Film Flow on a Moving Substrate Fig. E6.3.1 shows a coating experiment involving a flat photographic film that is being pulled up from a processing bath by rollers with a steady velocity U at an angle θ to the horizontal. As the film leaves the bath, it entrains some liquid, and in this particular

Question

Film Flow on a Moving Substrate

Fig. E6.3.1 shows a coating experiment involving a flat photographic film that is being pulled up from a processing bath by rollers with a steady velocity [latex]U[/latex] at an angle [latex] heta[/latex] to the horizontal. As the film leaves the bath, it entrains some liquid, and in this particular experiment it has reached the stage where: (a) the velocity of the liquid in contact with the film is [latex]v_{x}=U[/latex] at [latex]y=0[/latex], (b) the thickness of the liquid is constant at a value [latex]\delta[/latex], and (c) there is no net flow of liquid (as much is being pulled up by the film as is falling back by gravity). (Clearly, if the film were to retain a permanent coating, a net upward flow of liquid would be needed.)

Perform the following tasks:

Write down the differential mass balance and simplify it.
Write down the differential momentum balances in the [latex]x[/latex] and [latex]y[/latex] directions. What are the values of [latex]g_{x}[/latex] and [latex]g_{y}[/latex] in terms of [latex]g[/latex] and [latex] heta ?[/latex] Simplify the momentum balances as much as possible.
From the simplified [latex]y[/latex] momentum balance, derive an expression for the pressure [latex]p[/latex] as a function of [latex]y, ho, \delta, g[/latex], and [latex] heta[/latex], and hence demonstrate that [latex]\partial p / \partial x=0 .[/latex]

Assume that the pressure in the surrounding air is zero everywhere.

From the simplified [latex]x[/latex] momentum balance, assuming that the air exerts a negligible shear stress [latex] au_{y x}[/latex] on the surface of the liquid at [latex]y=\delta[/latex], derive an expression for the liquid velocity [latex]v_{x}[/latex] as a function of [latex]U, y, \delta[/latex], and [latex]\alpha[/latex], where [latex]\alpha= ho g \sin heta / \mu .[/latex]
Also derive an expression for the net liquid flow rate [latex]Q[/latex] (per unit width, normal to the plane of Fig. E6.3.1) in terms of [latex]U, \delta[/latex], and [latex]\alpha .[/latex] Noting that [latex]Q=0[/latex], obtain an expression for the film thickness [latex]\delta[/latex] in terms of [latex]U[/latex] and [latex]\alpha[/latex].
Sketch the velocity profile [latex]v_{x}[/latex], labeling all important features.

Accepted Answer

Assumptions and continuity. The following assumptions are reasonable:

The flow is steady and Newtonian, with constant density [latex] ho[/latex] and viscosity [latex]\mu[/latex].
The [latex]z[/latex] direction, normal to the plane of the diagram, may be disregarded entirely. Thus, not only is [latex]v_{z}[/latex] zero, but all derivatives with respect to [latex]z[/latex], such as [latex]\partial v_{x} / \partial z[/latex], are also zero.
There is only one nonzero velocity component, namely, that in the direction of motion of the photographic film, [latex]v_{x}[/latex]. Thus, [latex]v_{y}=v_{z}=0[/latex].
Gravity acts vertically downwards.

Because of the constant-density assumption, the continuity equation, [latex](5.48)[/latex], simplifies, as before, to:

[latex]\frac{\partial ho}{\partial t}+\frac{\partial\left( ho v_{x} ight)}{\partial x}+\frac{\partial\left( ho v_{y} ight)}{\partial y}+\frac{\partial\left( ho v_{z} ight)}{\partial z}=0,[/latex] (5.48)

[latex]\frac{\partial v_{x}}{\partial x}+\frac{\partial v_{y}}{\partial y}+\frac{\partial v_{z}}{\partial z}=0.[/latex] (E6.3.1)

But since [latex]v_{y}=v_{z}=0[/latex], it follows that:

[latex]\frac{\partial v_{x}}{\partial x}=0.[/latex] (E6.3.2)

so [latex]v_{x}[/latex] is independent of distance [latex]x[/latex] along the film. Further, [latex]v_{x}[/latex] does not depend on [latex]z[/latex] (assumption 2); thus, the velocity profile [latex]v_{x}=v_{x}(y)[/latex] depends only on [latex]y[/latex] and will appear the same for all values of [latex]x[/latex].

Momentum balances. With the stated assumptions of a Newtonian fluid with constant density and viscosity, Eqn. (5.73) gives the [latex]x[/latex] and [latex]y[/latex] momentum balances:

[latex]\begin{aligned} ho\left(\frac{\partial v_{x}}{\partial t}+v_{x} \frac{\partial v_{x}}{\partial x}+v_{y} \frac{\partial v_{x}}{\partial y}+v_{z} \frac{\partial v_{x}}{\partial z} ight) &=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^{2} v_{x}}{\partial x^{2}}+\frac{\partial^{2} v_{x}}{\partial y^{2}}+\frac{\partial^{2} v_{x}}{\partial z^{2}} ight)+ ho g_{x} \ ho\left(\frac{\partial v_{y}}{\partial t}+v_{x} \frac{\partial v_{y}}{\partial x}+v_{y} \frac{\partial v_{y}}{\partial y}+v_{z} \frac{\partial v_{y}}{\partial z} ight) &=-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^{2} v_{y}}{\partial x^{2}}+\frac{\partial^{2} v_{y}}{\partial y^{2}}+\frac{\partial^{2} v_{y}}{\partial z^{2}} ight)+ ho g_{y} (5.73)\end{aligned}[/latex]

