Fluids - Tarleton State University



Fluids

I. Pressure

Concept 1:

In a fluid, the forces are not located at a specific point like a rope pulling on a block. Instead, the forces are distributed across an area of the fluid. For this reason, it is easier to work with force per area instead of force!! This is also true for solid objects like beams or bridges when doing real world problems where forces are distributed across the object.

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A. Definition - P

The magnitude of pressure is defined as the force component perpendicular to an infinitesimally small surface area divided by the surface area.

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B. You should note that pressure has a direction (air pressure in a tire pushes outward on the tire and not inward). However, we don’t usually need the full power of vectors in the problems at this level (much like doing 1-D problems) since the force on any object placed in an ideal fluid is always perpendicular to the object’s surfaces. Thus, textbooks often use pressure when they mean the “magnitude of the pressure.”

Example: Bed of Nails Magic Trick

Most people think that a material tears/breaks when too great of a force is

applied to the object. Actually the material’s response is determined by both

the magnitude of the force and how the force is distributed. In other words, it

is the pressure that is important.

A layer of atoms in a material are held together by electrical forces between the atoms. A simple model would be a group of kids in a line holding hands as in “Rover-Red-Rover” shown as white circles below. The applied force can be modeled as a group of kids (six kids – black dots shown below) running into the line. Increasing the force would be like increasing the total number of kids running into the line. If the force is spread out (each of the kids runs into the line at a different place – first picture) the line may be able to handle the force. However, if the kids all rune into the line at the same place (second picture) then the same force may break through the line. A common example of this phenomena is floating. When you lay out in the water, you spread out your normal force, this means that each water molecule has to apply less force to support you. However, if you try and walk on water, your normal force is distributed upon a smaller area and the water molecules are unable to apply enough force to offset your weight before they are pushed aside (line breaks). This is the same principle behind the bead of nails in which you distribute your force across many nails.

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II. Density (actually mass density)

Concept 2: A fluid can change its shape depending on the vessel that it is placed into. This makes it difficult to talk about the amount of mass in a certain shape. Thus, we don’t talk about the mass of the free body, but the mass per volume (density) of the free body.

A. Definition - (

The density of a uniformly object is defined as the object’s mass divided by the object’s volume.

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Density is a useful concept even for solid objects!! The mass of a gold ball depends on the size of the ball. However, the density of gold is the same for any size ball. It is only a property of gold at a particular temperature and pressure. Thus, the densities of materials are tabulated in handbooks including the CRC and can be used to distinguish between different materials.

III. Increasing Pressure at A Depth

We know that the pressure applied by the water against the hull of a submarine at great depths is very large and can even crush the submarine if the submarine welds have the slightest flaw. We will use Newton’s 2nd Law to derive the following equation for this important example both as an example of how fluids are just an extension of the past work this semester and because it leads to Pascal’s principle.

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A fluid of density ( is shaded in the figure below. We now consider the free body diagram for a cylindrical section of fluid that has a cross sectional area A and thickness h.

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IV. Pascal’s Principle

A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

A. Submarine Example

The first term in our previous work demonstrates Pascal’s principle. The pressure applied by air on the surface of the water, P0, is transferred to the submarine at any depth. The second term is the increase in pressure due to the weight of the water above the submarine pushing down upon the submarine and the water surrounding the submarine.

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For the submarine, we see that the pressure is applied at all points around the submarine in order to ensure that the net external force is zero (i.e. it is not accelerating). We neglected this in our previous FBD work in order to simplify the drawing. For the submarine, the pressure would need to be greatest at the bottom since it must hold up the weight of the submarine and the submarine has a finite size. Our previous work suggests that this extra pressure must be the second term in our equation which only depends on the fluid!! This will lead us in the next section to Archimedes’s principle which was known hundreds of years before Pascal’s principle.

B. Pressure on a Vessel

If instead of a submarine, we consider an infinitesimally small unit of volume then the pressure must be the same on all sides of the volume element. This fact has some interesting consequences. Consider an infinitesimal volume element of fluid next to the wall of a container as shown below.

The pressure applied on the left side of the volume element by the container must be the same as the pressure applied on the right side of the element by the rest of the fluid since the volume element can not accelerate. By Newton’s third law, this also means that the volume element must apply the same pressure back up the container. Thus, our formula in section III also gives us the pressure on the walls of a vessel. Since the pressure on a dam increases at the bottom of the dam, the thickness of the dam must be larger at the bottom.

