Homework #4 Solution. μ 1. μ 2

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1 Homework #4 Solution 4.20 in Middleman We have two viscous liquids that are immiscible (e.g. water and oil), layered between two solid surfaces, where the top boundary is translating: y = B y = kb y = 0 μ 1 U This is a one-dimensional flow problem, where liquid will only move in the x-direction due to the influence of the top boundary. Hence, we will consider the x-component of the Navier-Stokes equation: ρ ( u x t + u u x x x + u u x y y + u u x z z ) = P x + μ u x ( 2 x u x + 2 u x z 2 ) + ρg x Since we only have velocity in the x-direction, gradients in the y-direction, there is no imposed pressure gradient, and gravity is perpendicular to the flow, we can dramatically simplify this equation: 0 = μ 2 u x This equation can be applied individually to both fluid regions. Hence, we need to solve the equations: 0 = μ 1 2 u x (1) 0 = 2 u x The superscript in the velocities denotes which region is described by the velocity. et s first consider the bottom layer of fluid, where we have the governing equation:

2 0 = d2 u x dy 2 This can be solved readily; integrating twice: Similarly, for the top fluid, dy = E u x (y) = Ey + F 0 = d2 u x (1) dy 2 u x (1) (y) = Cy + D We have four unknown constants, and hence we need a total of four boundary conditions. At the bottom of fluid 2, we have a solid boundary that is not moving. Assuming no-slip at this surface, we have the boundary condition: u x (0) = 0 (i) What about the boundary between fluids 1 and 2? We expect that the velocity profile across the entire channel should be a continuous function, and hence the velocities of the two fluids must match at their interface, which is located at y = kb: u x (1) (kb) = u x (kb) (ii) We also expect that the shear stresses imposed by fluid 1 on fluid 2 at the interface should equal the stresses imposed by fluid 2 on fluid 1 (i.e. stress should be a continuous function as well). This is known as continuity of stress! This implies, τ (1) xy (kb) = τ xy (kb) (iii) Finally, we can impose no-slip on fluid 1 at the top boundary, which gives is our fourth and final boundary condition: u x (1) (B) = U (iv) et s apply these to our equations. The velocity of the bottom fluid is given by: u x (y) = Ey + F Applying boundary condition (i), we find F = 0 According to boundary condition (iii),

3 μ 1 (1) τ (1) xy (kb) = τ xy (kb) dy y=kb = μ 1 C = E E = ) C dy y=kb We have already simplified our system and now we have the equations, u x (y) = ) Cy u x (1) (y) = Cy + D et s apply boundary condition (ii), which tells us that Applying the final boundary condition, u x (1) (kb) = u x (kb) CkB + D = ) CkB D = 1) CkB u x (1) (B) = CB + D = U D = U CB We now have two equations and two unknowns. Thus, we can solve, 1) CkB = U CB Now we can also determine D: C [ 1) kb + B] = U C = U B [ 1) k + 1] D = U CB D = U [1 1 1) k + 1 ] = 1) ku 1) k + 1

4 Therefore, we have the final equations for the velocity profiles in the two different fluid regions: u x (y) = 1 ) U [ 1) k + 1] ( y B ) u (1) x (y) 1 = U 1) k + 1 [( y B ) + (μ 1 1) k] 4.31 in Middleman Here we will ignore edge effects and assume that the flow is unidirectional. Because the flow is in the x-direction, we can start with the x-momentum Navier-Stokes equation: ρ ( u x t + u u x x x + u u x y y + u u x z z ) = P x + μ u x ( 2 x u x + 2 u x z 2 ) + ρg x In this problem, there is no x-component of the gravitational force, and we are looking for a steady-state solution, so the ρg x and time-derivative terms drop out. Furthermore, we only have a non-zero velocity in the x-direction, and a velocity gradient in the y-direction. Thus, any term with a velocity other than that in the x-direction, or a gradient in a direction other than the y-direction, drops out. We are simply left with: 0 = P x + μ 2 u x P x = μ 2 u x = C where C is some constant. Now that we have a governing equation for our flow, we need to specify boundary conditions and solve the above equation. We will only look at half of the flow domain, since the flow enters in the middle of the plates. y Q y = 2H x x = 0 x = /2 y = 0

5 We now have two separate ordinary differential equations to solve. We will first consider the pressure: Integrating: P x = C P = Cx + A Suppose the pressure at the inlet is P i and the pressure at the outlet is P o. Then we have the boundary conditions: P(x = 0) = P i P (x = 2 ) = P o Thus, A = P i. Applying the second boundary condition and solving for C, we find that: Thus, the pressure as a function of x is: C = 2(P o P i ) P = 2(P o P i ) x + P i Notice that all that really matters is the pressure difference. If we shift our pressures such that the pressure at the outlet P o is zero, then: The second ODE we have to solve is: C = 2ΔP P = 2ΔP x + ΔP μ d2 u x dy 2 = C d 2 u x dy 2 = C In the second line, we have absorbed the viscosity into our constant. Solving: We have the boundary conditions (no-slip): dy = C y + D u x = C y 2 + Dy + E 2

6 u x (y = 0) = 0 u x (y = 2H) = 0 The first B.C. immediately tells us that E = 0. We already know C from our solution of the ODE for the pressure gradient. Thus, we have to use the second boundary condition to solve for D. Doing this, we find that: D = C H = CH μ = 2ΔPH In the last expression, we substituted the expression for C we obtained previously. Therefore, our velocity profile is: u x = C y 2 + Dy + E 2 u x = ΔP y2 + 2ΔPH y = ΔP y(2h y) The volumetric flow rate can be obtained from integrating our velocity profile over the cross sectional area: Here, the integral bounds are: Q half = u x dydz 0 y 2H 0 z W Remember that we are only looking at half of the problem. We expect, by symmetry, for the other half to have an equal volumetric flowrate. Thus, the full volumetric flowrate is given by: Q = 2Q half = 2 ΔP y(2h y)dydz = 2ΔPW 2H (2Hy y 2 )dy 0 = 2ΔPW = 2ΔPW [Hy 2 y3 2H 3 ] 0 [4H3 8H3 3 ]

7 Solve for ΔP: = 8ΔPWH3 3 ΔP = 3Q 8WH 3 We can substitute this into our expression for the pressure as a function of x: P(x) = 2ΔP x + ΔP = 3Q 2x (1 8WH3 ) astly, to get the total pressure acting on one of the plates (both halves), we have to integrate: Where the integral bounds are: Thus, F = 2W F = 2 P(x)dxdz 0 x /2 0 0 z W /2 3Q 2x (1 8WH3 ) dx In the above expression, we have multiplied the integral by two to account for the fact that we are integrating only over half of the total plate surface. Evaluating, = 3Q /2 x2 [x 4WH3 ] 0 = 3Q2 16H 3

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