Plastics Engineering 3E Episode 13 docx

404 25 h 20 I f 15 8 d m s v 0. g 10 5: Q c 35 0 7 I I -I + . . Non-isothermi Analysis of polymer melt flow )thermal v 0 0.0002 0.0004 0.0006 O.OOO8 0.0001 0.0003 0.0005 0.0007 0.0009 Injection rate (@Is) 0.001 Fig. 5.28 Variation of Cavity Pressure Loss with Injection Rate may be seen that for the isothermal situation the pressure increases as the injection rate increases. However, in the situation where the melt freezes off as it enters the mould, the relationship is quite different. In this case the pressure is high at high injection rates, in a similar manner to the isothermal situation. In fact at extremely high injection rates the two curves would meet because the melt would enter the mould so fast it would not have time to be affected by the melt temperature. However, in the non-isothermal case the pressure is also high at low injection rates. This is because slow injection gives time for significant solidification of the melt and this leads to high pressures. It is clear therefore that in the non- isothermal case there is an optimum injection rate to give minimum pressure. In Fig. 5.28 this is seen to be about 3.0 x m3/s for the situation considered here. This will of course change with melt temperature and mould temperature since these affect the freeze-off time, tf, in the above equations. Some typical values for r)o and n are given in Table 5.2. The viscosity flow curves for these materials are shown in Fig. 5.17. To obtain similar data at other temperatures then a shift factor of the type given in equation (5.27) would have to be used. The temperature effect for polypropyl- ene is shown in Fig. 5.2. Example 5.15 During injection moulding of low density polyethylene, 15 kg of material are plasticised per hour. The temperature of the melt entering the Analysis of polymer melt flow 405 mould is 190°C and the mould temperature is 40°C. If the energy input from the screw is equivalent to 1 kW, calculate (a) the energy required from the heater bands (b) the flow rate of the circulating water in the mould necessary to keep its Solution The steady flow energy equation may be written as temperature at 40 f 2" q-W=Ah where q is the heat transfer per unit mass W is the work transfer per unit mass h is enthalpy (5.90) Enthalpy is defined as the amount of heat required to change the temperature of unit mass of material from one temperature to another. Thus the amount of heat required to change the temperature of a material between specified limits is the product of its mass and the enthalpy change. The enthalpy of plastics is frequently given in graphical form. For a perfectly crystalline material there is a sharp change in enthalpy at the melting point due to the latent heat. However, for semi-crystalline plastics the rate of enthalpy change with temperature increases up to the melting point after which it varies linearly with temperature increases up to the melting point after which it varies linearly with temperature as shown in Fig. 5.29. For amorphous plastics there is only a change in slope of the enthalpy line at glass transition points. Fig. 5.29 shows that when LDPE is heated from 20°C to 190°C the change in enthalpy is 485 Hkg. Temperature ("c) Fig. 5.29 Enthalpy Variation with Temperature 406 Analysis of polymer melt flow (a) In equation (5.90) the sign convention is important. Heat is usually taken as positive when it is applied to the system and work is positive when done by the system. Hence in this example where the work is done on the system by the screw it is regarded as negative work. So using (5.90) for a mass of 15 kg per hour 11 ( 60 60 q - (-4) = 15 485 x - x -) q = 1.02 kW The heater bands are expected to supply this power. in cooling the melt from 190°C to 40°C. (b) At the mould there is no work done so in terms of the total heat absorbed q = mAh (485 - 40) - x - 60 60 = 1.85 kW This heat must be removed by the water circulating in the mould at a rate, Q q = QAh or by definition of enthalpy q = QC,AT where C, is the specific heat (= 4.186 kJ/kg"C for water) and AT is the temperature change (= 4°C i.e. f2"C) 1.85 = Q x 4.186 x 4 Q = 0.11 kg/s = 0.11 litreds It is also possible to estimate the number of cooling channels required. If the thermal conductivity of the mould material is K then the heat removed through the mould per unit time will be given by K*A (AT) 4= where AT is the temperature between the melt and the circulating fluid and Y is the distance of the cooling channels from the mould A is the area through which the heat is conducted to the coolant. This is usually taken as half the circumference of the cooling channel multiplied by its length. Analysis of polymer melt flow Weight (g) Angle (e") 407 50 100 200 500 loo0 2000 0.57 1.25 2.56 7.36 17.0 42.0 KnDL(AT) 4= 2Y 2y9 L= KnD(AT) The K value for steel is 11.5 cal/m.s.T, so assuming that the cooling chan- nels have a diameter of 10 mm and they are placed 40 mm from the cavity, then 2 x 40 x 10-~ x 1.85 x io3 L= 11.5 x 4.2 x n x 10 x x (190 - 40) = 0.65 m If the length of the cavity is 130 mm then five cooling channels would be needed to provide the necessary heat removal. Flow rate (kg/min) Pressure drop (MN/m2) Bibliography Pearson, J.R.A. Mechanics of Polymer Processing, Elsevier Applied Science, London (1985) Throne, J.L. Plastics Process Engineering, Marcel Dekker, New York (1979) Fenner, R.T. Principles of Polymer Processing, Macmillan, London, (1979) Tadmor, Z. and Gogos, C.G. Principles of Polymer Processing, Wiley Interscience, New York Brydson, J.A. Flow Properties of Polymer Melts, George Godwin, London (1981) Grober, H. Fundamentals of Heat Transfer, McGraw-Hill, New York (1961) Cogswell, F.N. Polymer Melt Rheology, George Godwin, London (1981) Carreau, P.J., De Kee, D. and Domoux, M. Can. J. Chem Eng., 57 (1979) p. 135. Rao, N.S., Design Formulas for Plastics Engineers, Hanser, Munich (1991). Kliene, I., Marshall, D.J. and Friehe, C.A. J. SOC. Plasr. Eng., 21 (1965) p. 1299. Muenstedt, H. Kunsrsroffe, 68 (1978) p. 92. ( 1979) 0.21 0.4 0.58 0.8 1.3 2.3 1.8 3.0 4.0 5.2 7.6 12.0 Questions 5.1 In a particular type of cone and plate rheometer the torque is applied by means of a weight suspended on a piece of cord. The cord passes over a pulley and is wound around a drum which is on the same axis as the cone. There is a direct drive between the two. During a test on polythene at 190°C the following results were obtained by applying a weight and, when the steady state has been achieved, noting the angle of rotation of the cone in 40 seconds. If the diameter of the cone is 50 mm and its included angle is 170", estimate the viscosity of the melt at a shear stress of 104 N/m2. 5.2 Derive expressions for the velocity profile, shear stress, shear rate and volume flow rate during the isothermal flow of a power law fluid in a rectangular section slit of width W, depth H and length L. During tests on such a section the following data was obtained. 408 Analysis of polymer melt flow If the channel has a length of 50 mm, a depth of 2 mm and a width of 6 mm, establish the applicability of the power law to this fluid and determine the relevant constants. The density of the fluid is 940 kg/m3. 5.3 The viscosity characteristics of a polymer melt are measured using both a capillary rheometer and a cone and plate viscometer at the same temperature. The capillary is 2.0 mm diameter and 32.0 mm long. For volumetric flow rates of 70 x m3/s and 200 x m3/s, the pressures measured just before the entry to the capillary are 3.9 MNhZ and 5.7 MN/m*, respectively. The angle between the cone and the plate in the viscometer is 3" and the diameter of the base of the cone is 75 mm. When a torque of 1.18 Nm is applied to the cone, the steady rate of rotation reached is observed to be 0.062 rds. Assuming that the melt viscosity is a power law function of the rate of shear, calculate the percentage difference in the shear stresses given by the two methods of measurement at the rate of shear obtained in the cone and plate experiment. 5.4 The correction factor for converting apparent shear rates at the wall of a circular cylindrical capillary to true shear rates is (3n + 1)/4n, where n is the power law index of the polymer melt being extruded. Derive a similar expression for correcting apparent shear rates at the walls of a die whose cross-section is in the form of a very long narrow slit. A slit die is designed on the assumption that the material is Newtonian, using apparent viscous properties derived from capillary rheometer measurements, at a particular wall shear stress, to calculate the volumetric flow rate through the slit for the same wall shear stress. Using the correction factors already derived, obtain an expression for the error involved in this procedure due to the melt being non-Newtonian. Also obtain an expression for the error in pressure drop calculated on the same basis. What is the magnitude of the error in each case for a typical power law index n = 0.37? 5.5 Polyethylene is extruded through a cylindrical die of radius 3 mm and length 37.5 mm at a rate of 2.12 x m3/s. Using the flow curves supplied, calculate the natural time of the process and comment on the meaning of the value obtained. 5.6 Polythene is passed through a rectangular slit die 5 mm wide, 1 mm deep at a rate of 0.7 x m3/s. If the time taken is 1 second, calculate the natural time and comment on its meaning. 