Báo cáo hóa học: " Research Article Inequalities for Single Crystal Tube Growth by Edge-Defined Film-Fed Growth Technique" ppt

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2009, Article ID 732106, 28 pages doi:10.1155/2009/732106 Research Article Inequalities for Single Crystal Tube Growth by Edge-Defined Film-Fed Growth Technique Stefan Balint 1 and Agneta M. Balint 2 1 Department of Computer Science, Faculty of Mathematics and Computer Science, West University of Timisoara, Blv. V.Parvan 4, 300223 Timisoara, Romania 2 Faculty of Physics, West University of Timisoara, Blv. V.Parvan 4, 300223 Timisoara, Romania Correspondence should be addressed to Agneta M. Balint, balint@physics.uvt.ro Received 3 January 2009; Accepted 29 March 2009 Recommended by Yong Zhou The axi-symmetric Young-Laplace differential equation is analyzed. Solutions of this equation can describe the outer or inner free surface of a static meniscus the static liquid bridge free surface between the shaper and the crystal surface occurring in single crystal tube growth. The analysis concerns the dependence of solutions of the equation on a parameter p which represents the controllable part of the pressure difference across the free surface. Inequalities are established for p which are necessary or sufficient conditions for the existence of solutions which represent a stable and convex outer or inner free surfaces of a static meniscus. The analysis is numerically illustrated for the static menisci occurring in silicon tube growth by edge-defined film-fed growth EFGs technique. This kind of inequalities permits the adequate choice of the process parameter p.With this aim this study was undertaken. Copyright q 2009 S. Balint and A. M. Balint. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction The first successful Si tube growth was reported in 1. Also a theory of tube growth by E.F.G. process is developed there to show the dependence of the tube wall thickness on the growth variables. The theory concerns the calculation of the shape of the liquid-vapor interface i.e., the free surface of the meniscus and of the heat flow in the system. The inner and the outer free surface shapes of the meniscus Figure 1 were calculated from Young-Laplace capillary equation, in which the pressure difference Δp across a point on the free surface is considered to be Δp  ρ · g · H eff  constant, where H eff represents the effective height of the growth interface Figure 1. The above approximation of Δp is valid when H eff  h, where h is the height of the growth interface above the shaper top. Another approximation used in 1 is that the outer and inner free surface shapes are approximated by circular segments. With these relatively tight tolerances concerning the menisci in conjunction with the heat 2 Journal of Inequalities and Applications flow calculation in the system, the predictive model developed in 1 has been shown to be useful tool understanding the feasible limits of the wall thickness control. A more accurate predictive model would require an increase of the acceptable tolerance range introduced by approximation. The growth process was scaled up by Kaljes et al. in 2 to grow 15 cm diameter silicon tubes. It has been realized that theoretical investigations are necessary for the improvement of the technology. Since the growth system