hydrostatics - memorial university of newfoundland · forces on submerged surfaces • summary: –...

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Hydrostatics

ENGR 5961 Fluid Mechanics I: Dr. Y.S. Muzychka

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Introduction •  In Fluid Mechanics hydrostatics considers

fluids at rest: –  typically fluid pressure on stationary bodies and

surfaces, pressure measurements, buoyancy and flotation, and fluid masses under rigid body motion, to name a few.

–  We will consider a simple element of stationary fluid and examine the how the pressure, which is a scalar field, varies throughout the element.

–  A scalar field unlike a vector field, only has a magnitude associated with it at any point, and not direction. Pressure at any point is the same in all directions.


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Pressure Distribution

€

p = p(x,y,z)


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Pressure Distribution •  Pressure Gradient

–  If we examine the rate of change of pressure through the element in any direction we find (using a Taylor series approximation):

–  Similiarly, we find:

€

px = p

px+dx = p +∂p∂xdx

px − px+dx = p − p +∂p∂xdx

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

∂p∂xdx

€

px − px+dx = p − p +∂p∂xdx

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

∂p∂xdx

py − py+dy = p − p +∂p∂ydy

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

∂p∂ydy

pz − pz+dz = p − p +∂p∂zdz

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

∂p∂zdz

** We can also develop the Taylor expansion using a central Coordinate as given in the figure. For simplicity we choose point O to be located at the lower left corner. We get the same result.


Pressure Distribution •  Considering force balances across each

element displacement, we obtain:

•  The body force is the weight per unit volume:

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€

d F s = −

∂p∂x

dx⎛

⎝ ⎜

⎞

⎠ ⎟ dydzˆ i − ∂p

∂ydy

⎛

⎝ ⎜

⎞

⎠ ⎟ dxdzˆ j − ∂p

∂zdz

⎛

⎝ ⎜

⎞

⎠ ⎟ dxdy ˆ k

d F s = −

∂p∂x

ˆ i − ∂p∂y

ˆ j − ∂p∂z

ˆ k ⎛

⎝ ⎜

⎞

⎠ ⎟ dxdydz = −∇p⋅ dxdydz

€

d F g = ρ

g dxdydz


Pressure Distribution 6

•  For a stationary fluid, the force balance is strictly between pressure forces and body forces (weight):

•  The force balance after defining the change in pressure across a differential length is:

€

d F = d

F s + d

F g = 0

€

−∂p∂x

ˆ i + ∂p∂y

ˆ j + ∂p∂z

ˆ k ⎛

⎝ ⎜

⎞

⎠ ⎟ dxdydz

Pressure

+ ρ g dxdydzBody Force = 0

€

−∇p + ρ g = 0


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Pressure Distribution •  This is a vector equation, which must be expanded

into its three components:

•  If our co-ordinates are gravity oriented (positive z-axis upwards), then: €

−∂p∂x

+ ρgx = 0 x − direction

−∂p∂y

+ ρgy = 0 y − direction

−∂p∂z

+ ρgz = 0 z − direction

€

∂p∂x

= 0 x − direction∂p∂y

= 0 y − direction

∂p∂z

= −ρg z − direction


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Pressure Distribution •  We can extend this to fluids in rigid body

motion using Newton’s law F = ma, to obtain:

•  This form allows us to determine the effect of acceleration on fluid masses held in containers which are accelerating or decelerating. €

−∂p∂x

+ ρgx = ρax x − direction

−∂p∂y

+ ρgy = ρay y − direction

−∂p∂z

+ ρgz = ρaz z − direction


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Standard Atmosphere •  The pressure in the

atmosphere varies with elevation.

•  We can model the first 11 km (troposphere) using a simple linear relationship for the change in temperature and use ideal gas law.


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Standard Atmosphere •  Defining the following equations:

–  where R = 287.0 J/kgK (gas constant for air), β = 0.0065 K/m (lapse rate in the first 11 km), and To = 288.15 K (15 C) we obtain:

–  where pa = 101.3 kPa is the sea level (z = 0) pressure of the atmosphere.

€

−dpdz

+ ρg = 0

p = ρRTT(z) = To − βz

€

p(z) = pa 1−βzTo

⎛

⎝ ⎜

⎞

⎠ ⎟

gRβ

=101.3 1− 2.2558 ×10−5 z( )5.258


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Standard Atmosphere •  Sea level conditions of the Standard Atmosphere


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Example - 5 •  Calculate the absolute pressure you would experience

if you were at an altitude of 10 km and a depth of 10 km below the ocean surface. Place the final results in terms the number of atmospheres for each fluid.


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Manometers •  Variation of pressure within a fluid is frequently used for

making pressure measurements:

•  We desire a simple methodology to predict the pressure difference in these devices.

•  Modern devices use strain gages attached to membranes.


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Manometers •  Types of Manometers

–  Piezotube –  U-Tube (for vacuum pressures) –  U-Tube differential (for pressure difference) –  Inclined (for small pressure difference)

•  Manometers allow for accurate pressure measurement provided the manometer fluid’s density is accurately known.

