(Viscosity)
Planar Laminar flow
With solids, friction is the force that opposes motion between two surfaces pressed together. In fluid flow, viscosity is the force that opposes motion in and of a fluid. The easiest case to consider is laminar flow.
Laminar flow occurs when a fluid can be pictured as split into thin layers which slide smoothly over each other.
The thin layers (or laminas) are held back by viscous drag between the surfaces of the layers.
The thin layers (or laminas) are held back by viscous drag between the surfaces of the layers.
For example, if two flat solid plates are separated by a viscous fluid, an external force is needed to slide the top plate at constant speed over the fixed lower plate.
Viscous drag acts not only between the fluid and the upper plate but also between adjacent laminas of fluid. |  |
The velocity of the laminas differs by a small amount from the layers on each side as shown in the diagram.
The velocity decreases uniformly from the upper plate speed to zero at the lower plate.
The velocity decreases uniformly from the upper plate speed to zero at the lower plate.
Laminar flow of a fluid in a tube
Laminar flow in a tube can be thought of as co-axial tubes sliding past each other.
The velocity profile for the sliding tubes has the fluid moving slowly on the outside and quickly at the centre. |  |
Even though the fluid elements travel in a straight line, the flow is rotational because a small paddle wheel placed in the tube (anywhere but the exact centre) will rotate. The rotation is due to the difference in velocities.
Newton's law of Viscosity
Shear stress is the tangential force (in the plane of the lamina) divided by the area across which the force acts (the area vector is at right angles to the plane of the lamina).
Shear strain is the distance moved in the direction of the force divided by the perpendicular distance that separates the opposing forces producing the "twist". |  |
For solids, shear stress divided by shear strain gives an elastic modulus.
For viscous liquids, since the strain is increasing all the time, shear stress divided by the rate of shear strain gives the viscosity coefficient.
In particular, for the simple two flat plates geometry
The force depends directly on the area of the plates and the velocity gradient between the plates. This is known as Newton's Law of Viscosity. Force/Area has the units of pressure, i.e. Pascal (Pa). Velocity/distance has the units of s-1. Therefore, the viscosity coefficient has the unit of Pa.s. There is an older unit called the poise, 1 poise = 0.1 Pa.s.
Viscosity coefficients for some fluids
Fluid
|
Temperature
|
Coefficient
|
CO2
|
20°C
|
15 μPa.s
|
Air
|
20°C
|
18 μPa.s
|
Petrol
|
20°C
|
290 μPa.s
|
Water
|
90°C
|
320 μPa.s
|
Water
|
20°C
|
1 mPa.s
|
Blood
|
37°C
|
2 mPa.s
|
Motor Oil
|
20°C
|
0.03 Pa.s
|
Motor Oil
|
0°C
|
0.11 Pa.s
|
Glycerine
|
20°C
|
1.5 Pa.s
|
Typical Polymers*
|
Tg+20°C
|
10 GPa.s
|
Tg+10°C
|
100 GPa.s
| |
Tg
|
100 TPa.s
| |
Soda Lime Glass
|
800°C
|
1 MPa.s
|
515°C
|
100 TPa.s
|
*Tg: See below. The glass transition temperature is defined for the situation where there is a smooth transition from liquid to solid. At the glass transition temperature the material is considered to become solid.
Example F5
An airtrack supports a cart that rides on a thin cushion of air 1 mm thick and 0.04 m2 in area. The viscosity of air is 18 x 10-6 Pa.s. Find the force required to move the cart at a constant speed of 0.2 m.s-1
An airtrack supports a cart that rides on a thin cushion of air 1 mm thick and 0.04 m2 in area. The viscosity of air is 18 x 10-6 Pa.s. Find the force required to move the cart at a constant speed of 0.2 m.s-1
Answer F5
Viscosity at the Atomic Level
At the atomic level, viscous movement occurs when tightly bonded molecular units flow past each other. For this to occur, there has to beweak bonding between the units. This means that viscosity is important in materials with secondary bonding between molecular units. Also, materials with a large number of defects will have similarly weakened bonding.
Materials with secondary bonding:
Material
|
Bonding
|
polar liquid (water)
|
Hydrogen
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molten metals
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Ion/Electron
|
polymers
|
Van der Waals
|
Viscosity and Temperature
The bonds between atoms in the solid state are due to a large negative potential energy and a small positive kinetic energy.
(See Interatomic Potential Energy Function) |  |
At absolute zero (0K, -273°C) there is no thermal motion, i.e. no kinetic energy, to assist bond breaking.
As the temperature increases, atoms are given more and more kinetic energy so the total energy of the bond becomes less negative. The bonds between atoms weaken with increasing temperature.
As the temperature increases, atoms are given more and more kinetic energy so the total energy of the bond becomes less negative. The bonds between atoms weaken with increasing temperature.
When the potential energy and kinetic become comparable a solid can change phase into a liquid.
When the kinetic energy dominates then there can be a change of phase into a gas.
The phase changes are usually abrupt, but in some situations there can be a gradual change and a smooth transition between solid and liquid.
