Exercise 5: Exact Solutions to the Navier-Stokes Equations I

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1 Fluid Mechanics, SG4, HT009 September 5, 009 Exercise 5: Exact Solutions to the Navier-Stokes Equations I Example : Plane Couette Flow Consider the flow of a viscous Newtonian fluid between two parallel plates located at y = 0 and y = h. The upper plane is moving with velocity U. Calculate the flow field. Assume the following: Steady flow: Parallel, fully-developed flow: Two-dimensional flow: No pressure gradient: The streamwise Navier-Stokes equation is v = 0, w = 0, t = 0 p x i = 0 u i x = 0 z = 0 u t + ū )u = ρ p + ν u, can be simplified using the above assumptions. We get With the boundary conditions u y = 0 u y = A u = Ay + B. u0) = 0 B = 0, uh) = U A = U/h we finally obtain uy) = Uy h. Example : Plane Poiseuille Flow Channel Flow) Consider the flow of a viscous Newtonian fluid between two solid boundaries at y = ±h driven by a constant pressure gradient p = [ P, 0, 0]. Show that u = P µ h y ), v = w = 0. Navier-Stokes equations: Boundary conditions: ū t + ū )ū = ρ p + ν ū ū = 0. ūy = ±h) = 0

2 Y X Z Figure : Coordinate system for plane Poiseuille flow. We are considering stationary flow and thus ū t = 0. The constant pressure gradient implies ū = ūy). Changes of ū in x,z would require a changing pressure gradient in x,z. The continuity equation ū reduces to v = 0. The boundary condition vy = ±h) = 0 then implies y v = 0. Consider the spanwise z) component of the Navier-Stokes equations: w }{{} v y = ν w y w = c y + c =0 The boundary conditions wy = ) = wy = h) = 0 imply c = c = 0 and thus w = 0. We can conclude that ū = [uy),0,0]. Consider now the streamwise x) component of the Navier-Stokes equations: 0 = P ρ + ν u y u y = P νρ y + d {µ = ρν} uy) = P µ y + d y + d The boundary conditions at y = ±h give 0 = P µ h + d h + d and 0 = P µ h d h + d We can directly conclude that d = 0 and this gives d = P µ h. The solution is thus u = P µ h y ), v = w = 0. Energy Dissipation in Poiseuille Flow a) Calculate the dissipation function for the plane Poiseuille flow computed above, or in terms of the bulk velocity U u = P µ h y ), v = w = 0, u = 3U h h y ), v = w = 0.

3 The mass-flow rate through the channel is Q = udy = Uh. The dissipation function is defined as dissipation to heat due to viscous stresses) For incompressible flows, it can be re-written as where we used the fact that e ij ξ ij = 0. Φ = τ ij u i x j. Φ = τ ij u i x j = µe ij e ij + ξ ij ) = µe ij e ij, The deformation tensor for the Poiseuille flow becomes e ij = / Φ = µ [ b) Calculate the total dissipation for unit area φ = Φdy = ) u + y ) ] u = µ y u 0 y 0 u y ) u. y µ 3U ) h y dy = 6µU h. and therefore c) Write the mechanical energy equation for this flow. Integrate over the channel width and relate the total dissipation φ to the pressure gradient and the mass flux. The mechanical energy equation is obtained by multiplying the Navier-Stokes equations by u i the energy is ρ/)u i u i ). One gets ρ D ) Dt u p τ ij iu i = ρf i u i u i + u i. x i x j Considering the Poiseuille flow and re-writing the last term as u i τ ij x j = u iτ ij x j Φ, the energy equation reduces to 0 = up + y uτ xy) Φ. Integrating across the channel each term in the expression above, one obtains for the first term updy = P udy = QP, where Q is the flow rate. This term represents the work rate by pressure forces. The second term due to the no-slip boundary conditions. y uτ xy) dy = [uτ xy )] h = 0 3

4 The third term is the total dissipation φ = Φdy defined above. Summarising 0 = QP Φdy. One can check the results, using the expression for φ obtained in b). Just recall that Q = udy = Uh, and the pressure gradient can be expressed in terms of U as P = 3µU h. Therefore QP = 6µU /h = φ. Example 3: Poiseuille Flow Pipe Flow) Consider the viscous flow of a fluid through a pipe with a circular cross-section given by r = a under the constant pressure gradient P = p. Show that z u z = P 4µ a r ) u r = u θ = 0. R Z Figure : Coordinate system for Poiseuille flow. Use the Navier-Stokes equations in cylindrical coordinates see lecture notes) u r t + ū )u r u θ r = p ρ r + ν u r u r r ) u θ r θ u θ t + ū )u θ + u r u θ r = p ρr θ + ν u θ + u r r θ u ) θ r u z t + ū )u z = p ρ z + ν u z r r r u r) + r u θ θ + u z z = 0. We know that p p = 0 and θ r = 0 and can directly see that u r = u θ = 0 satisfys the two first equations. From the continuity equation we get u z z = 0 u z = u z r,θ) only. Considering a steady flow we get from the axial component of the Navier-Stokes equations ū )u z = u z u z z u z u z z = ρ P + ν u z. 4

