Wall δ Flow rate е Liquid film y www (a) Simplify the continuity equation and show that vx = x(y), i.e., it is a function of y only. др (b) O Simplify the momentum balance equation in the y-direction and show that there is no variation of pressure horizontally across the film. Since P₁y_53 = 0, i.e., dy = P atm" the pressure P др everywhere in the film is equal to Patm. This also means that =0. Ox (c) (d) (e) Simplify the momentum balance equation in the x-direction and show that =-pg. Solve the above equation with appropriate boundary conditions and show that V₁ = pg 28 (2 -(2y8 – y²). 2μ - Calculate the volumetric flowrate Q based on vx and show that the film thickness 1/3 3μQ 8 = pg
The following figure shows a “wetted-wall” column, in which a thin film of a reacting liquid
of viscosity μ flows steadily down a plane wall, possibly for a gas-absorption study. The volumetric
flow rate of liquid is specified as Q per unit width of the wall (normal to the plane of the diagram).
Assume that there is only one non-zero velocity component, vx , which does not vary in the x direction
and that the gas exerts negligible shear stress on the liquid film.
(a) Simplify the continuity equation and show that vx = vx(y), i.e., it is a function of y only.
(b) Simplify the momentum balance equation in the y-direction and show that dP/dy=0 i.e.,
there is no variation of pressure horizontally across the film. Since atmy
P =Patm everywhere in the film is equal to Patm. This also means that dP/dx=0
(c) Simplify the momentum balance equation in the x-direction and show that it equals the equation c in the image.
(d) Solve the above equation with appropriate boundary conditions and show that it equals the equation d in the image.
(e) Calculate the volumetric flowrate Q based on vx and show that the film thickness equals equation e in the image
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