SiO2 can form an amorphous glass (below left) or a crystal quartz (below right) depending on the speed of the temperature change.
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Above 1600° SiO2 is a liquid with its tetrahedral base units in random motion. If liquid SiO2 cools quickly then the tetrahedral units do not have time to move to their lowest energy configuration and the tetrahedrons are linked in random orientations by secondary bonds, thus forming a glass. If liquid SiO2 cools slowly then the tetrahedrons can jostle past each other into their lowest energy configurations and link with long range order to form a crystal.
The change to a crystal is abrupt with a sudden increase in the viscosity coefficient. The change to a glass is smooth with a gradual increase in viscosity. At some point there will be a large increase in viscosity and this is called the glass transition temperature (Tg in the above graph) and the material is then said to be a solid.
Polymers and Temperature
Polymers are long chains of atoms joined together with primary bonds but cross linked between chains with secondary bonds. Temperature has a stronger effect on the secondary bonds than the primary bonds.
At low temperatures, polymers like most materials, are brittle. There is an elastic region ending in brittle fracture.
At normal temperatures polymers have a non-linear elastic region with a smaller modulus (slope) and a yield point followed by limited plasticity before fracture. At warm temperatures the modulus decreases further and there is a larger easy glide region due to secondary bonds breaking and reforming. The chains tend to align with the applied stress. At high temperatures there is viscous flow. |  |
Heavily cross-linked polymers, like rubber, may have different properties such as non-linear elasticity and high yield points.
Viscosity and Time
Time is not usually a factor for solids, but provides opportunities for bond breaking when secondary bonds are important. If the atoms are moving apart at a particular instant of time then the probability of the bond between them breaking is enhanced.
Example F6
A glass slab at room temperature, with dimensions 140 mm × 60 mm × 20 mm, has a shear force of 8.4 kN across its largest faces. The viscosity of this glass at room temperature is 10+16 Pa.s and the average separation between molecules is 0.5 nm. Find the time for two neighbouring molecules to slide past each other.
A glass slab at room temperature, with dimensions 140 mm × 60 mm × 20 mm, has a shear force of 8.4 kN across its largest faces. The viscosity of this glass at room temperature is 10+16 Pa.s and the average separation between molecules is 0.5 nm. Find the time for two neighbouring molecules to slide past each other.
Answer F6
Note that this is independent of the atom spacing.
Drag Force and Terminal Velocity
An object moving through a viscous fluid has a resistive drag force exerted on it by the fluid. This prevents the object's velocity from increasing without limit (e.g. cars, boats) as it eventually reaches the maximum applied force. It means there is a terminal speed which is the maximum speed for the given conditions.
In general, 
, where x starts at 1 for low speeds and increases to 2 ,4 etc for high speeds.
, where x starts at 1 for low speeds and increases to 2 ,4 etc for high speeds.
The drag force is affected by shape and speed, through the drag coefficient, b and, v.
For a car at typical speeds drag force is given by:
 |
• CD is the drag coefficient produced
by the shape of the car. • A is the "front-on" cross-sectional area. • ρ is the density of the fluid. • v is the speed of the car. |
For most cars CD lies between 0.2 and 0.5 (the Model T Ford was 0.7).
Drag coefficients for some passenger vehicles
Vehicle (class)
|
CD
|
CD×A (m2)
|
VW Polo (class A)
|
0.37
|
0.636
|
Ford Escort (class B)
|
0.36
|
0.662
|
Opel Vectra (class C)
|
0.29
|
0.547
|
BMW 520i (class D)
|
0.31
|
0.649
|
Mercedes 300SE (class E)
|
0.36
|
0.785
|
Including drag, the resultant force on a car is given by:
The initial acceleration causes the speed to increase from zero, which in turn causes the acceleration to decrease. Equilibrium is achieved when the drag force equals the force the engine can provide, the acceleration goes to zero and a constant (terminal) speed is reached.
Stokes' Law and Terminal Speeds
Stokes calculated the drag force on a sphere at low speeds as 
, where η is the viscosity, r is the radius, and v, is the speed.
, where η is the viscosity, r is the radius, and v, is the speed.
Terminal speed for a sphere falling under gravity in a medium such as air or water.
As indicated before, when the acceleration goes to zero, the speed goes to a constant (terminal) value.
Terminal Speeds in Air
Object
|
Terminal Speed [m.s-1]
|
95% distance [m]
|
7kg shot put
|
145
|
2500
|
skydiver
|
60
|
430
|
baseball
|
42
|
210
|
tennis ball
|
31
|
115
|
basketball
|
20
|
47
|
ping-pong ball
|
9
|
10
|
1.5mm rain drop
|
7
|
6
|
parachutist
|
5
|
3
|
The "95% distance" is the distance to achieve 95% of the terminal speed.
Example F7
An oil drop has a density of 930 kg.m-3. The terminal velocity of a spherical drop of this oil falling in air at 20°C is 0.18 m.s-1. At 20°C, air density is 1.2 kg.m-3 and its viscosity is 18 μPa.s. Find the radius of the droplet.
