Ch2 Fluid Statics

FLUID STATICS

Introduction •

In many fluid problems fluid is at rest or moves as a rigid body

•

When fluid is at rest (hydrostatic condition), the pressure variation is due only to the weight of the fluid and may be calculated by integration. Important applications are: – pressure distribution in the atmosphere and oceans; – the design of manometer pressure instruments; – forces on submerged flat and curved surfaces – buoyancy on a submerged body – behaviour of floating bodies

•

If the fluid is moving in a rigid-body motion, the pressure also can be easily calculated, because the fluid is free of shear stress

Pressure at a Point •

Pressure is the normal force per unit area at a given point acting on a given plain within the fluid mass

•

How pressure at a point varies with the orientation of the plain?

•

Consider free body diagram obtained by removing a small triangular wedge of fluid from some arbitrary location within a fluid mass.

•

Since there are no shearing stresses, the only forces are due to pressure and weight.

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To generalize the case allow fluid element to move as a rigid body with nonzero acceleration

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Apply Newton’s second law to fluid element

Pressure at a Point

Pressure at a Point Equations of motion in y and z direction

 x y z ay 2  x y z  x y z  Fz  pz x y  ps x s cos    2   2 az

F

y

 p y x z  ps x s sin   

With  y   s cos  and  z   s sin  equations of motion reduce to

p y  ps   a y

y 2

p z  ps    a z   

z 2

At a point, as σx, σy and σz approach zero: ps  p y  pz

The pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present (Pascal’s law)

Basic Equation for Pressure Field •

How the pressure in a fluid at rest varies from point to point?

•

Consider a small rectangular fluid element

•

Two types of forces acting on this element are – surface forces due to pressure, and – a body force equal to the weight of the element

Basic Equation for Pressure Field •

How the pressure in a fluid at rest varies from point to point?

•

Consider a small rectangular fluid element

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Two types of forces acting on this element are – surface forces due to pressure, and – a body force equal to the weight of the element

•

Applying Newton’s second law of motion to the fluid element we obtain general equation of motion for a fluid in which there are no shearing stresses

p   kˆ   a

Pressure Variation in a Fluid at Rest For a fluid at rest a = 0 and

p   kˆ  0 Pressure does not depend on x or y and depends only on z

p   z For liquids or gases at rest the pressure gradient in the vertical direction at any point in a fluid depends only on the specific weight of the fluid at that point Pressure decreases as we move upward in a fluid at rest Specific weight does not necessarily be a constant, and for gases may vary with elevation

Pressure Variation. Incompressible Fluid For liquids constant specific weight is assumed. Pressure variation is obtained by direct integration

Pressure Variation. Incompressible Fluid For liquids constant specific weight is assumed. Pressure variation is obtained by direct integration



p2 p1

z2

dp    dz z1

p2  p1    z2  z1  p1   h  p2

- hydrostatic pressure distribution

Pressure head, h, is the height of a column of fluid that would give the specified pressure difference p  p2 h 1  For fluid with a free surface, pressure p at any depth h below the free surface: p   h  p0

Pressure Variation. Incompressible Fluid p1   h  p2

p   h  p0

Pressure in a homogeneous, incompressible fluid at rest depends on the depth of the fluid relative to some reference plane, and it is not influenced by the size or shape of the tank or container in which the fluid is held.

Fluid equilibrium in a container of arbitrary surface

Pressure Variation. Incompressible Fluid The transmission of pressure throughout a stationary fluid is the principle upon which many hydraulic devices are based (hydraulic jacks, lifts, presses, hydraulic controls on aircraft and other types of heavy machinery)

Transmission of fluid pressure

Pressure Variation. Compressible Fluid Density of gases can change significantly with changes in pressure and temperature. But specific weights of common gases are comparatively small, therefore pressure gradient in vertical direction is correspondingly small Thus, in problems involving gases in tanks, pipes, and so on effect of elevation changes on the pressure can be neglected. If variations in heights are large (thousands of feet) variation in specific weight must be accounted for. For gases

p   RT dp gp  dz RT p2 dp p2 g z2 dz p1 p  ln p1   R z1 T

To integrate last equation, temperature variation with elevation must be known. For isothermal condition  g  z2  z1   p  p1 exp    RT 0  