consists of a small die type 1mm width and athintubeorder of 200 μm wall thickness, the width of the melt/solid interface and the meniscus are accordingly very small. Therefore, it is essential to obtain accurate solution for the free surface of the meniscus, the temperature, and the liquid-crystal interface position in this tinny region. For single crystal tube growth by edge-defined film-fed growth E.F.G. technique, in hydrostatic approximation the free surface of a static meniscus is described by the Young- Laplace capillary equation 3: γ ·  1 R 1  1 R 2   ρ · g · z − p. 1.1 Here γ is the melt surface tension, ρ denotes the melt density, g is the gravity acceleration, 1/R 1 , 1/R 2 denote the main normal curvatures of the free surface at a point M of the free surface, z is the coordinate of M with respect to the Oz axis, directed vertically upwards, and p is the pressure difference across the free surface. For the outer free surface, p  p e  p m − p e g − ρ · g · H and for the inner free surface, p  p i  p m − p i g − ρ · g · H. Here p m denotes the hydrodynamic pressure in the meniscus melt, p e g ,p i g denote the pressure of the gas flow introduced in the furnace in order to release the heat from the outer and inner walls of the tube, respectively, and H denotes the melt column height between the horizontal crucible melt level and the shaper top level. When the shaper top level is above the crucible melt level, then H>0, and when the crucible melt level is above the shaper top level, then H<0 see Figure 1. To calculate the outer and inner free surface shapes of the static meniscus, it is convenient to employ the Young-Laplace 1.1 in its differential form. This form of the 1.1 can be obtained as a necessary condition for the minimum of the free energy of the melt column 3. For the growth of a single crystal tube of inner radius r i ∈ R gi , R gi  R ge /2 and outer radius r e ∈ R gi  R ge /2,R ge  the axi-symmetric differential equation of the outer free surface is given by z   ρ · g ·z −p e γ  1   z   2  3/2 − 1 r ·  1   z   2  · z  for r ∈  r e ,R ge  , 1.2 which is the Euler equation for the energy functional I e  z    R ge r e  γ ·  1   z   2  1/2  1 2 · ρ · g · z 2 − p e · z  · r · dr, z  r e   h e > 0,z  R ge   0. 1.3 Journal of Inequalities and Applications 3 z Inner gas flow Outer gas flow r i r e Tube Outer free surface Inner free surface Meniscus melt Shaper Capillary channel Crucible melt R gi R ge H H eff r h i h e 0 z i r z e r α c α c α g α g Figure 1: Axisymmetric meniscus geometry in the tube growth by E.F.G. method. The axi-symmetric differential equation of the inner free surface is given by z   ρ · g ·z −p i γ  1   z   2  3/2 − 1 r ·  1   z   2  · z  for r ∈  R gi ,r i  , 1.4 which is the Euler equation for the energy functional I i  z    r i R gi  γ ·  1   z   2  1/2  1 2 · ρ · g · z 2 − p i · z  · r · dr, z  R gi   0,z  r i   h i > 0. 