PLASTICS EXTRUSION (ΕΚΒΟΛΗ ΠΛΑΣΤΙΚΩΝ)

Σχετικά έγγραφα
MEΡΟ VI ΔΚΒΟΛΖ ΠΟΛΤΜΔΡΧΝ (POLYMER EXTRUSION)

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Section 9.2 Polar Equations and Graphs

ZLW Series. Single-stage Monoblock Centrifugal Pump ZL PUMP GROUP.,LTD

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

Homework 8 Model Solution Section

Finite Field Problems: Solutions

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Inverse trigonometric functions & General Solution of Trigonometric Equations

Group 30. Contents.

Section 8.3 Trigonometric Equations

Example Sheet 3 Solutions

Areas and Lengths in Polar Coordinates

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

EE512: Error Control Coding

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

Strain gauge and rosettes

Calculating the propagation delay of coaxial cable

derivation of the Laplacian from rectangular to spherical coordinates

UDZ Swirl diffuser. Product facts. Quick-selection. Swirl diffuser UDZ. Product code example:

The Simply Typed Lambda Calculus

Section 7.6 Double and Half Angle Formulas

Statistical Inference I Locally most powerful tests

Areas and Lengths in Polar Coordinates

2 Composition. Invertible Mappings

Approximation of distance between locations on earth given by latitude and longitude

RECIPROCATING COMPRESSOR CALCULATION SHEET ISOTHERMAL COMPRESSION Gas properties, flowrate and conditions. Compressor Calculation Sheet

Numerical Analysis FMN011

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Concrete Mathematics Exercises from 30 September 2016

Right Rear Door. Let's now finish the door hinge saga with the right rear door

Srednicki Chapter 55

RECIPROCATING COMPRESSOR CALCULATION SHEET

( y) Partial Differential Equations

Matrices and Determinants

ΜΜ917-Σχεδιασμός Ενεργειακών Συστημάτων

[1] P Q. Fig. 3.1

4.6 Autoregressive Moving Average Model ARMA(1,1)

Solutions to Exercise Sheet 5

Heat exchanger. Type WT. For the reheating of airflows in rectangular ducting PD WT 1. 03/2017 DE/en

TMA4115 Matematikk 3

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Second Order Partial Differential Equations

Chapter 7 Transformations of Stress and Strain

Other Test Constructions: Likelihood Ratio & Bayes Tests

6.4 Superposition of Linear Plane Progressive Waves

Second Order RLC Filters

C.S. 430 Assignment 6, Sample Solutions

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Differential equations

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Parametrized Surfaces

1 String with massive end-points

Example 1: THE ELECTRIC DIPOLE

Exercises to Statistics of Material Fatigue No. 5

Lifting Entry 2. Basic planar dynamics of motion, again Yet another equilibrium glide Hypersonic phugoid motion MARYLAND U N I V E R S I T Y O F

the total number of electrons passing through the lamp.

Shenzhen Lys Technology Co., Ltd

Tridiagonal matrices. Gérard MEURANT. October, 2008

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Instruction Execution Times

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

4 Way Reversing Valve

Capacitors - Capacitance, Charge and Potential Difference

Lecture 26: Circular domains

Αναερόβια Φυσική Κατάσταση

Introduction to Theory of. Elasticity. Kengo Nakajima Summer

PARTIAL NOTES for 6.1 Trigonometric Identities

Math 6 SL Probability Distributions Practice Test Mark Scheme

ST5224: Advanced Statistical Theory II

Math221: HW# 1 solutions

Gearmotor Data. SERIES GM9000: We have the GM9434H187-R1

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Bounding Nonsplitting Enumeration Degrees

Mean bond enthalpy Standard enthalpy of formation Bond N H N N N N H O O O

Fractional Colorings and Zykov Products of graphs

D Alembert s Solution to the Wave Equation

4 Way Reversing Valve

Homework 3 Solutions

Every set of first-order formulas is equivalent to an independent set

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11


RECIPROCATING COMPRESSOR CALCULATION SHEET

Lifting Entry (continued)

Forced Pendulum Numerical approach

4.4 Superposition of Linear Plane Progressive Waves

MS SERIES MS DESK TOP ENCLOSURE APPLICATION EXAMPLE FEATURE. Measuring instruments. Power supply equipments

CRASH COURSE IN PRECALCULUS

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Thi=Τ1. Thο=Τ2. Tci=Τ3. Tco=Τ4. Thm=Τ5. Tcm=Τ6

Μεταπτυχιακή διατριβή. Ανδρέας Παπαευσταθίου

wave energy Superposition of linear plane progressive waves Marine Hydrodynamics Lecture Oblique Plane Waves:

(1) Describe the process by which mercury atoms become excited in a fluorescent tube (3)

DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C

Transcript:

PLASTICS EXTRUSION (ΕΚΒΟΛΗ ΠΛΑΣΤΙΚΩΝ) 1

ΤΙ ΕΙΝΑΙ Η ΕΚΒΟΛΗ? ΜΙΑ ΑΠΌ ΤΙΣ ΚΥΡΙΕΣ ΔΙΕΡΓΑΣΙΕΣ ΣΤΗΝ ΒΙΟΜΗΧΑΝΙΑ ΠΟΛΥΜΕΡΩΝ ΣΥΝΕΧΗΣ ΔΙΕΡΓΑΣΙΑ ΜΕ ΜΕΓΑΛΗ ΕΥΕΛΙΞΙΑ ΟΣΟΝ ΑΦΟΡΑ ΤΟ ΤΕΛΙΚΟ ΠΡΟΙΟΝ ΣΥΧΝΑ ΕIΝΑΙ ΤΟ ΠΡΩΤΟ ΣΤΑΔΙΟ ΣΕ ΜΙΑ ΣΕΙΡΑ ΔΙΕΡΓΑΣΙΩΝ ΜΟΡΦΟΠΟΙΗΣΗΣ Rheology Extrusion Univ. Thessaly 2015

ΣΕ ΠΟΙΑ ΠΟΛΥΜΕΡΗ ΕΦΑΡΜΟΖΕΤΑΙ Primary Uses are Thermoplastics: LDPE, LLDPE, HDPE, ABS, PC, PS, Nylon, PVC, PP Melt Index and Density should be matched to application Some uses for Elastomers and Thermosets Important to watch age of material and processing conditions 3

ΕΙΔΗ ΕΚΒΟΛΗΣ ΠΟΛΥΜΕΡΩΝ Compounding Pellets for future use Blown Film Bags, film. Cast Film Plastic Food Packaging Sheet Foam Trays, packaging via thermoforming 4

ΕΙΔΗ ΕΚΒΟΛΗΣ ΠΟΛΥΜΕΡΩΝ Pipe and Tubing PVC Pipe; Garden Hoses Extrusion Coating Paper Milk Cartons with Plastic Coating Wire and Cable Coating Underground Cables Monofilament Fishing Line, Ropes 5

ΣΥΝ-ΕΚΒΟΛΗ (CO-EXTRUSION) Allows Opportunity for Several Layers with Different Properties All Extruders for Each Material Goes into Common Die Die Design Determines Division of Layers 6

The history of extrusion goes back to Archimedes and before BUT modern developments based on understanding of the physical phenomena are less than 50 years old. 7

Ο ΒΑΣΙΚΟΣ ΜΟΝΟΚΟΧΛΙΟΣ ΕΚΒΟΛΕΑΣ 8

Advantages of Single Screw: Low Cost Straightforward Design Reliability Disadvantages of Single Screw: Mixing is not very good (for some applications) 9

ΘΕΡΜΑΝΣΗ ΚΑΙ ΨΥΞΗ Heating Bring to startup temperature Maintain desired temperatures Cooling Water or Air Cooled To shutdown an extruder quickly To cool down when the polymer overheats To keep from bridging in the feed throat To keep from melting in the grooved feed 10