•  Typical fluids include: water (S.G. = 1), oil (S.G. 0.797-0.827), mercury (S.G. = 13.55).


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Manometers •  The basic equation is:

•  Separating variables and integrating we find:

•  More simply put, we define:

€

dpdz

= −ρg

€

dppo

p

∫ = −ρg dzzo

z

∫p − po = −ρg(z − zo) = ρg(zo − z)

€

p − po = ρghh = zo − z

€

h > 0 downh < 0 up


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Manometers •  Reference scale for manometry problems:


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Manometers •  We can use this simple approach to define the pressure

change (increases) as we move downwards through the fluid and (decreases) as we move upwards.

•  We also know, since the fluid is stationary that pressure along any horizontal line is constant. At any interface, we jump across.

€

pA + ρH2Og(10") − ρHgg(3")

Jump−Across

+ ρOilg(4")Jump−Across − ρHgg(5")

Jump−Across

− ρH2Og(8") = pB

€

p = po + ρgh


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Example - 6 •  Consider the system we used on the last slide:

Assuming that points A and B are connected to a pipe with flowing water in it, determine the actual pressure difference between A and B. Use the properties of fluids from your text.


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Example - 7 •  A mixture of crude oil (SG = 0.87) and water is

pumped into a settling tank that is 10 m high and vented to the atmosphere. After sometime a layer of oil 2 m thick forms at the top of the tank with 7.5 m of water below. Find: – The pressure at the bottom of the tank – Mean density of the combined fluids – Pressure gradient in the system


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Forces on Submerged Surfaces •  Consider the following system:


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Forces on Submerged Surfaces •  The pressure and force distribution is as follows:


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Forces on Submerged Surfaces •  We wish to resolve the pressure distribution and obtain

the following information: –  Resultant force (FR) –  Location of the Centroid of the surface (CofG) –  Location of the Center of Pressure (CofP)

•  This will allow us to make appropriate calculations for forces and moments required in systems.

•  We will also require some information about moments of area.

•  Later we will consider curved surfaces and develop an alternate approach for these systems.


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Forces on Submerged Surfaces •  The force on the plate is obtained by integrating the

pressure force over the surface:

•  We also know we can write the pressure at any point as:

where

•  Thus:

€

FR = pdAA∫

€

p = patm + ρgh

€

h = y sin(θ)

€

FR = patm + ρgy sinθ[ ]A∫ dA

€

FR = patmA + ρgsinθ yA∫ dA


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Forces on Submerged Surfaces •  The integral is the first moment of area about the x-axis,

which may be simplified according to:

•  Where yC is y coordinate of the location of the centroid of the surface.

•  Thus we may finally write:

•  Frequently, we do not need need to consider the absolute pressure, as the net force is due to gage pressure.

€

ydA = yCAA∫

€

FR = patmA + ρgsinθyCA = patm + ρghC( )Absolute Pressure at Centroid

A


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Forces on Submerged Surfaces •  Since the pressure varies with depth, the resultant

force does not act the centroid of the surface. •  To find the location of the resultant force, we need to

consider moments about the x-axis:

•  Combining gives:

€

yPFR = y patm + ρgy sinθ( )dAA∫ = patm y + ρgy 2 sinθ( )dAA∫

€

yPFR = patm ydAA∫

First−Moment

+ ρgsinθ y 2A∫ dA

Second −Moment

€

ydA = yCAA∫y 2dA = Ixx = Ixx,c + AyC

2A∫

€

yPFR = patm + ρghC( )yCA + ρgsin(θ)Ixx,c


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Forces on Submerged Surfaces •  Simplifying and solving for yP we get:

•  Ixx,c is the second moment about the centroid. It is tabulated for many shapes.

•  If we are dealing with gage pressure, which we frequently do, the above formula simplifies to give:

since for gage pressure FR = ρgycA. •  yP is the center of pressure.

€

yPFR = FR yC + ρgsin(θ )Ixx,c

€

yP = yC +ρgsin(θ )Ixx,c

FR

€

yP = yC +Ixx,cAyC


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Forces on Submerged Surfaces •  For surfaces which are not symmetric with respect to

the y-axis, we must also locate the C of P in the x-direction. The analysis is similar and leads to the following:

or

•  Here Ixy,c is product of inertia about the centroid. For symmetric surfaces, this parameter is equal to zero, for non-symmetric surfaces it is not zero!

€

xP = xC +ρgsin(θ )Ixy,c

FR

€

xP = xC +Ixy,cAyC


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Forces on Submerged Surfaces •  Summary:

–  Resultant force is the product of pressure at the C of G and the area of the surface.

