity of the polymer melt is 105 Pa-s, and the head pressure in the barrel is 4.0 MPa. What rotational speed of the screw is required to achieve a volumetric flow rate of 90 cm3/s? Solution: A = tan-1(p/(πD)) = tan-1(95/110π) = 15.37° Qd = 0.5π2 D2 N dc sinAcosA = 0.5π2(0.110)2(N)(7.0x10-3)sin 15.37 cos 15.37 = 106.8 N x 10-6 m3/s Qb = π(4x106)(0.110)(7x10-3)3(sin 15.37)2/12(105)(3.0) = 8.81 x 10-6 m3/s Qx = Qd - Qb = 106.8 N x 10-6 – 8.81 x 10-6 = 90 x 10-6 m3/s 106.8 N = 90.0 + 8.81 = 98.81 N = 98.81/106.8 = 0.9252 rev/s = 55.51 rev/min. 13.4 An extruder has a barrel diameter of 2.5 in and a length of 6.0 ft. The screw has a channel depth of 0.25 in, a flight angle of 20°, and rotates at 55 rev/min. The material being extruded is polypropylene. At the present settings, the volumetric flow rate of the polymer melt is 1.50 in3/sec and the head pressure is 500 lb/in2. (a) Under these operating characteristics, what is the viscosity of the polypropylene? (b) Using Figure 13.2, approximate the temperature in °F of the polypropylene. Solution: (a) Qd = 0.5π2 D2 Ndc sinA cosA = 0.5π2(2.52)(55/60)(0.25)sin 20 cos 20 Qd = 2.27 in3/sec Qb = Qd – Qx = 2.27 – 1.50 = 0.78 in3/sec η = pπDdc3 in2 A/(12 QbL) = 500π(2.5)(0.253)sin2(20)/(12(0.78)(6)(12)) = 0.011 lb-s/in2 (b) X-axis is Log scale. Therefore LOG(0.011) = -1.96 This is very close to –2, which is at the 10-2 hash mark on the y-axis. This yields about 410°F. 13.5 An extruder has diameter = 80 mm and length = 2.0 m. Its screw has a channel depth = 5 mm, flight angle = 18 degrees, and it rotates at 1 rev/sec. The plastic melt has a shear viscosity = 150 Pa-s. Determine the extruder characteristic by computing Qmax and pmax and then finding the equation of the straight line between them. Solution: Qmax = Qd = 0.5π2(0.08)2(1)(5x10-3)sin 18 cos 18 = 0.158 x 10-3(0.3090)(0.9510) = 46.4 x 10-6 m3/s pmax = 6π(0.08)(1)(2)(150)(cot 18)/(5x10-3)2 = 452.4(3.077)/25x10-6 = 55 x 106 Pa = 55 MPa Qx = 46.4 x 10-6 - (46.4x10-6/55)p Qx = 46.4 x 10-6 - 0.8436 x 10-6 p, where p has units of MPa 13.6 Determine the helix angle A such that the screw pitch p is equal to the screw diameter D. This is called the "square" angle in plastics extrusion - the angle that provides a flight advance equal to one diameter for each rotation of the screw. Solution: Assume flight land = zero. From Eq. (15.4), tan A = pitch/πD If pitch = D, then A = tan-1(1/π) = 17.66° 13.7 An extruder barrel has a diameter of 2.5 in. The screw rotates at 60 rev/min; its channel depth = 0.20 in, and its flight angle = 17.5°. The head pressure at the die end of the barrel is 800 lb/in2 and the length of the barrel is 50 in. The viscosity of the polymer melt is 122 x 10-4 lb-sec/in2. Determine the volume flow rate of the plastic in the barrel. Solution: Qd = 0.5π2(2.5)2(1)(.2)sin 17.5 cos 17.5 = 0.5(12.337)(0.3007)(0.9537) = 1.769 in3/sec Qb = π(800)(2.5)(.2)3(sin 17.5)2/12(122x10-4)(50) = 0.621 in3/sec Qx = 1.769 - 0.621 = 1.148 in3/sec. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 7 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 13.8 An extruder barrel has a diameter of 4.0 in and an L/D ratio of 28. The screw channel depth = 0.25 in, and its pitch = 4.8 in. It rotates at 60 rev/min. The viscosity of the polymer melt is 100 x 10-4 lb-sec/in2. What head pressure is required to obtain a volume flow rate = 150 in3/min? Solution: A = tan-1(pitch/πD) = tan-1(4.8/4π) = 20.9° Qd = 0.5π2(4)2(1)(0.25)sin 20.9 cos 20.9 = 19.74(0.3567)(0.9342) = 6.578 in3/sec = 394.66 in3/min Qx = Qd - Qb = 394.66 - Qd = 150 Qb = 394.66 - 150 = 244.66 in3/min = 4.078 in3/sec L = 4(28) = 112 in. Qb = πp(4)(0.25)3(sin 20.9)2/(12(100x10-4)(112)) = 4.078 0.0018592 p = 4.078 p = 2193.4 lb/in2 13.9 An extrusion operation produces continuous tubing with outside diameter = 2.0 in and inside diameter = 1.7 in. The extruder barrel has a diameter = 4.0 in and length = 10 ft. The screw rotates at 50 rev/min; it has a channel depth = 0.25 in and flight angle = 16°. The head pressure has a value of 350 lb/in2 and the viscosity of the polymer melt is 80 x 10-4 lb-sec/in2. Under these conditions, what is the production rate in length of tube/min, assuming the extrudate is pulled at a rate that eliminates the effect of die swell (i.e., the tubing has the same OD and ID as the die profile)? Solution: Qd = 0.5π2(4)2(50/60)(.25)sin 16 cos 16 = 16.45(0.2756)(0.9613) = 4.358 in3/sec Qb = π(350)(4)(.25)3(sin 16)2/(12(80x10-4)(120)) = 0.453 in3/sec Qx = 4.358 - 0.453 = 3.905 in3/sec. Ax = 0.25π(22 - 1.72) = 0.872 in2 vx = 3.905/0.872 = 4.478 in/sec = 22.39 ft/min. 13.10 Continuous tubing is produced in a plastic extrusion operation through a die orifice whose outside diameter = 2.0 in and inside diameter = 1.5 in. The extruder barrel diameter = 5.0 in and length = 12 ft. The screw rotates at 50 rev/min; it has a channel depth = 0.30 in and flight angle = 16°. The head pressure has a value of 350 lb/in2 and the viscosity of the polymer melt is 90 x 10-4 lb-sec/in2. Under these conditions, what is the production rate in length of tube/min, given that the die swell ratio is 1.25. Solution: Qd = 0.5π2(5)2(50/60)(.3)sin 16 cos 16 = 30.84(0.2756)(0.9613) = 8.171 in3/sec Qb = π(350)(5)(.3)3(sin 16)2/(12(90x10-4)(144)) = 0.725 in3/sec Qx = 8.171 - 0.725 = 7.446 in3/sec Die swell ratio applied to OD and ID: OD = 2(1.25) = 2.5, ID = 1.5(1.25) = 1.875 Ax = 0.25π(2.52 - 1.8752) = 2.1476 in2 vx = 7.446/2.1476 = 3.467 in/sec = 17.34 ft/min 13.11 An extruder has barrel diameter and length of 100 mm and 2.8 m, respectively. The screw rotational speed = 50 rev/min, channel depth = 7.5 mm, and flight angle = 17°. The plastic melt has a shear viscosity = 175 Pa-s. Determine: (a) the extruder characteristic, (b) the shape factor Ks for a circular die opening with diameter = 3.0 mm and length = 12.0 mm, and (c) the operating point (Q and p). Solution: Qmax = Qd = 0.5π2(.1)2(50/60)(7.5x10-3) sin 17 cos 17 = 308.4 x 10-6(0.2924)(0.9563) = 86.2 x 10-6 m3/s pmax = 6π(.1)(50/60)(2.8)(175)(cot 17)/(7.5x10-3)2 = 44.75 x 106 Pa =44.75 MPa Qx = 86.2 x 10-6 - 1.926 x 10-12 p, where p has units of Pa (b) Given: Dd = 3 mm, Ld = 12 mm. Ks = π(3 x 10-3)4/(128(175)(12 x 10-3)) = 0.9467 x 10-12 (c) 0.9467 x 10-12 p = 86.2 x 10-6 - 1.926 x 10-12 p 2.8727 x 10-12 p = 86.2 x 10-6 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 p = 30.0 x 106 Pa = 30 MPa Qx = 0.9467 x 10-12 (30 x 106) = 28.4 x 10-6 m3/s Check with extruder characteristic: Qx = 86.2 x 10-6 - 1.926 x 10-12 (30 x 106) = 28.4 x 10-6 m3/s. 13.12 For Problem 13.11, assume the material is acrylic. (a) Using Figure 13.2, determine the temperature of the polymer melt. (b) If the temperature is lowered 20°C, estimate the resulting viscosity of the polymer melt. (Hint: the y-axis of Figure 13.2 is a log scale, not linear). Solution: (a) When viscosity = 175 Pa-s, Log(175) = 2.243, and temperature is approximately 260°C. (b) At 240°C, Log(viscosity) is approximately 2.7 and viscosity = 102.7 = 500 Pa-s. (Note: due to the log scale, small changes in the estimate will result in large changes in viscosity. 