Chapter 2 HYDRAULICS Hydrostatics Fluid statics Fluid at

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Chapter 2 HYDRAULICS Hydrostatics

Chapter 2 HYDRAULICS Hydrostatics

 • • Fluid statics – Fluid at rest Hydrostatics – water at rest

• • Fluid statics – Fluid at rest Hydrostatics – water at rest » Act normal to the boundary » Independent of viscosity » No shear force Hydrostatics pressure and its measurement P = F/A • It is the same in all direction • should be normal to the surface 2

Pressure at a point in a Static fluid Small triangular prism of liquid at

Pressure at a point in a Static fluid Small triangular prism of liquid at rest ∑FH = 0 But from geometry so 3

Note: Px = 0 Py , g, Z Now take the limit as Thus,

Note: Px = 0 Py , g, Z Now take the limit as Thus, Since was arbitrary, pressure at a point in a static fluid is independent of orientation 4

Pressure variation in water Hydraulics Lecture notes 5

Pressure variation in water Hydraulics Lecture notes 5

 • Pressure variation is not a function of x or y. The total

• Pressure variation is not a function of x or y. The total differential equation The integration of the above differential gives the pressure variation in the vertical direction P + z = constant • Pressure increases linearly with decreasing elevation P 2 Z P 1 Z 2 Z 1 • Pressure is proportional to the depth below the free surface 6

Hydrostatics - Definition þ Hydrostatics is the study of liquid problems in which there

Hydrostatics - Definition þ Hydrostatics is the study of liquid problems in which there is no relative motion between liquid elements. i. e. When liquids are stationary or at rest. þ The discussion in this chapter focuses on Fluid pressure and its measurement.

Pressure at a point in a static fluid þ Pressure – the force exerted

Pressure at a point in a static fluid þ Pressure – the force exerted on a unit area. If the total force, F is uniformly distributed over an area A, then þ P= F/A [N/m 2] þ If the force is not uniformly distributed, the expression of P will be the average value. þ þ If the pressure varies from point to point, the pressure at any point is given by: þ P = d. F/d. A þ Where, d. F = the force acting on an elementary area, d. A.

Pressure at a point … cont’d þ Note: Ø A mass of fluid kept

Pressure at a point … cont’d þ Note: Ø A mass of fluid kept in a container/solid boundary exerts forces against the boundary surfaces. Ø The forces are always act in a direction normal to the surface. Because the fluid is at rest, there are no shear stresses in it.

Pressure at a point … cont’d The pressure at a point in a fluid

Pressure at a point … cont’d The pressure at a point in a fluid at rest is the same in all directions – Pascal’s Law. Proof: Consider a two dimensional small wedge shaped fluid element having unit width normal to plane of the paper. þ þ Let P 1, P 2 and P 3 be the average pressure acting on faces ab, ac and bc, respectively.

Pressure at a point … cont’d þ Since the wedge of fluid is in

Pressure at a point … cont’d þ Since the wedge of fluid is in equilibrium, P 1 = P 3 þ Similarly, þ If dz shrinks to a point and W becomes zero, thus at a point, P 2 = P 3 þ Therefore, the pressure at a point is independent of its orientation.

Pressure variation in a fluid at rest Consider a small parallelepiped fluid element of

Pressure variation in a fluid at rest Consider a small parallelepiped fluid element of size dxdydz at any point in a static mass of fluid. þ Since the fluid is at rest, the element is in equilibrium under the various forces acting on it. þ Let P be the pressure intensity at the midpoint of the element. þ

Pressure variation…cont’d

Pressure variation…cont’d

Pressure variation…cont’d With similar analysis in y direction, we can get: This means that

Pressure variation…cont’d With similar analysis in y direction, we can get: This means that there is no change in pressure with the horizontal distance.

