fluid mechanics (2014 2015) al azhar university

8/10/2019 Fluid Mechanics (2014 2015) Al Azhar University

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FLUID MECH NICS

PROF DR ANAS M ELMOLLA

DR AMIR M MOBASHER

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FLUID MECHANICS AL-AZHAR UNIVERSITYFACULTY OF ENGINEERING

Pof.Dr. Anas M. Elmolla & Dr. Amir M. Mobasher ii

FLUID MECHANICS

Contents

1.Units and Dimensions & Fluid Properties................ 2

2.Fluid Pressure and Hydrostatic Force.... 20

3.Fluid masses subjected to Acceleration ...... 37

4.Equilibrium of Floating Bodies.... 45

5.Kinematics of Fluid Flow.. 51

6.Flow through Pipes.. 82

7.Momentum Principle.. 109

8.Solved Problems...... 11

9.Assignments..... 184


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Pof.Dr. Anas M. Elmolla & Dr. Amir M. Mobasher 2

CHAPTER 1

UNITS AND DIMENSIONS & FLUID PROPERTIES

1.1Fluids :

Fluid can be defined as a substance which can deform continuously when

being subjected to shear stress at any magnitude. This includes any liquid

or gas.

,.

1.2Fluid mechanics :

Fluid mechanics is a division in applied mechanics related to the behaviour

of liquids or gases which is either in rest or in motion.

(. (

The study related to a fluid in rest or stationary is referred to fluid

statics , otherwise it is referred to as fluid dynamics

.


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1.3Applications Of Fluid Mechanics In Civil Engineering

Water pipelines:

Water distribution systems and Sewer systems:

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Dams and water control structures:

Rivers and manmade canals:

Coastal and Harbour structures:


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1.4Units And Dimensions :

a- Dimensions:

Length Time Mass ForceL T M F

Mass:is the property of a body of fluid that is a measure of its inertia or

resistance to a change in motion. It is also a measure of the quantity of

fluid.

Force (Weight):is the amount that a body weights, that is, the force with

which a body is attracted towards the earth by gravitation.

- Types of Dimensions:

MLT System

FLT System

Force (F) = Mass (M) x Acceleration (L/T2)

F = MLT-2

b- Units:

System Quantity

Length (L) Time (T) Mass (M) Force (F)

International (SI) M sec Kg N

French (c.g.s) Cm sec gm dyne

British (English) Foot (ft) sec slug Pound (Ib)

Kilogram weigth M sec Kg Kgw

1 N = 1 Kg x 1 m/s

2

1 dyne = 1 gm x 1 cm/s2

1 Ib = 1 slug x 1 ft/s2

- Some conversion factors:

a-Length (L):

1 ft = 12 inch 1 inch = 2.54 cm 1 ft = 0.3048 m

b-Mass (M):1 slug = 14.59 Kg


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c-Force (F):

1 Kgw= 9.81 N 1 gmw= 981 dyne 1 N = 105dyne

1 Ib = 4.44 N 1 Kgw= 2.205 Ib 1 Ib = 453.60 gm

Note: Acceleration due to gravity, g = 9.81 m/s2 = 32.20 ft/s2.

Example 1-1

Convert the following:

1.5Main Water properties:

1- Density ():

Density (): mass per unit volume

.

= M/ V = ML-3

(Dimension)

= Kg/m3=

gm/cm

3=

slug/ft

3(units)

water= 1000 Kg/m3= 1 gm/cm

3= 1.94 slug/ft

3

2- The specific weight () :

The specific weight () = weight per unit volume

= W/ V = FL-3 (Dimension)


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= g

= Kgw/m3= N/m

3=

dyne/cm

3=

Ib/ft

3(units)

water= 9810 N/m3

=981 dyne/cm

3= 62.42

Ib/ft

3

Water reaches a maximum density at 4C. It becomes less dense when

heated.

4. ,

Density of sea water about 4% more than that of fresh water. Thus, when

fresh water meets sea water without sufficient mixing, salinity increases

with depth.

%.4

Temperature (C) Density (, kg/m3

) Specific Weight (, N/m3

)

0(ice) 917 8996

0 (water) 999 9800

4 1000 9810

10 999 9800

20 998 9790

30 996 9771

40 992 9732

50 988 9692

60 983 9643

70 978 9594


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80 972 9535

90 965 9467

100 958 9398

Example 1-2

A cylindrical water tank (see the figure) is suspended vertically by its

sides. The tank has a 2-m diameter and is filled with 40C water to 1 m in

height. Determine the force exerted on the tank bottom.

3- Specific gravity (S.G), Relative density (R.D) :

Specific gravity (S.G): the ratio of the specific weight of any liquid

to that of water at 4C.

.4


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S.G = R.D = lquid/ water = lquid/ water (no units)

(for exampleS.G Hg= 13.60)

Example 1-3

Determine the specific weight, density and specific gravity of a liquid that

occupies a volume of 200 lit. , and weighs 178 kg. Will this fluid float on

the surface of an oil of specific gravity (0.8)? Provide results in SI units.

Solution

4- Surface Tension :

Surface tension (): A liquids ability to resist tension.

.

The surface tension () of a liquid is usually expressed in the units of force

per unit length.

At the interface between a liquid and a gas, i.e., at the liquid surface, and at

the interface between two immiscible (not mix-able) liquids, the out-of-

balance attraction force between molecules forms an imaginary surface

film, which exerts a tension force in the surface. This liquid property is

known as surface tension.

,)

(


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,

.

Molecular forces

:

Cohesion :Cohesion enables a liquid to resist tensile stress (inner

force between liquid molecules) .

Adhesion

:adhesion enables it to adhere to another body (attractionforce between liquids, and a solid surface).

.


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4-1 Capillarity

:

Capillary effect is the rise or fall of a liquid in a small-diameter tube. It

is caused by surface tension.

4-2 Water droplets

Dh

cos4

DP

4


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Example 1-4

Calculate the capillary effect in millimeters in a glass tube of 5 mm diam.,

when immersed in (i) water and (ii) in mercury (S.G. = 13.6). The

temperature of liquid is 20 C and the values of surface tension of water

and mercury at this temperature in contact with air are 0.0075 kg/m and

0.052 kg/m respectively. The contact angle for water = 0 and for mercury

= 130.

Solution

In water:

In mercury: = -1.97 mm.

Example 1-5

Calculate the internal pressure of a 25 mm diameter soap bubble if the

tension in the soap film is 0.5N/m.

Solution


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5- Viscosity :

Consider that oil fills the space between two parallel plates at a distance

ya part. A horizontal force Fis applied to the upper plate and moves

it to the right at velocity vwhile the lower plate remains stationary. Theshear force F is applied to overcome the oil resistance , and it must

be equal to because there is no acceleration involved in the process.

The proportionally constant, , is called dynamicviscosity of the fluid

the absolute viscosity of the fluid.


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The perviouse equation is commonly known at Newton's law of viscosity.

Most liquids abide by this relationship and are called Newtonian fluids.

Liquids that do not abide by this linear relationship are known as non-

Newtonian fluids. These include most house paints and blood.

