1. Introduction

When a liquid is applied to a spinning substrate, or if a pre-wetted substrate is spun, centrifugal forces act to drive any irregularities in the film thickness outward, away from the axis of rotation. The result is that the film becomes thinner and more uniform as the rotation proceeds. Consequently spin coating is used in such applications as coating magnetic storage discs, optical devices, and semiconductor wafers to obtain very thin but uniform films on flat substrates.

When the substrate is curved, centrifugal forces will act to produce a uniform layer only in horizontal regions where the substrate is perpendicular to the axis of rotation. But the coating layer may be very irregular in regions where the substrate is highly curved and the normal vector from the surface is not parallel to the axis of rotation.

In this work we will derive the lubrication form of the fluid mechanical equations for a thin liquid film sprayed on an arbitrarily curved, rotating, axisymmetric substrate. Our goal is to predict how the coating thickness changes with time as a function of the substrate geometry, the rotation rate, the rheological properties of the coating liquid, and the geometry and flux of the spray gun. We then discretize the equations and solve the partial differential equations governing the flow. Using an implicit method of solving the finite difference representation of the partial differential equations, we require a minimum of computer resources.

The theory we develop in this work is used to analyze one specific application: the spray/spin coating of the interior of aluminum beverage containers. When a

spinning can is spray painted, centrifugal forces help to cause a more uniform coating layer on the can substrate. Indeed this is the purpose of rotating the can a high spin rates while spray coating the interior. But centrifugal forces can also cause such coating irregularities as drop formation. In this work we will demonstrate how such parameters as the spray gun placement and the rotation rate contribute to how long the beverage can may remain in the spin phase of its coating process before this potential defect occurs.

## 2. Derivation of the evolution equation

Consider an arbitrarily curved, axisymmetric substrate with the parameter s representing arc length along the substrate. The parameter ϕ represents the circumferential angle and n is the distance normal to the substrate (see Figure 1). u is the velocity parallel to the substrate, in the s direction; v is the velocity normal to the substrate, in the n direction; and w the velocity in the circumferential direction ϕ. ω is the rotation rate around the z axis. Note that while the substrate is axisymmetric, the coating need not be. r is the distance from the can centerline to a given point ð Þ s, ϕ on the can substrate. For this geometry, the metric coefficients are listed in Table 1. The Appendix of a text book on fluid mechanics by Kee et al. [1] lists the momentum and continuity equations governing the motion of Newtonian fluids on orthogonal, curved substrates. From here we find that for our geometry the continuity equation is given by Figure 1.

$$\frac{\partial(ru)}{\partial s} + \frac{\partial(rv)}{\partial n} + \frac{\partial(w)}{\partial \phi} = \mathbf{0}.\tag{1}$$

Figure 1.

Curvilinear coordinate system. ϕ is the circumferential angle, eð Þ<sup>s</sup> is the unit vector parallel to the substrate, eð Þ <sup>n</sup> the unit vector perpendicular to the substrate, and r sð Þ is the distance from the centerline to a given point on the substrate. θ is the angle of the substrate with respect to the horizontal, ω is the rotation rate, and g is the acceleration of gravity.


Table 1.

Coordinates and their metric coefficients.

Numerical Simulation of the Spin Coating of the Interior of Metal Beverage Cans DOI: http://dx.doi.org/10.5772/intechopen.90381

If θ is the angle the substrate makes with the horizontal, and the gravity vector points in the negative z direction, and then if the can is set upright during the coating process, ρg sin θ is the force parallel to the substrate, and ρg cos θ is the force perpendicular to the substrate. The centrifugal force points in the r direction due to a rotation about the z axis; consequently ρωrsin θ is the centrifugal force on the coating perpendicular to the substrate and ρωr cos θ the force parallel to the substrate. When the coating layer is sufficiently thin and the Reynolds number is sufficiently small that we can neglect the inertial terms, the momentum equations governing the motion of the liquid coating reduce to

