stokes law experiment discussion

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Viscosity and Stoke’s Equation

One tube filled with water
One tube filled with glycerin
Teflon ball

Finding viscosity of a liquid by measuring velocity of small balls sinking in the tall tubes, and applying Stoke’s equation. Two long tubes are filled with fluids of different viscosities, one with water and the other with glycerin. Both tubes have two dark rings a meter apart.

Drop a ball from the top of the tube
Start the stopwatch when the ball passes the first ring (this time will be used to calculate the terminal velocity of the ball)
switch off the stopwatch when the ball passes the second ring
Record time
Retrieve ball with net
Repeat with a different ball
Ask students why Stoke;s Law will not work for water in this case.
Using the times recorded in glycerin find the terminal velocity and use that in stokes equation to find the viscosity of glycerin. Also, compare it to the given values in the table below.
Alternatively use the viscosity of glycerin to calculate the terminal velocity. Is it close to the value you found experimentally?

Explanation:

Viscosity is internal friction inside a fluid. Viscous forces oppose the motion of one portion of the fluid relative to another. Viscosity is denoted by η (eta) as the ratio of the shear stress (Force/Area) to the strain rate. Shear stress is the the friction between fluid particles sliding past each other. It is defined as the force tangent to the material surface divided by the area in which it acts. Strain rate is the rate of change of deformation or strain in a material. It is given as the velocity of the moving surface divided by l (the original length of the material). This equation is given below:

$stokes law experiment discussion$

The pascal second [Pa*s] is the SI unit for viscosity.

Viscosity is also very temperature dependent. For gases, when temperature increases the viscosity increases. For liquids, when temperature increase the viscosity decreases. For the purposes of our demonstration we will keep both tubes at room temperature(20° C) with little to no fluctuation in heat.

In glycerin the balls should fall with a constant velocity. In water the same balls will sink with non-uniform motion. In the water the flow is irregular and chaotic, otherwise known as turbulent flow.

Stokes law deals with small spherical objects that are moving through a fluid at small velocities. The reason why it has to be at small velocities is so that the flow is laminar and not turbulent. For that to happen we need to have a small Reynolds number.The viscosity can be determined from the velocity, mass and volume of the ball. Unfortunately for our purposes we cannot show stokes law with water, because the inertial forces dominate over the viscous forces. This means we have a large Reynolds number.

When the ball is moving through the fluid the force of gravity is acting down on it while the forces of buoyancy and drag are acting on it in the opposite direction. The figure below shows this:

Rearranged:

$stokes law experiment discussion$

To make the weight more relatable to the density we can use the relation below:

$stokes law experiment discussion$

Where ρ is the density of the ball and V is the volume of the ball (or the amount of water that it displaces.

The drag force is given below:

$stokes law experiment discussion$

This is called Stoke’s law.

Plugging everything into the equation we can then find our expression for Stokes law.

$stokes law experiment discussion$

Below we have a table with three different balls with different densities and size moving through the glycerin at slow velocities.

Written by Nick McCabe

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Stokes’ Law

What is stokes’ law.

Stokes’ law is a mathematical equation for the drag force experienced by small spherical particles passing through a viscous fluid medium. It deals with the resistive ( friction ) force applied to a body under the action of gravity as it is dropped into a fluid – liquid or air.

Historically, Stokes’ law has been named after Anglo-Irish physicist and mathematician George Gabriel Stokes after he derived the expression for drag force in 1851.

Stokes’ Law Equation

According to Stokes’ law, the drag force F d experienced by a spherical particle flowing through a viscous fluid is given by the following formula.

F d = 6πηrv

η is the viscosity of the fluid

r is the radius of the particle

v is the velocity of the particle relative to the fluid

SI Unit : Newton or N

Assumption of Stokes’ Law

Stokes’ law is valid under the following conditions.

Particles must be solid, smooth, and spherical.
Particles must be of uniform density.
Particles must be sufficiently large (>0.001mm) as compared to molecules of fluid so that the thermal (Brownian) motion of the fluid molecules does not affect the particles.
Particles should not interfere with each other during the fall.
The suspension should not have any turbulence, i.e., the flow must be laminar.

Derivation of Stokes’ Law

Stokes’ law can be derived using dimensional analysis. Consider the forces acting on a spherical particle as it sinks through a liquid column under the action of gravity. The viscous drag force F d acts on the particle is proportional to the following factors.

The radius of the sphere, r
The viscosity of the fluid, η
The velocity of the particle, v

This proportionality can be represented as,

F d ∝ η a r b v c

Replacing the proportionality sign with the equality sign,

F d = k η a r b v c

Where k is the dimensionless proportionality constant

Writing the above equation in terms of dimensions,

[MLT -2 ] = [ML -1 T -1 ] a [L] b [LT -1 ] c

Or, [MLT -2 ] = M a L -a+b+c T -a-c

Equating the superscripts of mass, length, and time on both sides.

1 = – a + b + c

– 2 = – a – c

Solving the above set of equations.

F d = k η r v

Experimentally, the value of k was found to be 6π . Hence,

F d = 6 πηrv

Stokes’ Law and Settling Velocity

Settling or terminal velocity is the maximum velocity attainable by an object as it falls through the fluid. It occurs when the sum of the forces due to drag and buoyancy is equal to the downward force due to gravity. There is no force on the object at the terminal velocity, and hence, the acceleration is zero. The object moves in a steady state.

By using the drag, buoyant, and gravitational forces, the particle’s terminal velocity is given by the following equation,

ρ s is the density of the particle

ρ f is the density of the fluid

Applications of Stokes’ Law

Stokes’ law is used in the falling-sphere viscometer, in which a sphere of known density descends a vertical tube. The terminal velocity can be measured by recording the time and the distance traveled. This device is used to measure the viscosity of a fluid.

Sedimentation is the propensity for particles in suspension to settle out of the fluid as deposits known as sediments. Stokes’ law is essential in understanding the sedimentation of small particles under the force of gravity. It can determine the sedimentation rate, which is the sediment’s terminal velocity.

Stokes’ law can explain the floatation of clouds. Since raindrops are smaller in size, their terminal velocities are small. Therefore, they remain suspended in the air in the form of clouds. As they increase their size, their terminal velocities increase, and they start falling in the form of rain.

When jumping off a plane, a parachute helps a person to slow down and land softly. The reason is that the person with the parachute achieves a terminal velocity.

In Millikan’s oil drop experiment , Stokes’ law is used to determine the electron ’s charge.

