Materials Tutorial 4 - Fluid Materials

 

Contents

Fluids

Pressure in a Fluid

Pressure and Hydraulic Systems

Pressure and Depth

Archimedes' Principle

Terminal Velocity in Liquids

Viscosity and Drag Force

Stokes' Law
Radius and Terminal Velocity

 

Fluids

A fluid material is one that adopts the shape of its container.  It cannot resist any shear forces that are applied to it.  It can be a liquid or a gas.  The general properties of a fluid are:

However there are significant differences between fluids, depending on the phase of the matter:

Some apparently solid materials are actually very viscous (gooey) fluids.  Glass is one such example:

Other examples include silicone putty.  You can roll it into a ball.  Then leave it a while and it flattens out into a disc.

 

Question 1

Explain why glass can be considered as a super-cooled liquid.

Answer

 

 

Pressure in a Fluid

You will have encountered pressure before.  It is defined as:

perpendicular force per unit area

The equation is:

[ p - pressure (N m-2); F - force (N); A - area (m2).]

 

The units for pressure are Pascals (Pa) or Newton per square metre (N m-2) where:

1 Pa = 1 N m-2

 

If the force is at an angle, we need to take the vertical component of the force.

 

 

 So our equation is changed to:

Question 2

A force of 250 N is applied at an angle of 40o to the horizontal onto a surface of 25 cm in diameter. 

 

Calculate the pressure.  Give your answer to an appropriate number of significant figures.

Answer

 

Pressure in a fluid acts equally in all directions.  This is called Pascal's Law.  This is why you are not crushed by atmospheric pressure.

 

 

Pressure and Hydraulic Systems

This has important implications in hydraulic systems.  Consider a very simple hydraulic system consisting of a master cylinder of area A1 and a slave cylinder of area A2.  This is shown in the diagram below:

 

 

Force F1 is applied to the master cylinder.  There is a resulting pressure, p.  Since pressure is the same throughout the system, pressure p acts on the slave cylinder to give force F2.  So we can write:

Simple rearrangement gives:

 

Question 3

A simple hydraulic system has a master cylinder of area 0.0015 m2, while the slave cylinder has an area of 0.055 m2

A force of 12 N is applied to the master cylinder.  Calculate the force from the slave cylinder.

Answer

 

The system above is a force multiplier.  The ratio of the forces = ratio of the areas.  This is shown below:

 

 

Question 4

Show that your answer to Question 3 is consistent with the statement about force multipliers.

Answer

 

Pressure and Depth

SCUBA divers will tell you how when they dive, the pressure acting on them from the water increases.  (SCUBA = Self Contained Underwater Breathing Apparatus).

 

Consider a cylinder of area A m2, filled with fluid of density r kg m-3 to a height of h m.  

 

 

The volume of the cylinder = area height = hA

The mass of the fluid in the cylinder = volume density = hAr

Remember that weight is a force, NOT the mass.

The weight of the fluid = mass gravitational field strength = hArg

 

Pressure = weight (force) area:

 

The areas cancel out to give:

p = hrg

 

 

Therefore the pressure is proportional to the depth in a liquid.  This equation works for liquids, since they are incompressible, so the density of a liquid remains the same throughout the body, however deep we go.  However this model does not work for a gas.  Gases are easily compressed, so the density of a gas changes with the height, something that pilots of aeroplanes need to be aware of.

 

Question 5

A SCUBA diver dives to a depth of 20 m.  Calculate the pressure acting on the diver at that depth.

Density of seawater = 1030 kg m-3.

Gravitational field strength = 9.81 N kg-1.

Answer

 

The Pascal is quite a small unit of pressure.  Engineers tend to use the atmospheric pressure unit, bar

1.0 bar = 1.0 105 Pa

 

Your answer should show that the pressure acting on the SCUBA diver is about 2 bar.

 

Archimedes' Principle

Any object that floats in a liquid has an upthrust that is equal to its weight.  Since the weight acts vertically downwards, the upthrust acts vertically upwards.  The idea is shown below:

 

The weight of water displaced is the same as the weight of the object.  If the upthrust is less than the weight, the object will sink.

 

When an object is totally immersed in the water, the volume of water displaced is the volume of the object.  You will have used this idea to find out the density of an irregular object.  See Materials Tutorial 1.