Standard Atmosphere Standard atmosphere is an idealized representation of mean condition in the earth’s atmosphere Properties for standard atmospheric condition at sea level are listed in Table Temperature profile for the U.S. standard atmosphere is shown on Figure

Since temperature variation is represented by a series of linear segments, equation



p2 p1

dp p2 g z2 dz  ln    p1 R z1 T

can be integrated to obtain pressure variation

Pressure Measurement (Figure) •

Pressure is designated as either absolute pressure of gage pressure

•

Absolute pressure is measured relative to a perfect vacuum (absolute zero pressure), gage pressure is measured relative to the local atmospheric pressure

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Negative gage pressure is referred to as a suction or vacuum pressure

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In this course pressure will be assumed to be gage pressures unless specifically designated absolute

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Pressure difference is independent on the reference, so that no special notation is required

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Pressure is measured in Pa (SI) or psf, psi (BG).

•

Pressure can also be expressed as a height of column of a liquid

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Barometer is used to measure atmospheric pressure

Manometry •

Manometers use vertical or inclined liquid columns to measure pressure.

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Three common types of manometers include – piezometer tube, – U-tube manometer – inclined-tube manometer

Piezometer Tube p   h  p0 p A   1h1

Disadvantages: - pA must be greater than p0 - h is limited - fluid in container must be liquid

U-Tube Manometer p A   2 h2   1h1

Advantage: gage fluid can be different from fluid in container

If fluid in container is gas p A   2 h2

For high pA mercury is used

For small pA water or other light liquids can be used

Differential Manometer p A  pB   2 h2   3 h3   1h1

p A  pB  h2   2   1   2.9 kPa

Inclined-Tube Manometer

p A  pB   2l2 sin    3 h3   1h1

for gases p A  pB   2l2 sin 

Mechanical and Electronic Pressure Measuring Devices •

Manometers have some disadvantages: – they are not well suited for measuring very high pressures, or pressure that are changing rapidly with time; – they require the measurement of one or more column heights, which is time consuming

•

Bourdon tube pressure gage uses a hollow, elastic, and curved tube to measure pressure;

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Aneroid barometer is used for measuring atmospheric pressure;

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Pressure transducer converts pressure into an electrical output: – pressure transducers using Bourdon tube – strain-gage pressure transducers – piezoelectric pressure transducers

Problem 2.38 An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 m as shown in Fig. P2.38. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates a differential reading of 735 mm Hg as illustrated. Based on these data what is the atmospheric pressure at the ocean surface?

Problem 2.38 An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 m as shown in Fig. P2.38. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates a differential reading of 735 mm Hg as illustrated. Based on these data what is the atmospheric pressure at the ocean surface? Solution Let

pa – absolute pressure inside shell patm – surface atmosphere pressure γsw – specific weight of seawater

Problem 2.38 An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 m as shown in Fig. P2.38. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates a differential reading of 735 mm Hg as illustrated. Based on these data what is the atmospheric pressure at the ocean surface? Solution Let

pa – absolute pressure inside shell patm – surface atmosphere pressure γsw – specific weight of seawater

Manometer equation patm  10 sw  0.36 sw  0.735 Hg  pa So that

patm  pa  10.36 sw  0.735 Hg kN  kN  kN       0.765 m   133 3   10.36 m   10.1 3   0.735 m   133 3  m  m  m      94.9 kPa

Problem 2.42 The manometer fluid in the manometer has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer.