The absolute viscosity has the dimension of force per unit area (stress)

times the time interval considered. It is usually measured in the unit of

poise.

1 poise =0.1 N sec/m2

6- Kinematic viscosity:

Kinematic viscosity, , is obtained by dividing the absolute viscosity by

the mass density of the fluid at the same temperature;

= /= M L-1

T-1

/ M L-3

= L2T

-1

The kinematic viscosity unit is cm2

/sec Stoke


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The absolute viscosities and the kinematic viscosities of pure water and air

are shown as functions of temperature in the next table.

.

,

.

Water Air

Temperature Absolute Kinematic Absolute Kinematic

(C) Viscosity Viscosity Viscosity Viscosity

N sec/m2

cm2

/sec N sec/m2

cm2

/sec

0 1.781 x l0-3

.785 x 10-6

1.717 x 10-5 .329 x 10-5

5 1.518 x l0-3

.519 x 10-6 1.741 x l0-5

.371 x l0-5

10 1.307 x 10- .306 x l0- 1.767x 10- .417 x l0-

15 1.139 x l0- 1.139 x l0- 1.793 x l0- .463 x 10-

20 1.002 x l0- 1.003 x l0- 1.817 x l0- .509 x 10-

25 0.890 x 10- 3.893 x 10- 1.840 x l0- .555 x10-

30 0.798 x l0-3

3.800 x 10-6 1.864 x 10-' .601 x 10-5

40 0.653 x 10- 3.658 x 10-6 1.910 x l0- .695 x 10-

50 0.547 x l0- 1553 x 10- 1.954 x l0- .794 x l0-

60 0.466 x 10- 3.474 x 10-6 2.001 x 10-5 .886 x l0-

70 0.404 x 10- 3.413 x 10- 2.044 x l0- .986 x 10-

80 0.354 x 10-3

3.364 x I0-6

2.088 x 10-5

-.087 x 10-5


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90 0.315 x l0-3

3.326 x l0-6

2.131 x l0-5

2.193 x l0-5

100 0.282 x 10- 1294 x 10- 2.174 x l0- -.302 x l0-

Example 1-6

A thin plate weighting 125gm takes 3.3 sec. to fall a distance of one meter

in a liquid with a specific gravity of 0.9 between two vertical boundaries

0.5 cm apart, the plate moves at a distance of 0.2 cm from one of them. If

the surface area of the plate is 1.5 m2. What are the dynamic viscosity of

the liquid in poise and the kinematic viscosity in stokes.

Solution


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7- Compressibility of liquids :

Compressibility (change in volume due to change in pressure) is inversely

proportional to its volume modulus of elasticity (Bulk Modulus of

Elasticity K).

Water is often considered incompressible, but it does have a finite, lowcompressibility.


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CHAPTER 2

FLUID PRESSURE AND HYDROSTATIC FORCE

2.1 Hydrostatic Pressure:

For liquids at rest, the pressure at all points in a horizontal plane is the

same.

2.2 Absolute and Gage Pressures

If the water body has a free surface that is

exposed to atmospheric pressure, Patm. Point A ispositioned on the free surface such that PA= Patm

(PB)

abs= P

A + h = P

atm+ h = absolute pressure


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The difference in pressure heads at two points in liquid at rest is always

equal to the difference in elevation between the two points.

Gage pressure: is the pressure measured with respect to atmospheric

pressure (using atmospheric pressure as a base).

Atmospheric pressure:

Patm = 10.33m high column of water = 10.33 = 10.33 x 1 x 9810 =

= 0.76m high column of Hg = 0.76 Hg

= 0.76 x 13.60 x 9810 =

= 101.3 x 103N/m

2(Pa) 1 x 10

5N/m

2(Pa) 1 bar

= 14.70 Ib/in2= 2116 Ib/ft

2

Absolute pressure:Pabs= Pgage+ Patm

If Pabsis negative it is called Vacum Pressure.

the water body has a free surface that is exposed to

atmospheric pressure, Patm

. Point A is positioned on

the free surface such that PA= Patm

Pressure head, h = P/

(PB/)

(PA/) = (h)


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2.3 Pascal's law :

Pascal's law states that the pressure intensity at a point in a fluid at rest is

the same in all directions.

Consider a small prism of fluid of unit thickness in the z-direction

contained in the bulk of the fluid as shown below. Since the cross-section

of the prism is equilateral triangle, P3is at an angle of 45 with the x-axis.

If the pressure intensities normal to the three surfaces are P1, P2, P3, as

shown then since:-

Force = Pressure x Area

Force on face AB = P1x (AB x I)

BC = P2x (BCx 1)

AC = P3x (ACx 1)

Resolving forces vertically:

P1x AB = P3 x AC cos

But AC cos = AB Therefore P1= P3Resolving forces horizontally:

P2x BC = P3 x AC sin

But AC sin = BC Therefore P2= P3

Hence P1= P2= P3,

In words: the pressure at any point is equal in all directions.

Pressure Variation in A static fluid

For a static fluid, pressure varies only with elevation within the fluid. This

can be shown by consideration of equilibrium of forces on a fluid element


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Newton's law (momentum principle) applied to a static fluid

F = ma = 0 for a static fluid

i.e., Fx= Fy= Fz= 0

Fz= 0

pdxdy p p

zdz dxdy gdxdydz ( )

0

p

zg

Basic equation for pressure variation with elevation

0y

p

0dxdz)dyy

pp(pdxdz

0Fy

0x

p

0dydz)dxx

pp(pdydz

0Fx

For a static fluid, the pressure only varies with elevation z and is constant

in horizontal xy planes.


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Pressure Variation for a Uniform-Density Fluid

p

zg

p z 2 1 2 1p p z z Alternate forms:

1 1 2 2p z p z p z

p z 0 0

i.e., p z pz

Pascal's Application:

A small force F1 applied to a piston with a smallarea produces a much larger force F2 on the larger

piston. This allows a hydraulic jack to lift heavy objects.

= constant for liquid

constant

constant (piezometric pressure)

gage

increase linearly with depth

constant (piezometric head)


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Example 2-1

The diameters of cylindrical pistons A and B are 3 cm and 20 cm,

respectively. The faces of the pistons are at the same elevation and the

intervening passages are filled with an incompressible hydraulic oil. A

force P of 100 N is applied at the end of the lever, as shown in the

figure. What weight Wcan the hydraulic jack support?

Solution

Balancing the moments produced by P and F, we obtain for the

equilibrium condition.

Thus,

From Pascal's law, the pressure PAapplied at A should be the same as thatof PBapplied at B. We may write


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2.4 Free Surface of Water

Free surface of liquid in a vessel may be subjected to:

- atmospheric pressure (open vessel) or,

- any other pressure that is exerted in the vessel (closed vessel).

2.5 Surface of Equal Pressure

- The hydrostatic pressure in a body of liquid varies with the vertical

distancemeasured from the free surface of the water body.

All points on a horizontal surfacein the liquid have the same pressure.

2.6 Manometers:

A manometer: Is a tube bent in the form of a U containing a fluid of

known specific gravity. The difference in elevations of the liquid surfaces

under pressure indicates the difference in pressure at the two ends.

The liquid used in a manometer is usually heavier than the fluids to be

measured. It must not mix with the adjacent liquids (i.e., immiscible

liquids).


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The most used liquids are:

- Mercury (specific gravity = 13.6),

- Water (sp. gr. = 1.00),

- Alcohol (sp. gr. = 0.9), and- Other commercial manometer oils of various specific gravities.

Two types of manometers:

1.An open manometer: has one end open to atmospheric pressure and

is capable of measuring the gage pressure in a vessel.

2. A differential manometer: connects each end to a different pressure

vessel and is capable of measuring the pressure difference between the two

vessels.

Inverted U-tube manometer is used when there is a small pressure

difference or a light liquid such as oil is used.


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Example 2-2

Determine the pressure difference P

Solution

The pressures at points 3 and 4 are, respectively,

So that,

and,


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2.7 Hydrostatic Force on a plane Surface

For a strip at depth h below the free surface:

The differential force dF acting on the element dA is:

Integrate both sides and note that and are constants,

Location of Total Hydrostatic Force (Centre of Pressure)

From the figure above, S is the intersection of the prolongation of the

submerged area to the free liquid surface. Taking moment about point S.

A.sin dydF

sin yhP

hF A..

ydydFFAA

.A.sinAsin first moment of area about point S

dFyyFP.


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Where,

Again from Calculus, is called moment of inertia denoted byI. Since our reference point is S,

Thus,

By transfer formula for moment of inertia the formula

for ypwill become or

From the figure above, , thus, the distance between

cgand cpis

A.sin dydF

hF A..


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Example 2-3

Determine F and yP

Solution


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2.8 Hydrostatic Forces on Curved Surfaces

For a curved surface, the pressure forces, being normal to the local area

element, vary in direction along the surface and thus cannot be added

numerically. The figure shows pressure distribution on a semi-cylindricalgate.

,

F,

curved surface

.

.

The resultant hydrostatic force is computed by considering the free-body

diagram of a body of fluid in contact with the curved surface, as illustrated

below.

A

B

A

BF

A

B

F

C

W

FCB

FAC

Free-body


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The steps involved in calculating the horizontal and vertical components of

the hydrostatic forceFare as follows:

Summation of forces in the horizontal direction FHgives

FH = FAC

where FAC is the hydrostatic force on plane surface AC. It acts

through the center of pressure of side AC.

FH curved surface

.

Summation of forces in the vertical direction FV gives

FV = W + FCB

where W is the weight of the fluid (acting through the center of

gravity) of the free-body diagram and FCBis the hydrostatic force

(acting through the centroid) on the surface CB.

FV curved

surface

free surface.

The line of action of FVis obtained by summing the moments

about any convenient axis.

The overall resultant force is found by combining the vertical and

horizontal components vectorialy:

The angle the resultant force makes to the horizontal is:

The position of Fis the point of intersection of the horizontal

line of action of FH and the vertical line of action of FV .

22

VH FFF

H

V

F

F1tan


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Example 2-4

Determine the total hydrostatic pressure and the center of pressure on the 3

m long, 2m high quadrant gate in the shown figure.

Solution

The horizontal component is equal to the hydrostatic pressure force on the

projection plane A/B.


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CHAPTER 3

FLUID MASSES SUBJECTED TO ACCELERATION

3.1 Fluid Mass Subjected to Horizontal Acceleration:

Consider a tank felled with liquid moves towards the right side with

a uniform acceleration a.

As the tank stats moving under the action of acceleration force, the

liquid does not remain in horizontal level.

The liquid surface falls down on the direction of motion and rise up

on the back side of the tank, the liquid surface makes angle with

the horizontal.

Consider the equilibrium of a fluid particle a lying on the inclined

free surface as shown in the figure.

The force acting on the liquid particle

are weight of the particle W = mg

acting vertically downwards,

accelerating force F acting toward

right and normal pressure P exerted

by the liquid.


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Acceleration force F = ma

Resolving horizontally P sin = F = ma (i)

Resolving vertically P cos = W = mg (ii)

Dividing Equation (i) to (ii)P sin / P cos = ma / mg = a/g

tan = a/g

uniform velocity

.

Now consider the equilibrium of the entire mass of the liquid. Let P1and

P2are the hydrostatic pressure on the back side and front side of the tank.

Net force P = P1- P2

By Newtons second law of motion,

P = ma

(P1-P2) = ma

Example 3-1

An open rectangular tank 3m long. 2.5m wide and 1.25m deep is

completely filled with water. If the tank is moved with an acceleration of

1.5m/s2, find the slope of the free surface of water and the quantity of

water which will spill out of the tank.


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Solution

Slope of the free surface of water

Quantity of water which will spill out of the tank

From the above figure we can see that the depth of water on the front side,

Quantity of water which will spill out of the tank,

3.2 Fluid Mass Subjected to Vertical Acceleration :

Consider a tank containing liquid and moving vertically upwards with a

uniform acceleration. Since the tank is subjected t an acceleration in the

vertical direction only, therefore the liquid surface will remain horizontal.

Now consider a small column of liquid of height h and area dA as shown

in the figure.


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The force acting on the liquid column s the weight of liquid column W

acting vertically downwards, acceleration force F and pressure force

exerted by the liquid on the column.ApplyingNewtonssecond law of motion,

Fy= m ay

P - W = m ay

p.dA - hdA = hdA / g. a

p.dA = hdA/ g .a + hdA

p = h (1+ a/g)

a ,

.

If the liquid is moving vertically downward with a uniform acceleration,

the pressure will be,

p = h (1 -a/g)

Example 3-2

An open rectangular tank 4m long and 2.5m wide contains an oil of

specific gravity 0.85 up to a depth of 1.5m. Determine the total pressure on

the bottom of the tank, when the tank is moving with an acceleration of

g/2m/s2 (i) vertically upwards (ii) vertically downwards.

Solution

(i) Total pressure on the bottom of the tank, when it is vertically upwards:

oil= 0.85 x 9810 = 8.34 KN/m3

p = h (1 + a/g) = 8.34x 1.50 (1 + g/2g) = 18.765 KN/m2

Total pressure force on the bottom of the tank,

P = 18.765 x (4 x 2.50) = 187.65 KN


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(ii) Total pressure on the bottom of the tank, when it is vertically

downwards:

p = h (1 - a/g) = 8.34 x 1.50 (1 - g/2g) = 6.255 KN/m2

Total pressure force on the bottom of the tank,

P = 6.255 x (4 x 2.50) = 62.55 KN

3.3 Accelerated along inclined plane:

Consider a tank open at top, containing a liquid and moving upwards along

inclined plane with a uniform acceleration as shown in the figure.

We know that when the tank starts moving, the liquid surface falls down

on the front side and rises up on the back side of the tank as shown in the

figure.


