chapter 1-3 fluid mechanics final

8/18/2019 CHAPTER 1-3 Fluid Mechanics Final

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CHAPTER 1

PROPERTIES OF FLUIDS

1.1 Introduction

Fluids mechanics deals with study of fluids-liquid and gases. The study can

be behavior of liquid fluids at rest (static) and in motion (dynamic). The study of

fluid mechanics is important because our life depend on them. The air we breathe,

flight of birds in air, the motion of fish in water, circulation of blood in veins of

human body, flow of oil and gas in pipelines, transportation of water in pipe, all

follow the principles of fluid mechanics. Engineers have applied these principles in

the design of dams, construction of ships, airplanes, turbo-machinery etc. Fluids in

motion are potential sources of energy and can be converted into useful wor to drive

a water turbine or windmill. The principles of fluid mechanics are also applied to

fluid power system in which pressured fluid is used to transmit power. !ydraulic

drives and controls have become more and important due to automation and

mechani"ation. Today, a very large part or modern machinery is controlled

completely or partly by fluid power.

Fluid can be defined as substance that has ability to flow. #ases e$pand

whereas liquids do not. %iquid have no shape of it&s own but rather tae the shape of

the container in which it is placed. That means if liquid or volume less than volume

of the container is poured into container the fluids will occupy a volume of the

container and will have a free surface. #ases e$pand and occupy full volume of the

container. #ases are compressible which means their volume changes with pressure

where as liquids are incompressible. 'ompressible flows are again divided into

subsonic and supersonic depending on gas velocity less or greater than sound velocity.

Their application is in et propulsion system, aircraft and rocets.

1.2 International Syste o! Units "SI#

n the te$t we shall use * units. The dimension in any system can be

considered as either primary or secondary dimensions. n the * units there are +

primary dimensions.


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a) rimary nits

Diension International Sy$ol Unit

/ass / 0ilogram (g)

%ength % /eter (m)

Time T *econd (s)Temperature 0 0elvin (0)

Electric 'urrent 1 1mpere (1)

b) *econdary nits

*econdary units is a combination of primary units such as 2ewton (2 or gm3s 4),

5oule (5 or 2m), 6att (6 or 2m3s) etc.

1.% S&eci!ic 'ei()t and ass density

Two important parameters that tend to indicate heaviness of the substance are

specific weight and mass density. The specific weight in the weight of substance per

unit volume and is commonly designated by #ree letter 7gamma& (γ ). n equation

form,

V

W

Volume

Weight

==γ

/ass density is the mass per unit volume of the substance. t commonly

designated by a #ree letter 7rho& ( ρ ). n the equation form,

V

m

Volume

mass== ρ

There e$ist an important relation between specific weight and mass density.

6eight of the substance w 8 mg

6eight of the substance3unit volume,

g V

mg

ρ γ ==

4


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For ideal gases, the density of gas is depended on the pressure and temperature of the

gas. The density can be obtained by the gas equation9

mRT PV =

or

RT P ρ =

where M gasmolar

t Coefficien gasUniversal

R

ℜ

== :

: :

Thus the specific weight is the product of mass density and acceleration due to

gravity.

n the * units, γ will be e$pressed in 23m; and < in g3m;. The values of specific

weight and mass density of water at different temperature are given in Table . and

Table .4 gives

Table . hysical roperties of 6ater

Temperature Specific

Weight

Mass

Density

Dynamic

Viscosity

Kinematic

Viscosity

Surface

tension

T γ ρ µ ν σ

(°C) (kN/m³) (kg/m³) N-s/m² m/s² N/m

0 9.81 1000 1.75 x 10 ;− 1.75 x 10 =− 0.0756

30 9.77 996 8.00 x 10 +− 1.02 x 10 >− 0.0712

60 9.65 984 4.60 x 10 +− 4.67 x 10 >− 0.0662

90 9.47 956 3.11 x 10 +− 3.22 x 10 >− 0.0608

the mass density for common fluids. From this table, other fluids can be compared

with water in terms of density and specific of weight.

;


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Table .4 hysical roperties of 'ommon fluids at *tandard 1tmospheric ressure

Fluids Specific

Gravity

Specific

Weight

Mass

Density

Dynamic

Viscosity

Kinematic

Viscosity

s γ ρ µ ν

- (kN/m³) (kg/m³) N-s/m² m 4 /s

Air 0.0012 11.8 1.20 1.81 x 10 ?− 1.51 x 10 ?−

Ammonia 0.830 8.31 829 2.20 x 10+ 2.65 x 10 >−

Glycerine 1.263 12.34 1258 950 x 10 ;− 7.55 x 10 +−

Kerosene 0.823 8.03 819 1.92 x 10 ;− 2.34 x 10 =−

Mercury 13.60 133.1 13570 1.56 x 10 ;− 1.14 x 10 >−

Methanol 0.79 7.73 788 5.98 x 10+ 5.58 x 10 >−

SAE 10 Oil 0.87 8.71 869 8.14 x 10 4− 9.36 x 10 ?−

SAE 30 Oil 0.89 8.71 888 4.40 x 10 ,− 4.95 x 10 +−

Turpentine 0.87 8.51 868 1.38 x 10 ;− 1.58 x 10 =−

Water 1.00 9.79 998 1.02 x 10 ;− 1.02 x 10 =−

Sea Water 1.03 10.08 1028 1.07 x 10 ;− 1.04 x 10 =−

1.* S&eci!ic (ra+ity

t is the ratio of specific weight of the substance to the specific weight of water

at +@'. 1 convenient method to measure sp. gravity is by means of a hydrometer. t

is dipped into the liquid and a calibrated scale gives the specific gravity. t should be

noted that specific gravity is a dimensionless number and its value for a particular

substance is the same regardless of the system of units. t is abbreviated as (s).

w

f

oC at water of weight specific

fluid of weight specific gravitySpecific

ρ

ρ ==

+ : : : : :

: : : :

The specific gravity can also be e$pressed as ratio of mass density of the substance to

mass density of water at +@'.

+


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E,a&le 1.1

1 tan of glycerol has a mass of 4AAg and volume of A.B?m ;. CetermineD

(a) 6eight of glycerol

(b) Censity

(c) *pecific weight

(d) *pecific gravity

Solution-

(a) From 2ewton&s %aw9

68 mg

Thus, 68 4AA $ B. 8 .>=2

(b) From equation (.4)

;3,4=?B?.A

,4AAmkg

V

m === ρ

(c) From equation (.;)

;

3;E.,4E,.B,4=; mkN g =×== ρ γ

(d) From equation (.+)

4=.,,AAA

,4=?===

W

s s ρ

ρ

1. /iscosity

?


