iidande forcesoncurvedsurfaces buoyancy

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Southern Methodist University Bobby B. Lyle School of Engineering CEE 2342/ME 2342 Fluid Mechanics Roger O. Dickey, Ph.D., P.E. II. HYDROSTATICS D. Hydrostatic Forces on Curved Surfaces

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Page 1: IIDandE ForcesOnCurvedSurfaces Buoyancy

Southern Methodist University

Bobby B. Lyle School of Engineering

CEE 2342/ME 2342 Fluid Mechanics

Roger O. Dickey, Ph.D., P.E.

II. HYDROSTATICS

D. Hydrostatic Forces on Curved Surfaces

E. Buoyant Force

Page 2: IIDandE ForcesOnCurvedSurfaces Buoyancy

Homework No. 7 Additional problems dealing

with hydrostatic pressure forces on submerged

planar surfaces.

Page 3: IIDandE ForcesOnCurvedSurfaces Buoyancy

Reading Assignment:

Chapter 2 Fluid Statics, Sections 2.10 and 2.11

Page 4: IIDandE ForcesOnCurvedSurfaces Buoyancy

D. Hydrostatic Forces on Curved Surfaces

For curved surfaces, it is usually convenient to

separately determine the horizontal and vertical

components of the hydrostatic pressure force,

and then compute the vector sum of the

components to obtain the resultant force by

applying the following principles of hydrostatics

and vector algebra:

Page 5: IIDandE ForcesOnCurvedSurfaces Buoyancy

1. the horizontal component of the hydrostatic

pressure force on any surface equals the force

exerted on a projection of the surface onto a

vertical plane.

2. the vertical component of the resultant

pressure force equals the weight of the vertical

column of liquid above a surface.

Page 6: IIDandE ForcesOnCurvedSurfaces Buoyancy

3. the magnitude of the resultant pressure force is then,

4. The line of action of the resultant force lies at an angle from the horizontal determined from,

22VHR FFF

FH

FVFR

H

V

H

V

FF

FF 1tan tan

Page 7: IIDandE ForcesOnCurvedSurfaces Buoyancy

An efficient problem solving algorithm for

applying these principles involves developing a

free-body diagram of the fluid volume enclosed

by the: (i) curved solid surface, (ii) horizontal

projection of the surface onto a hypothetical

vertical plane, and (iii) vertical projection of the

surface onto a hypothetical horizontal plane.

Page 8: IIDandE ForcesOnCurvedSurfaces Buoyancy

*Important Point –

A projection can be visualized as the “shadow”

cast by a 3-dimensional body onto a plane

surface.

Page 9: IIDandE ForcesOnCurvedSurfaces Buoyancy

Consider the example

shown in Figure 2.23

(b), p. 69 Modified:

Liquid element of volume has unit length perpendicular to the plane of the image.

h

hC1

hC2

VV

Area, A1

Area, A2

Page 10: IIDandE ForcesOnCurvedSurfaces Buoyancy

In the figure, hC1 is the depth to the centroid of

the horizontal rectangular area bounded by line

AB, and hC2 is the depth to the centroid of the

vertical rectangular area bounded by line AC.

Page 11: IIDandE ForcesOnCurvedSurfaces Buoyancy

All forces acting on the fluid element can be shown on a free-body diagram of the element as shown in Figure 2.23 (c), p. 69 Modified:

Horiz. pressure force on vertical plane surface bounded by AC

Vertical pressure force on horiz. plane surface bounded by AB

Horizontal &Vertical reaction forces from the tank wall on arc BC, acting through Point O

Weight acting through the center of gravity (CG) of the fluid element

Page 12: IIDandE ForcesOnCurvedSurfaces Buoyancy

Applying Newton’s second law in both the

horizontal and vertical directions for the static

fluid element:

Forces perpendicular to the plane of the sketch

counteract; regardless they do not contribute to

the resultant pressure force on the curved surface.

2

2

..

0FF

FF

maF

H

H

HorizHoriz

WFFWFF

maF

V

V

VertVert

1

1

..

0

0 0

Page 13: IIDandE ForcesOnCurvedSurfaces Buoyancy

Now, apply the basic principles of hydrostatics:

Substituting for F1, F2, and W in previous equations:VW

AhFAhF

C

C

(iii) (ii)

(i)

222

111

VAhFVAhF

CV

CV

11

11

22 AhF CH

Total volume of liquid above curved surface

FV equals the total weight of the vertical column of liquid above the curved surface, as expected

Page 14: IIDandE ForcesOnCurvedSurfaces Buoyancy

The resultant pressure force on the curved

portion of the tank wall is then equal in

magnitude, but opposite in direction to the

resultant of the reaction forces FH and FV as

shown in Figure 2.23 (d), p. 69 Modified:

H

V

FF1tan

FH

FVFR

Page 15: IIDandE ForcesOnCurvedSurfaces Buoyancy

Notice that a concurrent force system is formed by three coplanar forces intersecting at the common Point O:

• horizontal pressure force, F2, acting on the

vertical plane surface bounded by line AC

• the vector sum of the vertical pressure force on the horizontal plane surface bounded by line

AB and the fluid element weight, i.e., (F1 + W)

