Fiber spinning and draw resonance: theoretical background

Transcription

1 Fiber spinning and draw resonance: theoretical background dr. ir. Martien Hulsen March 17, Introduction The fiber spinning process is nicely described in the book by Dantig & Tucker ([1], Chapter 5, Exercise 12), as follows: The fiber spinning process is used to make all types of synthetic textile fibers (nylon, polyester, rayon, etc.). In the melt spinning version of the process, molten polymer is extruded through a die called a spinneret to create a thin cylinder. Far away from the spinneret, the fiber is wrapped around a drum, which pulls it away at a pre-determined take-off speed. The take-off speed is much higher than the extrusion speed, so the filament is stretched considerably in length, and decreases in diameter, between the spinneret and the take-up point. The filament also cools in between these two points, so that it is solid by the take-up point. In the actual process a bundle of many filaments are extruded and stretched together, but for analysis purposes we will consider a single filament. A 0 v 0 A() v() v D 0 r L The polymer swells slightly as it exits the die, and one usually starts the analysis from the point of maximum swell. Let be the distance along the fiber measured from this point, and let A() be the cross-sectional area of the filament (A = π R 2, where R is the radius). The total length of the spinline is L. The area A 0 and speed v 0 at the spinneret are specified, and the speed at the take-up point, v 0 D r, is also specified. D r, called the draw ratio, is an important operating parameter of the process. Because the filament is thin, and because it only sees small shear stresses from air drag, one can assume that the velocity v is independent of radial co-ordinate r. First we will analye the steady state process and want to find A() and v(). However it is known from the literature (see for example the paper of Denn [2]) that the filament becomes unstable for small disturbances for some critical value of D r. This phenomenon is called draw resonance. We will try to predict the critical value of D r by numerical methods. In particular we are interested in the influence of inertia since the speeds of commercial spinlines are quite high (up to 3000 m/min). 1

2 2 A model Our model for the spin line will be based on a one-dimensional approximation, which is valid for slender axisymmetric fibers, i.e. the diameter of the fiber is much smaller than the length: R/L 1. This approximation can be obtained by perturbation analysis. The same equations are obtained by assuming that the velocity v is constant in a crosssection of the fiber and thus only a function of and time t: v (, t). Note that we want to derive the unsteady equations and have introduced a dependence on time t. This is also necessary for the cross-sectional area: A(, t) (=π R 2 (, t)). We assume that we have isothermal conditions, that the fluid is Newtonian and the viscosity µ is constant up to the take-up 1. Furthermore we assume that the polymer has a constant density ρ and that air drag and gravity will be neglected. The equations we need for modeling the flow are the mass balance, the momentum balance and the constitutive equation (see also the book by Dantig & Tucker [1], in particular Chapter 2, Exercise 5 and Chapter 5, Exercise 12). R() v () By applying the integral form of the continuity equation (Eq. (2.32) of [1]) to the fixed control volume in the figure, we find in the limit 0: or (ρ A) (ρ Av ) = 0 (1) (Av ) = 0 (2) since we have assumed that ρ is a constant. By using the result of Example in Dantig & Tucker [1] we find that due to the continuity equation we have in the fiber v r = r v 2 (3) Note that v / does not depend on r. From Appendix B.3 of [1] we find for the non-ero components of D in the fiber D rr = 1 v 2 D θθ = 1 v 2 D = v (4) (5) (6) 1 These assumptions are invalid for commercial fiber spinning of melts, where non-isothermal conditions and viscoelasticity have a huge effect on the critical D r for draw resonance. However for the dynamically similar process of extrusion coating the assumptions are much better (see Denn [2]) 2

3 This is a uniaxial elongational flow having an stretch rate of v /. Using the Newtonian fluid model (see [1]) we find for the non-ero components of the stress-tensor σ in the fiber σ rr = p µ v σ θθ = p µ v σ = p + 2µ v where p is the pressure. Using the boundary condition σ rr = 0 we get (7) (8) (9) p = µ v (10) σ θθ = 0 (11) σ = 3µ v leaving σ as the only non-ero stress component. By applying the -component of the integral form of the momentum balance (Eq. (2.61) of [1]) to the fixed control volume in the figure, we find in the limit 0: or with Eq. (12) or by using Eq. (1) (ρ Av ) (ρ Av ) ρ A( v (12) (ρ Av2 ) = (Aσ ) (13) (ρ Av2 ) = (3Aµ v ) (14) + v v ) = (3Aµ v ) (15) Note, that Eq. (13) is still quite general with respect to the choice of the constitutive equation. Eq. (14) (and also Eq. (15)) is applicable for generalied Newtonian fluids when needed. The equations derived need to be supplemented with boundary and initial conditions. According to Denn [2] the following boundary conditions are appropriate A( = 0, t) = A 0 (16) v ( = 0, t) = v 0 (17) v ( = L, t) = v 0 D r (18) Note that for unsteady flows A( = L, t) is allowed to vary. In addition to the boundary conditions we need to give initial conditions for A and v on the interval (0, L). In summary we have derived two partial differential equations Eqs. (2) and (14) (or (15)) for two functions: A(, t) and v (, t) on the interval (0, L). These equations need to fulfill the boundary conditions Eqs. (16) (18) and initial conditions for A and v. 3

