1

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 work 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. Hydraulic

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

mechanization. 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. Gases expand

whereas liquids do not. Liquid have no shape of its own but rather take 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. Gases expand and occupy full volume of the

container. Gases are compressible which means their volume changes with pressure

where as liquids are incompressible. Compressible flows are again divided into

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

velocity. Their application is in jet propulsion system, aircraft and rockets.

1.2 International System of Units (SI)

In the text we shall use SI units. The dimension in any system can be

considered as either primary or secondary dimensions. In the SI units there are 4

primary dimensions.

2

a) Primary Units

Dimension International Symbol Unit Mass M Kilogram (kg)

Length L Meter (m) Time T Second (s)

Temperature K Kelvin (K) Electric Current A Ampere (A)

b) Secondary Units

Secondary units is a combination of primary units such as Newton (N or kgm/s2

1.3 Specific weight and mass density

),

Joule (J or Nm), Watt (W or Nm/s) etc.

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 Greek letter gamma ( ). In equation

form,

VW

VolumeWeight

==

Mass density is the mass per unit volume of the substance. It commonly

designated by a Greek letter rho ( ). In the equation form,

Vm

Volumemass

==

There exist an important relation between specific weight and mass density.

Weight of the substance w = mg

Weight of the substance/unit volume,

gVmg ==

3

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 equation;

mRTPV =

or

RTP =

where Mgasmolar

tCoefficiengasUniversalR ==___

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

gravity.

In the SI units, will be expressed in N/m3 and in kg/m3. The values of specific

weight and mass density of water at different temperature are given in Table 1.1 and

Table 1.2 gives

Table 1.1 Physical Properties of Water

Temperature Specific Weight

Mass Density

Dynamic Viscosity

Kinematic Viscosity

Surface tension

T (C) (kN/m) (Kg/m) N-s/m m/s N/m

0 9.81 1000 1.75 x 10 3 1.75 x 10 6 0.0756

30 9.77 996 8.00 x 10 4 1.02 x 10 7 0.0712

60 9.65 984 4.60 x 10 4 4.67 x 10 7 0.0662

90 9.47 956 3.11 x 10 4 3.22 x 10 7 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.

4

Table 1.2 Physical Properties of Common fluids at Standard Atmospheric Pressure

Fluids Specific Gravity

Specific Weight

Mass Density

Dynamic Viscosity

Kinematic Viscosity

s - (kN/m) (kg/m) N-s/m m 2 /s

Air 0.0012 11.8 1.20 1.81 x 10 5 1.51 x 10 5

Ammonia 0.830 8.31 829 2.20 x 10 4 2.65 x 10 7

Glycerine 1.263 12.34 1258 950 x 10 3 7.55 x 10 4

Kerosene 0.823 8.03 819 1.92 x 10 3 2.34 x 10 6

Mercury 13.60 133.1 13570 1.56 x 10 3 1.14 x 10 7

Methanol 0.79 7.73 788 5.98 x 10 4 5.58 x 10 7

SAE 10 Oil 0.87 8.71 869 8.14 x 10 2 9.36 x 10 5

SAE 30 Oil 0.89 8.71 888 4.40 x 10 1 4.95 x 10 4

Turpentine 0.87 8.51 868 1.38 x 10 3 1.58 x 10 6

Water 1.00 9.79 998 1.02 x 10 3 1.02 x 10 6

Sea Water 1.03 10.08 1028 1.07 x 10 3 1.04 x 10 6

1.4 Specific gravity

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

at 4C. A convenient method to measure sp. gravity is by means of a hydrometer. It

is dipped into the liquid and a calibrated scale gives the specific gravity. It 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. It is abbreviated as (s).

w

fo Catwaterofweightspecific

fluidofweightspecificgravitySpecific

==4_____

____

5

The specific gravity can also be expressed as ratio of mass density of the substance to

mass density of water at 4C.

Example 1.1

A tank of glycerol has a mass of 1200kg and volume of 0.95m3

(a) Weight of glycerol

. Determine:

(b) Density

(c) Specific weight

(d) Specific gravity

(a) From Newtons Law;

Solution:

W= mg

Thus, W= 1200 x 9.81 = 11.76kN

(b) From equation (1.2)

3/126595.0

1200 mkgVm

===

(c) From equation (1.3)

3/38.1281.91263 mkNg ===

(d) From equation (1.4)

6

26.110001265

===W

ss

1.5 Viscosity

Fluids offer resistance to shearing force. Viscosity is the property of the fluid

that determine amount of this resistance.

Consider a fluid in between two parallel plates Figure 1.1 where the upper

plate is moving with velocity V and lower plate is stationery. The distance between

the plates is y. The layer in contact with the upper plate is moving with velocity V

where as the layer is contact with lower plate which is fixed will have zero velocity.

The deformation of the fluid under the action of shear stress is assumed proportional

to the rate of change of velocity, may be expressed in the equation form,

Lower plate (fixed)Velocity = 0

The rate of change of velocity= [(v+dv)-v] / dy = dv/dy

v + dv

vdy

y

Upper plate (moving) Velocity = V

Figure 1.1

The displaced fluid due to shear stress acted and the shear stress is assumed

proportional with the velocity gradient. It can be illustrated in the equation;

7

yv

= (1.5)

Where,

= shear stress

= proportional coefficient

yv = velocity gradient

The shear force, FD

AFD =

acting on the lower plate surface is given by;

Where A= surface area of the lower plate

The unit of dynamic viscosity, is SI unit is (N-sm-2

or Pas). Kinematics viscosity

which is usually, denoted by the Greek letter nu (v) is determined by dividing

dynamic viscosity () by mass density of the fluid ( ). In the equation form;

= (1.6)

When the fluid is at rest the velocity gradient dv/dy is zero and therefore no shearing

force exists. The viscosity varies with temperature therefore values of for given

fluid are usually tabulated at various temperatures. There are experimental methods

to calculate viscosity. One such experimental method is Falling Sphere Viscometer.

