CP502 Advanced Fluid Mechanics Flow of Viscous Fluids Set 03

Continuity and Navier-Stokes equations for incompressible flow of Newtonian fluid ρ υ R. Shanthini 05 April 2012

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down an inclined plane under gravity Exercise 1: Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equations reduces to x y z θ direction of flow where u is the velocity in the x-direction, ρ is the density, μ is the dynamic viscosity, g is acceleration due to gravity, and θ is the angle of the plane to the horizontal. Solve the above equation to obtain the velocity profile u and obtain the expression for the volumetric flow rate for a flowing film of thickness h. Exercise 2: If there is another solid boundary instead of the free-surface at y = h and the flow occurs with no pressure gradient, what will be the volumetric flow rate? R. Shanthini 05 April 2012

} } y z x (1) (2) Step 1: Choose the equation to describe the flow Navier-Stokes equation is already chosen since the system considered is incompressible flow of a Newtonian fluid. x y z θ direction of flow Step 2: Choose the coordinate system Cartesian coordinate system is already chosen. Step 3: Decide upon the functional dependence of the velocity components } (1) Steady, fully developed flow and therefore no change in time and in the flow direction. Channel is not bounded in the z-direction and therefore nothing happens in the z-direction. } (2) R. Shanthini 05 April 2012

y z x Step 4: Use the continuity equation in Cartesian coordinates Flow geometry shows that v can not be a constant, and therefore we choose x y z θ direction of flow R. Shanthini 05 April 2012

The functional dependence of the velocity components therefore reduces to } y direction: v = 0 z direction: w = 0 x direction: u = function of (y) (3) Step 5: Using the N-S equation, we get x - component: y - component: z - component: R. Shanthini 05 April 2012

What was asked to be derived in Exercise 1 N-S equation therefore reduces to x - component: x y z θ direction of flow y - component: z - component: No applied pressure gradient to drive the flow. Flow is driven by gravity alone. Therefore, we get (4) x - component: y - component: What was asked to be derived in Exercise 1 p is not a function of z z - component: R. Shanthini 05 April 2012

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down an inclined plane under gravity Exercise 1: Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equation reduces to x y z θ direction of flow √done where u is the velocity in the x-direction, ρ is the density, μ is the dynamic viscosity, g is acceleration due to gravity, and θ is the angle of the plane to the horizontal. Solve the above equation to obtain the velocity profile u and obtain the expression for the volumetric flow rate for a flowing film of thickness h. Exercise 2: If there is another solid boundary instead of the free-surface at y = h and the flow occurs with no pressure gradient, what will be the volumetric flow rate? R. Shanthini 05 April 2012

(4) x y z Equation (4) is a second order equation in u with respect to y. Therefore, we require two boundary conditions (BC) of u with respect to y. h BC 1: At y = 0, u = 0 (no-slip boundary condition) θ BC 2: At y = h, (free-surface boundary condition) (5) Integrating equation (4), we get (6) Applying BC 2, we get Combining equations (5) and (6), we get (7) R. Shanthini 05 April 2012

y z x (8) (9) (10) h θ Integrating equation (7), we get Applying BC 1, we get B = 0 Combining equations (8) and (9), we get (10) R. Shanthini 05 April 2012

(10) x y z θ h Volumetric flow rate through one unit width fluid film along the z-direction is given by (11) R. Shanthini 05 April 2012

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down an inclined plane under gravity Exercise 1: Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equation reduces to x y z θ direction of flow √done where u is the velocity in the x-direction, ρ is the density, μ is the dynamic viscosity, g is acceleration due to gravity, and θ is the angle of the plane to the horizontal. Solve the above equation to obtain the velocity profile u and obtain the expression for the volumetric flow rate for a flowing film of thickness h. √done Exercise 2: If there is another solid boundary instead of the free-surface at y = h and the flow occurs with no pressure gradient, what will be the volumetric flow rate? R. Shanthini 05 April 2012

(4) y z x (12) (13) Equation does not change. BCs change. BC 1: At y = 0, u = 0 (no-slip boundary condition) h BC 2: At y = h, (free-surface boundary condition) θ u = 0 (no-slip boundary condition) (12) Integrating equation (4), we get (13) Integrating equation (12), we get Applying the BCs in (13), we get B = 0 and R. Shanthini 05 April 2012

y (14) z x (15) Therefore, equation (13) becomes Volumetric flow rate through one unit width fluid film along the z-direction is given by h θ (15) R. Shanthini 05 April 2012

Summary of Exercises 1 and 2 Free surface gravity flow Gravity flow through two planes x y z θ h x y z θ h (10) (14) (11) (15) Why the volumetric flow rate of the free surface gravity flow is 4 times larger than the gravity flow through two planes? R. Shanthini 05 April 2012