Pressure variation…cont’d Therefore, pressure is only a function of Z and basic equation of

Pressure variation…cont’d Therefore, pressure is only a function of Z and basic equation of fluid static. which is the

Pressure variation…cont’d For homogenous and incompressible fluids, is constant (so is ). Therefore, Where,

Pressure variation…cont’d For homogenous and incompressible fluids, is constant (so is ). Therefore, Where, c is the constant of integration and is equal to the value of P at z = 0. But since z is measured vertically downward, c implies the pressure at the free surface of the fluid (atmospheric pressure). Normally expressed as: For a point lying in the free surface of the fluid z= H+Zo and If Po is the atmospheric pressure at the free surface, then the constant of integration C = Po + (H+Zo)

Pressure variation…cont’d

Pressure variation…cont’d

Pressure variation…cont’d Substituting this value of C in the above equation gives Since the

Pressure variation…cont’d Substituting this value of C in the above equation gives Since the point lies in the fluid at a depth h below the free surface and for this point z = H+Zo-h, substituting for z in the above equation gives: For liquids γ is constant or

Pressure variation…cont’d The pressure at a point in a static mass of liquid depends

Pressure variation…cont’d The pressure at a point in a static mass of liquid depends only upon the vertical depth of the point below the free surface and the specific weight of the liquid, and it doesnot depend upon the shape and size of the bounding container. PA = P B = P C

Standard Atmosphere The atmospheric air exerts a normal pressure upon all surfaces in contact

Standard Atmosphere The atmospheric air exerts a normal pressure upon all surfaces in contact - atmospheric pressure. Atmospheric pressure varies with altitude. Atmospheric pressure is also known as barometric pressure as it is measured by a barometer At sea level atmospheric pressure = 101. 3 Kpa or 1. 013 bar or 10. 3 m of water or 760 mm of mercury – it is called standard atmospheric pressure

Standard Atmosphere Two common datum (zeros) for pressure measurement: 1. Absolute zero pressure. When

Standard Atmosphere Two common datum (zeros) for pressure measurement: 1. Absolute zero pressure. When pressure is measured above absolute zero, it is called absolute pressure. 2. The atmospheric pressure. When pressure is measured either above or below atmospheric pressure, it is called gage pressure. If the pressure of the fluid is below atmospheric pressure, it is called vacuum pressure or suction pressure or negative gage pressure.

Standard Atmosphere Pa= Patm + Pg Pa= Patm – Pv Pa = absolute Pressure

Standard Atmosphere Pa= Patm + Pg Pa= Patm – Pv Pa = absolute Pressure (+Ve) Patm = atm. Pressure Pg = gage pressure (+ve) Pv = vacuum pressure

Standard Atmosphere A device contains cells of air, water and glycerin. What are the

Standard Atmosphere A device contains cells of air, water and glycerin. What are the gage and absolute pressures at points 1 through 5 in KPa What are the equivalent columns of mercury (S. G 13. 6).

Standard Atmosphere P 3= P 2 - γg x 3= -6. 125 Kpa =

Standard Atmosphere P 3= P 2 - γg x 3= -6. 125 Kpa = 6. 125 Kpa (suction, vacuum) = 95. 175 Kpa ( abs ) The standard atmosphere = 101. 3 KPa (abs. ) P 5 = P 4 - γw x 2 = -25. 725 KPa = 75. 575 Kpa (abs) P 1 = 0 gage = 101. 3 KPa (abs) P 4 = P 3 P 2 = γg x 2. 5 m = 1. 25 x 9. 8 x 2. 5 = 24. 5 KPa = 125. 8 KPa (aps)

Gage A reads 200 Kpa. What is the height h of water ? What

Gage A reads 200 Kpa. What is the height h of water ? What should gage B read ? Pair gage = 180 -101. 3= 78. 7 Kpa Pair + γw h + γ hg x 0. 8 = 200 Kpa 78. 7 + 9. 8 h + (13. 6 x 9. 8) x 0. 8 = 200 h = 1. 50 m PB = 78. 7 + γ w (1. 5 + 0. 8) = 101. 24 Kpa

Pressure Measurement • By balancing the liquid column (whose pressure is to be found