Problem 2.42 The manometer fluid in the manometer has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer.

a

a

Problem 2.42 The manometer fluid in the manometer has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer.

a

a

Solution Initially

Finally

p A  2 H 2O  2 gf  1 H 2O  pB

(1)

p A   2  a   H 2O   2  2a   gf   1  a   H 2O  pB

 2

Subtract (2) from (1)

p A  p A  a H 2O  2a gf  a H 2O  pB  pB

Increment a

p  p   p  p   a  1.03 ft

New differential reading

B

B



A

2  H 2O   gf

A



h  2  2a  4.06 ft

Problem 2.44 The inclined differential manometer contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain a brine (SG = 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, θ

Problem 2.44 The inclined differential manometer contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain hi a brine (SG = 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the a inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, θ Solution p A  pB increases to p A  pB

b

Problem 2.44 The inclined differential manometer contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain hi a brine (SG = 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the a inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, θ Solution p A  pB increases to p A  pB

b

For final configuration

p A   hi  a   br   a  b sin    CCl4   hi  b sin    br  pB

or

p A  pB   br   CCl4

Differential reading along the tube From (1) or



  a  b sin   0

a b sin  p A  pB   br   CCl4  p A  pB sin  h  br   CCl4

h 









  h sin    0 

(1)

Problem 2.44 The inclined differential manometer contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain hi a brine (SG = 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the a inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, θ Solution  p A  pB sin  h   br   CCl4 





With p A  pB  0.1 psi for h  12 in. sin   0.466

Thus

  27.8o

b

Hydrostatic Force on a Plane Surface •

When a surface is submerged in a fluid, forces develop on it due to fluid

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We are interested in the direction, location and magnitude of those forces

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For fluids at rest force is perpendicular to the surface

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Pressure varies linearly with depth if fluid is incompressible

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For horizontal surface force calculation is straightforward

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When determining the resultant force on an area, the effect of atmospheric pressure often cancels

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If submerged surface is inclined determination of the resultant force is more involved

Hydrostatic Force on Inclined Plane Surface •

Consider plane surface with area of arbitrary shape submerged in the fluid and inclined with respect to fluid surface

Hydrostatic Force on Inclined Plane Surface •

Consider plane surface with area of arbitrary shape submerged in the fluid and inclined with respect to fluid surface

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Assume that fluid surface is open to atmosphere

Hydrostatic Force on Inclined Plane Surface •

Consider plane surface with area of arbitrary shape submerged in the fluid and inclined with respect to fluid surface

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Assume that fluid surface is open to atmosphere

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Let the plane in which the surface lies intersect the free surface at 0 and make an angle θ with this surface

Hydrostatic Force on Inclined Plane Surface •

Consider plane surface with area of arbitrary shape submerged in the fluid and inclined with respect to fluid surface

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Assume that fluid surface is open to atmosphere

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Let the plane in which the surface lies intersect the free surface at 0 and make an angle θ with this surface

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Define coordinate system so that 0 is the origin and y is directed along the surface

Hydrostatic Force on Inclined Plane Surface •

Consider plane surface with area of arbitrary shape submerged in the fluid and inclined with respect to fluid surface

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Assume that fluid surface is open to atmosphere

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Let the plane in which the surface lies intersect the free surface at 0 and make an angle θ with this surface

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Define coordinate system so that 0 is the origin and y is directed along the surface

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We wish to determine the direction, location, and magnitude of the resultant force acting on one side of this area due to the liquid in contact with the area.

Hydrostatic Force on Inclined Plane Surface •

The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution

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The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area (details)

Hydrostatic Force on Inclined Plane Surface •

The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution

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The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area

FR   hc A •

The resultant fluid force does not pass through the centroid of the area but is always below it

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The point through which the resultant fluid force acts is called the center of pressure. Coordinates, yR and xR, of the center of pressure are:

Hydrostatic Force on Inclined Plane Surface •

The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution

•

The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area

FR   hc A •

The resultant fluid force does not pass through the centroid of the area but is always below it

•

The point through which the resultant fluid force acts is called the center of pressure. Coordinates, yR and xR, of the center of pressure are:

I yR  xc  yc yc A

xR 

I xyc yc A

 xc

Hydrostatic Force on Inclined Plane Surface •

If the submerged area is symmetrical with respect to an axis passing through the centroid and parallel to either the x or y axes, the resultant force must lie along the line x = xc , since Ixyc is identically zero in this case

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As yc increases the center of pressure moves closer to the centroid of the area..