Limitations of Stokes’ Law

Some colloidal particles of the same mass fall slower than others due to the difference in the shapes of particles.
Many fast-falling particles may drag finer particles down along with them.
Particles greater than 0.08 mm in diameter settle quickly and cause turbulence
Dropping the Ball (Slowly) – Galileo.phys.virginia.edu
Stokes’ Law – Courseware.cutm.ac.in
Viscosity – Iitr.ac.in
Stokes’ Law Examples – Ocw.tudelft.nl
Stokes’ Law and Terminal Velocity – Schoolphysics.co.uk

Article was last reviewed on Thursday, February 2, 2023

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Stokes' law and terminal velocity

When any object rises or falls through a fluid it will experience a viscous drag, whether it is a parachutist or spacecraft falling through air, a stone falling through water or a bubble rising through fizzy lemonade. The mathematics of the viscous drag on irregular shapes is difficult; we will consider here only the case of a falling sphere. The formula was first suggested by Stokes and is therefore known as Stokes' law. Consider a sphere falling through a viscous fluid. As the sphere falls so its velocity increases until it reaches a velocity known as the terminal velocity. At this velocity the frictional drag due to viscous forces is just balanced by the gravitational force and the velocity is constant (shown by Figure 2). At this speed: Viscous drag = 6πηrv = Weight = mg The following formula can be proved (see dimensional proof)

schoolphysics Stokes' Law animation

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Fluid Mechanics: Stokes' Law and Viscosity Measurement Laboratory Investigation No. 3

Related Papers

Milan Shrestha

Using a photo-gated digital time interval measuring clock, acrylic spheres dropped in a vertical tube filled with castor oil such that they move through the oil under gravity, the viscosity of the oil is determined by applying the Stoke's law. The value of viscosity obtained is compared with the standard value for castor oil. The experimental setup is now available in two models. Of the two models, the one explained in this paper, which can be interfaced with a computer is accurate up to the third decimal resolution for time measurement, provides reasonably accurate value of viscosity.

OBJECT • To measure coefficient of the dynamic viscosity of the glycerine and ricine oil with a Stokes viscometer. • Evaluate the error of measurements. • Compare your results to the accepted value. THEORY Viscosity is a measure of the resistance of a fluid which is being deformed by either shear stress or tensile stress. Viscosity describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. For example, high-viscosity felsic magma will create a tall, steep stratovolcano, because it cannot flow far before it cools, while low-viscosity mafic lava will create a wide, shallow-sloped shield volcano. All real fluids (except superfluids) have some resistance to stress and therefore are viscous, but a fluid which has no resistance to shear stress is known as an ideal fluid or inviscid fluid. Viscosity coefficients can be defined in two ways: Dynamic viscosity, also absolute viscosity, the more usual one. The SI physical unit of dynamic viscosity is the pascal-second (Pa·s), (equivalent to N·s/m 2 , or kg/(m·s)). The usual symbol for dynamic viscosity used by mechanical and chemical engineers — as well as fluid dynamicists — is the Greek letter mu (µ). Kinematic viscosity is the dynamic viscosity divided by the density (units m 2 /s).

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3.2 Stokes’ law: examples

The Basics of Transport Phenomena

Home Courses The Basics of Transport Phenomena Course materials Lectures 3.2 Stokes’ law: examples

Course subject(s) 3. Drag force

EXAMPLE 3.2A: CALCULATING DRAG FORCE WITH STOKES’ LAW (ELEMENTARY)

Imagine a sphere with fluid flowing around it. Can you calculate its drag force, using Stokes’ Law? And, what has Reynolds Number got to do with it?

Example 3.2A: Calculating drag Force with Stokes' Law (Elementary)

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EXAMPLE 3.2B: SINKING TIME OF PLANKTON (MEDIUM)

Consider a small micro-organism living in the ocean.

At some point, it dies and will slowly fall at its terminal velocity to the ocean floor. We assume that the micro-organism is a sphere with a diameter of 10 μ m.

It has a density that is slightly higher than that of sea water. The ocean has a depth of H = 1 km.

1. What is its terminal velocity? 2. How long does it take for the organism to reach the ocean floor?

EXAMPLE 3.2C: SETTLING TANK FOR OIL/WATER SEPARATION (ADVANCED)

During the production of crude oil, water is often produced along with natural gas. The result is a homogeneous water-in-oil emulsion, with water droplets settling at the bottom of the tank. One of the simplest methods to separate this emulsion is to design a huge settling tank and wait until the water and oil are separated by gravity.

Generally, an important question is: How long does it take to separate the emulsion? All water droplets can be represented by spheres with an average diameter of dwater = 50 µm. Furthermore, we know that:

– The viscosity of water is µwater = 10-3 Pa·s and the density of water is equal to ρwater = 1000 kg/m3

– The viscosity of crude oil is µoil = 5.10-3 Pa·s and the density is ρoil = 800 kg/m3

– The gravity constant is g = 9.81 m/s2

– The height of the tank is 4 meter.

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Forces and Motion

Practical investigation into Viscosity in liquids (Stokes Law).

Practical investigation into Viscosity in liquids(Stokes Law)

Introduction

When dealing with fluid/mechanical systems, it is important to know what affects the rate of descent of an object through a liquid.

There are many factors that affect the descent of an object through a liquid such as:

1) Temperature of the liquid

2) Mass* of object

3) Size/surface area of object

4 Viscosity of liquid

5) Angle of descent

Temperature

I would like to investigate the correlation between temperature and time of descent. Reading suggests that the colder the liquid the longer it will take for the object to reach the bottom.

Mass*& Surface area/size

Gravity accelerates at 9.81 ms-1 independent of mass. Hence increasing the mass will not affect the experiment of surface area. Thus using an object of various sizes it would be possible to investigate the proportionately of size on the descent of the object.

I feel it is important to investigate the affects of how a more viscous liquid would impede the progress of an object descending through a liquid. Therefore I have included this factor into my investigation.

Angle of descent

I would like to observe the affects of the object descending at an angle. Such at sediment in a bottle is there a way in which bottles should be stored that may hasten descent?

To investigate the rate of decent of an object falling through a liquid (simulated by a ball bearing) and investigate some of the factors that will affect this.

I have completed this investigation by using many simple experiments to reach a firm conclusion on the rate of decent of a ball bearing through a liquid.

Size/surface area of ball bearing

The first set of experiments was to show the speed of five differently sized ball bearings descending through water. All five ball bearings seemed to reach their terminal velocity at the same timed interval.

Viscosity of liquid

The second set of experiments was to show the affects of a more viscous liquid (engine oil) on the decent of a ball bearing. Using the same method as the experiment for surface area, the ball bearing took longer to descend through the liquid.

On the third experiment I varied the temperature of the oil. The difference was at a higher temperature the ball bearing descended faster through the liquid.

Finally the fourth experiment compared angles of which the ball bearing descended through a liquid. The results showed a greater angle to the vertical reduced the speed of decent through the liquid.

Initial apparatus

The following apparatus was used to complete the investigation:

2.2 metre plastic tube (colourless) Diameter 0.03 metre

900ml Water 900ml Engine oil

Five ball bearings with diameter and masses as follows:

Class Mass (g) Diameter (mm) Very small 0.12 2.96 Small 0.88 6.98 Medium 4.07 9.98 Large 8.96 13.98 Very large 16.69 16.9

Clamp and clamp stand.

Tape measure

Bunsen to heat oil (in a pan).

This is a preview of the whole essay

Each experiment shall be repeated 3 times and an average calculated to plot a graph.

Method and Results tables

Experiment 1

For the first experiment, I decided to measure the speed of a ball bearing descending through water.

My final decision on how to complete the experiment was to time the ball bearing at equally spaced intervals. At first, the obvious way to do this was to time at every 10cm interval. However this proved very difficult to do without some sought of mechanical or electrical aid to take the measurement. Therefore I settled for 20cm intervals but timed from a 10cm starting point. This worked particularly well and produced the results that follow this explanation.