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C. Example – Hydraulic Lift

A useful application of Pascal’s principle is the hydraulic lift. Consider an enclosed fluid shown below. A downward pressure of P is applied to a piston of area A1. This pressure is undiminished through the fluid (Pascal’s principle) and pushes upward on a second piston on area A2.

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Using the definition of pressure, we have

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Thus, we see that the upward force on the second piston can be made large than the downward force F1 by the ratio of the areas of the two pistons.

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This is the technique in which an auto mechanic can push down on a small piston and raise a car. It might seem sensible to think that the forces on the pistons would be the same. However, the force is increased because there are many more fluid atoms pushing on the larger area piston than on the smaller piston!!

V. Archimedes’s Principle

Any object completely or partially submerged in a fluid is buoyed up by a force with magnitude equal to the weight of the fluid displaced by the object.

Consider an object of mass mo that is partially or totally submerged like the one shown below with its free body diagram.

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The weight of the object can be related to the density of the object, (o, and the object’s total volume V by

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Using Archimedes’s principle, the buoyancy force is related to density of the fluid (f, and the volume of the fluid displaced (this is the portion of the volume of the object which is submerged) Vs.

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Substituting into our Newton II work we have

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Replacing mo in terms of the object’s density, we have

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Case I: Floating (Partially Submerged) Object

In this case the object is in equilibrium so Ay = 0. This gives us the following condition:

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This says that the fraction of the volume of the object that is submerged is equal to the ratio of the density of the object to the density of the fluid. Since the largest fraction of the volume that can be submerged is 100%, we also see that the density of the object must be less than the density of the fluid if the object is to float..

Case II: Totally Submerged Object

In this case the volume of water displaced is equal to the volume of the object.

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Putting this condition in our Newton II results, we have

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If the object’s density if greater than the fluid then the object will accelerate downward (sink) until it hits the bottom and an additional force is added to our free body diagram. If the object’s density is less than the density of the fluid, the object will accelerate upward to the surface and will become only partially submerged as in case II. If the density of the object and the fluid is the same then the object will be in equilibrium and remain submerged at this depth.

VI. Continuity Equation

For an incompressible fluid the volume of fluid passing through any size cross section of a pipe must be the same due to conservation of mass. This condition can also be stated in a more useful form as follows:

The product of the cross section and the component of the velocity of the fluid perpendicular to the cross sectional area is a constant.

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This conserved quantity is called flux which is Latin for “water flow.”

Proof: Consider a fluid flowing through a pipe of varying cross sectional area as shown below.

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An amount of fluid of mass M enters the pipe on the left at speed v1. Since the amount of fluid in the pipe can not increase, the same mass of fluid must be pushed out on the right side of the pipe at speed v2.

We can write the mass of the volume of fluid entering the left hand side over a time t as

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We can write the mass of the volume of fluid exiting the right hand side over a time t as

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Equating our two results gives us the continuity equation as

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Q.E.D.

The continuity equation explains many interesting everyday events including the fact that when water flows through a reduced cross sectional area it will travel at a higher speed. This is why we reduce the area of a garden hose to make a sprayer and why fast moving rapids occur when a river becomes shallow!!! It is just a simple consequence of conservation of atoms!!!

VII. Bernoulli’s Equation

Bernoulli’s equation is a consequence of applying conservation of energy to a fluid.

Consider a fluid flowing through a pipe as shown below.

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The work done on the fluid by an external force F1 in displacing the fluid by (x1 is

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The work done on the fluid by an external force F2 as the fluid is displaced by (x2 is

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Assuming that there are no other un-conservative forces acting on the fluid, we have from conservation of mechanical energy

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The kinetic energy of the fluid at position 1 is given by

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The kinetic energy of the fluid at position 2 is given by

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The gravitational potential energy at point 1 is given by

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The gravitational potential energy at point 2 is given by

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Substituting our results into our conservation of mechanical energy equation and performing some algebra gives us Bernoulli’s equation.

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Bernoulli’s Equation

According to Bernoulli’s equation a faster moving fluid applies a lower pressure upon an object than a slower moving fluid.

This and other interpretations of Bernoulli’s equation will allow us to explain many interesting physical phenomena including curve balls, lift on air plane wings, and why fluids flow down hill. We will discuss some of these in class. What is important to remember is that no matter what we call it, we are still simply analyzing systems using energy analysis!!!

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Single Contact Force

Distributed Force

High Pressure, Same Force

Low Pressure, Same Force

F

F

h

A

x

y

P

P

P

P

P

P

Same height

x

y

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v1

v2

y1

y2

F1

F2

(x1

(x2

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