$$\begin{aligned} \frac{\partial p}{\partial t} &= -\rho \text{g} \sin \theta + \rho a^2 r \cos \theta + \mu [2u\_x + v\_m + u\_{nn}] + \frac{\mu}{r} \left[ 2r\_i u\_i + w\_{\phi} \right] + \frac{\mu}{r^2} \left[ w\_{\phi \phi} - w\_{\phi} - 2r\_i (w\_{\phi} + u) \right], \\\\ \frac{\partial p}{\partial t} &= \rho \text{g} \cos \theta + \rho a^2 r \sin \theta + \mu \left\{ \left[ v\_{\mu} + u\_{\mu} + 2r v\_{m} + w\_{n\phi} \right] + \frac{\left[ 2r\_i v\_i + 2r\_i u\_{\mu} + 2w\_{n} - v\_{\phi \phi}/r \right]}{r} \right\}, \\\\ \frac{1}{r} \frac{\partial p}{\partial \phi} &= \mu \left\{ \left[ w\_{\mu} + v\_{\phi \mu} + w\_{\theta} r\_i \right] + \frac{\left[ u\_{\phi \phi} + w\_i + r w\_{m} + 2w\_{\phi \phi} + 2u\_{\phi} r\_i^2 \right]}{r} \right\}. \end{aligned}$$

Here p is the pressure, ρ is the density of the liquid, and μ is the viscosity of the liquid. Meyers et al. [2] found that the Coriolis forces were of the same magnitude as the inertial forces. Because we have neglected the inertial forces on the grounds that the Reynolds number is small, we shall also neglect Coriolis forces in our analysis. We assume that the coating layer obeys the no-slip boundary condition at the surface of the substrate, n ¼ 0

$$u(\mathfrak{s}, n=0, \phi) = v(\mathfrak{s}, n=0, \phi) = w(\mathfrak{s}, n=0, \phi) = \mathbf{0}.\tag{2}$$

If F, a function of s, n, ϕ, and time t, is always zero at the free surface, then it must be given by the coating thickness h minus the value of the coordinate n

$$F(s, n, \phi, t) = h(s, \phi, t) - n = 0. \tag{3}$$

We define eð Þ<sup>s</sup> as the unit vector in the direction parallel to the substrate s, eð Þ <sup>n</sup> as the unit vector in the direction perpendicular to the substrate n, and eð Þ <sup>ϕ</sup> as the unit vector in the circumferential direction ϕ. From Eq. (3) we find that the unit vector perpendicular to the coating surface, denoted by ^f, is defined by

$$\hat{\mathbf{f}} = \frac{\nabla F}{|\nabla F|} = \frac{h\_s \mathbf{e}^{(\mathbf{s})} - \mathbf{e}^{(\mathbf{n})} + \left(h\_{\phi}/r\right)\mathbf{e}^{(\phi)}}{\sqrt{\left(h\_s\right)^2 + 1 + \left(h\_{\phi}/r\right)^2}}.$$

At the free surface, the substantial derivative of F must be zero. Consequently the kinematic condition on the free surface is given by

$$\frac{DF}{Dt} = \mathbf{0} = h\_t - \imath h\_t + \upsilon + w\left(\frac{h\_\phi}{r}\right).$$

Employing the substitution

$$
\mathcal{L} = \sqrt{(h\_s)^2 + \mathbf{1} + \left(\frac{h\_\phi}{r}\right)^2},
$$

the mean curvature of the free surface for this geometry is given by

Methods for Film Synthesis and Coating Procedures

$$\kappa = -\nabla \cdot \hat{\mathbf{f}} + \frac{\mathbf{1}}{R^{(\text{sub})}} = -\frac{r\_s h\_s + r h\_\infty}{r \mathcal{L}} + \frac{h\_s^2 h\_\infty}{\left(\mathcal{L}\right)^3} - \frac{h\_\phi \phi}{r^2 \mathcal{L}} + \frac{h\_\phi^2 h\_{\phi\phi}}{r^4 \left(\mathcal{L}\right)^3} + \frac{\mathbf{1}}{R^{(\text{sub})}},\tag{4}$$

at <sup>n</sup> <sup>¼</sup> <sup>h</sup>. Here <sup>R</sup>ð Þ sub is the radius of curvature of the substrate. Adding the curvature of the free surface relative to the substrate with the curvature of the substrate, given by 1=Rð Þ sub , to obtain the total curvature of the free surface has been shown to be valid by Schwartz and Weidner [3] as long as hs ≪ 1 and hκ ≪ 1.

