1. Lec 25: Viscosity, Bernoulli equation

2. For next time:
Read: § 11-4 to
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 note conversion factor in caption.

15. 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.

16. 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

17. 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

18. 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:

19. Fluid mechanics
The result is or

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

21. Fluid Mechanics
With v = constant (incompressible) so

22. 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.

23. 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:

24. 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.

25. 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.

26. 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

27. Fluid Mechanics or
These are forms of the Bernoulli equation

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

29. 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.)

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

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

32. Fluid Mechanics The second term is the velocity or dynamic pressure.
For flows where the elevation z is approximately constant,

33. 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

34. 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.

35. 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.

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

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

38. 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.

39. 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:

40. 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 .

41. 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

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