Now that I had my first set of results, I thought I would introduce a variable of five differently sized ball bearings. This enabled me to investigate the surface area, mass* and how this effected the rate of descent.

The results of experiment 1 are as follows:

Very Small Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.19 0.21 0.20 0.200 10-30. 0.17 0.18 0.20 0.183 20-40 0.18 0.19 0.20 0.190 30-50 0.20 0.18 0.19 0.190

Small Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.24 0.25 0.24 0.243 10-30. 0.20 0.21 0.21 0.206 20-40 0.18 0.18 0.19 0.183 30-50 0.17 0.19 0.18 0.180 40-60 0.18 0.18 0.18 0.18 50-70 0.17 0.18 0.19 0.18

Medium Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.32 0.33 0.32 0.323 10-30. 0.28 0.28 0.29 0.283 20-40 0.23 0.23 0.22 0.227 30-50 0.19 0.21 0.20 0.200 40-60 0.21 0.20 0.20 0.203 50-70 0.20 0.20 0.21 0.203

Large Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.26 0.28 0.28 0.273 10-30. 0.23 0.22 0.20 0.217 20-40 0.20 0.21 0.20 0.203 30-50 0.20 0.19 0.20 0.197 40-60 0.20 0.20 0.20 0.200 50-70 0.20 0.19 0.2 0.197

Very Large Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.33 0.32 0.30 0.317 10-30. 0.28 0.29 0.27 0.280 20-40 0.27 0.26 0.26 0.263 30-50 0.23 0.24 0.25 0.240 40-60 0.22 0.23 0.23 0.227 50-70 0.22 0.22 0.22 0.220

Experiment 1.2

The second part of the first experiment measured the time taken for five differently sized ball bearings to descend through water. The purpose of this part of the experiment was to make it more clear how surface area and mass* affects the rate of descent.

At this point and introduction of a sixth ball was used. This ball was 'almost' the same diameter as the tube it descended through (22mm in diameter) & (29.02g). Using this ball bearing, I could investigate more deeply into 'up thrust' that wasn't so noticeable with the smaller ball bearings.

The results of the experiments are as follows:

Ball bearing Time taken (s) to descend 1 metre through water 1 2 3 Average Very small 1.92 1.96 1.94 1.940 Small 1.39 1.42 1.38 1.397 Medium 0.29 1.28 1.28 0.950 Large 1.27 1.30 1.29 1.287 Very large 1.64 1.62 1.63 1.630 Extra large 3.60 3.55 3.59 3.580

On investigation this experiment was not valid as it changed the parameters of the original investigation i.e. the liquid was forced under pressure between the ball bearing and tube.

Experiment 2

The second experiment had all the variables the same apart from the viscosity of the liquid that was changed. The liquid used for this experiment was engine oil.

Although it is easier to measure the rate of descent through engine oil because it is slower, the measured interval remained at 20cm.

The results are as follows:

Very small Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 1.89 1.86 1.88 1.877 10-30. 1.83 1.77 1.80 1.800 20-40 1.76 1.75 1.78 1.763 30-50 1.76 1.79 1.77 1.773 40-60 1.77 1.78 1.77 1.773

Small Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.86 0.84 0.85 0.850 10-30. 0.69 0.67 0.68 0.680 20-40 0.62 0.62 0.62 0.620 30-50 0.62 0.60 0.61 0.610 40-60 0.61 0.62 0.61 0.613

Medium Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.45 0.49 0.46 0.467 10-30. 0.38 0.37 0.39 0.380 20-40 0.39 0.36 0.38 0.377 30-50 0.39 0.38 0.36 0.377 40-60 0.37 0.38 0.37 0.373

Large Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.92 0.90 0.91 0.910 10-30. 0.77 0.77 0.76 0.767 20-40 0.48 0.49 0.49 0.487 30-50 0.48 0.46 0.46 0.467 40-60 0.49 0.48 0.47 0.480

Very large Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 1.52 1.52 1.53 1.523 10-30. 0.68 0.69 0.67 0.680 20-40 0.71 0.71 0.70 0.707 30-50 0.72 0.70 0.71 0.710 40-60 0.71 0.70 0.69 0.700

Experiment 2.2

The second part of experiment 2 investigates the time taken for five various sized ball bearings to descend through engine oil. This allows me to investigate the affect of viscosity on the surface area of a ball bearing.

The results are as follows.

Ball bearing Time taken (s) to descend 1 metre through engine oil 1 2 3 Average Very small 13.50 13.58 13.54 13.540 Small 5.02 5.05 5.02 5.030 Medium 3.38 3.39 3.37 3.380 Large 3.58 3.60 3.57 3.583 Very large 5.48 5.49 5.48 5.483

Experiment 3

Experiment 3 varied the temperature of the liquid that a ball bearing descended through. I used engine oil for this experiment, as this would prove to have the most change in viscosity when heated. The medium sized ball bearing was used to complete this experiment however all other variables remained the same.

The results are shown below:

Temperature Time taken (s) to descend 1 metre through engine oil 1 2 3 Average 35 2.67 2.65 2.63 2.650 40 2.27 2.29 2.30 2.287 45 2.20 2.21 2.19 2.200 50 2.13 2.17 2.15 2.150 55 1.90 1.91 1.92 1.910 60 1.88 1.72 1.81 1.803

Experiment 4

The fourth and final experiment was to measure the rate of descent of a ball bearing through a liquid at varied angles to the horizontal. I decided to take measurements at every 10o to the vertical.

This experiment would then allow me to investigate and observe if there is any change to the rate of descent caused by the applied angle. I only required taking measurements up to 80o as at 90o the ball bearing would not descend.

The results are shown below.

Angle Time taken (s) to descend 1 metre through engine oil 1 2 3 Average 10 1.69 1.82 1.66 1.723 20 1.87 1.94 1.90 1.903 30 2.20 2.15 2.17 2.173 40 2.46 2.37 2.39 2.407 50 2.76 2.71 2.72 2.730 60 2.92 2.96 2.90 2.927 70 3.54 3.50 3.52 3.520 80 6.15 6.10 6.30 6.183

The next section that starts on the following page displays graphically the results and other relevant information of the experiments mentioned in this report.

A number of conclusions can be drawn from my investigative work into the rate of descent of a ball bearing through a liquid and the factors that affect this.

1) I have found that one of the major factors effecting the ball bearings descent is the size of the ball bearing. From the discussion I mentioned the way in which the liquid flows around the ball bearing as it descends. This proved along with my experiments that the larger the ball bearing the slower the rate of descent. This occured in both engine oil and water.

2) The second major factor was the viscosity of the liquid in which the ball bearing descends. The more viscous liquid made the ball bearing descend at a slower rate.

I can also conclude that the temperature of the liquid in which the ball bearing descends through can increase or decrease the rate of descent. With a higher temperature the liquid becomes less viscous and the ball bearing descends at a faster rate.

3) I have found that the angle in which the ball bearing descends through will decrease the speed of when it will reach the bottom. However as I mentioned in the discussion an extra force was acting upon this and therefore made this experiment invalid.

4) The final conclusion to be drawn from my investigation, is that the ball bearings seemed to reach their terminal velocity in the same timed interval. For experiment 1 it was 40-60 cm and for experiment 2 this was also 40-60. Therefore I would be able to conclude with a third liquid that it may be possible that the liquid does not effect the point in which a ball bearing reaches its terminal velocity.

