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Wasit University- College of Engineering- Civil dept.

Fluid Mechanic

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Unit 1

Introduction

Prior to fluid mechanics, statics, and dynamics, was taken, involve solid

mechanics. Mechanics is the field of science focused on the motion of material

bodies. Mechanics involves force, energy, motion, deformation, and material

properties. When mechanics applies to material bodies in the solid phase, the

discipline is called solid mechanics. When the material body is in the gas or

liquid phase, the discipline is called fluid mechanics. In contrast to a solid, a

fluid is a substance whose molecules move freely past each other. More

specifically, a fluid is a substance that will continuously deform [that is, flow

under the action of a shear stress]. Alternatively, a solid will deform under the

action of a shear stress but will not flow like a fluid. Both liquids and gases are

classified as fluids.

This lecture notes introduces fluid mechanics by describing gases, liquids, and

the continuum assumption. This lecture notes also presents an approach for

using units and primary dimensions in fluid mechanics calculations.

Liquids and Gases

Liquids and gases differ because of forces between the molecules. As shown in

the figure 1.1, a liquid will take the shape of a container whereas a gas will

expand to fill a closed container. The behavior of the liquid is produced by

strong attractive force between the molecules. This strong attractive force also

explains why the density of a liquid is much higher than the density of gas. A


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gas is a phase of material in which molecules are widely spaced, molecules

move about freely, and forces between molecules are minuscule, except during

collisions. Alternatively, a liquid is a phase of material in which molecules are

closely spaced, molecules

move about freely, and there are strong attractive forces between molecules.

Figure 1.1

Civil Engineering Fluid Mechanics

Why are we studying fluid mechanics on a Civil Engineering course? The

provision of adequate water services such as the supply of potable water,

drainage, sewerage is essential for the development of industrial society. It is

these services which civil engineers provide. Fluid mechanics is involved in

nearly all areas of Civil Engineering either directly or indirectly. Some

examples of direct involvement are those where we are concerned with

manipulating thefluid:

Sea and river (flood) defenses;

Water distribution / sewerage (sanitation) networks;

Hydraulic design of water/sewage treatment works;


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Dams;

Irrigation;

Pumps and Turbines;

Water retaining structures.

And some examples where the primary object is construction - yet analysis of

the fluid mechanics is essential:

Flow of air around buildings;

Bridge piers in rivers;

Ground-water flow.

Dimensions and Units,

The dimensions and units that are used in fluid mechanics will be presented.

This information is essential for understanding most aspects of fluid mechanics.

Dimensions

A dimension is a category that represents a physical quantity such as mass,

length, time, momentum, force, acceleration, and energy. To simplify matters,

engineers express dimensions using a limited set that are called primary

dimensions. Table 1.1 lists one common set of primary dimensions. Secondary

dimensions such as momentum and energy can be related to primary

dimensions by using equations. For example, the secondary dimension “force”


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is expressed in primary dimensions by using Newton’s second law of motion,

F= ma. The primary dimensions of acceleration are L/T2, so

In Eq. (1.1), the square brackets mean “dimensions of.” This equation reads “the

primary dimensions of force are mass times length divided by time squared.”

Note that primary dimensions are not enclosed in brackets.

Units

While a dimension expresses a specific type of physical quantity, a unit assigns

a number so that the dimension can be measured. For example, measurement of

volume (a dimension) can be expressed using units of liters. Similarly,

measurement of energy (a dimension) can be expressed using units of joules.

Most dimensions have multiple units that are used for measurement. For

example, the dimension of “force” can be expressed using units of newtons,

pounds-force, or dynes.

The primary units of the SI system are shown in the table 1.1 below:

Table 1.1


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There are many derived units all obtained from combination of the above

primary units. Those most used are shown in the table 1.2 below:

Table 1.2

Unit Systems

In practice, there are several unit systems in use. The International System of

Units (abbreviated SI from the French “Le Système International d'Unités”) is

based on the meter, kilogram, and second. There are other systems in common


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use in the United States. The U.S. Customary System (USCS), sometimes called

English Engineering System, uses the pound-mass (lbm) for mass, the foot (ft)

for length, the pound-force (lbf) for force, and the second (s) for time. The

British Gravitational (BG) System is similar to the USCS system that the unit of

mass is the slug. To convert between these systems, the relationships are


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Unit 2

Fluid Properties

A fluid has certain characteristics by which its physical condition may be

described. These characteristics are called properties of the fluid. This lecture

notes introduces material properties of fluids.

Mass Density, 𝝆

Mass density is defined as the ratio of mass to volume at a point, given by

𝜌 =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑=

𝑚

∀ (2.1)

Mass density has units of kilograms per cubic meter (kg/m3) or pounds-mass per

cubic foot (lbm/ ft3). The mass density of water at 4°C is 1000 kg /m

3 (62.4 lbm

ft3), and it decreases slightly with increasing temperature. The mass density of

air at 20°C and standard atmospheric pressure is 1.2 kg/m3 (0.075 lbm/ft

3), and

it changes significantly with temperature and pressure. Mass density, often

simply called density, is represented by the Greek symbol 𝝆 (rho).

Specific volume, 𝒗𝒔

Specific volume is the reciprocal of the density, that is the volume occupied by

unit mass of fluid, given by

𝑣𝑠 =1

𝜌


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Specific Weight, 𝜸

The gravitational force per unit volume of fluid, or simply the weight per unit

volume, is defined as specific weight. It is given the Greek symbol 𝛾

(gamma).

𝛾 =𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑=

𝑊

∀ (2.2)

Water at 20°C has a specific weight of 9790 N/m3 (or 62.4 lb/ft

3). In contrast,

the specific weight of air at 20°C and standard atmospheric pressure is 11.8

N/m3. Specific weight and density are related by:

Specific Gravity, SG

The ratio of the specific weight of a given fluid to the specific weight of water

at the standard reference temperature 4°C is defined as specific gravity, SG:

The specific weight of water at atmospheric pressure is 9790 N m3. The specific

gravity of mercury at 20°C is

Because specific gravity is a ratio of specific weights, it is dimensionless and

therefore independent of the system of units used.


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Perfect Gas Law

The perfect gas law relates important thermodynamic properties, and is often

used to calculate density. One form of the law is

Where p is the absolute pressure, ∀ is the volume, n is the number of moles, Ru

is the universal gas constant (the same for all gases), and T is absolute

temperature. Absolute pressure, is referred to absolute zero. The universal gas

constant is 8312m. N/kg.mol.K in the SI system and 1545 ft-lb/ lbm.mol-°R in

the traditional system. Eq. (2.4) can be rewritten as

𝑝 =𝑛𝑀𝑅𝑢

∀𝑀𝑇

where M is the molecular weight of the gas. The product of the number of

moles and the molecular weight is the mass of the gas. Thus nM/∀is the mass

per unit volume, or density. The quotient Ru /M is the gas constant, R. Thus a

second form of the ideal gas law is

To determine the mass density of a gas, solve Eq. (2.5) for 𝜌:


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Viscosity

The property of viscosity is important to engineering practice because it leads to

significant energy loss when moving fluids contact a solid boundary, or when

different zones of fluid are flowing at different velocities.

Viscosity, 𝝁

Viscosity (also called dynamic viscosity, or absolute viscosity) is a measure of a

fluid’s resistance to deformation under shear stress. For example, crude oil has a

higher resistance to shear than does water. Crude oil will pour more slowly than

water from an identical beaker held at the same angle. The symbol used to

represent viscosity is 𝝁 (mu). To understand the physics of viscosity, it is useful

to refer back to solid mechanics and the concepts of shear stress and shear

strain.

Shear stress, 𝝉, (tau), is the ratio of force/area on a surface when the force is

aligned parallel to the area. Shear strain is a change in an interior angle of a

cubical element, that was originally a right angle. The shear stress on a material

element in solid mechanics is proportional to the strain, and the constant of

proportionality is the shear modulus:

In fluid flow, however, the shear stress on a fluid element is proportional to the

rate (speed) of strain, and the constant of proportionality is the viscosity:

Now consider the laminar motion of a real fluid along a solid boundary as in

Fig. 2.1. Observations show that, while the fluid has a finite velocity, V, at any


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finite distance from the boundary, there is no velocity at the boundary. Thus, the

velocity increases with increasing distance from the boundary. These facts are

summarized on the velocity profile, which indicates relative motion between

adjacent layers. a velocity gradient that becomes less steep ( dV/dy becomes

smaller) with distance from the boundary has a maximum shear stress at the

boundary, and the shear stress decreases with distance from the

Fig.2.1

The shear stress (shear force per unit area) is

From Eq. (2.6) it can be seen that the viscosity 𝝁 is related to the shear stress

and velocity gradient.


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A common unit of viscosity is the poise, which is 1 dyne-s/ cm2 or 0.1 N.s/m

2.