Pressure Measurement • By balancing the liquid column (whose pressure is to be found out) by the same or another column. These are also called tube gauges to measure the pressure • By balancing the liquid column (whose pressure is to be found out) by the spring or dead weight. These are called mechanical gauges to measure the pressure. • Tube gages to measure fluid pressure are • Piezometer tube • Manometer • Mercury Barometer • Device used to measure atmospheric pressure

Mercury Barometer

Mercury Barometer

Pressure Measurement Piezometer Tube – Simplest form of manometer, – used for measuring moderate

Pressure Measurement Piezometer Tube – Simplest form of manometer, – used for measuring moderate pressure – The height to which the liquid rises up in the tube gives the pressure head directly Manometer – Improved form of a piezometer tube – Can measure comparatively high pressure and negative pressure – Types are: • • Simple manometer, Mercury barometer U – tube manometer and Inclined-tube manometer Micromanometer, Differential manometer, and Inverted differential manometer

Piezometer tube • consist of a vertical tube • open at the top •

Piezometer tube • consist of a vertical tube • open at the top • Attached to a container in which the pressure is required

Piezometer tube Positive Pressure Negative Pressure PA +h 1 γw = h 2 γm

Piezometer tube Positive Pressure Negative Pressure PA +h 1 γw = h 2 γm PA + h 1 γ w + h 2 γ m = 0 PA = h 2 γ m - h 1 γ w PA = -h 1 γw - h 2 γm

U – tube manometer Widely used to measure the difference in pressure between two

U – tube manometer Widely used to measure the difference in pressure between two containers or two points in a given system. • We can use one of the two ways of manometer reading techniques, • Surface of equal pressure (SEP) • Step by step procedure (SS) • Start at one end and write the pressure there • Add the change in pressure there • + If next meniscus is lower • - If next meniscus is higher P(2) = P(3) by SEP

Differential manometer – In many cases only the difference between two pressures is desired,

Differential manometer – In many cases only the difference between two pressures is desired, and for this purpose differential manometers may be used A and B at the same level and containing same liquid A and B at different level and containing different liquid

Differential manometer Inverted U-tube manometer is used for measuring pressure differences in liquids. P(2)

Differential manometer Inverted U-tube manometer is used for measuring pressure differences in liquids. P(2) = P (3) SEP

Inverted differential manometer – Used for measuring difference of low pressures, where accuracy is

Inverted differential manometer – Used for measuring difference of low pressures, where accuracy is the prime consideration PA – γ h 1 = PB - γ' h 3 - γ‘ h 2

Inclined – tube Manometer • • Used to measure small pressure changes in fluids

Inclined – tube Manometer • • Used to measure small pressure changes in fluids The differential reading is measured along the inclined tube • If pipes A and B contain a gas then the contribution of gas column is always negligible

Mechanical Gauges – For measuring high pressure – Measuring pressure in boilers or other

Mechanical Gauges – For measuring high pressure – Measuring pressure in boilers or other pipes, where tube gauges can not be conveniently used. – Types of gauges are: Dead weight pressure gauge Bourdon's tube pressure gauge Diaphragm Pressure gauge

Advantage and Limitations of manometer • Manometer does not have to be calibrated against

Advantage and Limitations of manometer • Manometer does not have to be calibrated against any standard ( Adv. ) • Measure very small pressure differences, it cannot be used conveniently for large pressure differences ( Lim. ) • Some liquids are unsuitable for use because they do not form well-defined menisci. ( Lim. ) • Unsuitable for measuring fluctuating pressures – Slow response ( Lim. )

Examples Calculate the pressure difference between A and B for the setup shown in

Examples Calculate the pressure difference between A and B for the setup shown in the Figure. PA + 1. 7γw - 1. 0γHg + 0. 56γw - 0. 76γHg - 0. 25γw = PB

Examples Determine the pressure at A, B, C and D in kpa PA =

Examples Determine the pressure at A, B, C and D in kpa PA = γw (4 m) = 9. 8 x 4 = 39. 2 KPa PB PC = = PD PA PB + (3 γw) x 3 39. 2 + 9 x 9. 8 = 127. 4 KPa PB + (3γw) 2 = 98. 10 KPa 1 Pascal (Pa) = KM/m 2

Examples If a pressure in a tank is 345 kpa, find the equivalent pressure

Examples If a pressure in a tank is 345 kpa, find the equivalent pressure head of : 1 - Water 2 -Mercuy 3 -Oil with S. G. 0. 92

Hydrostatic Force on plane surfaces 1. Horizontal surfaces • • The elemental forces Pd.