Any clarification? R. Shanthini 05 April 2012

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down a vertical plane under gravity Exercise 3: A viscous film of liquid draining down the side of a wide vertical wall is shown in the figure. At some distance down the wall, the film approaches steady conditions with fully developed flow. The thickness of the film is h. Assuming that the atmosphere offers no shear resistance to the motion of the film, obtain an expression for the velocity distribution across the film and show that x y z h where ν is the kinematic viscosity of the liquid, Q is the volumetric flow rate per unit width of the plate and g is acceleration due to gravity. R. Shanthini 05 April 2012

Workout Exercise 3 in 5 minutes! R. Shanthini 05 April 2012

Oil Skimmer Example An oil skimmer uses a 5 m wide x 6 m long moving belt above a fixed platform (q = 60º) to skim oil off of rivers (T = 10ºC). The belt travels at 3 m/s. The distance between the belt and the fixed platform is 2 mm. The belt discharges into an open container on the ship. The fluid is actually a mixture of oil and water. To simplify the analysis, assume crude oil dominates. Find the discharge and the power required to move the belt. r = 860 kg/m3 m = 1x10-2 Ns/m2 U h l y x g R. Shanthini 05 April 2012 30º

Oil Skimmer Discharge = ? N-S equation reduces to x - component: y - component: z - component: No applied pressure gradient to drive the flow. Flow is driven by gravity alone. Therefore, we get (16) x - component: y - component: p is not a function of z z - component: R. Shanthini 05 April 2012

(16) (17) (18) Sign changes in the equation BC 1: At y = 0, u = 0 (no-slip boundary condition) BC 2: At y = h, (free-surface boundary condition) u = U (no-slip boundary condition) (17) Integrating equation (16), we get (18) Integrating equation (17), we get Applying the BCs in (18), we get B = 0 and R. Shanthini 05 April 2012

(19) (20) Therefore, equation (18) becomes Volumetric flow rate through one unit width fluid film along the z-direction is given by (20) per unit width of the belt R. Shanthini 05 April 2012

Power = Force x Velocity [N·m/s] Oil Skimmer Power Requirements = ? How do we get the power requirement? What is the force acting on the belt? Equation for shear? Power = Force x Velocity [N·m/s] Shear force (t · L · W) t = m(du/dy) R. Shanthini 05 April 2012

Evaluate t = m(du/dy) at the moving belt (19) At the moving belt = 19.21 N/m2 R. Shanthini 05 April 2012

Power = shear force at the belt * L * W * U = (19.21 N/m2) (6 m) (5 m) (3 m/s) = 1.73 kW To reduce the power requirement, decrease the shear force R. Shanthini 05 April 2012

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid over a porous plate sucking the fluid Exercise 4: An incompressible, viscous fluid (of kinematic viscosity ν) flows between two straight walls at a distance h apart. One wall is moving at a constant velocity U in x-direction while the other is at rest as shown in the figure. The flow is caused by the movement of the wall. The walls are porous and a steady uniform flow is imposed across the walls to create a constant velocity V through the walls. Assuming fully developed flow, show that the velocity profile is given by Also, show that (i) u approaches Uy/h for small V, and (ii) u approaches for very large Vh/ν. x y z U V u v h R. Shanthini 05 April 2012

Step 1: Choose the equation to describe the flow done Step 2: Choose the coordinate system done Step 3: Decide upon the functional dependence of the velocity components Steady, fully developed flow and therefore no change in time and in the flow direction. Channel is not bounded in the z-direction and therefore nothing happens in the z-direction. } (1) Step 4: Use the continuity equation in Cartesian coordinates x y z U V u v h R. Shanthini 05 April 2012

The functional dependence of the velocity components therefore reduces to } y direction: v = V z direction: w = 0 x direction: u = function of (y) (2) Step 5: Using the N-S equation, we get x - component: y - component: z - component: R. Shanthini 05 April 2012

(3) v y u z x N-S equation therefore reduces to x - component: y - component: z - component: No applied pressure gradient to drive the flow. Flow is caused by the movement of the wall. Therefore, we get (3) x - component: x y z U V u v h R. Shanthini 05 April 2012

(3) Equation (3) is a second order equation in u with respect to y. Therefore, we require two boundary conditions (BC) of u with respect to y. BC 1: At y = 0, u = 0 (no-slip boundary condition) BC 2: At y = h, u = U (no-slip boundary condition) Integrating equation (3), we get (4) (5) Integrating equation (4), we get Applying the BCs in equation (5), we get (6) R. Shanthini 05 April 2012 (7)

(8) v y u z x From equations (6) and (7), we get Substituting the above in equation (5), we get (8) x y z U V u v h R. Shanthini 05 April 2012

(8) v y u z x Could you recognize the above profile? (i) For small V, expand exp(Vy/ν) and exp(Vh/ν) using Taylor series as follows: For small V, we can ignore the terms with power. We then get x y z U V u v h Could you recognize the above profile? R. Shanthini 05 April 2012

(8) For very large Vh/ν, exp(Vh/ν) goes to infinity. Therefore. Divide equation (8) by exp(Vh/ν). We then get For very large Vh/ν, exp(-Vh/ν) goes to zero. Therefore, we get x y z U V u v h R. Shanthini 05 April 2012