5.7 In a plunger type injection moulding machine the torpedo has a length of 30 mm and a diameter of 23 mm. If, during the moulding of polythene at 170°C (flow curves given), the plunger moves forward at a speed of 50 mm/s estimate the pressure drop along the torpedo. The barrel diameter is 25 mm. Analysis of polymer melt flow 409 5.8 The exit region of a die used to extrude a plastic section is 10 mm long and has the cross-sectional dimensions shown below. If the channel is being extruded at the rate of 3 dmin calculate the power absorbed in the die exit and the melt temperature rise in the die. Flow curves for the polymer melt are given in Fig. 5.3. The product pCp for the melt is 3.3 x 106. 5.9 The exit region of a die used to extrude a plastic channel section is 10 mm long and has the dimensions shown below. If the channel is being extruded at the rate of 3 dmin. calculate the power absorbed in the die exit, and the dimensions of the extrudate as it emerges from the die. The flow curves in Fig. 5.3 may be used. All dimensions in mm 5.10 During extrusion blow moulding of 60 mm diameter bottles the extruder output rate is 46 x m3/s. If the die diameter is 30 mm and the die gap is 1.5 mm calculate the wall thickness of the bottles which are produced. The flow curves in Fig. 5.3 should be used. 5.11 Polyethylene is injected into a mould at a temperature of 170°C and a pressure of 100 MN/m2. If the mould cavity has the form of a long channel with a rectangular cross-section 6 mm x 1 mm deep, estimate the length of the flow path after 1 second. The flow may be assumed to be isothermal and over the range of shear rates experienced ( lo3 - 16 s-l) the material may be considered to be a power law fluid. 5.12 Repeat the previous question for the situation in which the mould temperature is 60°C and the freeze-off temperature is 128°C. What difference would it make if it had been assumed that the material was Newtonian with a viscosity of 1.2 x Id Ns/m*. 5.13 During the blow moulding of polypropylene bottles, the parison is extruded at a temper- ature of 230°C and the mould temperature is 50°C. If the wall thickness of the bottle is 1 mm and the bottles can be ejected at a temperature of 120°C estimate the cooling time in the mould. 5.14 An injection moulding is in the form of a flat sheet 100 mm square and 4 mm thick. The melt temperature is 23OoC, the mould temperature is 30°C and the plastic may be ejected from the mould at a centre-line temperature of 90°C. If the runner design criterion is that it should be ejectable at the same instant as the moulding, estimate the required runner diameter. The thermal diffusivity of the melt is 1 x lo-' m2/s. 5.15 For a particular polymer melt the power law constants are = 40 kN.s"/m and n = 0.35. If the polymer flows through an injection nozzle of diameter 3 mm and length 25 mm at a rate of 5 x 10-~ m3/s, estimate the pressure drop in the node. 5.16 Polythcne at 170'C is used to injection mould a disc with a diameter of 120 mm and thickness 3 nun. A sprue gate is used to feed the material into the centre of the disc. If the 410 Analysis of polymer melt flow injection rate is constant and the cavity is to be filled in 1 second estimate the minimum injec- tion pressure needed at the nozzle. The flow curves for this grade of polythene are given in Fig. 5.3. m3, the melt temperature is 170"C, the mould temperature is 50°C and rectangular gates with a land length of 0.6 mm are to be used. If it is desired to have the melt enter the mould at a shear rate of Id s-I and freeze-off at the gate after 3 seconds, estimate the dimensions of the gate and the pressure drop across it. It may be assumed that freeze-off occurs at a temperature of 114°C. The flow curves in Fig. 5.3 should be used. m3/s. Using the flow curves provided and assuming the power law index n = 0.33 over the working section of the curves, calculate the total pressure drop through the die. Also estimate the dimen- sions of the extruded tube. 5.17 During the injection moulding of a polythene container having a volume of 4 x 5.18 Polyethylene at 170°C passes through the annular die shown, at a rate of 10 x 5.19 A polythene tube of outside diameter 40 mm and wall thickness 0.75 mm is to be extruded at a linear speed of 15 mm/s. Using the 170°C polythene flow curves supplied, calculate suitable die exit dimensions. 