However I can conclude that the size of the ball bearing and also the mass does not effect where it reaches its terminal velocity. As you already know, if two objects of the same size but with different masses are dropped from the same height they will descend and hit the ground at the same time. It is only air resistance that will affect the descent if the objects size is slightly different. I can relate this to my experiments in finding the terminal velocity of the ball bearings through the liquid, and therefore explain why the occurance happened with only a slight varience with the very large ball bearings.

Practical investigation into Viscosity in liquids (Stokes Law).

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Learn How to Determine the Coefficient of Viscosity Using Stoke’s Method

Last Updated on August 29, 2023 by Nourhan Essam

Imagine that you have a cup with a hole in the bottom. If you pour honey or glycerin into the cup, you will find that the cup drains very slowly. But in the case of water, the cup will drain much more quickly. That happens because “viscosity” is higher in the case of honey or glycerin when compared to other liquids’ viscosities.

Viscosity is one of the essential parameters that describe the physicochemical properties of liquids. It is a measurement of a fluid’s resistance to flow or change in shape at a given rate

In this article, we will discuss what is meant by viscosity, the concept of coefficient viscosity, how to determine the coefficient of viscosity using Stoke’s method, and more.

Table of Contents

What is Viscosity?

Viscosity can be defined as a force multiplied by a time divided by an area. It is a measure of a fluid’s resistance to flow and describes the internal friction of a fluid that is moving.

For further understanding, when you put some drops of honey on one side on an inclined surface and some water droplets on the other side and observe the flow of fluid. You will notice that water flows much faster than honey, which flows very slowly and cannot be moved easily. This happens due to the viscosity of the material.

Notice that a high viscosity of fluid resists motion because its molecular makeup gives it a variety of internal friction. However, the low-viscosity fluid flows easily because its molecular makeup results in little or no friction in case of movement.

Gases have viscosity like fluids but it is difficult to be noticed in ordinary circumstances.

Viscosity depends on factors related to the fluid’s state, such as:

The fluid temperature
Deformation rate
pressure

Note: The dependence on these factors may be negligible in certain cases. For example, the Newtonian fluid’s viscosity does not vary with the deformation rate.

Coefficient of Viscosity

To understand what is meant by the coefficient of viscosity, you should understand the meaning of “The viscous grad ient” first.

Viscous Gradient

The viscous gradient is the distinction inside the velocity among the fluid’s adjacent layers If more force is applied by the upper layer to move forward the more will be the viscous gradient.

The viscous gradient is the difference in the velocity between the fluid’s adjacent layers. If more force is applied by the upper layer to move forward.

It is given by v/x, where v is the velocity difference and x is the difference of distance between the two layers.

Now, let’s define the coefficient of viscosity:

The coefficient of viscosity is the ratio of the shearing stress to the velocity gradient of the fluid (η).

The coefficient of viscosity is given by the following relation:

η = F . d / A .ⅴ

F is the force required to maintain a unit velocity gradient between two parallel layers of liquid of a unit area.
d is the distance between the two layers of liquid skidding over each other
A is the area
ⅴ is the velocity.

The coefficient of viscosity is a measure of how easily or difficult a fluid flows under the effect of external forces, and it plays a very important role in many industrial, scientific, and everyday applications.

Examples of Coefficient of Viscosity

Water : The coefficient of viscosity for water at 20°C is nearly 0.001 Pa·s (pascal-seconds) or 1 centipoise (cP). This is considered a relatively low viscosity that allows water to flow easily and is one of the reasons why it is commonly used for viscosity comparisons.
Honey : The coefficient of viscosity for honey can range from 10 to 20 Pa·s or higher, depending on factors such as the honey type and the temperature. This high viscosity gives honey its thick and slow-flowing characteristic.
Motor Oil : The coefficient of viscosity for motor oil has a wide range, from 20 to 100 centistokes (cSt) or higher. Motor oil is designed to have a higher viscosity compared to water and this ensures effective lubrication of engine components. Higher-viscosity oils have higher coefficients of viscosity.
Air : Gases also have a coefficient of viscosity. The coefficient of viscosity for air at atmospheric pressure or room temperature is approximately 0.000018 Pa·s or 0.018 millipoise (mP). The viscosity of gases is considerably lower than that of fluids and this low viscosity allows air to flow easily.
Molten Glass : The coefficient of viscosity for molten glass can range from around 10^2 to 10^8 Pa·s, indicating a high resistance to flow. And this depends on its temperature and composition.

The Coefficient of Viscosity Using Stoke’s Method

Stoke’s method or Stoke’s law is a mathematical equation that expresses the velocities of the small spherical objects in a fluid medium and says that any object when rises or falls through a fluid it will experience a frictional or drag force because of the fluid.

In other words, Stoke’s law states that the force of viscosity on a small sphere moving through a viscous fluid is given by the following equation:

Where: F is the frictional force acting on the interface between the fluid and the particle. μ is the dynamic viscosity r is the radius of the spherical object V is the flow velocity relative to the object

Importance of Stoke’s Law

Stoke’s law is important in explaining the formation of clouds, the falling from a parachute, why the larger rain droplets hurt more than the smaller ones, and more.

Millikan depended on the concept of Stokes’ law in his oil-drop experiment, which was used to determine the electronic charge.

PraxiLabs, the 3D virtual lab solution, provides students with access to realistic biology, chemistry, and physics labs and enriches their understanding with a variety of informational and educational content .. Pick the best plan for you now!

Stoke’s Law Derivation

The viscous force acting on a sphere is directly proportional to the following factors:

Coefficient of viscosity (η) as F ∝ η a …(i)

The radius of the sphere (r) as F ∝ r b …(ii)

The velocity of the object (v) as F ∝ v c …(iii)

Combining the above three equations we get,

F ∝ η a r b v c

Removing the proportionality sign and adding the proportionality constant k we get.

F = k η a r b v c ……(1)

Now, equating the dimensions of parameters on either side of equation (1)

[MLT –2 ] = [ML –1 T –1 ] a [L] b [LT -1 ] c

Simplifying the above equation,

[MLT –2 ] = [M a ⋅ L –a+b+c ⋅ T –a–c ] ……(2)

Equating the superscripts of mass, length, and time respectively from equation (2), we get

a = 1 ….(a)

–a + b + c = 1 ……(b)

a + c = 2 ……(c)

from (b), (c)

c = 1 ……(d)

from (a) & (b) in (d), we get

–1 + b + 1 = 1

b = 1 ……(e)

From (a), (d), and (e)

F = k η r v

For any spherical body, the value of k was experimentally obtained to be 6π.

Thus, the viscous force on a spherical body falling through a liquid is given by the equation, F = 6πηrv

Determination of the Coefficient of Viscosity by Stoke’s Method Experiment From PraxiLabs

PraxiLabs provides a virtual lab simulation for the experiment (Determination of the Coefficient of Viscosity by Stoke’s Method)

Learning Objectives

By the end of the determination of the coefficient of viscosity using Stoke’s method experiment, the students should be able to:

Understand the definition of the viscosity.
Detect the CGS and MKS (SI) units for measuring the coefficient of viscosity.
Explain the origin of the viscosity in both liquids and gases.
Estimate the temperature effect on the coefficient of viscosity for fluids and enumerate the other different factors that could change the viscosity of a liquid.
Set up an experiment to measure the coefficient of viscosity of liquids.