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Let,

= Inclination of the plane with the horizontal

= Angle, which the liquid surface makes with the horizontal, and

a = Acceleration of the tank

Acceleration force F = ma

Resolving horizontally P sin = FH= F cos= m a cos (i)

Resolving vertically P cos = W+ FV= mg + m a sin (ii)

Dividing Equation (i) to (ii)

P sin / P cos = ma cos/ mg + m a sin

tan = a cos/ g + a sin

tan = aH/(g+aV)

If the liquid is moving downward,

tan = aH/(g-aV)

where,aH= horizontal component of the acceleration

aV= vertical component of the acceleration

Example 3-3

A rectangular box containing water is accelerated at 3 m/s2upwards on an

inclined plane 30 degree to the horizontal. Find the slope of the free liquid

surface.

Solution

aH= 3 cos 30 = 2.598 m/s2

aV= 3 sin 30 = 1.50 m/s2

tan = 2.598/(9.81+1.50) = 0.2297

= 12.940


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3.4 Uniform Rotation about Vertical Axis:

When at rest, the surface of mass of liquid is horizontal at PQas shown in

the figure. When this mass of liquid is rotated about a vertical axis at

constant angular velocity radian per second, it will assume the surfaceABCwhich is parabolic. Every particle is subjected to centripetal force or

centrifugal force

CF = m2xwhich produces centripetal

acceleration towards the center of rotation. Other forces that acts are

gravity force W = mgand normal force N.

Where tan is the slope at the surface of paraboloid at any distance x from

the axis of rotation.

From Calculus, y = slope, thus


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For cylindrical vessel of radius r revolved about its vertical axis, the

height hof paraboloid is:

=


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CHAPTER 4

EQUILIBRIUM OF FLOATING BODIES

4.1 Principle of Buoyancy

The general principle of buoyancy is expressed in the Archimedes

principle, which is stated as follows:

For an object partially or completely submerged in a fluid, there is a net

upward force (buoyant force) equal to the weight of the displaced fluid.

,"

" " "

.

Weight of floating body (W) = Weight of liquid displaced (Fb)

W = Fb

b. Vb= w . Vsub

S.G.w. LBD = w . LBhsub

hsub =S.G.D (in this example)

Where Vsubis the displaced volume (see the figures below).


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4.2 Stability of Floating Bodies

Any floating body is subjected by two opposing vertical forces. One is the

body's weight Wwhich is downward, and the other is the buoyant force

Fbwhich is upward. The weight is acting at the center of gravity G

and the buoyant force is acting at the center of buoyancy B.

W

G

,"

"

Fb

B. ,

Center of Bouyancy"

" :


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It is the point of application of the force of buoyancy on the body.

It is always the center of gravity of the volume of fluid displaced.

Wand Fbare always equal and if these forces are collinear, the body

will be in upright position as shown below.

The body may tilt from many causes like wind or wave action causing the

center of buoyancy to shift to a new positionB/ as shown below.

Metacenter M:is the intersection point of the axis of the body and the

line of action of the buoyant force.

If Mis above G, Fband Wwill produce a righting moment

RMwhich causes the body to return to its neutral position, thus

the body is stable.

If M is below G, the body becomes unstablebecause of the

overturning moment OMmade by Wand Fb.

If Mcoincides with G, the body is said to be just stable which

simply means critical neutral.


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:

M

G

:

:

M

G

,WFb.

:

M

G ,WFb.

:

M

G ,WFb

,

.

The value of righting moment or overturning moment is given by

The distance MG is called metacentric height.

Metacentric hight MG:is the distance between the centre of gravity of

a floating body ant the metacenter.

Use (-) if Gis above Band (+) if G is below B.

Note that: Mis always above B.

Value of MB:


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Assume that the body is rectangular at the top view and measures Bby

Lat the waterline when in upright position.

The moment due to the shifting of the buoyant force is equal to themoment due to shifting of wedge.

For small value of , tan sin and note that 1/12 LB3= I, thus,


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The formula above can be applied to any section.

Where

v = volume of the wedge either immersion or emersion

s = horizontal distance between the center of gravity of the wedges = angle of tilting

I = moment of inertia of the waterline section of the body

Some remarks about the weight of an object

The weight of an object in a fluid medium refers to the tension in the

spring when the object is attached to a spring balance.

The weight (of the object) registered by the spring balance depends on

the medium in which it is measured. [See the illustration below.]

The weight commonly referred to in daily use is the weight in air.


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CHAPTER 5

KINEMATICS OF FLUID FLOW

5.1 Kinematics :

Kinematics of fluid describes the fluid motion and its consequences

without consideration of the nature of forces causing the motion.

(

, )

5.2 Difference between open-channel flow and the pipe flow:

Pipe flow

:

The pipe is completely filled with the fluid being transported.

The main driving force is likely to be a pressure gradient along the pipe.

.

Open-channel flow

:

Fluid flows without completely filling the pipe.

Gravity alone is the driving force, the fluid flows down a hill.

,

.


54/215



5.3 Types of Flow

:

Steady and Unsteady flow

:

The flow parameters such as velocity (v), pressure (P) and density (r) of a

fluid flow are independent of time in a steady flow. In unsteady flow they

are independent.

Uniform and non-uniform flow

:

A flow is uniform if the flow characteristics at any given instant remain

the same at different points in the direction of flow, otherwise it is termed

as non-uniform flow.

.

.

Examples of flow types:

For a steady flow 0ooo ,z,yx

tv

0ooo ,z,yx

tvFor an unsteady flow

0ot

svFor a uniform flow

For a non-uniform flow 0ot

sv


55/215



Laminar and turbulent flow

)(

:

Laminar flow)(

The fluid particles move along smooth well defined path or streamlines

that are parallel, thus particles move in laminas or layers, smoothly gliding

over each other

Turbulent flow

The fluid particles do not move in orderly manner and they occupy

different relative positions in successive cross-sections.

There is a small fluctuation in magnitude and direction of the velocity of

the fluid particles

Transitional flow

The flow occurs between laminar and turbulent flow

.

,


56/215




57/215



Reynolds Experiment:

Reynold performed a very carefully prepared pipe flow experiment.

Reynold found that transition from laminar to turbulent flow in a pipe

depends not only on the velocity, but only on the pipe diameter and theviscosity of the fluid.

Reynolds number is used to check whether the flow is laminar or

turbulent. It is denoted by Rn. This number got by comparing inertial

force with viscous force.

Where

V: mean velocity in the pipe [L/T]

D: pipe diameter [L]

: density of flowing fluid [M/L3]

ForcesViscous

ForcesInertialVDVDRn


58/215



: dynamic viscosity

[M/LT]

: kinematic viscosity

[L2/T]

"

"

."

" . "

"

The Kind of flow depends on value of Rn

If Rn< 2000 the flow is Laminar

If Rn> 4000 the flow is turbulent

If 2000 < Rn< 4000 it is called transition flow.

Laminar Vs. Turbulent flows

.

.

:

"

"

.)