If the atmospheric pressure is zero, then the tensor equation relating the change in pressure across the free surface due to surface tension is given by

$$\left[\pi\_{\hat{\eta}} + p\delta\_{\hat{\eta}}\right]\hat{f}\_j = -\sigma\kappa\hat{f}\_i. \tag{5}$$

Here σ is the surface tension of the liquid, and τij is the stress tensor in the liquid at the free surface. For our geometry, the components of the stress tensor are given by

$$\begin{aligned} \tau\_{\tt s} &= 2\mu u\_{\tt s}, & \tau\_{\tt m} &= 2\mu v\_{\tt n}, & \tau\_{\tt \phi \phi} &= \frac{2\mu}{r} \left[w\_{\phi} + \mu r\_{\tt s}\right], \\\\ \tau\_{\tt m} &= \tau\_{\tt n} = \mu \left[v\_{\tt s} + u\_{\tt n}\right], & \tau\_{n\phi} &= \tau\_{\phi n} = \mu \left[w\_{n} + \frac{v\_{\phi}}{r}\right], \\\\ \tau\_{s\phi} &= \tau\_{\phi \ast} = \mu \left[\frac{u\_{\phi}}{r} + w\_{\tt s} - \frac{\omega r\_{\tt s}}{r}\right]. \end{aligned}$$

If we sum the three equations found in Eq. (5) over j ¼ s, n, ϕ for each i ¼ s, i ¼ n, and i ¼ ϕ, we have the following three equations:

$$-\sigma \kappa h\_s = p h\_s + \tau\_{ss} h\_s + \tau\_{sn} + \tau\_{s\phi} \frac{h\_{\phi}}{r},\tag{6}$$

$$-\sigma \kappa = p + \tau\_{ns} h\_s + \tau\_{nn} + \tau\_{n\phi} \frac{h\_\phi}{r},\tag{7}$$

$$-\sigma \kappa \frac{h\_{\phi}}{r} = p \frac{h\_{\phi}}{r} + \tau\_{\phi \flat} h\_{s} + \tau\_{\phi \flat \mathfrak{u}} + \tau\_{\phi \flat \phi} \frac{h\_{\phi}}{r},\tag{8}$$

all evaluated at the free surface n ¼ h. Because we have assumed that the pressure in the air is zero, the pressure in the liquid is positive if κ is negative. If we eliminate the pressure p and the curvature of the free surface κ from the three Eqs. (6)–(8), we find the following equations are valid at the free surface n ¼ h:

$$
\tau\_{sn} \left( \mathbf{1} - h\_s^2 \right) + (\tau\_{sr} - \tau\_{nn}) h\_s - \tau\_{n\phi} \frac{h\_s h\_\phi}{r} + \tau\_{s\phi} \frac{h\_\phi}{r} = \mathbf{0},\tag{9}
$$

$$
\tau\_{n\phi} \left( \mathbf{1} - \left[ \frac{h\_{\phi}}{r} \right]^2 \right) + \tau\_{\phi\phi} \frac{h\_{\phi}}{r} + \tau\_{s\phi} h\_s - \tau\_{ss} h\_s \frac{h\_{\phi}}{r} - \tau\_{ns} \frac{h\_s h\_{\phi}}{r} = \mathbf{0}. \tag{10}
$$

We now scale the dependent and independent variables with various characteristic lengths of the substrate geometry. These include h0, a characteristic coating thickness, Rc the maximum radius of the substrate, u<sup>0</sup> a characteristic velocity, and α a small parameter representing the ratio of a typical length perpendicular to the substrate to a typical length parallel to the substrate. Denoting dimensionless variables by bars, we have

Numerical Simulation of the Spin Coating of the Interior of Metal Beverage Cans DOI: http://dx.doi.org/10.5772/intechopen.90381

$$\overline{\phi} = \frac{aR\_{\epsilon}\phi}{h\_0}, \qquad \overline{s} = \frac{as}{h\_0}, \quad \overline{n} = \frac{n}{h\_0}, \quad \overline{w} = \frac{w}{u\_0}, \quad \overline{u} = \frac{u}{u\_0}, \quad \overline{v} = \frac{v}{au\_0},$$

$$\overline{r} = \frac{r}{R\_{\epsilon}}, \quad \overline{t} = \frac{au\_0t}{h\_0}, \qquad \overline{\kappa} = R\_{\epsilon}\kappa, \quad \overline{p} = \frac{ph\_0}{u\_0\mu}, \quad \overline{Q} = \frac{Q}{u\_0h\_0}.$$

If Q is the flux, with units of length cubed over time, then the flux in the s and ϕ directions are defined by

$$
\overline{Q}^{(5)} = \int\_0^{\overline{h}} \overline{u} \, d\overline{n}, \qquad \overline{Q}^{(\overline{\phi})} = \int\_0^{\overline{h}} \overline{w} \, d\overline{n}.
$$

