Hydro Static Forces

8/3/2019 Hydro Static Forces

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HYDROSTATIC FORCES ON SUBMERGED PLANES

When a fluid is in contact with a surface is exerts a normal force on the surface. The walls of

reservoirs, sluice gates, flood gates, oil and water tanks and the hulls of ships are exposed to theforces exerted by fluids in contact with them. The fluids are generally under static condition. For

the design of such structures it is necessary to determine the total force on them. It is alsonecessary to determine the point of action of this force. The point of action of the total force is

known as centre of pressure or pressure centre. From the basic hydrodynamic equation it isknown that the force depends on the pressure at the depth considered.

The total force is obtained by integrating the basic equation over the area

From the definition of centre of gravity or centroid

The integral over the area is nothing but the second moment or the moment of inertia of the area

about the axis considered. Thus there is a need to know the centre of gravity and the moment of

inertia of areas.

CENTROID AND MOMENT OF INERTIA OF AREAS

In the process of obtaining the resultant force and centre of pressure, the

determination of first and second moment of areas is found necessary andhence this discussion. The moment of the area with respect to theyaxis canbe obtained by summing up the moments of elementary areas all over thesurface with respect to this axis as shown in Fig.

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The integral has to be taken over the area. Ifmomentsare taken with respect to a parallel axis at adistance ofk

from theyaxis equation can be written as

Fig.1As kis a constant, it is possible to choose a value ofx = k,

such that the moment about the axis is zero. The moment about the axis through the centre of

gravity is always zero.

The point of intersection of these centroidal axes is known as the centroid of the area. It can be

shown that the moment of the area about any line passing through the centroid to be zero. With

reference to the Fig. the second moment of an area about they axis.Iy is defined as

Considering an axis parallel toy axis through the centroid and taking the second moment of thearea about the axis and calling it asIG, wherex is the distance from the axis and the centroid.

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The moment of inertia of an area about any axis is equal to the sum of themoment of inertia about a parallel axis through the centroid and the productof the area and the square of the distance between this axis and centroidal

axis. These two equations are used in all the subsequent problems.The second moment is used in the determination of the centre of pressure for plane areas

immersed in fluids.The product of inertia is defined as

It can be shown that whenever any one of the axes is an axis of symmetryfor the area, Ixy= 0.The location of the centre of gravity, moment of inertia through the centroidIG and moment of inertia about edge Iedge (specified) for some basic shapes

are given in Table

Centre of Gravity and Moment of Inertia for some typical shapes

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FORCE ON AN ARBITRARILY SHAPED PLATE IMMERSED IN A LIQUID

Case 1 : Surface exposed to gas pressure : For plane surface, force = area pressure Thecontribution due to the weight of the gas column is negligible. The resultant acts at the centroid

of the area as the pressure at all depths are the same.

Case 2 : Horizontal surface at a depthy.P= y and asy is ve, force =Ay in whichy may

also be expressed as head of the fluid. The resultant force acts vertically through the centroid ofthe area, Here also the pressure at all locations are the same.

Case : Plane inclined at angle with horizontal. Refer Fig.

FIG-2

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This equation is extensively used in the calculation of total force on a surface. Equations andthose given in Table 3.1 are used to obtain the location of the centroid. The following important

conclusions can be drawn from this equation.

1. h equals the pressure at the centroid. The total force thus equals the product of areaand the pressure at the centroid.

2. A special case of the situation is a vertical surface where = 90 and sin = 1. and soh =y in this case.

CENTRE OF PRESSURE FOR AN IMMERSED INCLINED PLANE

The centre of pressure is determined by taking moments of the force on elementary areas withrespect to an axis (say O in Fig.2) and equating it to the product of the distance of the centre of

pressure from this axis and the total force on the area

For surfaces with an axis of symmetry, the centre of pressure will lie on that axis. In other casesxcpandycpare calculated by the use of moments.

With reference to the Fig.2, let CP (xcp,ycp) be the centre of pressure.

The area element considered here being dxdy. Referring to the Fig.1

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IG is the moment of inertia along the centroidal axis and y is the location of centroid along they

direction.

If the plane is vertical, then,y = h the depth to the centroid.

In case the height hcpis required instead ofycp, (along the plane) then substitutingycp= hcp/sin andy = h / sin Substituting in equation (3.3

hcp/ sin = [IG/{(h / sin ) A}] + h/sin

there fore hcp = [(IGsin2 / h A)] + hThis is the general equation when the depth of the centre of pressure is required in the case ofinclined planes. If = 90 (vertical surface), then sin2 = 1. These equations are fairly simple, the

main problem being the calculation of the moment of inertia for odd shapes.

CENTRE OF PRESSURE FOR IMMERSED VERTICAL PLANES

Case 1: A rectangle of width b and depth d, the side of length bbeing horizontal.

Case 2: A circle of diameterd.

Case 3: A triangle of height h with base b, horizontal and nearer the free surface.