The viscosity of water at 20°C is one centipoise (10–2

poise) or 10–3

N.s/m2. The

unit of viscosity in the traditional system is Ib.s/ft2.

Kinematic Viscosity,

Many equations of fluid mechanics include the ratio 𝜇

𝜌. Because it occurs so

frequently, this ratio has been given the special name kinematic viscosity. The

symbol used to identify kinematic viscosity is 𝑣 (nu). Units of kinematic

viscosity are m2/s, as shown.

The units for kinematic viscosity in the traditional system are ft2/s.

Temperature Dependency

The effect of temperature on viscosity is different for liquids and gases. The

viscosity of liquids decreases as the temperature increases, whereas the viscosity

of gases increases with increasing temperature. To understand the mechanisms

responsible for an increase in temperature that causes a decrease in viscosity in

a liquid, it is helpful to rely on an approximate theory that has been developed

to explained the observed trends (1). The molecules in a liquid form a structure

with “holes” where there are no molecules, as shown in Fig. 2.2. Even when the

liquid is at rest, the molecules are in constant motion, but confined to cells. The

cell structure is caused by attractive forces between the molecules. The cells


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may be thought of as energy barriers. When the liquid is subjected to a rate of

strain and thus caused to move, as shown in Fig. 2.2, there is a shear stress, 𝜏,

imposed by one layer on another in the fluid. This force/area assists a molecule

in overcoming the energy barrier, and it can move into the next hole. The

magnitude of these energy barriers is related to viscosity, or resistance to shear

deformation. At a higher temperature the size of the energy barrier is smaller,

and it is easier for molecules to make the jump, so that the net effect is less

resistance to deformation under shear. Thus, an increase in temperature causes a

decrease in viscosity for liquids. An equation for the variation of liquid

viscosity with temperature is

where C and b are empirical constants that require viscosity data at two

temperatures for evaluation.

Fig. 2.2

As compared to liquids, gases do not have zones to which molecules are

confined by intermolecular bonding. Gas molecules are always undergoing

random motion. If this random motion of molecules is superimposed upon two

layers of gas, where the top layer is moving faster than the bottom layer,


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periodically a gas molecule will randomly move from one layer to the other. As

the gas temperature increases, more of the molecules will be making random

jumps. Just highly mobile gas molecules have momentum, which must be

resisted by the layer to which the molecules jump. Therefore, as the temperature

increases, the viscosity, or resistance to shear, also increases.

Newtonian versus Non-Newtonian Fluids

Fluids for which the shear stress is directly proportional to the rate of strain are

called Newtonian fluids. Because shear stress is directly proportional to the

shear strain dV/dy, a plot relating these variables (see Fig. 2.3) results in a

straight line passing through the origin. The slope of this line is the value of the

dynamic (absolute) viscosity. For some fluids the shear stress may not be

directly proportional to the rate of strain; these are called non-Newtonian

fluids.

Fig. 2.3


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EXAMPLE 2.1 CALCULATING VISCOSITY OF LIQUID AS A FUNCTION OF

TEMPERATURE

The dynamic viscosity of water at 20°C is and the viscosity at 40°C is

Estimate the viscosity at 30°C.

Problem Definition

Situation: Viscosity of water is specified at two temperatures.

Find: The viscosity at 30°C by interpolation.

Properties:

a) Water at 20°C,

b) Water at 40°C,

Plan

1. Linearize Eq. (2.9) by taking the logarithm.

2. Interpolate between the two known values of viscosity.

3. Solve for and b in this linear set of equations.

4. Change back to exponential equation, and solve for 𝜇 at 30°C.

Solution

1. Logarithm of Eq. (2.9)

2. Interpolation

3. Solution for and b

4. Substitution back in exponential equation

At 30°C


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EXAMPLE 2.2MODELING A BOARD SLIDING ON A LIQUID LAYER

A board 1 m by 1 m that weighs 25 N slides down an inclined ramp (slope 20°) with a

velocity of 2.0 cm/s. The board is separated from the ramp by a thin film of oil with a

viscosity of 0.05 N.s/m2. Neglecting edge effects, calculate the space between the board and

the ramp.

Problem Definition

Situation: A board is sliding down a ramp, on a thin film of oil.

Find: Space (in m) between the board and the ramp.

Assumptions: A linear velocity distribution in the oil.

Properties: Oil,

Sketch:

Plan

1. Draw a free body diagram of the board, as shown in “sketch.”

• For a constant sliding velocity, the resisting shear force is equal to the component of weight

parallel to the inclined ramp.

• Relate shear force to viscosity and velocity distribution.

2. With a linear velocity distribution dV/dy, can everywhere be expressed as ΔV/Δy where ΔV

is the velocity of the board, and Δy is the space between the board and the ramp.

3. Solve for Δy.


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EXAMPLE 2.3 MODELING A VELOCITY PROFILE OF A LIQUID

LAYER


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SOLUTION


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Bulk Modulus of Elasticity

The bulk modulus of elasticity, E, is a property that relates changes in pressure

to changes in volume (e.g., expansion or contraction)

where dp is the differential pressure change, d∀is the differential volume

change, and ∀ is the volume of fluid. Because is negative for a positive

dp, a negative sign is used in the definition to yield a positive Ev. The elasticity

is often called the compressibility of the fluid.

Surface Tension

Surface tension, 𝜎 (sigma), is a material property whereby a liquid at a material

interface, usually liquid-gas, exerts a force per unit length along the surface.

According to the theory of molecular attraction, molecules of liquid

considerably below the surface act on each other by forces that are equal in all

directions. However, molecules near the surface have a greater attraction for

each other than they do for molecules below the surface because of the presence

of a different substance above the surface. This produces a layer of surface

molecules on the liquid that acts like a stretched membrane. Because of this

membrane effect, each portion of the liquid surface exerts “tension” on adjacent

portions of the surface or on objects that are in contact with the liquid surface.

This tension acts in the plane of the surface, and is given by:


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where L is the length over which the surface tension acts.

Surface tension for a water–air surface is 0.073 N/m at room temperature. The

magnitude of surface tension decreases with increasing temperature. The effect

of surface tension is illustrated for the case of capillary action (rise above a

static water level at atmospheric pressure) in a small tube (Fig. 2.4). Here the

end of a small-diameter tube is inserted into a reservoir of water, and the

characteristic curved water surface profile occurs within the tube. The relatively

greater attraction of the water molecules for the glass rather than the air causes

the water surface to curve upward in the region of the glass wall. Then the

surface tension force acts around the circumference of the tube, in the direction

indicated. It may be assumed that the contact angle 𝜃 (theta) is equal to 0° for

water against glass. The surface tension force produces a net upward force on

the water that causes the water in the tube to rise above the water surface in the

reservoir.

Fig. 2.4

Surface tension forces for several different cases are shown in Fig. 2.5. Case (a)

is a spherical droplet of radius r. The surface tension force is balanced by the

internal pressure.


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Case (b) is a bubble of radius r that has internal and external surfaces and the

surface tension force acts on both surfaces, so

Case (c) is a cylinder supported by surface-tension forces. The liquid does not

wet the cylinder surface. The maximum weight the surface tension can support

is

where L is the length of the cylinder.

Case (d) is a ring being pulled out of a liquid. This is a technique to measure

surface tension. The force due to surface tension on the ring is


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Fig. 2.5

EXAMPLE 2.5 CAPILLARY RISE IN A TUBE

To what height above the reservoir level will water (at 20°C) rise in a glass tube, such as that

shown in Fig. 2.4, if the inside diameter of the tube is 1.6 mm?

Problem Definition

Situation: A glass tube of small diameter placed in an open reservoir of water induces

capillary rise.

Find: The height the water will rise above the reservoir level.

Sketch: See Figure 2.4.

Properties: Water (20 oC), 𝜎=0.073N/m, 𝛾 =9790 N/m3.

Plan

1. Perform a force balance on water that has risen in the tube.

2. Solve for Δh

Solution

1. Force balance: Weight of water (down) is balanced by surface tension force (up).


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Because the contact angle θ for water against glass is so small, it can be assumed to be 0°;

therefore cosθ=1. Therefore:

2. Solve for Δh

Vapor Pressure

The pressure at which a liquid will vaporize, or boil, at a given temperature, is

called its vapor pressure. This means that boiling occurs whenever the local

pressure equals the vapor pressure. Vapor pressure increases with temperature.

