iidande forcesoncurvedsurfaces buoyancy
TRANSCRIPT
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
Homework No. 7 Additional problems dealing
with hydrostatic pressure forces on submerged
planar surfaces.
Reading Assignment:
Chapter 2 Fluid Statics, Sections 2.10 and 2.11
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:
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.
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
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.
*Important Point –
A projection can be visualized as the “shadow”
cast by a 3-dimensional body onto a plane
surface.
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
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.
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
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
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
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
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
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
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.
(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+
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.
*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.
Refer to Handout II.D. Hydrostatic Forces on
Curved Surfaces Example Problems.
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.
Archimedes Screw Pumps
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:
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.
For example, consider the pressure distribution on the surface of a tethered buoy submerged in seawater:
*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.
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.
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.
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.
Homework No. 8 Hydrostatic pressure forces on submerged curved surfaces, and buoyant force.