We can eliminate the pressure from the s, n, and ϕ momentum equations using identities for partial differentiation of functions:

$$\begin{split} \left(\overline{p}\_{i}\right)\_{\overline{n}} - \left(\overline{p}\_{\overline{n}}\right)\_{i} = 0 &= \mu \overline{u}\_{\overline{nm}} + a\mu \left(\frac{-\overline{u}\overline{\rho}\_{\overline{\theta}\overline{n}}}{\overline{r}^{2}} - \frac{2\overline{u}\overline{\rho}\overline{r}}{\overline{r}^{2}} + \frac{\overline{w}\_{\overline{n}}}{\overline{r}}\right) \\ &+ a\left[ -\frac{\rho\mathbf{g}\sin\theta\theta\_{\overline{i}}}{R\_{c}} + \rho a^{2}\overline{r}\_{\overline{r}}\sin\theta + \frac{\rho a^{2}\cos\theta\theta\_{\overline{i}}}{R\_{c}} \right] + O\left(a^{2}\right), \end{split} \tag{11}$$

$$\left(\overline{p}\_{\overline{\theta}}\right)\_{\overline{n}} - \left(\overline{p}\_{\overline{n}}\right)\_{\overline{\theta}} = 0 = \mu\overline{w}\_{\overline{nm}}\overline{r} + O\left(a^{2}\right). \tag{12}$$

The scaled form of the continuity equation is given by

$$\frac{\partial}{\partial \overline{\mathfrak{T}}} (\overline{r}\overline{u}) + \frac{\partial}{\partial \overline{n}} (\overline{r}\overline{v}) + \frac{\partial}{\partial \overline{\phi}} (\overline{w}) = 0. \tag{13}$$

At the substrate, n ¼ 0, we have the scaled representation of the no-slip boundary conditions:

$$
\overline{u}\left(\overline{\mathfrak{s}}, \overline{n} = 0, \overline{\mathfrak{\phi}}, \overline{t}\right) = a\overline{w}\left(\overline{\mathfrak{s}}, \overline{n} = 0, \overline{\mathfrak{\phi}}, \overline{t}\right) = \overline{w}\left(\overline{\mathfrak{s}}, \overline{n} = 0, \overline{\mathfrak{\phi}}, \overline{t}\right) = 0. \tag{14}
$$

The scaled form of the kinematic condition is shown to be

$$
\overline{h}\_{\overline{\mathfrak{r}}} = \overline{u}\overline{h}\_{\overline{\mathfrak{r}}} - \overline{v} + \overline{w}\frac{\overline{h}\_{\overline{\phi}}}{\overline{r}},
\tag{15}
$$

at n ¼ h. The boundary conditions, Eq. (9) and Eq. (10), are given by

$$
\overline{u}\_{\overline{\pi}} + \mathcal{O}(a^2) = \mathbf{0}, \qquad \overline{w}\_{\overline{\pi}} + \mathcal{O}(a^2) = \mathbf{0}, \tag{16}
$$

at n ¼ h.

Instead of transforming the equations representing the pressure discontinuity across the liquid interface, given by Eq. (9) and Eq. (10), we will eliminate the pressure term in Eqs. (9) and (10) using the momentum equations and the identities for the partial differentiation of implicit functions. Eq. (7) gives the pressure in the liquid at the free surface <sup>p</sup> <sup>s</sup>, <sup>n</sup> <sup>¼</sup> <sup>h</sup>, <sup>ϕ</sup>, <sup>t</sup> � �. The momentum equations give us pn, ps , and <sup>p</sup><sup>ϕ</sup> in the liquid. The directional derivatives of <sup>p</sup>, along with <sup>n</sup> <sup>¼</sup> <sup>h</sup> <sup>s</sup>, <sup>ϕ</sup>, <sup>t</sup> � �, in the s and ϕ directions, respectively, are

Methods for Film Synthesis and Coating Procedures

$$\begin{aligned} \left[\overline{p}\_{\tilde{\tau}} + \overline{p}\_{\overline{n}} \overline{h}\_{\overline{i}}\right] \vert\_{\overline{n} = \overline{h}} &= \frac{\partial}{\partial \overline{s}} \ \overline{p}(\overline{s}, \overline{n} = \overline{h}, \overline{\phi}, \overline{t}), \\\\ \left[\overline{p}\_{\overline{\phi}} + \overline{p}\_{\overline{n}} \overline{h}\_{\overline{\phi}}\right] \vert\_{\overline{n} = \overline{h}} &= \frac{\partial}{\partial \overline{\phi}} \ \overline{p}(\overline{s}, \overline{n} = \overline{h}, \overline{\phi}, \overline{t}) \ .\end{aligned}$$