Note that there are two ways to boil a liquid. One way is to raise the

temperature, assuming that the pressure is fixed. For water at 14.7 psia, this can

be accomplished by increasing the temperature of water at sea level to 100°C,

thus reaching the temperature where the vapor pressure is equal to the same

value. However, boiling can also occur in water at temperatures much below

100°C , if the pressure in the water is reduced to the vapor pressure of water

corresponding to that lower temperature. For example, the vapor pressure of

water at 10°C is 0.178 psia (approximately 1% of standard atmospheric

pressure). Therefore, if the pressure in water at 10°C is reduced to 0.178 psia,

the water boils. Such boiling often occurs in localized low-pressure zones of

flowing liquids, such as on the suction side of a pump. When localized low-

pressure boiling does occur in flowing liquids, vapor bubbles start growing in

local regions of very low pressure and then collapse in regions of higher


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pressure downstream. This phenomenon, which is called cavitation, can cause

extensive damage to fluids systems.

Solved Problems

Problem 2.1 Calculate the density and specific weight of nitrogen at an

absolute pressure of 1 MPa and a R = 297 J/kg/K at a temperature of 40oC .

Solution

Perfect gas law

The temperature in absolute units is T =273 + 40 = 313 K.

The specific weight is

Problem 2.2 Find the density, kinematic and dynamic viscosity of crude oil in

traditional units at 100oF. use


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Solution

The density of water at standard condition is 1.94 slugs/ft3, so the density of

crude oil is 0.86 × 1.94 =1.67 slugs/ft3 or 1.67 × 32.2 =53.8 lbm/ft3.

The dynamic viscosity is ρν = 1.67 × 6.5 × 10−5

=1.09 × 10−4

lbf·s/ft2.

Problem 2.3 Two parallel glass plates separated by 0.5 mm are placed in water

at 20oC. The plates are clean, and the width/separation ratio is large so that end

effects are negligible. How far will the water rise between the plates?

The surface tension at 20oC is 7.3×10−2 N/m

Solution

The weight of the water in the column h is balanced by the surface tension

force.

whdρg = 2wσ cos θ

where w is the width of the plates and d is the separation distance. For water

against glass, cos θ = 1. Solving for h gives

Problem 2.4 Air at 15oC forms a boundary layer near a solid wall. The velocity

distribution in the boundary layer is given by


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where U = 30 m/s and δ = 1 cm. Find the shear stress at the wall (y = 0).


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Unit 3

Fluid Static

This unit begins mechanics of fluids in depth by introducing many concepts

related to pressure and by describing how to calculate forces associated with

distributions of pressure. This chapter is restricted to fluids that are in

hydrostatic equilibrium.

Pressure is defined as the ratio of normal force to area at a point. The pressure

often varies from point to point. For example, pressure acting on the water tank

wall will vary at different locations on the wall. Pressure is a scalar quantity;

that is, it has magnitude only. Units for pressure give a ratio of force to area.

Newtons per square meter of area, or Pascal (Pa), is the SI unit. The USC units

include psi, which is pounds-force per square inch, and psf, which is pounds-

force per square foot.

Absolute Pressure, Gage Pressure, and Vacuum Pressure

Absolute pressure is referenced to regions such as outer space, where the

pressure is essentially zero because the region is devoid of gas. The pressure in

a perfect vacuum is called absolute zero, and pressure measured relative to this

zero pressure is termed absolute pressure. When pressure is measured relative

to prevailing local atmospheric pressure, the pressure value is called gage

pressure. For example, when a tire pressure gage gives a value of 300 kPa (44

psi), this means that the absolute pressure in the tire is 300 kPa greater than

local atmospheric pressure. To convert gage pressure to absolute pressure, add

the local atmospheric pressure. When pressure is less than atmospheric, the


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pressure can be described using vacuum pressure. Vacuum pressure is defined

as the difference between atmospheric pressure and actual pressure.

Figure 3.1 provides a visual description of the three pressure scales. Gage,

absolute, and vacuum pressure can be related using equations labeled as the

“pressure equations.”

Fig. 3.1

Pressure Variation with Elevation

Hydrostatic Differential Equation

The hydrostatic differential equation is derived by applying force equilibrium to

a static body of fluid. To begin the derivation, isolate a small cylindrical body,

and then sketch a free-body diagram (FBD) as shown in Fig. 3.2. The

cylindrical body is oriented so that its longitudinal axis is parallel to an arbitrary


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𝑙 direction. The body is Δ𝑙 long, ΔA in

cross-sectional area, and inclined at an

angle 𝛼 with the horizontal.

Apply force equilibrium in the 𝑙 direction:

Simplify and divide by the volume of the body Δl ΔA to give

From Fig. 3.2, the sine of the angle is given by

Combining the previous two equations and letting approach zero gives

The final result is

Equation (3.1) means that changes in pressure correspond to changes in

elevation.

Hydrostatic Equation

The hydrostatic equation is used to predict pressure variation in a fluid with

constant specific weight (constant density). Integrating Eq. (3.1) will give

Where, z: is elevation, (vertical distance above a fixed reference point - datum),

pz; piezometric pressure.

Fig. 3.2


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Dividing Eq. (3.2) by 𝛾 gives

Where, h : piezometric head. Since h is constant in Eq. (3.3),

Multiplying Eq. (3.4a) by 𝛾 gives

In Eq. (3.4b), letting and letting gives

The hydrostatic equation is given by either Eq. (3.4a), (3.4b), or (3. c). These

three equations are equivalent because any one of the equations can be used to

derive the other two. Piezometric pressure and head are related by

When hydrostatic equilibrium prevails in a body of fluid of constant density,

then h will be constant at all locations.


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Pressure Measurements

Four scientific instruments for measuring pressure: the barometer, Bourdon-

tube gage, piezometer and manometer, transducer will described.

1. Barometer: it is an instrument that is used to measure atmospheric

pressure. The most common types are the mercury barometer. A mercury

barometer is made by inverting a mercury-filled tube in a container of

mercury as shown in Fig. 3.3. Atmospheric pressure will push the

mercury up the tube to a height h. The mercury barometer is analyzed by

applying the hydrostatic equation:Thus, by measuring h, local

atmospheric pressure can be determined using Eq. (3.6).

2. Bourdon-Tube Gage

A Bourdon-tube gage measures pressure by sensing the deflection of a coiled

tube. The tube has an elliptical cross section and is bent into a circular arc, as

shown in Fig. 3.4.When atmospheric pressure (zero gage pressure) prevails,

the tube is undeflected. When pressure is applied to the gage, the curved tube

tends to straighten, thereby actuating the pointer to read a positive gage

pressure.


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3. Piezometer

A piezometer is a vertical tube, usually transparent, in which a liquid rises in

response to a positive gage pressure. Pressure in the pipe pushes the water

column to a height h, and the gage pressure at the center of the pipe is

𝑝 = 𝛾ℎ. The piezometer has several advantages: simplicity, direct

measurement and accuracy. However, a piezometer cannot easily be used for

measuring pressure in a gas, and a piezometer is limited to low pressures

because the column height becomes too large at high pressures.

4. Manometer

A manometer, is a device for measuring pressure by raising or lowering a

column of liquid. In the case shown, positive gage pressure in the pipe

pushes the manometer liquid up a height ∆ℎ. To use a manometer, the height

of the liquid in the manometer related to pressure.

The general equation for the pressure difference measured by the

manometer is:

where 𝛾𝑖 and hi are the specific weight and deflection in each leg of

the manometer. It does not matter where one starts; that is, where one

defines the initial point 1 and final point 2.


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Change in Piezometric Head for Pipe Flow

For Fig. 3.7, apply the manometer equation (3.7) between points 1 and 2:


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Fig. 3.7


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Solved Problems

Problem 3.1 A tank that is open to the atmosphere contains a 1.0-m layer of

oil (ρ = 800kg/m3) floating on a 0.5m layer of water (ρ = 1000 kg/m

3).

Determine the pressure at elevations A, B, C, and D. Note that B is midway

between A and C.

Solution At a horizontal interface of two fluids,

pressure will be constant across the interface.

Thus the pressure in the oil at A equals the pressure in the air (atmospheric

pressure). pA = patm= 0 kPa (gage)

Since the oil layer is a static fluid of constant density, the piezometric

pressure is constant

pA + γoilzA = pB + γoilzB = pC + γoilzC = constant ……………….(1)

where z denotes elevation. Let zA = 0, zB = −0.5 m, zC = −1.0 m. Then, Eq.

(1) becomes pA = pB + γoil (−0.5) = pC + γoil (−1.0)

So pB = pA + γoil (0.5)= patm + (800) (9.81) (0.5) /1000= 3.92 kPa (gage)

Similarly, pC = pA + γoil (1.0)= patm + (800) (9.81) (1.0) /1000= 7.85 kPa

At elevation C, pressure in the oil equals pressure in the water. Since the

piezometric pressure in the water is constant, we can write

pC + γwater zC = pD + γwater zD

or, pD = pC + γwater (zC − zD)= 7.85 + (1000) (9.81) (0.5)/1000= 12.8 kPa


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Problem 3.2 A U-tube manometer contains kerosene, mercury and water, each

at 70 oF. The manometer is connected between two pipes (A and B), and the

pressure difference, as measured between the pipe centerlines, is

pB −pA = 4.5 psi. Find the elevation difference z in the manometer.