 

Archimedes' principle states:

Any body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced

 

We can explain Archimedes' Principle using the relationship between pressure and depth.  Consider a uniform object of area A and thickness t at a depth of d in a liquid of density r:

 

 

The pressure on the top surface is:

 Since force = pressure area, we can work out the downwards force on the top surface:

 

Pressure on the bottom surface is:

 

Since force = pressure area, we can work out the upwards force on the bottom surface:

 

 

The upwards force on the object is in the opposite direction to the downwards force, therefore:

 

 

The volume of the object is tA.  This volume is the same as the volume of liquid that is displaced.  The mass of liquid displaced is volume density, so the mass of liquid displaced is tAr.  Therefore the weight displaced is tArg.  We have seen how the depth term has cancelled out.  This is consistent with the density of the water being the same whatever the depth, due to liquids being incompressible.

 

Archimedes' principle can be applied to gases.  However the upthrust from the air is small enough to be negligible.

 

Question 6

A diver is salvaging a spherical cannon ball from the wreck of an ancient ship.  The cannon ball has a diameter of 10 cm and is made of iron of density 7900 kg m-3.  The density of seawater is 1030 kg m-3.

What is the force needed to lift the cannon ball:

(a) on land;

(b) under the water?

 

Does the depth of the wreck matter?

 

Acceleration due to gravity = 9.81 m s-2.

Answer

 

 

Terminal Velocity in Liquids

It is easy to measure terminal velocity in a liquid, since the terminal speed is low.  This experiment is one of the core practicals:

 

 

When a ball bearing is dropped into a viscous liquid, it almost immediately reaches its terminal speed, and measuring it is simply a matter of timing the motion between two fixed points a known distance apart.  You will also have to:

When the ball bearing drops into the liquid (usually glycerine), there are two forces:

Since the weight is greater than the upthrust, the ball bearing will accelerate.  The velocity-time graph will look like this:

 

At terminal speed, the upwards forces of upthrust and drag and the downwards force of the weight are balanced

 

 

The upthrust is the same as the weight of fluid displaced by Archimedes' principle.  Therefore, if the weight is greater than the upthrust, the object will accelerate downwards until the drag balances the difference between the weight and the upthrust.  This is true of all fluids, for example air, or water, or chocolate.

 

Question 7

A ball bearing of diameter 2.50 mm is made of steel.  It is released into glycerine and falls.

(a) What is the upthrust?

(b) Show that the acceleration is about 8 ms-2.

 

The density of this grade of steel is 8050 kg m-3;

The density of glycerine is 1260 kg  m-3;

Acceleration due to gravity is 9.81 m s-2.

Answer

 

You will see that the initial acceleration is not 9.81 m s-2.   This is because upthrust is a force acting in the opposite direction to the weight.  However you would not be able to time the ball bearing as it fell through the glycerine if its acceleration were 8 m s-2.  Some other force is acting.  This is the drag force.

 

Viscosity and Drag Force

Viscosity has a working definition of:

 

the quantity that describes a fluid's resistance to flow.

 

There are more formal definitions, but we won't use them here.  Viscosity is temperature dependent.  If you heat up a viscous material like oil, it becomes less viscous, i.e. more runny.  Cold car engines use more petrol because the oil is more viscous, so more energy is needed to circulate it.

 

When the ball bearing is falling at terminal velocity, the resultant force = 0 (Newton I).  The weight is balanced by the upthrust and the drag force.

 

Weight - (Upthrust + Drag) = 0

 

Question 8

Use answers from Question 7 to work out what the drag force is at terminal velocity.  What do you notice?

Answer

 

Stokes' Law

We can calculate the drag force by Stokes' Law.  It was worked out by George Gabriel Stokes (1819 - 1903).  Consider a ball falling  through a viscous liquid at terminal velocity.  There are the three forces acting on it:

Stokes' Law is given by:

 

[Fd - drag force (N); r - radius of the sphere (m); v - terminal speed (m s-1 ).]

 

The strange looking symbol, h, is "eta", a Greek lower case letter long 'ē', the Physics Code for the coefficient of the viscosity of a fluid. 

 

The units for h are N s m-2.  An alternative SI unit is Pascal seconds (Pa s).

 

For air, h = 1.8 10-5 N s m-2.  For glycerine, h = 0.950 N s m-2.

 

This relationship only works when the fluid flow around the object is laminar (smooth).  It does not work if the flow is turbulent.

 

Question 9

Use your answer to Question 8 to calculate the terminal velocity of the ball bearing in Question 7, as it falls through the glycerine.

Comment on your answer.

For glycerine, h = 0.950 N s m-2.

Answer

Question 10

A student carelessly omits the upthrust in answering Questions 7, 8, and 9.  What is the effect on the answer for the terminal velocity?

