Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering...

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Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322 Chapter 2 Chapter 2
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Page 1: Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322.

Water Pressure and Pressure Force (Revision)

The Islamic University of Gaza

Faculty of Engineering

Civil Engineering Department

Hydraulics - ECIV 3322

Chapter 2Chapter 2

Page 2: Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322.

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2.1 Free Surface of Water

• A horizontal surface upon which the pressure is constant every where.

• Free surface of water in a vessel may be subjected to:

- atmospheric pressure (open vessel) or,

- any other pressure that is exerted in the vessel (closed vessel).

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2.2 Absolute and Gage Pressures

• Atmospheric pressure is approximately equal to a 10.33-m-high column of water at sea level.

• Any object located below the water surface is subjected to a pressure greater than the atmospheric pressure (P > Patm).

Let:

dA = cross-sectional area of the prism. the prism is at rest. So, all forces acting upon it must be in equilibrium in all directions.

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Notice that:

• If the two points are on the same elevation, h = 0 PA=PB

• In other words, for water at rest, the pressure at all points in a horizontal plane is the same.

Equilibrium in x- direction: Fx = PA dA – PB dA + L dA sin = 0

PB – PA = h The difference in pressure between any two points in still water is always equal to: the product of the specific weight of water () and the difference in elevation between the two points (h).

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Pressure gages: are usually designed to measure pressures above or below the atmospheric pressure.

Gage pressure: is the pressure measured with respect to

atmospheric pressure (using atmospheric pressure as a base).

Absolute pressure: Pabs = Pgage + Patm

Pressure head, h = P/

If the water body has a free surface that is exposed to atmospheric pressure, Patm. Point A is positioned on the free surface such that PA= Patm

(PB )abs= PA + h = Patm + h = absolute pressure

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Notice that:• Any change in pressure at point B would cause an equal

change at point A, because the difference in pressure head between the two points must remain constant = h.

Pascal's law : A pressure applied at any point in a liquid at rest is

transmitted equally and undiminished in all directions to every other point in the liquid.

This principle has been made use of in the hydraulic jacks that lift heavy weights by applying relatively small forces.

The difference in pressure heads at two points in water at rest is always equal to the difference in elevation between the two points. (PB /) – (PA /) = (h)

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Example 2.1

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2.3 Surface of Equal Pressure• The hydrostatic pressure in a body of water varies with the

vertical distance measured from the free surface of the water body.

• All points on a horizontal surface in the water have the same pressure.

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2.4 Manometers

A manometer

Is a tube bent in the form of a U containing a fluid of known specific gravity. The difference in elevations of the liquid surfaces under pressure indicates the difference in pressure at the two ends.

Two types of manometers:

1. An open manometer: has one end open to atmospheric pressure and is capable of measuring the gage pressure in a vessel.

2. A differential manometer: connects each end to a different pressure vessel and is capable of measuring the pressure difference between the two vessels.

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• The liquid used in a manometer is usually heavier than the fluids to be measured. It must not mix with the adjacent liquids (i.e., immiscible liquids).

• The most used liquids are: - Mercury (specific gravity = 13.6), - Water (sp. gr. = 1.00), - Alcohol (sp. gr. = 0.9), and - Other commercial manometer oils of various specific

gravities.

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A simple step-by-step procedure for pressure computation

Step1: Make a sketch of the manometer system approximately to scale.

Step 2: Draw a horizontal line at the level of the lower surface of the manometer liquid, M. The pressure at points 1 and 2 must be the same since the system is in static equilibrium.