ΕΝΔΟ-ΚΟΧΛΙΑ ΘΕΡΜΑΝΣΗ ΚΑΙ ΨΥΞΗ Cartridge Heaters to heat from both sides Fluid Heating and Cooling to control melt temperature to prevent melting in the feed zone to increase pressure generation in feed 11

ΕΠΙΠΛΕΟΝ ΕΞΟΠΛΙΣΜΟΣ ΣΥΣΤΗΜΑΤΑ ΤΡΟΦΟΔΟΣΙΑΣ Gravimetric versus RPM-based Type of hopper ΠΙΝΑΚΑΣ ΕΛΕΓΧΟΥ ΠΑΡΑΚΟΛΟΥΘΗΣΗ ΛΕΙΤΟΥΡΓΙΑΣ ΑΝΤΛΙΕΣ (GEAR PUMPS) ΣΥΣΤΗΜΑΤΑ ΜΕΤΑΔΟΣΗΣ ΚΙΝΗΣΗΣ 12

ΕΠΙΠΕΔΕΣ ΚΕΦΑΛΕΣ ΕΚΒΟΛΗΣ (FLAT EXTRUSION DIES) 13

ΣΠΙΡΑΛ ΚΕΦΑΛΕΣ ΕΚΒΟΛΗΣ (SPIRAL EXTRUSION DIES) 14

ΛΕΙΤΟΥΡΓΙΚΑ ΧΑΡΑΚΤΗΡΙΣΤΙΚΑ ΤΟΥ ΚΟΧΛΙΑ Ζώνη τροφοδοσίας Ζώνη Συμπίεσης (τήξη) Ζώνη Άντλησης τήγματος L/D Ratio Flighted Length Outer Diameter of Screw Compression Ratio Feed Depth Metering Depth 15

ΓΕΩΜΕΤΡΙΚΑ ΧΑΡΑΚΤΗΡΙΣΤΙΚΑ ΤΟΥ ΚΟΧΛΙΑ Βάθος καναλιού γωνία ελίκωσης 16

Η ΖΩΝΗ ΜΕΤΑΦΟΡΑΣ ΣΤΕΡΕΩΝ (ΖΩΝΗ ΤΡΟΦΟΔΟΣΙΑΣ Solids Conveying Zone) How the solid pellets convey???? Barrels: rough surface (sometimes intentionally grooved) Screws: smooth (polished) surface 17

Darnell and Mol (1956) developed and isothermal model that relates the mass flow rate to the ratio of outlet to inlet pressure: M s HWpv H P fs 2H k ln 1 f z P f W b o b b sin sin arcsin 1 1 s 2 2 2 f k fsk 1 f 2 s F r =W*dz*P*f s z L sin tan( ) 1afs W2H z Pz Poexp fb f 2 1 12 s a W H 18

Mass Flow Rate of the solid bed as a function of the ratio f s /f b : Ms[kg/hr] 4500 4000 3500 3000 2500 2000 1500 1000 500 0 0.000 0.111 0.250 0.429 0.667 fs/fb Ο ρυθμός μεταφοράς των στερεών σε σχέση με το λόγο fs/fb. Max (M) when f s is small and f b is large 19

ΗΖΩΝΗΤΗΞΕΩΣ (MELTING ZONE) 20

21

22

Barrier SCREWS 23

BASIC EXTRUDER ANALYSIS DESIGN OF A MELT SCREW PUMP so somehow we must generate pressure from stress (from the fluid viscosity). In drag flow we have no pressure generation: 24

THIS DEVICE IS A PUMP BECAUSE IT GENERATES PRESSURE!!! Let s see how this principle can be put into practical use.by a (conceptional design): shallow channel of finite length covered by an infinite moving plate..but this does not represent a practical solution!! 25

Now, let s do what Einstein called a Gedanken (i.e. thought) construction of a practical device. twist and turn the shallow channel convert infinite plate into a barrel 26