Hydrostatic Force on plane surfaces 1. Horizontal surfaces • • The elemental forces Pd. A acting on A Subject to constant pressure F = ∫pd. A = p∫d. A = p. A is all parallel. • Normal to the surface. • To find line of action (pressure centre) of the resultant, the moment of area about Y-axis P - Constant - distance from the y- axis to the resultant. - distance to the centroid of the area. Forces on horizontal plane surface Hence, for a horizontal area subjected to static fluid pressure, the resultant passes through the centroid of the area.

2. 6 Hydrostatic Force on Surfaces • Forces on Submerged Plane Surfaces: For design

2. 6 Hydrostatic Force on Surfaces • Forces on Submerged Plane Surfaces: For design purposes, it is essential to calculate the magnitude, direction and location of total forces on surfaces submerged in a liquid. A) Horizontal Plane Surfaces The surface is at a depth h below the free surface of the liquid. Since every point on the surface is at the same depth, the pressure is constant over the entire plane surface. i. e. Therefore, Acting at the centroid of the surface

2. 6 Hydrostatic Force on Surfaces B) Force on a vertical plane surface: Here,

2. 6 Hydrostatic Force on Surfaces B) Force on a vertical plane surface: Here, since the depth of liquid varies from point to point on the surface, the pressure is not constant over the entire surface. Therefore, the total pressure force on the entire surface is: where is the first moment of the area about 0 -0

2. 6 Hydrostatic Force on Surfaces For a vertical plane surface, the CP does

2. 6 Hydrostatic Force on Surfaces For a vertical plane surface, the CP does not coincide with the centeroid of the area. Since the pressure increases with depth, the CP lies below the centroid of the surface area. This position can be determined as follows. Taking the moment of F about O-O, The total force on the strip is O-O is ---- (1) and its moment about Thus, the sum of the moments of the force on all the strips, --- (2) Equating (1) and (2), but is the second moment of inertia.

2. Inclined surface • Pressure on a plane inclined at angle from the horizontal

2. Inclined surface • Pressure on a plane inclined at angle from the horizontal • The line of action of the resultant force has its piercing point in the surface at point called the pressure centre , with coordinates (xp, yp)

For design purposes, it is essential to calculate the magnitude, direction and location of

For design purposes, it is essential to calculate the magnitude, direction and location of total forces on surfaces submerged in a liquid. d. F = P d. A = γh d. A = γ y sin α d. A = γ sin α y d. A where is the first moment of the area about 0 46

The moment d. M of the force d. F about 0 is d. M

The moment d. M of the force d. F about 0 is d. M = d. F y = γ d. A h y = γ d. A y sin α y = γ sin α y 2 d. A This may be integrated over the area and set equal to Fyp is the second moment of the area about 0 yp acts at the center of pressure which is a distance e below the center of gravity such that 47

But, since , because IG is positive, this shows that center of pressure is

But, since , because IG is positive, this shows that center of pressure is below the center of gravity IG-the moment of inertia of the plane with respect to its own centroid,

where IC. G = moment of inertia about the horizontal axis x-x through the

where IC. G = moment of inertia about the horizontal axis x-x through the center of gravity. 49

Hydrostatic Force on Surfaces

Hydrostatic Force on Surfaces

Hydrostatic Force on Surfaces

Hydrostatic Force on Surfaces

Hydrostatic Force on Surfaces

Hydrostatic Force on Surfaces

The concept of pressure prism • The resultant hydrostatic force • Line of action

The concept of pressure prism • The resultant hydrostatic force • Line of action on a plane surface • Prismatic volume – its base the given surface area and altitude at any point of the base given by P = h Force acting on the element area A is F = h A = d d - an element of volume of the pressure prism After integrating, F = the volume of the pressure prism equals the magnitude of the resultant force acting on one side of the surface Xp and Yp are distances to the centroid of the pressure prism Hence the line of action of the resultant passes through the centroid of the pressure prism.