Assuming the depth ofCG to bePm in all the cases. h =y in the case

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COMPONENT OF FORCES ON IMMERSED INCLINED RECTANGLES

Consider a case of a rectangle ofa d, with side dinclined at to the horizontal, immersed in a

fluid with its centroid at a depth ofh m. For this case it can be shown that (i) The horizontal

component of the resultant force equals the force on the vertical projection of the area and(ii) The vertical component equals the weight of the fluid column above this area.The net force acting perpendicular to the area is given by = Ah

The horizontal component equals = A h sin

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FORCES ON CURVED SURFACES

(i) Vertical forces : The vertical force on a curved surface is given by the weight of the liquidenclosed by the surface and the horizontal free surface of the liquid. The force acts along the

centre of gravity of the volume. In case there is gas pressure above the surface, the force due togas pressure equals the product of horizontal projected area and the gas pressure and acts at thecentroid of the projected area. If the other side of the surface is exposed to the same gas pressure,

force due to the gas pressure cancels out. This applies to doubly curved surfaces and inclined

plane surfaces.

(ii) Horizontal forces: The horizontal force equals the force on the projected area of the curvedsurface and acts at the centre of pressure of the projected area.

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The value can be calculated using the general equation.F= A h,

whereA is the projected area and h is the depth of the centroid of the area.These two statements can be proved as indicated below. Refer to Fig. 3 The volume above the

surface can be divided into smaller elements. At the base of each element, the vertical

force equals the weight of the smallelement. Thus the total vertical forceequals the sum of theweights of all the elements or theweight of theliquid enclosed between the area andthe horizontal surface. Consider an

imaginary vertical surface AB. The

element between AB and the

surface AB is in equilibrium. AB gives the projected vertical area. The

horizontal force on this area due to liquid

FIG 3

pressure should equal the horizontal force on the curved surface for the

volumeAB AB to be in equilibrium. Hence the horizontal force equals theforce on the projected area due to liquid pressure

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Problem 3.6.In a water reservoir, the vertical gate provided for opening is a semicircular plate

of dia 3 m with diameter horizontal and at the water level. Determine the total pressure and its

point of action if water levelis up to the top edge of the gate.

Total force on the gate = Ah

h = centre of gravity of the semicircular surface 2D/ 3= 2 3/3 = 0.6366 m

Total force = 1000 9.81 ( 32/4 2) 0.6366 = 22072 N = 22.072 kNIbase= D4/ 128 (about the diameter)

Depth centre of pressure =Ibase/Ah

= ( 34/128)(2 4/ 32) (1/0.6366)

= 0.8836 m

Problem 3.7.A water tank is provided with a gate which has a shape of a quadrant of

a circle of 3 m radius. The gate is positioned in such a way that one straight edge of it ishorizontal. Determine the force acting on the gate due to water and its point of action if the tank

is filled with water upto 2 m above the edge.

Distance of centre of gravity of the gate from the top edge = 2D/3 = 4r/3

Total pressure on the gate = h A= 1000 9.81 [2 + 4 3/3 ] [ 32/4] = 226976 N

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LAMINAR & TURBULENT FLOWS

Characteristics Of Turbulent Flow

The most important characteristic of turbulent motion is the fact that velocity andpressure at a point fluctuate with time in a random manner.

Fig. 32.1 Variation of horizontal components of velocity for laminar and turbulent flows at a

point P

The mixing in turbulent flow is more due to these fluctuations. As a result we can see

more uniform velocity distributions in turbulent pipe flows as compared to the laminarflows .

Fig. 32.2 Comparison of velocity profiles in a pipe for (a) laminar and (b) turbulent flows

Turbulence can be generated by -

1. frictional forces at the confining solid walls2. the flow of layers of fluids with different velocities over one another

The turbulence generated in these two ways are considered to be different.

Turbulence generated and continuously affected by fixed walls is designated as wall turbulence

, and turbulence generated by two adjacent layers of fluid in absence of walls is termed as free

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turbulence . One of the effects of viscosity on turbulence is to make the flow more

homogeneous and less dependent on direction.

Turbulence can be categorised as below -

Homogeneous Turbulence: Turbulence has the same structurequantitatively in all parts of the flow field.

Isotropic Turbulence: The statistical features have no directional

preference and perfect disorder persists.

Anisotropic Turbulence: The statistical features have directional

preference and the mean velocity has a gradient.

Homogeneous Turbulence : The term homogeneous turbulence implies that the velocity

fluctuations in the system are random but the average turbulent characteristics are

independent of the position in the fluid, i.e., invariant to axis translation.

Consider the root mean square velocity fluctuations

, ,

In homogeneous turbulence, the rms values ofu', v'and w'can all be different, but each value

must be constant over the entire turbulent field. Note that even if the rms fluctuation of anycomponent, say u's are constant over the entire field the instantaneous values of u necessarily

differ from point to point at any instant.

Isotropic Turbulence: The velocity fluctuations are independent of the axis of reference,

i.e. invariant to axis rotation and reflection. Isotropic turbulence is by its definitionalways homogeneous . In such a situation, the gradient of the mean velocity does not

exist, the mean velocity is either zero or constant throughout.

In isotropic turbulence fluctuations are independent of the direction of reference and

= = or

It is re-emphasised that even if the rms fluctuations at any point are same, their instantaneous

values necessarily differ from each other at any instant.

Turbulent flow is diffusive and dissipative . In general, turbulence brings about better

mixing of a fluid and produces an additional diffusive effect. Such a diffusion is termed

as "Eddy-diffusion ".( Note that this is different from molecular diffusion)

At a large Reynolds number there exists a continuous transport of energy from the freestream to the large eddies. Then, from the large eddies smaller eddies are continuously

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formed. Near the wall smallest eddies destroy themselves in dissipating energy, i.e.,

converting kinetic energy of the eddies into intermolecular energy.

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Hydro Static Forces

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