1.5 The state of the arts at the time 1993-1994, concerning the dependence of the shape of the meniscus free surface on the pressure difference p for small and large bond numbers, in the case of the growth of single crystal rods by E.F.G. technique, are summarized in 4. According to 4, for the general differential equation 1.2, 1.4 describing the free surface of a liquid meniscus, there are no complete analysis and solution. For the general equation only numerical integrations were carried out for a number of process parameter values that were of practical interest at the moment. Later, in 2001, Rossolenko shows in 5 that the hydrodynamic factor is too small to be considered in the automated single crystal tube growth. Finally, in 6 the authors present theoretical and numerical study of meniscus dynamics under axi-symmetric and asymmetric configurations. In 6 the meniscus free surface is approximated by an arc of 4 Journal of Inequalities and Applications constant curvature, and a meniscus dynamics model is developed to consider meniscus shape and its dynamics, heat and mass transfer around the die-top and meniscus. Analysis reveals the correlations among tube wall thickness, effective melt height, pull-rate, die-top temperature, and crystal environmental temperature. In the present paper the shape of the inner and outer free surfaces of the static meniscus is analyzed as function of p, the controllable part of the pressure difference across the free surface, and the static stability of the free surfaces is investigated. The novelty with respect to the considerations presented in literature consists in the fact that the free surface is not approximated as in 1, 6, by an arc with constant curvature, and the pressure of the gas flow introduced in the furnace for releasing the heat from the tube wall is taken in consideration. The setting of the thermal conditions is not considered in this paper. 2. Meniscus Outer Free Surface Shape Analysis in the Case of Tube Growth Consider the differential equation z   ρ · g ·z −p e γ  1   z   2  3/2 − 1 r ·  1   z   2  · z  for R gi  R ge 2 ≤ r ≤ R ge , 2.1 and α c ,α g such that 0 <α c <π/2 −α g ,α g ∈ 0,π/2. Definition 2.1. A solution z  zx of the 2.1 describes the outer free surface of a static meniscus on the interval r e ,R ge R gi  R ge /2 <r e <R ge  if possesses the following properties: a z  r e −tanπ/2 −α g , b z  R ge −tan α c , and c zR ge 0andzr is strictly decreasing on r e ,R ge . The described outer free surface is convex on r e ,R ge  if in addition the following inequality holds: d z  r > 0 ∀r ∈ r e ,R ge . Theorem 2.2. If there exists a solution of 2.1, which describes a convex outer free surface of a static meniscus on the closed interval r e ,R ge , then the following inequalities hold: − γ · π/2 −  α c  α g  R ge − r e · cos α c  γ R ge · sin α c ≤ p e ≤−γ · π/2 −  α c  α g  R ge − r e · sin α g  ρ · g ·  R ge − r e  · tan  π 2 − α g   γ r e · cos α g . 2.2 Journal of Inequalities and Applications 5 Proof. Let zr be a solution of 2.1, which describes a convex outer free surface of a static meniscus on the closed interval r e ,R ge  and αr−arctanz  r. The function αr verifies the equation α   r   p e − ρ · g · z  r  γ · 1 cos α  r  − 1 r · tan α  r  2.3 and the boundary conditions α  r e   π 2 − α g ,α  R ge   α c . 2.4 Hence, there exists r  ∈ r e ,R ge  such that the following equality holds: p e  −γ · π/2 −  α c  α g  R ge − r e · cos α  r    ρ · g · z  r    γ r  · sin α  r   . 2.5 Since z  r > 0onr e ,R ge , z  r is strictly increasing and αr−arctan z  r is strictly decreasing on r e ,R ge , therefore the values of αr are in the range α c ,π/2 − α g  and the following inequalities hold: α c ≤ α  r   ≤ π 2 − α g , sin α g ≤ cos α  r   ≤ cos α c , sin α c ≤ sin α  r   ≤ cos α g , −tan  π 2 − α g  ≤ z   r  ≤−tanα c , ρ · g ·  R ge − r  · tan α c ≤ ρ · g · z  r  ≤ ρ · g ·  R ge − r  · tan  π 2 − α g  . 