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y

v

∂

∂8 velocity gradient

The shear force, FC acting on the lower plate surface is given by9

! " ×= τ

6here 18 surface area of the lower plate

The unit of dynamic viscosity, I is * unit is (2-sm -4 or as). 0inematics viscosity

which is usually, denoted by the #ree letter 7nu& (v) is determined by dividing

dynamic viscosity (I) by mass density of the fluid ( ρ ). n the equation form9

ρ

µ ν = (.=)

6hen the fluid is at rest the velocity gradient dv#dy is "ero and therefore no shearing

force e$ists. The viscosity varies with temperature therefore values of I for given

fluid are usually tabulated at various temperatures. There are e$perimental methodsto calculate viscosity. Jne such e$perimental method is Falling *phere iscometer.

n this method a sphere of nown diameter is dropped into a liquid. Ky determining

the time required for the sphere to fall through a certain distance, its terminal velocity

(v) can be calculated. The stoes equation can be written as

−= ,

,E

4

ρ

σ

ν

gd v

6here d 8 diameter of sphere

L 8 sphere density

< 8 fluid density

M 8 inematics density

>


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1 number of viscometers are available in the maret. These viscometers are electronic

devices with digital panel and measured viscosity most of the liquids such paint,

lubrication oil, polymer compound, chemical compositions etc.

Fluids obeying 2ewton&s law of viscosity (equation .=) and for which I has a

constant value are nown as 2ewton&s fluids. /ost common fluids such as air, water

and oil come under this category for which shear stress in proportional to velocity

gradient. The fluids that do not obey 2ewton&s law of viscosity are nown as non-

2ewtonian fluids such as human blood, lubrication oils, molten rubber and sewage

sludge etc.

1 general relation between shear stress and velocity gradient for non-2ewton&s

fluids may be written as9

n

dy

dv $

+=τ (.>)

6here 1 and K are constants. Kased on the value of power inde$ 7n& non N

2ewtonian fluids are classified as9

seudoplastic (such as mil, cement, clay) nO

Kingham-plastic (such as sewage sludge, toothpaste) n8

Cilatent (such as lubrication oil, butter, printing in) nP

1 2ewtonian fluid is a special case of nonN2ewtonian fluid for which 1 8 A and

power inde$ n 8 .

The dynamic viscosity of various fluids at various temperatures is shown in Figure

.4.


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Figure .4 Cynamic viscosities versus Temperature

E,a&le 1.2

The viscosity of a fluid is to be measured by a viscometer constructed of two >?cm

long concentric cylinders as shown in Figure E.4. The outer radius of the inner

cylinder is ?cm, and the gap between the two cylinders is A.4cm. The inner cylinder

is rotated at 4AArpm, and the torque is measured to be A.2m. Cetermine the viscosity

of the fluid.

Figure E.4

B

1 b s o l u t e v i s c o s i t y

, 2 . s 3 m 4


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Solution-

The shear force, R% y

v ! " π µ τ 4×

∂

∂=×=

6here shear force can be calculated by

N R

Tor&ue ! " ;;.?

,?.A

E.A===

8 4QRf 8 4(;.+4)(A.?)(4AA)3=A 8 ;.+ m3s

smt

uV

y

v3=>.4=,=

AA,4.A

A,+.;=

−=

−=

∂∂

Thus, the dynamic viscosityD

43AA4B.A

)=>.4=,=)(>?.A)(,?.A(4

;;.?m Ns==

π

µ

1.0 Co&ressi$ility and ul 3odulus

'onsider a mass of fluid m whose initial pressure and volume is and

respectively. %et the fluid be compressed by application of force such that final

pressure is Sd and volume reduced to Nd. !ence, change in pressure is dp and

change in volume is Nd. olumetric strain in defined as change in volume divided

by original volume and is Nd3. The bul modulus denoted by and is defined as

change in pressure to volumetric strain9

8 'hanges in pressure3olume strain

or

dV dP V k −= (.)

A


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%et mass of fluid is mD

m 8


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8 (.+)

(ii) For an adiabatic process where no heat is allowed to enter or leave during

compression the relation between pressure and density is given by

.const P

=γ

ρ (.?)

1fter differentiation will give9

γ ρ ρ

P

d

dP = (.=)

where 8 ratio of specific heats at constant pressure and at constant volume or 8

'3'

1gain substitute d3d< into equation (.) will give9

8 (.>)

The ratio of adiabatic bul modulus is equal to the ratio of specific heat of fluid as

constant pressure to that at constant volume. For liquids is almost equal to one, but

for gases the difference is large for e$ample for air 8 .+.

1.4 3ac) no. and Co&ressi$ility

/ach. 2o. is defined as ratio of velocity of flow (v) to local velocity of sound

(a) and is a measure of compressibility effects.

a

v M = (.)

The velocity of propagation of sound waves in a fluid, flow is e$pressed as

4


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ρ d

dP a = or

ρ d

dP a =4 (.B)

*ubstituting the value of d3dρ in equation (.) we get

4ak ρ = (.4A)

*ubstituting value of a in equation (.4) we get

k

v M

ρ 4= (.4)

For liquids the bul modulus is large and velocities small and, hence, /ach. 2o. is

negligible or effect of compressibility is neglected. #as velocities are high bul

modulus is low, and hence, /ach. no. is high and compressibility cannot be neglected.

#ases can only be treated as incompressible if pressure changes are small and /ach.

no. is less than A.;.

1.5 Sur!ace tension

The molecules of the liquid are attracted by the molecules of the same liquid

by a force nown as 7'ohesion&. This force eeps the molecules bonded together.

The force of attraction between molecules of two different liquids that do not

mi$ each other or between liquids molecules and solid boundary containing the liquid

or between molecules or liquid on side and molecules of air (or gas) on the other side

is nown as 7adhesion&.

;


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Figure .;

Figure .; is shown a molecule of liquid at the surface is acted on by

imbalance cohesion and adhesive forces giving rise to surface tension. t is

commonly denoted by #ree letter, 7sigma& (σ) and is defined as force per unit length

of the surface. n the equation, it can be written as

%

! =σ (.44)

The units of L in * units will be 23m. n many engineering problems surface

tension forces are very small compared with other forces acting on the fluid and may

therefore be neglected. !owever, surface tension can cause serious errors in capillary

effects particularly in manometer.

For a droplet or a half bubble, the surface tension effect can be illustrated by

analy"ing a free-body diagram as shown in Figure .+.