In this experiment, we will need:

Cylindrical glass jar – Steel balls of different radii – Highly viscous liquids – Ruler – Stopwatch.

The experiment of determination of the coefficient of viscosity depends on measuring the fall-terminal speed of a metal sphere that falls in a glass jar filled with a viscous fluid, as a function of the sphere’s radius. Hence the viscosity coefficient of the liquid can be readily determined.

PraxiLabs provides properties of matter virtual labs to students and educators that include a set of hands-on experiments designed to support students’ understanding of the principles of physics. Subscribe now!

Theoretical Background

How to Determine the Coefficient of Viscosity Using Stoke's Method

Consider a steel ball of radius (r) is released from the top of a glass jar filled with a viscous fluid with viscosity coefficient (η). The instant the ball hits the liquid’s surface, it stops momentarily and then begins to accelerate downward under the effect of gravity.

At the same time, the so-called viscous force ( F v ) begins to oppose the motion of the ball, as it depends directly on the speed of the ball (see eq.(3)). As time goes on, the viscous force (which is also called the drag force or Stoke’s force) increases in magnitude until it reaches a value together with the buoyant force that balances the weight of the ball.

Theory In such a case, the ball moves with a constant speed called the terminal velocity ( v o or sedimentation velocity). Before reaching the bottom of the jar, the ball decelerates and eventually comes to a stop. This means that within a certain length of the jar away from its ends (both are marked on the glass jar with two marks A and B), the ball moves with a uniform speed, under the effect of three balanced forces:

1- The Ball Weight

m ball is the mass of the ball

g is the acceleration due to gravity

V ball is the volume of the ball.

And ρ ball is the density of the ball’s material.

2- Upthrust force (or the buoyant force)

Upthrust force is the force exerted by the fluid on the immersed body and acting upward. Recall that the buoyant force equals the weight of the immersed body:

Where ρ liquid is the density of the liquid’s material. However, since the volume of immersed body V ball equals the volume of displaced liquid V liquid ,

3- Viscous force

Viscous force acting the ball due to the viscosity of the liquid ? , acts upward,

However, when the ball moves with the terminal velocity vo , the net force acting on the ball is zero; then

Effect of the Jar’s Wall

To take into account the proximity of the jar’s wall which affects the motion of the ball, the terminal speed vo can be calculated to the first approximation using the relation

Where R is the inner radius of the jar and v is the measured velocity.

Effect of Temperature

The viscosity coefficient depends on the fluid’s temperature. As the temperature increases, the coefficient of the viscosity decreases for liquids and increases for gases.

This can be interpreted according to the origin of viscosity in both. For liquids, the viscosity originates from the frictional force between adjacent layers when they move past each other, therefore increasing temperature, and causes the separation distance between layers to increase and the viscosity decrease.

On the other hand, in gases, the viscosity originates from the collisions made by the gas molecules with each other within the same layer. Increasing the temperature causes these collisions to increase and the viscosity increases as well.

The steps of the Experiment

In PraxiLabs virtual lab of the determination of Coefficient of Viscosity Using Stoke’s Method experiments, you will do the following steps to conduct the experiment successfully

Choose a liquid from the drop list, such as glycerin.
Select the ball’ radius from the slide bar.
Adjust the distance between the two movable marks “A” and “B” (10 – 90 cm).
Click on the “Drop Ball” button to drop the solid ball inside the jar. When the ball reaches the mark “A”, Click the “Start” button to start the stopwatch.
When the ball (inside the red circle) reaches the mark “B”, click the “Stop” button. 6. Click the “Record” button to record this time.
Click the “New Trial” button to start over with the same ball if you want.
Repeat the previous steps for all the available ball’ radii.
The experiment has ended, an Excel sheet will be downloaded that contains the data you’ve recorded.
You will calculate r (m) and r 2 (m2) / the apparent speed of the ball / the terminal speed of the ball.
Plot a relation between vo and r 2 , which should yield a straight line that passes through the origin, then calculate the slope of the line.

11. Finally, Calculate the viscosity coefficient of the chosen liquid using the following relation:

Now, you can easily conduct (determination of the Coefficient of Viscosity by Stoke’s Method Experiment) on your own for free. Create your free account and try it now !

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Stoke law determend the viscosity of liquid

In an experiment to determine the viscosity by the Stoke method. When we throw the ball into the liquid it will reach a certain mark and start recording the time. This mark what determines its location, such that it is 5 cm further from the top edge, why not more or less. And the bottom mark. What determined their location and why

fluid-dynamics

$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$ – Community Bot Commented Nov 3, 2021 at 18:46
$\begingroup$ @Community I think the question is pretty clear, actually. $\endgroup$ – Gert Commented Nov 3, 2021 at 19:00

When terminal velocity has been reached then Stokes Law tells us that:

$$(\rho_p-\rho_f)g\frac43 \pi R^3=6\pi R\mu v$$

Knowing mass $m$ and velocity $v$ (and radius $R$ ) then allows calculating viscosity $\mu$ .

But prior to reaching terminal velocity the ball's equation of motion is:

$$(\rho_p-\rho_f)g\frac43 \pi R^3\ddot{y}-6\pi R\mu \dot{y}=0$$

where $y$ is the vertical direction.

Solving that equation allows to determine the amount of displacement $y$ that's needed before the ball's velocity has become constant (i.e. terminal).

Those who designed the experiment for you determined that distance was about $5\mathrm{cm}$ .

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Stokes’s law

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Academia - Stokes Law Presentation

Stokes’s law , mathematical equation that expresses the drag force resisting the fall of small spherical particles through a fluid medium. The law, first set forth by the British scientist Sir George G. Stokes in 1851, is derived by consideration of the forces acting on a particular particle as it sinks through a liquid column under the influence of gravity . In Stokes’s law, the drag force F acting upward in resistance to the fall is equal to 6 πrηv , in which r is the radius of the sphere , η is the viscosity of the liquid, and v is the velocity of fall.

The force acting downward is equal to 4 / 3 πr 3 ( d 1 − d 2 ) g , in which d 1 is the density of the sphere, d 2 is the density of the liquid, and g is the acceleration due to gravity. At a constant velocity of fall called the terminal velocity , the upward and downward forces are in balance. Equating the two expressions given above and solving for v therefore yields the required velocity expressed as v = 2 / 9 ( d 1 − d 2 ) gr 2 / η .

Stokes’s law finds application in several areas, particularly with regard to the settling of sediment in fresh water and to measurements of the viscosity of fluids. Because its validity is limited to conditions in which the motion of the particle does not produce turbulence in the fluid, however, various modifications have been set forth.

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Stoke’s Law

Stoke’s Law: Observe a raindrop falling from a height if you look closely you will notice that the speed of all the raindrops is constant and even though it falls from a height under the influence of gravity its velocity seems constant. These questions are answered using Stoke’s law

Stoke’s law was first proposed by the famous English scientist George Gabriel Stokes . This law tells us that the velocity of any object increases gradually when it is dropped in any medium until it reaches a constant velocity then that object proceeds without changing its velocity.