Noting that [latex]g_{x}=-g \sin heta[/latex] and [latex]g_{y}=-g \cos heta[/latex], these momentum balances simplify to:

[latex]\frac{\partial p}{\partial x}+ ho g \sin heta=\mu \frac{\partial^{2} v_{x}}{\partial y^{2}},[/latex] (E6.3.3)

[latex]\frac{\partial p}{\partial y}=- ho g \cos heta.[/latex] (E6.3.4)

Integration of Eqn. (E6.3.4), between the free surface at [latex]y=\delta[/latex] (where the gauge pressure is zero) and an arbitrary location [latex]y[/latex] (where the pressure is [latex]p[/latex] ) gives:

[latex]\int_{0}^{p} d p=- ho g \cos heta \int_{\delta}^{y} d y+f(x),[/latex] (E6.3.5)

so that:

[latex]p= ho g \cos heta(\delta-y)+f(x) .[/latex] (E6.3.6)

Note that since a partial differential equation is being integrated, a function of integration, [latex]f(x)[/latex], is again introduced. Another way of looking at it is to observe that if Eqn. (E6.3.6) is differentiated with respect to [latex]y[/latex], we would recover the original equation, [latex](\mathrm{E} 6.3 .4)[/latex], because [latex]\partial f(x) / \partial y=0[/latex].

However, since [latex]p=0[/latex] at [latex]y=\delta[/latex] (the air/liquid interface) for all values of [latex]x[/latex], the function [latex]f(x)[/latex] must be zero. Hence, the pressure distribution:

[latex]p=(\delta-y) ho g \cos heta,[/latex] (E6.3.7)

shows that [latex]p[/latex] is not a function of [latex]x[/latex].

In view of this last result, we may now substitute [latex]\partial p / \partial x=0[/latex] into the [latex]x[/latex]- momentum balance, Eqn. (E6.3.3), which becomes:

[latex]\frac{d^{2} v_{x}}{d y^{2}}=\frac{ ho g}{\mu} \sin heta=\alpha,[/latex] (E6.3.8)

in which the constant [latex]\alpha[/latex] has been introduced to denote [latex] ho g \sin heta / \mu[/latex]. Observe that the second derivative of the velocity now appears as a total derivative, since [latex]v_{x}[/latex] depends on [latex]y[/latex] only.

A first integration of Eqn. (E6.3.8) with respect to [latex]y[/latex] gives:

[latex]\frac{d v_{x}}{d y}=\alpha y+c_{1} .[/latex] (E6.3.9)

The boundary condition of zero shear stress at the free surface is now invoked:

[latex] au_{y x}=\mu\left(\frac{\partial v_{y}}{\partial x}+\frac{\partial v_{x}}{\partial y} ight)=\mu \frac{d v_{x}}{d y}=0.[/latex] (E6.3.10)

Thus, from Eqns. (E6.3.9) and (E6.3.10) at [latex]y=\delta[/latex], the first constant of integration can be determined:

[latex]\frac{d v_{x}}{d y}=\alpha \delta+c_{1}=0, \quad ext { or } \quad c_{1}=-\alpha \delta.[/latex] (E6.3.11)

A second integration, of Eqn. (E6.3.9) with respect to [latex]y[/latex], gives:

[latex]v_{x}=\alpha\left(\frac{y^{2}}{2}-y \delta ight)+c_{2} .[/latex] (E6.3.12)

The second constant of integration, [latex]c_{2}[/latex], can be determined by using the boundary condition that the liquid velocity at [latex]y=0[/latex] equals that of the moving photographic film. That is, [latex]v_{x}=U[/latex] at [latex]y=0[/latex], yielding [latex]c_{2}=U[/latex]; thus, the final velocity profile is:

[latex]v_{x}=U-\alpha y\left(\delta-\frac{y}{2} ight) .[/latex] (E6.3.13)

Observe that the velocity profile, which is parabolic, consists of two parts:

A constant and positive part, arising from the film velocity, [latex]U[/latex].
A variable and negative part, caused by gravity, which reduces [latex]v_{x}[/latex] at increasing distances [latex]y[/latex] from the film and eventually causes it to become negative.

Exactly how much of the liquid is flowing upwards, and how much downwards, depends on the values of the variables [latex]U, \delta[/latex], and [latex]\alpha[/latex]. However, we are asked to investigate the situation in which there is no net flow of liquid - that is, as much is being pulled up by the film as is falling back by gravity. In this case:

[latex]Q=\int_{0}^{\delta} v_{x} d y=\int_{0}^{\delta}\left[U-\alpha y\left(\delta-\frac{y}{2} ight) ight] d y=U \delta-\frac{1}{3} \alpha \delta^{3}=0,[/latex] (E6.3.14)

giving the thickness of the liquid film as:

[latex]\delta=\sqrt{\frac{3 U}{\alpha}} .[/latex] (E6.3.15)

The velocity profile for this case of [latex]Q=0[/latex] is shown in Fig. E6.3.2.

Question 6.3: Film Flow on a Moving Substrate Fig. E6.3.1 shows a coating ...

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