Solution Beginning at location B and adding pressure

differences until location A is reached gives

pB + (2 + z) γwater − zγHg − 2γkero = pA

Rearranging pB − pA = 2(γkero − γwater) + z (γHg − γwater) ……………(1)

Looking up values of specific weight and substituting into Eq. (1) gives

[4.5 × 144] lb/ft2 = [h(2 ft) (51 − 62.3) lb/ft

3 + (z ft) (847 − 62.3)lb/ft

3]

So, z =(648 + 22.6)/784.7= 0.855 ft

Problem 3.3 A container, filled with water at 20oC, is open to the atmosphere

on the right side. Find the pressure of the air in the enclosed space on the left

side of the container.

Solution The pressure at elevation 2

is the same on both the left and right side.

p2 = patm + γ (0.6 m)= 0+(9.81 kN/m3)(0.6 m)= 5.89 kPa

Since the piezometric head is the same at elevations 1 and 2

p1/γ+ z1 =p2/γ+ z2

so, p1 = p2 + γ (z2 − z1)= (5.89 kPa) + (9.81 kN/m3)(−1.0 m)= −3.92 kPa


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38

Problem 3.4 (Differential-Manometers)


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39

Forces on Plane Surfaces (Panels)

This section explains how to represent hydrostatic pressure distributions on one

face of a panel with a resultant force that passes through a point called the

center of pressure. This information is relevant to applications such as dams,

gates and water tanks.

Pressure Distribution

A description of the pressure at all points along a plane surface of arbitrary

shape surface is called a pressure distribution. As shown in Fig. 3.7 the

pressure distribution is linear and act normal to the surface. The pressure

distribution acting on an area dA exerted force will be;

The resultant force F passes through a point called the center of pressure (CP).

For a uniform pressure distribution, the CP is located at the centroid of area of

the panel. Notice that the CP is located below the centroid of area.

Magnitude of Resultant Hydrostatic Force

To derive an equation for the resultant force on a panel under hydrostatic

loading, sum forces using an integral. The situation is shown in Fig. 3.7. Line

AB is the edge view of a panel submerged in a liquid. The plane of this panel

intersects the horizontal liquid surface at axis 0-0 with an angle α . The distance

from the axis 0-0 to the horizontal axis through the centroid of the area is given

by ý The distance from 0-0 to the differential area dA is y. The pressure on the

differential area is,

Where is the average pressure


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40

Fig. 3.7

The differential force is

The hydrostatic force on one side of a panel of arbitrary shape (e.g., rectangular,

round, elliptical) submerged in liquid is given by the product of area and

pressure at the centroid of area.


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41

Line of Action of the Resultant Force

The torque due to the resultant force F will balance the torque due to the

pressure distribution.

The 2nd moments of area about a line through the centroid of some common

shapes are given in table 3.1.


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42

Table 3.1


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43

Solved Problems

Problem 3.5 A rectangular gate of dimension 1 m

by 4 m is held in place by a stop block at B. This

block exerts a horizontal force of 40 kN and a vertical

force of 0 kN. The gate is pin-connected at A, and the

weight of the gate is 2 kN. Find the depth h of the water.


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44

Solution

A free-body diagram of the gate is shown in Fig.

where W is the weight of the gate, F is the equivalent

force of the water, and r is the length of the moment

arm. Summing moments about A gives

The equivalent force of the water is


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45

Substituting Eqs. (2) and (3) into Eq. (1) gives

Problem 3.6 A container is formed by joining two plates, each 4 ft long with a

dimension of 6 ft in the direction normal to the paper. The plates are joined by a

pin connection at A and held together at the top by two steels rods (one on each

end). The container is filled with concrete (S = 2.4) to a depth of 1.5 ft. Find the

tensile load in each steel rod.

Solution

A free-body diagram of plate ABC is shown in Fig.

Summing moments about A


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46

The geometry of plate ABC is shown in Fig.

To find the distance x2, note that portion BC of the plate is above the surface of

the concrete. Thus use values for a plate of dimension 3 ft by 6 ft.

From equation (1)


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48

Problem 3.8


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49

Problem 3.9


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50

Forces on Curved Surfaces

This section explains how to calculate forces on surfaces that have curvature.

These calculation is a first step for the design of components such as tanks,

pipes, and curved gates. On a curved surface the forces pdA on individual

elements differ in direction, so a simple summation of them may not be made. It

is easier to sum forces for the free body shown in the Fig. 3.8.

Fig. 3.8

The lower sketch in Fig. 3.8b shows the resultant force acting on the curved

surface F acting on the free body. Using the FBD and summing forces in the

horizontal direction shows that

The line of action for the force FAC is through the center of pressure for side AC,

as in the force acting on submerged plane area, and designated as yp. The

vertical component of the equivalent force is

Where W: is the weight of the fluid in the free body

FCB: is the force on the side CB.


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51

The force FCB acts through the centroid of surface CB, and the weight acts

through the center of gravity of the free body. The line of action for the vertical

force may be found by summing the moments about any convenient axis.

If atmospheric pressure prevails above the free surface (Fig.3.9a) and on the

outside of surface AB, then force caused by atmospheric pressure cancels out

and equilibrium gives F=Ɣ ∀ ABCD =W↓ 3.16

When the pressure distribution on a curved surface comes from the liquid

underneath (Fig.3.9b), the equivalent force on upper face of curved surface is

given by

F=Ɣ ∀ ABCD =W↑ 3.17

Fig.3.9


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52


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53

Solved Problems

Problem 3.10 A sector gate, of radius 4 m and length5m, controls the flow of

water in a horizontal channel. For the (equilibrium) conditions shown in Fig.,

determine the total resultant force on the gate.


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54

Solution

1. The horizontal component is the force that would be exerted by the water

on a vertical projection of the curved surface. The depth d of this

projection is

d=4 sin 30◦ = 2 m

and its centroid is 1 +d/2 = 2 m below the free surface.

Therefore Horizontal force Fx = 𝛾ℎ𝐴=9810×2 (2 × 5) m2= 196.2k N

2. The total vertical component of force=weight of imaginary water above

the gate up to water surface

AB = (4 − 4 cos 30◦) m = 0.536 m

Fv = 𝛾ℎ𝐴=9810 × 1(0.536 × 5)=26.291kN

W=𝛾∀=9810[(𝜋 × 42) ×30

360− (

1

2× 2 × 4 cos 30°)] × 5 = 35.546𝑘𝑁

Fy=Fv+W=26.291+35.546=61.837kN


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55

yp= 2 + 5×23/12

2×(2×5)= 2.167𝑚

4. The center of gravity of the imaginary fluid above the gate up to water

surface may be found by taking moments about BC. It is 0.237 m to the left of

BC.

𝐹 = √196.22 + 61.832 = 205.71𝑘𝑁

at an angle, tan-1

(61.83/196. 2)= 17.5◦ to the horizontal

Problem 3.11. This tainter gate is pivoted

at O and is 30 ft long.

Calculate the horizontal and vertical

components of force on the face of the gate.


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56

Problem 3.12

Problem 3.13


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57

The Buoyant Force Equation

Fig. 3.10

A buoyant force is defined as the upward force that is produced on a body that

is totally or partially submerged in a fluid. Buoyant forces are significant for

most problems as surface ships, sediment transport in rivers. In in Fig. 3.10a

shown, consider a body ABCD submerged in a liquid of specific weight Ɣ. The

pressures acting on the lower portion of the body create an upward force equal

to the weight of liquid needed to fill the volume above surface ADC.

The upward force is

As shown by Fig. 3.10a, pressures acting on the top surface of the body create a

downward force equal to the weight of the liquid above the body:

Subtracting the downward force from the upward force gives the buoyant force

FB acting on the body:

Hence, the buoyant force (FB) equals the weight of liquid that would be needed

to occupy the volume of the body.


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58

The body that is floating as shown in Fig. 3.10b. Pressure acts on curved surface

ADC causing an upward force equal to the weight of liquid that would be

needed to fill volume VD (displaced volume). The buoyant force is given by

Hence, the buoyant force equals the weight of liquid that would be needed to

occupy the volume VD. We can write a single equation for the buoyant force:

Stability of Immersed Bodies

When a body is completely immersed in a liquid, its stability depends on the

relative positions of the center of gravity of the body and the centroid of the

displaced volume of fluid, which is called the center of buoyancy.

If the center of buoyancy is above the center of gravity (Fig. 3.11a), any

tipping of the body produces a righting couple, and consequently, the

body is stable.