5.20 The exit region of a die used to blow plastic film is shown below. If the extruder output is 100 x m3/s of polythene at 170°C estimate the total pressure drop in the die between points A and C. Also calculate the dimensions of the plastic bubble produced. It may be assumed that there is no inflation or draw-down of the bubble. Flow data for polythene is given in Fig. 5.3. angle = 2" C Melt Melt 5.21 A polyethylene moulding material at 170°C passes along the channel shown at a rate of m3/s. Using the flow curves given and assuming n = 0.33 calculate the pressure. drop 4 x along the channel. Analysis of polymer melt flow 41 1 I all dimensions in mm 1 5.22 A power law plastic is injected into a circular section channel using a constant pressure, (a) the flow is isothermal (b) the melt is freezing off as it flows along the channel. 5.23 A polymer melt is injected into a circular section channel under constant pressure. What is the ratio of the maximum non-isothermal flow length to the isothermal flow length in the same time for (a) a Newtonian melt and (b) a power law melt with index. n = 0.3. 5.24 A power law fluid with the constants 90 = 104 Ndm2 and n = 0.3 is injected into a circular section channel of diameter 10 mm. Show how the injection rate and injection pressure vary with time if. P. Derive an expression for the flow length assuming that (a) the injection pressure is held constant at 140 MN/m2 (b) the injection rate is held constant at The flow in each case may be considered to be isothermal. 5.25 Polyethylene at 170°C is used to injection mould a flat plaque measuring 50 mm x 10 mm x 3 mm. A rectangular gate which is 4 mm x 2 mm with a land length of 0.6 mm is situated in the centre of the 50 mm side. The runners are 8 mm diameter and 20 mm long. The material passes from the barrel into the runners in 1 second and the pressure losses in the nozzle and sprue may be taken as the same as those in the runner. If the injection rate is fixed at m3/s, estimate (a) the pressure losses in the runner and gate and (b) the initial packing pressure on the moulded plaque. Flow curves are supplied. 5.26 A lace of polyethylene is extruded with a diameter of 3 mm and a temperature of 190°C. If its centre-line must be cooled to 70°C before it can be granulated effectively, calculate the required length of the water bath if the water temperature is 2OoC. The haul-off speed is 0.4 m/s and it may be assumed that the heat transfer from the plastic to the water is by conduction only. 5.27 Using the data in Tables 5.1 and 5.2, calculate the flow lengths which would be expected if the following materials were injected at 100 MN/mz into a wide rectangular cross-section channel, 1 mm deep. Materials - LDPE, polypropylene, polystyrene, PVC, POM, acrylic, polycarbonate, nylon 66 and ABS. Note that the answers will give an indication of the flow ratios for these materials. The flow should be assumed to be non-isothermal. 5.28 It is desired to blow mould a cylindrical plastic container of diameter 100 mm and wall thickness 2.5 mm. If the extruder die has an average diameter of 40 mm and a gap of 2 mm, calculate the output rate needed from the extruder. Comment on the suitability of an inflation pressure in the region of 0.4 MN/m2. The density of the molten plastic may be taken as 790 kg/m3. Use the flow curves in Fig. 5.3. 5.29 During the blow moulding of polyethylene at 170°C the parison is 0.4 m long and is left hanging for 1 second. Estimate the natural time for the process and the amount of sagging which occurs. The density of the melt may be taken at 730 kg/m3. m3/s. 412 Analysis of polymer melt flow 5.30 The viscosity, t). of plastic melt is dependent on temperature, T, and pressure, P. The variations for some common plastics are given by equations of the form t)/t)~ = 1PAP and t)lt)~ = leA7 where AT = T - TR(OC), AP = P - PR (htN/m2), and the subscript R signifies a reference value. Spica1 values of the constants A and B are given below. A (x 10-3) B (x10-3) Acrylic Polypropylene LDPE Nylon Acetal -28.32 -7.53 -11.29 -12.97 -7.53 9.54 6.43 6.02 4.22 3.89 During flow along a particular channel the temperature drops by 40°C and the pressure drops by 50 MN/mz. Estimate the overall change in viscosity of the melt in each case. Detennine the ratio of the pressure change to the temperature change which would cause no change in viscosity for each of the above materials. APPENDIX A - Structure of Plastics A.l Structure of Long Molecules Polymeric materials consist of long chain-like molecules. Their unique struc- tural configuration affects many of their properties and it is useful to consider in more detail the nature of the chains and how they are built up. The simplest polymer to consider for this purpose is polyethylene. During the polymerisa- tion of the monomer ethylene, the double bond (see Fig. A.l) is opened out enabling the carbon single bonds to link up with neighbouring units to form a long chain of CH2 groups as shown in Fig. A.2. This is a schematic represen- tation and conceals the fact that the atoms are jointed to each other at an angle as shown in Fig. A.3. Fig. A.l Ethylene monomer HHHHH Ill11 IIIII -c - c -c -c -c- HHHHH Fig. A.2 Polyethylene molecule 413 [...]