fluid statics - university of memphis statics.pdf · 2 3 fluid statics another type of manometer...

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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 = hρg

Wednesday, September 5, 2012

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Fluid Statics 3

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.


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.

The pressure at point B

( ) ( )2 1 2BP P g a g hρ ρ= + +


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Fluid Statics 5

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

= + += + ++ + = + +


Fluid Statics 6

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

− = + + − +− = + + − −− = + −− = −


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Fluid Statics 7

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

− = + + − +− = + + − −− = + −− = −


Fluid Statics 8

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


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Fluid Statics 9

Fluid Statics ¢ We can start with a generalized object

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


Fluid Statics 10

Fluid Statics


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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.

Notice the orientation of the y and z axis.


Fluid Statics 12

Fluid Statics ¢ On any differential area, dA, a

differential force is developed by the pressure acting on that area.


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Fluid Statics 13

Fluid Statics ¢ The differential force is labeled as dRf.


Fluid Statics 14

Fluid Statics ¢ The pressure acting on the differential area

is a function of how deep below the surface the differential area is.


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Fluid Statics 15

Fluid Statics ¢  If the distance along the z-axis (the sloped

surface) to the differential area is z, then the depth can be determined as z sin θ


Fluid Statics 16

Fluid Statics ¢ So the differential force on dA can be

written as dRf = ρ g z sin θ dA


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Fluid Statics 17

Fluid Statics ¢ The total force acting on the submerged

surface will be the sum of all the differential forces


Fluid Statics 18

Fluid Statics ¢ Summing over the area of the submerged

surface we can use an integral form.

dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA


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Fluid Statics 19

Fluid Statics ¢  Integrating on the left and bringing the

constants out on the right we have


Rf = ρ( ) g( )sinθ z( )dAA∫


Fluid Statics 20

Fluid Statics ¢ From statics, the term on the right integral is

the first moment of the area about what would be the x-axis (into the page)




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Fluid Statics 21

Fluid Statics ¢ By definition, the centroid of the are along

the z-axis would be



zc =z( )dA

A∫dA

A∫zcA = z( )dA

A∫


Fluid Statics 22

Fluid Statics ¢ Substituting into the force expression



Rf = ρ( ) g( )sinθ zcA( )

zc =z( )dA

A∫dA

A∫zcA = z( )dA

A∫


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Fluid Statics 23

Fluid Statics ¢  It is important to remember that the distance

to the centroid is measured from the fluid surface along the slope



Rf = ρ( ) g( )sinθ zcA( )


Fluid Statics 24

Fluid Statics


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Fluid Statics 25

Fluid Statics ¢ The zc is the distance to the centroid of the

submerged area, it is not the location of the line of action of the force acting on the area


Fluid Statics 26

Fluid Statics ¢ We will use another variable, zr, to position

the location of the resultant force acting normal to the surface.


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Fluid Statics 27

Fluid Statics ¢  If we sum the moments of the differential

forces about the x-axis, we can write that as

dMaboutxA∫ = z ρ( ) g( )A∫ zsinθ( )dA


Fluid Statics 28

Fluid Statics ¢ The left side is the total moment and we can

again take constants outside the integral on the right

Maboutx = ρ( ) g( ) sinθ( ) z2A∫ dA


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Fluid Statics 29

Fluid Statics ¢ The integral term on the right is the second

moment of the area about the x-axis which in this text is written as Ixx


Maboutx = ρ( ) g( ) sinθ( ) Ixx


Fluid Statics 30

Fluid Statics ¢  If the submerged surface is a common

shape or one that can be broken up into common shapes, it may be more convenient to locate the second moment about the centroidal axis and then use the parallel axis theorem to find the moment about the x-axis.


Maboutx = ρ( ) g( ) sinθ( ) Ixx


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Fluid Statics 31

Fluid Statics ¢  If you can use the parallel axis theorem,

then Maboutx = ρ( ) g( ) sinθ( ) z2

A∫ dA

Maboutx = ρ( ) g( ) sinθ( ) IxxIxx = Ixxc + Azc

2


Fluid Statics 32

Fluid Statics ¢ Calculating the moment using the resultant

force and the distance to its line of action


Maboutx = ρ( ) g( ) sinθ( ) IxxIxx = Ixxc + Azc

2

zrRf = ρ( ) g( ) sinθ( ) Ixxc + Azc2( )zr ρ( ) g( ) sinθ( ) zc( )A = ρ( ) g( ) sinθ( ) Ixxc + Azc2( )


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Fluid Statics 33

Fluid Statics ¢ Manipulating

zr ρ( ) g( ) sinθ( ) zc( )A = ρ( ) g( ) sinθ( ) Ixxc + Azc2( )zr zc( )A = Ixxc + Azc

2( )zr =

Ixxc + Azc2( )

zc( )A

zr =Ixxczc( )A + zc


Fluid Statics 34

Fluid Statics


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Fluid Statics 35

Homework Problem 4-1


Fluid Statics 36



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Fluid Statics 37



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 39

Fluid Statics ¢ We can start with a generalized object

submerged in a fluid


Fluid Statics 40

Fluid Statics ¢ The generalized object has a uniform

thickness that we can say is 1 unit


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Fluid Statics 41

Fluid Statics ¢  It has an arbitrary cross section


Fluid Statics 42

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 43

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

the fluid


Fluid Statics 44

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 45

Fluid Statics ¢ Again, our positive direction will be to go

deeper into the fluid.


Fluid Statics 46

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 47

Fluid Statics ¢ The area of the differential element will be dz

by dy ¢ We will label that dA


Fluid Statics 48

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 49

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

below the surface of the water.


Fluid Statics 50

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 51

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 52

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 53

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 54

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 55

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

along the plate (y)


Fluid Statics 56

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 57

Fluid Statics ¢  h is then related to y as

sinh y θ=


Fluid Statics 58

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 59

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 60

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 61

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 62

Fluid Statics ¢ Substituting the second expression into the first

( )0 sinA

F P gy dAρ θ= +∫


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Fluid Statics 63

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 64

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 65

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 66

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 67

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 68

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 69

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 70

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 71

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 72

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 73

Fluid Statics ¢  We can also generate the moment of all the

differential forces and sum these moments

CP Cy P A


Fluid Statics 74

Fluid Statics ¢  The sum of these differential moments is

A

CP C

yPdA

y P A∫


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Fluid Statics 75

Fluid Statics ¢  These two moment expressions have to be equal

so we have

CP CAyPdA y P A=∫


Fluid Statics 76

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 77

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 78

Fluid Statics ¢  Manipulating the expression

( ) ( )( )

20

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 79

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.



Fluid Statics 80

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


0 sinC xx CP CP y A g I y P Aρ θ+ =


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Fluid Statics 81

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

area of the surface

0 sinP P gyρ θ= +


fluid statics - university of memphis statics.pdf · 2 3 fluid statics another type of manometer...

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