CP502 Advanced Fluid Mechanics Flow of Viscous Fluids and Boundary Layer Flow Lectures 3 and 4

R. Shanthini 25 Aug 2010 Continuity and Navier-Stokes equations for incompressible flow of Newtonian fluid ρ υ

R. Shanthini 25 Aug 2010 Steady, incompressible flow of Newtonian fluid in an infinite channel with stationery plates - fully developed plane Poiseuille flow Navier-Stokes equation will be used since the system considered is an incompressible flow (Assumption 1) of a Newtonian fluid (Assumption 2) Step 2: Choose the coordinate system Cartesian coordinate system is chosen Step 1: Choose the equation to describe the flow Fixed plate Fluid flow direction 2a x y z

R. Shanthini 25 Aug 2010 Continuity and Navier-Stokes equations for incompressible flow of Newtonian fluid in Cartesian coordinates Continuity: Navier-Stokes: x - component: y - component: z - component:

R. Shanthini 25 Aug 2010 Steady, incompressible flow of Newtonian fluid in an infinite channel with stationery plates - fully developed plane Poiseuille flow Step 3: Decide upon the functional dependence of the velocity components } y direction: v = function of (t, x, y, z) z direction: w = function of (t, x, y, z) x direction: u = function of (t, x, y, z) (1) Fixed plate Fluid flow direction 2a x y z

R. Shanthini 25 Aug 2010 Assumption (3): steady flow and therefore no change in time Assumption (4): infinite channel and therefore nothing happens in z-direction Assumption (5): fully developed flow and therefore no change in the flow direction (that is, x-direction) Using the above 3 assumptions, we reduce (1) to the following: } y direction: v = function of (y) z direction: w = 0 x direction: u = function of (y) (2)

R. Shanthini 25 Aug 2010 Fixed plate Fluid flow direction 2a x y z Step 4: Use the continuity equation in Cartesian coordinates Flow geometry shows that v can not be a constant, and therefore we choose

R. Shanthini 25 Aug 2010 } y direction: v = 0 z direction: w = 0 x direction: u = function of (y) (3) The functional dependence of the velocity components therefore reduces to Step 5: Using the N-S equation, we get x - component: y - component: z - component:

R. Shanthini 25 Aug 2010 N-S equation therefore reduces to Ignoring gravitational effects, we get p is not a function of z (4) x - component: y - component: z - component: x - component: y - component: z - component: p is not a function of y p is a function of x only

R. Shanthini 25 Aug 2010 Rewriting (4), we get (5) u is a function of y only and therefore LHS is a function of y only p is a function of x only and is a constant and therefore RHS is a function of x only LHS = left hand side of the equation RHS = right hand side of the equation function of (y) = function of (x)Therefore (5) gives,= constant It means = constant That is, pressure gradient in the x-direction is a constant.

R. Shanthini 25 Aug 2010 Rewriting (5), we get where is the constant pressure gradient in the x-direction Since u is only a function of y, the partial derivative becomes an ordinary derivative. (6) Therefore, (5) becomes (7)

R. Shanthini 25 Aug 2010 Integrating (6), we get (8) where C 1 and C 2 are constants to be determined using the boundary conditions given below: Substituting the boundary conditions in (8), we get Fixed plate Fluid flow direction 2a x y and Therefore, (8) reduces to Parabolic velocity profile } no-slip boundary condition

R. Shanthini 25 Aug 2010 Fixed plate Moving plate at velocity U Fluid flow direction 2a x y Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow where C 1 and C 2 are constants to be determined using the boundary conditions given below: Substituting the boundary conditions in (8), we get and Therefore, (8) reduces to (8) } no-slip boundary condition Parabolic velocity profile is super imposed on a linear velocity profile

R. Shanthini 25 Aug 2010 Fixed plate Moving plate at velocity U Fluid flow direction 2a x y Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow If zero pressure gradient is maintained in the flow direction, then ∆ p = 0. Therefore, we get Linear velocity profile Point to remember: - Pressure gradient gives parabolic profile - Moving wall gives linear profile

R. Shanthini 25 Aug 2010 Steady, incompressible flow of Newtonian fluid in a pipe - fully developed pipe Poisuille flow Fixed pipe z r Fluid flow direction 2a φ Workout on your own

R. Shanthini 25 Aug 2010 Steady, incompressible flow of Newtonian fluid between a stationary outer cylinder and a rotating inner cylinder - fully developed pipe Couette flow Step 2: Cylindrical polar coordinate is chosen Step 1: Continuity and Navier-Stokes equations are used Step 3: functional dependence of the velocity components are determined φ aΩaΩ a b r z direction: u = function of (t, r,, z) z direction: u z = function of (t, r,, z) r direction: u r = function of (t, r,, z)

R. Shanthini 25 Aug 2010 Assumptions: - Steady flow no time dependence - fully developed no change in the axial (z)-direction - axi-symmetric flow no change in the tangential ( )-direction φ aΩaΩ a b r z direction: u = function of (r) z direction: u z = 0 r direction: u r = function of (r) Flow condition: no flow in the axial (z-) direction is assumed.

R. Shanthini 25 Aug 2010 Step 4: Use the continuity equation Flow geometry shows that velocity component in the r-direction can not be a constant divided by r in order to satisfy the “no flow through the walls” boundary conditions, and therefore we choose u r = 0 u r = constant/r or u r = 0 φ aΩaΩ a b r z direction: u = function of (r) z direction: u z = 0 r direction: u r = 0

R. Shanthini 25 Aug 2010 Step 5: Using the N-S equation, ignoring the gravity, we get Radial component: Tangential component: Axial component: direction: u = function of (r) z direction: u z = 0 r direction: u r = 0

R. Shanthini 25 Aug 2010 Radial component of the N-S equation reduces to (9) Can you identify the centrifugal force??? which gives Axial component of the N-S equation reduces to p is not a function of z

R. Shanthini 25 Aug 2010 = function of r only = f(r) (say) we get which gives and Pressure takes different values, which is the same point in the tangential direction. Since pressure is continuous, it is impossible. (10) Tangential component of the N-S equation reduces to which gives Starting from, Therefore, we conclude f(r) = 0, which gives (11)

R. Shanthini 25 Aug 2010 (12) Combining (10) and (11), we get Radial component was (9) There are two unknowns and two equations. Solve (12) to get the tangential velocity component Then solve (9) to get the pressure p u

R. Shanthini 25 Aug 2010 Integrating (12) gives (13) where C 1 and C 2 are constants to be determined using the no-slip boundary conditions given below: φ aΩaΩ a b r z Substituting the boundary conditions in (13) gives and Therefore (13) could be written as (14)

R. Shanthini 25 Aug 2010 Let us introduce the following non-dimensional variables: (14) Using the non-dimensional, (14) is reduced to the following: (15) This simple circular flow exists only at low rotational speed of the inner cylinder. As the rotational speed increases, this simple steady flow is replaced by another steady flow in which the space between cylinders is filled with (Taylor) vortices.