Plugging the pressure at the interface, found by Eq. (7), into the momentum equations yields

$$\begin{split} \left. \mu \overline{u}\_{\overline{m}} \right|\_{n=h} &= \lambda \sin \theta - \Gamma (\overline{r} \cos \theta) + a \left[ \frac{\Lambda}{H} \overline{s}\_{\overline{t}} + \lambda \sin \theta \overline{h}\_{\overline{t}} + \Gamma \overline{r} \cos \theta \overline{h}\_{\overline{t}} \right] \\ &+ a \mu \left[ \frac{2 \overline{r} \overline{u}}{\overline{r}^2} + \frac{\overline{w}\_{\phi}}{\overline{r}} - \frac{2 \overline{w}\_{\overline{n}}}{\overline{r}^2} \overline{h}\_{\overline{t}} \right] + O \left( \alpha^2 \right), \end{split} \tag{17}$$

$$\mu \overline{w}\_{\overline{m}} \vert\_{\overline{n}=h} = a \left[ \frac{\Lambda}{H} \frac{\overline{\kappa}\_{\overline{\phi}}}{\overline{r}} + \lambda \cos \theta \overline{h}\_{\overline{\phi}} + \Gamma \overline{r} \sin \theta \overline{h}\_{\overline{\phi}} \right] + a \mu \frac{\overline{w}\_{\overline{\tau}}}{r} + O(a^2), \tag{18}$$

where

$$
\lambda = \frac{\rho \text{g} \text{R}\_{\text{c}}^2}{\mu u\_0}, \qquad \Gamma = \frac{\rho a o^2 \text{R}\_{\text{c}}^3}{\mu u\_0}, \qquad \Lambda = \frac{\sigma a}{\mu u\_0}, \qquad H = \frac{h\_0}{R\_{\text{c}}}.
$$

Here the κ terms have been assumed to be of order Oð Þ1 , while Eq. (4) would indicate that they are <sup>O</sup> <sup>α</sup><sup>2</sup> ð Þ. In order to bring the surface tension terms into the <sup>O</sup>ð Þ <sup>α</sup> problem, we are in effect assuming that <sup>Λ</sup> is <sup>O</sup> <sup>1</sup>=α<sup>2</sup> ð Þ. Others, including Benney [4] and Atherton and Homsy [5], have used a similar "specific promotion" for similar thin film problems.

We now expand the scaled velocities in a regular perturbation series expansion in powers of the small parameter α:

$$\begin{aligned} \overline{u} &= \overline{u}^{(0)} + a\overline{u}^{(1)} + \dots a^m \overline{u}^{(m)}, \\\\ \overline{v} &= \overline{v}^{(0)} + a\overline{v}^{(1)} + \dots a^m \overline{v}^{(m)}, \\\\ \overline{w} &= \overline{w}^{(0)} + a\overline{w}^{(1)} + \dots a^m \overline{w}^{(m)}. \end{aligned}$$

We then use boundary conditions to determine the appropriate constants of integration and solve for <sup>u</sup>ð Þ <sup>0</sup> , <sup>v</sup>ð Þ <sup>0</sup> , and <sup>w</sup>ð Þ <sup>0</sup> and <sup>u</sup>ð Þ<sup>1</sup> , <sup>v</sup>ð Þ<sup>1</sup> , and <sup>w</sup>ð Þ<sup>1</sup> . The <sup>O</sup>ð Þ<sup>1</sup> velocities are given by

$$
\overline{u}^{(0)}\left(\overline{\mathfrak{s}}, \overline{n}, \overline{\phi}\right) = A\_0 \left[\frac{\overline{n}^2}{2} - \overline{h}\overline{n}\right], \qquad \overline{w}^{(0)} = \mathbf{0}, \tag{19}
$$

where A<sup>0</sup> ¼ λ sin θ � Γr cos θ. We now determine the order α velocities

