The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Hydraulics - ECIV 3322
Chapter 2
Water Pressure and Pressure Force (Revision)
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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Equilibrium in x- direction:
Fx = PA dA – PB dA + g L dA sin q = 0
PB – PA = g h
The difference in pressure between any
two points in still water is always equal to:
the product of the specific weight of water
(g) and the difference in elevation between
the two points (h).
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.
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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 + g h = Patm + g h = absolute pressure
 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/g
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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 /g) – (PA /g) = D(h)
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.
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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
gM.h = gW.y + PA
PA = gM.h - gW.y
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A simple step-by-step procedure for pressure computation
b) For a differential manometers
P2 = P1
gM.h + gw .(y - h) + PB = gW.y + PA
DP = PA – PB = h (gM - gw )
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Example 2.2
Determine the pressure
difference DP
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 (q ), 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.
h
h
horizontal view
projection of AB on
the dam surface
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• For a strip at depth h below the free surface:
P  γ h  γ y sin θ
dF  γ y sin θ.dA
• The total pressure force over the surface:
F   dF   γ y sin θ.dA  γ sin θ.A. y
A
F  γ .h.A
A
Where:
y   y dA A is the distance measured from the x-axis to the
A
centroid (C.G.) of the plane
The total hydrostatic pressure force on any submerged plane
dF  γ y sin θ.dA
surface is equal to the product of the surface area and the
pressure acting at the centroid (C.G.) of the plane surface.
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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 = g 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`).
YP 
 y dF
A
F

2
y
 dA
A
Ay
Ix
I o  Ay 2 I o



y
Mx
Ay
Ay
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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:
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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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F
x
0
FH  FA' B
• 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.
F
y
0
FV  WAA'  WABA'
• 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
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the mass of fluid.
Resultant force
• The overall resultant force is found by combining the
vertical and horizontal components vectorialy:
F  FH2  FV2
• The angle the resultant force makes to the horizontal is:
 FV
q  tan 
 FH
1



• 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 .
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Pressure distribution on a semi-cylindrical gate
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