Stokes Law

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  • 02.05.2020
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Viscosity of Fluid Flow Stokes’s Law

to apply Stokes formula when solving experiment, calculation and qualitative problems

Learning Objective for the lesson

Viscosity
Velocity gradient
Shearing stress
Absolute viscosity
Kinematic viscosity
Resistive force
Cohesion or Intermolecular forces
Molecular collisions

Vocabulary

DEMONSTRATION ON FLUID FLOW SPEED

Which liquid would flow faster?

Video on Liquid Viscosities

https://www.youtube.com/watch?v=uXTjOpmlyZw

It is the quantity that describes a fluid’s resistance to flow.
A fluid with a higher viscosity flows slowly; while a fluid with low viscosity flows fast.

Viscosity

We can say:
The water is less viscous than oil.
The water is thinner than oil.
Vice versa, we can say:
The oil is more viscous than water.
The oil is thicker than water.

Comparing Fluid Viscosity

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Absolute Viscosity
Kinematic Viscosity

Types of Viscosity

It is sometimes called dynamic viscosity or simple viscosity.
It is the ratio of the shearing stress to the velocity gradient of the fluid.
It uses the symbol “eta” ( η ).

Absolute Viscosity

How do scientists and engineers determine the viscosity of fluids?

Suppose that we have two plates separated by a layer of liquid (d) with a certain viscosity (η).

d

A horizontal force is then applied to the moving plate.

F

d

The force required to move the upper plate distorts a portion of the liquid from its original shape.

A

d

Which layer of liquid has the highest velocity? Lowest velocity?

Absolute Viscosity

It is the ratio of the shearing stress to the velocity gradient of the fluid.
It uses the symbol “eta” ( η ).

𝜂𝜂= 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑠𝑠ℎ𝑒𝑒𝑎𝑎𝑟𝑟𝑖𝑖𝑛𝑛𝑔𝑔 𝑠𝑠𝑡𝑡𝑟𝑟𝑒𝑒𝑠𝑠𝑠𝑠 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑣𝑣𝑒𝑒𝑙𝑙𝑜𝑜𝑐𝑐𝑖𝑖𝑡𝑡𝑦𝑦 𝑔𝑔𝑟𝑟𝑎𝑎𝑑𝑑𝑖𝑖𝑒𝑒𝑛𝑛𝑡𝑡 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡

𝜂𝜂= 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑠𝑠ℎ𝑒𝑒𝑎𝑎𝑟𝑟𝑖𝑖𝑛𝑛𝑔𝑔 𝑠𝑠𝑡𝑡𝑟𝑟𝑒𝑒𝑠𝑠𝑠𝑠 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑣𝑣𝑒𝑒𝑙𝑙𝑜𝑜𝑐𝑐𝑖𝑖𝑡𝑡𝑦𝑦 𝑔𝑔𝑟𝑟𝑎𝑎𝑑𝑑𝑖𝑖𝑒𝑒𝑛𝑛𝑡𝑡 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡

𝜂𝜂= 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑠𝑠ℎ𝑒𝑒𝑎𝑎𝑟𝑟𝑖𝑖𝑛𝑛𝑔𝑔 𝑠𝑠𝑡𝑡𝑟𝑟𝑒𝑒𝑠𝑠𝑠𝑠 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑣𝑣𝑒𝑒𝑙𝑙𝑜𝑜𝑐𝑐𝑖𝑖𝑡𝑡𝑦𝑦 𝑔𝑔𝑟𝑟𝑎𝑎𝑑𝑑𝑖𝑖𝑒𝑒𝑛𝑛𝑡𝑡 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡

𝜂𝜂= 𝐹 𝐴 Δ𝑣 Δ𝑑 𝐹 𝐴 𝐹𝐹 𝐹 𝐴 𝐴𝐴 𝐹 𝐴 𝐹 𝐴 Δ𝑣 Δ𝑑 Δ𝑣 Δ𝑑 Δ𝑣𝑣 Δ𝑣 Δ𝑑 Δ𝑑𝑑 Δ𝑣 Δ𝑑 𝐹 𝐴 Δ𝑣 Δ𝑑

𝐹=𝜂𝐴 Δ𝑣 Δ𝑑

For a fixed speed v, the force required to move the upper plate is equal to:

𝑭=𝜼 𝑨𝒗 𝒅

Relationship between quantities

An increase in fluid viscosity, increases the force required.

𝜂𝜂≈𝐹𝐹

An increase in the contact area of the plate, increases the force required.

𝐴𝐴≈𝐹𝐹

An increase in velocity, increases the force required.

𝑣𝑣≈𝐹𝐹

An increase in the liquid layer, decreases the force required.

1 𝑑 1 1 𝑑 𝑑𝑑 1 𝑑 ≈𝐹𝐹

Absolute Viscosity

Its SI unit is Pascal second [ Pa·s ]
But its most common unit is dyne second per square centimeter [ dyne·s·cm-2 ] or poise [P].