If the center of gravity is above the center of buoyancy (Fig. 3.11c), any

tipping produces an increasing overturning moment, thus causing the

body to turn through 180°.

Finally, if the center of buoyancy and center of gravity are coincident,

the body is neutrally stable—that is, it lacks a tendency for righting or for

overturning, as shown in Fig. 3.11b.

Fig. 3.11


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59

Stability Floating Bodies

The stability for floating bodies than for immersed bodies is very important

because the center of buoyancy may take different positions with respect to the

center of gravity, depending on the shape of the body and the position in which

it is floating. When the center of gravity G is above the center of buoyancy C

(center of displaced volume) for floating body, the body will be stable and

equilibriu. The reason for the change in the center of buoyancy for the ship is

that part of the original buoyant volume, as shown in Fig.3.12by the wedge

shape AOB, is transferred to a new buoyant volume EOD. Because the buoyant

center is at the centroid of the displaced volume, it follows that for this case the

buoyant center must move laterally to the right. The point of intersection of the

lines of action of the buoyant force before and after heel is called the

metacenter M, and the distance GM is called the metacentric height.

If GM is positive—that is, if M is above G,the body is stable

If GM is negative, the body is unstable.

Fig.3.12

Consider the prismatic body shown in Fig. 3.12, which has taken a small angle

of heel 𝛼. First evaluate the lateral displacement of the center of buoyancy CC’,

then it will be easy by simple trigonometry to solve for the metacentric height

GM or to evaluate the righting moment.

The righting couple =𝑊. 𝑀𝐺. sin 𝛼


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60

Where : W weight of body and 𝛼 angle of heel.

By similar triangle EOD and C’CM: ∆𝑦

𝑏/2=

𝐶′𝐶

𝐶𝑀 find CM

𝐺𝑀 = 𝑀𝐶 − 𝐺𝐶


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62

Solved Problems

Problem 3.14: An 18-in. diameter concrete cylinder (SG = 2.4) is used to raise

a 60-ft long log to a 45o angle. The center of the log is pin-connected to a pier at

point A. Find the length L of the concrete cylinder.


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63

Solution

A free-body diagram is

where BL and BC are the buoyant forces on the log and concrete, respectively.

Similarly, WL and WC represent weight.

Summing moments about point A


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64

Problem 3.15 As shown, a cube (L=60mm ) suspended in carbon tetrachloride

is exactly balanced by an object of mass m1=700g. Find the mass m2 of the

cube.

Equilibrium of accelerated fluid masses

If a body of fluid is moved at a constant velocity, then it obeys the equations

derived earlier for static equilibrium.

If a body of fluid is accelerated such that, after some time, it has adjusted so that

there are no shearing forces, there is no motion between fluid particles, and it


Fluid Mechanic

65

moves as a solid block, then the pressure distribution within the fluid can be

described by equations similar to those applying to static fluids.

Uniform vertical acceleration:

If a body of fluid is accelerated uniformly vertically upward, the forces acting

on a small vertical element(Fig. 3.13), area A and length h, with a uniform

acceleration a in vertical direction:

But Newton's 2nd law gives: 𝐹 = 𝑚𝑎

𝐹𝑦 = 𝑝A − γhδA = (ρ h δA)a

Divide the equation by 𝛿𝐴:

∴ 𝑝 = 𝛾ℎ +𝛾

𝑔ℎ𝑎

OR ∴ 𝑝 = 𝛾ℎ(1 +𝑎

𝑔 ) .....3.21

Thus, the mass of fluid accelerated upward increase the pressure at any point

ratio of 𝑎

𝑔 .

Therefore if a=g the pressure will be :

∴ 𝑝 = 𝛾ℎ (1 +𝑔

𝑔 ) = 2𝛾ℎ) .....3.22

Also, If a=0 the pressure will be hydrostatic pressure:

∴ 𝑝 = 𝛾ℎ (1 +0

𝑔 ) = 𝛾ℎ

If the acceleration downward the pressure will decrease in with ratio of 𝑎

𝑔 :

∴ 𝑝 = 𝛾ℎ (1 −𝑎

𝑔 )) .....3.23

In case of a mass of fluid is decelerated uniformly vertically upwardwith a=g,

the pressure at any point:


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66

∴ 𝑝 = 𝛾ℎ (1 −𝑔

𝑔 ) = 0

So, at any point of free falling mass of fluid the pressure equal to zero.

In case of a>g; ∴ 𝑝 = −𝛾ℎ (1 −𝑎

𝑔 )) .....3.24

For all previous cases the liquid have horizontal surface.

Uniform horizontal acceleration:

If a container of fluid is accelerated uniformly and horizontally with, then the

slope of the isobars within the fluid is given by:

𝜃 = 𝑡𝑎𝑛−1 𝑎

𝑔

The free surface of a liquid is normally taken as a line of constant

pressure - or isobar - and the equation above gives the surface slope of an

accelerated fluid.

Consider element (1) in Fig.3.14 directly under surface of liquid. The resultant

of forces 𝛿𝐹acting on element a equal to mass (𝛿𝑊

𝑔) times acceleration a

according 2nd

Newton low.

𝛿𝐹 =𝛿𝑊

𝑔a ……..3.25

There are two forces acting on the element, the 1st is the pressure on bottom

surface which is normal to surface and the 2nd

is element weight 𝛿𝑊 acting

vertically downward. So,

𝛿𝐹 = 𝛿𝑊 tan 𝜃 ……….3.26

Equating equations 3.25 and 3.26 produce ,


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67

𝛿𝑊𝑡𝑎𝑛𝜃 =𝛿𝑊

𝑔𝑎

∴ 𝑡𝑎𝑛𝜃 =𝑎

𝑔

The pressure at any point in liquid equal to 𝑝 = 𝛾ℎ.

Fig.3.14

Example 3.13: A vessel of 4m with, 2m hight and 6m length rest on horizontal

plan with water having depth 1m. the acceleration of vessel 2m/s2.Find the

pressure on points C & D in figure shown and the total pressure on rear wall.


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68

Example: 3.14


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69

Example3.15: A vessel cross sectional area 3m *1m rest on inclined plan with

liquid as shown in figure. the acceleration of vessel 2m/s2.Find what should the

acceleration along the inclined plane so that the free surface at the edge touches

the point c. assume the sides of vessel are high enough so that no liquid spills.

Uniform rotation about a vertical axis:

If liquid is placed in a container and rotated about a vertical axis at a constant

angular velocity, then after some time it will move as a solid body with no

shearing of the fluid. Under such conditions the liquid is said to be moving as a

"forced vortex" with velocity 𝑣 = 𝜔𝑟where r radius from the axis and 𝜔

angular velocity. [This contrasts with "free-vortex" motion in which the fluid

velocity varies inversely with distance from the axis of rotation.]


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70

With 𝜔 constant, all fluid experiences an acceleration 𝑎 = 𝜔2𝑟 (centrifugal)

directed towards the axis of rotation. The liquid surface will concave downward

(parabolic revolution) and for equilibrium of a typical small horizontal element.

The resultant of forces 𝛿𝐹acting on element a equal to mass (𝛿𝑊

𝑔) times

acceleration a according 2nd

Newton low.

𝛿𝐹 =𝛿𝑊

𝑔a ……..3.27

There are two forces acting on the element, the 1st is the pressure on bottom

surface which is normal to tangent of parabolic surface at that point and the 2nd

is element weight 𝛿𝑊 acting vertically downward. So,

𝛿𝐹 = 𝛿𝑊 tan 𝜃 ……….3.28

Equating equations 3.27 and 3.28 produce ,

𝛿𝑊𝑡𝑎𝑛𝜃 =𝛿𝑊

𝑔𝑎

∴ 𝜃 = 𝑡𝑎𝑛−1 𝑎

𝑔 , 𝑤ℎ𝑒𝑟𝑒 𝑎: 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎 = 𝜔2𝑟

𝑡ℎ𝑒𝑛, 𝑡𝑎𝑛𝜃 = 𝜔2𝑟

𝑔 𝑜𝑟, 𝜃 = 𝑡𝑎𝑛−1

𝜔2𝑟

𝑔 ….3.29

The radius r in equation 3.29 variable, therefore θ will be variable vector. So

equation 3.29 represents the slope of tangent of parabolic surface at that point.

Mathematically slope of tangent represent the 1st derivative of vertical axis y to

horizontal axis x.