In this method a sphere of known diameter is dropped into a liquid. By determining

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

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

= 1

18

2

gdv

8

Where d = diameter of sphere

= sphere density

= fluid density

= kinematics density

A number of viscometers are available in the market. 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 Newtons law of viscosity (equation 1.6) and for which has a

constant value are known as Newtons fluids. Most 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 Newtons law of viscosity are known as non-

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

sludge etc.

A general relation between shear stress and velocity gradient for non-

Newtons fluids may be written as;

n

dydvBA

+= (1.7)

Where A and B are constants. Based on the value of power index n non

Newtonian fluids are classified as;

Pseudoplastic (such as milk, cement, clay) n1

A Newtonian fluid is a special case of nonNewtonian fluid for which A = 0 and

power index n = 1.

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

1.2.

9

-20 0 20 40 60 80 100 120

0.50.40.30.2

0.1

0.060.040.030.02

0.016432

1 x 10 -3

6432

1 x 10-4

6432

1 x 10-5

5

Temperature, oC

Castor Oil

GlyserinSAE 30 oil

SAE10 oil

Crude oil (SG 0.86)

Kerosene

Aniline Mercury

Carbon tetrachloride

Ethyl alcoholWater

Gasoline (SG 0.68)Benzene

Helium

Carbon DioxideAir

Hydrogen

Figure 1.2 Dynamic viscosities versus Temperature

u=0

t

R

Example 1.2

The viscosity of a fluid is to be measured by a viscometer constructed of two 75cm

long concentric cylinders as shown in Figure E1.2. The outer radius of the inner

cylinder is 15cm, and the gap between the two cylinders is 0.12cm. The inner cylinder

is rotated at 200rpm, and the torque is measured to be 0.8Nm. Determine the viscosity

of the fluid.

Figure E1.2

Abs

olut

e vi

scos

ity

, N .

s/m

2

10

The shear force,

Solution:

RLyvAFD 2

==

Where shear force can be calculated by

NR

TorqueFD 33.515.08.0

===

V= 2Rf = 2(3.142)(0.15)(200)/60 = 3.14 m/s

smt

uVyv /67.2616

0012.0014.3=

=

=

Thus, the dynamic viscosity:

2/0029.0)67.2616)(75.0)(15.0(2

33.5 mNs==

1.6 Compressibility and Bulk Modulus

Consider a mass of fluid m whose initial pressure and volume is P and V

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

pressure is P+dP and volume reduced to VdV. Hence, change in pressure is dp and

change in volume is dV. Volumetric strain in defined as change in volume divided

by original volume and is dV/V. The bulk modulus denoted by k and is defined as

change in pressure to volumetric strain;

k= Changes in pressure/Volume strain

or

11

dVdPVk = (1.8)

Let mass of fluid is m:

m = V (1.9)

After differentiation of equation (1.9):

0=+ VddV

ddVV

= (1.10)

Substitute V/dV into equation (1.8), thus;

ddPk = (1.11)

From equation (1.11) the value of k is dependent on the relationship between pressure

and density. For liquids, changes of density with pressure are small and Bulk

modulus k is high. These liquids can be considered incompressible. However, for

gases the compressibility is so large that value of k is not a constant but proportional

to pressure. For gases relation between pressure and mass density in obtained from

characteristic equation of a gas and particular relation between pressure and density is

established depending on type of compression process.

(i) For an isothermal process where the temperature is maintained constant the

characteristic equation is written as

.constPddP

==

.constP =

(1.12)

.constPddP

==

(1.13)

12

Substitute dP/d into equation (1.6) gives

k= P (1.14)

(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

.constP =

(1.15)

After differentiation will give;

P

ddP

= (1.16)

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

CP/C

1.7 Mach no. and Compressibility

V

Again substitute dP/d into equation (1.11) will give;

k= P (1.17)

The ratio of adiabatic bulk 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 example for air = 1.4.

Mach. No. is defined as ratio of velocity of flow (v) to local velocity of sound

(a) and is a measure of compressibility effects.

avM = (1.18)

13

The velocity of propagation of sound waves in a fluid, flow is expressed as

ddPa = or

ddPa =2 (1.19)

Substituting the value of dP/d in equation (1.11) we get

2ak = (1.20)

Substituting value of a in equation (1.12) we get

kvM

2

= (1.21)

For liquids the bulk modulus k is large and velocities small and, hence, Mach. No. is

negligible or effect of compressibility is neglected. Gas velocities are high bulk

modulus is low, and hence, Mach. no. is high and compressibility cannot be

neglected. Gases can only be treated as incompressible if pressure changes are small

and Mach. no. is less than 0.3.

1.8 Surface tension

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

by a force known as Cohesion. This force keeps the molecules bonded together.

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

mix 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 known as adhesion.

14

Liquid

Vapor

B

A

Figure 1.3

Figure 1.3 is shown a molecule of liquid at the surface is acted on by

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

commonly denoted by Greek letter, sigma () and is defined as force per unit length

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

LF

= (1.22)

The units of in SI units will be N/m. In many engineering problems surface

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

therefore be neglected. However, 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

analyzing a free-body diagram as shown in Figure 1.4.

PR

2R

2R

2

Figure 1.4

The pressure force exerted in the droplet is given by

15

2RPF =

The force due to surface tension is

RF 2=

The pressure force and tension must be balance each other;

RRP 22 =

RP 2= (1.23)

1.9 Capillarity

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

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

cohesive force of the liquid molecules and adhesion of liquid surface to solid glass

surface.