Pressure Diagram

Pressure Diagram

Practical application of pressure force & center of pressure It is very important for

Practical application of pressure force & center of pressure It is very important for hydraulic structures design such as dams, gates and tanks. Example: What should be the minimum weight of the dam not to be overturn by the force of water ponded on its upstream. Take moment about the toe of the dam (about O),

3. Curved surface • Elemental forces vary in direction and not parallel- vector addition

3. Curved surface • Elemental forces vary in direction and not parallel- vector addition Case 1. Liquid is resting on top of a curved base The element of fluid ABC is equilibrium (as the fluid is at rest).

Horizontal Forces • Force on CB = 0 • FAC must equal and be

Horizontal Forces • Force on CB = 0 • FAC must equal and be in the opposite direction to the horizontal resultant force • RH on the curved surface. • AC is the projection of the curved surface AB onto a vertical plane • RH = Resultant force on the projection of the curved surface onto a vertical plane. RH acts horizontally through the centre of pressure of the projection of the curved surface onto an vertical plane.

Vertical Forces RV= weight of liquid vertically above the curved surface and extending up

Vertical Forces RV= weight of liquid vertically above the curved surface and extending up to the free surface • act vertically downward through the centre of gravity of the mass of fluid

Resultant Force Vectorial summation of vertical and horizontal forces And acts through O at

Resultant Force Vectorial summation of vertical and horizontal forces And acts through O at an angle of . The position of O is the point of integration of the horizontal line of action of and the vertical line of action of

Forces on curved surfaces (also applies to plain surface) Pv = Ph = the

Forces on curved surfaces (also applies to plain surface) Pv = Ph = the weight of the liquid above the surface the force exerted on the vertical projection of the curved surface.

Examples A rectangular gate 2 m wide is hinged at point B and rests

Examples A rectangular gate 2 m wide is hinged at point B and rests against a smooth wall at point A as shown. Calculate: a) The pressure force on the gate due to the sea Water (S. G. 1. 034) b) The force exerted by the wall at point A. c) The location of the center of pressure. (neglect weight of gate)

Moment about B = 0

Moment about B = 0

Case 2. When liquid is below the curved surface Example - curved sluice gate

Case 2. When liquid is below the curved surface Example - curved sluice gate Calculation of the forces acting from the fluid below is very similar to when the fluid is above.

Horizontal Forces • The horizontal reaction force which is equal and in the opposite

Horizontal Forces • The horizontal reaction force which is equal and in the opposite direction to the pressure force on the vertical plane A'B. FA’B RH = Resultant force on the projection of the curved surface onto a vertical plane.

Vertical Forces • If the curved surface were removed and the area it were

Vertical Forces • If the curved surface were removed and the area it were replaced by the fluid, the whole system would be in equilibrium. • Thus the force required by the curved surface to maintain equilibrium is equal to that force which the fluid above the surface would exert - i. e. the weight of the fluid. Rv =Weight of the imaginary volume of fluid vertically above the curved surface.

Resultant Force Vectorial summation of vertical and horizontal forces And acts through O at

Resultant Force Vectorial summation of vertical and horizontal forces And acts through O at an angle of . The position of O is the point of integration of the horizontal line of action of and the vertical line of action of

Examples The tank shown in the figure contains oil and water as shown find

Examples The tank shown in the figure contains oil and water as shown find the resultant force on side ABC which is 1. 2 m wide. Convert the oil height to equivalent pressure

Moment about C = 1. 566 m from C or 3. 234 below A

Moment about C = 1. 566 m from C or 3. 234 below A

Examples Find the moment M at O to hold the gate closed. Determine P

Examples Find the moment M at O to hold the gate closed. Determine P on the upper surface of fluid i. e the tank is pressurized under Vacuum

Moment about 0 = F 1 x 0. 848 - Fw x 1. 131

Moment about 0 = F 1 x 0. 848 - Fw x 1. 131 = 17. 96 x 0. 848 - 9. 979 x 1. 13 = 3. 944 N. m counter clockwise Moment required to hold the gate is 3. 944 N. m clockwise.

