atthe top of the oxygenator to slowly flow along the plastic bag's walls downward under theinfluence of gravity. Blood and Air were in direct contact, thus supporting a very efficient masstransport of O2 from Air to blood and transport of CO2 from blood to Air. Later, in the mid 20thcentury, scientists and engineers replaced these bags with gas permeable membranes, whichfacilitated gas transport between blood and Air. Nevertheless, the mass transport coefficient inmembrane-supported devices was lower than in the Direct Blood oxygenator. Consider isothermal, steady, unidirectional laminar flow of Blood down the wall of the direct contact oxygenator as illustrated in Figure 2a. The thin film of blood of approximate thickness8 = 165 [um] is flowing due to gravity only. The Shear Stress - Shear Rate data for blood at 25°C is presented in Figure 2b. a) (30 points) Using the data presented in Figure 2b determine the coefficients of the Shear Rate - Shear Stress relationship for Blood. b) (10 points) Apply the continuity equation for this application. What is the conclusion? c) (30 points) Develop a mathematical model [differential equation(s) + boundary conditions]that will represent the flow of Blood in the film along the walls of the plastic bag. Start from the conservation of momentum equations (Navier-Stokes) and show your work for the simplification. d) (50 points) Solve the mathematical model developed in (c) and obtain an algebraic expression that will represent the velocity profile u,(y) of Blood. e) (30 points) Develop the expression for the volumetric flow rate (Q) of blood in the Direct Blood oxygenator. Determine the volumetric flow rate of blood in Q(=)mL/min] if the width of the bag is W = 1.25 [m]. f) (10 points) Make a graph u (y) versus y; use Excel. g) (30 points) If the exponent 'n' in the solutions obtained in parts (d) and (e) is set to n =1,(and n= u) do your solutions reduce to i) velocity profile and ii) volumetric flow rate that could be obtained for a Newtonian fluid. Check and show all your work. Assumptions: The flow is assumed to be fully developed, isothermal, unidirectional and laminar. Momentum transfer between Blood and Air is negligible. One could assume that bloodди,is a non-Newtonian Power-Law fluid: T, =-nдуAlso, ignore entrance and exit effects of Blood flow. State any additional assumption. \begin{aligned} &\text { Momentum equation in } x \text { direction: }\\ &\rho\left[\frac{\partial u_{x}}{\partial t}+u_{x} \frac{\partial u_{x}}{\partial x}+u_{y} \frac{\partial u_{x}}{\partial y}+u_{z} \frac{\partial u_{x}}{\partial z}\right]=-\left[\frac{\partial \tau_{x x}}{\partial x}+\frac{\partial \tau_{x y}}{\partial y}+\frac{\partial \tau_{x z}}{\partial z}\right]+\rho g_{x} \end{aligned} \begin{aligned} &\text { Momentum equation in } y \text { direction: }\\ &\rho\left[\frac{\partial u_{y}}{\partial t}+u_{x} \frac{\partial u_{y}}{\partial x}+u_{y} \frac{\partial u_{y}}{\partial y}+u_{z} \frac{\partial u_{y}}{\partial z}\right]=-\left[\frac{\partial \tau_{y x}}{\partial x}+\frac{\partial \tau_{y y}}{\partial y}+\frac{\partial \tau_{y z}}{\partial z}\right]+\rho g_{y} \end{aligned} \begin{aligned} &\text { Momentum equation in z direction: }\\ &\rho\left[\frac{\partial u_{z}}{\partial t}+u_{x} \frac{\partial u_{z}}{\partial x}+u_{y} \frac{\partial u_{z}}{\partial y}+u_{z} \frac{\partial u_{z}}{\partial z}\right]=-\left[\frac{\partial \tau_{z x}}{\partial x}+\frac{\partial \tau_{z y}}{\partial y}+\frac{\partial \tau_{z z}}{\partial z}\right]+\rho g_{z} \end{aligned}

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