Answer

 

When you are measuring the terminal velocity of ball bearings in a viscous fluid, you need to have a start point some distance below the surface of the glycerine.

 

Question 11

Use your answer to Question 9 to give an estimate of the distance travelled by the ball bearing as it accelerates to terminal velocity.  What is your assumption?

Answer

 

So the start line a couple of centimetres below the surface will ensure that the ball bearing is travelling at the terminal velocity.  However the Stokes' Law equation does not lend itself easily to graphical analysis.  Let's look at this more closely:

 

The temptation is to say that:

 

There is a problem in that the drag force is not constant.  If we use a ball bearing of half the diameter, the weight goes down by 8 times.  Using the answers to the questions above, the weight of a ball bearing of diameter 1.25 mm will give these results:

 

Diameter / 10-3 m

Weight / N

Upthrust / N

Drag Force / N

2.50

6.46 10-4 N

1.01 10-4 N

5.45 10-4 N

1.25

8.08 10-5 N

1.27 10-5 N

6.81 10-5 N

 

Therefore the drag force will be eight times less.  So instead of the terminal velocity being doubled when the radius is halved, it is reduced by four times.  Let's look at this further.

The 6ph is a constant, which we will call k.  So the relationship between drag force and terminal velocity becomes:

 

 

If the radius is halved, the force goes down eight times.  So the new terminal velocity v' is given by:

 

 

Therefore we need to look at a new relationship between the radius and the terminal velocity.

 

 

Relationship between Radius and Terminal Velocity

I have written two arguments here to establish the relationship between the radius of a ball bearing and its terminal velocity.  The first does not take into account the upthrust (which is not on the syllabus for some boards).  The second does take into account the upthrust.

 

Argument without upthrust

Upthrust is not on the syllabus for some of the boards.  The derivation of this equation is an approximation, as we are not taking into account the upthrust.  The resulting uncertainty is relatively low, compared with the uncertainties of the other measurements made in the experiment.  If you want to see the effect of upthrust on the argument, please go to the next section.

 

Consider a ball bearing of mass m made of material of density r being dropped into a viscous fluid of viscosity h.  The gravity field is g (= 9.81 m s-2 as you aren't likely to take the apparatus elsewhere).

 

Weight = density volume g

 

We can write this as:

 

So we can bring in the Stokes' Law equation in by writing:

 

 

Cancelling out gives us:

And rearranging gives:

 

Then we take the square root to show the equation in the form it's usually presented:

 

 

We can rearrange to make v the subject:

 

 

Therefore we can plot a graph of v against r2:

 

The gradient will be:

 

So if we know the density of the ball-bearing, we can work out the coefficient of the viscosity.

 

Question 12

A ball bearing of diameter 2.50 mm is made of steel.  It is released into glycerine and falls.

 

Calculate the terminal velocity.

 

The density of this grade of steel is 8050 kg m-3;

For glycerine, h = 0.950 N s m-2;

Acceleration due to gravity is 9.81 m s-2.

Answer

 

Argument with upthrust

The derivation of this equation takes into account the upthrust.

 

Consider a ball bearing of mass m made of material of density r being dropped into a viscous fluid of viscosity h, and density s.  The gravity field is g (= 9.81 m s-2 as you aren't likely to take the apparatus elsewhere).  By Archimedes, the upthrust is the weight of the fluid displaced = volume of the object density of the fluid

 

The strange looking symbol s, is sigma, a Greek lower case letter 's'.

 

We can write the weight as:

We can write the upthrust as:

 

We know that the drag force = weight - upthrust:

 

We can rewrite this by substituting:

We can write this as:

 

 

So we can bring in the Stokes' Law equation in by writing:

 

 

Cancelling out gives us:

 

 

And rearranging gives:

 

 

The usual form of this equation is the square root:

 

This rearranges to make v the subject:

 

Therefore we can plot a graph of v against r2:

 

The gradient will be:

 

So if we know the density of the ball-bearing and the viscous fluid, we can work out the coefficient of the viscosity.  The term (r - s) is simply the difference between the densities of the solid ball bearing and the viscous liquid.

 

Question 13

A ball bearing of diameter 2.50 mm is made of steel.  It is released into glycerine and falls.

 

Calculate the terminal velocity.

 

The density of this grade of steel is 8050 kg m-3;

The density of glycerine is 1260 kg  m-3;

For glycerine, h = 0.950 N s m-2;

Acceleration due to gravity is 9.81 m s-2.

 

Answer

 

The study of fluid flow is not easy, and is part of university level study at second year.