•

Centroidal coordinates and moments of inertia for some common areas are provided

Example 2.6 The 4-m-diameter circular gate is located in the inclined wall of a large reservoir containing water (γ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would have to be applied to the shaft to open the gate

Example 2.6 The 4-m-diameter circular gate is located in the inclined wall of a large reservoir containing water (γ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would gave to be applied to the shaft to open the gate Solution (a) Magnitude of the force FR   hc A  1.23 MN

Location xR 

I xyc yc A

 xc  0

since area is symmetrical, and center of pressure must lie along diameter A-A

Example 2.6 The 4-m-diameter circular gate is located in the inclined wall of a large reservoir containing water (γ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would gave to be applied to the shaft to open the gate Solution Moment of inertia

 R4 I xc  4 and yR 

I xc  yc  11.6 m yc A

distance below the shaft to the center of pressure yR  yc  0.0866 m

Example 2.6 The 4-m-diameter circular gate is located in the inclined wall of a large reservoir containing water (γ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would gave to be applied to the shaft to open the gate Solution (b) Moment

M

c

0

M  FR  yR  yc   1.07  105 N m

Example 2.7 A large fish-holding tank contains seawater (γ = 64.0 lb/ft3) to a depth of 10 ft. To repair some damage to one corner of the tank, a triangular section is replaced with a new section as illustrated. Determine the magnitude and location of the force of the seawater on this triangular area.

Example 2.7 A large fish-holding tank contains seawater (γ = 64.0 lb/ft3) to a depth of 10 ft. To repair some damage to one corner of the tank, a triangular section is replaced with a new section as illustrated. Determine the magnitude and location of the force of the seawater on this triangular area. Solution Force magnitude FR   hc A  2590 lb

Force location 3 ba 81 4 I xc   ft 36 36 I yR  xc  yc  9.06 ft yc A I xyc

ba 2 81   b  2d   ft 4 72 72

xR 

I xyc yc A

 xc  0.0278 ft

Pressure Prism •

Pressure prism is a geometric representation of the hydrostatic force on a plane surface

•

Consider pressure distribution along a vertical wall of tank with liquid (Fig. a). Average pressure , pav, occurs at the depth h/2, and resultant force on area A = bh  h A   2

FR  pav A   

Pressure Prism •

Volume, shown on Fig. b is called the pressure prism

•

Magnitude of the resultant fluid force is equal to the volume of the pressure prism and passes through its centroid (details) FR  volume 

1 h   h   bh      A 2  2

Pressure Prism •

Above-mentioned is valid for surfaces that do not extend up to the fluid surface. In this case, the cross section of the pressure prism is trapezoidal

•

Specific values can be obtained by decomposing pressure prism into two parts, ABDE and BCD. Thus FR  F1  F2

Pressure Prism •

Location of FR can be determined by summing moments about some convenient axis: FR yR  F1 y1  F2 y2

Example 2.8 A pressurized tank contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate bolted to its side. When the pressure gage on the top of the tank reads 50 kPa, what is the magnitude and location of the resultant force on the attached plate? The outside of the tank is at atmospheric pressure.

Solution

Solution Resultant force: F1   ps   h1  A  24.4 kN  h2  h1   A  0.954 kN 2  

F2   

FR  F1  F2  25.4 kN

Solution Vertical location of the resultant force: FR yO  F1  0.3 m   F2  0.2 m 

yO  0.296 m

Note that the air pressure ised in the calculation of the force was gage pressure

Hydrostatic Force on a Curved Surface •

Development of a free-body diagram of a suitable volume of fluid can be used to determine the resultant fluid force acting on a curved surface

•

For example, consider curved section BC of the open tank, which has a unit length perpendicular to the plane of slide

•

Find resultant fluid force acting on this section

Hydrostatic Force on a Curved Surface •

Develop free-body diagram

•

Determine magnitude and location of forces F1 and F2 using relationships for planar surfaces

•

Weight acts through the center of gravity of fluid contained within the volume

•

Forces FH and FV represent components of the force that the tank exerts on the fluid

Hydrostatic Force on a Curved Surface •

For this force system to be in equilibrium: FH  F2

•

FV  F1  W

Magnitude of resultant force FR 

 FH 

2

  FV 

2

Hydrostatic Force on a Curved Surface •

Resultant force FR passes through the point O, which can be located by summing moments about appropriate axis

•

Resultant force of the fluid acting on the curved surface BC is equal and opposite in direction to that obtained from the free-body diagram

Example 2.9 The 6-ft-diameter drainage conduit is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall.

