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Lec 25 : Viscosity, Bernoulli equation

Lec 25 : Viscosity, Bernoulli equation. For next time: Read: § 11-4 to 11-12. Outline: Viscosity Bernoulli equation Examples Important points: Know what a Newtonian fluid is and how to calculate shear forces Understand when you can apply the Bernoulli equation

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Lec 25 : Viscosity, Bernoulli equation

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  1. Lec 25: Viscosity, Bernoulli equation

  2. For next time: • Read: § 11-4 to 11-12. • Outline: • Viscosity • Bernoulli equation • Examples • Important points: • Know what a Newtonian fluid is and how to calculate shear forces • Understand when you can apply the Bernoulli equation • Know how to use different forms of the Bernoulli equation to solve problems

  3. Viscosity • Consider a stack of copy paper laying on a flat surface. Push horizontally near the top and it will resist your push. F

  4. Viscosity • Think of a fluid as being composed of layers like the individual sheets of paper. When one layer moves relative to another, there is a resisting force. • This frictional resistance to a shear force and to flow is called viscosity. It is greater for oil, for example, than water.

  5. Shearing of a solid (a) and a fluid (b) The crosshatching represents (a) solid plates or planes bonded to the solid being sheared and (b) two parallel plates bounding the fluid in (b). The fluid might be a thick oil or glycerin, for example.

  6. Shearing of a solid and a fluid • Within the elastic limit of the solid, the shear stress  = F/A where A is the area of the surface in contact with the solid plate. • However, for the fluid, the top plate does not stop. It continues to move as time t goes on and the fluid continues to deform.

  7. Shearing of a fluid • Fluids are broadly classified in terms of the relation between the shear stress and the rate of deformation of the fluid. • Fluids for which the shear stress is directly proportional to the rate of deformation are know as Newtonian fluids.

  8. Shearing of a fluid • Engineering fluids are mostly Newtonian. Examples are water, refrigerants and hydrocarbon fluids (e.g., propane). • Examples of non-Newtonian fluids include toothpaste, ketchup, and some paints.

  9. Shearing of a fluid • Consider a block or plane sliding at constant velocity u over a well-oiled surface under the influence of a constant force Fx. • The oil next to the block sticks to the block and moves at velocity u. The surface beneath the oil is stationary and the oil there sticks to that surface and has velocity zero.

  10. Shearing of a fluid • No-slip boundary condition--The condition of zero velocity at a boundary is known in fluid mechanics as the “no-slip” boundary condition.

  11. Shearing of a fluid

  12. Shearing of a fluid • It can be shown that the shear stress  is given by • The term du/dy is known as the velocity gradient and as the rate of shear strain. • The coefficient is the coefficient of dynamic viscosity, .

  13. Coefficient of dynamic viscosity • Intensive property. • Dependent upon both temperature and pressure for a single phase of a pure substance. • Pressure dependence is usually weak and temperature dependence is important. • Can be found in Figure 9-10--note conversion factor in caption.

  14. TEAMPLAY • Determine the force to slide, at a speed of 0.5 m/s, two blocks of 1.0 m square separated by 2 cm with SAE 10W-30 oil and determine the same force if the blocks are separated by water. Assume that the temperature is 40 C.

  15. Shearing of a fluid • And we see that for the simple case of two plates separated by distance d, one plate stationary, and the other moving at constant speed V

  16. Shearing of a fluid • Two concentric cylinders can be used as a viscometer to measure viscosity For the inner cylinder, The torque is T=FR, V=R, and A=2RL. So at the inner cylinder, F=2 R2 L/d

  17. Fluid Mechanics • The text obtains the Bernoulli equation from momentum considerations, as will most fluid mechanics courses. We will obtain it from the first law of thermodyamics. • Consider the following equation for steady-state flow:

  18. Fluid mechanics • The result is • or

  19. Fluid mechanics • On a mass-specific basis • And rearranging the enthalpy terms

  20. Fluid Mechanics • With v = constant (incompressible) • so

  21. Fluid Mechanics • The term (u2 –u1) will come up later as a ‘head loss’ term, usually treated with experimental data. • It represents losses due to friction as the fluid flows. • Often ‘frictionless’, adiabatic flow is assumed and (u2 –u1) as well as q disappear.