1 𝑃𝑎∙𝑠=10 𝑃

It is the ratio of the viscosity of a fluid to its density.
It uses the symbol “nu” ( ν ).


Its SI unit is meter square per second [m2/s].
Its most common unit is centimeter square per second [ cm2/s ] or stokes [St].

Kinematic Viscosity

𝜈𝜈= 𝜂 𝜌 𝜂𝜂 𝜂 𝜌 𝜌𝜌 𝜂 𝜌

1 𝑐𝑚2/𝑠=1 𝑆𝑡

Stokes’ Law and Terminal Velocity

A falling body accelerates due to the force of gravity.
However, the presence of air yields two opposing forces:
Bouyant force
Resistive force (Air friction)

Terminal Velocity

Fb

Fr

Fg

Since air friction increases as speed increases, the resultant force acting on the body will eventually be equal to zero.
At that instance, the body moves at its maximum constant velocity or terminal velocity.

Terminal Velocity

Fg

Fb

Fr

If we assume that the falling body is a small sphere with radius r falling slowly at a speed of v through a fluid of viscosity η, its resistive force can be determined by:

Stokes’s Law

𝑭 𝒓 =𝟔𝝅𝜼𝒓𝒗

Fb

Fr

Fg

where:
Fr is the resistive force
η is the viscosity of the fluid
r is the radius of the sphere
v is the velocity of the falling sphere

Deriving the Terminal Velocity

𝐹 𝑔 𝐹𝐹 𝐹 𝑔 𝑔𝑔 𝐹 𝑔 = 𝐹 𝑏 𝐹𝐹 𝐹 𝑏 𝑏𝑏 𝐹 𝑏 + 𝐹 𝑟 𝐹𝐹 𝐹 𝑟 𝑟𝑟 𝐹 𝑟

𝑚 𝑠𝑝ℎ𝑒𝑟𝑒 𝑚𝑚 𝑚 𝑠𝑝ℎ𝑒𝑟𝑒 𝑠𝑠𝑝𝑝ℎ𝑒𝑒𝑟𝑟𝑒𝑒 𝑚 𝑠𝑝ℎ𝑒𝑟𝑒 𝑔𝑔= 𝜌 𝑓𝑙𝑢𝑖𝑑 𝜌𝜌 𝜌 𝑓𝑙𝑢𝑖𝑑 𝑓𝑓𝑙𝑙𝑢𝑢𝑖𝑖𝑑𝑑 𝜌 𝑓𝑙𝑢𝑖𝑑 𝑉 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉𝑉 𝑉 𝑠𝑝ℎ𝑒𝑟𝑒 𝑠𝑠𝑝𝑝ℎ𝑒𝑒𝑟𝑟𝑒𝑒 𝑉 𝑠𝑝ℎ𝑒𝑟𝑒 𝑔𝑔+6𝜋𝜋𝑟𝑟𝜂𝜂𝑣𝑣

𝜌 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉 𝑠𝑝ℎ𝑒𝑟𝑒 𝑔= 𝜌 𝑓𝑙𝑢𝑖𝑑 𝑉 𝑠𝑝ℎ𝑒𝑟𝑒 𝑔+6𝜋𝑟𝜂𝑣

𝜌 𝑠 𝑉 𝑠 𝑔= 𝜌 𝑓 𝑉 𝑠 𝑔+6𝜋𝑟𝜂𝑣

𝜌 𝑠 4 3 𝜋 𝑟 3 𝑔= 𝜌 𝑓 4 3 𝜋 𝑟 3 𝑔+6𝜋𝑟𝜂𝑣

4 3 𝜋 𝑟 3 𝑔 𝜌 𝑠 − 𝜌 𝑓 =6𝜋𝑟𝜂𝑣

𝒗= 𝟐𝝅 𝒓 𝟐 𝒈 𝟗𝜼 𝝆 𝒃𝒂𝒍𝒍 − 𝝆 𝒂𝒊𝒓

Fb

Fr

Fg

The block of ice (temperature 0°C) shown below is drawn over a level surface lubricated by a layer of water 0.10 mm thick. Determine the magnitude of the force F needed to pull the block with a constant speed of 0.50 m/s. At 0°C, the viscosity of water has the value of 1.79 x 10-3 N·s/m2.

Problem Solving

The viscous force on an oil drop is measured to be equal to 3.0 x 10-15 N when the drop is falling through with a speed of 4.5 x 10-4 m/s. If the diameter of the drop is 5.0 x 10-6 m, what is the viscosity of air?

Problem Solving

oil drop

Using the relationship:
𝑣𝑣=𝑘𝑘 𝑑 2 𝑑𝑑 𝑑 2 2 𝑑 2
On our next lesson, we will find out the value of k and its consistency by dropping marbles of different diameters into a container of oil.

Viscosity Experiment

THANK YOU!

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