Step 3: a) For open manometers P2 = P1

M.h = W.y + PA

PA = M.h - W.y

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A simple step-by-step procedure for pressure computation

b) For a differential manometers

P2 = P1

M.h + w .(y - h) + PB = W.y + PA

P = PA – PB = h (M - w )

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Example 2.2

Determine the pressuredifference P

Solution:

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Single-reading manometer A differential manometer installed in a flow - measured system

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2.5 Hydrostatic Force on a Flat Surface• The area AB of the back face of a dam inclines at an angle (), and,• X - axis lies on the line at which the water free surface intersects with

the dam surface,• Y - axis running down the direction of the dam surface.

horizontal view projection of AB on the dam surface

h

h

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θsinγγ yhP

Aθ.sinγ dydF

• For a strip at depth h below the free surface:

• The total pressure force over the surface:

ydydFFAA

θ.A.sinγAθ.sinγ .A.γ hF

The total hydrostatic pressure force on any submerged plane

surface is equal to the product of the surface area and the

pressure acting at the centroid (C.G.) of the plane surface.

Where:

is the distance measured from the x-axis to the centroid (C.G.) of the plane

Aθ.sinγ dydF

AdAyyA

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Notes:

• Pressure forces acting on a plane surface are distributed over every part of the surface.

• They are parallel and act in a direction normal to the surface.

• They can be replaced by a single resultant force F = h`A. acting normal to the surface.

• The point on the plane surface at which this resultant force acts is known as the center of pressure (C.P.).

• The center of pressure of any submerged plane surface is always below the centroid of the surface (Yp > Y`).

yyA

I

yA

yAI

M

I

yA

dAy

F

dFy

Y oo

x

xAAP

2

2

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The centroid, area, and moment of inertia with respect to the centroid of some common geometrical plane surfaces are given below.

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Example 2.3

For the vertical trapezoidal gate,Determine F and YP

Solution:

Page 21: Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322.

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Example 2.3

Determine F and YP

Solution:

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2.6 Hydrostatic Forces on Curved Surfaces

• The hydrostatic force on a curved surface can be best analyzed by

resolving the total pressure force on the surface into its horizontal and

vertical components.

• Then combine these forces to obtain the resultant force and its direction.

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• FH = Resultant force on the projection of the curved surface onto a vertical plane.

• FH acts horizontally through the centre of pressure of the projection of the curved surface onto a vertical plane.

• We can use the pressure diagram method to calculate the position and magnitude of the resultant horizontal force on a curved surface.

BAH FF '0 xF

'' ABAAAV WWF 0 yF

• FV = The resultant vertical force of a fluid above a curved surface equal to the weight of fluid directly above the curved surface.

• It acts vertically downward through the centre of gravity of the mass of fluid.

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Resultant force• The overall resultant force is found by combining the

vertical and horizontal components vectorialy:

• The angle the resultant force makes to the horizontal is:

• The position of F is the point of intersection of the horizontal line of action of FH and the vertical line of action of FV .

22VH FFF

H

V

F

F1tan

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Pressure distribution on a semi-cylindrical gate

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2.7 Buoyancy

• When a body is submerged or floating in a static fluid, the resultant force exerted on it by the fluid is called the buoyancy force.

• The buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward.

• This force will act vertically upward through the centroid of the volume of fluid displaced, known as the centre of buoyancy.

Archimedes' Principle

The weight of a submerged body is reduced by an amount equal

to the weight of the liquid displaced by the body.

BodybyDisplacedFluidb VF

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Buoyancy of a submerged body

dAhhVF BodybyDisplacedFluidV )( 12

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• The stability of a floating body is determined by the relative positions of the center of gravity of body G and the center of buoyancy B.

• The body is in equilibrium if:

- W must equal Fb (buoyancy force)

- G and B lie on the same vertical line.

2.8 Flotation Stability

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• When the body is displaced through an angle , the center of buoyancy move from B to B` and a turning moment is produced.

• M (metacentre) is the point at which the line of action of the upthrust Fb intersects the vertical line through G.

• GM known as the metacentric height.

• Then: GMW generatedMoment

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2.8 Flotation Stability

The equilibrium of a body may be: • Stable Equilibrium : if when displaced returns to

equilibrium position.

• Unstable Equilibrium: if when displaced it returns to a new equilibrium position.

• Neutral Equilibrium

The stability depends upon whether, when given a small displacement, it tends to return to the equilibrium position, move further from it or remain in the displaced position.

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