Plastics Engineering 3E Episode 12 ppsx

Analysis of polymer melt flow SO PQ= pQ x C, x AT AT=- P PCP 369 (5.64) where p is the density of the fluid and Cp is its specific heat. 5.10 Experimental Methods Used to Obtain Flow Data In Section 5.1 1 design examples relating to polymer processing will be illus- trated. In these examples the flow data supplied by material manufacturers will be referred to, so it is proposed in this section to show how this data may be obtained. The equipment used to obtain flow data on polymer melts may be divided into two main groups. (a) Rotational Viscometers - these include the cone and plate and the concen- tric cylinder. (b) Capillary Viscometer - the main example of this is the ram extruder. Cone and Plate Viscometer In this apparatus the plastic to be analysed is placed between a heated cone and a heated plate. The cone is truncated and is placed above the plate in such a way that the imaginary apex of the cone is in the plane of the plate. The angle between the side of the cone and the plate is small (typically 4"). The cone is rotated relative to the plate and the torque. T, necessary to do this is measured over a range of rotational rates, 6. Refemng to Fig. 5.14 Area of annulus Force = 2nn. dr Torque = 217r'r. dr = 217r. dr SO Total Torque, T = 2rcr't . dr I 0 Also shear strain, Strain rate, T = ($) 3rR3t XI-86 y=-=-=- h ra! ct .e y=- a! (5.65) (5.66) 370 Analysis of polymer melt flow Molten plastic Heated plate [fixed I Fig. 5.14 Cone and Plate Viscometer t And since viscosity q=- Y 3T a 9=-*- 2n~3 b (5.67) The disadvantage of this apparatus is that it is limited to strain rates in the region 10 to 1 s-' whereas in plastics processing equipment the strain rates are in the order of 10 s3-l to IO4 s-'. Concentric Cylinder Viscometer In this apparatus the polymer melt is sheared between concentric cylinders. The torque required to rotate the inner cylinder over a range of speeds is recorded so that viscosity and strain rates may be calculated. Torque, T = 2nRL R t = 2nR2Lt (5.68) Referring to Fig. 5.15 Strain rate, Viscosity, (5.69) (5.70) Analysis of polymer melt flow 37 1 Hwted - concentric cylinders, Molten /plastic Fig. 5.15 Concentric Cylinder Viscometer As in the previous case, this apparatus is usually restricted to relatively low strain rates. Ram Extruder In this apparatus the plastic to be tested is heated in a barrel and then forced through a capillary die as shown in Fig. 5.16. Normally the ram moves at a constant velocity to give a constant volume flow rate, Q. From this it is conventional to calculate the shear rate from the Newtonian flow expression. p=- 4Q nR3 Since it is recognised that the fluid is Non-Newtonian, this is often referred to as the apparent shear rate to differentiate it from the true shear rate. If the pressure drop, P, across the die is also measured then the shear stress, t, may be calculated from PR This leads to a definition of apparent viscosity as the ratio of shear stress to apparent shear rate A plot of apparent viscosity against shear rate produces a unique flow curve for the melt as shown in Fig. 5.3. Occasionally this information may be based on the true shear rate. As shown in Section 5.4(a) this is given by 372 Analysis of polymer melt flow r Piston 1/ xtrudote k Fig. 5.16 Section Through Ram Extruder However, this then means that the true shear rate must always be used in the flow situation being analysed. Experience has shown that this additional complexity is unnecessary because if the Newtonian shear rate is correlated with flow data which has been calculated using a Newtonian shear rate, then no error is involved. Note that rotational viscometers give true shear rates and if this is to be used with Newtonian based flow curves then, from above, a correction factor of (4n/3n + 1) needs to be applied to the true shear rate. The ratio (3n + 1)/4n is called the Rabinowitsch Correction Factor and it is used to convert Newtonian shear rates to true shear rates. There are two other points worth noting about this test. Firstly the flow data is produced using a capillary die so that its use on channels of a different geometry would require a correction factor. However, in most cases of practical interest, the factor is not significantly different from 1 and so there is no justification for the additional complication caused by its inclusion. Secondly, the pressure drop, P, in the above expression is the pressure drop due to shear flow along the die. If a pressure transducer is used to record the Analysis of polymer melt flow 373 pressure drop as shown in Fig. 5.16, then it will also pick up the pressure losses at the die entry. This problem may be overcome by carrying out further tests using either a series of dies