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Civil Engineering Hydraulics Mechanics of Fluids

Pressure and Fluid Statics

The fastest healing part of the body is the tongue.

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Common Units

  In order to be able to discuss and analyze fluid problems we need to be able to understand some fundamental terms commonly used

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Common Units

  The most used term in hydraulics and fluid mechanics is probably pressure

  Pressure is defined as the normal force exerted by a fluid per unit of area   The important part of that definition is

the normal (perpendicular) to the unit of area

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Common Units

  The Pascal is a very small unit of pressure so it is most often encountered with a prefix to allow the numerical values to be easy to display

  Common prefixes are the Kilopascal (kPa=103Pa), the Megapascal (MPa=106Pa), and sometimes the Gigapascal (GPa=109Pa)

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Common Units

 A bar is defined as 105 Pa so a millibar (mbar) is defined as 10-3 bar so the millibar is 102 Pa

The word bar finds its origin in the Greek word báros, meaning weight.

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Common Units

 Standard atmospheric pressure or "the standard atmosphere" (1 atm) is defined as 101.325 kilopascals (kPa).

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Common Units

 This "standard pressure" is a purely arbitrary representative value for pressure at sea level, and real atmospheric pressures vary from place to place and moment to moment everywhere in the world.

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Common Units

 Pressure is usually given in reference to some datum  Absolute pressure is given in

reference to a system with no pressure whatsoever (a vacuum)

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Common Units

 Pressure is usually given in reference to some datum  Gage pressure, the more commonly

used pressure, is the difference between local atmospheric pressure and the absolute pressure of the system being measured

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Common Units

  If the gage pressure of the system being measured is less than local atmospheric pressure, the pressure may be termed a vacuum pressure

 This does not imply that it has no pressure, just that the pressure is less then local atmospheric

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Common Units

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Common Units

 To be the most precise, when giving pressure, you should state if it is an absolute or a gage pressure

 There is a difference

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Pressure at a point

 What can appear as non-intuitive is that the pressure at any point in a fluid is the same in all directions

  It would seem to make more sense if the pressure was greater on the top of the point than on the bottom and not at all on the sides but remember we are talking about a point.

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Pressure at a point

 The statics (yes I said statics) of the situation can be used to define just what is happening

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Pressure at a point

 We can start with a fluid at rest and therefore at static equilibrium

  In that fluid, we can pick a segment with a triangular cross section and a unit depth, into the fluid or into the page, with a thickness of 1 unit

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Pressure at a point

The shape was a thickness of 1 in the y-direction.

Remember the pressure is defined as the force exerted normal to the surface.

Since we have the FBD of this wedge, we are showing forces action on the wedge.

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Pressure at a point

We need to include one more force on this FBD and that is the weight of this wedge of fluid.

W

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Pressure at a point

We can start by writing our expressions for force balances along each axis. W

( )( ) ( )( )( )

( )( ) ( )( )( )

2 3

1 3

01 1 cos

01 1 sin

z

z

x

x

FF W P x P lFF P z P l

θ

θ

=

= − + Δ −

=

= Δ −

∑∑∑∑

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Pressure at a point The weight of the wedge can be found by solving for the volume of the wedge and then using the mass density and gravity to find the weight.

W

( )( ){ }1 12

W Vg

W g x z

ρ

ρ

=

= Δ Δ

ρ is the mass density of the fluid and g is the universal gravitational constant.

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Pressure at a point Substituting for W in our initial expressions we have

W

Since the left side of all the expressions are equal to 0, we can divide both sides by 1 and get rid of the 1’s in both expressions.

( )( ){ } ( )( ) ( )( )( )

( )( ) ( )( )( )

2 3

1 3

01 1 1 1 cos2

01 1 sin

z

z

x

x

F

F g x z P x P l

FF P z P l

ρ θ

θ

=

= − Δ Δ + Δ −

=

= Δ −

∑∑∑∑

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Pressure at a point Now we have

W

In the second expression, the term l sin θ is equal to Δz so we have

( )( ){ } ( ) ( )( )

( ) ( )( )

2 3

1 3

10 - - cos2

0 sin

g x z P x P l

P z P l

ρ θ

θ

= Δ Δ + Δ

= Δ −

( )( ){ } ( ) ( )( )

( ) ( )

2 3

1 3

1 3

10 - - cos2

0

g x z P x P l

P z P zP P

ρ θ= Δ Δ + Δ

= Δ − Δ∴ =

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Pressure at a point

W

In the first expression, the term l cos θ is equal to Δx so we have

( )( ){ } ( ) ( )( )

