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EDEXCEL NATIONAL CERTIFICATE/DIPLOMA
PRINCIPLES AND APPLICATIONS of FLUID MECHANICS UNIT 13
NQF LEVEL 3
OUTCOME 2 - HYDROSTATICS
TUTORIAL 2 - IMMERSED SURFACES
CONTENT
Be able to determine the forces acting in hydrostatic systems
Hydraulic devices : devices e.g. hydraulic jack, hydraulic press, hydraulic braking system; system parameters e.g. cylinder dimensions, input and output forces, internal pressure, input and output motions
Immersed surfaces : surfaces e.g. retaining walls of tanks and reservoirs, lock and sluice gates, immersed rectangular and circular inspection covers and hatches; system parameters e.g. surface dimensions, depth of immersion, hydrostatic pressure and thrust, position of centre of pressure
Before you start you should make sure that you fully understand first and second moments of area and basic calculus. If you are not familiar with this, you should do tutorial found on www.freestudy.co.uk before proceeding.
Let’s start this tutorial by revising the fundamental properties of liquids.
1. FORCES ON SUBMERGED SURFACES
1.1 TOTAL FORCE
Consider a vertical area submerged below the surface of liquid as shown.
Fig.
The area of the elementary strip is dA = B dy
You should already know that the pressure at depth h in a liquid is given by the equation: p = gh is the density and h the depth.
In this case, we are using y to denote depth so p = gy
The force on the strip due to this pressure is dF = p dA = B g y dy
The total force on the surface due to pressure is denoted R and it is obtained by integrating this expression between the limits of y 1 and y 2.
It follows that
y y R ρgB
2 1
2 2
This may be factorised.
2
y y y y R ρgB^2
(y 2 - y 1 ) = D so B(y 2 - y 1 ) = BD =Area of the surface A
(y 2 + y 1 )/2 is the distance from the free surface to the centroid y.
It follows that the total force is given by the expression
R = gA y
The term A y is the first moment of area and in general, the total force on a submerged
surface is
R = g x 1st moment of area about the free surface.
SOLUTION
For a given width B, the area is a rectangle with the free surface at the top edge.
h B
h B
1 moment
2 moment h
h 2 momentofareaabout thetopedgeisB
h 1 momentofareaabout thetopedgeisAy B
A Bh 2
h y
2
3
st
nd
3 nd
2 st
2h h
It follows that the centre of pressure is h/3 from the bottom.
The total force is R = gAy = gBh^2 /2 and for a unit width this is gh^2 /
The moment bout the bottom is R x h/3 = (gh^2 /2) x h/3 = gh^3 /
SELF ASSESSMENT EXERCISE No. 1
- A vertical retaining wall contains water to a depth of 20 metres. Calculate the turning moment about the bottom for a unit width. Take the density as 1000 kg/m^3. (13.08 MNm)
- A vertical wall separates seawater on one side from fresh water on the other side. The seawater is 3.5 m deep and has a density of 1030 kg/m^3. The fresh water is 2 m deep and has a density of 1000 kg/m^3. Calculate the turning moment produced about the bottom for a unit width. (59.12 kNm)
Let's look at a harder example next.
WORKED EXAMPLE No. 2
The diagram shows a hinged circular vertical hatch diameter D that flips open when the water level outside reaches a critical depth h. Show that for this to happen the hinge
must be located at a position x from the bottom given by the formula
8h-4D
8 h-5D 2
D x
Given that the hatch is 0.6 m diameter, calculate the position of the hinge such that the hatch flips open when the depth reaches 4 metres.
Fig. 3 SOLUTION
The hatch will flip open as soon as the centre of pressure rises above the hinge creating a clockwise turning moment. When the centre of pressure is below the hinge, the turning moment is anticlockwise and the hatch is prevented from turning in that direction. We must make the centre of pressure at position x.
8h- 4D
8h-5D 2
D
8h-4D
8h-4D-D 2
D
8h-4D
D
D
x
16h-8D
D
D
D
16h-8D
D
x -
D
h-
2
D
16 h-
D
y h- 16 y
D
x h-
y h-x 16 y
D
Equateforh
y 16 y
D
y 4
πD
y 4
πD 64
πD
A y
I Ay h
about thesurface firstmomentofarea
secondmomentofarea h
h h-x 2
D
y h-
2 2
2 2
2
2 2
2 4 2 2 gg
Putting D = 0.6 and h = 4 we get x = 0.5 m
