Ρ gz c 1 r and p rz 1 2 ρω 2 r 2 c 2 z both of

• Homework Help
• 12
• 100% (26) 26 out of 26 people found this document helpful

This preview shows page 2 - 7 out of 12 pages.

−ρ gz + C 1 r ( ) and p r,z ( ) = 1 2 ρω 2 r 2 + C 2 z ( ) . Both of these results are satisfied by a pressure given by p r,z ( ) = −ρ gz + 1 2 ρω 2 r 2 + C . Now set the pressure to atmospheric at vertical heights h(r),
p 0 = −ρ gh r ( ) + 1 2 ρω 2 r 2 + C . Since p 0 and C are both constant, this equation can only be satisfied if h r ( ) = 1 2g ω 2 r 2 + C 3 where C 3 is an unknown constant, but p 0 = C 3 + C = C 4 . To determine the remaining constant C, use the fact the total volume of the water cannot change. Therefore 2 π h r ( ) rdr = π R 2 0 R H and 1 2g ω 2 r 2 + C 3 rdr = 1 2 R 2 0 R H so 1 8g ω 2 R 4 + 1 2 C 3 R 2 = 1 2 R 2 H yielding C 3 = H 1 4g ω 2 R 2 and h r ( ) = H + ω 2 2g r 2 R 2 2 . The surface shape is a parabola, with the lowest point at r=0 and the highest at r=R. 2. The figure below shows flow of a layer of Newtonian fluid with viscosity μ and constant thickness H down a plane inclined at an angle θ . This gravity-driven flow is steady and unidirectional in the x- direction. Calculate the velocity profile v X (y), the flow rate Q per unit width and the viscous stress τ at the plane surface y=0. YX
and hence d 2 v x dy 2 = ρ g μ sin θ . Integrating twice yields v x y ( ) = ρ g 2 μ sin θ y 2 + C 1 y + C 2 . As boundary conditions, we use the no-slip condition at y=0, and assume that the air exerts negligible shear stress on the liquid, because the viscosity of air is very low compared to any liquid. Then v x = 0 at y=0 and μ dv x dy = 0 at y=H . The no-slip condition yields C 2 =0 and the shear stress condition yields the equation μ ρ g μ sin θ y + C 1 = 0 at y=H or C 1 = ρ g μ sin θ H . The velocity profile is therefore v x y ( ) = ρ g μ sin θ Hy y 2 2 . The flow rate per unit width is Q = v x y ( ) 0 H dy = ρ g μ sin θ Hy y 2 2 dy 0 H or Q = ρ g μ sin θ H 3 2 H 3 6 = ρ gH 3 3 μ sin θ . The shear stress at y=0 is
τ yx y = 0 = μ dv x dy y = 0 = ρ gHsin θ . 3. In the slot-coater shown below, a coating liquid is extruded through a die and onto a solid sheet, the “web,” moving with velocity U in the x-direction. The pressure where the slot meets the web is P 1 and the gap between the web and the die is h. There is a forward-flow region of length L 1 and also a liquid- filled region of length L 2 in which the tendency for backflow is balanced by the movement of the web. The flow is fully developed, steady and unidirectional in most of the slot, and gravitational effects are negligible. The coating thickness far from the coater is h , and any evaporation or cooling that takes place has a negligible effect on the coating density. It may be assumed that P 0 , P 1 , U, h and L 1 are known. a) Derive an expression for the velocity profile v x (y) in the forward flow region. b)
• • • 