How Does Film Thickness Affect Mass Flow Rate in Laminar Liquid Boundary Layers?

[painfive]

Here is the problem: (From sabersky, problem 8.9)

Vapor condenses on a vertical surface to form a liquid film. The film moves under gravity and forms a laminar liquid boundary layer. Derive an expression for the mass flow rate dm/dt as a function of the local film thickness $\delta$. Neglect any velocity components in the y-direction. (the positive x direction is down and y points away from the solid surface)

Answer: dm/dt=$\rho g \delta^3/3 \nu$, where $\rho$ is the density and $\nu$ is the kinematic viscosity.

I'm really stuck on this one. The continuity equation is useless because there must be vapor condensing on to the film (or else dm/dt would be constant). Assuming this vapor has no momentum, I was able to use the momentum equation to get:

$$\frac{\partial}{\partial x} \int_0^\delta \rho u^2 dy =-\tau_0+\rho g$$

where $\tau_0$ is the shear force at the solid surface. Since g and $\nu$ appear not as a sum but as a product in the final answer, I assume there must be another equation relating them. Can anyone help me out here?

 

[Jhero]

Dear sabersky,

Thank you for your question. I understand your confusion and I will do my best to help you derive the expression for the mass flow rate.

First, let's review the problem. We have a vertical surface where vapor is condensing to form a liquid film. This film moves under gravity and forms a laminar liquid boundary layer. We are asked to derive an expression for the mass flow rate dm/dt as a function of the local film thickness \delta, neglecting any velocity components in the y-direction.

To solve this problem, we will use the conservation of mass principle. This principle states that the mass entering a control volume must equal the mass leaving the control volume, plus the rate of change of mass within the control volume.

Let's consider a small control volume of width dx and height \delta, as shown in the figure below:

<img src="" alt="control volume">

The mass entering this control volume in the x-direction is given by \rho u \delta dx, where \rho is the density and u is the velocity in the x-direction. The mass leaving the control volume is given by \rho (u+\frac{\partial u}{\partial x} dx) (\delta+\frac{\partial \delta}{\partial x} dx). The rate of change of mass within the control volume is given by \frac{\partial}{\partial t}(\rho \delta dx).

Using the conservation of mass principle, we can write the following equation:

\rho u \delta dx = \rho (u+\frac{\partial u}{\partial x} dx) (\delta+\frac{\partial \delta}{\partial x} dx) + \frac{\partial}{\partial t}(\rho \delta dx)

Simplifying and neglecting higher order terms, we get:

\rho u \delta = \rho u \delta + \delta \frac{\partial u}{\partial x} + u \frac{\partial \delta}{\partial x} + \frac{\partial \delta}{\partial t}

Since we are neglecting any velocity components in the y-direction, we can assume that u is only a function of x. Therefore, \frac{\partial u}{\partial y} = 0 and the equation becomes:

\frac{\partial}{\partial x} \int_0^\delta \rho u^2 dy = \