2.6 Equality 2.5 and inequalities 2.6 imply inequalities 2.2. Corollary 2.3. If r e  R ge /n with 1 <n<2 · R ge /R gi  R ge , then inequalities 2.2 become − γ · π/2 −  α c  α g  R ge · n n − 1 · cos α c  γ R ge · sin α c ≤ p e ≤−γ · π/2 −  α c  α g  R ge · n n − 1 · sin α g  ρ · g ·R ge ·  n − 1  n · tan  π 2 − α g   γ R ge · n · cos α g . 2.7 6 Journal of Inequalities and Applications Corollary 2.4. If n → 2 · R ge /R gi  R ge ,thenr e → R gi  R ge /2 and 2.7 becomes − 2 ·γ · π/2 −  α c  α g  R ge − R gi · cos α c  γ R ge · sin α c ≤ p e ≤−2 ·γ · π/2 −  α c  α g  R ge − R gi · sin α g  ρ · g ·  R ge − R gi  2 · tan  π 2 − α g   2 · γ R gi  R ge · cos α g . 2.8 If n → 1,thenr e → R ge and p e →−∞. Theorem 2.5. Let n be such that 1 <n<2 ·R ge /R gi  R ge .Ifp e satisfies the inequality p e < −γ · π/2 −  α c  α g  R ge · n n − 1 · cos α c  γ R ge · sin α c , 2.9 then there exists r e in the closed interval R ge /n, R ge  such that the solution of the initial value problem z   ρ · g ·z −p e γ ·  1   z   2  3/2 − 1 r ·  1 z   2  · z  for R gi  R ge 2 <r≤ R ge z  R ge   0,z   R ge   −tan α c , 0 <α c < π 2 − α g ,α g ∈  0, π 2  2.10 on the interval r e ,R ge  describes the convex outer free surface of a static meniscus. Proof. Consider the solution zr of the initial value problem 2.10. Denote by I the maximal interval on which the function zr exists and by αr the function αr−arctan z  r defined on I. Remark that for αr the equality 2.3 holds. Since z   R ge  > 0,z   R ge   −tan α c < 0,z   R ge  > −tan  π 2 − α g  , 2.11 there exists r  ∈ I,0<r  <R ge such that for any r ∈ r  ,R ge  the following inequalities hold: z   r  > 0,z   r  ≤−tanα c ,z   r  ≥−tan  π 2 − α g  . 2.12 Let r ∗ be defined by r ∗  inf  r  ∈ I | 0 <r  <R ge such that for any r ∈  r  ,R ge  inequalities  2.12  hold  . 2.13 It is clear that r ∗ ≥ 0 and for any r ∈ r ∗ ,R ge  inequalities 2.12 hold. Journal of Inequalities and Applications 7 From 2.12 and 2.13 it follows that z  r is strictly increasing and bounded on r ∗ ,R ge . Therefore z  r ∗  0lim r →r ∗ ,r>r ∗ z  r exists and satisfies −tan  π 2 − α g  ≤ z   r ∗  0  ≤−tanα c . 2.14 Moreover, since zr is strictly decreasing and possesses bounded derivative on r ∗ ,R ge , zr ∗  0lim r →r ∗ ,r>r ∗ zr exists too, it is finite, and satisfies 0 <  R ge − r ∗  · tan α c ≤ z  r ∗  0  ≤  R ge − r ∗  · tan  π 2 − α g  < ∞. 2.15 We will show now that r ∗ >R ge /n and z  r ∗  0−tanπ/2 − α g . In order to show that r ∗ >R ge /n we assume the contrary, that is, that r ∗ ≤ R ge /n . Under this hypothesis we have α  r ∗  0  − α  R ge   −α   r   ·  R ge − r ∗    − p e γ  ρ · g ·z  r   γ  sin α  r   r   · R ge − r ∗ cos α  r   >  π/2 −  α c  α g  R ge · n n − 1 · cos α c − 1 R ge · sin α c  ρ · g ·z  r   γ  sin α  r   r   · R ge − r ∗ cos α  r   > π 2 −  α c  α g  2.16 for some r  ∈ r ∗ ,R ge . Hence αr ∗  0 >π/2 − α g . This last inequality is impossible, since according to the inequality 2.14, we have αr ∗  0 ≤ π/2 − α g . Therefore, r ∗ , defined by 2.14,satisfiesr ∗ >R ge /n. In order to show that z  r ∗  0−tanπ/2 − α g  we remark that from the definition 2.14 of r ∗ it follows that at least one of the following three equalities holds: z   r ∗  0   −tan α c ,z   r ∗  0   −tan  π 2 − α g  ,z   r ∗  0   0. 