Figure .+

The pressure force e$erted in the droplet is given by

+


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4 R P ! π =

The force due to surface tension is

σ π R ! 4=

The pressure force and tension must be balance each other9

σ π π R R P 44 =

R P

σ 4= (.4;)

1.6 Ca&illarity

f a small diameter glass tube is inserted into water through a free surface the

water will rise in the tube. This phenomenon is nown as capillarity and is caused by

cohesive force of the liquid molecules and adhesion of liquid surface to solid glasssurface.

The rise in level of the capillarity tube will depend on L and angle of contact, U

as shown in Figure .?.

Figure .?

?


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%ength of line of contact of the liquid with the tube 8 Qd

ertical component of the surface tension force 8 (Qd).L.cosU

6eight of column of liquid, 68 γ π

hd 4

+

Thus, for the equilibrium of surfaces tension and gravity forces requires as

γ π

θ σ π hd d 4

+cos =

d h

γ

θ σ cos+= or

d P

θ σ cos+= (.4+)

'onsequently, when one does not wish a meniscus to rise appreciably in a

tube, a large value of diameter is chosen. t is believed that trees, even very tall ones,

send water to their highest branches by means of capillarity effects. !ence, capillary

passages must be e$tremely fine. n water and certain other liquids that e$hibit

capillarity the meniscus is 7concave&. These liquids wet the glass and angle of contact

U is less than BA@. n some other liquids such as mercury the meniscus is conve$. The

liquids do not wet the solid surface and angle of contact U is more than BA@. #lass

tubes are commonly used in manometer and capillary action is a serious source of

error in reading levels in such tubes. They should have as large a diameter as is

conveniently possible to minimi"e errors due to capillarity.

1.17 /a&or Pressure

'avitation is given to the phenomenon that occurs at the solid boundaries of

liquid streams when the pressure of the liquid is reduced to vapor pressure of the

liquid at the prevailing temperature (Table .;). 1ny attempt to reduce the pressure

still further merely causes the liquid to vapori"e more quicly and clouds of vapor

bubble form. The bubbles of vapor formed in the region of cavitation move

downstream to a region

=


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of higher pressure where they collapse (see Figure .?). t is repeated formation and

collapse of vapor bubbles which can have damaging effects upon the walls of the

solid surface. The actual time between formation and collapse may not be more than

3AA of a second, but dynamic force caused by this phenomenon may be very severe.

t is only a matter of having enough bubbles formed over a sufficient period of time

for the destruction of the metal begins. 'avitation may occur in pumps, turbines,

hydrofoils, propellers, and in venture-meters. n the case of turbines, cavitation is

most liely to occur on the blade surfaces near the tail race where as for pumps it is

most liely to occur at inlet to the impeller. 'avitation can also occur if a liquid

contains dissolved air or gases, since solubility of gases in liquid decreases as the

pressure is reduced. #as and air bubbles will be released as vapor bubble with the

same damaging effects. 'are should be taen to avoid cavitation as far as possible but

if this proves impracticable, than the parts liely to be affected by cavitations should

be constructed of especially resistant metals such as stainless steel.

Table .; *aturation vapor pressure of water

Temperature (V') apor ressure (a)

A =?

A 4;A

4A 4;+A

+A >+AA

=A 4AAAA

A +>+AA

AA A;AA

>


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Figure .? 'avitation phenomena inside the no""le

Pro$les

. Cetermine the density of air, hydrogen, and carbon dio$ide at an absolute

pressure of ;AA23mW and a temperature of ;.X'.

(;.;? g3m;, A.4;; g3m;, ?.A g3m;)

4. 'alculate the specific weight and density of air at absolute pressure of ++? a

and a temperature of ; X'.

(+.B= g3m;, +.B 23m;)

;. f the volume of liquid decreases by A.4Y for an increase of pressure from ==>

23mW to ?=B= 23mW, calculate the bul modulus of elasticity of the liquid.

(+,++.?/a)

+. 1 soap bubble ? mm in diameter has an internal pressure in e$cess of outside

pressure of 4.A= $ A-4 a, calculate the tension in the soap film.

(A.;;;23m)

?. f the pressure inside a water droplet is A.4 a in e$cess of e$ternal pressure,

and given surface tension of water in contact with air at 4AX' is equal to A.A>;=

23m, determine the diameter of the droplet.

(.+>4mm)

=. 1ir is introduced a no""le into a tan of water to from a stream of bubbles. f

the bubbles are intended to have diameter of 4 mm, determine the pressure of air


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at the no""le e$ceed that of surrounding water (given tension of water 8 A.A>;=

23m).

(+>.423m4)

>. The air in an automobile tyre is at 4.B+; /a absolute at 4=.=X'. 1ssuming no

change in the volume of air if the temperature rises to rises to =4.4X', determine

the air pressure.

(;.+B/a)

. 1 gas occupying a volume of ;AA liters at a certain temperature and pressure of

A.;+ 23mmW is compressed isothermally to ?A liters. 'alculate the initial and

final bul modulus of elasticity.

(A.;+23mm4, A.=23mm4)

B. 1t a certain point in a fluid, the shear stress in A.44 23mW and the velocity

gradient is A.=> sec. f the mass density of the fluid is 4B; g3m ;, determine

the inematic viscosity.

(=.;> $ A-+m43s)

A. 1 liquid flows between two fi$ed parallel boundaries. The velocity distribution

near to lower wall is given in the following tableD

y (mm) v (m3s)

.A .AA

4.A .BB

;.A 4.B

+.A ;.AA

?.A ;.AA

Cetermine the ma$imum and minimum shear stresses (The dynamic viscosity of

fluid is A.A?as).

(?A23m4, A)

. Two plates are arranged as in Figure Z in the liquid. The top plate is moving

with the velocity of A.?m3s and the middle plate is moving with the velocity of

4m3s in opposite direction. The area of both plates area A.4?m 4. lot the velocity

profile on all surfaces and determine the force acting on the middle plate (Tae

the viscosity of liquid is A.Aas).

B


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Figure Z

(;.>?2)

4. /ercury does not adhere to a glass surface, so when a glass tube immersed in

a pool of mercury, the meniscus is depressed, as a shown in Figure Z4. The

surface of mercury is A.?+ 23m and the angle contact is +AX'. 'alculate the

depression distance in a mm glass tube.

Figure Z4

(.mm)

;. The vapor pressure of water at AAX' is A 23mW, because water boils under

these conditions. The vapor pressure of water decreases appro$imately linearly

with decreasing temperature at a rate of ;. 23mW3X'. 'alculate the boiling

temperature of water at an altitude of ;AAA m, where the atmospheric pressure is

=B 23mW absolute.