13.13 Consider an extruder in which the barrel diameter = 4.5 in and length = 11 ft. The extruder screw rotates at 60 rev/min; it has channel depth = 0.35 in and flight angle = 20°. The plastic melt has a shear viscosity = 125 x 10-4 lb-sec/in2. Determine: (a) Qmax and pmax; (b) the shape factor Ks for a circular die opening in which Dd = 0.312 in and Ld = 0.75 in; and (c) the values of Q and p at the operating point. Solution: (a) Qmax = 0.5π2(4.5)2(1)(0.35)sin 20 cos 20 = 34.975(0.342)(0.9397) = 11.24 in3/sec pmax = 6π(4.5)(1)(132)(0.0125)(cot 20)/(0.35)2 = 3139 lb/in2 (b) Given: Dd = 0.312 in., Ld = 0.75 in. Ks = π(0.312)4/128(0.0125)(0.75) = 0.024808 (c) From (a), Qx = Qmax - (Qmax/pmax)p = 11.24 - 0.003581p From (b), Qx = 0.024808p Combining,.024808p = 11.24 -.003581p 0.02839p = 11.24 p = 395.9 lb/in2 Qx = 11.24 - 0.003581(395.9) = 9.82 in3/sec 13.14 An extruder has a barrel diameter = 5.0 in and length = 12 ft. The extruder screw rotates at 50 rev/min; it has channel depth = 0.30 in and flight angle = 17.7°. The plastic melt has a shear viscosity = 100 x 10-4 lb-sec/in2. Find: (a) the extruder characteristic, (b) the values of Q and p at the operating point, given that the die characteristic is Qx = 0.00150 p. Solution: (a) Qmax = 0.5π2(5)2(50/60)(0.3)sin 17.7 cos 17.7 = 30.84(0.3040)(0.9527) = 8.93 in3/sec pmax = 6π(5)(50/60)(144)(0.01)(cot 17.7)/(0.3)2 = 3937.6 lb/in2 Qx = Qmax - (Qmax/pmax)p = 8.93 - 0.002268p (b) Given: die characteristic Qx = 0.0015p Qx = 8.93 - 0.002268p = 0.0015p 0.00377p = 8.93 p = 2370 lb/in2 Qx = 8.93 - 0.002268(2370) = 3.55 in3/sec 13.15 Given the data in Problem 13.14, except that the flight angle of the extruder screw is a variable instead of a constant 17.7°. Use a spreadsheet calculator to determine the value of the flight angle that maximizes the volumetric flow rate Qx. Explore values of flight angle between 10° and 20°. Determine the optimum value to the nearest tenth of a degree. Solution: The author’s spreadsheet computations returned an optimum value of 13.5°. 13.16 An extruder has a barrel diameter of 3.5 in and a length of 5.0 ft. It has a screw channel depth of 0.16 in and a flight angle of 22°. The extruder screw rotates at 75 rev/min. The polymer melt has a shear viscosity = 65 x 10-4 lb-sec/in2 at the operating temperature of 525°F. The specific gravity of the polymer is 1.2 and its tensile strength is 8000 lb/in2. A T-shaped cross section is extruded at a rate of Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 0.11 lb/sec. The density of water is 62.5 lb/ft3. (a) Find the equation for the extruder characteristic. (b) Find the operating point (Q and p), and (c) the die characteristic that is indicated by the operating point. Solution: (a) Qmax = 0.5π2 D2 Ndc sinA cosA = 0.5π2(3.5)2(75/60)(0.16)sin 22 cos 22 = 4.199 in3/sec pmax = 6πDNLη cot A/dc2 =6π(3.5)(75/60)(60)(0.0065)(cot 22)/(0.16)2 = 989.8 lb/in2 Qx = Qmax - (Qmax/pmax)p = 4.199 - 0.004242p (b) Given: T-shaped cross section extruded at 0.14 lb/sec. Density of polymer ρ = specific gravity of polymer x ρwater =1.2(62.4 lb/ft3) = 75 lb/ft3 Convert to lb/in3: ρ= 75 lb/ft3 /(123 in3/ft3) = 0.0433 lb/in3 Qx = 0.11/0.0433 = 2.540 in3/sec. 2.540 = 4.199 - 0.004242p 0.004242p = 4.199 – 2.540 = 1.659 p = 391.1 lb/in2 (c) Qx = Ks p Ks = Qx/p = 2.540/391.1 = 0.00649 Qx = 0.00649 p