ANALYSIS OF INCLINED & CURVED SUBMERGED SURFACES (2)

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ANALYSIS OF ANALYSIS OF INCLINED & CURVED INCLINED & CURVED

SUBMERGED SUBMERGED SURFACESSURFACES

PRESENTED BY: Kuldeep Thakur

ANALYSIS OF INCLINED & CURVED SUBMERGED SURFACES

Hydrostatic Thrusts

Buoyant Force

External

Forces

1.(A)1.(A)Hydrostatic Thrusts on a Hydrostatic Thrusts on a

Submerged Inclined Submerged Inclined SurfaceSurface

Hydrostatic Thrusts on Submerged inclined Plane Surface

When a body is submerged in a fluid, and fluid & body both are at rest, so hydrostatic pressure acts on the whole surface of the body.

Due to the existence of hydrostatic pressure in a fluid mass, a normal force is exerted on any part of a solid surface which is in contact with a fluid. The individual forces distributed over an area give rise to a resultant force.

The resultant force FR is acting perpendicular to the plane since no shear force is present when the fluid is at rest

FR has a line of action that passes through the point (xcp , ycp), which is called the

center of pressure.

The differential force dF acting on dA is given by

dF = ρgh dA

The magnitude of the resultant force can be obtained by integrating the differential force over the whole area

The integral represents the first moment of the area about the x axis, which is equal to

where yc is the y coordinate of the centroid of the plane surface.

From trigonometry, hc = yc sinθ

where hc is the vertical distance from the fluid surface to the centroid of the plane surface, then the resultant force is simplified to

FR = ρghcA

The center of pressure, (xcp,ycp) can be obtained by summing moments about the y and x axis, respectively. First, by equating the sum of moments of all pressure forces about the x axis to the moment of the resultant force:

where is the second moment of the area or the area moment of inertia (Ix) about the x axis.

Parallel Axis TheoremParallel Axis Theorem--

The moment of inertia about any axis parallel to that axis through the center of mass is

given by

According to the parallel axis theorem, the moment of inertia can also be written as

Ix = Ix' + Ayc2

where Ix' is the second moment of the area with respect to the centroidal axis, which is parallel to the x axis.

Hence, the center of pressure coordinate ycp is given by

Similarly, xcp

1.(B)Hydrostatic Thrusts on Hydrostatic Thrusts on

Submerged Curved Submerged Curved SurfaceSurface

Hydrostatic Thrusts on Submerged Curved

Surfaces

On a curved surface, the direction of the normal changes from point to point, and hence the pressure forces on individual elemental surfaces differ in their directions. Therefore, a scalar summation of them cannot be made. . Instead, the resultant thrusts in certain directions are to be determined and these forces may then be combined vectorially

Surfaces

Surfaces the magnitude and the line

of action of the resultant force FR exerted on the surface can be best derived by splitting the force into its horizontal and vertical components.

The x-component of the resultant force FRx is the normal force acting on the vertical projection of the curved surface (i.e., surface AE). This force FRx passes through the center of pressure for the projected area AE.

Surfaces

The y-component of the resultant force FRy is the weight of the liquid directly above the curved surface.

Note that this volume can be either real or imaginary.

Here , the volume is real since the liquid actually

occupies this volume.

Surfaces

In this case the y-component of the resultant force is given by the weight of the liquid in a volume above surface AB (i.e., volume ABC); the volume in this case is an imaginary one since volume ABC is not occupied by an actual liquid.

In both cases the force, FRy ,passes through the center of gravity of volume ABCD.

Surfaces

If the gravitational acceleration is assumed to be constant and the fluid is incompressible, then the center of gravity is the same as the centroid of the fluid volume.

Pressure forces are always perpendicular to the surface AB (i.e., the normal stresses).

Since all points on a circle have a normal passing through the center of a circle, the resultant force FR has to pass through point E.