$$\overline{u}^{(1)} = A\_0 \frac{2\overline{r}\_5}{\overline{r}^2} \left[ \frac{\overline{n}^4}{24} - \frac{\overline{h}\overline{n}^3}{6} + \frac{\overline{h}^2 \overline{n}^4}{4} - \frac{\overline{h}^3 \overline{n}}{6} \right] + B\_0 \left[ \frac{\overline{n}^2}{2} - \overline{h}\overline{n} \right] + C\_0 \left[ \overline{h}^2 \overline{n} - \frac{\overline{h}\overline{n}^2}{2} \right], \quad \text{(20)}$$

$$
\overline{w}^{(1)} = D\_0 \left[ \frac{\overline{n}^2}{2} - \overline{h}\overline{n} \right],
\tag{21}
$$

with the constants B0, C0, and D<sup>0</sup> given by

Numerical Simulation of the Spin Coating of the Interior of Metal Beverage Cans DOI: http://dx.doi.org/10.5772/intechopen.90381

$$B\_0 = \left[\frac{\Lambda}{H}\overline{\kappa}\_{\tilde{\tau}} + \dot{\lambda}\cos\theta\overline{h}\_{\tilde{\tau}} + \Gamma\overline{r}\sin\theta\overline{h}\_{\tilde{\tau}}\right],$$

$$C\_0 = [-\dot{\lambda}\sin\theta\theta\_{\tilde{\tau}} + \Gamma\overline{r}\_{\tilde{r}}\sin\theta + \Gamma\overline{r}\cos\theta\theta\_{\tilde{\tau}}],$$

$$D\_0 = \left[\frac{\Lambda}{H}\frac{\overline{\kappa}\_{\tilde{\phi}}}{\overline{r}} + \dot{\lambda}\cos\theta\frac{\overline{h}\_{\tilde{\phi}}}{\overline{r}} + \Gamma\overline{r}\sin\theta\frac{\overline{h}\_{\tilde{\phi}}}{\overline{r}}\right].$$

Using the appropriate boundary conditions, the order α velocity in the normal direction n may be determined by integrating the continuity equation. But if we limit our consideration to the order α problem, we can write the evolution equation in divergence form, negating the necessity of determining vð Þ<sup>1</sup> . Integrating Eq. (13) between the boundaries n ¼ 0 and n ¼ h, and employing Leibniz's rule, we have the kinematic condition

$$\overline{h}\_{\overline{i}} = -\frac{1}{\overline{r}} \left[ \frac{\partial \left( \overline{r} \overline{Q}^{(\overline{i})} \right)}{\partial \overline{s}} + \frac{\partial \overline{Q}^{(\overline{\phi})}}{\partial \overline{\phi}} \right].$$

The evolution equation, to order α, is then determined to be

$$
\begin{split}
\overline{h}\_{\overline{i}} &= -\frac{\mathbf{1}}{\overline{r}} \frac{\partial}{\partial \overline{s}} \left[ \left[ (-\boldsymbol{\lambda}\sin\theta + \Gamma \overline{r}\cos\theta) - a \left( +\frac{\boldsymbol{\Lambda}}{H} \overline{\kappa} + \dot{\lambda}\cos\theta \overline{h}\_{\overline{r}} - \Gamma \overline{r}\sin\theta \overline{h}\_{\overline{i}} \right) \right] \overline{r} \frac{\overline{h}^{3}}{3} \right] \\ &+ a(-\boldsymbol{\lambda}\sin\theta \theta\_{\overline{i}} + \Gamma \overline{r}; \sin\theta + \Gamma \overline{r}\cos\theta \theta\_{\overline{i}}) \overline{r} \frac{\overline{h}^{4}}{8} + a(-\boldsymbol{\lambda}\sin\theta + \Gamma \overline{r}\cos\theta) \frac{\overline{r}}{\overline{r}} \frac{\overline{h}^{5}}{3} \right] \\ &- \frac{\mathbf{1}}{\overline{r}} \frac{\partial}{\partial \overline{\phi}} \left[ a \left( -\frac{\boldsymbol{\Lambda}}{H} \frac{\overline{\kappa}\_{\overline{\phi}}}{\overline{r}} - \boldsymbol{\lambda}\cos\theta \frac{\overline{h}\_{\overline{\phi}}}{\overline{r}} - \Gamma \overline{r}\sin\theta \frac{\overline{h}\_{\overline{\phi}}}{\overline{r}} \right) \overline{\frac{h}{3}} \right].
\end{split}
$$