2.17 Since z  r ∗  0 <z  r ≤−tan α c for any r ∈ r ∗ ,R ge  it follows that the equality z  r ∗  0 −tan α c is impossible. Hence, we obtain that at r ∗ at least one of the following two equalities holds: z  r ∗  00, z  r ∗  0−tanπ/2 −α g .Weshownowthattheequalityz  r ∗  00 is impossible. For that we assume the contrary, that is, z  r ∗  00. Under this hypothesis, 8 Journal of Inequalities and Applications from 2.12 we have: p e  g · ρ · z  r ∗  0   γ r ∗ · sin α  r ∗  0  >g· ρ ·  R ge − r ∗  · tan α c  γ R ge · sin α c > γ R ge · sin α c − γ · π/2 −  α c  α g  R ge · n n − 1 · cos α c >p e 2.18 what is impossible. In this way we obtain that the equality z  r ∗  0−tanπ/2 − α g  holds. For r e  r ∗ the solution of the initial value problem 2.8 on the interval r e ,R ge  describes a convex outer free surface of a static meniscus. Corollary 2.6. If for p e the following inequality holds: p e < −2 · γ · π/2 −  α c  α g  R ge − R gi · cos α c  γ R ge · sin α c , 2.19 then there exists r e in the interval R gi  R ge /2,R ge  such that the solution of the initial value problem 2.10 on the interval r e ,R ge  describes a convex outer free surface of a static meniscus. Corollary 2.7. If for 1 <n  <n<2 · R ge /R gi  R ge  the following inequalities hold: − γ · π/2 −  α c  α g  R ge · n  n  − 1 · sin α g  ρ · g · R ge · n  − 1 n  · tan  π 2 − α g   γ R ge · n  cos α g <p e < −γ · π/2 −  α c  α g  R ge · n n − 1 · cos α c  γ R ge · sin α c , 2.20 then there exists r e in the interval R ge /n, R ge /n   such that the solution of the initial value problem 2.10 on the interval r e ,R ge  describes a convex outer free surface of a static meniscus. The existence of r e and the inequality r e ≥ R ge /n follows from Theorem 2.5. The inequality r e ≤ R ge /n  follows from Corollary 2.3. Remark 2.8. The solution of the initial value problem 2.10 is convex at R ge i.e., z  R ge  > 0 if and only if p e < γ R ge · sin α c . 2.21 Journal of Inequalities and Applications 9 That is because z  R ge  > 0 if and only if α  R ge −z  R ge  · cos 2 α c < 0, that is, p e γ − sin α c R ge < 0 ⇐⇒ p e < γ R ge · sin α c . 2.22 Moreover, if p e <γ/R ge · sin α c ,thesolutionzr of the initial value problem 2.8 is convex everywhere i.e., z  r > 0forr ∈ I, 0 <r≤ R ge . That is because the change of convexity implies the existence of r  ∈ I, 0 <r  <R ge such that αr   >α c ,zr   > 0andp e  ρ ·g ·zr   γ/r  · sin αr   >γ/R ge · sin α c , what is impossible. Theorem 2.9. If a solution z 1  z 1 r of 2.1 describes a convex outer free surface of a static meniscus on the interval r e ,R ge R gi  R ge /2 <r e <R ge , then it is a weak minimum for the energy functional of the melt column: I e  z    R ge r e  γ ·  1   z   2  1/2  1 2 · ρ · g · z 2 − p e · z  · r · dr z  r e   z 1  r e  ,z  R ge   z 1  R ge   0. 2.23 Proof. Since 2.1 is the Euler equation for 2.23,itissufficient to prove that the Legendre and Jacobi conditions are satisfied in this case. Denote by Fr, z, z   the function defined as F  r, z, z    r ·  1 2 · ρ · g · z 2 − p e · z  γ ·  1   z   2  1/2  . 2.24 It is easy to verify that we have ∂ 2 F ∂z 2  r · γ  1   z   2  3/2 > 0. 2.25 Hence, the Legendre condition is satisfied. The Jacobi equation  ∂ 2 F ∂z 2 − d dr  ∂ 2 F ∂z ∂z    · η − d dr  ∂ 2 F ∂z 2 · η    0 2.26 in this case is given by d dr ⎛ ⎜ ⎝ r · γ  1   z   2  3/2 · η  ⎞ ⎟ ⎠ − ρ · g · r ·η  0. 