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

as shown in Figure 1.5.

BA

h

Parallel Plates

F

Tube

BA

dL

Figure 1.5

16

Length of line of contact of the liquid with the tube = d

Vertical component of the surface tension force = (d)..cos

Weight of column of liquid, W= hd 24

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

hdd 24

cos =

dh

cos4

= or d

P cos4= (1.24)

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

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

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

passages must be extremely fine. In water and certain other liquids that exhibit

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

is less than 90. In some other liquids such as mercury the meniscus is convex. The

liquids do not wet the solid surface and angle of contact is more than 90. Glass

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 minimize errors due to capillarity.

1.10 Vapor Pressure

Cavitation 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. Any attempt to reduce the pressure still further

merely causes the liquid to vaporize more quickly and clouds of vapor bubble form.

The bubbles of vapor formed in the region of cavitation move downstream to a region

17

of higher pressure where they collapse (see Figure 1.5). It 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

1/100 of a second, but dynamic force caused by this phenomenon may be very severe.

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

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

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

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

most likely to occur at inlet to the impeller. Cavitation can also occur if a liquid

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

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

same damaging effects. Care should be taken to avoid cavitation as far as possible but

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

be constructed of especially resistant metals such as stainless steel.

Figure 1.5 Cavitation phenomena inside the nozzle

18

Problems

1. Determine the density of air, hydrogen, and carbon dioxide at an absolute

pressure of 300kN/m and a temperature of 38.8C.

2. Calculate the specific weight and density of air at absolute pressure of 445 kPa

and a temperature of 38 C.

3. If the volume of liquid decreases by 0.2% for an increase of pressure from 6867

kN/m to 15696 kN/m, calculate the bulk modulus of elasticity of the liquid.

4. A soap bubble 51 mm in diameter has an internal pressure in excess of outside

pressure of 2.06 x 10-2 kPa, calculate the tension in the soap film.

5. If the pressure inside a water droplet is 0.2 kPa in excess of external pressure,

and given surface tension of water in contact with air at 20C is equal to 0.0736

N/m, determine the diameter of the droplet.

6. Air is introduced a nozzle into a tank of water to from a stream of bubbles. If

the bubbles are intended to have diameter of 2 mm, determine the pressure of air

at the nozzle exceed that of surrounding water (given tension of water = 0.0736

N/m).

7. The air in an automobile tyre is at 2.943 MPa absolute at 26.6C. Assuming no

change in the volume of air if the temperature rises to rises to 62.2C, determine

the air pressure.

8. A gas occupying a volume of 300 liters at a certain temperature and pressure of

0.34 N/mm is compressed isothermally to 150 liters. Calculate the initial and

final bulk modulus of elasticity.

9. At a certain point in a fluid, the shear stress in 0.22 N/m and the velocity

gradient is 0.167 sec. If the mass density of the fluid is 1293 kg/m3, determine

the kinematic viscosity.

19

10. A liquid flows between two fixed parallel boundaries. The velocity distribution

near to lower wall is given in the following table:

y (mm) v (m/s)

1.0 1.00 2.0 1.99 3.0 2.98 4.0 3.00 5.0 3.00

Determine the maximum and minimum shear stresses (The dynamic viscosity of

fluid is 0.05Pas).

11. Two plates are arranged as in Figure Q11 in the liquid. The top plate is moving

with the velocity of 0.5m/s and the middle plate is moving with the velocity of

2m/s in opposite direction. The area of both plates area 0.25m2

3mm

3mm

2m/s

0.5m/s

. Plot the velocity

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

the viscosity of liquid is 0.01Pas).

Figure Q11

12. Mercury 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 Q12. The

surface of mercury is 0.514 N/m and the angle contact is 40C. Calculate the

depression distance in a 1 mm glass tube.

20

40o

dMercury

Figure Q12

13. The vapor pressure of water at 100C is 101 kN/m, because water boils under

these conditions. The vapor pressure of water decreases approximately linearly

with decreasing temperature at a rate of 3.1 kN/m/C. Calculate the boiling

temperature of water at an altitude of 3000 m, where the atmospheric pressure is

69 kN/m absolute.

-oooOOOooo-

21

CHAPTER 2

HYDROSTATIC PRESSURE

2.1 Introduction A fluid at rest is characterized by absence of relative motion between adjacent

fluid layers. Under such a condition, the velocity gradient is zero and there is no

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

rest do exert forces on the solid boundary. Knowledge of force variation or more

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

There are so many practical examples of fluids at rest such as water retained

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

The object 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

In 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 2.1.

1

2

h

h2

h1

depth

Figure 2.1 Static pressure increased linearly with the depth

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

given as

22

ghPP atm += (2.1)

by neglecting Patm

ghP =

equation 2.1 becomes

(2.2)

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

a example, for point 1 and 2 in Figure 1 can be written as:

11 ghP = (2.3)

22 ghP = (2.4)

Thus, the pressure different between points 1 and 2 can be written as:

hgPPP == 12 (2.5)

where h=h2-h

A B C DF

E

1

The fluid pressure at rest is constant along the horizontal line. In 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 2.2). 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. It also contributes to

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

Figure 2.2 Pressures are equal for all points along horizontal plane

23

Example 2.1

A tank is connected to a vertical tube is filled with water (= 9810N/m2

(a) Absolute pressure at levels A, B, C, D, E, and F

). Determine:

(b) Gage pressure at levels A, B, C, D, E, and F

0.7

1.8

0.9

0.4 D

C

B

A

Water

E

F

Patm

Figure E2.1

(a) Absolute pressure

Solution:

From equation 2.1, it can be written as

Point A: PA= Patm + hA, where hA= 0

= 101.3(103) + 9810(0)

= 101.3kPa

Point B: PB= Patm + hB, where hB= 0.7m

= 101.3(103) + 9810(0.7)

= 108.2kPa

Point C: PC= Patm + hC, where hC= 2.5m

= 101.3(103

Point D: P

) + 9810(2.5)

= 125.8kPa

D= Patm + hD, where hD= 3.4m

24

= 101.3(103) + 9810(3.4)

= 134.7kPa

Point E: PE= PC since hE = hC

Thus, PE= 125.8kPa

Point F: PF= Patm + hF, where hF= 3.8m

= 101.3(103) + 9810(3.8)

= 138.6kPa

(b) Gage pressure

PgA= PA - Patm = 0kPa

PgB= PB - Patm = 6.87kPa

PgC= PC - Patm = 24.5kPa

PgD= PD - Patm = 33.4kPa

PgE= PE - Patm = 24.5kPa

PgF= PF - Patm

F1 F2

21A1 A2

Plunger or piston

Hydraulic fluid,

= 37.3kPa

Now consider a simple hydraulic jack in Figure 2.3:

Figure 2.3 Hydraulic Jack

25

Using equilibrium principle, the pressure exerted by the fluid on a plunger or a

piston is written as

AFP = (2.6)

From which it follows that 1

11 A

FP = and 2

22 A

FP =

Since the pressure at points 1 and 2 is equal (i.e. same level), we have

2

2

1

1

AF

AF

= (2.7)

which can be used to solve for F2 if F1, A1 and A2 are given, or vice versa.

Example 2.2:

For a hydraulic jack as shown in Figure E2.2, determine the weight that could be

lifted if the 400N input force F1

F1 F2

21A1 A2

Plunger or piston

Hydraulic fluid,

is applied to the plunger. The diameters of the

plunger and the piston are10mm and 75mm, respectively

Figure E2.2

26

1

212 A

AFF =

Solution:

From equation 2.7 above;

where 2521 10855.74/)01.0( mA==

2522 1042.44/)075.0( mA==

Therefore, F2

2.2 Pressure and head

= 22.5kN

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

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

to measure, either a gage, vacuum or absolute pressure. A 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.

Patm

Pgage

Pvacuum

P absolute

P absolute

P absolute = 0

Figure 2.4 Type of pressure

27

2.3 Mercury Barometer

A typical mercury barometer is shown in Figure 2.5 and used to measure

atmospheric pressure. It 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. It can be assumed that the space between the mercury and the end

of the tube contains a vacuum with zero pressure.

At 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 Patm. Thus, the atmospheric pressure can be calculated as

Patm

P = 0

hP atm

= gh (2.8)

Figure 2.5

2.4 The Bourdon Gauge

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

robust and simple to use. A 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 fixed to the

frame as shown in Figure 2.6.

28

X

X

Section at X-X Pointer

Flattened phosphor-bronze tube

Figure 2.6

When the pressure inside the tube exceeds 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 linkage to a pointer moving over a scale. Zero reading is of

course obtained when the pressure inside the tube equals the local atmospheric

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

may be made.

2.5 Manometric 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.

2.5.1 Simple U-tube Manometer

29

h1

h2

C

D

B

Patm

AP

1

2 Figure 2.7

Consider the U-tube manometer connected via a small hole to a pipe (Figure

2.7) carrying a fluid of density 1 at pressure PA (which is to be measured).

Let the open end of the U-tube be subjected to atmospheric pressure, Patm.

At the common surface B-C with the configuration as shown in the diagram

we have :

PA + 1gh1 = PB = PC = PD + 2gh2

P

,

or

A + 1gh1 = PD + 2gh2 (2.9)

Now PA is the pressure to be measured (P) and PD = Patm.

Thus P Patm = (2h2 - 1h1)g (2.10)

2.5.2 Differential Manometer

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

as shown in Figure 2.8.

PA + 1gh1 = PD + 3gh3 + 2gh2,

Or differential pressure is given by

30

PA PD = (3h3 + 2h2 - 1h1

h1

h2

C

B

D

12

A

h3 3

)g (2.11)

Figure 2.8

2.5.3 Inverted U-tube Manometer

Another type of differential manometer as shown in Figure 2.9

PB + 1gh1 = PA

PB + 2gh2 = PC

PC + 3gh3 = PD

or PD 3gh3 2gh2 + 1gh1 = PA (2.12)

or PA PD = (1h1 2h2 3h3)g (2.13)

31

h1

h2

C

B

D

1

2

h3

3

A

Figure 2.9

2.5.4 An Inclined Manometer

An inclined manometer as shown in Figure 4 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 L.

h Zero levely

P1

AIR

Diameter, D

Fluid densityDiameter of tube, d

LP2

Figure 2.10 Inclined Manometer

Applying the equal level-equal pressure principle, we have

P1 = PGas

But P1 = P2 + h and h= L sin. The above equation then becomes,

32

( ) sin21 LPP += (2.14)

h

A B

air, air

Pair

Patm

Mercury, sg = 13.6

Example 2.3

U-tube manometer containing a mercury (sg= 13.6) as a working fluid is connected to

a tank that contains air as shown in Figure E2.3. The other end of the manometer is

exposed to the atmosphere. Determine the pressure in the tank if h= 0.4m.

Figure E2.3

Solution:

Since the density of air is neglected, the pressure at point A will correspond to

the pressure of air in the tank.

