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Civil Engineering Hydraulics Mechanics of Fluids

Fluid Statics

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Fluid Statics 2

Pressure Variation with Depth

While the arrows might look strange, remember that the pressure is exerted normal to any surface the fluid is in contact with.

Even thought the arrows are in different directions, the pressures at each point shown, other then H, has the same magnitude.

p = h!g

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Fluid Statics 3

Review Problem

Fluid Statics 4

Another Type of Manometer The figure shown below is a typical case of using a manometer to measure the pressure drop along a conduit.

If the fluid were at rest, the pressure at points 1 and 2 would be the same.

However, since the fluid is moving, we lose pressure due to friction so the pressure at point 1 is higher than the pressure at point 2.

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Fluid Statics 5

Another Type of Manometer The figure shown below is a typical case of using a manometer to measure the pressure drop along a conduit.

The pressure at point B

( ) ( )2 1 2BP P g a g hρ ρ= + +

Fluid Statics 6

Another Type of Manometer Since point A and B are in the same fluid and the fluid isn’t moving, PA is equal to PB.

( )( ) ( )

( ) ( ) ( )

1 1

2 1 2

1 1 2 1 2

A

B

P P g a hP P g a g hP g a h P g a g h

ρρ ρ

ρ ρ ρ

= + += + ++ + = + +

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Fluid Statics 7

Another Type of Manometer If we rearrange terms to determine the difference in pressure at points 1 and 2 we get.

( ) ( ) ( )

( )

1 2 1 2 1

1 2 1 2 1 1

1 2 2 1

1 2 2 1

P P g a g h g a hP P ga gh ga ghP P gh ghP P gh

ρ ρ ρρ ρ ρ ρρ ρρ ρ

− = + + − +− = + + − −− = + −− = −

Fluid Statics 8

Another Type of Manometer So by knowing the density of the fluids and the relative heights in the manometer, we are able to determine the pressure drop along the condiut.

( ) ( ) ( )

( )

1 2 1 2 1

1 2 1 2 1 1

1 2 2 1

1 2 2 1

P P g a g h g a hP P ga gh ga ghP P gh ghP P gh

ρ ρ ρρ ρ ρ ρρ ρρ ρ

− = + + − +− = + + − −− = + −− = −

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Fluid Statics 9

Fluid Statics

 We will be looking at systems that are in static equilibrium so we can use our fundamentals of force and moment balances to look at the systems

Fluid Statics 10

Fluid Statics  We can start with a generalized object

with a rectangular cross section with a vertical orientation submerged in a fluid

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Fluid Statics 11

Fluid Statics  Then we can generalize to a surface

with any type of cross section and any orientation.

Fluid Statics 12

Fluid Statics

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Fluid Statics 13

Fluid Statics

Fluid Statics 14

Fluid Statics The gas tank of an automobile is sketched in a profile view in Figure 2.16. The lower edge of a semicircular plug is located 1 cm from the tank floor. The tank is filled to a height of 20 cm with gasoline and pressurized to 130 kPa. Calculate the force exerted on the semicircular plug. Assume that gasoline properties are the same as those for octane.

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Fluid Statics 15

Homework Problems

Fluid Statics 16

Homework Problems

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Fluid Statics 17

Homework Problems

Additional Slides

The following are slides that I have used in previous semesters and develop the class material in a slightly different manner.

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Fluid Statics 19

Fluid Statics  We can start with a generalized object

submerged in a fluid

Fluid Statics 20

Fluid Statics  The generalized object has a uniform

thickness that we can say is 1 unit

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Fluid Statics 21

Fluid Statics   It has an arbitrary cross section

Fluid Statics 22

Fluid Statics  We can start by taking a parallel at the surface

of the object to the surface of the fluid.

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Fluid Statics 23

Fluid Statics  This will be our y-axis  The positive direction will be to go deeper into

the fluid

Fluid Statics 24

Fluid Statics  From the point of intersection of our y-axis and

the surface of the fluid we will generate a z-axis

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Fluid Statics 25

Fluid Statics  Again, our positive direction will be to go

deeper into the fluid.

Fluid Statics 26

Fluid Statics  We can look at any differential element in the

fluid and find the pressure on that element

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Fluid Statics 27

Fluid Statics  The area of the differential element will be dz

by dy  We will label that dA

Fluid Statics 28

Fluid Statics  To see what the pressure on that differential

area is, we will need to find the depth of that differential area below the surface of the fluid.

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Fluid Statics 29

Fluid Statics  We can look at a differential area at a depth h

below the surface of the water.