Injection Molding 13.17 Compute the percentage volumetric contraction of a polyethylene molded part, based on the value of shrinkage given in Table 13.1. Solution: S = 0.025 for polyethylene from Table 13.1. Volumetric contraction = 1.0 - (1 -.025)3 = 1.0 - 0.92686 = 0.07314 = 7.314% Note that we are not using the parameter S from Table 13.1 in the way it was intended to be used. Its intended use is to compute the oversized dimension of a mold cavity in injection molding. Instead, we are using the shrinkage term to calculate the amount of (volumetric) reduction in size of the part after the polymer is injected into the cavity. In fact, a slightly different shrinkage parameter value may apply in this case. 13.18 The specified dimension = 225.00 mm for a certain injection molded part made of ABS. Compute the corresponding dimension to which the mold cavity should be machined, using the value of shrinkage given in Table 13.1. Solution: S = 0.006 for ABS from Table 13.1. Dc = 225.00 + 225.00(0.006) + 225.00(0.006)2 = 225.00 + 1.35 + 0.0081 = 226.36 mm. 13.19 The part dimension for a certain injection molded part made of polycarbonate is specified as 3.75 in. Compute the corresponding dimension to which the mold cavity should be machined, using the value of shrinkage given in Table 13.1. Solution: S = 0.007 for polycarbonate from Table 13.1. Dc = 3.75 + 3.75(0.007) + 3.75 (0.007)2 = 3.75 + 0.0263 + 0.0002 = 3.7765 in. 13.20 The foreman in the injection molding department says that a polyethylene part produced in one of the operations has greater shrinkage than the calculations indicate it should have. The important dimension of the part is specified as 112.5 ±0.25 mm. However, the actual molded part measures 112.02 mm. (a) As a first step, the corresponding mold cavity dimension should be checked. Compute the correct value of the mold dimension, given that the shrinkage value for polyethylene is 0.025 (from Table 13.1). (b) What adjustments in process parameters could be made to reduce the amount of shrinkage? Solution: (a) Given: S = 0.025, Dc = 112.5 + 112.5(.025) + 112.5(.025)2 = 115.383 mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 10 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 (b) Adjustments to reduce shrinkage include: (1) increase injection pressure, (2) increase compaction time, and (3) increase molding temperatures. 13.21 An injection molded polyethylene part has a dimension of 2.500 in. A new material, polycarbonate, is used in the same mold. What is the expected corresponding dimension of the polycarbonate molding? Solution: For polyethylene the shrinkage is 0.025 in/in (from Table 13.1). Die Cavity = Dc = Dp + DpS + DpS2 = 2.500 + 2.500(0.025) + 2.500(0.025)2 = 2.564 in For polycarbonate, the shrinkage is 0.007 in/in Part dimension = Dc/(1 + S + S2) = 2.564/(1 + 0.007 + 0.0072) = 2.546 in Other Molding Operations and Thermoforming 13.22 The extrusion die for a polyethylene parison used in blow molding has a mean diameter of 18.0 mm. The size of the ring opening in the die is 2.0 mm. The mean diameter of the parison is observed to swell to a size of 21.5 mm after exiting the die orifice. If the diameter of the blow molded container is to be 150 mm, determine (a) the corresponding wall thickness of the container and (b) the wall thickness of the parison. Solution: (a) rs = Dp/Dd = 21.5/18.0 = 1.194 tm = tp Dp/Dm = rs td Dp/Dm = (1.194)(2.0)(21.5)/150.0 = 0.342 mm (b) tp = rs td = (1.194)(2.0) = 