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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

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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.

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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.

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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?

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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

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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?

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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 /

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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.

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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)

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Total Pressure force on submerged plane

Ah

yAF

sin

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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

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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

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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.

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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.

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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.

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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

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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.

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Determine the force F required to hold the cone in the

position shown in Fig. X3.8.5. Assume the cone is

weightless

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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)

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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.

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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)

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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)

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