(

Laminar flows characterized by: low velocities

small length scales

high kinematic viscosities

Rn< Critical R

n

Viscous forces are dominant

Turbulent flows characterized by high velocities

large length scales

low kinematic viscosities

Rn> Critical R

n

Inertial forces are dominant


59/215



Example 5-1

40 mm diameter circular pipe carries water at 20oC. Calculate the largest

flow rate (Q) which laminar flow can be expected.

Solution

Streamlines and Streamtubes:

Streamline ""

:

A curve that is drawn in such a way that it is tangential to the velocity

vector at any point along the curve. A curve that is representing the

direction of flow at a given time. No flow across a stream line.

.

Streamtube""

:

A set of streamlines arranged to form an

imaginary tube. Ex: The internal surface of a

pipeline.

.

sec/1028.6)04.0(4

05.0. 352 mAVQ

sec/05.02000101

)04.0(2000

6 mV

VVDRn

CTat o20101 6

mD 04.0


60/215



Compressible And Incompressible Flows:

Incompressible Flowis a type of flow in which the density () is constant

in the flow field.

Compressible Flow is the type of flow in which the density of the fluid

changes in the flow field.

Ideal and Real Fluids:

a- Ideal Fluids

It is a fluid that has no viscosity, and incompressible

Shear resistance is considered zero

Ideal fluid does not exist in nature e.g. Water and air are assumed ideal

b- Real Fluids

It is a fluid that has viscosity, and compressible

It offers resistance to its flow e.g. All fluids in nature

,.

Rotational And Irrotational Flows:

Rotational flow is the type of flow in which the

fluid particles while flowing along stream-lines also

rotate about their own axis.

Ir-rotational flowis the type of flow in which the

fluid particles while flowing along stream-lines donot rotate about their own axis.

One, Two And Three Dimensional Flows:

One-dimensional flowis the type of flow in which flow parameters (such

as velocity, pressure, depth etc.) are a function of time and one space

coordinate only.

v = f(x, t)


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e.g. Flow through a straight uniform diameter pipe

The flow is never truly 1 dimensional, because viscosity causes the fluid

velocity to be zero at the boundaries.

Two-dimensional flow is the type of flow in which flow parameters

describing the flow vary in two space coordinates and time.

v = f(x, y, t)

Streamlines in two-dimensional flow are curved lines on a plane and are

the same on all parallel planes. An example is flow over a weir which

typical streamlines can be seen in the figure below.

Three-dimensional flowis the type of flow in which the flow parameters

describing the flow vary in three space coordinates and time.

v = f(x, y, z, t)

Although in general all fluids flow three-dimensionally, with pressures and

velocities and other flow properties varying in all directions, in many cases

the greatest changes only occur in two directions or even only in one.


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5.4 Volume flow rateDischarge :

The discharge is the volume of fluid passing a given cross-section per unit

time.

5.5 Mean Velocity:

It is the average velocity passing a given cection.

The velocity in the pipe is not constant across the cross section. Crossing

the centreline of the pipe, the velocity is zero at the walls increasing to a

maximum at the centre then decreasing symmetrically to the other wall.

This variation across the section is known as the velocity profile or

distribution.

5.6 Continuity equation for Incompressible Steady flow

:

Cross section and elevation of the pipe are varied along the axial direction

of the flow.


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)(.. '22'11 massfluidfluxmassdVoldVol

Conservation law of mass

Mass enters the

control volumeMass leaves the

control volume

QVAVAdt

dSA

dt

dSA

dtdVol

dtdVol

.......

..

22112

21

1

'22'11

QVAVA 2211 ..Continuity equation for

Incompressible Steady flow


64/215



,

.

Apply Newtons Second Law:

Fxis the axial direction force exerted on the control volume by the wall of

the pipe.

5.7 Energy Head in Pipe Flow

Water flow in pipes may contain energy in three basic forms:

1- Kinetic energy ,

2- potential energy ,

3- pressure energy .

Consider the control volume:

In time interval dt:Water particles at sec.1-1 move to sec. 1`-1` with velocity V1.

t

VMVM

dt

VdMaMF

12

xxx WFAPAPF

2211

)(.

)(.

)(.

.

12

12

12

zzz

yyy

xxx

VVQF

VVQF

VVQF

rateflowmassQtMbut

)(. 12

VVQF

Conservation of

moment equation


65/215



Water particles at sec.2-2 move to sec. 2`-2` with velocity V2.

To satisfy continuity equation:

The work done by the pressure force:

-ve sign because P2is in the opposite direction to distance traveled ds2

The work done by the gravity force:

The kinetic energy:

The total work done by all forces is equal to the change in kinetic energy:

Dividing both sides by gQdt

dtVAdtVA .... 2211

dtVAPdsAP ..... 111111

dtVAPdsAP .....222222

. on section 1-1

. on section 2-2

)(...2

1.

2

1.

2

1 21

2

211

2

1

2

2 VVdtVAVMVM

Bernoulli Equation

Energy per unit weight of water

OR: Energy Head

).(.. 2111 zzdtVAg

)(..2

1

).(......

2

1

2

22121 VVdtQzzdtQgdtQPdtQP

22

2

21

1

2

1

22z

P

g

Vz

P

g

V


66/215



Notice that:

In reality, certain amount of energy loss (hL) occurs when the water

mass flow from one section to another.

The energy relationship between two sections can be written as:

22

2

22

2

zP

g

VH

Kinetichead

Elevationhead

Pressurehead

Energyhead

= + +

11

2

112 z

P

g

VH

LhzPg

VzPg

V 222

21

1

2

1

22

Energy head and Head loss in pipe flow


67/215



5.8 Flow Through Orifices and Mouthpieces :

5.8.1Small Orifice (Under Constant Head) :

Consider large tank filled with a liquid with a small orifice located in the

wall at a depth H below the free surface, the ratio of jet area Aoat vena

contractato area of orifice A is symbolized by coeff icient of contraction

Cc

The coefficient of contraction Cc varies from 0.613 to 0.69

The Bernoullis equation applied between point 1 & 2:

Because of loss of energy in friction, the actual velocity Vais always less

than the theoretical one and hence Cv


68/215



The actual discharge passing the orifice is the product of the actual

velocity at the vena contracta and the area of the jet, thus

The coefficient of discharge Cd is defined as the ratio of the actual

discharge to the theoretical discharge and is expressed by

5.8.2Large Rectangular Orifice :

Large Orifice Discharging Free:


69/215



Consider a rectangular orifice of width B and height D, discharging under

a head H1above its top as shown in the Figure.

The flow through an elementary strip of area B.dh is

Integration of the above equation between limits H1and H2gives

Large Orifice Partially Submerged :

The discharge in this case is considered in two parts:


70/215



Where Q1is the discharge for the free area and Q2is the discharge for the

fully submerged area As.

Where, H is the difference in elevation of water surface. For convenience

the two coefficients of discharge may be assumed the same.

Large Orifice Completely Submerged

:

Applying Bernoullis equation betweenpoint 1 and point 2:


71/215



Where, H is the difference in elevation of water surface, A is the orifice

area and Cdis the coefficient of discharge.