  22. Fluid Mechanics • The work term w would normally be work done on the fluid by a compressor, fan, or pump or done by the fluid in a turbine. • For example, for frictionless flow in the absence of kinetic energy or potential energy changes:

  23. Fluid Mechanics • If work is done on the fluid by a pump, the work w will be negative, and P2 will be greater than P1 • If work is done by the fluid, as it passes through a liquid turbine, for example, then P2 will be less than P1 because w is positive.

  24. Fluid Mechanics • Return to the complete equation and think of the case of the pump. The work term can be that for the pump or fan to overcome friction in a piping or duct system. • For now let us assume the flow is frictionless and set w = 0.

  25. Fluid Mechanics • In the world of fluid mechanics, somewhat differently than for thermodynamics, density is used more often than specific volume. • We are considering incompressible fluids, so

  26. Fluid Mechanics • or • These are forms of the Bernoulli equation

  27. Fluid Mechanics • Bernoulli equation • Each of the terms has units of energy per unit mass. For adiabatic, no workinteractions, incompressible, frictionless, and steady flow, the Bernoulli equation says the energy content of the fluid [along a streamline] is a constant.

  28. Fluid Mechanics • The p/ term is just the pv term, the old flow work or flow energy term. • Energy can be traded between the flow energy (p/), kinetic energy (V2/2), and potential energy (gz), but the total energy of the flow will not change [along a streamline]. (Remember it is frictionless, adiabatic, steady, and no work is being done on it.)

  29. Fluid Mechanics • The Bernoulli equation can also be expressed in terms of pressures:

  30. Fluid Mechanics • P is called the static pressure and would be measured as shown in a fluid flow: Flow direction

  31. Fluid Mechanics • The second term is the velocityor dynamic pressure. • For flows where the elevation z is approximately constant,

  32. Fluid mechanics • The second term, and thus the velocity, can be obtained from a measurement of the static pressure and the stagnation pressure as shown. • Thus Stagnation pressure Static pressure Flow direction

  33. TEAMPLAY • A point in a flow where a fluid comes to rest (V=0) is known as a “stagnation point.” The pressure there is the stagnation pressure. • Use the Bernoulli equation and predict the stagnation pressure on the leading edge of a sailplane wing soaring at 40 mph at an altitude of 2,000 ft where the static pressure is 13.7 psia and the temperature is 60°F.

  34. Fluid Mechanics • Pitot-static tube--a device somewhat similar to the previous one used in measuring the velocity of aircraft. • Its operation is based on the Bernoulli equation and the velocity in the equation is the velocity of the aircraft.

  35. Fluid mechanics • The third term in the previous form of the equation is known as the elevation pressure.

  36. Fluid mechanics • It is also possible to write the equation in terms of elevation--often called “head”.

  37. Fluid Mechanics--example problem courtesy of Dr. Dennis O’Neal • A tank of water has a small nozzle at its base as shown. Find the velocity in ft/sec and the volumetric flow rate in ft3/sec from the nozzle.

  38. Fluid Mechanics • Assume the jet is cylindrical and that the pressure is atmospheric as soon as it leaves the nozzle. Apply Bernoulli’s equation between a point 1 on the surface of the tank and a point 2 at the nozzle exit:

  39. Fluid Mechanics • The pressure at point 1 and at point 2 is just that of the atmosphere, so P1 = P2. • At point 1 the height is z1 = H and at point 2 the height is z2 = 0. If the size of the top of the tank is very large compared to the outlet area, then (V1)2 is significantly less than (V2)2 .

  40. Fluid Mechanics • The Bernoulli Equation becomes • but the terms P1/g and P2/g are equal and cancel because the pressures are the same, and what is left is

  41. Fluid Mechanics • And so the exit velocity is • the discharge flow rate is

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