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Chapter 2. Fluid Statics

Fluid statics is concerned with the balance of forces which stabilize fluids at

rest. In the case of a liquid , as the pressure largely changes according to

its height, it is necessary to take its depth into account. Furthermore, even in

the case of relative rest (e.g. the case where the fluid is stable relative to its

vessel even when the vessel is rotating at high speed), the fluid can be

regarded as being at rest if the fluid movement is observed in terms of

coordinates fixed upon the vessel.

Pressure

When a uniform pressure acts on a flat plate of area A and a force F pushes the

plate, then the pressure p is :

p = F/A

When the pressure is not uniform, the pressure acting on the minute area ∆A is

expressed by the following equation:

Characteristics of pressure

The pressure has the following three characteristics.

1. The pressure of a fluid always acts perpendicular to the wall in contact with the

fluid.

2. The values of the pressure acting at any point in a fluid at rest are equal

regardless of its direction. Imagine a minute triangular prism of unit width in a

fluid at rest as shown in Fig. 1 Let the pressure acting on the small surfaces

dA1 , dA2 , and dA be p1 , p2 and p respectively. The following equations

are obtained from the balance of forces in the horizontal and vertical

directions:

Fig.11

dF ∆F

2

The weight of the triangle pillar is doubly infinitesimal, so it is omitted. From

geometry, the following equations are obtained:

Therefore, the following relation is obtained:

Pressure of fluid at rest In general, in a fluid at rest the pressure varies according to the depth. Consider a

minute column in the fluid as shown in Fig. 2. Assume that the sectional area

is dA and the pressure acting upward on the bottom surface is p and the

pressure acting downward on the upper surface (dz above the bottom surface)

is p + (dp/dz)dz. Then, from the balance of forces acting on the column, the

following equation is obtained:

Since p is constant for liquid, the following equation

ensues:

When the base point is set at zo below the upper

surface of liquid as shown in Fig. 3 and po is the

pressure acting on that surface, then p = po when

z = zo, so :

Thus it is found that the pressure inside a liquid increases in proportion to the

depth.

Fig.2

Fig.3

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Hydrostatic Pressure in Gases: Gases are compressible , with density

nearly proportional to pressure. Thus density must be considered as a variable in

the above

equation :

Separate the variables and integrate between points 1 and 2:

The integral over z requires an assumption about the temperature

variation T(z) . One common approximation is the isothermal atmosphere ,

where T=To :

In the spatial cases the gases become incompressible fluids (ρ=constant)

when the pressure and temperature at absolute values so that the above

integral become :

Ln(p2/p1)=-k( z2 – z1 )

where k = g/(RT)=1.139×10-4 m-1 , for z2 – z1 = 1 m and at 27 oC

p2/p1 = 0.9989 , if p1 = 100 kN/m2 then p2 = 99.989 kN/m2 ∆P=0

It can be neglected the head pressure of the gases .

Manometry :The technique of pressure measurement by employing the

hydrostatic law i.e., dp/dz=ρg is known as manometry and the devises used for

measurement are called manometers

1-single tube manometers : the tube is attached to the point at it is lower and left

open to the atmosphere at the upper end . Since the column of liquid in the

tube is at rest , the gage pressure at the lower end of the tube must be

balanced by the hydrostatic pressure due to the

column of liquid in it :

p = po + ρ g h

P

α

h h cosα

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This manometer may serve to measure the pressure above the atmospheric

pressure For a small gage pressure , it is better to use an inclined single

tube.

2- U-tube Manometers : Consists of two tubes joined at one end to form a U-

shape tube . This manometer can be employed to measure the pressure above

and below atmospheric in any fluid , liquid or gas .

Generally, the pressure gages divided to:

1- Tube gage:

i- Piezometer (small positive),

ii- Manometer (simple (small positive and negative), micro, differential and

invert of differential (large positive and negative))),

2-Mechaniical gage (Bourdon tube), c-type, spiral, twist and helical.

Gas under pressure gh'mP=ρ

h'

Liquid under vacuum

P=-ρgh

h

p

h

Liquid under pressure

P=ρgh

Gas under vacuum

ghmρ-P=

h

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Bourdon gage

3-Piezoelectric transducers, also called solid state pressure transducers, work on

the principle that an electric potential is generated in a crystalline substance

when it is subjected to mechanical pressure.

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For a fluid under a very high pressure , consider the U-tube , multi-fluid

manometer as shown in below figure .If we first label all inter mediate points

between A & a, the total pressure difference can be expressed in terms of a

series of intermediate terms as following :

PA+ ρ1g(ZA – Z1 ) - ρ2g (Z2 – Z1 ) - ρag ( Za –Z2 ) = Pa

PA-Pa= - ρ1g(ZA – Z1 )+ ρ2g (Z2 – Z1 )+ ρag ( Za –Z2 )

For a multiple-fluid manometer connected to the two chambers A and B

the difference in pressure between two chambers A and B.

