Electrospinning Process Model - Lab | MATE 460, Lab Reports of Materials science

Download Electrospinning Process Model - Lab | MATE 460 and more Lab Reports Materials science in PDF only on Docsity! Engineering Computational Laboratory Fall 2000 ELECTROSPINNING PROCESS MODEL Terry Shetter Delia Garcia Department of Materials Engineering Drexel University ABSTRACT The emerging technology of electrospinning has found the interest of many because of the breakdown that occurs during processing. In this paper, a model has been used to describe this behavior and the major factors that affect it have been varied. The factors that most affect this process are viscosity, surface tension and electric current. INTRODUCTION Polymer breakdown in electrospinning has attracted much attention. The problem is attributed to electrohydrodynamics in the capillary tube. In this paper, an analytical model of steady state electrospinning in a single jet regime is invoked and the effects of secondary factors are studied. This model was chosen to supplement a senior design project so that the student would better understand what factors affected the electrospinning process. Included is a brief description of the background, problem, and discussion about the results that were attained by using a differential equation for the jet radius. BACKGROUND INFORMATION Thin polymer fibers can be made by electrospinning. This process involves ejecting a charged polymer solution from a capillary tube and applying an external electric field to elongate and accelerate the polymer. The polymer is deposited on a substrate and dried and/or can be chemically treated to become a thin fiber. Electrospinning is an emerging technology and is not understood well. The parameters that control the process are hydrostatic pressure in the capillary tube and the external electric field. The viscosity, conductivity, dielectric permeability, and surface tension of the polymer solution also affect the process. Rayleigh instabilities can describe the breakdown of the polymer solutions. However, this is not typically expected in electrospinning because polymers are highly viscous and exhibit non-rheological behavior. STATEMENT OF THE PROBLEM An asymptotic model of the electrospinning process is considered to evaluate and control the diameters of the polymer fibers. Engineering Computational Laboratory Fall 2000 NUMERICAL SIMULATION Equations taking into consideration the steady state flow of an infinite viscous jet, linear momentum balance, and electric stress tensor (three-dimensional) were reduced to a one- dimensional problem by averaging physical quantities over the jet cross-section. The following differential equation is used to model the jet radius:
 d dz R N R N R N dR dzw E R m
          





4 1 1 2 1 2 1 In this equation, Rbar = R/Ro is the dimensionless jet radius, Ro is the normalization constant, zbar = z/zo is the dimensionless axial coordinate and zo is the normalization length. The dimensionless Weber number Nw describes the ration of inertia forces to surface tension in the jet: Nw =
Q2/22 Ro 3S The dimensionless parameter NE, reciprocal of the Euler number, describes the ration of inertial forces to electrostatic field pressure: NE = 4o
Q4/2 J2 Ro 6 The effective Reynolds number NR for the fluid characterized by the power-law describes the ration of inertia forces to viscous forces: NR = Q2
/22 Ro 4 [6EJ Ro 2/ Q2
]m where:
: fluid density S: coefficient of surface tension E : electric field o: permittivity of vacuum Q: volumetric flow rate J : electric current The constants in these equations are:  = 2.3 x 103 cP S= 50 mN/m Ro = 45 m E = 40 kV/m J = 95 nA Q = 15 l/s The boundary condition is Rbarzbar = 0 = 1. The method of solution used for this analysis is Euler Backward. This method for the analysis was chosen because the differential equation for the jet radius is a non-linear equation. For the Euler Backward method, the generalized equation Engineering Computational Laboratory Fall 2000 original parameters 0 0.2 0.4 0.6 0.8 1 1.2 0.00E+00 5.00E+05 1.00E+06 1.50E+06 2.00E+06 2.50E+06 3.00E+06 3.50E+06 4.00E+06 4.50E+06 z Figure 1. The original parameters in the mathematical model. Engineering Computational Laboratory Fall 2000 Excel w=-10 0 0.2 0.4 0.6 0.8 1 1.2 0 200 400 600 800 1000 1200 z r Nr=2500 0 0.2 0.4 0.6 0.8 1 1.2 0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03 5.00E+03 6.00E+03 z r Engineering Computational Laboratory Fall 2000 Nw=0.1 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 200 250 300 350 z r Nw=1000 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 200 250 300 350 z r Engineering Computational Laboratory Fall 2000 Nr=2200, r=0.1 0.00E+00 2.00E-02 4.00E-02 6.00E-02 8.00E-02 1.00E-01 1.20E-01 0.00E+00 5.00E+04 1.00E+05 1.50E+05 2.00E+05 2.50E+05 3.00E+05 3.50E+05 4.00E+05 z r Nr=2500, r=0.1 0 0.02 0.04 0.06 0.08 0.1 0.12 0 50 100 150 200 250 300 350 z r Engineering Computational Laboratory Fall 2000 z=10 0 0.02 0.04 0.06 0.08 0.1 0.12 0 50 100 150 200 250 300 350 z r Engineering Computational Laboratory Fall 2000 Maple > restart; > diff(R(z)^(-4)+(Nw*R(z))^(-1)-Ne^(-1)*R(z)^2-Nr^(- 1)*(diff(R(z),z)^(-2))^(-1/2),z)-1; > solve(%,diff(R(z),z$2)); > EQ1:=diff(R(z),z$2)=%; > subs(diff(R(z),z)=W2,rhs(EQ1)); > AAA:=subs(R(z)=R2,%); > EQA:=R2-R1=dz*W2; EQB:=W2-W1=dz*AAA; > R2_NEW:=solve(EQA,R2); > subs(R2=%,EQB); > EQ_NEW:=%; > W1:=-1000.: R1:=.1:dz:=10.^(-7):Nr:=0.1:Nw:=10:Ne:=10: z:=0: for i from 1 to 300 do W2_NEW:=fsolve(EQ_NEW, W2); R1:= evalf(subs(W2=W2_NEW, R2_NEW)); W1:= evalf(W2_NEW); z:=evalf(z)+dz; dz1:=abs(0.1*W1/subs(W2=W2_NEW, R2=R1,AAA)); dz2:=abs(0.1*R1/W1); dz:=min(dz1,dz2); # if (abs(W1)<10^(-10)) then dz:=dz2; else dz:=dz1; fi; # dz:=min(dz, 10); print(z,R1,W1); od: