seminar 3 4
TRANSCRIPT
-
7/28/2019 Seminar 3 4
1/33
Seminar 3-4
Variation of pressure in static fluid 3.2
Pressure expressed as height of fluid 3.3
Absolute and Gage Pressure 3.4
Measurement of Pressure 3.5
Center of Pressure 3.7
Force on Curved Surface 3.8
Bouyancy and stability of submerged andfloating objects 3.9
-
7/28/2019 Seminar 3 4
2/33
Pressure at a Point Assumptions:
Fluid is at rest or
There is no relative motion, i.e. shearing stresses areabsent
= specific weight
h = vertical distance from the free water surface to thepoint
At a point the pressure is equal in all directions
p is actually the difference between the pressure at thepoint in question and the overlying atmospheric pressure
(taken as zero reference or gage pressure)
hp
)( 11 zzpp This can be easily developed to calculate
the difference in pressure between two
elevations within and incompressible fluid.
-
7/28/2019 Seminar 3 4
3/33
3.2.4 Where an underground oil pipeline crosses under a stream in a gully, it is 68 ft
deeper than on either side. When the oil ( s = 0.88) is not flowing, what is the oil
pressure in the line under the stream, if it is 32 psi at each side of the gully?
A pressure gage at elevation 4.8 m on the side of a storage tank containing oilreads 34.7 kPa. Another gage at elevation 2.2 m reads 57.5 kPa. Compute the
specific weight, density, and specific gravity of the liquid.
-
7/28/2019 Seminar 3 4
4/33
A bubble 4 in below the water surface contains 2x10-7 lb of air. If the temperature
is 60o F and the barometric pressure is 14.7 pisa, calculate the diameter of the
bubble. Ignore the partial pressure of water vapor inside the bubble.
Hint 1: what does perfect gas law tell you?
Hint 2: What is the air pressure in terms of the atmosphere and its location below thewater surface?
-
7/28/2019 Seminar 3 4
5/33
-
7/28/2019 Seminar 3 4
6/33
What is the gage pressure when we are at vapor pressure in a
pressure pipe system?
Absolute = Gage + atmospheric
Gage = Absolute - atmospheric
m
mkN
mkN
m
kNm
kNpressurecatmospheri
106.10
792.9
96.98
96.983.101338.2
3.101
3
2
2
2
-
7/28/2019 Seminar 3 4
7/33
A gage is connected to a tank in which the
pressure of the fluid is 42 psi above atmospheric
(see Figure). If the absolute pressure of the fluidremains unchanged but the gage is in a chamber
where the air pressure is reduced to a vacuum of
25 in Hg, what reading in psi will then be
observed?
-
7/28/2019 Seminar 3 4
8/33
If the atmospheric pressure is 780 mb abs and a gage attached to a tank reads
330 mm Hg vacuum, what is the absolute pressure within the tank?
0.1 /
-
7/28/2019 Seminar 3 4
9/33
In the manometer (see figure) liquid A is 8.4 kN/m3
and liquid B is 12.4 kN/m3. If the pressure at B is
207 kPa. If the pressure at B minus the pressure at Ais 145 kPa what would be the manometer reading x?
Express all pressure heads in terms of liquid B.
-
7/28/2019 Seminar 3 4
10/33
On a submerged object
The total pressure force acting on a
submerged object can be developed two
ways
1. By integrating the pressure distribution acting
perpendicular to and over the area of the
submerged object, or
2. By developing the force components, hydrostaticforce and weight force (typically applied to
curved surfaces)
-
7/28/2019 Seminar 3 4
11/33
Total Pressure force on submerged plane
Ah
yAF
sin
-
7/28/2019 Seminar 3 4
12/33
Where is the distance measured from the x-
axis to the centroid (center of gravity) along
the plane surface.