𝑆𝑜, 𝑡𝑎𝑛𝜃 = 𝑑𝑦

𝑑𝑥 ……………………….3.30

From equations 3.29 & 3.30,

𝑑𝑦

𝑑𝑥=

𝜔2𝑟

𝑔 ……………………….3.31

The result of integration of equation 3.31 is,


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71

𝑦 =𝜔2𝑟2

2𝑔+ 𝐶 …………………3.32

Equation 3.32 is parabolic equation and C is constant which can be equal to

zero when x (at the concave of parabola) and y equal to zero . So equation 3.32,

can be in form,

𝑦 =𝜔2𝑟2

2𝑔 ………………….3.33

Equation 3.33 can be express in the following form:

𝑦 =𝑣2

2𝑔 𝑤ℎ𝑒𝑟𝑒 , 𝑣 = 𝜔𝑟

In the vertical direction, the usual expression for pressure at any point in a static

fluid holds:

𝑝 = 𝛾ℎ

Where h: distance from the fluid surface to that point

Fig. 3.15


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72

Example3.16: the open circular tank of 2.6m high and 0.6m diameter contain

2.0m water. If cylinder is rotated about its axis, what is the maximum angular

velocity so that no water spills? Find the difference in total pressure at the

bottom and at the side s of cylinder due to rotation?

Example:3.17


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73

Unit 4

Kinematic of Fluid Motion

The force balance between pressure and weight in a static fluid was presented in

unit 3, which led to an equation for pressure variation with depth. This unit deal

with behaviour of fluid like velocity, acceleration and flow pattern (flow

classification). The fluid dynamic deal with fluid having accelerated movement

and there is relative acceleration between fluid particles. So, fluid dynamic

problems can be solved by applying one or more of the following equations:

1. Continuity equation , which enable to find the increasing of flow velocity

due to decreasing of flow cross section area.

2. Energy equation, which depend on energy conservation law and enable to

calculate the pressure, velocity and pressure head.

3. Momentum equation, which enable to find the force required to

accelerate the fluid mass depending on 2nd

Newton law of movement;

𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑜𝑓 𝑒𝑥𝑒𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠

Definition

Explain of some of fluid kinematic concept will help to understand of the three

equations.

Stream line: is an imaginary drawn through a flow field such that tangent to

line at any point on line indicates the direction of velocity vector at an instant.


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74

Stream tube: is an imaginary tube like boundary bounded by a number of

stream lines. No flow across through stream tube, flow should pass through the

ends of tube only as close conduit.

Fig. 4.1

Acceleration in fluid flow:

Acceleration is defined as the rate of change of velocity.

𝑎 =𝑑𝑉

𝑑𝑡

The change may be in magnitude, in direction or in all of them. The variation of

magnitude of velocity with respect to time only is called temporal acceleration

(local acceleration). The rate of change of velocity with respect to time in

direction of stream line in a point is called tangential acceleration and its

magnitude given by 𝑑𝑉𝑠

𝑑𝑡 . similarly, the normal acceleration with respect to time

would be 𝑑𝑉𝑛

𝑑𝑡 . The victorial summation of tangential and normal components of

acceleration gives local acceleration. The rate of change of velocity through

distance 𝑑𝑆 between two points on stream line at instance is called convective

acceleration. The tangential acceleration is given by

𝑎𝑠 = 𝑑𝑉

𝑑𝑆 𝑑𝑆

𝑑𝑡= 𝑣

𝑑𝑉

𝑑𝑆


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75

Flow Patterns

Uniform and non-uniform flow:

In a uniform flow, the velocity does not change along a fluid path and the fluid

paths are straight and parallel as shown in Fig. 4.2 for flow in a pipe.

In nonuniform flow, the magnitude or the direction of velocity changes along a

fluid path and

Steady or unsteady flow

Steady flow occurs in a system when none of the variables involved changes

with time; if any variable changes with time, the condition of unsteady flow

exist. Unsteady flow exists while the valve across pipe line is being adjusted;

with the valve opening fixed, steady flow occurs.

Combining the above we can classify any flow in to one of four types:


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76

Steady uniform flow. Conditions do not change with position in the stream or

with time. An example is the flow of water in a pipe of constant diameter at

constant velocity.

Steady non-uniform flow. Conditions change from point to point in the

stream but do not change with time. An example is flow in a tapering pipe

with constant velocity at the inlet - velocity will change as you move along

the length of the pipe toward the exit.

Unsteady uniform flow. At a given instant in time the conditions at every

point are the same, but will change with time. An example is a pipe of

constant diameter connected to a pump pumping at a constant rate which is

then switched off.

Unsteady non-uniform flow. Every condition of the flow may change from

point to point and with time at every point. An example is surface waves in

an open channel.

Laminar and Turbulent Flow

Laminar flow is a state of flow in which adjacent fluid layers move smoothly

with respect to each other. Laminar flow in a pipe has a smooth, parabolic

velocity distribution as shown in Fig. 4.4a.

Turbulent flow is an unsteady flow characterized by intense cross-stream

mixing. A turbulent flow. The mixing is apparent as the plume widens and

disperses. An instantaneous velocity profile for turbulent flow in a pipe is

shown in Fig. 4.4 b.


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77

One-Dimensional and Multi-Dimensional Flows

Fig. 4.5a shows the velocity distribution for an axisymmetric flow in a circular

duct. The flow is uniform, so the velocity does not change in the flow direction

(z). The velocity depends on only one dimension, namely the radius r, so the

flow is one-dimensional.

Fig.4.5b shows the velocity distribution for uniform flow in a square duct. In

this case the velocity depends on two dimensions, namely x and y, so the flow is

two-dimensional.

Fig.4.5c also shows the velocity distribution for the flow in a square duct but

the duct cross-sectional area is expanding in the flow direction so the velocity

will be dependent on z as well as x and y. This flow is three-dimensional.

Fig. 4.5


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78

Equation of Continuity

The application of the principle of conservation of mass to fluid flow in a

stream tube results in the "equation of continuity' expressing the continuity of

the flow from point to point along the streamtube. If the cross-sectional areas

and average velocities at sections 1 and 2 in the streamtube of Fig. 31 are

designated by A1, A2, V1and V2, respectively, the .quantity of fluid passing

section 1 per unit of time will be expressed by A1V1, and the mass of fluid

passing section 1 per unit of time will be 𝐴1𝑉1𝜌1. Similarly, the mass of fluid

passing section 2 will be𝐴2𝑉2𝜌2, Obviously, no fluid mass is being created or

destroyed between sections 1 and 2, and therefore

𝐴1𝑉1𝜌1 = 𝐴2𝑉2𝜌2 ……4.1

Fig. 4.6

Thus the mass of fluid passing any point in a streamtube per unit of time is the

same.

If this equation is multiplied by g, the acceleration due to gravity, there

results giving the equation of continuity in terms of weight.

𝐴1𝑉1𝑊1 = 𝐴2𝑉2𝑊2 ……4.2

The product will be found to have units of N/s and is termed the "weight rate of

flow" or "weight flow."


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79

For liquids, and for gases when pressure and temperature changes are

negligible, W1=W2 , resulting in

𝐴1𝑉1 = 𝐴2𝑉2 = 𝑄 ……4.3

indicating that for fluids of constant- density the product of cross-sectional area

and velocity at any point in a stream-tube will be the same. This product Q, is

designated as the "rate of flow" or "flow" and will have units of m3/s.

Energy Equation (Bernoulli’s Equation):

Energy equation depends on energy conservation law and enables to calculate

the pressure, velocity and pressure head. Flowing of a fluid subjected to

unbalanced forces or stresses. The motion continues as long as unbalanced

forces are applied. Consider a small element of fluid aligned along a streamline

(Fig. 4.7). It has a cross sectional area dA, pressure is assumed uniform across

its ends dA, and the local velocity is defined v.

Fig. 4.7

Applying Newton's laws of motion to the flow through the element along the

streamline (the force (in the direction of flow along the streamline) = mass x

acceleration).

F=m.a ……………………………………………….…4.4


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80

ds

dv v dA ds = g ds dA pdAds)dA

ds

dp(p sin …….. 4.5

dividing equation 4.5 by ρgdA. ds and defining -dz = ds sinθ , we have that:

0 =dz + g

vdv +

dp

…………………………..………4.6

The equation 4.6 is a form of Euler's equation, and relates p, v, and z in flow

field.

Equation 4.6 can be written in integral form:

∫𝛿𝑝

𝛾+ ∫ 𝛿𝑧 + ∫

𝑣𝛿𝑣

𝑔= 𝐶 ……………………………….. 4.7

where C is a constant of integration. If the equation 4.7 integrated between

section 1 and 2 in Fig. , so, C will be zero.