Examples Ph = Pv = horizontal hydrostatic pressure force on the vertical projection of

Examples Ph = Pv = horizontal hydrostatic pressure force on the vertical projection of the curved surface of the gate vertical buoyant force which is equal in magnitude to the weight of the fluid mass displaced

Tensile Stress in Pipes • Fluid pressure induces tensile stress in pipes. Consider unit

Tensile Stress in Pipes • Fluid pressure induces tensile stress in pipes. Consider unit width of tube For equilibrium In most of cases of pipes under high pressure, cp is assumed to coincide with the center of the pipe and this gives

 • For known value of the tensile force T, the tensile stress per

• For known value of the tensile force T, the tensile stress per meter length of pipe is: If is the allowable stress for the pipe material, the necessary thickness of pipe can be computed from

3. 3 Buoyancy and Stability of Floating and Submerged Bodies • Completely submerged or

3. 3 Buoyancy and Stability of Floating and Submerged Bodies • Completely submerged or floating (partially submerged) bodies in a fluid – net upward force • The tendency for an immersed body to be lifted up in the fluid, due to an upward opposite to the action of gravity – buoyancy . • Resultant fluid force – Buoyant force • Line of action of the force – center of buoyancy • The magnitude of the buoyant force is determined from Archimede’s principle “When a body is immersed in a fluid either wholly or partially, it is buoyed or lifted up by a force, which is equal to the weight of the fluid displaced by the body. ”

 • The principle of floatation - “The weight of a body floating in

• The principle of floatation - “The weight of a body floating in a fluid is equal to the buoyant force, which in turn is equal to the weight of the fluid displaced by the body. ” • For immersed body – If Fb > W then the body will rise until its weight equals the buoyant force. – If W > Fb then the body will tend to move down ward and it may finally sink. • The tendency for the body to return to the original upright position after it has been displaced slightly - stability of a submerged or a floating body.

Generally there are three state of equilibrium position: stable, neutral and unstable. • Stable

Generally there are three state of equilibrium position: stable, neutral and unstable. • Stable equilibrium position - when displaced, it returns to its equilibrium position. • Unstable equilibrium exists when a body tends to continue movement after a slight displacement. • Neutral equilibrium exists when a body remains in its displaced position.

 • Note that as long as the center of gravity falls below the

• Note that as long as the center of gravity falls below the center of buoyancy, the body is in a stable equilibrium position with respect to small rotations. Stability of submerged bodies • A completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position.

 • For floating bodies the stability problem is more complicated, since as the

• For floating bodies the stability problem is more complicated, since as the body rotates the location of center of buoyancy may change. • When M is above G, the body is stable; • when M is below G, it is unstable; and • when M is at G, it is in neutral equilibrium. Stability of a floating body, Stable configuration • A floating body can be stable even though the center of gravity lies above the center of buoyancy. • Metacentre is defined as the point of intersection between the axis of the floating body passing through the points B and G and a vertical line passing through the new center of buoyancy B’. • GM – Metacentric height

 • The position of the metacentre relative to the position of the center

• The position of the metacentre relative to the position of the center of gravity of a floating body determines the stability condition of the floating body. Stability of floating bodies , Unstable configuration When the metacentre lies above the center of gravity, G When the metacentre lies below Unstable Stable the center of gravity, G

 • There are two methods, which may be used to determine the metacentric

• There are two methods, which may be used to determine the metacentric height of a floating body. 1. Experimental method Note: For small angles tan = sin = radians Taking moment about the axis z-z But,

2. Analytical method is sometimes known as the metacentric radius Where, I is moment

2. Analytical method is sometimes known as the metacentric radius Where, I is moment of inertia of the body at the liquid surface about its longitudinal axis ; V is volume of the displaced fluid where

Thank you

Thank you