The pressure force acts at the center of
pressure which is below the centroid due to
the non uniform distribution of pressure along
the plane surface.
y
-
7/28/2019 Seminar 3 4
13/33
Mx is the first moment of the area with respect tothe x axis
Io is the second moment around the x-axis or the
moment of inertia Ic is the moment of inertia of the area around an axis
passing through its centroid and parallel to the x axis
yyA
IY
yAII
yA
IMIY
cp
co
o
x
op
2
-
7/28/2019 Seminar 3 4
14/33
-
7/28/2019 Seminar 3 4
15/33
-
7/28/2019 Seminar 3 4
16/33
A rectangular plate 5 ft by 4 ft is at an angle of 30o with the horizontal,
and the 5-ft side is horizontal. Find the magnitude of the force on one
side of the plate and the depth of its center of pressure when the top
edge is a) at the water surface; and b) 1 ft below the surface.
-
7/28/2019 Seminar 3 4
17/33
The common type of irrigation head gate
shown in Fig P3.15 is a plate that slides over
the opening to a culvert. The coefficient of
friction between the gate and its sliding waysis 0.6. Find the force required to slide open
this 600-lb gate if it is set a) vertically; b) on a
2:1 slope (n = 2), as is common.
-
7/28/2019 Seminar 3 4
18/33
Forces on Curved Surfaces
The hydrostatic force on a curved surface is bestdetermined by resolving the total pressure force onthe surface into its horizontal and verticalcomponents. (noting that the hydrostatic componentacts normal to a submerged surface.
-
7/28/2019 Seminar 3 4
19/33
1. The horizontal component of the total hydrostaticpressure force on any surface is always equal to thetotal pressure on the vertical projection of thesurface. The resultant force of the horizontalcomponent can be located through the center ofpressure of this projection
-
7/28/2019 Seminar 3 4
20/33
2. The vertical component of the total hydrostaticpressure force on any surface is always equal to theweight of the entire water column above thesurface extending vertically to the free water
surface. The resultant force of the verticalcomponent can be located through the centroid ofthis column.
-
7/28/2019 Seminar 3 4
21/33
Determine the force F required to hold the cone in the
position shown in Fig. X3.8.5. Assume the cone is
weightless
-
7/28/2019 Seminar 3 4
22/33
a) Find the horizontal and vertical forces per foot of width acting on the Tainter
gate shown in Fig. P3.2. b) Locate the horizontal force and indicate the line of
action of the vertical force without actually computing its location. C) Locate the
vertical force (hint consider the resultant)
-
7/28/2019 Seminar 3 4
23/33
Buoyancy
The weight of a submerged body is reduced by an
amount equal to the weight of the liquid displaced
by the body (Archimedes Principle).
Thus, a buoyant force is created equal to the volumeof fluid displaced is created by partially or fully
submerged objects.
This force acts vertically upward through the centroid
of the displaced volume.
-
7/28/2019 Seminar 3 4
24/33
A rectangular block of uniform material and length L =
800 mm, width b = 300 mm, and depth d = 50 mm, is
floating in a liquid. It assumes the position shown in
Fig. P3.28 when a uniform vertical load of 20 N/m isapplied at P. A) Find the weight of the block. B) If the
load is suddenly removed, what is the righting
moment before the block starts to move? (Hint: refer
also to Fig 3.19)
-
7/28/2019 Seminar 3 4
25/33
-
7/28/2019 Seminar 3 4
26/33
-
7/28/2019 Seminar 3 4
27/33
-
7/28/2019 Seminar 3 4
28/33
-
7/28/2019 Seminar 3 4
29/33
-
7/28/2019 Seminar 3 4
30/33
FVABottom = (0.8 * 9790)*[(1m*6m*1.414m)+(1/4**(0.707m)2*1m)+(1/2*(0.707m)
2*1m]
= 74.19 kN
Net vertical force (fluid A) = 60.1 kN 74.19 kN = -11.18 kN (or upward)
Alternate vertical force due to displaced volume
FAby = (0.8 * 9790) * 1m * [(3/4 * PI()*(0.707m)2
+ 1/2* (0.707m)2]
= -11.18 kN (equivalent to vertical force balance)
NET VERTICLE FORCE ON CYLINDER:
FV = -11.18 57.7 + 30.7 = -38.11 kN (or upward force)
-
7/28/2019 Seminar 3 4
31/33
-
7/28/2019 Seminar 3 4
32/33
-
7/28/2019 Seminar 3 4
33/33