PA = PAir (1)

Applying equal level-equal pressure principle, we have

PA = PB (2)

But

33

PB = Patm + Hg h (3)

Where Hg= sg (Hg at 4deg.) = 13.6 (9810) = 133416N/m

)4.0(13341)10(100 3 +=+= hPP Hgatmudara

2

Substitute (3) and (2) into (1) will give

kPaPudara 4.153=

Example 2.4

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

pressurized tank containing gas and water as shown Figure E2.4. Pressure gage

attached to the tank reads 80 kPa, determine the gage pressure in the pipe. (Given h1=

40cm, h2= 84cm, h3= 57cm, h4

Gas

Water

h 1 = 40 cmh 2 = 84 cmh 3 = 57 cmh 4 = 45 cm

Mercury ,sg = 13 .6

A

h3Oil , sg = 0.8

h1

h2

h4

= 45cm)

Figure E2.4

From equal level-equal pressure,

DCB PPP ==

But, 3hPP HgED +=

Where )( 2hPP wFE +=

And kPaPP gasF 80==

34

And also 4hPP oilBA +=

Substitute (2), (3) and (4) into (1);

432 )( hhhPP oilHgwatergasA +++=

PA = 80(103) + 9810(0.84) +13.6(9810)(0.57)+ 0.8(9810)(0.45)

= 167kPa.

Example 2.5

Figure E2.5 show a three-fluid manometer containing oil (sg= 0.82), mercury (sg=

13.6) and water used to measure large pressure differences. Determine the pressure

difference between A and B. Given h1= 45cm, h2= 70cm, h3= 25cm and h4

Water

Gas

A

h4

h3

Oil, sg = 0.82

Gasoline,Sg = 0.7 Mercury,

Sg = 13.6

B

h2

h1

= 20cm.

Neglect the air density.

Figure E2.5

DCA PPP ==

Solution:

but 11 hPhPP WAWDE +=+=

35

and 1hPPP WAEF +==

Also 2hPP HgGF +=

but GH PP =

Also 43 hhPP oilgasolinHB ++=

Substituting 1, 2, 3, 4, 5, 6 and 7 give

432 hhhhPP oilgasolinWHgBA =

or )( 432 hsghsghhsgPP oilgasolinHgWBA =

= (9810)[(13.6)(0.7)-(0.45)-0.7(0.25)-0.82(0.2)]

= 85.7kPa.

Problems

1 The Crosby gage tester shown in the figure is used to calibrate or to test

pressure gages. When the weights and the piston together weigh 89.0 N, the

gage being tested indicates 179kPa. If the piston diameter is 20mm, what is

the percentage error exists in the gage?

Weight

Piston

Air

Oil

Figure Q1

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

is reduced to 10% of atmospheric pressure. The inner radius is 15 cm, and the

36

outer radius is 15.5 cm. The seal is located half way between the inner and

other radius. If the atmospheric pressure is 100kPa, what force is required to

pull the shells apart?

3. If exactly 20 bolts of 2.5-cm diameter are needed to hold the air chamber

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

needed at B-B? Here D = 50 cm and d = 25 cm.

Figure Q3

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

weights A and B, respectively, one above the other where A > B

Liquid B

Liquid A

. Which

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

through the liquids?

p p p p

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

Figure Q4

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

has an inside diameter of 1 mm (d = 1.0 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

37

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

the right tube. State your main reason or assumption for making your choice.

d 3d

Figure Q5

6. A tank is fitted with a manometer on the side, as shown. The liquid in the

bottom of the tank and in the manometer has a specific gravity (s) of 3.0, the

depth of this bottom liquid is 20 cm. A 10-cm layer of water lies on top

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

7.

Water

Liquid

h = ?10 cm

20 cm

Figure Q6

7. Determine the gage pressure in pipe A.

Mercury

0.5 m

1.0 m

Water

1.3 m

A

Oil (SG = 0.90)

38

Figure Q7

8. Considering the effects of surface tension, estimate the gage pressure at the

center of pipe A.

10 cm

Glass tube ( 1mm ID. 4 mm OD)

Water level in tube

A

Figure Q8

9. What is the pressure at the center of pipe B?

50 cm

10 cm

4

350 cm

B

= 10 kN/m3

= 20 kN/m3

Figure Q9

39

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

Water

A

150 cm

Oil (SG=0.8)

Mercury (SG =13.6)

90 cm

Water

30 cm

30 cm

Figure Q10

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

of 1.01. Determine the pressure difference in Pa between two points A and B at the

same level at the base of the legs when the difference in water level h is 75mm.

40

CHAPTER 3

FLUID STATICS

3.1 Introduction

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

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

forces acting on submerged surfaces. We shall also examine problems involving

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

ships and submarines.

In this chapter, the submerged surfaces are divided into the following types; a)

straight horizontal and vertical surfaces, b) straight inclined surfaces, and c) curved

surfaces.

The analysis of hydrostatic force on submerged surfaces assumes the following

conditions.

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

stress for fluids at rest.

2. Pressure varies linearly with depth for incompressible fluid.

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

pressure

3.2 Horizontal and vertical surface

Any given depth h, the resultant fluid force FR

FRh hFR

P=gh

on the horizontal and vertical

surfaces may be represented as shown in Figure 3.1.

Figure 3.1

41

Magnitude of the resultant force FR

PAFR =

at the bottom of the tank, Figure 3.1 is given

by

(3.1)

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

vertical

Surface. Figure 5.8(b), we have

PAFR 21

= (3.2)

3.3 Inclined Surface

The surface tilted at an angle from the horizontal is shown in Figure 3.2. The

pressure variation and hence the resultant hydrostatic force FR

pcFR

O

AP

CXC

XP y

ycyp

hchp

, on the surface can be

presented as shown in Figure 3.2.