An oil drop has a density of 930 kg.m-3. The terminal velocity of a spherical drop of this oil falling in air at 20°C is 0.18 m.s-1. At 20°C, air density is 1.2 kg.m-3 and its viscosity is 18 μPa.s. Find the radius of the droplet.
Answer F7
Poiseuille's Law and Laminar flow in a tube
The French prounciation of this "law" is something like "Pwasweeyer".
Consider a solid cylinder of viscous fluid, (viscosity η), flowing inside a hollow cylindrical pipe of length, L, and internal radius, R, as shown below. The flow is driven by a pressure difference, ΔP, and can be modelled as a number of thin co-axial cylinders flowing past each other. There will be a stationary thin cylinder at the outer edge and the maximum speed cylinder will be at the centre. The velocity profile will be parabolic.
The volume flux (flow rate) is given by:
|  |
In this course you will not be asked to derive this formula. For those interested.
Note that this depends directly on the pressure difference and inversely on the length. It also depends on the 4th power of the radius, so that will be a dominant factor.
• The larger the pressure difference, the greater the flux.
• The larger the cross-sectional area, the greater the flux.
• The shorter the length, the greater the flux.
• The larger the pressure difference, the greater the flux.
• The larger the cross-sectional area, the greater the flux.
• The shorter the length, the greater the flux.
Note: blood is a viscous fluid but it does not follow Poiseulle's equation because it has platelets in its plasma.
Example F8
A small pipe has an inner radius of 4 mm. A fluid with a viscosity of 4 x 10-3 Pa.s flows through it at a rate of 10-6 m3.s-1. Find the pressure difference across a 2 m length of the pipe.
A small pipe has an inner radius of 4 mm. A fluid with a viscosity of 4 x 10-3 Pa.s flows through it at a rate of 10-6 m3.s-1. Find the pressure difference across a 2 m length of the pipe.
Answer F8
Reynold's number and Turbulent flow
In turbulent flow the fluid paths change abruptly and unpredictibly with time.
This picture on the right shows water flows from a tap that are laminar on the left and turbulent on the right.
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This picture on the right shows cigarette smoke rising. It starts as laminar flow and then changes to turbulent flow.
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Poiseuille's law only holds for laminar flow. For turbulent flow you need to find experimentally what rules apply to the specific situation.
Reynold's number
Reynold's number represents the ratio of driving force to viscous force.
From Newton's law of motion, the definition of density and the equation of continuity, the driving force on the fluid is given by:
From Newton's law of viscosity, the viscous drag force is given by:
The ratio of these gives Reynold's number:
In turbulent flow, the driving force will dominate and in laminar flow the viscous force will dominate.
Thus Reynold's number 
as a ratio of these, gives an indication as to whether a fluid flow is laminar or turbulent.
as a ratio of these, gives an indication as to whether a fluid flow is laminar or turbulent.
A "rule of thumb" for laminar vs turbulent flow.
NR< 2000
|
Laminar flow
|
2000 <NR< 3000
|
Unstable, may flip between laminar and turbulent
|
NR > 3000
|
Turbulent flow
|
Example F9
A pipe with diameter 300mm has water (density 1000 kg.m-3, viscosity 1 mPa.s), flowing through it. Find Reynold's number for the flow when there is a:
(a) flow speed of 3 mm.s-1
(b) flow speed of 30 mm.s-1
(a) flow speed of 3 mm.s-1
(b) flow speed of 30 mm.s-1
Answer F9
A ten-fold increase in speed changes the flow from laminar to turbulent.
Summarising:
Laminar flow occurs when a fluid can be pictured as split into thin layers which slide smoothly over each other.
| |
Laminar flow can be in planes or cylinders
| |
Newton's law of Viscosity:
|  |
Viscosity is important in materials with secondary bonding between tightly bound molecular units.
| |
The bonds between atoms weaken with increasing temperature.
| |
Time provides opportunities for bond breaking.
| |
Viscous Drag force for a car at typical speeds
| |
Stokes' Law:
| |
Terminal speed for a sphere falling under gravity:
|  |
Poiseuille's Law:
| |
Reynold's number:
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DO YOU KNOW
1. An ideal liquid has zero viscosity and zero compressibility.
2.Viscosity is due to transport of momentum.
3. With increase in temperature, the viscosity of liquid decreases but viscosity of gases increases.
4. The reciprocal of viscosity is called fluidity.
Poiseuille's Law Derivation
Consider a solid cylinder of fluid, of radius r inside a hollow cylindrical pipe of radius R.
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The driving force on the cylinder due to the pressure difference is:
The viscous drag force opposing motion depends on the surface area of the cylinder(length L and radius r):
 |
In an equilibrium condition of constant speed, where the net force goes to zero.
We know empirically that the velocity gradient should look like this:
 |
At the centre
• r=0 •  • v is at its maximum.
At the edge
• r=R • v=0 |
From the velocity gradient equation above, and using the empirical velocity gradient limits, an integration can be made to get an expression for the velocity.
Which has a parabolic form as expected.
Now the equation of continuity giving the volume flux for a variable speed is:
Substituting the velocity profile equation and the surface area of the moving cylinder:
Poiseuille's equation