Solution Develop free-body diagram and determine forces F1   hc A  281 lb Weight   vol  441 lb FH  F1  281 lb

FV =Weight  441 lb

Solution Magnitude of the resultant force FR =

 FH    FV   523 lb 2

2

The force the water exerts on the conduit wall is equal, but opposite in direction, to the forces FH and FV Note, the line of action of the resultant force passes through the center of the conduit.

Buoyancy, Flotation, and Stability

Archimedes’ Principle A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces

Archimedes’ Principle Magnitude of buoyant force

Archimedes’ Principle Magnitude of buoyant force FB   V

Archimedes’ Principle Buoyant force passes through the centroid of the displaced volume. The point through which the buoyant force acts is called the center of buoyancy

Archimedes’ Principle Buoyant force passes through the centroid of the displaced volume. The point through which the buoyant force acts is called the center of buoyancy Summing moments of forces with respect to axis passing through D

FB yc  F2 y1  F1 y1  Wy2 on substitution for forces:

Vyc  VT y1   VT  V  y2

If a body is immersed in a fluid in which γ varies with depth, such as in a layered fluid, the buoyant force passes through the center of gravity of the displaced volume

Example 2.10 A spherical buoy has a diameter of 1.5 m, weighs 8.50 kN, and is anchored to the sea floor with a cable. Although the buoy normally floats on the surface, at certain times (waves, rising tide) the water depth increases so that the buoy is completely immersed as illustrated. For this condition what is the tension of the cable?

Example 2.10 A spherical buoy has a diameter of 1.5 m, weighs 8.50 kN, and is anchored to the sea floor with a cable. Although the buoy normally floats on the surface, at certain times (waves, rising tide) the water depth increases so that the buoy is completely immersed as illustrated. For this condition what is the tension of the cable? Solution

FB   V  17.85 kN

T  FB  W  9.35 kN

Stability Submerged or floating bodies can be either in a stable or unstable position. Stable equilibrium – when displaced body returns to its equilibrium position Unstable equilibrium – when displaced body moves to a new equilibrium position

Stability of Completely Immersed Bodies For a completely submerged body with a center of gravity below the center of buoyancy rotation from equilibrium position will create a restoring couple. If center of gravity is above the center of buoyancy, resulting couple will cause the body to overturn and move the a new equilibrium position

Stability of a completely immersed body

Stability of Floating Bodies When floating body rotates the location of the center of buoyancy may change

Stability of a floating body- stable configuration

Stability of Floating Bodies When floating body rotates the location of the center of buoyancy may change

Stability of a floating body- unstable configuration

Pressure Variation in a Fluid with Rigid-Body Motion Even though a fluid may be in motion, if it moves as a rigid body there will be no searing stresses present For such a fluid the general equation of motion

In component form

Pressure Variation in a Fluid with Rigid-Body Motion Even though a fluid may be in motion, if it moves as a rigid body there will be no searing stresses present For such a fluid the general equation of motion p   kˆ   a

In component form 

p   ax x



p   ay y



p     az z

Consider two classes of problems; rigid-body uniform motion, and rigid-body rotation

Pressure Variation in a Fluid with Rigid-Body Motion Linear Motion Consider an open container of a liquid that is translating along a straight path with a constant acceleration. Apply general equation of motion

Linear acceleration of a liquid with a free surface

Pressure Variation in a Fluid with Rigid-Body Motion Linear Motion Consider an open container of a liquid that is translating along a straight path with a constant acceleration. Apply general equation of motion ay dz  Slope of line of constant pressure, dp = 0, is given by dy g  az

Linear acceleration of a liquid with a free surface

Pressure Variation in a Fluid with Rigid-Body Motion Linear Motion If ay=0, az≠0, fluid surface will be horizontal, but pressure distribution is not hydrostatic dp     g  az  dz

Linear acceleration of a liquid with a free surface

Example 2.11 The cross section for the fuel tank of an experimental vehicle is shown in figure. The rectangular tank is vented to the atmosphere, and a pressure transducer is located in its side as illustrated. During testing of the vehicle, the tank is subjected to a constant linear acceleration, ay . (a) Determine an expression that relates ay and the pressure (in lb/ft2) at the transducer for a fuel with a SG = 0.65. (b) What is the maximum acceleration that can occur before the fuel level drops below the transducer?

