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Lec 25: Viscosity, Bernoulli equation

FLUI D MECHANI CS

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• 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

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Viscosity

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

F

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Shearing of a fluid

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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, .

dy

du

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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.

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TEAMPLAYTEAMPLAY

• 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.

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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

d

Vμ

dy

duμτ

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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=2RL. So at the inner cylinder,

F=2 R2 L/d

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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:

e

2e

eei

2i

iicvcv gz

2hmgz

2hmWQ

dt

dE vv

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Fluid mechanics

• The result is

• or

12

21

22

12cv zzg22

hhmWQvv

e

2e

eei

2i

iicv gz2

hmgz2

hmWQ0vv

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Fluid mechanics

• On a mass-specific basis

• And rearranging the enthalpy terms

12

21

22

12 zzg22

hhwq vv

12

21

22

121122 22)( zzguuvPvPwq

vv

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Fluid Mechanics

• With v = constant (incompressible)

• so

)PP(vvPvP 121122

12

21

22

1212 zzg22

)PP(vw)uu(q vv

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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.

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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:

)PP(vw 12

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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.

)PP(vw 12

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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.

0zzg22

)PP(v 12

21

22

12 vv

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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

1v

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Fluid Mechanics

0zzg22

)PP(12

21

2212

vv

•or

ttanconsgz

2

Pgz

2

P2

222

1

211

vv

•These are forms of the Bernoulli equation

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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.

ttanconsgz

2

P 2

v

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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.)

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Fluid Mechanics

• The Bernoulli equation can also be expressed in terms of pressures:

ttanconsgz2

1P 2 v

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Fluid Mechanics

ttanconsgz2

1P 2 v

•P is called the static pressure and would be measured as shown in a fluid flow:

Flow direction

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Fluid Mechanics

• The second term is the velocity

or dynamic pressure.

• For flows where the elevation z is approximately constant,

2

2

1V

stagnation

2

P PressureStagnation

PressureDynamicPressureStaticvρ2

1P

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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

Flow direction

Static pressure

staticstagnation2 PPρV

2

1

Stagnation pressure

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TEAMPLAYTEAMPLAY

• 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.

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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.

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Fluid mechanics

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

constant212

P gz v

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Fluid mechanics

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

constant

2

2P

zg g

v

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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.

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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:

2

222

1

211 z

g2g

Pz

g2g

P

vv

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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 .

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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

0

g2g

PH0

g

P 2221

v

gH22 V

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Fluid Mechanics

• And so the exit velocity is

• the discharge flow rate is

sec/ft1.32)ft16)(sec/ft2.32(2 22 V

3in ft ftin sec secft

2

22 2 2

41 132.1 2.8

4 4 12Q A d

V V