–  It acts a the C of P which is below the C of G. –  The location of the C of P requires the second moment about

the centroid. –  If the plate is not symmetric with respect to the y-axis, we

need to locate the C of P in the x-direction as well. –  Frequently we use gage pressure in the resultant force. –  Taking the coordinate system at the C of G rather than the

free surface we have:

€

xP = −ρgsinθIxy,cpC A

€

yP = −ρgsinθIxx,cpC A

€

FR = ρghcA = pCA


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Example - 8 •  Find the force on the submerged gate shown in the

figure (we will sketch in class). The gate is 5 ft into the page. The specific weight of water in BG units is 64.2 lbf/ft3. Compute the force at the wall and at the reaction forces at the hinge.


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Curved Surfaces •  Forces on curved surfaces are also important, but we

utilize a different approach, than in plane surfaces, since integration is more complex.


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Curved Surfaces •  The horizontal force component is calculated for a

virtual vertical surface using:

•  The vertical force component is found to be the weight of the fluid over the surface:

•  The line of action of the vertical weight is through the centroid of the volume of fluid above the curved surface. This can be difficult to find but not impossible.

€

FH = ρghcA = pCA

€

FV = ρghdAzAz∫ = ρgdV =

V∫ ρgV


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Example - 9 •  Calculate the hydrostatic force on the curved surface

of the dam (sketched in class). The dam level is 24 ft and is described by the parabolic curve:

where x0 = 10 ft and y0 = 24 ft. Assume the specific weight of water is in BG units is 64.2 lbf/ft3.

€

y = y0xx0

⎛

⎝ ⎜

⎞

⎠ ⎟

2


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Buoyancy and Stability •  Floating bodies or suspended bodies in a fluid are

another common application of fluid statics. •  The buoyant force is due to the hydrostatic pressure

distribution around a submerged or partially submerged body:

•  Archimedes principle:

€

FB = ρgVdisplaced


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Buoyancy and Stability •  Proof is simple (referring to the figure on the last slide):

but

so

that is the buoyant force equals the weight of fluid displaced.

€

dFz = (ρgh2 − ρgh1)dA = ρg h2 − h1( )dA

€

h2 − h1( )dA = dV

€

FB = dFz∫ = ρgdV∫ = ρgVdisplaced

€

FB > FG Body FloatsFB = FG Neutrally BuoyantFB < FG Body Sinks


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Example - 10 •  It is desired to make a floatation device from steel in

the shape of a sphere. The device is to have an outer diameter of 30 cm. How thick should the vessel wall be to assure floatation in fresh water. What happens if the sphere is then placed in oil S.G. = 0.8 and seawater S.G. 1.025? Assume that the steel has a density of 7854 kg/m3. What about the contents inside, if any?


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Example - 11 •  A helium weather balloon is designed to carry a payload of

250 [kg] (carriage, instruments, balloon fabric etc.). The balloon has a nearly spherical shape with a diameter of 12 [m] and is filled with helium gas having a density ρ = 1.66 x 10-1 [kg/m3 ]. The air temperature of the standard atmosphere varies linearly with altitude up to 11 km. If the balloon is constructed of a material which prevents significant heat transfer from the inside of the balloon to the atmospheric air outside, i.e. assume constant temperature inside the balloon. Determine the altitude to which the balloon will rise, i.e find z. Hint: consider the air as an ideal gas.


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Buoyancy and Stability •  Stability of floating bodies is quite important. The

relative locations of the center of gravity of a body and its center of buoyancy determine stability.


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Buoyancy and Stability •  In general, the buoyant force and gravitational force of

a body in equilibrium are inline. •  Keep in mind that the buoyant force is located at the

center of buoyancy which is the centroid of the submerged portion of the body. Its location changes as a body is displaced, the center of mass of the body does not move.

•  As a result, if the body is displaced in such a manner that they are no longer in line, a moment arises which may restore the body to its position (stable) or overturn the body (unstable).


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Buoyancy and Stability


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Buoyancy and Stability •  In general, we can define the Metacentric height of a

floating body as:

•  If we find that:

•  The analysis leading to the above equation is based on the assumption of a small displacement occuring.

€

MG =Io

Vsubmerged

−GB

€

MG Metacentric HeightGB Separation of CofG and CofBIo Moment of Inertia of Waterline

Vsubmerged Volume of Submerged Portion

€

MG > 0 StableMG < 0 Unstable


Example - 12 •  A floating rectangular barge has a displacement in water of

H, with a total height of 2H, a width 2L, and a length B (into the page B >> 2L). Assuming (in this case) that the barge has a weight such that its CofG is always at the waterline, determine the ratio of L/H which ensures stability for small displacements.

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Example - 13 •  A 4 m wood log having a 15 cm diameter is

to be weighted at one end in an attempt to provide vertical stability. The specific weight of the wood is 0.5 that of water. A 20 kg cylindrical lead weight sleeve of length 30 cm is added to one end. Assuming that the weight has negligible volume as compared with the volume of the log, will the log float vertical or will it find another stable orientation, i.e. find the meta-centric height. Hint, you will need to find the CofG of the weighted log using a simple moment (pivot) analysis.

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hydrostatics - memorial university of newfoundland · forces on submerged surfaces • summary: –...

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