Me305_Fall2011_Sec04_HW1_solutions

Me305_Fall2011_Sec04_HW1_solutions - FALL 2011/Section...

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Unformatted text preview: FALL 2011/Section 04/ME MIDDLE EAST TECHNICAL UNIVERSITY DEPARTMENT OF MECHANICAL ENGINEERING ME 305 FLUID MECHANICS I HOMEWORK 1 SOLUTIONS 3. A fluid between two very long parallel plates is heated in a way that its viscosity decreases linearly from 0.8 Pa s at the lower plate to 0.4 Pa s at the upper plate. The spacing between the two plates is 0.5 mm. The upper plate moves steadily at a velocity of 10 m/s, in a direction parallel to both plates. The pressure is constant everywhere, the fluid is Newtonian and assumed incompressible. Neglect gravitational effects. a) Obtain the fluid velocity u as a function of y , i.e. u ( y ), where y is the vertical axis perpendicular to the plates. Plot the velocity profile across the gap between the plates using an appropriate software (hand- plotting is not acceptable). b) Calculate the value of the shear stress at the upper plate. Show the direction of the shear stress (i) on the plate, and (ii) on the top surface of the fluid element adjacent to the plate. Mark the directions with respect to the direction of the flow Solution: Assumptions: Flow parallel to plates 1-D flow Constant pressure Newtonian, incompressible fluid Neglect gravitational effects a) Taking an infinitesimal fluid element and applying force balance (assuming one dimensional flow) the force equilibrium in x direction yields U =10 m/s μ (y) h = 0.5 mm x y dx dy fluid element μ dx dy dy dτ τ + τ dx pdy pdy gG g¡ = 0 → G = constant (1) Then, Newton’s law of viscosity yields G = ¢ g£ g¡ = constant (2) The viscosity is changing linearly with respect to y. Thus, it can be expressed in the form ¢ = ¤¡ + ¥ where A , B are constants....
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Me305_Fall2011_Sec04_HW1_solutions - FALL 2011/Section...

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