∫𝛿𝑝

𝛾

𝑝2

𝑝1+ ∫ 𝛿𝑧

𝑧2

𝑧1+ ∫

𝑣𝛿𝑣

𝑔

𝑉2

𝑉1= 0 ……………………..4.8

Where, p1 and p2 represent to pressure in points 1 and 2, z1 and z2 represent

elevation from datum of sections 1 and 2 and V1 and V2 represent velocities of

point 1 and 2 respectively. For incompressible fluid 𝛾 constant, so, the result of

equation 4.8 is,

𝑃2−𝑃1

𝛾+ 𝑍2 − 𝑍1 +

𝑉22−𝑉1

2

2𝑔= 0 ……………………4.9

∴ 𝑃1

𝛾+ 𝑍1 +

𝑉12

2𝑔=

𝑃2

𝛾+ 𝑍2 +

𝑉22

2𝑔 ……………4.10

Equation 4.10 called Bernoulli’s equation (for frictionless, steady flow). All of

terms of Bernoulli’s equation having dimension of length (L) or dimension of

energy times dimension of weight( FL/F). The elevation head represent the

potential energy per unit weight as below,


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81

∴ 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑑 = 𝑧

The velocity head represent the kinetic energy per unit weight as below,

∴ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑 =𝑉2

2𝑔

The pressure head represent the pressure energy per unit weight as below,

∴ 𝑝𝑟𝑒𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 =𝑝

𝛾

The sum of elevation, velocity and pressure heads for ideal steady

incompressible flow is constant for all point in stream line,

𝑃

𝛾+ 𝑧 +

𝑉2

2𝑔= 𝐻 = 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑

the sum of elevation and pressure heads called piezometric head which

represent the manometric height of liquid from datum,

𝑃

𝛾+ 𝑧 = ℎ = 𝑝𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑

Hydraulic and Energy Grade Lines

The energy grade line (EGL) shows the height of the total Bernoulli constant

𝑃

𝛾+ 𝑧 +

𝑉2

2𝑔= 𝐻 = 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑. The EGL has constant height.

The hydraulic grade line (HGL) shows the height corresponding to elevation

and pressure head 𝑃

𝛾+ 𝑧 = ℎ = 𝑝𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑, that is, the EGL minus the

velocity head V2/2g. The HGL is the height to which liquid would rise in a

piezometer tube

In an open-channel flow the HGL is identical to the free surface of the

water.


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82

The EGL will drop slowly due to friction losses and will drop sharply

due to a substantial loss (a valve or obstruction) or due to work extraction

(to a turbine).

The EGL can rise only if there is work addition (as from a pump or

propeller).

The HGL generally follows the behavior of the EGL with respect to

losses or work transfer, and it rises and/or falls if the velocity decreases

and/or increases.

Example 4.1: A flow of water from a reservoir to a pipe of different diameters

shown in Figure below. Calculate 1) the discharge and velocity at each pipe, 2)

the pressure in each pipe and 3) the energy and hydraulic grade lines.


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83

Example 4.2: A flow of water from a closed reservoir with interior pressure of

50 kPa to a pipe of different diameters shown in figure below. Calculate 1) the

discharge and velocity at each pipe, 2) the pressure in each pipe and 3) the

energy and hydraulic grade lines.

Example 4.3: A pipe gradually tapers from 0.6m at A to 0.9m at point B. the

elevation difference between A and B is 3m. Find pressure head and pressure at

point B if the pressure head at A is 15m and velocity at A is 2m/s. Assume the

frictionless flow.


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84

Applications of Bernoulli’s equation:

1) Flow through orifice

a) With constant head

When an open tank fill with liquid and drains through a port at the

bottom of the tank. The elevation of the liquid in the tank is constant

above the drain. The drain port is at atmospheric pressure. The flow is

steady, viscous effects are unimportant and velocity at liquid surface is

zero. The Bernoulli equation between points 1 and 2 on streamline:

2

2

22

1

2

11 z + g 2

v +

g

p = z +

g 2

v +

g

p

p1 = p2 because the pressure at the outlet and the tank surface are the

same (atmospheric).

The velocity at the tank surface zero, then:

g

V)zz(

2000

2

2

21

gHzzgV 2)(2 212

𝑄𝑡ℎ𝑒𝑜 = 𝑉2𝐴2

𝑄𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶𝑑 × 𝑄𝑡ℎ𝑒𝑜

𝑤ℎ𝑒𝑟𝑒 𝐶𝑑 𝑖𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 (0.64)

Fig.4.8


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85

If the tank is closed with interior pressure of Po then Bernoulli’s

equation can be expressed as:

g

Vzz

Po

20)(0

2

221

)(22 HP

gV o

b) With variable head (time for the tank to empty)

For cylindrical tank, the tank cross sectional area is A. In a time dt the

level falls by dH

We have an expression for the discharge from the tank

gHACQ dact 22

This discharge out of the orifice is the same as the flow in the tank so,

gHACdt

dHA d 22

Integrating between the initial level, h1, and final level, h2, gives the

time it takes to fall this height:

2

1

2

1

225.0

t

t

d

H

H

dtA

gACdHH

tA

gACH

dH

H

2

225.0 2

1

tA

gACHH

d

22

21


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86

Example 4.4; Determine the water discharge through standard orifice of 0.5 cm

diameter under water head of 0.9m. use Cd=0.61.

Example 4.5: Determine the actual discharge for orifice in Figure.


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87

Example 4.6: Find the velocity at A in Fig. The barometer reading is

750mmHg.

Example 4.7: A cylindrical tank filled with water to the top. The tank is

2m height and have an opening of 20mm diameter at the bottom base.

Determine the time required for the water to reach height of 3m, 2m, 1m

and 0.


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88

2) Syphons

Example 4.9: Find the flowrate and the pressure of syphon shown in fig.


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89

3) Pitot Tube

Two piezometers, one as normal and one as a Pitot tube within the pipe can

be used as shown below to measure velocity of flow.

The tube opening is directed upstream so that the fluid flows into the

opening untile the pressure builds up in the tube sufficiently to with stand

the impact of velocity against it. Directly in front of opening the fluid is at

rest. The pressure at 1 and 2 is known from the liquid column in the tubes.

By applying Bernoulli’s eq. between points 1 and 2 produce,

𝑃1

𝛾+

𝑣12

2𝑔=

𝑃2

𝛾 →

𝑣12

2𝑔=

𝑃2

𝛾−

𝑃1

𝛾= ℎ2 − ℎ1 = ∆ℎ

𝑣 = √2𝑔∆ℎ

𝑣 = √2𝑔∆ℎ(𝑆𝐺𝑚

𝑆𝐺− 1)

𝑆𝐺𝑚: monomeric liquid SG


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90

Example 4.10: Determine velocity for pitot tube shown in Fig. below with

Δh=0.3m.

4) Venturi Meter

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

rapidly converging section which increases the velocity of flow and hence

reduces the pressure. It then returns to the original dimensions of the pipe by

a gently diverging ‘diffuser’ section.


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91

Apply Bernoulli along the streamline from point 1 to point 2

2

2

22

1

2

11 z + g 2

v +

g

p = z +

g 2

v +

g

p

Substituting and rearranging gives

Or 𝑄𝑡ℎ𝑒𝑜 = 𝐴1√2𝑔(

𝑃1−𝑃2𝛾

+𝑧1−𝑧2)

(𝐴1𝐴2

)2−1= 𝐴1√

2𝑔(𝑃1−𝑃2

𝛾+𝑧1−𝑧2)

(𝐷1𝐷2

)4−1

Actual discharge takes into account the losses due to friction, include a

coefficient of discharge (Cd ≈ 0.9)

𝑄𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶𝑑𝐴1√2𝑔(

𝑃1 − 𝑃2

𝛾+ 𝑧1 − 𝑧2)

(𝐷1

𝐷2)4 − 1


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92

In terms of the manometer readings

𝑃1 + 𝛾𝑧1 − 𝛾𝑚∆ℎ − 𝛾(𝑧2 − ∆ℎ) = 𝑃2

𝑃1 − 𝑃2 + 𝛾𝑧1 − 𝛾𝑧2 = 𝛾𝑚∆ℎ − 𝛾∆ℎ = ∆ℎ(𝛾𝑚 − 𝛾)

𝑃1 − 𝑃2

𝛾+ 𝑧1 − 𝑧2 = ∆ℎ(

𝛾𝑚

𝛾− 1)

Giving 𝑄𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶𝑑𝐴1√2𝑔∆ℎ(

𝛾𝑚𝛾

−1)

(𝐷1𝐷2

)4−1

This expression does not include any elevation terms. (z1 or z2) When used

with a manometer, the Venturimeter can be used without knowing its angle.

Example 4.11: A horizontal venturimeter with inlet and throat diameters

200mm and 100mm respectively is used to measure flow of water. The pressure

at inlet is 176.58kPa and the vacuum pressure at the throat is 300mm Hg .

Determine the rate of flow. Take Cd=0.98.


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93

Example 4.12: A horizontal venturimeter with inlet and throat diameters

300mm and 150mm respectively is used to measure flow of water. The reading

of differential manometer connected to the inlet and the throat is 200mm

mercury. Take Cd=0.98.Determine the rate of flow.