5 But we know that u z z Integrate once in r gives and integrating again we get = 0 from the continuity equation. We get u z = r r ru z r ) = P µ r ru z r ) = P µ r r u z r = P µ r + c u z r = P µ r + c r, u z = P 4µ r + c lnr) + c using the boundary conditions u z r = 0) < c = 0. We also have u z = 0 at r = a and this gives c = P a 4µ and we finally get u z = P 4µ a r ). Example 4: Asymptotic Suction Boundary Layer Calculate the asymptotic suction boundary layer, where the boundary layer over a flat plate is kept parallel by a steady suction V 0 through the plate. Assumptions: Two-dimensional flow: Parallel, fully-developed flow: Steady flow: Momentum equations: Normal momentum equation gives Boundary conditions: Continuity gives z = 0, w = 0 x = 0 t = 0 u t + uu x + vu y = ρ v t + uv x + vv Streamwise momentum equation at y p ) x + ν u x + u y y = p ) ρ y + ν v x + v y p y = 0 y = 0 : u = 0, v = V 0 y : u U u x + v y = 0 v = V 0 U V 0 y = p ρ x + ν U y 5

6 Resulting streamwise momentum equation p x = 0 Characteristic equation u V 0 y = ν u y u y = V 0 u ν y λ = V 0 ν λ λ = 0,λ = V 0 ν uy) = A + Be V 0 y/ν With the boundary conditions at y = 0 and y = we get uy) = U e V0y/ν). Example 5: Flow on an Inclined Plate Two incompressible viscous fluids flow one on top of the other down an inclined plate at an angle α see figure 3). They both have the same density ρ, but different viscosities µ and µ. The lower fluid has depth h and the upper h. Assuming that viscous forces from the surrounding air is negligible and that the pressure on the free surface is constant, show that [ u y) = h + h )y ] g sinα) y. ν h h Y X Figure 3: Coordinate system for flow down an inclined plate. Make the ansatz ū = [u y),0,0] and ū = [u y),0,0]. The continuity equation u x + v y = 0 gives v = 0 v = c and the boundary condition at y = 0 give v = 0. y Layer : Layer : N S ē y : 0 = p ρ y g cosα) p = ρg cosα)y + f x) N S ē x : 0 = ρ f x) + ν d u dy + g sinα) f x) = c N S ē y : 0 = p ρ y g cosα) p = ρg cosα)y + f x) N S ē x : 0 = ρ f x) + ν d u dy + g sinα) f x) = c 6

7 The pressure at the free surface y = h + h is p 0 : p 0 = ρg cosα)h + h ) + f x) f = p 0 + ρgh + h )cosα) f = 0 The pressure is continuous at y = h : p 0 + ρgh cosα) = ρgh cosα) + f x) f = p 0 + ρgh + h )cosα) f = 0 This gives the pressure: p y) = p y) = py) = ρg cosα)y + p 0 + ρg cosα)h + h ) We now have two momentum equations in x: And four boundary conditions: d u 0 = ν + g sinα) ) dy d u 0 = ν + g sinα) ) dy BC: No slip on the plate: u 0) = 0 du BC: No viscous forces on the free surface: µ dy = 0 y=h+h du BC3: Force balance at the fluid interface: µ du dy = µ y=h dy BC4: Continous velocity at the interface: u y=h = u y=h y=h ) du dy = g ν y sinα) + c u = g ν y sinα) + c y + c ) du dy = g ν y sinα) + c u = g ν y sinα) + c y + c BC c = 0 BC µ g ν h + h )sinα) + c ) = 0 c = g ν h + h )sinα) BC3 µ g ν y sinα)+c ) = µ g ν y sinα)+c ) {µ = νρ} c = µ µ c = g ν h +h )sinα) BC4 This gives us the velocities: g h sinα) + g h + h )sinα)h = g h sinα) + g h + h )sinα)h + c ν ν ν ν ) h c = g sinα) h + h )h ) ν ν u y) = g sinα) ν u y) = g sinα) ν u y) = g ν y sinα) + g ν h + h )sinα)y u y) = g sinα) ν [h + h )y y ] y + g sinα) ) h h + h )y + g sinα) ν h + h )h ) ν ν [h + h )y ] [ ] ) h y + g sinα) h + h )h ν ν The velocity in layer does depend on h but not on the viscosity in layer. This is because the depth is important for the tangential stress boundary condition at the interface, unlike the viscosity. There is no acceleration of the upper layer and thus the tangential stress must be equal to the gravitational force on the upper layer which depends on h but not on ν. 7

MA3D1 Fluid Dynamics Support Class 5 - Shear Flows and Blunt Bodies

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