(B.>o')

-oooJJJooo-

4A


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CHAPTER 2

H8DROSTATIC PRESSURE

2.1 Introduction

1 fluid at rest is characteri"ed by absence of relative motion between adacent

fluid layers. nder such a condition, the velocity gradient is "ero and there is no

shear stress, therefore, viscosity of fluid has no effect on fluids at rest. Kut fluids at

rest do e$ert forces on the solid boundary. 0nowledge of force variation or more

appropriately pressure variations in a static fluid is important to an engineer.

There are so many practical e$amples of fluids at rest such as water retained

by a dam, an overhead tan supplying water to the public, gas or fuel in a tan truc.

The obect of this chapter is to measure pressure variations in a static fluid to discuss

pressure and pressure measurement and laws of fluid pressure.

2.2 Hydrostatic Pressure

n the gravity environment, the static pressure in the fluid is proportionally increased

linearly with the depth and always acted perpendicular or normal to the surface as

shown in Figure 4..

Figure 4. *tatic pressure increased linearly with the depth

The hydrostatic pressure at every location with the depth, h from the free surface is

given as

4


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gh P P atm ρ += (4.)

by neglecting atm equation 4. becomes

gh P ρ = (4.4)

and this is called gage pressure that always use in the calculation of static pressure. 1s

a e$ample, for point and 4 in Figure can be written asD

,, gh P ρ = (4.;)

44 gh P ρ = (4.+)

Thus, the pressure different between points and 4 can be written asD

h g P P P ∆=−=∆ ρ ,4 (4.?)

where [h8h4-h

The fluid pressure at rest is constant along the hori"ontal line. n other words, the

pressures for all points with similar depth have same magnitude, and it is independent

of the shape or cross section area of the fluid container (see Figure 4.4). This may be

stated as the equal level-equal pressure principle that forms the basis for many

pressure-measuring devices such as barometer and manometer. t also contributes to

the operation of a hydraulic ac that a small input force creates a larger output force.

Figure 4.4 ressures are equal for all points along hori"ontal plane

44


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E,a&le 2.1

1 tan is connected to a vertical tube is filled with water (8 BA23m 4). CetermineD

(a) 1bsolute pressure at levels 1, K, ', C, E, and F

(b) #age pressure at levels 1, K, ', C, E, and F

Figure E4.

Solution-

(a) 1bsolute pressure

From equation 4., it can be written as

oint 1D 18 atm S h1, where h18 A

8 A.;(A;) S BA(A)

8 A.;a

oint KD K8 atm S hK, where hK8 A.>m

8 A.;(A;) S BA(A.>)

8 A.4a

oint 'D '8 atm S h', where h'8 4.?m

8 A.;(A;) S BA(4.?)

8 4?.a

oint CD C8 atm S hC, where hC8 ;.+m

4;


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8 A.;(A;) S BA(;.+)

8 ;+.>a

oint ED E8 ' since hE 8 h'

Thus, E8 4?.a

oint FD F8 atm S hF, where hF8 ;.m

8 A.;(A;) S BA(;.)

8 ;.=a

(b) #age pressure

g18 1 - atm 8 Aa

gK8 K - atm 8 =.>a

g'8 ' - atm 8 4+.?a

gC8 C - atm 8 ;;.+a

gE8 E - atm 8 4+.?a

gF8 F - atm 8 ;>.;a

2ow consider a simple hydraulic ac in Figure 4.;D

Figure 4.; !ydraulic 5ac

4+


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sing equilibrium principle, the pressure e$erted by the fluid on a plunger or a

piston is written as

! P = (4.=)

From which it follows that,

,,

! P = and

4

44

! P =

*ince the pressure at points and 4 is equal (i.e. same level), we have

4

4

,

,

!

! = (4.>)

which can be used to solve for F4 if F, 1 and 14 are given, or vice versa.

E,a&le 2.2-

For a hydraulic ac as shown in Figure E4.4, determine the weight that could be

lifted if the +AA2 input force F is applied to the plunger. The diameters of the

plunger and the piston areAmm and >?mm, respectively

Figure E4.4

4?


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Solution-

From equation 4.> above9

,

4,4

! ! =

where4?4

, ,AE??.>+3)A,.A( m −×== π

4?44 ,A+4.++3)A>?.A( m −×== π

Therefore, F4 8 44.?2

2.2 Pressure and )ead

There are several types of pressure-measuring devices available. Cevices such as a

barometer, Kourdon gage, and manometer are among commonly used instruments

to measure, either a gage, vacuum or absolute pressure. 1 pressure gage measures

a pressure relative to the local atmospheric pressure when the pressure is above the

atmospheric pressure whereas a vacuum gage is used when the pressure is below

the atmospheric pressure.

Figure 4.+ Type of pressure

4=


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2.% 3ercury aroeter

1 typical mercury barometer is shown in Figure 4.? and used to measure

atmospheric pressure. t consists of a vertical closed glass tube with a column of

mercury inside. The barometer is constructed so as to avoid having any trapped air at

the end of the tube. t can be assumed that the space between the mercury and the end

of the tube contains a vacuum with "ero pressure.

1t the bottom of the column, mercury is contained in a small reservoir in a

small reservoir. The pressure acting on the surface of the reservoir is atmospheric

pressure atm. Thus, the atmospheric pressure can be calculated as

P atm ( ρ gh (4.)

Figure 4.?

2.* T)e ourdon 9au(e

This is most common type of pressure gauge which is compact, reasonably

robust and simple to use. 1 curved tube of elliptical cross-section is closed at one end

is free to move, but the other end-through which the fluid enters is rigidly fi$ed to the

frame as shown in Figure 4.=.

4>


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Figure 4.=

6hen the pressure inside the tube e$ceeds outside pressure (usually

atmospheric), the cross-section tends to become circular, thus causing the tube

uncurve slightly. The movement of the free end of the tube is transmitted by a

suitable mechanical linage to a pointer moving over a scale. \ero reading is of course obtained when the pressure inside the tube equals the local atmospheric

pressure. Ky using tubes of appropriate stiffness, gauges for a wide range of pressure

may be made.

2. 3anoetric Pressure

The barometer analysis shows that vertical columns of liquid can be used to

measure pressure. The science of this measurement is called manometer.

There are different types of manometers with varying degrees of sensitivity which

embody the principle already derived and used for pressure measurement.

4


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2..1 Si&le U:tu$e 3anoeter

Figure 4.>

'onsider the -tube manometer connected via a small hole to a pipe (Figure

4.>) carrying a fluid of density ρ at pressure 1 (which is to be measured).

%et the open end of the -tube be subected to atmospheric pressure, atm.

1t the common surface K-' with the configuration as shown in the diagram

we have D

4B


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1 S ρgh 8 K 8 ' 8 C S ρ4gh4,

or

1 S ρgh 8 C S ρ4gh4 (4.B)

2ow 1 is the pressure to be measured () and C 8 atm.

Thus N atm 8 (ρ4h4 - ρh)g (4.A)

2..2 Di!!erential 3anoeter

This is used to measure the pressure differential between two fluid reservoirs

as shown in Figure 4..

1 S ρgh 8 C S ρ;gh; S ρ4gh4,

Jr differential pressure is given by

1 N C 8 (ρ;h; S ρ4h4 - ρh)g (4.)

Figure 4.

2..% In+erted U:tu$e 3anoeter

;A


31/70

1nother type of differential manometer as shown in Figure 4.B

K S ρgh 8 1

K S ρ4gh4 8 '

' S ρ;gh; 8 C

or C N ρ;gh; N ρ4gh4 S ρgh 8 1 (4.4)

or 1 N C 8 (ρh N ρ4h4 N ρ;h;)g (4.;)

Figure 4.B

2..* An Inclined 3anoeter

1n inclined manometer as shown in Figure + is used to achieve a greater

accuracy and sensitivity in pressure measurement. This is because the slight

pressure change could cause a noticeable change in %.

;


32/70

Figure 4.A nclined /anometer

1pplying the equal level-equal pressure principle, we have

8 #as

Kut 8 4 S h and h8 % sinU. The above equation then becomes,

( )θ γ sin4, % P P += (4.+)

E,a&le 2.%

-tube manometer containing a mercury (sg8 ;.=) as a woring fluid is connected to

a tan that contains air as shown in Figure E4.;. The other end of the manometer is

e$posed to the atmosphere. Cetermine the pressure in the tan if h8 A.+m.

Figure E4.;

Solution-

;4


33/70

*ince the density of air is neglected, the pressure at point 1 will correspond to

the pressure of air in the tan.

1 8 1ir ()

1pplying equal level-equal pressure principle, we have

1 8 K (4)

Kut

K 8 atm S !g h (;)

6here !g8 sg (!g at +deg.) 8 ;.= (BA) 8 ;;+=23m4

*ubstitute (;) and (4) into () will give

)+.A(,;;+,),A(,AA ; +=+= h P P )g atmudara γ

kPa P udara +.,?;=∴

E,a&le 2.*

The manometer is used to measure the pressure in the pipe and is connected to the

pressuri"ed tan containing gas and water as shown Figure E4.+. ressure gage

attached to the tan reads A a, determine the gage pressure in the pipe. (#iven h8

+Acm, h48 +cm, h;8 ?>cm, h+8 +?cm)

;;


34/70

Figure E4.+

From equal level-equal pressure,

"C $ P P P ==

Kut, ;h P P )g * " γ +=

6here )( 4h P P w ! * γ +=

1nd kPa P P gas ! EA==

1nd also +h P P oil $ γ +=

*ubstitute (4), (;) and (+) into ()9

+;4 )( hhh P P oil )g water gas γ γ γ +++=

1 8 A(A;) S BA(A.+) S;.=(BA)(A.?>)S A.(BA)(A.+?)

8 =>a.

E,a&le 2.

Figure E4.? show a three-fluid manometer containing oil (sg8 A.4), mercury (sg8

;.=) and water used to measure large pressure differences. Cetermine the pressure

;+


35/70

difference between 1 and K. #iven h8 +?cm, h48 >Acm, h;8 4?cm and h+8 4Acm.

2eglect the air density.

Figure E4.?

Solution-

"C P P P ==

but ,, h P h P P W W " * γ γ +=+=

and ,h P P P W * ! γ +==

1lso 4h P P )g + ! γ +=

but + ) P P =

1lso +; hh P P oil gasolin ) $ γ γ ++=

*ubstituting , 4, ;, +, ?, = and > give

+;4 hhhh P P oil gasolinW )g $ γ γ γ γ −−−=−

or )( +;4 h sg h sg hh sg P P oil gasolin )g W $ −−−=− γ

;?


36/70

8 (BA)](;.=)(A.>)-(A.+?)-A.>(A.4?)-A.4(A.4)^

8 ?.>a.

Pro$les

The 'rosby gage tester shown in the figure is used to calibrate or to test

pressure gages. 6hen the weights and the piston together weigh B.A 2, the

gage being tested indicates >Ba. f the piston diameter is 4Amm, what is

the percentage error e$ists in the gage_

Figure Z

];=.Y^

4. Two hemispheric shells are perfectly sealed together and the internal pressure

is reduced to AY of atmospheric pressure. The inner radius is ? cm, and the

outer radius is ?.? cm. The seal is located half way between the inner and

other radius. f the atmospheric pressure is AAa, what force is required to

pull the shells apart_

]=.?2^

;. f e$actly 4A bolts of 4.?-cm diameter are needed to hold the air chamber

together at 1-1 as a result of the high pressure within, how many bolts will be

needed at K-K_ !ere C 8 ?A cm and d 8 4? cm.

;=


37/70

Figure Z;

]? bolts^

+. The reservoir shown in the figure contains two immiscible liquids of specific

weights and $, respectively, one above the other where P $. 6hich

graph depicts the correct distribution of gage pressure along a vertical line

through the liquids_

p p p p

(a) (b) (c) (d)

Figure Z+

]b^

?. This manometer contains water at room temperature. The glass tube on the left

has an inside diameter of mm (d 8 .A mm). The glass tube on the right is

three times as large. For these conditions, the water surface level in the left

tube will be a) higher than the water surface level in the right tube, b) equal to

the water surface level in the right tube, c) less than the water surface level in

the right tube. *tate your main reason or assumption for maing your choice.

;>


38/70

Figure Z?

]a^

=. 1 tan is fitted with a manometer on the side, as shown. The liquid in the

bottom of the tan and in the manometer has a specific gravity (s) of ;.A, the

depth of this bottom liquid is 4A cm. 1 A-cm layer of water lies on top bottom

liquid. Find the position of the liquid surface in the manometer.

Figure Z=

];.;;cm^

>. Cetermine the gage pressure in pipe 1.

Figure Z>

]4B.>a^

;


39/70

. 'onsidering the effects of surface tension, estimate the gage pressure at the

center of pipe 1.

Figure Z

]=Ba^

B. 6hat is the pressure at the center of pipe K_

Figure ZB

]-.Aa^

A. Find the pressure at the center of pipe 1.

;B


40/70

Figure ZA

]B.+>a^

. The top of an invented -tube manometer is filled with an oil of specific gravity

of .A. Cetermine the pressure difference in a between two points 1 and K at the

same level at the base of the legs when the difference in water level h is >?mm.

]>.;=a^

+A


41/70

CHAPTER %

FLUID STATICS

%.1 Introduction

n the previous chapter it was noted that the hydrostatic pressure parts of fluid static.

n this chapter we shall develop equations to calculate the magnitude and location of

forces acting on submerged surfaces. 6e shall also e$amine problems involving

ability of floating bodies. *uch analysis of fluid helps in the design of dams, gates,

ships and submarines.

n this chapter, the submerged surfaces are divided into the following types9 a) straight

hori"ontal and vertical surfaces, b) straight inclined surfaces, and c) curved surfaces.

The analysis of hydrostatic force on submerged surfaces assumes the following

conditions.

. Force is always perpendicular to the surface since there is no shear

stress for fluids at rest.

4. ressure varies linearly with depth for incompressible fluid.

;. The resultant fluid force passes through the point called the center of

pressure

%.2 Hori;ontal and +ertical sur!ace

1ny given depth h, the resultant fluid force FR on the hori"ontal and vertical

surfaces may be represented as shown in Figure ;..

Figure ;.

+


42/70

/agnitude of the resultant force FR at the bottom of the tan, Figure ;. is given

by

P ! R = (;.)

where 1 is the area of the surface upon which the pressure is acting. For the

vertical

*urface. Figure ?.(b), we have

P ! R

4

,= (;.4)

%.% Inclined Sur!ace

The surface tilted at an angle U from the hori"ontal is shown in Figure ;.4. The

pressure variation and hence the resultant hydrostatic force FR , on the surface can be

presented as shown in Figure ;.4.

Figure ;.4

+4


43/70

/agnitude FR can be written as

P ! C R = (;.;)

where ' is a pressure at the centroid surface (point ') and is written asD

C oC gh P P ρ += (;.+)

This pressure is equivalent to the average pressure on the surface.

we can see that

a) the magnitude of the force is independent of the angle U,

b) it is perpendicular to the surface, and

c) it passes through the point of application called the center of pressure

(point ).

n many cases, the pressure at point J is the atmospheric pressure and may be

ignored in the analysis. This simplifies equation (;.+) to

C C gh P ρ = (;.?)

2oticed that when U 8 BA@ , the surface becomes vertical and when U 8 A@ , the

surface becomes hori"ontal.

%.%.1 Center o! Pressure

2e$t we need to determine the line of action of the resultant force FR distance y and

hence its location below the free surface, hp. t is found that T action y is given by

C

- P

y

. y = (;.=)

where - . is the moment of inertia of the area with respect to the $ a$is. sing

parallel a$is4

C - - y . . += , equation (;.=) can be rewritten as

+;


44/70

C

- C P

y

. y y += (;.>)

where -

. is the moment of inertia of the area with respect to the centroid a$is. -

.

commonly used shapes is given in Figure ;.;. Jbserve that the center pressure

below the centroid ' of the surface since A3 >C - y . ,

E$pressing y8 h3sinU and y'8 h'3sinU, we obtain from equation ;.>

C

C P h

. hh +=

6here θ 4

sin - . . = . n some te$tboos, the location of the centroid of the submerge

surface from the fluid free surface, C h is denoted as h and y' as y .

Figure ;.; shows the centroid y for the simple shape obects and Figure ;.+ shows

the second moment of area - . for the simple shapes.

++


45/70

+?


46/70

Figure ;.;


47/70

/ a 0 ;4, −=

6here 1, 14, and 1; are the areas and $, a, and b are the centroid for the

above shapes respectively.

2. A,ial 3oent o! Inertia o! an Area

The a$ial moment of inertia of an area is the summation of the a$ial momentsof inertia of the elements.

d y . 0 ∫ = 4

d 0 . y ∫ = 4

%. Parallel A,is T)eore

The parallel a$is theorem states that the a$ial moment of inertia of an area

about any a$is equals the a$ial moment of inertia of the area about a parallel

a$is through the centroid of the area plus the product of the area and the

square of the distance between the two parallel a$es.

+>


48/70

4 m . .

0 0 += ′

4 n . . y y += ′

%.* 3oent o! a Force

The resultant force FR and its location y or h, are determined, thus the calculation of

the moment needed to overcome the resulting moment due to this force about certain

point. 'onsider a force F acting perpendicular to the body at point K as shown in

Figure ;.+.

Figure ;.+

This force tends to rotate the body about point 1 in a counter-clocwise direction. Thetendency of a force to rotate the body is called the moment of a force about that point.

The magnitude of this moment about point 1 is given byD

!d M =

+


49/70

where d is the perpendicular distance from point 1 to the point of application of the

force. The direction of the moment is indicated using either clocwise or counter-

clocwise. n this case, /1 is in the counter-clocwise direction Figure ;.+.

E,a&le %.1-

'onsider a rectangular gate 1K hinged along 1 to support the water pressure as shown

in Figure E;.. CetermineD

(a) The resultant hydrostatic force e$erted on the gate 1K

(b) The center of pressure

(c) The force acting on the stopper at K

The gate width is ;m.

Figure E;.

Solution-

(a) Resultant force

FR 8 h' 1 ()

where 18 =(;) 8 m4 and h' 8 + S ; 8 >m

From equation ()9

+B


50/70

FR 8 BA (>) () 8 .4+/2

(b) %ocation of center of pressure

C

C P h

. hh +=

where+4;4 ?+)BA(sin)=)(;(

,4

,sin m . . - =°== θ

mh P +;.>)>)(,E(

?+> =+=

(c) The force acting at the stopper K

Taing moment at 1,

)+( −×=× P R $ h ! l !

MN ! $

>,.A=

+;.;4+.,=

×=

E,a&le %.2-

Cetermine the friction coefficient required to hold the dam from moving as shown in

Figure E.4. The normal force of dam is ?A/23m.

Figure E.4

?A


51/70


52/70

6 8 the weight of the enclosed volume supported by the curved 1K and

that 6 8


53/70

44

vh R ! ! ! +=

where Fv8 F S 6 and Fh8 F4

For each force components9

h ! C γ =,

where 18 4.+(;) 8 >.4m4 and h'8 ;.=m

kN ! ;.4?+)4.>)(=.;(BE,A, ==∴

V gV W γ ρ ==

6here;44 ?>.,;);()+.4(

4

,

+

,ml r V === π π

kN W 4.,;;)?>.,;(BE,A ==∴

h ! C γ =4

6here 18 4.+(;) 8 >.4m; and h'8 ;.= S .4 8 +.m

kN ! ;;B)4.>)(E.+(BE,A4 ==∴

Therefore, Fv8 4?+.; S ;;.4 8 ;>.+2

Fh8 F48 ;;B2

1nd

FR 8 (;>.+4 S ;;B4)34 8 ?+.2

E,a&le %.*

?;


54/70

Cetermine the force per unit width , required to hold the gate as shown in Figure

E;.+.

Figure E;.+

Solution-

!ydrostatic force acting on the control volume9

mkN h ! ) 3=4.,B)4)(,(BE,A === γ

Ky taing moment at hingeD

mkN P 3>.,??.4

=4.,B4=

×=

%.0 uoyancy= Floatation and Sta$ility

6hen a body is completely submerged or floating in a fluid, the resultant fluid

force acting in an upward direction on the body is called the /uoyancy force1 This

force tends to lift the body upward and its e$istence is due to the fact that () the

fluid pressure increases with depth, and (4) the pressure force acting from below is

larger than the pressure force acting from above.

%.0.1 uoyancy and Floatation

'onsider a body submerged completely in a fluid as shown in Figure ?.4.

The resultant force on the bottom surface of the body is greater than the

resultant force on the top surface of the body.

?+


55/70

Figure ;.= Kuoyancy force acting on a submerged body

The difference between these two forces is the buoyant force which gives

the net upward force. This buoyant force will pass through the point called

center of /uoyancy or the centroid of the displaced volume, C K which

happened to be at the same point as the center of gravity of the body, # in

the case of a completely submerged body. 6riting force balance on the

body, we have

h h h ! ! ! f f f top/ottom $ γ =γ −γ =−= ,4

!owever, the term h is basically the volume of the fluid body (or volume

of the displaced fluid by the body). E$pressing this volume as ( h, we

may write the equation above as

V ! f $ γ =

where V is volume of the displaced fluid1 Thus, we conclude that the

buoyant force acting on the body is equal to the weight of the fluid

displaced by the body and therefore, proportional to the density of the fluid.

n the case of a floating body, Figure ;.>, the weight of the entire body must

be equal to the buoyant force, which is the weight of the fluid whose volume

is equal to the volume of the su/merged portion of the floating body.

??


56/70

Fi(ure %.4 Kuoyancy force acting on a floating body

Thus, for a floating body in static equilibrium, we may write

W ! $ = or total /ody su/ f V V γ =γ

where total V 8 total volume of the body or volume of the entire body

su/V 8 volume of the submerged, portion of the body, which is equal to

the

volume of the displaced fluid (Figure ;.>(c))

Rewriting the above equation as

f

/ody

f

/ody

total

su/

V

V

ρρ=

γ γ =

6e observe that the body is completely submerged when the density ratio is

equal to that is when the density of the body is equal to the fluid density.

6e can conclude that a body immersed in a fluid will

() Rise to the surface of the fluid and float when the density of the bodyis less than the fluid density,

(4) Remain at rest at any point in the fluid when its density is equal to the

fluid density, and

(;) *in to the bottom when the density of the body is greater than the

fluid density.

?=


57/70

Fi(ure %.5 *ituations of a body immersed in a fluidD float, remain at rest, or

sin depending on the density of the body relative to the fluid density

E,a&le %.

1 cuboids has the si"e of b`h`w is floating in the water ( air 8 B.23m;) as

shown in Figure E;.?. Cetermine the portion of body that above the water

surface, a if 5asad8 ;.?23m;, b8 =m, h8 m dan w8 ?m.

Figure E;.?

Solution-

w/hV w/ 2ah3 V

total

su/

=−=

From the relationship,

f

/ody

f

/ody

total

su/

V

V

ρ

ρ=

γ

γ =

−=γ γ

−= BEAA;?AA

,E, 23 ha f

/ody

?>


58/70

a ( 4 1 5 6 m

.+)=.?4;E,.A4A(

E,.B)=.?4;E,.A4A()=.?4;)(E,.B4.,( sma =

×+××+−×

=

?


59/70

%.0.2 Sta$ility o! Floatin( and Iersed odies

*tability is an important issue for a floating and immersed body such as in

the design of a ship, submarine and barge (Figure ;.B).

Figure ;.B 1 barge used for transportation at 0uala 0urau

n the vertical direction under static equilibrium, the weight and the buoyant

force on a floating or immersed body will balance each other, and such body

is said to be vertically sta/le (vertical stability).

For the rotational stability, the condition depends upon the relative locationof the center of gravity # of the body and the center of buoyancy ' K, Figure

;.A. 1 floating or immersed body is stable if the point # is below point ' K,

Figure ;.B(a). nder this condition, the body will return to its original stable

position due to the restoring moment or couple produced by the body.

?B


60/70

Figure ;.A *tability of an immersed body (a) stable with restoring couple

as shown in (d), (b) neutral, and (c) unstable with overturning

couple as shown in (e).

!owever, a floating body will still be stable even if # is above 'K. *ee

Figure ;.. This is because the body will still produce the restoring moment

since the centroid of the displaced volume is now shifted to the side to point

'K. The lines of action of the buoyancy force before and after rotation will

meet at the point called the meta7center C /1

The distance between # and '/ is called the metacentric height h/ and is

used as a measure of sta/ility for a floating body. The larger it is, the more

stable the floating body will be. 2ote that the floating body is unstable if the

point '/ is below point #.

Figure ;. *tability of an immersed body

To determine whether the floating body is stable or not is given by thefollowing equationsD

=A


61/70

V

. C C $ M =−

and for the body is in stable condition when it rotate at certain angle resulting

from the reversed moment, it must has the following conditionsD

A>−= + M M C C h *table

A=−= + M M C C h 2eutral

A


62/70

olume of displacement, ;;

+E>E,A4?

,A?AAAmV =

×=

1nd, mV

.