2.27 10 Journal of Inequalities and Applications For 2.27 the following inequalities hold: r · γ  1   z   2  3/2 ≥ r · γ · cos 3  π 2 − α g   r · γ · sin 3 α g , −ρ · g · r ≤ 0. 2.28 Hence, the equation  η  · r · γ · sin 3 α g    0 2.29 is a Sturm type upper bound for 2.277. Since every nonzero solution of 2.29 vanishes at most once on the interval r e ,R ge , the solution ηr of the initial value problem d dr ⎛ ⎜ ⎝ r · γ  1   z   2  3/2 · η  ⎞ ⎟ ⎠ − ρ · g · r ·η  0, η  r e   0,η   r e   1 2.30 has only one zero on the interval r e ,R ge 7. Hence the Jacobi condition is satisfied. Definition 2.10. A solution z  zr of 2.1 which describes the outer free surface of a static meniscus is said to be stable if it is a weak minimum of the energy functional of the melt column. Remark 2.11. Theorem 2.9 shows that if z  zr describes a convex outer free surface of a static meniscus on the interval r e ,R ge , then it is stable. Theorem 2.12. If the solution z  zr of the initial value problem 2.10 is concave (i.e., z”r < 0) on the interval r e ,R ge R gi  R ge /2 <r e <R ge , then it does not describe the outer free surface of a static meniscus on r e ,R ge . Proof. z  r < 0onr e ,R ge  implies that z  r is strictly decreasing on r e ,R ge . Hence z  r e  > z  R ge −tan α c > −tanπ/2 −α g . Theorem 2.13. If p e >γ/R ge ·sin α c and there exists r e ∈ R gi R ge /2,R ge  such that the solution of the initial value problem 2.10 is the outer free surface of a static meniscus on r e ,R ge , then for p e the following inequalities hold: γ R ge · sin α c <p e ≤ ρ · g ·  R ge − r e  · tan  π 2 − α g   γ r e · cos α g . 2.31 Proof. Denote by zr the solution of the initial value problem 2.10 which is assumed to represent the outer free surface of a static meniscus on the closed interval r e ,R ge .Letαr be defined as in Theorem 2.2.forr ∈ r e ,R ge . Since p e >γ/R ge · sin α c , we have z  R ge  −1/cos 2 α c  · α  R ge  < 0. Hence α  R ge  > 0 and therefore αr <αR ge α c for r<R ge , r [...]... National Authority for Research supporting this research under the Grant ID 354 no 7/2007 References 1 L Erris, R W Stormont, T Surek, and A S Taylor, “The growth of silicon tubes by the EFG process,” Journal of Crystal Growth, vol 50, no 1, pp 200–211, 1980 2 J P Kalejs, A A Menna, R W Stormont, and J W Hutchinson, “Stress in thin hollow silicon cylinders grown by the edge-defined film-fed growth technique,”... of Inequalities and Applications Since z Rgi > 0, tan αc > 0, z Rgi there exists r ∈ I, Rgi < r < Rgi inequalities hold: z r > 0, z Rgi < tan π − αg , 2 3.12 Rge /2 such that for any r ∈ r , Rgi the following z r > 0, z r < tan π − αg 2 3.13 Let r ∗ be defined by r ∗ sup r ∈ I | Rgi < r < Rgi Rge such that for any r ∈ Rgi , r 2 inequalities 3.13 hold 3.14 It is clear that r ∗ ≤ Rgi Rge /2 and for. .. of Crystal Growth, vol 104, no 1, pp 14–19, 1990 3 R Finn, Equilibrium Capillary Surfaces, vol 284 of Grundlehren der Mathematischen Wissenschaften, Springer, New York, NY, USA, 1986 4 V A Tatarchenko, Shaped Crystal Growth, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1993 5 S N Rossolenko, “Menisci masses and weights in Stepanov EFG technique: ribbon, rod, tube, ” Journal of Crystal Growth, ... therefore the following inequalities hold: arctan z r is strictly π − αg , 2 sin αg ≤ cos α r ≤ cos αc , αc ≤ α r ≤ sin αc ≤ sin α r ρ · g · r − Rgi · tan αc ≤ ρ · g · z r 3.5 3.6 ≤ cos αg , ≤ ρ · g · r − Rgi · tan Equality 3.5 and inequalities 3.6 imply inequalities 3.2 π − αg 2 Journal of Inequalities and Applications Corollary 3.3 If ri −γ · 13 m · Rgi with 1 < m < Rgi Rge /2 · Rgi , then inequalities. .. 