... described also apply to other polymers and the structures of several of the more common polymers are given below 416 structure of Plastics / - -\ I \ Fig AS@) Random conformation of carbon backbone on molecular chain C H Fig A.6 Chain branch i polyethylene n structure of Plastics 417 Polypropylene H H H H I I I I -c-c-c-c-c-cI I I I H H I I I I H CH3 H CH3 H CH3 Polyvinyl Chloride (PVC) H H H H H H... becoming available and a mass of spherulites will start to grow The solid polymer will then consist of a large number of small spherulites 423 Structure of Plastics Fig A 12 Qpical illustration of spherulites Twisted tddcd chain growth from nucleus Fig A .13 Structure of spherulite The ease with which a polymer will form into crystalline regions depends on the structure of the molecular chain It can be seen,... carried out Equally, a polymer may be crystalline but optical measurements will show no signs of orientation Orientation can be introduced into plastics such as polyethylene and polypropylene (both semi-crystalline) by cold drawing at room temperature Other brittle plastics such as polymethyl methacrylate and polystyrene (both amorphous) cannot be cold drawn but can be drawn at elevated temperatures APPENDIX... 4n2EI AL2 W h3 12 4n2EWh3 n2E h = 12whL2 = 3(i) so L For many plastics u,/E = 30 x to 35 x for short-term loading so For longer term loading, E will decrease more than uc so uJE will get larger This will cause the acceptable h/L to decrease (2.1 1) K~EI K~EI Pc = -so a = , L 2 AL2 n2d2 Now for circular rod, cc = 16L2 = /0.005 x 16 x 1502 n2 = 13. 5 mm ... of this is polypropylene which in the atactic form is an amorphous material of little commercial value but in the isotactic form is an extremely versatile large tonnage plastic material Structure of Plastics 419 H H H H X H H I I I I I I I -c-c-c-c-c-c-c1 1 1 1 1 1 1 X H X H H H X (3 At.ctic H H X H H H X I l I I I I I -c-c-c-c-c-c-c1 X 1 1 H H H t X l H ~ ~ H syndidretie (b) H H H H H H H 1 1 1 1... Alternating -A - B - A - B - A - B- (2) Random A - A - B - A - A - A -B -B - A- (3) Block -A - A - A - B - B - B - B - B - B -A - A- B I (4)Graft - A - A - A - A - A - A - A I I B B I I B I B 420 Structure of Plastics (01 Atactic Polyprapylene Ibl lsotactlc Polypropylene Fig A.8 Polypropylene structures A.3 Arrangement of Molecular Chains A picture of an individual molecular chain has been built up as a long... crystalline fashion This is illustrated in Fig A.9 During the 1940’s it was proposed that partially crystalline polymers consisted of regions where the molecular chains where gathered in an ordered Structure of Plastics 42 1 Fig A.9 Crystalline structure of polyethylene fashion whereas adjacent regions had a random distribution of chains The crystalline regions were considered to be so small that an individual... imperfect folds, chain entanglements etc, are regarded as the diffuse (amorphous) regions viewed in X-ray diffraction studies As an alternative it has been suggested that crystalline 422 Structure of Plastics Crystalline wbn Fig A.10 Fringed micelle model 10 Fig A 11 Folded chain model (folded chains) and amorphous (random chains) regions can exist in a similar manner to that proposed in the fringed... tends to produce larger spherulites than fast cooling It is believed that the spherulites g o in all directions from a central nucleus, by the twisting rw of the folded chain platelets as shown in Fig A .13 The size of a spherulite will be limited by the growth of adjacent spherulites If the polymer melt is cooled very quickly it may undercool, i.e remain molten at a temperature below its melting point... the same length The length of each chain depends on a series of random events during the polymerisation process One chain may grow rapidly in a region with an abundant supply of monomer 415 Structure of Plastics whereas other chains stop growing prematurely as the supply of monomer dries up This means that a particular sample of synthetic polymer will not have a unique value for its molecular weight Instead . Mechanics of Polymer Processing, Elsevier Applied Science, London (1985) Throne, J.L. Plastics Process Engineering, Marcel Dekker, New York (1979) Fenner, R.T. Principles of Polymer Processing,. Carreau, P.J., De Kee, D. and Domoux, M. Can. J. Chem Eng., 57 (1979) p. 135 . Rao, N.S., Design Formulas for Plastics Engineers, Hanser, Munich (1991). Kliene, I., Marshall, D.J. and. HHHHH Ill11 IIIII -c - c -c -c -c- HHHHH Fig. A.2 Polyethylene molecule 413 414 Structure of Plastics U w H Fig. A.3 Polyethylene molecule In all the groups along the chain,