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Mechanical Sciences - II

Mechanical Sciences - II. Chapter I Fluids & Fluid Properties. F. V. Recap: Fluids. The upper plate keeps on moving, and the fluid keeps on deforming. No equilibrium state – but an equilibrium velocity V of the plate.

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Mechanical Sciences - II

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  1. Mechanical Sciences - II

  2. Chapter I Fluids & Fluid Properties

  3. F V Recap: Fluids • The upper plate keeps on moving, and the fluid keeps on deforming. No equilibrium state – but an equilibrium velocity V of the plate. • The equilibrium velocity V increases as the shear force F increases. • This means that mere deformation of fluid does not build up resisting forces. • Since there is an equilibrium velocity (more the velocity larger the force), a resisting force is developing for sure. • The resisting force is, thus, dependent not on deformation, but on deformation rates.

  4. Recap: Fluids A fluid does not resist shear force by acquiring an equilibrium strain, but by acquiring an equilibrium rate of strain. A fluid deforms continually under the action of a shear force, but at a rate determined by the magnitude of the shear force and the fluid properties. Vijay Gupta

  5. Fluids A fluid at rest (or in uniform motion) cannot sustain any shear force. Vijay Gupta

  6. Pressure Imagine a small surface immersed in a fluid: • A fluid at rest (or in uniform motion) cannot apply a shear force on this surface. • The only possible force is, thus, normal to the surface, whatever be the orientation of the surface. • This force is compressive, and depends on the area of the surface.

  7. Pressure • The compressive force acting on a surface immersed in a fluid is expressed as a force per unit area and is termed pressure. • Thus, pressure is measured as force per unit area, and its units are N/m2. • It is given the name Pascal (Pa). • Typical atmospheric pressure is about 105 Pa.

  8. Pressure at a Point Horizontal force balance:px = pn Vertical force balance: py = pn A δl δy θ B O δx Pascal law: Pressure at a point is same in all directions Vijay Gupta

  9. Gauge and Vacuum Pressures Actual Pressure at a point Gauge Pressure Absolute pressure Atmospheric pressure -ve gauge, or Vacuum pressure Actual Pressure at another point Absolute pressure Absolute zero pressure

  10. Stresses in Fluids in Motion When a fluid is in motion, there may exist shear stresses as well.

  11. Rate of Deformation y γ x Rate of deformation = Moving with velocity Vo Stationary Deformation in time δt = γ Vijay Gupta

  12. Newton Law of Viscosity For the parallel-plate geometry shown earlier: Shear stresses in fluids are proportional to the rates of shear strain: The constant of proportionality μis termed as viscosity, and is a material property Vijay Gupta

  13. Effect of Viscosity Vo Penetration of the effect of motion of the bottom plate due to action of viscosity.

  14. Effect of Viscosity Penetration depth increases with time. Action of viscosity is like a process of diffusion. The viscous action of the fluid resists the motion of the lower plate. At the steady-state, the fluid attempts to drag the upper plate to the right.

  15. Dimensionsand Units of Viscosity Therefore, Dimensions of viscosity: Units: kg/ms [= Pa.s] centipoise = 10-3Pa.s Vijay Gupta

  16. Typical Values of Viscosity Vijay Gupta

  17. y + + x Sign Convention A positively (negatively) oriented stress on a face with normal in a positive (negative) coordinate direction is termed a positive stress.

  18. y - + x Sign Convention A negatively (positively) oriented stress on a face with normal in a positive (negative) coordinate direction is termed a positive stress.

  19. Surface Tension Water strider Vijay Gupta

  20. Surface Tension Vijay Gupta

  21. L Surface Tension Force F ~ L F = σ L For water in contact with air, σ = 7.56X10-2N/m

  22. Compressibility Change in pressure = dp Decrease in volume = - dV Fractional decrease in volume = -dV/V Compressibility = Units?

  23. Compressibility in perfect gas For iso-thermal process: pV = C gives V = C/p Then, dV/dp = - C/p2= - V/p Therefore, β = - (1/V)(dV/dp) = 1/p

  24. Compressibility in perfect gas For adiabatic process: pVγ = C gives V = Kp1/γ Then, dV/dp = -K(1/γ)/p1/γ - 1= -(1/γ)V/p Therefore, β = - (1/V)(dV/dp) = 1/γp Sinceγ > 1, adiabatic compressibility is less than the iso-thermal compressibility. Why?

  25. Non-Newtonian Effects Rod-Climbing: An example of Weissenberg effect. A rod is rotating is a dish of a viscoelastic liquid and the liquid climbs up the rod. The fluid is high-molecular weight  PIB in low-molecular weight polyisobutylene.

  26. Non-Newtonian Effects Die swell: Die swell for a 2.0% aqueous solution of polyacrylamide. The liquid fall under gravity on exiting the capillary.

  27. Non-Newtonian Effects Open Siphon: Extensional-viscosity effect The open-siphon effect of a 0.75% aqueous solution of PolyoxWSR 301. Only the slightest spilling will part empty the beaker.

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