having different lengths or a die with a very short (theoretically zero) length. In the former case, the pressure drops for the various lengths of die may be extrapolated to give the pressure drop for entry into a die of zero length. In the second case this pressure is obtained directly by using the so-called zero length die. This is then subtracted from the measured pressure loss, PL, on the long die being considered, so that (5.71) In addition, if the swelling of the extrudate is measured in each of these two tests then the swelling ratio using the long die will be BSR and the swelling ratio using the short die will be BER (see Section 5.6). Using equation (5.44) and (5.47) this enables the shear and tensile components of the recoverable strains to be calculated and from them the shear and tensile moduli. From this relatively simple test, therefore, it is possible to obtain complete flow data on the material as shown in Fig. 5.3. Note that shear rates similar to those experienced in processing equipment can be achieved. Variations in melt temperature and hydrostatic pressure also have an effect on the shear and tensile viscosities of the melt. An increase in temperature causes a decrease in viscosity and an increase in hydrostatic pressure causes an increase in viscosity. Wically, for low density polyethlyene an increase in temperature of 40°C causes a vertical shift of the viscosity curve by a factor of about 3. Since the plastic will be subjected to a temperature rise when it is forced through the die, it is usually worthwhile to check (by means of Quation 5.64) whether or not this is significant. Fig. 5.2 shows the effect of temperature on the viscosity of polypropylene. A change in pressure from atmospheric (= 0.1 MN/m2) to about 100 MN/m2 (a pressure likely to be experienced during processing) causes a vertical shift of the viscosity curve by a factor of about 4 for LDPE. This effect may be important during processing because the material can be subjected to large changes in pressure in sections such as nozzles, gates, etc. However, it should be noted that in some cases the increase or decrease in pressure results in, or is associated with, an increase or decrease in temperatures so that the net effect on viscosity may be negligible. Other factors such as the use of additives also have an effect on the shape of the flow curves. Flame retardants, if used, tend to decrease viscosity whereas pigments tend to increase viscosity. Fig. 5.17 shows flow curves for a range of plastics. Melt Flow Rate (also known as Melt Flow Index) The Melt Flow Rate Test is a method used to characterise polymer melts. It is, in effect, a single point ram extruder test using standard testing conditions (BS 374 Analysis of polymer melt flow IO000 -1m E N I 10 1 66( 290" C) PVC (1 70°C) PMMA POM(21 OOC) 100 1000 10000 100000 Shear rate (s-1) Fig. 5.17 Viscosity Curves for a Range of Plastics (230°C unless otherwise shown) Oil -filled t hermomete wells Weight w 'r=%il I -Barrel 'Heater , lnsulati Polymei on . melt Fig. 5.18 Diagram of Apparatus for Measuring Melt Flow Index Analysis of polymer melt flow 375 2782) as illustrated in Fig. 5.18. The polymer sample is heated in the barrel and then extruded through a standard die using a standard weight on the piston, and the weight (in gms) of polymer extruded in 10 minutes is quoted as the melt flow rate (MFR) of the polymer. Flow Defects When a molten plastic is forced through a die it is found that under certain conditions there will be defects in the extrudate. In the worst case this will take the form of gross distortion of the extrudate but it can be as slight as a dullness of the surface. In most cases flow defects are to be avoided since they affect the quality of the output and the efficiency of the processing operation. However, in some cases if the flow anomaly can be controlled and reproduced, it can be used to advantage - for example, in the production of sheets with matt surface finish. Flow defects result from a combination of melt flow properties, die design and processing conditions but the exact causes and mechanisms are not completely understood. The two most common defects are (a) Melt Fracture When a polymer melt is flowing through a die, there is a critical shear rate above which the extrudate is no longer smooth. The defect may take the form of a spiralling extrudate or a completely random configuration. With most plastics it is found that increasing the melt temperature or the WD ratio of the die will increase the critical value of shear rate. It is generally believed that the distortion of the extrudate is caused by slip-stick mechanism between the melt and the die wall due to the high shear rates. If there is an abrupt entry to the die then