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Question 4 (Unit 8) - 25 marks Consider the steady, laminar flow of two liquids, A and B, with viscosities MA = u and B = 2, respectively, between infinite parallel plates at % = a, as shown in the diagram below. The plate at % = a is fixed, while the plate at 2 = -a moves with constant velocity -Vi, where V > 0. The liquids do not mix, and each forms a layer of depth a. There is an applied pressure gradient acting on both liquids, given by Vp = - Ci (where C > 0 is constant), and the effects of gravitation can be assumed to be negligible. z a fixed plate liquid A pressure gradient Vp = Ci PA u interface pressure gradient x liquid B Vp = Ci ub = 2 -Vi -a moving plate (a) Assuming that each fluid is of constant viscosity and is Newtonian, what other assumption must be made about the fluids in order to apply Equations (2.8) on page 179 of Unit 8? Given the problem statement, why is this additional assumption reasonable? [2] In order to apply the equations on page 179, we must assume that the density is constant. This is reasonable because both fluids are labelled as liquids where, generally, changes in density are negligible. Another assumption is that the flow is fully-developed due to the plates being infinite. (b) The following assumptions are to be made. Write down the mathematical consequences of each of them. 1. The flow is two-dimensional. 2. There is no variation in the direction into the page. 3. The flow is steady. 4. There is no variation of velocity parallel to the plates. Hence write down the continuity equation for either of the two liquids. [3] 1) The velocity field has no y-component so u = U1 + U1 k and = 0 2) There is no change in the y-direction so = 0 3) There is no change with respect to time = at 4) There is no change in u with respect to so = is = ôx = The continuity equation states V . u = 0 which means + From assumptions 2 and 4 above, we know that our = = 0. This simplifies the continuity equation for these liquids = (c) Show that the fluid velocities UA in liquid A and u B in liquid B are given by UA = (z) i and ub = UB (z) i. State the boundary conditions at the upper and lower plates, and at the interface % = 0. (Note that the shear stress must vary continuously between the plates.) 7] For both fluids, the fluid velocity field is From assumption 1, U2 = 0, therefore the velocity field From assumption 2, ay of = 0, therefore the velocity field is u = a From assumption 3, at a = 0, therefore the velocity field is = From assumption 4, dui = ous ax = 0, therefore the velocity field is = x ou3 From the continuity equation, we know that = 0. This means that = constant. Because the ôz plates form boundaries, we know that = 0 and because the fluid cannot pass the boundary, if U3 is a constant, it must be 0 at all points in the fluid. This means U3(2) K = 0. We are left with UA = and ub = The boundary condition at the upper plate is UA = 0 for Z = a The boundary condition at the lower plate is UB = - V for Z = - a The interface boundary condition is UA = ub and TA = TB for Z = 0 (d) Write down the .--components of the Navier-Stokes equations for UA and UB. Solve them, using the boundary conditions from part (c), to show that - - - - , = [13]

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