- #1
painfive
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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 [itex]\delta[/itex]. 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=[itex]\rho g \delta^3/3 \nu [/itex], where [itex]\rho[/itex] is the density and [itex]\nu[/itex] 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:
[tex]\frac{\partial}{\partial x} \int_0^\delta \rho u^2 dy =-\tau_0+\rho g[/tex]
where [itex]\tau_0[/itex] is the shear force at the solid surface. Since g and [itex]\nu[/itex] 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?
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 [itex]\delta[/itex]. 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=[itex]\rho g \delta^3/3 \nu [/itex], where [itex]\rho[/itex] is the density and [itex]\nu[/itex] 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:
[tex]\frac{\partial}{\partial x} \int_0^\delta \rho u^2 dy =-\tau_0+\rho g[/tex]
where [itex]\tau_0[/itex] is the shear force at the solid surface. Since g and [itex]\nu[/itex] 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?