Figure 3.2

42

Magnitude FR

APF CR =

can be written as

(3.3)

where PC

CoC ghPP +=

is a pressure at the centroid surface (point C) and is written as:

(3.4)

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 ,

b) it is perpendicular to the surface, and

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

(point P).

In many cases, the pressure at point O is the atmospheric pressure and may be

ignored in the analysis. This simplifies equation (3.4) to

CC ghP = (3.5)

Noticed that when = 90, the surface becomes vertical and when = 0, the

surface becomes horizontal.

3.3.1 Center of Pressure

Next we need to determine the line of action of the resultant force FR distance yP and hence its location below the free surface, hp. It is found that T action yP

C

XP Ay

Iy =

is given

by

(3.6)

43

where XI is the moment of inertia of the area with respect to the x axis. Using

parallel axis 2CXX AyII += , equation (3.6) can be rewritten as

C

XCP Ay

Iyy += (3.7)

where XI is the moment of inertia of the area with respect to the centroid axis. XI

commonly used shapes is given in Figure 3.3. Observe that the center pressure P

below the centroid C of the surface since 0/ >CX AyI ,

Expressing yP= hP/sin and yC= hC

CCP Ah

Ihh +=

/sin, we obtain from equation 3.7

Where 2sinXII = . In some textbooks, the location of the centroid of the submerge

surface from the fluid free surface, Ch is denoted as h and yC y as .

Figure 3.3 shows the centroid y for the simple shape objects and Figure 3.4 shows

the second moment of area XI for the simple shapes.

44

45

Figure 3.3

1. First Moment of Area Note:

It is used to determine the centroid for a complex shape. It can be calculated using

the following method:

46

xa b

A1

A2A3

= -

bAaAxA 321 =

Where A1, A2, and A3

y

xy

x

dA

are the areas and x, a, and b are the centroid for the

above shapes respectively.

2. Axial Moment of Inertia of an Area

The axial moment of inertia of an area is the summation of the axial moments

of inertia of the elements.

dAyI x = 2

dAxI y = 2

3. Parallel Axis Theorem

The parallel axis theorem states that the axial moment of inertia of an area

about any axis equals the axial moment of inertia of the area about a parallel

axis through the centroid of the area plus the product of the area and the

square of the distance between the two parallel axes.

47

x

y

0

x

y

n

m

2AmII xx +=

2AnII yy +=

3.4 Moment of a Force

The resultant force FR and its location yP or hP

M=Fd

AC

F

Bd

AC=

, are determined, thus the calculation of

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

point. Consider a force F acting perpendicular to the body at point B as shown in

Figure 3.4.

Figure 3.4

This force tends to rotate the body about point A in a counter-clockwise direction.

The tendency of a force to rotate the body is called the moment of a force about that

point. The magnitude of this moment about point A is given by:

FdM A =

48

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

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

clockwise. In this case, MA is in the counter-clockwise direction Figure 3.4.

(a) The resultant hydrostatic force exerted on the gate AB

Example 3.1:

Consider a rectangular gate AB hinged along A to support the water pressure as

shown in Figure E3.1. Determine:

(b) The center of pressure

(c) The force acting on the stopper at B

The gate width is 3m.

Water

=9810N/m3h

l

h = 10 ml =6 m

A

B

Figure E3.1

(a) Resultant force

Solution:

FR = hC A (1)

where A= 6(3) = 18m2 and hC = 4 + 3 = 7m

From equation (1);

49

FR

(b) Location of center of pressure

= 9810 (7) (18) = 1.24MN

CCP Ah

Ihh +=

where 4232 54)90(sin)6)(3(121sin mII X ===

mhP 43.7)7)(18(547 =+=

(c) The force acting at the stopper B

Taking moment at A,

)4( = PRB hFlF

MNFB 71.0643.324.1

=

=

Water

10m

60 F=

Example 3.2:

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

Figure E1.2. The normal force of dam is 50MN/m.

Figure E1.2

50

mkNAhF /8.42460sin/)10)(5(9810 ===

Solution:

Hydrostatic force per unit width acting on the dam surface:

For force acts along horizontal component:

mkNFFH /9.36730cos8.42430cos ===

Therefore, the friction factor required to hold the dam;

0074.050

109.367 3=

=

3.5 Curved Surface

For the curved surface as shown in Figure 3.5, the pressure variation and hence the

resultant hydrostatic force FR

A

B B

A

F2

D

FVFR

Fh

W

PG

F1

FR

P

on the surface can be represented as shown.

Figure 3.5

Let Fh = horizontal force exerted on the fluid mass

Fv F

= vertical force exerted on the fluid mass

1 = resultant force of the fluid exerted on the surface AD

51

F2

2FFh =

= resultant force of the fluid exerted on the surface BD

W = the weight of the enclosed volume supported by the curved AB and

that W = gV where V is volume of the enclosed liquid block, and it passes

through the centroid of this volume.

The force balances under static equilibrium and obtain the following equations:

WFFv += 1

The magnitude of the total resultant fluid force acting on the curved surface is given

by:

22vhR FFF +=

Example 3.3:

A water channel with a 3m long quarter-circular section AB of radius 2.4m is

designed as shown in Figure E3.3. Determine the total resultant force exerted by

water on section AB. Given air= 9.81kN/m3

2.4 m

3.6 m

A

B

WATER

.

Figure E3.3

Solution:

Total resultant force is given by

52

22vhR FFF +=

where Fv= F1 + W and Fh= F

AhF C=1

2

For each force components;

where A= 2.4(3) = 7.2m2 and hC

kNF 3.254)2.7)(6.3(98101 ==

= 3.6m

VgVW ==

Where 322 57.13)3()4.2(21

41 mlrV ===

kNW 2.133)57.13(9810 ==

AhF C=2

Where A= 2.4(3) = 7.2m3 and hC

kNF 339)2.7)(8.4(98102 ==

= 3.6 + 1.2 = 4.8m

Therefore, Fv= 254.3 + 133.2 = 387.4kN

Fh= F2= 339kN

And

FR= (387.42 + 3392)1/2

= 514.8kN

53

P

Hinge

2m

2mWater

0.5m

Example 3.4

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

E3.4.

Figure E3.4

mkNAhFH /62.19)2)(1(9810 ===

Solution:

Hydrostatic force acting on the control volume;

By taking moment at hinge:

mkNP /7.155.2

62.192=

=

3.6 Buoyancy, Floatation and Stability

When 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 buoyancy force. This

force tends to lift the body upward and its existence is due to the fact that (1) the

fluid pressure increases with depth, and (2) the pressure force acting from below is

larger than the pressure force acting from above.

3.6.1 Buoyancy and Floatation Consider a body submerged completely in a fluid as shown in Figure 5.12.

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

resultant force on the top surface of the body.

54

h

FB

Ftop= fh1A

Fbottom= fh2A

h1h2

Figure 3.6 Buoyancy 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 buoyancy or the centroid of the displaced volume, CB

hAAhAhFFF ffftopbottomB === 12

which

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

the case of a completely submerged body. Writing force balance on the

body, we have

However, the term hA is basically the volume of the fluid body (or volume

of the displaced fluid by the body). Expressing this volume as V = hA, we

may write the equation above as

VF fB =

where V is volume of the displaced fluid. 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.

In the case of a floating body, Figure 3.7, 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 submerged portion of the floating body.

55

FB

VsubFB

(a) (b) (c)

Vtotalmg

FBFB=mg

mg

Figure 3.7 Buoyancy force acting on a floating body

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

WFB = or totalbodysubf VV =

where totalV = total volume of the body or volume of the entire body

subV = volume of the submerged, portion of the body, which is equal to

the

volume of the displaced fluid (Figure 3.7(c))

Rewriting the above equation as

f

body

f

body

total

sub

VV

=

=

We observe that the body is completely submerged when the density ratio is

equal to 1 that is when the density of the body is equal to the fluid density.

We can conclude that a body immersed in a fluid will

(1) Rise to the surface of the fluid and float when the density of the body

is less than the fluid density,

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

fluid density, and

56

(3) Sink to the bottom when the density of the body is greater than the

fluid density.

body < f : floating

body < f : suspended

body < f : sinking

Figure 3.8 Situations of a body immersed in a fluid: float, remain at rest, or

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

Example 3.5

A cuboids has the size of bhw is floating in the water ( air= 9.81kN/m3) as

shown in Figure E3.5. Determine the portion of body that above the water

surface, a if Jasad= 3.5kN/m3

h

water

a

w

, b= 5m, h= 8m dan w= 6m.

Figure E3.5

Solution:

wbhVwb)ah(V

total

sub

=

=

From the relationship,

f

body

f

body

total

sub

VV

=

=

57

=

=

98003500181 )(ha

f

body

a = 5 . 1 4 m

Note: Observe that from equation above, the portion above the water

surface will be less if the density of the body is higher

Example 3.6

Balloon has a mass of 20kg and the diameter of 10m is filled with the helium

gas (Figure E3.6). The density of helium gas is 0.81kg/m3. Determine the

acceleration of the balloon after it releases from the ground (Take air density

as 1.2kg/m3

10m

Helium

).

Figure E3.6

Solution:

The forces in equilibrium position as in equation below:

amgmF TTB =

)20()20(

VgVVa

He

Heair

++

= , where 33 6.523534 mV ==

Thus, the acceleration is

58

2/07.4)6.52381.020(

81.9)6.52381.020()6.523)(81.92.1( sma =+

+=

3.6.2 Stability of Floating and Immersed Bodies

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

the design of a ship, submarine and barge (Figure 3.9).

Figure 3.9 A barge used for transportation at Kuala Kurau

In 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 stable (vertical stability).

For the rotational stability, the condition depends upon the relative location

of the center of gravity G of the body and the center of buoyancy CB, Figure

3.10. A floating or immersed body is stable if the point G is below point CB,

Figure 3.9(a). Under this condition, the body will return to its original stable

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

59

(a) stable (b) neutral (c) unstable

w

FBCBG

w

FB

CBG

w

FBCBG

(d) Restoring Couple

Restoring Moment

w

FB

Overturning Moment

(e) Overturning Couple

w

FB

Figure 3.10 Stability 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).

However, a floating body will still be stable even if G is above CB. See

Figure 3.11. 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

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

meet at the point called the meta-center CM.

The distance between G and CM is called the metacentric height hM and is

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

stable the floating body will be. Note that the floating body is unstable if the

point CM

W

G

CB

W

CM

CB

FB

G

Restoring Couple

hM

is below point G.

Figure 3.11 Stability of an immersed body

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

following equations:

60

VICC BM =

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

from the reversed moment, it must has the following conditions:

0>= GMM CCh Stable

0== GMM CCh Neutral

0

61

Volume of displacement, 33

48781025

105000 mV ==

And, mVICC BM 2.14878

5840===

Thus, metacentric height, mxhM 344.05sin03.0

5sin===

And the vertical distance between CG and CB

mxCC BM 857.05tan03.02.1

5tan)( ==

is

3.7 Liquids in Relative Equilibrium

If 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 take 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.

3.7.1 Horizontal Acceleration

Consider a particle O of mass m on the free surface of the liquid as in Figure

3.11.

A C B

a

W

F

R

O

Figure 3.12

62

Since the particle is at rest relative to the tank, it will have the same

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

ag

WmaF ==

where W= weight of particles.