Surfaces

The magnitude of the resultant force is then determined by FR = (FRx + FRy)

The direction of the resultant force is given by

tan θ = (FRy / FRx)

2.2.Buoyant Force on a Buoyant Force on a Submerged bodySubmerged body

Buoyant Force on a Submerged body

Buoyancy- When a body is either

wholly or partially immersed in a fluid, a lift is generated due to the net vertical component of hydrostatic pressure forces experienced by the body.

This lift is called the buoyant force and the phenomenon is called buoyancy.

Hydrostatic pressure forces act on the entire surface of the body.

Therefore the buoyant force FB on the entire submerged body is given by-

Where is the total volume of the submerged

Archimedes Principle-

The buoyant force on a submerged body-

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of liquid displaced by the body, and acts vertically upward through the centroid of the displaced volume.

Thus the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

The buoyant force on a partially immersed body-

According to Archimedes principle, the buoyant force of a partially immersed body is equal to the weight of the displaced liquid.

Therefore the buoyant force depends upon the density of the fluid and the submerged volume of the body.

For a floating body in static equilibrium and in the absence of any other external force, the buoyant force must balance the weight of the body.

3.3.External forces on the External forces on the Inclined and Curved Inclined and Curved

SurfacesSurfaces

External forces on Inclined and Curved Surfaces

When the fluid or body or both are in motion at different velocities, some external forces also act on the body (plane, inclined, curved) along with the pressure forces.

The magnitude of these external forces depend upon the angle which the body makes with the direction of flow of fluid.

The total force exerted by the fluid on the body is perpendicular of the surface of the body. Thus the total force is inclined to the direction of the motion.

The total force can be resolved in two components, one in direction of motion and other perpendicular to the direction of motion.

Determining drag is important in many engineering applications, such as the design of automobiles, airplanes, submarines and buildings.

DRAG- The component of the total force in the direction of

motion is called DRAG and denoted by FD. Drag (D) consists of both friction and pressure drag,

and it is often expressed in terms of a drag coefficient (CD) as

D = CD ρAV2/2

LIFT- The component of the total force in the direction

perpendicular to the direction of motion is known as LIFT and denoted by FL. Lift force occurs only when the axis of the body is inclined to the direction of fluid flow. If the axis of the body is parallel to the direction of fluid flow, lift force is zero. In that case only Drag force acts.

L = CL ρAV2/2

Frictional Drag: this is the drag between the moving fluid and relativity stationary surface. May be referred to as viscous drag This value is dependent/sensitive on the Reynolds number.

Pressure Drag: this is the drag as a result of a pressure drop - wake. Pressure drag is mildly dependent on the Reynolds number, but generally not sensitive to it. Pressure drag is highly sensitive to body shape.

For both Frictional and Pressure drag.... It's important to note that they are both VERY dependent on fluid viscosity. That is, in inviscid (viscosity = 0) flow, both of drop to 0.

AoA: Angle of attack - draw a line in parallel to fluid flow. Now draw a line from the leading edge of a body to the trailing edge. Measure the angle between these two lines. Streamlined Body: Drag losses are primarily due to viscous losses. Streamlined bodies of the same thickness are significantly more aerodynamic than a bluff body. Some of the drag coefficients for flow over two-dimensional and three-dimensional bodies are summarized in the following tables.

ShapeArea,

ReynoldsNumb

Drag CoefficientCD

D2 > 104 1.05

πD2/4 > 104

L/D =0.5 => 1.10L/D =1.0 =>

0.93L/D =2.0 =>

0.83L/D =4.0 =>

πD2/4 > 104

θ =10o => 0.30θ =30o => 0.55θ =60o => 0.80θ =90o => 1.15

DRAG COEFFICIENTS FOR THREE DIMENSIONAL BODIES-

ShapeArea,

ReynoldsNumber,

Drag Coefficient

bD** > 104 2.0

bD > 104 2.1

bD< 105

>5×105

1.2 (Laminar)0.3

(Turbulent)

bD > 104 1.2

bD > 104 1.0

Drag Coefficient for

Two-Dimensional Body-

(**Note: b is the length

into the page)

THANK YOUTHANK YOU

ANALYSIS OF INCLINED & CURVED SUBMERGED SURFACES (2)

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