( ) ( )

2 3

1 3

1 3

10 - - cos2

0

g x z P x P l

P z P zP P

ρ θ= Δ Δ + Δ

= Δ − Δ∴ =

0 = -!g 12

!x( ) !z( )"#$

%&'+ P2 !x( ) - P3 !x( )

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Pressure at a point

W

Dividing all the terms by Δx we have

( )( ){ } ( ) ( )2 3

1 3

10 - -2

g x z P x P x

P P

ρ= Δ Δ + Δ Δ

=

0 = -!g 12

!z( )"#$

%&'+ P2 - P3

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Pressure at a point

W Remember that we are talking about pressure at a point in the fluid.

We can reduce our wedge to a point by allowing the Δx and Δz dimensions to approach 0.

As Δz approaches 0, the weight term also approaches 0.

This reduces our expression to 0 = -!g 1

2!z( )"

#$

%&'+ P2 - P3

0 = P2 - P3P1 = P3(P2 = P3 = P1

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Pressure at a point

W Notice that the value for θ was not critical for our derivation and the density of the fluid did not enter into our calculation at the end.

At any point in a fluid, the magnitude of the pressure is the same in all directions. 2 3

1 3

2 3 1

0 -P PP PP P P

==

∴ = =

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Pressure at a point

W Note that this statement is made about any point, it not made about any two points having the same pressure. That is a different problem and not covered by the assumptions that we just made.

2 3

1 3

2 3 1

0 -P PP PP P P

==

∴ = =

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Pressure Variation with Depth   If we consider a fluid with a constant

density over a depth  We can consider most gasses as

having a constant density with depth over reasonable depths and most liquids also

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Pressure Variation with Depth  We can start by remembering that

pressure is the force exerted per unit of area.

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Pressure Variation with Depth

h

d We can start with a cylinder of diameter d and find the pressure at some depth h in the cylinder.

The fluid has a mass density of ρ and the pressure at the top of the cylinder is patm (atmospheric pressure).

The pressure at the depth h is going to be the sum of the patm and the pressure exerted by the weight of the fluid about the depth h.

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Pressure Variation with Depth

h

d The weight of the fluid can be determined by taking the volume of the fluid and multiplying that by the product of the mass density ρ and the gravitational constant g.

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4

dV h

dF V g h g

π

πρ ρ

=

= =

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Pressure Variation with Depth

h

d The area that the force is acting normal to is the cross sectional area of the cylinder.

2

2

2

4

4

4

dV h

dF V g h g

dA

π

πρ ρ

π

=

= =

=

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Pressure Variation with Depth

h

d So the added pressure of the fluid is the force exerted by the fluid divided by the area it is acting over.

F =V!g = "d2

4h!g

A = "d2

4

p = FA

=

!d24h!g

"d24

= h!gThe development in the text uses z for the depth of the fluid rather than h.

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Pressure Variation with Depth

h

d We can check for dimensional consistency.

p = h!gmass! length

time2length2

= length! masslength3

! lengthtime2

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Pressure Variation with Depth

h

d The pressure is not dependent on the area.

If we assume that density remains constant with depth and that the gravitational constant also remains constant with depth the pressure becomes a linear function of depth.

As you go deeper in a fluid, the pressure increases linearly.

p = h!g

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Pressure Variation with Depth

In your text, the symbol z is used for depth because h is often also used as a variable in thermodynamics.

We may use both depending on the problem. The important thing is to take the variable as measuring the depth below the surface in the fluid.

p = h!g

Pressure Variation with Depth

  In the previous slide, we made two assumptions  The first was that we were starting from 0

gage pressure at the top of the fluid which was the reference for our depth measurement

 The second was that the density of the fluid did not change with depth

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p = h!g

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Pressure Variation with Depth

 A more formal expression for change of pressure with changing depth would be

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p2 ! p1 = !gdz =z1

z2

" !gdh =h1

h2

" ! dz =z1

z2

" ! dhh1

h2

"

Pressure Variation with Depth

 This would allow for a fluid which might change density as a function of depth

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p2 ! p1 = !gdz =z1

z2

" !gdh =h1

h2

" ! dz =z1

z2

" ! dhh1

h2

"

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Problem

pressure increases downward in a given fluid and decreases upward

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Problem

pressure increases downward in a given fluid and decreases upward

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Reading

  Sections 2-2 and 2-3

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Homework

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Homework

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Homework