The dimensional evolution equation is found to be given by

$$\begin{split} h\_{l} &= -\frac{1}{\mu r} \frac{\partial}{\partial t} \left[ \left( -\rho \text{g} \sin \theta + \rho \alpha^{2} r \cos \theta - \sigma \kappa\_{i} - \rho \text{g} \cos \theta h\_{i} - \rho \alpha^{2} r \sin \theta h\_{i} \right) r \left( \frac{h^{3}}{3} \right) \right] \\ &- \frac{1}{\mu r} \frac{\partial}{\partial t} \left[ \left( -\rho \text{g} \sin \theta \theta\_{i} + \sigma r\_{i} \sin \theta + \rho \alpha^{2} r \cos \theta h\_{i} \right) r \left( \frac{h^{4}}{8} \right) \\ &+ \frac{r\_{s}}{r} \left( -\rho \text{g} \sin \theta + \rho \alpha^{2} r \cos \theta \right) \left( \frac{h^{5}}{3} \right) \right] \\ &- \frac{1}{\mu r} \frac{\partial}{\partial \phi} \left[ \left( -\sigma \frac{\kappa\_{\phi}}{r} - \rho \text{g} \cos \theta \frac{h\_{\phi}}{r} - \rho \alpha^{2} r \sin \theta \frac{h\_{\phi}}{r} \right) \left( \frac{h^{3}}{3} \right) \right]. \end{split} \tag{22}$$

### 3. Nondimensionalization

We shall employ the maximum radius of the substrate as our length scale, <sup>r</sup>max <sup>¼</sup> Rc, and scale time with the quantity <sup>t</sup> <sup>∗</sup> <sup>¼</sup> <sup>3</sup>μRc=σ. We include the following substitutions:

Methods for Film Synthesis and Coating Procedures

$$\hat{h} = \frac{h}{R\_c}, \quad \hat{r} = \frac{r}{R\_c}, \quad \hat{s} = \frac{s}{R\_c}, \quad \hat{\kappa} = \kappa R\_c, \quad \hat{\phi} = \phi, \quad \hat{t} = \frac{t}{t^\*}.$$

Substituting these scale factors in the dimensional evolution equation, Eq. (22), leads to the nondimensional evolution equation

$$\begin{split} \dot{h}\_{i} &= -\frac{1}{\dot{r}} \frac{\partial}{\partial \dot{\theta}} \left[ \left( -\dot{\lambda}\_{2} \sin \theta + \Gamma\_{2} \dot{r} \cos \theta - \dot{\kappa}\_{i} - \dot{\lambda}\_{2} \cos \theta \dot{h}\_{i} - \Gamma\_{2} \dot{r} \sin \theta \dot{h}\_{i} \right) \dot{r} \dot{h}^{3} \right] \\ &- \frac{1}{\dot{r}} \frac{\partial}{\partial \dot{\theta}} \left[ \left( +\dot{\lambda}\_{2} \sin \theta \theta\_{i} - \Gamma\_{2} \dot{r}\_{i} \sin \theta + \Gamma\_{2} \dot{r} \cos \theta \dot{\kappa}\_{i} \right) \dot{r} \left( \frac{3 \dot{h}^{4}}{8} \right) \right] \\ &- \frac{1}{\dot{r}} \frac{\partial}{\partial \dot{\theta}} \left[ \left[ -\dot{\lambda}\_{2} \sin \theta + \Gamma\_{2} \cos \theta \right] \frac{\dot{r}\_{i}}{\dot{r}} \left( \dot{\lambda}^{5} \right) \right], \\ &- \frac{1}{\dot{r}} \frac{\partial}{\partial \dot{\theta}} \left[ \left[ -\frac{\dot{\kappa}\_{\dot{\phi}}}{\dot{r}} - \dot{\lambda}\_{2} \cos \theta \frac{\dot{h}\_{\dot{\phi}}}{\dot{r}} - \Gamma\_{2} \dot{r} \sin \theta \frac{\dot{h}\_{\dot{\phi}}}{\dot{r}} \right] \dot{h}^{3} \right], \end{split} \tag{23}$$

in nondimensional units. Here λ<sup>2</sup> and Γ<sup>2</sup> are given by

$$
\lambda\_2 = \frac{\rho \mathbf{g} R\_c^2}{\sigma}, \qquad \Gamma\_2 = \frac{\rho a \rho^2 R\_c^3}{\sigma}.
$$

The geometry of the substrate is delineated by a schematic drawing which gives us the radius of each circular arc, Ri, the subtended angle φ<sup>i</sup> of each element, and the length of each straight segment (see Figure 2). Consequently there is a discontinuity in the curvature between adjacent elements, but the slope is always continuous. In order to use the lubrication approximations, the coating thickness must

Figure 2. Substrate made up of straight segments and circular arcs of radius R<sup>i</sup> and subtended angle φi.