This article will explore Stoke’s law, its definition, formula, examples, applications, conditions, etc.

Table of Content

What is Stoke’s Law?

Stoke’s law definition, stoke’s law mathematical expression, stokes law formula, importance of stoke’s law, stoke’s law derivation, limitations of stoke’s law, negative density difference in stoke’s equation, high content of dispersed solids, dielectric constant, brownian movement, conditions for stoke’s law, stoke’s law applications, velocity of raindrops, terminal velocity, terminal velocity formula, stoke’s law examples.

Stoke’s Law is an equation that expresses the velocity of small spherical particles in a fluid medium. The law is established by taking into account the forces acting on a certain particle as it falls through a liquid column under the influence of gravity.

Stokes Law state that, “There is a drag force working on the spherical body which is following in a column and the upward drag force on the body ultimately equals the gravitational force then the body drops with a constant velocity.”

The force that retards a sphere passing through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity. Sir George G. Stokes, an English physicist, stated the viscous drag force acts on a body dropping in a liquid, this force is denoted using the letter ‘F’.

In Stokes’s law, the drag force F acting upward in resistance to the fall is equal to 6πrηv , in which r is the radius of the sphere, η is the viscosity of the liquid, and v is the velocity of the fall.

F is the drag force acting on the object,
η (Greek letter eta) is the dynamic viscosity of the fluid,
r is the radius of the spherical object, and
v is the velocity of the object relative to the fluid.

Stokes’s law gives us a mathematical equation that tells us about the drag force acting on a spherical particle when passing through a liquid under the influence of gravity. Stokes Law Formula is stated as,

F = 6πηrv where F is the frictional force η is the viscosity of the liquid r is the radius v is the velocity of the flow

Following are the importance of Stoke’s Law:

Millikan employs this rule in his oil-drop experiment to determine the electric charge. Working of a parachute is based on the principle of Stoke’s Law. Cloud formation and floating are explained using Stoke’s Law.

The viscous force acting on a sphere is directly proportional to the following factors:

Coefficient of viscosity (η) as F ∝ η a …(i)

Radius of the sphere (r) as F ∝ r b …(ii)

Velocity of the object (v) as F ∝ v c …(iii)

Combining the above three equations we get,

F ∝ η a r b v c

Removing the proportionality sign and adding the proportionality constant k we get.

F = k η a r b v c ……(1)

Now, equating the dimensions of parameters on either side of equation (1)

[MLT –2 ] = [ML –1 T –1 ] a [L] b [LT -1 ] c

Simplifying the above equation,

[MLT –2 ] = [M a ⋅ L –a+b+c ⋅ T –a–c ] ……(2)

Equating the superscripts of mass, length, and time respectively from equation (2), we get

a = 1 ….(a)

–a + b + c = 1 ……(b)

a + c = 2 ……(c)

from (b), (c)

c = 1 ……(d)

from (a) & (b) in (d), we get

–1 + b + 1 = 1

b = 1 ……(e)

from (a), (d), and (e)

F = k η r v

For any spherical body, the value of k was experimentally obtained to be 6π.

Thus, the viscous force on a spherical body falling through a liquid is given by the equation,

Hence, this is the required relation of Stoke’s Law.

Various limitations of Stoke’s Law are discussed below in the article,

Stokes’ equation is invalid if the density difference in the equation is negative that is when the particles are lighter than the dispersion medium. This results in floatation or creaming most commonly seen in emulsion systems.

When the solid content of a suspension is high, Stokes’ equation may not show the real sedimentation rate. High solid content imparts additional viscosity to the system, which must be taken into consideration if the correct rate of settling is to be determined. The equation contains only the viscosity of the medium.

The dielectric constant, which is not used in Stokes’ equation, is a significant parameter in many contexts. The electrical potential between two charges is inversely proportional to the medium’s dielectric constant. As a result, the zeta potential is affected by the medium’s dielectric constant. The assumption is that if an automobile has a low dielectric constant, the double layer is several times thicker than in an aqueous medium, resulting in different zeta potential and, therefore, a different setting. These considerations are critical for using non-aqueous vehicles such as sesame oil, corn oil, and chlorofluorocarbon propellant (in aerosol suspension).

Another aspect that can affect the precision of Stokes’ equation results is Brownian movement , which is a spontaneous (zigzag) movement of particles floating in a fluid caused by collisions with fast-moving atoms or molecules in the gas or liquid. Sedimentation is mitigated to some degree by Brownian migration. This results in a significant difference between the real rate of sedimentation and the rate measured using Stokes’ equation.

Conditions under which Stoke’s Law is valid are,

The fluid through which the body moves must have an infinite extension. The body is perfectly rigid and smooth. There is no slip between the body and the fluid. The motion of the body does not give rise to turbulent motion. Hence, the motion is streamlined. The size of the body is small, but it is larger than the distance between the molecules of the liquid. Thus, the medium is homogeneous and continuous for such a body.

Raindrops do not reach extremely high speeds during their free fall. If this does not occur, a person walking in the rain will be injured. This is because the viscous drag in air opposes the velocity of raindrops as they descend due to gravity. The drop reaches a terminal velocity when the viscous force equals the force of gravity. As a result, raindrops reaching the earth have low kinetic energy .

When we jump out of an aeroplane, a parachute assists us in landing safely on the ground. Because in this case, the person falls with g acceleration due to gravity, but the acceleration decreases due to viscous drag in the air until the person attains terminal velocity. The person then drops at a constant speed and opens his parachute close to the ground at a predetermined period, allowing him to land safely near his destination.

It is used to calculate the charge of an electron. (The oil drop method of Millikan)

When a body falls through a viscous fluid, it produces relative motion between its different layers. As a result, the body experiences a viscous force that tends to retard its motion. As the velocity of the body increase, the viscous force (F = 6 π η r v) also increases. A stage is reached when the weight of the body becomes just equal to the sum of the upthrust and viscous force. Then no net force acts on the body, and it begins to move with a constant velocity. So, here comes the concept of Terminal Velocity.

The maximum constant velocity acquired by a body while falling through a viscous medium is called its Terminal Velocity.

Various force acting on solid immersed in liquid

As the speed of an object increases, so does the drag force acting on it, which also depends on the substance it is passing through (for example air or water). At some speed, the drag or force of resistance will equal the gravitational pull on the object (buoyancy is considered below). At this point, the object stops accelerating and continues falling at a constant speed called the terminal velocity (also called settling velocity).

The formula for terminal velocity (v t ) is,

v t = 2r 2 × (ρ−σ)g / 9η where ρ is the mass density of the sphere σ is the mass densities of the fluid r is the radius of the raindrop

Reynolds Number Streamline Flow Bernoulli’s Principle

Example 1: A raindrop of radius 0.3 mm falls through the air with a terminal velocity of 1 m/s. The viscosity of air is 18 × 10 -5 Poise. Find the viscous force on the raindrop.