Example 2.11 The cross section for the fuel tank of an experimental vehicle is shown in figure. The rectangular tank is vented to the atmosphere, and a pressure transducer is located in its side as illustrated. During testing of the vehicle, the tank is subjected to a constant linear acceleration, ay . (a) Determine an expression that relates ay and the pressure (in lb/ft2) at the transducer for a fuel with a SG = 0.65. (b) What is the maximum acceleration that can occur before the fuel level drops below the transducer? Solution (a) Slope of the surface ay dz  dy g change in depth ay z1   0.75 ft g

or

z1   0.75 ft 

ay g

Pressure at the transducer

p   h    0.5 ft  z1   20.3  30.4

ay g

Example 2.11 The cross section for the fuel tank of an experimental vehicle is shown in figure. The rectangular tank is vented to the atmosphere, and a pressure transducer is located in its side as illustrated. During testing of the vehicle, the tank is subjected to a constant linear acceleration, ay . (a) Determine an expression that relates ay and the pressure (in lb/ft2) at the transducer for a fuel with a SG = 0.65. (b) What is the maximum acceleration that can occur before the fuel level drops below the transducer? Solution (b) Maximum acceleration

  ay 

0.5 ft   0.75 ft  

or



 ay 

max



g

 

max



2g 3

for a standard acceleration of gravity

a 

y max

 21.5 ft/s 2

Note:

p1  p2

Problem 2.92 An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi/hr. As the truck slows uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration?

Problem 2.92 An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi/hr. As the truck slows uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration? Solution Acceleration a

V 55  5.280 ft/mi    16.1 ft/s 2 t 5  3600 s/hr 

Slope dz a y   0.5 dy g

Pressure Variation in a Fluid with Rigid-Body Motion Rigid-Body Rotation A fluid contained in a tank that rotates with a constant angular velocity about an axis will rotate as a rigid body p   r 2 r

Pressure gradients (details):

Differential pressure

Rigid-body rotation of a liquid in a tank

dp 

p p dr  dz r z

p 0 

or

p   z

dp   r 2 dr   dz

Pressure Variation in a Fluid with Rigid-Body Motion Rigid-Body Rotation Along a surface of constant pressure dp = 0, so that

dz r 2  dr g

and equation for surfaces of constant pressure is

 2r 2 z  constant 2g

Surfaces of constant pressure are parabolic

Pressure distribution in a rotating liquid

Pressure Variation in a Fluid with Rigid-Body Motion Rigid-Body Rotation Pressure distribution  2 r 2 p   z  constant 2

Pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical direction

Pressure distribution in a rotating liquid

Example 2.12 It has been suggested that the angular velocity, ω, of a rotating body or shaft can be measured by attaching an open cylinder of liquid, as shown in figure, and measuring with some type of depth gage the change in the fluid level, H – h0 , caused by the rotation of the fluid. Determine the relationship between this change in fluid level and the angular velocity.

Example 2.12 It has been suggested that the angular velocity, ω, of a rotating body or shaft can be measured by attaching an open cylinder of liquid, as shown in figure, and measuring with some type of depth gage the change in the fluid level, H – h0 , caused by the rotation of the fluid. Determine the relationship between this change in fluid level and the angular velocity. Solution Height, h, of the free surface

 2r 2 h  h0 2g Initial volume of fluid Vi   R 2 H

Volume of rotating fluid dV  2 rhdr V  2 

R

0

  2r 2   2 R 4 r  h0 dr    R 2 h0 4g  2g 

Example 2.12 It has been suggested that the angular velocity, ω, of a rotating body or shaft can be measured by attaching an open cylinder of liquid, as shown in figure, and measuring with some type of depth gage the change in the fluid level, H – h0 , caused by the rotation of the fluid. Determine the relationship between this change in fluid level and the angular velocity. Solution Volume of fluid in tank remains constant Vi  V

