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L ECTURE 5 Properties Of Fluids-Cont. By Dr. Mohamed Fekry 2 nd Sem.1434.

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Presentation on theme: "L ECTURE 5 Properties Of Fluids-Cont. By Dr. Mohamed Fekry 2 nd Sem.1434."— Presentation transcript:

1 L ECTURE 5 Properties Of Fluids-Cont. By Dr. Mohamed Fekry 2 nd Sem.1434

2 P ROPERTIES O F F LUIDS Density (  ) and Specific Volume (v) Specific Gravity (SG) Specific Weight (  ) Density of ideal gas Coefficient of Compressibility (k) Coefficient of Volume Expansion (b) Viscosity (  ) Surface Tension (  )& Capillary Effect (h) 2

3 3 - When two solid bodies in contact move relative to each other, a friction force develops at the contact surface in the direction opposite to motion. - The situation is similar when a fluid moves relative to a solid or when two fluids move relative to each other. - It appears that there is a property that represents the internal resistance of a fluid to motion or the “fluidity,” and that property is the viscosity. The force a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity.

4 4 Real Experiment: One moving shaft inside another hollow shaft filled with oil. In the case the thickness of oil film is equal to: L = Ri − Rs Results: Mention the relationships of the following parameters. F = applied force A = contact area L = thickness of the fluid layer V = the velocity of the moving plate Viscosity measures a fluids ability to resist shear stress Viscosity (  )

5 Hypothetical Experiment : 5 Where: v = Velocity of the moving plate (varies linearly from 0 at the stationary surface to maximum at the contact surface between the moving plate and oil). A = Contact area between the moving plate and oil. F = Applied force to the moving plate. L = Thickness of oil layer

6 6 L & A = constants F & v v & A = constants F & L L & v = constants F & A Where μ is the proportional constant.

7 7 By observation: F = μA v/L Where F is the external applied force. Summation of forces: Σ F x = 0 Driving forces – Resisting forces = 0 F − F R = 0 ↔ F = F R = τ A F\A = μ v/L τ = μ v/L Assumptions: Small gap thickness v is not too large v/L = Δ v /Δ L = Slope of the velocity distribution (assuming linear distribution)

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10 10 The ideal fluid, with no viscosity (μ = 0), falls on the horizontal axis, while the true elastic solid plots along the vertical axis. A plastic that sustains a certain amount of stress before suffering a plastic flow corresponds to a straight line intersecting the vertical axis at the yield stress. There are certain non-Newtonian fluids in which μ varies with the rate of deformation. These are relatively uncommon in engineering usage. Typical non-Newtonian fluids include paints, printer’s ink, gels, sludge and slurries, and certain plastics.

11 Consider a fluid layer of thickness within a small gap between two concentric cylinders, such as the thin layer of oil in a journal bearing. The gap between the cylinders can be modeled as two parallel flat plates separated by a fluid. Noting that torque is T= FR (force times the moment arm, which is the radius R of the inner cylinder in this case), the tangential velocity is V =ωR (angular velocity times the radius), and taking the wetted surface area of the inner cylinder to be A=(2πRL) by disregarding the shear stress acting on the two ends of the inner cylinder, torque can be expressed as where L is the length of the cylinder and n is the number of revolutions per unit time, which is usually expressed in rpm (revolutions per minute). Note that the angular distance traveled during one rotation is 2π rad, and thus the relation between the angular velocity in rad/min and the rpm is ω = 2π n.. Therefore, two concentric cylinders can be used as a viscometer, a device that measures viscosity. 11

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