5.8.3Mouthpieces or Short Tubes

External Mouthpieces


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Internal (Re-entrant) Mouthpieces


73/215



5.8.4Discharge Under Varying Head :

Discharge through an orifice or over a notch under falling head is a case of

unsteady flow, since the velocity of flow will be decreasing with time.

Consider a tank of area Afitted with an orifice of area a. The inflow to the

tank is q, which is supposed to be constant. Let at any instant of time the

liquid surface is at a distance habove the orifice and in time dtthe level

falls by dh.

The discharge through the orifice

During time dt the volume of fluid in the tank is reduced by A.dh. We can,

therefore, write

The time t required to lower the fluid level from H1to H2is


74/215



If there be inflow into the tank q = 0, we can, therefore, write

Assume a cross sectional area of tank A is constant, so

5.8.5Flow Over Notch :

Discharge Over A Rectangular Notch

:

Let,

H = Height of the water above the crest of the weir

L = Length of the weir and

Cd= Coefficient of discharge


75/215



Let us consider a horizontal strip of water of thickness dh at a depth h from

the water surface as shown in figure.

Area of the strip =L.dh

The total discharge, over the weir, may be found out by integrating the

above equation within the limits 0 and H.

Example 5-2

A weir of 8m long is to be built across a rectangular channel to discharge a

flow of 9m3 /s. If the maximum depth of water on the upstream side of

weir is to be 2m, what should be the height of the weir? Adopt Cd= 0.62.

Solution

Therefore height of weir should be = 2.0 - 0.72 = 1.28 m.

ghdhLCdq d 2..

H

d ghdhLCQ

0

2..

H

d dhhgLCQ0

.2.

2

3

2.

3

2HgLCQ d

2

3

81.92862.03

29 Hxxx

mH 72.0


76/215



Discharge Over Triangular V Notch

:

If the angle of the "V" is then the Area of the strip b is

So the discharge is

Example 5-3

A right-angled V-notchwas used to measure the discharge of a centrifugal

pump. If the depth of water at V-notch is 200mm, calculate the discharge

over the notch in liters per minute. Assume Cd= 0.62.

Solution

2tan)(2

hHb

dhhhHgCQ

H

d .)(2

tan2.20

H

d hHhgCQ

0

2

5

2

3

52

52

2tan2.2

2

5

2tan2.

15

8HgCQ d

smxxxQ /026.02.02

90tan81.9262.0

15

8 325

min/1560/26 litersslitersQ


77/215



Discharge Over A Cippoletti TrabizoidalWeir:

Let us split up the trapezoidal weir into a rectangular weir and a

triangularnotch.

So total discharge,

21 QQQ

2

5

2

3

2tan2.

15

82.

3

2HgCHgLCQ dd


78/215



5.8.6Discharge through a Venturimeter :

The Venturi meter is a device for measuring discharge in a pipe. It

consists of a rapidly converging section which increases the velocity of

flow and hence reduces the pressure. It then returns to the originaldimensions of the pipe by a gently diverging diffuser section. By

measuring the pressure differences the discharge can be calculated.

Let d1 = diameter at the inlet (section 1)

p1= pressure at section 1

v1= velocity at section 1

A1= area at section1

d2, p2, v2, A2are the corresponding values at the throat (section 2)

Applying Bernoullisequations at sections 1 and 2, we get

As pipe is horizontal z1=z2


79/215



Where h = ( p1-p2)/ g difference of pressure heads at sections 1 and 2.

From the continuity equation at sections 1 and 2, we obtain

Note that the above expression is for ideal condition and is known as

theoretical discharge. Actual discharge will be less than theoreticaldischarge

Cdis the coefficient of venturimeter and its value is always less then 1


80/215



Expression of h given by differential U-tube manometer:

Case 1: The liquid in the manometer is heavier than the liquid flowing

through the pipe:

Case 2: The liquid in the manometer is lighter than the liquid flowing

through the pipe:

5.8.7Orifice meter :

Orifice meter: is a device used for measuring the rate of flow of a fluid

flowing through a pipe.

It is a cheaper device as compared to venturimeter. This also works onthe same principle as that of venturimeter.

It consists of flat circular plate which has a circular hole, in concentric

with the pipe. This is called orifice.

The diameter of orifice is generally 0.5 times the diameter of the pipe

(D), although it may vary from 0.4 to 0.8 times the pipe diameter.


81/215



Let d1 = diameter at the inlet (section 1)



A1= area at section1

d2, p2, v2, A2are the corresponding values at the throat (section 2)

Applying Bernoullis equations at sections 1 and 2, we get

Let A0is the area of the orifice.

By continuity equation, we have


82/215



Thus, discharge,

If Cdis the co-efficient of discharge for orifice meter, which is defined as

Hence,

The coefficient of discharge of the orifice meter is much smaller than thatof a venturimeter.

5.8.8Pitot tube :

Orifice meter:is a device used for measuring the velocity of flow at any

point in a pipe or a channel.

Principle: If the velocity at any point decreases, the pressure at that point

increases due to the conservation of the kinetic energy into pressureenergy.


83/215



In simplest form, the pitot tube consists of a glass tube, bent at right

angles.

.

Let p1= pressure at section 1




H = depth of tube in the liquidh = rise of liquid in the tube

above the free surface

Applying Bernoullis equations at sections 1 and 2, we get

This is theoretical velocity. Actual velocity is given by

Cvis the coefficient of pitot-tube.


84/215



CHAPTER 6

FLOW THROUGH PIPES

6.1 Introduction:

Any water conveying system may include the following elements:

pipes (in series, pipes in parallel)

elbows

valves

other devices

If all elements are connected in series, the arrangement is known as a

pipeline. Otherwise, it is known as a pipe network.

6.2 Flow Through A Single Pipe:

6.2.1Calculation of Head (Energy) Losses:

When a fluid is flowing through a pipe, the fluid experiences some

resistance due to which some of energy (head) of fluid is lost.


85/215



:

,

.

6.2.1.1 Losses of Head due to Friction

Energy loss through friction in the length of pipeline is commonly

termed the major loss hf

This is the loss of head due to pipe friction and to the viscous

dissipation in flowing water.

The resistance to flow in a pipe is a function of:

The pipe length, L

The pipe diameter, D

The mean velocity, V

The properties of the fluid ()

The roughness of the pipe, (the flow is turbulent).


86/215



Several formulas have been developed in the past. Some of these

formulas have faithfully been used in various hydraulic engineering

practices.

Darcy-Weisbach formula

The Hazen -Williams Formula

The Manning Formula

The Chezy Formula

Darcy-Weisbach Equation

Friction Factor: (F)

For Laminar flow: (Rn< 2000) [depends only on Reynolds number

and not on the surface roughness]

For turbulent flow in smooth pipes (e/D = 0) with 4000 < Rn< 105is

The thickness of the laminar sublayer decrease with an increase in

Rn.