Given, Radius of the raindrop, r = 0.3mm = 0.03 cm. Terminal velocity, v = 1 m/s =100 cm/s. Viscosity of air, η = 18 × 10 -5 Poise. According to Stokes law, force of viscosity on rain drop is F = 6π η r v = 6 × 3.142 × 18 × 10 -5 Poise × 0.03 cm × 100 cm/s = 1.018 × 10 -2 dyne

Example 2: A solid metal ball falling in a liquid column attains a terminal velocity of 3 m/s. Find the viscosity of the liquid if the radius of the ball is r = 7 cm and its density is d l = 1000kg/m 3 and the density of the metal ball is d s = 7050 kg/m 3 (Take g = 10 m/s 2 )

Given, Radius of Sphere r = 3 cm = 0.03 m Density of Sphere d s = 7050 kg/m 3 Density of liquid d l = 1000 kg/m 3 Terminal Velocity = 5 m/s v t = 2r 2 × (ρ−σ)g / 9η Substituting the values in the terminal velocity equation, 3 = 2(0.03) 2 × (7050 -1000)10 / 9η 3 = (6×10 -4 × 60500) / 9η 3 = 36.3/9η 9η = 12.1 η = 12.1/9 = 1.344 kg/ms

Example 3: The speed of water in a water stream is 18 km/h near the surface. If the stream is 10 m deep, then determine the Shearing stress between the surface layer and the bottom layer (coefficient of viscosity of water, η=10 −3 Pa s).

Since, velocity of water at the bottom of the river is 0 m/s, Therefore, dv = 18 km/h = (18 × 5/18) m/s = 5 m/s Also, dx = 10 m and η=10 −3 Pa s Force of viscosity, F = η A (dv/dx) Therefore, Shearing stress = F / A = η (dv/dx) = 10 −3 Pa s × (5 m/s / 10 m) = 5 ×10 −2 N m -2

Important Physics Related links:

Torricelli’s Law
Radius Of Gyration Definition
Heat Pump And Refrigerator
Translational Motion Definition
Vernier Caliper Parts
Uses Of Metamorphic Rocks
How To Construct A Bridge?
Work Done Definition
Formula Of Magnetic Moment

Summary – Stoke’s Law

Stoke’s Law is like a guidebook explaining why objects like raindrops fall at a steady speed instead of speeding up non-stop. Proposed by George Gabriel Stokes, this law dives into the world where tiny spherical objects slow dance with fluids, revealing why they eventually settle into a graceful, constant descent. Picture a raindrop tumbling down, initially picking up speed due to gravity’s pull.

According to Stoke’s Law, as it falls, the liquid around it starts to push back, gradually increasing until this push matches gravity’s pull. When these forces balance out, our raindrop glides smoothly at a constant pace, hitting the ground gently instead of with a splash. This isn’t just about raindrops; Stoke’s Law has its fingers in many pies, from parachutes fluttering to safety, clouds wafting across the sky, to the nitty-gritty of scientific experiments like measuring an electron’s charge.

It’s a fascinating peek into how objects interact with the world on a microscopic level, transforming seemingly chaotic descents into harmonious falls.

Stokes law – FAQs

Define stoke’s law.

The equation which gives the drag force acting on the body if it is in free fall in any liquid column is called Stoke’s Law. It is used to find the terminal velocity of an object.

When do we use Stoke’s Law?

Stoke’s Law is used to determine the terminal velocity of any object. This law also works in explaining the working of parachutes, cloud formation and others.

When is Stoke’s Law valid?

Stoke’s Law is valid only when the velocity of the object is constant under influence of the gravitational force which is balanced by the drag force of the liquid.

What are the Applications of Stoke’s Law?

Various applications of Stoke’s Law are, Measuring the Viscosity of Fluid Finding the Terminal Velocity of an object, etc.

What is the Force on the Water Drop falling with a Constant Velocity?

When a drop falls at a constant velocity, the viscous force is balanced by its weight i.e. the force of gravity. Therefore, the resulting force is zero .

What is Terminal Velocity?

Terminal velocity is the velocity of an object when the net force acting on the object is zero. In the case of raindrops, the mass of the drop acts downward while the viscous force of the air acts upward when these two force balance each other the velocity attained by the raindrop is called the Terminal Velocity.

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Experience in modernizing a laboratory study of stokes' law

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Published: November 1968
Volume 11 , pages 138–140, ( 1968 )

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Literature cited

L. D. Landau and E. M. Lifshitz, Fluid Mechanics [in Russian], Moscow, 1954.

A. I. Kitaigorodskii, Introduction to Physics [in Russian], Fizmatgiz, 1959.

Syslo Mieczyslaw, “Uogolnione pravo Stokesa dla pryl spójnych,” Zecz. nauk. Acad. górn.-hutn., no. 94, 1964.

E. V. Zolotykh, “Theoretical and experimental study of viscosimeters,” Trudy institutov Komiteta standartov mer i izmeritel'nykh priborov, no. 75, 135, Moscow, 1964.

Google Scholar

T. Maxwothy, “Accurate measurements of sphere drag at low Reynolds numbers,” J. Fluid Mech. vol. 23, part 2, pp. 369–372, 1965.

Daniel Nahshol, “Measuring viscosity by Stoke's law,” American J. of Physics, vol. 33, no. 9, pp. 657, 658, 1965.

G. Barr, Monograph of Viscosimetry [Russian translation], Moscow-Leningrad, 1938.

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Translated from Izvestiya VUZ. Fizika, Vol. 11, No. 11, pp. 150–152, November, 1968.

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Lisenkov, N.A. Experience in modernizing a laboratory study of stokes' law. Soviet Physics Journal 11 , 138–140 (1968). https://doi.org/10.1007/BF00816087

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Received : 11 April 1968

Issue Date : November 1968

DOI : https://doi.org/10.1007/BF00816087

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Navier-Stokes equations

From cfd-wiki.

The Navier-Stokes equations are the basic governing equations for a viscous, heat conducting fluid. It is a vector equation obtained by applying Newton's Law of Motion to a fluid element and is also called the momentum equation . It is supplemented by the mass conservation equation, also called continuity equation and the energy equation . Usually, the term Navier-Stokes equations is used to refer to all of these equations.

Equations Boundary conditions Derivation Derivation of continuity equation Derivation of momentum equations forces and stresses deformation and rotation Newtonian Fluids Stokes Hypothesis substitution Other formulation Derivation of the energy equation Existence and uniqueness History References External links

The instantaneous continuity equation (1), momentum equation (2) and energy equation (3) for a compressible fluid can be written as:

	(1)
	(2)
	(3)

For a Newtonian fluid, assuming Stokes Law for mono-atomic gases, the viscous stress is given by:

(4)

Where the trace-less viscous strain-rate is defined by:

(5)

$q_j$

(6)

$Pr$

(7)

To close these equations it is also necessary to specify an equation of state. Assuming a calorically perfect gas the following relations are valid:

(8)

$\gamma$

(9)

Note that the corresponding expression (15) for Favre averaged turbulent flows contains an extra term related to the turbulent energy.

$\mu$

Boundary conditions

Derivation of continuity equation.