Fluid Statics 30

Fluid Statics  Remember that the pressure is the same in all

directions so if we know how the pressure varies with depth (and we do) we can find the pressure on this differential area

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Fluid Statics 31

Fluid Statics  The differential area is at a depth h so the

pressure on that area is the pressure at the surface plus the pressure from the column of fluid above the differential area

Fluid Statics 32

Fluid Statics   If the fluid surface is exposed to the

atmosphere, then the pressure at a depth h is equal to Ph = Patm + ρgh

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Fluid Statics 33

Fluid Statics  We can generalize and say that the pressure at

the surface is P0, then the pressure at a depth h is equal to Ph = P0 + ρgh

Fluid Statics 34

Fluid Statics  Since we are going to sum the forces

generated by the fluid on the top of the surface, we need to translate this pressure expression into one along the surface

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Fluid Statics 35

Fluid Statics  We can do this by writing h in terms of distance

along the plate (y)

Fluid Statics 36

Fluid Statics  By the way we set up the axis, the distance

along the place from the origin (y) is the hypotenuse of a right triangle with h as one of the sides

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Fluid Statics 37

Fluid Statics   h is then related to y as

sinh y θ=

Fluid Statics 38

Fluid Statics  So our expression for the pressure on any

differential area can be rewritten as

0 sinP P gyρ θ= +

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Fluid Statics 39

Fluid Statics   In order to find the total force exerted on the

plate, we need to sum up all the forces exerted on all the differential areas that make up the surface

0 sinP P gyρ θ= +

Fluid Statics 40

Fluid Statics  The force on any differential area is equal to the

pressure on that area times the area

0 sin

dA

P P gyF PdA

ρ θ= +=

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Fluid Statics 41

Fluid Statics  The force on the entire surface is equal to the

some of all the forces all over the surface

0 sindAA A

F F PdA

P P gyρ θ

= =

= +∫ ∫

Fluid Statics 42

Fluid Statics  Substituting the second expression into the first

( )0 sinA

F P gy dAρ θ= +∫

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Fluid Statics 43

Fluid Statics  Manipulating the expressions we have

( )( ) ( )0

0

sin

sinA

A A

F P gy dA

F P dA gy dA

ρ θ

ρ θ

= +

= +

∫∫ ∫

Fluid Statics 44

Fluid Statics  Manipulating the expressions we have

( ) ( )0

0

sin

sinA A

A

F P dA gy dA

F P A g ydA

ρ θ

ρ θ

= +

= +

∫ ∫∫

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Fluid Statics 45

Fluid Statics  From statics, hopefully you remember, ŷ is the

distance to the y centroid of the surface

0 sinA

A

yA ydA

F P A g ydAρ θ

=

= +

∫∫

)

Fluid Statics 46

Fluid Statics  You text uses a different symbol for the centroid

so we will adopt this also

0 sin

c A

A

y A ydA

F P A g ydAρ θ

=

= +

∫∫

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Fluid Statics 47

Fluid Statics  So our expression for the magnitude of the

force acting on the surface is

( )0 sin cF P A g y Aρ θ= +

Fluid Statics 48

Fluid Statics  From our triangle developed earlier, the depth

to the centroid is equal to yc sin(θ)  We will label this hc

0 cF P A gh Aρ= +

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Fluid Statics 49

Fluid Statics  The pressure terms are the pressure at a depth

equal to the depth of the centroid of the surface

( )0

0

c

c C

F P A gh AF A P gh AP

ρρ

= += + =

PC is the pressure at the depth of the centroid of the surface

Fluid Statics 50

Fluid Statics  We know what the magnitude of the force is

that is generated by the pressure but for analysis we would need to also need to know where the equivalent point force would be located.

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Fluid Statics 51

Fluid Statics   This time we will use equivalent moments to find the

location of the line of action of the equivalent force   We will call this distance yCP

Fluid Statics 52

Fluid Statics   The moment of the equivalent force about the origin

(point O) is then equal to

CP Cy P A

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Fluid Statics 53

Fluid Statics   We can also generate the moment of all the

differential forces and sum these moments

CP Cy P A

Fluid Statics 54

Fluid Statics   The sum of these differential moments is

A

CP C

yPdA

y P A∫

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Fluid Statics 55

Fluid Statics   These two moment expressions have to be equal

so we have

CP CAyPdA y P A=∫

Fluid Statics 56

Fluid Statics   Now we can replace the expression for the pressure

at any y and we have

( )0 sin CP CAy P gy dA y P Aρ θ+ =∫

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Fluid Statics 57

Fluid Statics   Manipulating the expression

( )( ) ( )

0

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sin

sin

CP CA

CP CA A

y P gy dA y P A

yP dA gy dA y P A

ρ θ

ρ θ

+ =

+ =

∫∫ ∫

Fluid Statics 58

Fluid Statics   Manipulating the expression

( ) ( )( )

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20

sin

sin

CP CA A

C CP CA

P y dA g y dA y P A

P y A g y dA y P A

ρ θ

ρ θ

+ =

+ =

∫ ∫∫

Remember that the integral of ydA over the area is the same as the first moment of the area which is why we can substitute yCA for the integral

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Fluid Statics 59

Fluid Statics   The integral of y2 dA over the area is the second

moment of the area about the x axis.   Remember that the y distance is measured from the

origin, which in this case is the intersection of the x and y axis.

Remember that the integral of ydA over the area is the same as the first moment of the area which is why we can substitute yCA for the integral

Fluid Statics 60

Fluid Statics   We represent this as the moment of inertia of the

area about the x axis and give it the label Ixx

  So we can substitute this into our expression

Remember that the integral of ydA over the area is the same as the first moment of the area which is why we can substitute yCA for the integral

0 sinC xx CP CP y A g I y P Aρ θ+ =

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Fluid Statics 61

Fluid Statics  We can do this by taking the integral over the

area of the surface

0 sinP P gyρ θ= +