27

Now, let s (conceptually) unwind the channel, and turn it into.a CHANNEL between two flat plates (assume the screw is stationary and THE BARREL ROTATES): The barrel moves with V b =πdn where N rotational speed of screw (e.g. RPM) and z the downchannel direction. The down channel velocity component is: V bz =V b cosθ=πdncosθ and: L=z cosθ Recall the FLAT PLATE EQUATIONS for drag flow with an opposing pressure flow: Q VHW 2 3 H dp 12 dz 28

Use the helical geometry of the channel: Q N = revs per second (rpm/60) of screw 1 2 D 2 2 HN sin cos 3 DH 12 sin 2 P L 29

If we take into account the leakage flow rate from the small clearance (δ) between the barrel and the screw: Q L 2 2 3 D 12e P tan L in our analysis we neglect this term ~ 0 Q 1 2 D 2 2 3 DH HN sin cos 12 sin 2 P L 2 2 3 D 12e P tan L NOTE: 1. If there is no pressure build up (e.g. no constriction of flow at the end of the extruder), the output would be maximum, i.e. drag flow only: 1 2 2 Q max D HN sin cos 2 2. If the end is closed, Q=0 and we may equate drag and pressure flow which gives the MAXIMUM POSSIBLE PRESSURE: 3 1 2 2 DH 2 P 6DLN D HN sin cos sin P max 2 2 12 L tan μέγιστη παροχή μέγιστη πτώση πίεσης Since μ is large for polymer melts, extremely large (AND VERY DANGEROUS!!!) pressures can develop. 30

MATCHING OF DIE AND SCREW DESIGN CHARACTERISTICS We have seen that for Newtonian fluids Q versus ΔP are linear for both extruder (pressure build up) and die (pressure drop). The equation can be plotted as follow: ΣΗΜΕΙΟ ΛΕΙΤΟΥΡΓΙΑΣ ΕΚΒΟΛΕΑ 31

For the extruder: 1 2 2 Q max D HN sin cos 2 P max 2 6DLN tan Careful..!! L is the length of the MELTING ZONE ONLY! L For the DIE (κεφαλή) the pressure drop vs flow rate can be obtained by the usual equations: P mh P 2mR ( 2n1 ) L F 1 1 2 n Q W ( 3n1 ) Q LC 1 3 n n n z 32 2H y r x

All the previous are straightforward for NEWTONIAN FLUID ΙΝ ΤΗΕ EXTRUDER. But what happens if the fluid behaves in a NON NEWTONIAN manner??? e.g. POWER LAW mode there is not a simple closed form solution to solve problems of combined pressure and drag flow! So we will use the.equivalent VISCOSITY (ισοδύναμο ιξώδες) ( back of the envelope calculations): ref velocity gap V H 2RN H m n1 e.g. ref velocity gap V H 2 100mm 2mm 80 60 s 1... s 1 ref 734Pa s 0. 5 15000 ref 0. 5 The above APPROXIMATIONS apply ONLY IN THE EXTRUDER!!! 33

Power Requirement (κατανάλωση ισχύος εκβολέα) of the extruder: The power required by the turning screw is needed to: 1. Raise the temperature from room temperature to extrusion temperature in the die (room temperature~20 o C die temp ~ 200 o C). 2. Melt the polymer (heat of fusion). m Q ή( kg / hr ) 3. Pump the molten polymer. P o QC p T T QH pq out in Ισχύς= αύξηση Τ + τήξη + άντληση f Συνεισφορά τριών όρων! Motor power (ισχύς κινητήρα)p m : e: motor EFFICIENCY P 1 e m P o P o 21. 37kW e 0.7 34

And.some calculations 1. Torque (ροπή) with power calculated from previous equation: P o F V F P / o V T o F R T o P V o R T o PO N ( D 2H ) 60 ( D 2H ) 2 T o PO N 2 60 2. Power and Torque from.shear stress (από διατμητική τάση): P o F V P o wall Area V screw wall m n Area ( D 2H ) V screw L s N ( D 2H ) 60 L sin P o... L s 35

36

37

38