C C $ M 4.,+E>E

?E+A

===−

Thus, metacentric height, m 0

h M ;++.A?sin

A;.A

?sin===

1nd the vertical distance between '# and 'K is

m 0

C C $ M E?>.A

?tan

A;.A4.,

?tan

)( =−=−−

%.4 Li>uids in Relati+e E>uili$riu

f a vessel containing a liquid is at rest or moving with constant velocity the liquid is

not affected by the motion of the container, but if the container is given

continuously acceleration this will be imparted to the liquid which will tae up a

new position and come to rest with respect to its container and come to rest

relative to the vessel. The liquid is in relative equilibrium and is at rest with

respect to its container. There is no relative motion of the particles of the fluid

and therefore no shear stress. Fluid pressure is everywhere normal to the surface

on which it acts.

%.4.1 Hori;ontal Acceleration

'onsider a particle J of mass m on the free surface of the liquid as in Figure

;..

=4


63/70

Figure ;.4

*ince the particle is at rest relative to the tan, it will have the same

acceleration a, and will be subected to an accelerating force, F

a g

W ma ! ==

where 68 weight of particles.

The accelerating force, F is the resultant of the weight 6 of the particle acting

vertically downward and the pressure force, R acting normal to the free

surface due to surrounding fluid.

For equilibrium F8 6 tanθ, where θ is the angle of the free surface to the

hori"ontal.

Thus,

g

a=θ tan

and is constant for all points on the surface.

%.4.2 /ertical Acceleration

1s the acceleration is vertical the free surface will remain hori"ontal. 'onsider

a vertical prism of height h (Figure ;.;) e$tending from the free surface to $

and let the pressure intensity at $ be .

=;


64/70

Figure ;.;

1ccelerating force at $, F8 Force due to pressure N weight of prism

8 1 - ρgh1

Ky 2ewton&s second %awD

F ? ass acceleration

ah ! ×= ρ

Thus,

ah gh P ×=− ρ ρ

or

+= g

a gh P , ρ

%.4.% Forced /orte,

The liquid in the vessel is rotated with the vessel at the same angular velocity,

ω. 1 particle on the free surface will be in equilibrium under the action of its

weight 6 (Figure ;.+), the centrifugal accelerating force, F acting

hori"ontally and the fluid reaction R.

=+


65/70

Figure ;.+

For any point at radius $ and a height y from the lowest point J, if θ is theangle of inclination of the water surface to the hori"ontal,

W

!

d0

dy ==θ tan

For a constant value of ω, F will vary with $, since the centrifugal acceleration

is ω4$ dan F8 (63g)ω4$.

The surface angle therefore varies and,

g

0

d0

dy 4

tan ω

θ ==

ntegrating will give,

malar g

0d0

g

0 y

0

+== ∫ 444

A

4ω ω

f y is measured from 1K, y8 A when $8 A and

g

0 y

4

44ω

=

The water surface is therefore a parabolic revolution.

E,a&le %.5

1 tan containing water moves hori"ontally with a constant linear acceleration

a of ;m3s4. The tan is ;m long and the depth of water when the tan is at rest

is .?m. 'alculateD

(a) The angle of water surface when the tan is .?m

=?


66/70

(b) The ma$imum pressure intensity on the bottom

(c) The minimum pressure intensity on the bottom

Solution-

#ivenD a8 ;m3s4, h8 .?m

(a) The angle of water surface to hori"ontal, θ

o

g

a,>tan

, == −θ

(b) The depth at 1 (ma$imum pressure),

mhhho

B=.,,>tan?.,?.,tan =+=+= θ

(c) The depth at K (minimum pressure),

mhhho

$ A+.,,>tan?.,?.,tan =−=+= θ

Pro$les

. 'onsider the two rectangular gates shown in the figure. They are both the

same si"e, but one 3+ate 1) is held in place by a hori"ontal shaft through

its midpoint and the other 3+ate K) is cantilevered to a shaft at its top. 2ow

consider the torque required to hold the gates in places as ! is increase.

'hoose the valid statement(s)D a) T increases with !. b) T $ increases with

!. c) T does not change with !. d) T $ does not change with !.

Figure Z

]b and c^

4. Find the force of the gate on the bloc.

==


67/70

Figure Z4

]A+.+2^

;. 2eglecting the weight of the gate, determine the force acting on the hinge of

the gate.

Figure Z;

]4.;/2^+. The rectangular gate measures =m by +m and is pin-connected at point 1. f

the surface on which the gate rests at 1 is frictionless. 6hat is the reaction at

1_ 2eglect the weight of the gate.

Figure Z+

]??>2^

?. 1 4m $ 4m gate is installed at the end of water reservoir, as shown, and is

hinged at the top. The gate hinge is # m below the reservoir water surface.

The gate is connected to a rectangular tan of water which is 4 m wide (into

=>


68/70

the paper) and filled with = m of water. The weight to the tan is negligible.

!ow long (%) would the tan have to be open the gate_

Figure Z?].;m^

=. The triangular gate 1K' is pivoted at the bottom edge 1' and closes a

triangular opening 1K' in the wall of the tan. The opening is + m wide (6 8

+ m) and B m high (! 8 B m). The depth d of water in the tan is A m.

Cetermine the hydrostatic force on the gate and the hori"ontal force required

at K to hold the gate closed.

Figure Z=

];4;.>2^

>. Estimate the depth d needed for the rectangular gate to automatically open if

the weight 68 =A2 as shown in Figure Z>. The gate is +m high and 4m

wide. 2eglect the weight of the gate.

=


69/70

Figure Z>

];.4+m^

. For the plane rectangular gate (% $ 6 in si"e), Figure (a), what is the

magnitude of the reaction at 1 in terms of w and the dimensions % and 6_ For

the cylindrical gate, Figure B(b), will the magnitude of the reaction of 1 be

greater than, less than, or the same as that for the plane gate_ 2eglect the

weight of the gates.

(a) (b)

Figure

]A.?Aw6l4^

B. The floating platform shown is supported at each corner by a hollow sealed

cylinder m in diameter. The platform itself weighs ;A2 in air, and each

cylinder weighs .A2 per meter of length as in Figure ZB. 6hat total

cylinder length % is required for the platform to float m above the water surface_ 1ssume that the specific weight of the water is A,AAA 23m. The

platform is square in plan view.

Figure ZB

=B


70/70

]4.4+m^

A. The coffee cup in Figure ZA is removed from the drag race, placed on a

turntable, and rotated about its central a$is until a rigid-body mode occurs.

CetermineD

(a) the angular velocity which will cause the coffee to ust reach the lip of the

cup

(b) the gauge pressure at point 1 for this condition

(Tae the density of coffee as AAg3m;)

Figure ZA

];=.>rad3s, A>.Ba^

>A

chapter 1-3 fluid mechanics final

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