4 3 Scaling We apply the following scaling: = L v = v v 0 A = A A 0 t = t L v 0 and find from Eq. (13) and (14) where (A v ) = 0 (19) ( (A v Re ) (A v 2 = 3 (A v ) (20) Re = ρv 0L µ is the Reynolds number. An alternative for Eq. (20) is (using (15)): (21) ReA ( v v ) v = 3 (A v ) (22) The boundary conditions Eqs. (16) (18) scale like A ( = 0, t ) = 1 (23) v ( = 0, t ) = 1 (24) v ( = 1, t ) = D r (25) In the following we will drop the superscript () where possible. 4 Steady state For the steady state solution we can derive an analytical solution. We use the dimensionless form Eqs. (19) (20) with ()/ = 0: d (Av ) = 0 (26) d ( Re Av 2 3A dv ) = 0 (27) where we have dropped the superscript (). This means that Av = 1 (constant flow rate) (28) Re Av 2 3A dv = C 1 (constant momentum flux) (29) 4

5 where C 1 is a constant, still to be determined. We have already used Eqs. (23) and (24) to determine that the constant flow rate is 1 in the dimensionless form. After some basic calculus we finally find that v = C 1 (C 1 + Re) exp( C 1 /3) Re where we already used the boundary condition Eq. (24). The constant C 1 still needs to be determined using the boundary condition at = 1, Eq. (25). For Re = 0 (no inertia) the analytical solution is easy: and also (30) C 1 = 3 ln D r (31) A = Dr (32) v = Dr (33) 3A dv = Aσ = C 1 (34) or: the force in the fiber is a constant. 5 Linear stability To assess the stability of the steady state solution we want to know whether this solution is stable for (infinitesimal) small perturbations. This is called a linear stability analysis. A nice introduction to this field of scientific research with an emphasis on flow problems is Drain [3]. The unsteady equations are or after some rearranging v (Av ) = 0 (35) ( v v ) ReA + v = 3 (A v ) (36) = v v + 3 Re = v A v 1 A v + 3 Re (37) 2 v 2 (38) To determine the linear stability we split the unsteady solution A and v as follows A(, t) = Ā() + A (, t) (39) v (, t) = v () + v (, t) (40) where Ā(), v () are the steady state solutions of A and v, respectively. Since we do a linear stability analysis we assume that A and v are small. After substitution of Eqs. (39) and (40) into Eqs. (37) and (38) and using that a. Ā() and v () fulfill the steady state equations b. for a linear analysis higher-order terms (like v / A ) can be neglected 5

6 we find that = a 1 A + a 2 + a 3v + a v 4 v = b 1 A + b 2 + b 3v + b 4 The coefficients a i and b i are functions of : v + b 5 (41) 2 v 2 (42) a 1 = d v (43) a 2 = v (44) a 3 = d Ā (45) a 4 = Ā (46) b 1 = 3 1 d Ā Re Ā 2 b 2 = 3 1 d v Re Ā d v b 3 = d v b 4 = v d Ā Re Ā b 5 = 3 Re The boundary conditions can easily be derived from the boundary conditions for A and v (Eqs. (23) (25)): (47) (48) (49) (50) (51) A ( = 0, t) = 0 (52) v ( = 0, t) = 0 (53) v ( = 1, t) = 0 (54) Note that we have a time-dependent problem now that also requires an initial condition. In linear stability analysis it is usual to seek for solutions of the following form (Drain [3]): A = e λ j t  j () (55) j=1 v = e λ j t ˆv j () (56) j=1 which is called a superposition of normal modes and is basically a separation of variables technique. Note, that λ j,  j, ˆv j are possibly complex valued and that in Eq. (55) and (56) we should have taken the real part of the right-hand side, for example A = 1 2 ( e λ j t  j () + e λ j t  j ()) (57) j=1 6