1 0.75 0.5 0.075 0.0751 0.0754 0.0755 r m Figure 5: α versus r for pe −1000; −900; −800 Pa ×10−4 10 8 −1000 Pa pe z m 6 4 pe 2 0 0.075 0.0751 0.0752 −2000 Pa pe −2500 Pa 0.0753 0.0754 0.0755 r m Figure 6: z versus r for pe −2500; −2000; −1000 Pa 4 Numerical Illustration Numerical computations were performed for a Si cylindrical tube for the following data: Rge 75.5 · 10−3 m , ρ 74.5 · 10−3 m , Rgi1... condition For the same numerical data inequality 2.20 becomes −2580 < pe < −1179.443 Pa −2000; −2500 Pa Figures Integration of 2.10 illustrates the above phenomenon for pe 6 and 7 and also the fact that the condition is not necessary see pe −1000 Pa For the same numerical data inequality 2.21 becomes pe < 4.768 Pa Integration of 2.10 illustrates the above phenomenon for pe 2; 3; 4 Pa Figures 8 and 9 For. .. Figures 18 and 19 see pi −500 Pa For the same numerical data inequality 3.23 becomes pi < −4.8322 Pa Integration of 3.10 illustrates the above phenomenon for pi −7; −6; −5 Pa Figures 20 and 21 For the considered numerical data inequality 3.24 becomes −9.664 Pa < pi < −5 Pa there exists ri ∈ Rgi , m · Rgi such −4.827 Pa Integration shows that for pi 26 Journal of Inequalities and Applications 1.75... theoretical predictions This test is not the subject of this paper For the considered numerical data, inequality 2.7 becomes −1179.443 Pa ≤ pe ≤ −194.682 Pa 4.2 Integration of 2.10 shows that for pe −1100; −1000 Pa there exists re ∈ Rge /n, Rge such that the solution is a convex outer free surface of a static meniscus on re , Rge , but for Journal of Inequalities and Applications 21 1.75 1.5 pe 3 Pa 1.25 α... relations hold: − arctan z r and remark that for r ∈ I ∩ Rge /n, Rge the following pe ρ · g · z r sin α r 1 · − − cos α r γ γ r α r ≥ ρ · g · Rge n − 1 n n · tan αc − − Rge γ ·n Rge ρ · g · Rge n − 1 1 · · tan αc cos α r γ ·n ≥ 0 2.35 Hence: z r − 1/cos2 α r · α r ≤ 0 for r ∈ I ∩ Rge /n, Rge 3 Meniscus Inner Free Surface Shape Analysis in the Case of Tube Growth Consider now the differential equation... 0.075 r m Figure 14: z versus r for pi −1000; −900; −550 Pa Moreover, for pe 5; 10; 100 Pa it is not anymore the outer free surface of a static meniscus Figures 10 and 11 Hence, inequality 2.32 is not a sufficient condition For the same numerical data inequality 2.33 becomes 4.768 Pa ≤ pe ≤ 72.5 Pa 5; 10 Pa Figure 10 the solution of 2.10 is not We have already obtained that for pe anymore the outer free . Corporation Journal of Inequalities and Applications Volume 2009, Article ID 732106, 28 pages doi:10.1155/2009/732106 Research Article Inequalities for Single Crystal Tube Growth by Edge-Defined Film-Fed Growth. solution for the free surface of the meniscus, the temperature, and the liquid -crystal interface position in this tinny region. For single crystal tube growth by edge-defined film-fed growth E.F.G.. its differential form. This form of the 1.1 can be obtained as a necessary condition for the minimum of the free energy of the melt column 3. For the growth of a single crystal tube of inner