$V_{m}$

Applying the Reynolds transport theorem and divergence theorem one obtains:

$0= \frac {dM}{dt}= \int_{V_{m}} \frac {\partial \rho}{\partial t} dV + \int_{S_m} \rho \boldsymbol{u}\cdot\boldsymbol{n} \, dS = \int_{V_{m}} \left( \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \cdot \left( \rho \boldsymbol{u} \right) \right) dV$

(1)

at all points of the fluid. For an incompressible fluid the change rate of density is zero. One can simplify (1) to:

$\boldsymbol{\nabla} \cdot \boldsymbol{u} = 0.$

Derivation of momentum equations

All is started from Newton's law:

FORCE is equal MASS time ACCELERATION

(2)

and which is applied to the fluid parsel

Expansion, rotation and deformation of a fluid parsel

forces and stresses

(3)

$\textbf{F}$

(4)

There are two types of forces: body(mass) forces and surface forces. Body forces act on the entire control volume. The most common body force is that due to gravity. Electromagnetic phenomena may also create body forces, but this is a rather specialized situation.

Surface forces act on only surface of a control volume at a time and arise due to pressure or viscous stresses.

$dx,dy,dz$

At the moment we assume this parallelepiped isolated from the rest of the fluid flow , and consider the forces acting on the faces of the parallelepiped.

Let the left forward top of a parallelepiped lies in a point O

$x$

(6)

(7)

(8)

(9)

The force due to the stress is the product of the stress and the area over which it acts.

(10)

(11)

(12)

(13)

(14)

deformation and rotation

(15)

(16)

(18)

(19)

(20)

(21)

(22)

Newtonian Fluids

$\tau$

Newton postulated (since then experimentally verified) that the shear force or shear stress needed to deform the fluid was linearly proportional to the velocity gradient:

$v / h= \partial u / \partial y$

Therefore, Newton postulated:

Fluids that have a linear relationship between stress and strain rate are called Newtonian fluids . This is a property of the fluid, not the flow. Water and air are examples of Newtonian fluids, while blood is a non-Newtonian fluid.

Stokes Hypothesis

Stokes extended Newton's idea from simple 1-D flows (where only one component of velocity is present) to multidimensional flows. He developed the following relations, collectively known as Stokes relations

$\lambda$

substitution

Other formulation.

Cauchy's equation of motion

The equation of motion for a Newtonian fluid is obtained by constitutive equation into Cauchy's equation to obtain

$e_{ij}$

If the viscous effects are negligible, we obtain the Euler equation

Derivation of the energy equation

we follow Wesseling (2001)

$V(t)$

Let heat diffusion be governed by Fourier's law:

$k$

By application of the transport theorem and substitution we obtain

External links

Navier-Stokes equations at mathworld.com
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PDF Experiment No. 16 Viscosity
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8Experiment 6: Viscosity (Stoke's Law)Viscosity is a property of uids (liquids and gases) which determines how much resistance is experienced by a. object trying to move through the uid. In this experiment we will use Stoke's Law and the concept of terminal velocity. ne the viscosity of glycerin.ObjectiveUse Stoke's Law to derive an equation relat.
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This is called Stoke's law. Where η is the viscosity of the fluid, r is the radius of the ball and is the terminal velocity of the ball in that particular fluid. Plugging everything into the equation we can then find our expression for Stokes law. Rearranging the equation above we can then find an expression for the viscosity of the fluid.
Stokes' law
In fluid dynamics, Stokes' law is an empirical law for the frictional force - also called drag force - exerted on spherical objects with very small Reynolds numbers in a viscous fluid. [1] ... The school experiment uses glycerine or golden syrup as the fluid, and the technique is used industrially to check the viscosity of fluids used in ...
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What is Stokes' Law. Stokes' law is a mathematical equation for the drag force experienced by small spherical particles passing through a viscous fluid medium. It deals with the resistive force applied to a body under the action of gravity as it is dropped into a fluid - liquid or air.Historically, Stokes' law has been named after Anglo-Irish physicist and mathematician George Gabriel ...
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Assuming an air temperature of -50°C and an altitude of 35,000 feet, the density of the air is about 10-3 g/cm3 and the viscosity is something like 1.5 x 10-4 poise. Substituting the various values into Equation 3.21, we find that the boundary-layer thickness is a few hundredths of a millimeter.
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this video illustrates an experiment for using stokes law to find coeficient of viscosity. please Kindly consider donating to help support the continued prod...
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Stokes' law shows that the frictional drag (F) is directly proportional to the weight of the sphere; in other words F is proportional to r 3.The formula for viscosity shows that the terminal velocity (v) is proportional to the radius squared; v is greater for a larger sphere than for a smaller one of the same material.
Fluid Mechanics: Stokes' Law and Viscosity Measurement Laboratory
Stokes' Law is written as, Fd = 6pmVd where Fd is the drag force of the fluid on a sphere, m is the fluid viscosity, V is the velocity of the sphere relative to the fluid, and d is the diameter of the sphere. Using this equation, along with other well-known principle of physics, we can write an expression that describes the rate at which the ...
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The experimental setup has been designed to use inexpen-. sive, like chronometers and glassware, or ubiquitous (like the. balance) laboratory materials. The setup directly measures. the force ...
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EXAMPLE 3.2C: SETTLING TANK FOR OIL/WATER SEPARATION (ADVANCED) During the production of crude oil, water is often produced along with natural gas. The result is a homogeneous water-in-oil emulsion, with water droplets settling at the bottom of the tank. One of the simplest methods to separate this emulsion is to design a huge settling tank and ...
Practical investigation into Viscosity in liquids (Stokes Law)
The first set of experiments was to show the speed of five differently sized ball bearings descending through water. All five ball bearings seemed to reach their terminal velocity at the same timed interval. Viscosity of liquid. The second set of experiments was to show the affects of a more viscous liquid (engine oil) on the decent of a ball ...
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Importance of Stoke's Law. Stoke's law is important in explaining the formation of clouds, the falling from a parachute, why the larger rain droplets hurt more than the smaller ones, and more. Millikan depended on the concept of Stokes' law in his oil-drop experiment, which was used to determine the electronic charge.
fluid dynamics
In an experiment to determine the viscosity by the Stoke method. When we throw the ball into the liquid it will reach a certain mark and start recording the time. This mark what determines its location, such that it is 5 cm further from the top edge, why not more or less. And the bottom mark. What determined their location and why. fluid-dynamics.
Stokes' law, viscometry, and the Stokes falling sphere clock
This article is part of the theme issue 'Stokes at 200 (part 2)'. In 1851, Stokes derived [1] the drag force on a spherical body in a fluid, now called Stokes' law, Fd 6π μRv, where R is the radius of the sphere, is the flow. v velocity and the constant μ is the fluid viscosity. Stokes' paper was 'On the effect of the internal ...
Stokes's law
The law, first set forth by the British scientist Sir George G. Stokes in 1851, is derived by consideration of the forces acting on a particular particle as it sinks through a liquid column under the influence of gravity. In Stokes's law, the drag force F acting upward in resistance to the fall is equal to 6 πrηv, in which r is the radius ...
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Stoke's Law Examples. Example 1: A raindrop of radius 0.3 mm falls through the air with a terminal velocity of 1 m/s. The viscosity of air is 18 × 10-5 Poise. Find the viscous force on the raindrop. Solution: Given, Radius of the raindrop, r = 0.3mm = 0.03 cm. Terminal velocity, v = 1 m/s =100 cm/s.
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Navier-Stokes equations

Boundary conditions

Derivation of momentum equations

forces and stresses

deformation and rotation