the tensilelshear stress history which the melt experiences is also considered to contribute to the problem. (b) Sharkskin Although this defect is also a visual imperfection of the extrudate it is usually differentiated from melt fracture because the defects are perpendicular to the flow direction rather than helical or irregular. In addi- tion, experience has shown that this defect is a function of the linear output rate rather than the shear rate or die dimensions. The most probable mechanism of sharkskin relates to the velocity of skin layers of the melt inside and outside the die. Inside the die the skin layers are almost stationary whereas when the extrudate emerges from the die there must be a rapid acceleration of the skin layers to bring the skin velocity up to that of the rest of the extrudate. This sets up tensile stresses in the melt which can be sufficient to cause fracture. 5.11 Analysis of Flow in Some Processing Operations Design methods involving polymer melts are difficult because the flow behaviour of these materials is complex. In addition, flow properties of the melt are usually measured under well defined uniform conditions whereas unknown effects such as heating and cooling in processing equipment make service conditions less than ideal. However, sufficient experience has been gathered 376 Analysis of polymer melt flow using the equations derived earlier that melt flow problems can be tackled quantitatively and with an accuracy which compares favourably with other engineering design situations. In this section a number of polymer processing methods will be analysed using the tools which have been assembled in this chapter. In some cases the heating and cooling of the melt may have an important effect on the flow behaviour and methods of allowing for this are developed. Example 5.2 In a plunger-type injection moulding machine the torpedo has a length of 40 mm, a diameter of 23 mm and is supported by three spiders. If, during moulding of polythene at 170°C, the plunger moves forward at a speed of 15 mm/s, estimate the pressure drop along the torpedo and the shear force on the spiders. The barrel diameter is 25 mm. Solution Assume that the flow is isothermal. Volume flow rate, Q = area x velocity IC = -(25 10-~)~ 15 10-~ 4 = 7.36 x m3/s. The gap between the torpedo and the barrel may be considered as a rectangular slit with T = (IC x 24 x 10F3)m and H = 1 x 10 m-3. 6Q 6 x 7.36 x TH2 - IC x 24 x So apparent strain rate, = ~ - x = 585 S-' From Fig. 5.3, at this strain rate, r] = 400 Ns/m2. so T = r]p = 400 x 585 = 2.34 x lo5 N/m2 2Lt 2 x 40 x x 2.34 x 105 p=-= = 18.7 MN/m2 H 1 x 10-3 The force on the spider results from (a) the force on the torpedo due to the pressure difference across it and (b) the force on the torpedo due to viscous drag. Cross-sectional area of torpedo = nr2 So force due to pressure = Rr2P = ~(11.5 x 10-3)2 x 18.7 x lo6 = 7.8 kN 377 Analysis of polymer melt flow Surface area of torpedo = RDL Viscous drag force = RDL~ = n(23 x = 0.68 kN So Total Force = 8.48 kN x 1OW3)(2.34 x lo5) Example 5.3 The output of polythene from an extruder is 30 x m3/s. If the breaker plate in this extruder has 80 holes, each being 4 mm diameter and 12 mm long, estimate the pressure drop across the plate assuming the material temperature is 170°C at this point. The flow curves in Fig. 5.3 should be used. Solution Assume the flow is isothermal. Flow rate through each hole So Apparent shear rate, = i, = m3/s 30 x 4Q 80 nR3 4 x 30 x x go x 23 x 10-9 59.6 s-l From Fig. 5.3 the shear stress, t, is 1.2 x 16 N/m2 and since PR 2L t=- 2 12 10-3 x 1.2 x 105 2 x 10-3 P= = 1.44 m/m2 Example 5.4 Eight polypropylene mouldings, each weighing 10 g are to be moulded using the runner layout shown in Fig. 5.19. If the injection time is 2 seconds and the melt temperature is 210°C calculate the pressure at each cavity if the injection pressure at the sprue is 80 MN/m2. The density of the polypropylene is 909 kg/m3 and the volume of the sprue is 5000 mm3. Assume that the flow is isothermal and ignore the pressure losses at comers. Solution Firstly it is necessary to determine the volume flow rate through the runners and gates. The total volume of the runners, gates and cavities may be calculated as follows: Volume of 8 cavities = 8 x 10 x 10-3/909 = 8.8 x = 88 x lo3 mm3 m3 378 Analysis of polymer melt flow 1Wmm 5 mm Gate 3 mm wide, 2 mm deep, 1.6mm long Fig. 5.19 Runner Layout, Example 5.4 Volume of long runners = 4 x nR2 x L = 4 n(3)2 100 = 11.3 x 103 mm3 Volume of short runners = 8 x ~(2.5)~ x 50 = 7.85 x lo3 mm3 Volume of sprue = 5 io3 m3 Total volume = 112.2 103 m3 The fill time is 2 seconds so the total volume flow rate, Q, is given by 112.2 x 103 Q= = 56.1 x lo3 mm3/s = 56.1 x m3/s To get the pressure loss in the runners and gates. The flow rate in the long runner is Q/4, hence 4Q RR~ 4 x ~(3)~ 4 x 56.1 x lo3 (i) For long runners i, = - = = 661 S-' [...]