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

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

surface due to surrounding fluid.

For equilibrium F= W tan, where is the angle of the free surface to the

horizontal.

Thus,

ga

=tan

and is constant for all points on the surface.

3.7.2 Vertical Acceleration

As the acceleration is vertical the free surface will remain horizontal. Consider

a vertical prism of height h (Figure 3.13) extending from the free surface to x

and let the pressure intensity at x be P.

P

x

BA

a Prism Cross sectional

Area = a

h

Figure 3.13

63

Accelerating force at x, F= Force due to pressure weight of prism

= PA - ghA

By Newtons second Law:

F = mass acceleration

ahAF =

Thus,

ahAghAPA =

or

+=

gaghP 1

3.7.3 Forced Vortex

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

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

weight W (Figure 3.14), the centrifugal accelerating force, F acting

horizontally and the fluid reaction R.

ow

A

Fy

D

B

Axis

of r

otat

ion

x

Figure 3.14

64

For any point at radius x and a height y from the lowest point O, if is the

angle of inclination of the water surface to the horizontal,

WF

dxdy

==tan

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

is 2x dan F= (W/g)2

gx

dxdy 2tan ==

x.

The surface angle therefore varies and,

Integrating will give,

malargxdx

gxy

x

+== 222

0

2

If y is measured from AB, y= 0 when x= 0 and

gxy

2

22=

The water surface is therefore a parabolic revolution.

Example 3.8

A tank containing water moves horizontally with a constant linear acceleration

a of 3m/s2

(a) The angle of water surface when the tank is 1.5m

. The tank is 3m long and the depth of water when the tank is at rest

is 1.5m. Calculate:

(b) The maximum pressure intensity on the bottom

65

(c) The minimum pressure intensity on the bottom

Solution:

Given: a= 3m/s2

(a) The angle of water surface to horizontal,

, h= 1.5m

o

ga 17tan 1 ==

(b) The depth at A (maximum pressure),

mhhh oA 96.117tan5.15.1tan =+=+=

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

mhhh oB 04.117tan5.15.1tan ==+=

Problems

1. Consider the two rectangular gates shown in the figure. They are both the

same size, but one (Gate A) is held in place by a horizontal shaft through

its midpoint and the other (Gate B) is cantilevered to a shaft at its top. Now

consider the torque 1' required to hold the gates in places as H is increase.

Choose the valid statement(s): a) TA increases with H. b) TB increases with

H. c) TA does not change with H. d) TB

does not change with H.

Shaft

Gate BGate A

HWater

Atmospheric Pressure

HWater

Atmospheric Pressure

Shaft

Figure Q1

2. Find the force of the gate on the block.

66

Block

Pivot

4 m x 4 m gate

Water

10 m

2 m 2 m

Figure Q2

3. Neglecting the weight of the gate, determine the force acting on the hinge of

the gate.

3 m

9 mAtmospheric pressure On this side of gate

9 m x 9 m gatewater

hinge

Figure Q3

4. The rectangular gate measures 6m by 4m and is pin-connected at point A. If

the surface on which the gate rests at A is frictionless. What is the reaction at

A? Neglect the weight of the gate.

Hinge

Atmospheric Pressure

Stop

A

Water6 m

3 m

30o

Figure Q4

5. A 12m x 12m gate is installed at the end of water reservoir, as shown, and is

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

The gate is connected to a rectangular tank of water which is 12 m wide (into

67

the paper) and filled with 6 m of water. The weight to the tank is negligible.

How long (L) would the tank have to be open the gate?

Figure Q5

6. The triangular gate ABC is pivoted at the bottom edge AC and closes a

triangular opening ABC in the wall of the tank. The opening is 4 m wide (W =

4 m) and 9 m high (H = 9 m). The depth d of water in the tank is 10 m.

Determine the hydrostatic force on the gate and the horizontal force P required

at B to hold the gate closed.

dH

W

Water

T = 20oC

View D-DView E-E

A,C

E

A

B

C

E

D

D

P

Figure Q6

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

the weight W= 60kN as shown in Figure Q7. The gate is 4m high and 2m

wide. Neglect the weight of the gate.

68

5m

Water4m d

W

Hinge

Atmospheric pressure

Gate stop Figure Q7

8. For the plane rectangular gate (L x W in size), Figure 8(a), what is the

magnitude of the reaction at A in terms of w

Water Water

L

HingeHinge1/4L

A

45 o

A

B B

1/4L

L Sin 45 o

Smooth Boundary

Rectangular Gate

and the dimensions L and W?

For the cylindrical gate, Figure 9(b), will the magnitude of the reaction of A be

greater than, less than, or the same as that for the plane gate? Neglect the

weight of the gates.

(a) (b)

Figure 8

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

cylinder 1 m in diameter. The platform itself weighs 30kN in air, and each

cylinder weighs 1.0kN per meter of length as in Figure Q9. What total

cylinder length L is required for the platform to float 1 m above the water

surface? Assume that the specific weight of the water is 10,000 N/m. The

platform is square in plan view.

10 m

Diameter = 1 m

Floating Platform

Weight = 30kN

L = ?1 m

Figure Q9

69

10. The coffee cup in Figure Q10 is removed from the drag race, placed on a

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

Determine:

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

cup

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

(Take the density of coffee as 1100kg/m3

r0

3cm3cm

3cm

7cm

z

)

Figure Q10

PROPERTIES OF FLUIDSHYDROSTATIC PRESSUREFLUID STATICS3.3.1 Center of PressureExample 3.5

If 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 take up a new...Figure 3.12Figure 3.13F = mass ( accelerationFigure 3.14Example 3.8