Fluid Statics - University of Memphis Statics.pdf · 2 3 Fluid Statics Another Type of Manometer...
Transcript of 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
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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.
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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
= + += + ++ + = + +
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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
− = + + − +− = + + − −− = + −− = −
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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
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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.
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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.
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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
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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
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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
dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA
Rf = ρ( ) g( )sinθ z( )dAA∫
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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)
dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA
Rf = ρ( ) g( )sinθ z( )dAA∫
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Fluid Statics 21
Fluid Statics ¢ By definition, the centroid of the are along
the z-axis would be
dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA
Rf = ρ( ) g( )sinθ z( )dAA∫
zc =z( )dA
A∫dA
A∫zcA = z( )dA
A∫
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Fluid Statics 22
Fluid Statics ¢ Substituting into the force expression
dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA
Rf = ρ( ) g( )sinθ z( )dAA∫
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
dRfA∫ = ρ( ) g( )A∫ zsinθ( )dA
Rf = ρ( ) g( )sinθ z( )dAA∫
Rf = ρ( ) g( )sinθ zcA( )
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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
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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
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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( ) z2A∫ dA
Maboutx = ρ( ) g( ) sinθ( ) Ixx
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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( ) z2A∫ dA
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
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Fluid Statics 32
Fluid Statics ¢ Calculating the moment using the resultant
force and the distance to its line of action
Maboutx = ρ( ) g( ) sinθ( ) z2A∫ dA
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
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Fluid Statics 34
Fluid Statics
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Fluid Statics 35
Homework Problem 4-1
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Fluid Statics 36
Homework Problem 4-2
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Fluid Statics 37
Homework Problem 4-3
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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
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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
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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
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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.
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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
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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.
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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
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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
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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)
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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=
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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= +
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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
= =
= +∫ ∫
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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
= +
= +
∫∫ ∫
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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
=
= +
∫∫
)
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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= +
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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
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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
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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
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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=∫
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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
20
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
+ =
+ =
∫∫ ∫
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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.
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 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
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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