... involving plastics are demonstrated Fouriers equation for non-steady heat flow in one dimension, x , is a2T 1 aT a at ax2 where T is temperature and a! is the thermal diffusivity defined as the ratio of thermal conductivity, K,to the heat capacity per unit volume a=- K PCP where p is density and C p is specific heat Most materials manufacturers supply data on the thermal diffusivity of their plastics. .. predictions of freeze-off time for the data in the above Example Example 5 .12 Estimate the heat transfer from the die to the melt as it passes through the die land in Example 5.5 Solution Volume of die land = RDHL = (r 260 0.7 4 10-91m3 260 0.7 4 x 10-9 Residence time = 109.6 x = 2.08 x s at Fourier number = X2 1 0 - ~ 2.08 x (0.35 x 10- 312 = 0.017 Fo = 1 Analysis of polymer melt flow 396 From Fig 5.23 it... the mean effective pressure (MEP) by the projected area of the moulding, as illustrated in Chapter 4 For plastics other than polypropylene it would be necessary to produce a similar set of curves using equation (5.88) and the Power Law data in Table 5.2 Table 5.2 Qpical Power Law Parameters for Plastics Material Temperature ("C) LDPE PVC Polypropylene Acetal (POM) Polystyrene ABS PMMA Polycarbonate... angle is taken as 8" H4 = H3 + Ltan8" = 1.28 At + 40 tan 8" = 6.9 mm C3 )j= 494.5 s-1 6 = 1/3(494.5) tan 4" = 11.52 s - ~ At C4 y = - - 6Q TH: - 7~ 6 x 109.6 x = 16.9 s - ~ x 260 x 10-3 x (6.9 x 10- 312 i = 1/3(16.9)tan4" = 0.39 s-l Analysis of polymer melt flow 382 As before (Z,"] psM =A [l 2n tana from Fig 5.3, t3 = 2.1 x lo5 Nm-2 at y = 494.5 2.1 x 1 6 = 0.6tan4" s-l [' (z)] 1.28 - = 3.36 MNlm2... and C p is specific heat Most materials manufacturers supply data on the thermal diffusivity of their plastics but in the absence of any information a value of 1 x lov7 m2/s may be used for most thermoplastics (see Table 1.8 and Table 5.1) Solutions to Fourier’s equation are in the form of infinite series but are often more conveniently expressed in graphical form In the solution the following dimensionless... dominant factor The above analysis was for conduction heat transfer (Bi + 00) When the plastic moulding is taken out of the mould we need to check the value of Bi In this case Bi = 20 x 1.5 x 0.25 10-3 = 0 .12 Analysis of polymer melt flow 394 The dimensionless temperature gradient is Fig 5.24 shows the modified ATIF, graph for a flat sheet for different values of Bi In this case 1/Bi = 8.33 So from Fig 5.24... with the mould but it takes 90 seconds to cool a further 30°C when it emerges out into the air This highlights the much faster cooling rate achieved in the mould Some typical moulding data for a range of plastics are given in Table 5.1 Note that the de-moulding temperature will be generally about 30°C below the Freeze-off temperature in order to ensure that the moulding is sufficiently solid for handling... dimensions will be the same as those of the tube 6Q Apparent shear rate, i/ = TH2 384 Analysis of polymer melt flow but volume flow rate, Q = VTH where V is the velocity of the plastic melt p = -6V H 6 x 20 - 120 s-1 =-1 from Fig 5.3, t = 1.3 x 1 N/m-2 and G = 3 x 104 N/m-2 6 So recoverable shear strain, YR, is given by t )/R=-= G 1.3 x 1 6 = 4.33 3x104 Assuming that swelling results from shear effects only... injection rate would decrease Note that for a Newtonian fluid, n = 1, so for the isothermal case, equation (5.82) becomes (9) 1I2 L = 0.408 H and for the non-isothermal case, equation (5.86) becomes (?) 112 L = 0.13 H For the non-isothermal cases the freeze-off time, t f , may be estimated by the method described in Example 5.1 1 Example 5.14 A power law fluid with constants ~0 = 1.2 x 104 Ndm2 and n =... between parallel plates Q= (w) n+l TVoH Now for the disc, T = 2nr and substituting for VOfrom (5.27) Now letting P I = 0 at R = R 1 and P2 = P at R2 = r now P = PO at r = 0 So for the conditions given = 12. 42 MN/m2 401 Analysis of polymer melt flow Note that if POis substituted back into the expression for P then the expression used earlier (Chapter 4) to calculate the mould clamping force is obtained . Volume of sprue = 5 io3 m3 Total volume = 112. 2 103 m3 The fill time is 2 seconds so the total volume flow rate, Q, is given by 112. 2 x 103 Q= = 56.1 x lo3 mm3/s = 56.1. whereas pigments tend to increase viscosity. Fig. 5.17 shows flow curves for a range of plastics. Melt Flow Rate (also known as Melt Flow Index) The Melt Flow Rate Test is a. 100 1000 10000 100000 Shear rate (s-1) Fig. 5.17 Viscosity Curves for a Range of Plastics (230°C unless otherwise shown) Oil -filled t hermomete wells Weight w 'r=%il