• the resultant reaction force of the tank wall on

the fluid element, FR

Page 16: IIDandE ForcesOnCurvedSurfaces Buoyancy

Concurrent force

system – F2, (F1+W),

and FR acting on the

fluid element while

intersecting at the

common Point O:

F2

y

x

F1+W

FR

CG

FV

FH

Page 17: IIDandE ForcesOnCurvedSurfaces Buoyancy

The xy-coordinates for Point O, (xO, yO), are

determined as follows:

(i) There are only two, statically balanced

horizontal forces pressure force F2 and

reaction force FH hence they must be co-

linear. Therefore, yO equals the y-coordinate

of the pressure center, yR , for the horizontal

pressure force, F2, acting on the vertical plane

surface bounded by AC. That is, yO = yR.

Page 18: IIDandE ForcesOnCurvedSurfaces Buoyancy

(ii) Based on static equilibrium, xO can be

determined by summing the moments of all

vertical forces about the z-axis

perpendicular to the xy-plane, and then

setting the sum equal to zero;

V

CGCO

CGCOV

V

FWxxFx

WxxFxF

M

11

11 0

0+

Page 19: IIDandE ForcesOnCurvedSurfaces Buoyancy

Where, xC1 is the x-coordinate of the centroid

of the horizontal plane surface bounded by line

AB, and xCG is the x-coordinate of the center

of gravity of the fluid element, Point CG.

Page 20: IIDandE ForcesOnCurvedSurfaces Buoyancy

*Important Point –

Use of a free body diagram of the fluid element,

and the principles enumerated above, are

generally applicable to any surface submerged in a

fluid. Thus, the problem solving algorithm

described here may also be applied as an

alternative method for determination of the

resultant pressure force on inclined planar

surfaces.

Page 21: IIDandE ForcesOnCurvedSurfaces Buoyancy

Refer to Handout II.D. Hydrostatic Forces on

Curved Surfaces Example Problems.

Page 22: IIDandE ForcesOnCurvedSurfaces Buoyancy

E. Buoyant Force

Archimedes (287-212 B.C.) was an ancient Greek

scientist and mathematician who established the

principles of buoyancy and flotation, and he

invented a type of low-head/high-volume pump

now called the Archimedes Screw. Remarkably,

his principles and ideas are largely unchanged

and widely applied to this day.

Page 23: IIDandE ForcesOnCurvedSurfaces Buoyancy

Archimedes Screw Pumps

Page 24: IIDandE ForcesOnCurvedSurfaces Buoyancy

Buoyant force on a submerged or floating body has a magnitude equal to the weight of the fluid displaced by the body. Buoyant force always acts vertically upward with a line of action through the centroid of the displaced fluid volume, which is called the center of buoyancy.

Archimedes’ Principle, and associated corollaries, follow directly from the principles of hydrostatics dealing with pressure forces on submerged surfaces:

Page 25: IIDandE ForcesOnCurvedSurfaces Buoyancy

The upward action of buoyant force is due to

greater pressures acting on the underside of a

body compared to those on the upper side. These

pressure differences are the result of the

hydrostatic distribution; simply put, the

underside of a submerged or floating body is at

greater depth, hence it experiences greater

pressure than the upper side.

Page 26: IIDandE ForcesOnCurvedSurfaces Buoyancy

For example, consider the pressure distribution on the surface of a tethered buoy submerged in seawater:

Page 27: IIDandE ForcesOnCurvedSurfaces Buoyancy

*Important Points –

1. A body floating partially submerged at the

surface of a liquid penetrates into the liquid

until it reaches the specific depth where the

weight of the displaced fluid exactly equals the

weight of the body. This is why the gross

weight of a ship is referred to as the ship’s

displacement.

Page 28: IIDandE ForcesOnCurvedSurfaces Buoyancy

2. Bodies submerged in a liquid have an apparent

weight that is less than their actual weight in air

because the vertical buoyant force counteracts

all or part of the weight. Differences between

weight measurements of a body in air, and

while the body is submerged in a liquid are

sometimes used to determine the volume of

irregularly shaped bodies.

Page 29: IIDandE ForcesOnCurvedSurfaces Buoyancy

3. For completely submerged incompressible bodies, i.e., bodies whose volume does not change with changing pressure, the buoyant force exerted by a an incompressible fluid is independent of submergence depth. In this case, the pressure difference between top and bottom remains constant because the vertical distance between them remains the same. For example, buoyant force on a fully submerged submarine remains constant regardless of depth – submarines change depth by changing their weight via ballast tanks.

Page 30: IIDandE ForcesOnCurvedSurfaces Buoyancy

4. For bodies whose volume depends upon the surrounding fluid pressure, i.e., compressible bodies, the buoyant force varies with submergence depth. For example, a scuba diver’s air bubbles expand due to decreasing hydrostatic pressure as they rise through seawater. Larger volume further increases the buoyant force causing bubbles to accelerate as they rise. They often burst, forming many smaller bubbles before reaching the surface.

Page 31: IIDandE ForcesOnCurvedSurfaces Buoyancy

Homework No. 8 Hydrostatic pressure forces on submerged curved surfaces, and buoyant force.