Question

A belt moves upward at velocity V, dragging a film of viscous liquid of constant thickness h. Near the belt, the film moves upward due to no slip. At its outer edge, the film moves downward due to gravity. - liquid dynamic viscosity: u density:P belt 1- Using the continuity equation, demonstrate that there is no flow in the x-direction. Clearly state all assumptions and boundary conditions. 2- Assuming zero shear stress at the outer film edge, derive an expression for the y-component of the velocity. Clearly state all assumptions and boundary conditions. 3- For what value of V is there no net flow either up or down?

hi i am in a fluid dynamics class and need some help with the question attached. please be specific and write out all steps so i know how to do the problem.

Answer 1

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(a) The assumption of parallel flow, 11 = w = 0 and y = V(x), satisfies conti- nuity and makes the x- and z-momentum equations irrelevant. We are left with the y-momentum equation: I dy dva pu-+-+w- Ох ду og + U 21 22 Xo д. oy? There is no convective acceleration, and the pressure gradient is negligible due to the free surface. We are left with a second-order linear differential equation for v(x): dv_P8 x+C PE+ dm2 = P8 1 u 2 "7++ C2 (b) d'y pg Integrate: C Integrate again: V = . dx² u dx u At the free surface, x = h, T = u(dv/dx) = 0, hence C1 =-pghlu. At the wall, v = V = C2. The solution is ob V = V - + u 2u (c) The average velocity is found by integrating the distribution v(x) across the film: Since livavg = Q per unit depth into the paper, there is no net up-or-down flow when V = pgh?|(34) Ans.(c)