25

22

8

2

Dg

QLF

g

V

D

LFhL

Where:F is the friction factorL is pipe lengthDis pipe diameterQis the flow ratehLis the loss due to friction

nR

64F

4/1

316.0

nRF


87/215



Moody diagram

A convenient chart was prepared by Lewis F. Moody and commonly

called the Moody diagram of friction factors for pipe flow, There are

4 zones of pipe flow in the chart:

A laminar flow zone where F is simple linear function of Rn

A critical zone (shaded) where values are uncertain because

the flow might be neither laminar nor truly turbulent

A transition zone where F is a function of both Rnand relative

roughness

A zone of fully developed turbulence where the value of F

depends solely on the relative roughness and independent of the

Reynolds Number


88/215




89/215



Typical values of the absolute roughness (e) are given in the next table

Example 6-1

The water flow in Asphalted cast Iron pipe (e = 0.12mm) has a diameter20cm at 20

oC. Is 0.05 m

3/s. determine the losses due to friction per 1 km

Solution


90/215



Hazen-Williams


91/215



Manning Formula

This formula has extensively been used for open channel design, it is also

quite commonly used for pipe flows


92/215



The Chezy Formula

Where C = Chezy coefficient

It can be shown that this formula, for circular pipes, is equivalent to

Darcys formula with the value for

F is Darcy Weisbeich coefficient

The following formula has been proposed for the value of C:

n is the Manning coefficient

Example 6-2

New Cast Iron (CHW= 130, n = 0.011) has length = 6 km and diameter =

30cm. Q= 0.32 m3/s, T=30

o. Calculate the head loss due to friction.

Solution

V C R S h 1 2 1 2/ /

2

4

C

V

D

Lhf

F

gC

8

hR

n

S

nSC

)00155.0

23(1

100155.023


93/215



Hazen-William Method

Manning Method

6.2.1.2 Minor losses

:

It is due to the change of the velocity of the

flowing fluid in the magnitude or in direction

[turbulence within bulk flow as it movesthrough and fitting]

The minor losses occurs du to :

Valves

Tees, Bends

Reducers

And other appurtenances

It has the common form

minor compared to friction losses in long pipelines but, can be the

dominant cause of head loss in shorter pipelines.

33332030130

6000710

710

8521

8748521

85 21

8748521

m..

.h

Q

DC

L.h

.

..f

.

..

HW

f

m

.

...h

D

nQLh

.f

.f

47030

32001106000310

3.10

335

2

335

2

2

22

22 gA

Qk

g

Vkh LLm


94/215



Losses due to Sudden contraction :

A sudden contractionin a pipe usually causes a marked drop in pressure

in the pipe due to both the increase in velocity and the loss of energy to

turbulence.

.


95/215



Head Loss Due to Gradual Contraction

:

Head losses due to pipe contraction may be greatly reduced by introducing

a gradual pipe transiti onknown as a confusor

Losses due to Sudden Enlargement :

Note that the drop in the energy line is much larger than in the case of a

contraction

g

V'k'h cc

2

2

2

g

VVh

L 2

2

21


96/215



Head Loss Due to Gradual Enlargement

:

Head losses due to pipe enlargement may be greatly reduced by

introducing a gradual pipe transitionknown as a diffusor

Loss due to pipe entrance:

General formula for head loss at the entrance of a pipe is also expressed in

term of velocity head of the pipe

g

VV'k'h EE

2

2

2

2

1

g

VKh entent

2

2


97/215



Head Loss at the Exit of a Pipe:

Head Loss Due to Bends in Pipes :

h V

gL

2

2

gVkh bb2

2


98/215



6.3 Compound Pipe flow :

When two or more pipes with different diameters are connected together

head to tail (in ser ies) or connected to two common nodes (in parallel).

The system is called compound pipe flow

6.3.1Flow Through Pipes in Series :

Pipes of different lengths and different diameters connected end to end (in

series) to form a pipeline

Discharge:The discharge through each pipe is the same

Head loss:The difference in liquid surface levels is equal to the sum of

the total head loss in the pipes:

Where

332211 VAVAVAQ

LBBB

AAA hz

g

VPz

g

VP 22

22

Hhzz LBA

4

1

3

1 j

mj

i

fiL hhh

g

VK

g

VK

g

VK

g

VK

g

V

D

LFh exitenlcent

i

i

i

iiL

22222

2

3

2

2

2

2

2

13

1

2


99/215



6.3.2Flow Through Parallel Pipe s

If a main pipe divides into two or more branches and again join together

downstream to form a single pipe, then the branched pipes are said to be

connected in parallel (compound pipes). Points A and B are called nodes.

Discharge:

Head loss: the head loss for each branch is the same

3

1

321

i

iQQQQQ

321 fffL hhhh

g

V

D

LF

g

V

D

LF

g

V

D

LF

222

2

3

3

33

2

2

2

22

2

1

1

11


100/215



Example 6-3

In the figure shown two new cast iron pipes in series, D1= 0.6 m, D2= 0.4

m length of the two pipes is 300m, level at A =80 m, Q = 0.5 m3/s (F1=

0.017, F2= 0.018). There are a sudden contraction between Pipe 1 and 2,

and Sharp entrance at pipe 1, and Sharp outlet at pipe 2.

1-Find the water level at B, and draw T.E.L & H.G.L

2-If the two pipes are connected in parallel to each other, and the

difference in elevation is 20 m. Find the total flow.

Solution

1,27.0,5.0 exitcent

KKK


101/215



Pipes are connected in parallel

Example 6-4

1-Determine the flow rate in each pipe (F=0.015)

2-Also, if the two pipes are replaced with one pipe of the same length

determine the diameter which give the same flow.


102/215



Solution

If the two pipes are replaced with one pipe of the same length


103/215



6.4 Pipe line with negative Pressure (syphon phenomena)

:

It is a long bent pipe which is used to transfer liquid from a reservoir at a

higher elevation to another reservoir at a lower level when the two

reservoirs are separated by a hill or high ground

Occasionally, a section of the pipeline may be raised to an elevation that is

above the local HGL .


104/215



Negative pressure exists in the pipelines wherever the pipe line is raised

above the hydraulic gradient line (between P & Q)

The negative pressure at the summit point can reach theoretically -10.33

m water head (gauge pressure) and zero (absolute pressure). But in the

practice water contains dissolved gasses that will vaporize before -10.33 m

water head which reduces the pipe flow cross section.

Example 6-5

Syphon pipe between two pipe has diameter of 20cm and length 500m as

shown. The difference between reservoir levels is 20m. The distance

between reservoir A and summit point S is 100m. Calculate the flow in the

system and the pressure head at summit. F=0.02


105/215



Solution


106/215



6.5 Three-reservoirs problem :

This system must satisfy:

The quantity of water brought to junction J is equal to the

quantity of water taken away from the junction:

All pipes that meet at junction J must share the same pressure at

the junction.

6.5.1Types of three-reservoirs problem:

Type 1:

Given the lengths , diameters, and materials of all pipes involved;D1, D2, D3, L1, L2, L3, and e or F

Given the water elevation in any two reservoirs,Z1and Z2 (for example)

Given the flow rate from any one of the reservoirs,Q1 or Q2 or Q3

Determine the elevation of the third reservoir Z3 (for example) andthe rest of Qs.