7 where () is the complex conjugate. However, the solutions for λ j,  j and ˆv j are always either real or come in complex pairs and therefore it is easier to work with (55) and (56) and combine the complex pairs afterwards to a real valued solution. If we substitute Eqs. (55) and (56) into the linearied equations (41) and (42) and considering each j independently we find that λ j,  j and ˆv j, j = 1,..., are the solution of the following system of equations in the unknowns A and v 2 λa = a 1 A + a 2 + a v 3v + a 4 (58) λv = b 1 A + b 2 + b v 3v + b 4 + b 2 v 5 2 (59) excluding the trivial solution A = v = 0. This is called an eigenvalue problem. The solutions λ j, j = 1,..., are called the eigenvalues and  j, ˆv j the corresponding eigenfunctions. The steady state solution Â, ˆv is stable if Re(λ j ) < 0 for all eigenvalues and unstable if Re(λ j ) > 0 for at least one eigenvalue, because that mode will grow in time like e λ j t = e (Re(λ j )+iim(λ j ))t = e Re(λ j )t ( cos(im(λ j )t) + i sin(im(λ j )t) ) (60) which will grow exponentially in time if Re(λ j ) > 0. In order to determine the linear stability of the steady state fiber spinning solution we need to find the eigenvalues λ j and determine the sign of Re(λ j ). 6 Numerical approach The eigenvalue problem given by Eqs. (58) and (59) cannot be solved by analytical means. Therefore, we have to use a numerical method. We will use the following approach: a. define the values of of the unknowns functions A and v in Eqs. (58) and (59) in a finite number of points on the interval [0, 1]. b. assume that the equations are valid in these discrete points and use finite difference formulas for the derivatives. c. solve the resulting matrix eigenvalue problem by a standard library routine from LAPACK or NAG. If the number of points is increased it can be expected that the solution to the discretied system converges to the solution of the continuous eigenvalue system. We will use a uniform grid i = ih, i = 0, 1,..., n with the grid sie h = 1/n. In each point i we have two unknowns A i and v i. Since A 0 = v 0 = v n = 0 due to the boundary conditions, we will end up with m = 2n 1 degrees of freedom in the matrix system. 2 Note that we use A and v for the unknown functions of the system, but these are different from A and v functions in Eqs. (37) (38). 7

8 For the difference formulas we will use central differencing where possible: a du ( i) a( i ) u( i+1) u( i 1 ) 2h a d2 u 2 ( i) a( i ) u( i+1 ) u( i ) h = a( i ) u i+1 u i 1 2h u( i) u( i 1 ) h h (61) = a( i ) u i+1 2u i + u i 1 h 2 (62) Both approximations lead to global errors that are of second-order, i.e. O(h 2 ). At the end point n we cannot use central differencing, because of the lack of a point outside of the interval [0, 1]. Therefore, we apply one-sided differencing (or upwind) at the last point: a du ( n) a( n ) u n u n 1 (63) h The local error is O(h 2 ). However we make this error only once, and it therefore matches nicely with the global error of the central difference schemes for the interior points. The discretied eigenvalue system can now be written as λũ = S ũ or (S λī)ũ = 0 (64) for ũ = 0. Here ũ is the vector of unknowns of length m and S the system matrix of sie m m. Problems 1. Verify Eqs. (1) (15). 2. Explain that the boundary conditions Eqs. (16) (18) represent the physical process of fiber spinning. 3. How many dimensionless numbers determine the fiber spinning process in the model presented here? 4. Explain why in early experiments the onset of draw resonance was found to be only dependent on the draw ration D r and to be independent of the length L and the flow rate A 0 v Verify the steady state solution Eq. (30). Hint: use dx (ax + b)(px + q) = 1 px + q ln bp aq ax + b 6. Is the force in the fiber in the steady state solution for Re = 0 also constant as in the case Re = Verify the linearied system Eqs. (41) (51). Why can terms like v / A be neglected? 8

9 8. An alternative approach to studying the linear stability is discretiing the system Eqs. (41) (42) directly and integrate the equations forward in time. How would you choose the initial conditions for A and v? How can an instability be detected? 9. How can the limiting case Re = 0 (or Re 1) be handled? References [1] J.A. Dantig and C.L. Tucker. Modeling in Materials Processing. Cambridge University Press, Cambridge, [2] M.M. Denn. Fiber spinning. In J.R.A. Pearson and S.M. Richardson, editors, Computational Analysis of Polymer Processing, London, Applied Science Publishers. [3] P.G. Drain. Introduction to Hydrodynamics Stability. Cambridge University Press, Cambridge,