PA+ ρ1g (ZA – Z1 ) - ρ2g (Z2 – Z1 ) + ρ3g (Z2 – Z3 ) - ρ4g (Z B – Z3 ) = PB

PA- PB = - ρ1g (ZA – Z1 ) + ρ2g (Z2 – Z1 ) - ρ3g (Z2 – Z3 ) + ρ4g (Z B – Z3 )

Manometer Example:

Given the indicated manometer,

determine the gage pressure at A.

ρgw = 9810 N/m3

ρgA = 0.83*9810 = 6807 N/m3

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ρgair = 11.8 N/m3

With the indicated points labeled on the manometer, we can write

PA+ ρ1g(ZA – Z1 ) - ρ2g (Z2 – Z1 ) - ρag ( Za –Z2 ) = Pa

PA-Pa= - ρ1g(ZA – Z1 )+ ρ2g (Z2 – Z1 )+ ρag ( Za –Z2 )

Neglect the contribution due to the air column. Substituting values, we obtained:

PA-Pa= - (6807 ×0.1) + (9810 × 0.18) = 1085 .1 N/m2

Buoyancy Fluid pressure acts all over the wetted surface of a body floating in a fluid, and the

resultant pressure acts in a vertical upward direction. This force is called

buoyancy. The buoyancy of air is small compared with the gravitational force of

the immersed body, so it is normally ignored. Suppose that a cube is located in

a liquid of density ρ as shown in below Figure.

For the vertical direction, where the atmospheric pressure is Po, the force F1

acting on the upper surface A is expressed by the following equation:

F1= ( Po + ρgh1 )A

The force F2 acting on the lower surface is

F2= ( Po + ρgh2)A

So, when the volume of the body in the liquid is V, the resultant force F from the

pressure acting on the whole surface of the body is

F = F2 – F1 = ρg ( h2 – h1)A

F= ρgVb

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Pressure distribution in rigid body motion

We will now consider an extension of our static fluid analysis to the case of

rigid body motion, where the entire fluid mass moves and accelerates

uniformly (as a rigid body). The container of fluid shown below is accelerated

uniformly up and to the right as shown.

Fist law of motion a m =ƩFx

dm =ρdx dy

ax = - (1/ρ)

= dm xa × 1

g dm - = dm ya × 1

)ρ(1/–g - = ya

+ dy dp =

P = -ρ ax x - ρ( g + ay ) y + C

)y / (g + a xa -tanθ =

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Example: The tank of liquid in the

right figure accelerates to the right with

the fluid in rigid-body motion. Compute

ax in m/s2. Find the gage pressure at point

A if the fluid is glycerin at 20 oC .

Solution: (a) The slope of the liquid gives us the acceleration:

y

x

ag

a

tan

at x=0 and y=0 then C=Po=0

Rotational motion Let us study the height of the water surface in the case where a cylindrical vessel

filled with liquid is rotating at constant angular velocity ω. The movement at

constant angular velocity like this is sometimes called gyrostatics, where the

liquid surface poses a concave free surface. Then let us take cylindrical

coordinates (r,θ,z) . Consider a minute element of mass m on the equipressure

plane .The forces acting on it are mg due to the gravitational acceleration g

in the vertical direction and mrω2 due to the centripetal acceleration rw2 in the

horizontal direction.

y

x A(-100,-15)

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If z = ho at r = 0, c = ho, and the following equation is obtained :

The free surface is now a rotating parabolic surface.

Example : A 16-cm-diameter open cylinder

27 cm high is full of water. Find the central

rigid-body rotation rate for which (a) one third

of the water will spill out; and (b) the bottom

center of the can will be exposed.

Solution: (a) One-third will spill out if the resulting parabolic surface is 18

cm deep:

(at z=27cm, r=8cm then z-ho=27-9=18cm)

b) The bottom is barely exposed if the parabolic surface is 27 cm deep:

(at z=27cm, r=8cm then z-ho=27-0=27cm)

ω

ω2 R2 ω2 (0.08)2

ω= 23.5 rad /s

ω2 (0.08)2

ω= 28.8 rad /s

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Problems

1. What is the water pressure on the sea bottom at a depth of 6500 m? The specific

gravity of sea water is assumed to be 1.03.

2. Obtain the pressure p at point A in Fig.1 (a) , (b) and (c).

3. Obtain the pressure difference p1 - p2 in Fig.2 (a) and (b) as shown blow .

4. An iceberg of specific gravity 0.92 is floating on the sea with a specific gravity

of 1.025. If the volume of the iceberg above the water level is 100 m3, what is the

total volume of the iceberg?

5. A cylindrical vessel of radius ro filled with water to height h is rotated around

the central axis, and the difference in height of water level is h'. What is the

rotational angular velocity? Furthermore, assuming ro = 10 cm and h = 18 cm,

obtain ω when h' = 10 cm and also the

number of revolutions per minute n when the cylinder bottom begins to appear.

Fig.1

Fig.2

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Answer

1. 6.57 107 pa

2. (a) P = Po + PgH , (b) P = Po – PgH , ( c) P = Po + ρ'gH' - ρgH

4. 976 m3

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