Q3= Q

1+ Q

2 Flow Direction????

This types of problems can be solved by simply using:

Bernoullis equationfor each pipe Continuity equationat the junction.


107/215



Example 6-6

In the following figure determine the flow in pipe BJ & pipe CJ. Also,

determine the water elevation in tank C

Solution


108/215



Type 2:

given the lengths , diameters, and materials of all pipes involved;

D1, D2, D3, L1, L2, L3, and e or F

given the water elevation in each of the three reservoirs,

Z1, Z2, Z3

determine the discharges to or from each reservoir,

Q1, Q2,and Q3.

1- First assume a piezometric surface elevation,P, at the junction.

2-This assumed elevation gives the head losses hf1, hf2, and hf3

3-From this set of head losses and the given pipe diameters, lengths,

and material, the trail computation gives a set of values for

discharges Q1, Q2,and Q3.

4-If the assumed elevation P is correct, the computed Qs should

satisfy:

5-Otherwise, a new elevationPis assumed for the second trail.

The computation of another set ofQs is performed until the above

condition is satisfied

Example 6-7

In the following figure determine the flow in each pipe.

This types of problems are most convenientlysolved by trail and error


109/215



Solution


110/215




111/215



CHAPTER 7

MOMENTUM PRINCIPLE

7.1 Development of the Momentum Equation :

It is based on the law of conservation of momentum principle, which states

that the net force acting on a f luid mass is equal to the change in

momentum of f low per unit time in that direction .the force acting on

fluid mass m isgiven by the Newton`s second law of motion

F = m x a

Where a is the acceleration acting in the same direction as force F

But

m is constant and can be taken inside the differential

This equation is known as the momentum principle

This equation can be written as:

F dt = d(mv)

Which is known as the impulsemomentum equationand states that:

The impulse of a force acting on a fluid of mass in short interval of time is

equal to the change of momentum (d(mv)) in the direction of force.


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7.2 Force exerted by a flowing fluid on a PipeBend

The impulse-momentum equation is used to determine the resultant

force exerted by a flowing fluid on a pipe bend.

Consider two section (1) and (2), as shown in Fig

Let,

V1= Velocity of flow at section (1),

P1= Pressure intensity at section (1),

A1 = Area of cross-section of pipe at section (1) and ,

V1, P1, A1 Corresponding values of velocity, pressure and area at

section (2).

Let Fx and Fy be the components of the forces exerted by the

flowing fluid on the bend in x and ydirection respectively.

Then the force exerted by the bend on the fluid in the x and y

direction will be equal to Fx and Fybut in the opposite direction.

The momentum equation in x direction is given by:


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P1A1P2A2cos

Fx= (Mass per sec) (change of velocity)

P1A1P2A2cos Fx= Q (V2cos - V1)

Fx= Q (V1 - V2cos ) + P1A1P2A2cos

Similarly the momentum equation in y direction gives

0P2A2sin

Fy=

Q (V2sin

- 0)Fy= Q (- V2sin )P2A2sin

Now the resultant force FRacting on the bend

And the angle made by the resultant force with horizontal direction is

given by

inoutext QVQVF )()(


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7.3 Applications of the Linear Momentum Equation:

A jet of fluid deflected by an object puts a force on the object. This

force is the result of the change of momentum of the fluid and can

happen even though the speed (magnitude of velocity) remainsconstant. If a jet of water has sufficient momentum, it can tip over

the block that deflects it.

The same thing can happen when a garden hose is used to fill

sprinkles.

Similarly, a jet of water against the blades of a Pelton wheel turbine

causes the turbine wheel to rotate.


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SOLVED PROBLEMS


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CHAPTER 1

UNITS AND DIMENSIONS & FLUID PROPERTIES

Problem 1-1: Identify the dimensions and units for the following

engineering quantities and terms (in British, and SI units): Density (),specific weight (),specific gravity (S.G.), discharge (Q), surface tension(), shear stress (), pressure intensity (p), pressure head (p/), dynamicviscosity (),kinematic viscosity (), Linear momentum, angular velocity(),Reynolds number (Rn= VD/),Froude number (Fn= V/gY).

Solution

Problem 1-2: Convert the following:a) A discharge of 20 ft

3/min. to lit/sec.

b) A force of 10 poundals to dynes.

c) A pressure of 30 lb / inch2to gm/cm

2.

d) A specific weight of 62.4 lb / ft3to kg/lit.

e) A density of 7 gm / cm3to slug / ft

3.

f) A dynamic viscosity of 25 dyne.sec/ cm2to lb.sec/ ft

2.

g) A dynamic viscosity of 10 gm / cm.sec to slug / ft.sec


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Solution

Problem 1-3: Determine the specific weight, density and specific gravity

of a liquid that occupies a volume of 200 lit., and weighs 178 kg. Will thisliquid float on the surface of an oil of specific gravity (0.8)? Provide

results in SI units.

Solution


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Since S.G of fluid (0.89) is higher than SG of oil (0.80) so the fluid will

not float over the oil (Higher S.G. means denser fluid)

Problem 1-4: Calculate the capillary rise in mm in a glass tube of 6 mmdiameter when immersed in (a) water, and (b) mercury, both liquids being

at C020. Assume to be 73103N/m for water and 0.5 N/m for mercury.

The contact angle for water and mercury are zero and 1300respectively.

Solution

Problem 1-5: Calculate the internal pressure of a 25 mm diameter soap

bubble if the surface tension in the soap film is 0.5 N/m.

Solution

Problem 1-6: The velocity distribution of a viscous liquid ( = 0.9 N.s/m

2) over a fixed boundary is approximately given by: v = 0.68y - y2 in

which y is the vertical distance in meters, measured from the boundary and

v is the velocity in m/s. Determine the shear stress at the boundary and at

the surface (at y = 0.34m), sketch the velocity and shear stress profiles forthe given flow.


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Solution

Problem 1-7: A square plate of 60 cm side slides down a plane inclined at30

0to the horizontal at a uniform speed of 10 cm/s. The plate weighs 3 kg

and a thin oil film of thickness 1.5 mm fills the spacing between the plate

and the plane. Find the viscosity of the filling oil and the shear at both

surfaces.

Solution


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Problem 1-8: A 25 mm diameter steel cylinder falls under its own weight

at a uniform rate of 0.1 m/s inside a tube of slightly larger diameter. Castor

oil of viscosity 0.25N.s/m2 fills the spacing between the tube and the

cylinder. Find the diameter of the tube. (specific weight of steel = 7.8t/m

3).

Solution

Problem 1-9: A rotating-cylinder viscometer consists of two concentric

cylinders of diameters 5.0 cm and 5.04 cm, respectively. Find the viscosity

of the tested oil which fills the gap between both cylinders to a height of

4.0 cm when a torque of 2 x 105dyne.cm is required to rotate the inner

cylinder at 2000 rpm.


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Solution


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Problem 1-10: A flat circular disk of radius 1.0 m is rotated at an angular

velocity of 0.65 rad/s over a fixed surface. Find the power required to

rotate the disk if an oil f

fluid mechanics (2014 2015) al azhar university

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