2.388 mm 13.23 A parison is extruded from a die with outside diameter = 11.5 mm and inside diameter = 7.5 mm. The observed die swell is 1.25. The parison is used to blow mold a beverage container whose outside diameter = 112 mm (a standard size 2-liter soda bottle). (a) What is the corresponding wall thickness of the container? (b) Obtain an empty 2-liter plastic soda bottle and (carefully) cut it across the diameter. Using a micrometer, measure the wall thickness to compare with your answer in (a). Solution: (a) Dd = (11.5 + 7.5)/2 = 9.5 mm, and td = (11.5 - 7.5)/2 = 2.0 mm tm = (1.25)2(2.0)(9.5)/112 = 0.265 mm (= 0.010 in) (b) Measured value should be close to calculated value. Some wall thicknesses are less. 13.24 A blow-molding operation is used to produce a bottle with a diameter of 2.250 in and a wall thickness of 0.045 in. The parison has a thickness of 0.290 in. The observed die swell ratio is 1.30. (a) What is the required diameter of the parison? (b) What is the diameter of the die? Solution: (a) Dp = tmDm/tp = (0.045)(2.250)/0.290 = 0.349 in (b) Dd = Dp/rs = 0.349/1.30 = 0.268 in 13.25 An extrusion operation is used to produce a parison whose mean diameter = 27 mm. The inside and outside diameters of the die that produced the parison are 18 mm and 22 mm, respectively. If the minimum wall thickness of the blow-molded container is to be 0.40 mm, what is the maximum possible diameter of the blow mold? Solution: Dd = (22 + 18)/2 = 20 mm, and td = (22 - 18)/2 = 2 mm rs = 27/20 = 1.35 Rearranging Eq. (13.22) in text, Dm = rsd3tdDd/tm = (1.35)2(2)(20)/(0.40) = 182.25 mm 13.26 A rotational molding operation is to be used to mold a hollow playing ball out of polypropylene. The ball will be 1.25 ft in diameter and its wall thickness should be 3/32 in. What weight of PP powder should be loaded into the mold in order to meet these specifications? The specific gravity of the PP grade is 0.90, and the density of water is 62.4 lb/ft3. Solution: Density ρ = specific gravity of polymer x ρwater = 0.90(62.4 lb/ft3) = 56.2 lb/ft3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 11 Solutions for Fundamentals of Modern Manufacturing, 4/e (published by Wiley)  MPGroover 2010 03-27-09, 04-28-09 Convert to lb/in3: ρ = 56.2 lb/ft3 /(1728 in3/ft3) = 0.0325 lb/in3 Volume = π(Do3 - Di3)/6 = 0.16667π[(1.25x12)3 - (1.25x12 - 3/16)3] = 10.91 in3 Weight W = (10.91)(0.0325) = 0.355 lb. 13.27 The problem in a certain thermoforming operation is that there is too much thinning in the walls of the large cup-shaped part. The operation is conventional pressure thermoforming using a positive mold, and the plastic is an ABS sheet with an initial thickness of 3.2 mm. (a) Why is thinning occurring in the walls of the cup? (b) What changes could be made in the operation to correct the problem? Solution: (a) As the starting flat sheet is draped over the convex cup-shaped mold, the portion contacting the base of the cup experiences little stretching. However, the remaining portions of the sheet must be stretched significantly to conform to the sides of the cup. Hence, thinning in these sides results. (b) The problem could be solved by either: (1) fabricating a negative mold to replace the current positive mold, since a negative mold will distribute the material more uniformly and result in approximately equal thinning throughout the sheet; or (2) prestretch the sheet as in Figure 13.38 in the text. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 12