Numerical Simulation of the Spin Coating of the Interior of Metal Beverage Cans DOI: http://dx.doi.org/10.5772/intechopen.90381

always be much smaller than a typical substrate length, Riφ<sup>i</sup> or li. At t ¼ 0 we assume that there is already a very thin paint layer on the substrate and always use a sufficiently small time step such that ^ <sup>h</sup>ð Þ ^s,^<sup>t</sup> is never zero. Consequently we do not have to introduce the effects of a contact angle in our analysis. We not only assume that there is no ϕ variation in the substrate but that the coating is also axisymmetric. Consequently we only have to solve for the coating height as a function of s and t.

### 4. Application of coating layer

We assume that the coating layer is laid down over time by a spray gun that emits a fan of gas that is directed toward the can substrate. Thus the accumulation of coating with time due to this fan is a function of <sup>s</sup> and <sup>t</sup>: <sup>∂</sup>hfanð Þ <sup>s</sup>, <sup>t</sup> <sup>=</sup>∂t. The spray sector is bounded by the angles Δφ and Δψ as shown in Figure 3. We assume that the subtended angle Δψ is significantly smaller than the angle Δφ so that if the spray fan geometry is an ellipse, it will have an eccentricity near 1.0.

If Qfan tot is the total flux from the fan, and q the flux in an elemental area of the fan, then

$$Q\_{\rm tot}^{\rm fan} = \int\limits\_{\rho\_1}^{\rho\_2} \int q(\rho, \mu) d\rho d\mu \dots$$

If the flux does not vary significantly in the ψ direction, then the flux is constant in the elemental shaded sector in Figure 3. Consequently the flux per unit distance across the thin dimension of the fan is given by qð Þ φ =ð Þ ϵυ , where ϵ ¼ Δψ. When the width of the spray fan in the ψ direction is less than one-half the circumference 2πr sð Þ wetted by an element of the fan (see Figure 4), the average flux deposited per revolution is given by

$$(q/\epsilon \nu)(\epsilon \nu/(2\pi r(s))) = q/(2\pi r(s)).\tag{24}$$

When expressed as a function of ∂hfan=∂t, Eq. (24) becomes

Figure 3. Assumed spray pattern.

Methods for Film Synthesis and Coating Procedures

Figure 4.

Beverage can and spray pattern geometry. Here υ is the distance from the orifice of the spray gun to a given point on the can substrate.

$$q d\rho = \frac{\partial h^{\text{fan}}}{\partial t} ds 2\pi r(s).$$

From Figure 4 we have the relation

sin γ ¼ υdφ=ds,

For a sufficiently thin fan, where the point on the substrate is sufficiently far from the centerline r ¼ 0, the circumferential-average rate of increase of height is given by

$$\frac{\partial h^{\text{fan}}}{\partial t} = q(\rho) \sin \chi / (2\pi r(s)\nu). \tag{25}$$

When we are near the centerline, this formula must be modified to

$$\frac{\partial h^{\rm fan}}{\partial t} = [\mathbf{1}/\nu(\boldsymbol{s})][q(\boldsymbol{s})/(\boldsymbol{\epsilon}\nu(\boldsymbol{s}))]\sin\gamma(\boldsymbol{s}) \cdot \mathbf{Min}[(\boldsymbol{\epsilon}\nu(\boldsymbol{s})/2\pi r(\boldsymbol{s}), \mathbf{1}/2], \tag{26}$$

to account for the fact that here the substrate is constantly being reached by the spray fan.

We also assume that there is a secondary "gas" which is uniform in the interior of the can and results in a constant ∂hgas=∂t. If 0<ζ <100 represents the percentage of secondary gas, then ∂hgas=∂t from this source is given by

$$\frac{\partial h\_{\rm gas}^{\rm fan}}{\partial t} = \frac{\zeta}{100} \frac{Q\_{\rm tot}}{\mathcal{A}\_{\rm tot}},$$

Numerical Simulation of the Spin Coating of the Interior of Metal Beverage Cans DOI: http://dx.doi.org/10.5772/intechopen.90381

where Atot is the total area of the interior of the can. We also must include the effects of geometric shading. For example, the dome blocks some flux from the spray gun from reaching the inner wall of the moat.