 2 R 4 R H    R 2 h0 4g 2

Thus

 2 R2 H  h0  4g Note: relationship between the change in depth and speed is not linear one

END OF LECTURE

Supplementary slides

Forces on an arbitrary wedge-shaped element of fluid

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Resultant surface force

 Fs  p x y z Weight of the fluid element

 W kˆ   x y z kˆ

Surface and body forces acting on small fluid element

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Variation of temperature with altitude in the U.S standard atmosphere back

Pressure Measurement

Graphical representation of gage and absolute pressure

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Pressure Measurement

patm   h  pvapor

Mercury barometer

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Liquid-filled Bourdon pressure gages for various pressure ranges

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Internal elements of Bourdon gages. The “C-shaped” Bourdon tube is shown on the left, and the “coiled spring” Bourdon tube for high pressure of 1000 psi and above is shown on the right

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Pressure transducer which combines a linear variable differential transformer (LVDT) with a Bourdon gage

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Pressure and resultant hydrostatic force developed on the bottom of an open tank

Resultant force

FR  pA

If atmospheric pressure act on both sides of the bottom the resultant force on the bottom is simply due to the liquid in the tank Since pressure is constant and uniformly distributed over the bottom, the resultant force acts through the centroid of the area

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Magnitude of Resultant force At given depth, h, force acting on dA is dF=γ hdA and is perpendicular to the surface Magnitude of the resultant force

Magnitude of Resultant force At given depth, h, force acting on dA is dF=γ hdA and is perpendicular to the surface Magnitude of the resultant force

FR    hdA    y sin  dA A

A

For constant γ and θ

FR   sin   ydA A

In terms of first moment of the area



A

ydA  yc A

Resultant force

FR   hc A

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Location of Resultant force Moment of resultant force must equal moment of distributed pressure force

Location of Resultant force Moment of resultant force must equal moment of distributed pressure force

FR yR   ydF    sin  y 2 dA A

Since

A

FR   Ayc sin 

yR 



A

then

y 2 dA yc A

or in term of moment of inertia, Ix

yR 

Ix yc A

By use of parallel axis theorem

I x  I xc  Ayc2 where Ixc is the moment of inertia with respect to axis passing through centroid and parallel to x axis. Finally:

yR 

I xc  yc yc A

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Location of Resultant force Moment of resultant force must equal moment of distributed pressure force By summing moments about y axis we get

Location of Resultant force Moment of resultant force must equal moment of distributed pressure force By summing moments about y axis we get

FR xR    sin  xydA A

and

xR

 

A

xydA yc A



I xy yc A

Using parallel axis theorem we get

xR 

I xyc yc A

 xc

where Ixyc is the product of inertia with respect to orthogonal coordinate system passing through centroid of the area and formed by translation of the x-y coordinate system

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Geometrical Properties of Some Common Shapes

back to example 2.6

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Parallel Axis Theorem for Moment of Inertia I x  I xc  Ayc2

Moment of inertia of an area Ix with respect to x axis is equal to the moment of inertia Ixc with respect to centroidal axis x’ parallel to the x axis, plus the product Ayc2 of the area A and of the square of the distance yc between the two axes

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Parallel Axis Theorem for Product of Inertia I xy  I xyc  Axc yc

Product of inertia with respect to an orthogonal set of axes (x-y coordinate system) is equal to the product of inertia with respect to an orthogonal set of axes parallel to the original set and passing through the centroid of the area, plus the product of the area and the x and y coordinates of the centroid of the area

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1 3 1 3 I xc  ba  bh 12 12

I xyc  0

h yc  2

b xc  2

bh3 2 1 h 2 yR    h 12 h bh 2 3

xR  xc 

b 2 back

Pressure variation. Linear Motion Pressure gradients

p 0 x

p   ay y

Change in pressure between two closely spaced points or

p     g  az  z p p dp  dy  dz y z

dp    a y dy    g  az  dz

Slope of line of constant pressure

Linear acceleration of a liquid with a free surface

ay dz  dy g  az

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