Example 4.13: in vertical pipe conveying oil of SG 0.8, two pressure gage

have been installed at A and B where the diameters are 160mm and 80mm

respectively. A is 2m above B. the pressure difference PA-PB=9.81kPa.

calculate the flowrate. If the pressure gage at A and B is replaced by mercury

manometer, calculate the difference of level of mercury.


Fluid Mechanic

94

Example 4.14: Find the discharge of water flowing through a pipe 300mm

diameter placed in inclined position where venturimeter is inserted, having a

throat diameters of 150mm. The reading of differential manometer with liquid

of SG 0.6 connected to the inlet and the throat is 300mm.. Take Cd=0.98.

5) Notches and s

A weir is a vertical barrier in the side of a tank or reservoir. The liquid is flow

over the weir with free surface. It is a device for measuring discharge. It is used

as both a discharge measuring device and a device to raise water levels. There

are many different designs of weir depending on the shape of weir opening. It

may be rectangular, trapezoidal or triangular weir. The weirs also classified

according its crest width to sharp crested weir, broad crested weir and ogee

weir. The sharp crested rectangular weir may be contracted when the length of


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95

weir opening is less than the channel width or suppress when the length of weir

is equal to channel width.

Fig. 4.9

Weir Assumptions

velocity of the fluid approaching the weir is small so we can ignore kinetic

energy.

The velocity in the flow depends only on the depth below the free surface.

A General Weir Equation

Consider a horizontal strip of width b, depth h below the free surface

velocity through the strip,

discharge through the strip,


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96

Integrating from the free surface, h = 0, to the weir crest, h = H, gives the total

theoretical discharge

This is different for every differently shaped weir or notch. We need an

expression relating the width of flow across the weir to the depth below the free

surface

Rectangular Weir

The width does not change with depth so,

Substituting this into the general weir equation gives

To get the actual discharge we introduce a coefficient of discharge, Cd, to

account for losses at the edges of the weir and contractions in the area of flow,

Example 4.15: Water enters the Millwood flood storage area via a rectangular

weir when the river height exceeds the weir crest. For design purposes a flow

rate of 0.163 m3/s over the weir can be assumed

1. Assuming a height over the crest of 200mm and Cd=0.2, what is the necessary

width, B, of the weir?

2. What will be the velocity over the weir at this design?


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97

‘V’ Notch Weir

The relationship between width and depth is dependent on the angle of the “V”.

The width, b, depth, h, from the free surface relationshipis

So the discharge is

The actual discharge is obtained by introducing a coefficient of discharge


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98

Example 4.16: Water is flowing over a 90o ‘V’ Notch weir into a tank with a

cross-sectional area of 0.6m2. After 30s the depth of the water in the tank is

1.5m. If the discharge coefficient for the weir is 0.8, what is the height of the

water above the weir?

6) Flow under the gates:

Fig. 4.10

The sluice gate is a hydraulic structure for regulate discharge in open channels.

It is cross the flow section which increases the velocity of flow at down stream

of the gate and hence reduces the pressure. Apply Bernoulli along the streamline

from sction 1 to section 2 shown in Fig. 4.10.


Fluid Mechanic

99

𝑦1 +𝑉1

2

2𝑔= 𝑦2 +

𝑉22

2𝑔

∴ 𝑦1 − 𝑦2 =𝑉2

2

2𝑔−

𝑉12

2𝑔

For unit length of gate, the continuity equation can be written as:

𝑞 = 𝑦1 × 1 × 𝑉1 = 𝑦2 × 1 × 𝑉2

𝑉1 =𝑦2

𝑦1𝑉2

2𝑔(𝑦1 − 𝑦2) = 𝑉22 − 𝑉2

2 (𝑦2

𝑦1)

2

= 𝑉22 [1 − (

𝑦2

𝑦1)

2

]

𝑉2 = √2𝑔(𝑦1−𝑦2)

1−(𝑦2𝑦1

)2 Or 𝑉1 = √

2𝑔(𝑦1−𝑦2)

(𝑦1𝑦2

)2

−1

Example 4.17: Flowing of water passing under the gate. The upstream water

depth is 4m and q=8m3/s/m.l. Determine the depth of water downstream the

gate and the velocity of flow?


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100

7) Cavitation

Cavitation is the phenomenon that occurs when the fluid pressure is reduced to

the local vapor pressure and boiling occurs. Cavitation typically occurs at

locations where the velocity is high as shown in Fig. 4.8. Under such conditions

vapor bubbles form in the liquid, grow, and then collapse, producing shock

waves, noise, and dynamic effects that lead to decreased equipment

performance and, frequently, equipment failure. When the flow velocity is

increased, the minimum pressure is reach the local vapor pressure. When the

zone of bubble formation is extended, the entire vapor pocket may

intermittently grow and collapse, producing serious vibration problems which

damage the centrifugal pump impeller and produce erosion in a spillway tunnel.

Fig 4.8

Example 4.18: Water flowing through a horizontal pipe as shown in Fig. 4.8.

the barometric pressure is 650mmHg and vapor pressure for the water is 27kPa

abs. The depth of water in the tank is 7m. Determine the maximum discharge

without cavitation occurred?


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101

Example 4.19: The discharge of water flowing through a pipe of diameter

200mm is 0.44m3/s. The barometric pressure is 100kPa abs and the vapor

pressure of water is 3.77kPa abs. The elevation difference between the

contraction and point 2 is 3m. Determine the contraction diameter to be the

cavitation to be not occurred.


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102

8) Nozzle flow

Fig. 4.9

The pressure of all points of liquid jet outside the nozzle is equal to

atmospheric pressure. So, it will be equal to zero when the atmospheric

pressure is the reference pressure. Then, the Bernoulli’s equation between

points 1 and 2 in Fig. 4.9 can be written as:

The velocity components for Vn are:

𝑉𝑥 = 𝑉𝑛 cos 𝜃 and 𝑉𝑦 = 𝑉𝑛 sin 𝜃

Therefore, for point 1: 𝑉1𝑥 = 𝑉𝑛 cos 𝜃 and 𝑉1𝑦 = 𝑉𝑛 sin 𝜃

For point 2: 𝑉2𝑥 = 𝑉𝑛 cos 𝜃 and 𝑉2𝑦 = 0, so, 𝑉2 = 𝑉𝑛 cos 𝜃,

which mean that the horizontal velocity component is constant along the

nozzle jet path.

∴ 𝑉1𝑥

2

2𝑔+

𝑉1𝑦2

2𝑔= ℎ +

𝑉22

2𝑔

(𝑉𝑛 𝑐𝑜𝑠 𝜃)2

2𝑔+

(𝑉𝑛 𝑠𝑖𝑛 𝜃)2

2𝑔= ℎ +

(𝑉𝑛 𝑠𝑖𝑛 𝜃)2

2𝑔

Therefore the maximum height of nozzle jet will be: ℎ =(𝑉𝑛 sin 𝜃)2

2𝑔


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103

The vertical component of velocity Vy is varied along the nozzle jet path

as below:

𝑉𝑦 = 𝑉𝑛 sin 𝜃 − 𝑔. 𝑡

The horizontal and vertical axes(x,y) of any point along the nozzle jet

path as below:

𝑦 = 𝑉𝑛 𝑠𝑖𝑛 𝜃 . 𝑡 −1

2𝑔𝑡2

𝑥 = 𝑉𝑥. 𝑡 = 𝑉𝑛 cos 𝜃. 𝑡

at the highest point of jet Vy reaches to zero, so:

ℎ =1

2𝑔𝑡2 → 𝑡2 = √

2ℎ

𝑔

and the horizontal distance of highest point determined as:

𝑋 = 𝑉𝑥. 𝑡2 = 𝑉𝑛 𝑐𝑜𝑠 𝜃. 𝑡2

Example 4.20: Determine the vertical and horizontal distance of highest point

of water jet from nozzle with velocity of 20m/s. Also, find the diameter of the

jet at the highest point if the diameter of nozzle is 2cm. The jet is inclined at 60o

with horizontal. Neglect air resistance.

Solution: ℎ =(𝑉𝑛 𝑐𝑜𝑠 𝜃)2

2𝑔 =

(20 sin 60)2

2×9.81=15.29m

𝑡2 = √2ℎ

𝑔= √

2 × 15.29

9.81= 1.766𝑠𝑒𝑐

𝑉𝑥 = 𝑉𝑛 𝑐𝑜𝑠 𝜃 = 20 cos 60 = 10m/s

𝑋 = 𝑉𝑥. 𝑡2 = 10 × 1.766 = 17.66𝑚

Q=Vn ×Anozzle=20.π/4(0.02)2=6.28×10

-3m

3/s


Fluid Mechanic

104

A2=Q/Vx=6.28/10=6.28×10-4

m2

D=0.028m=2.8cm

Example 4.21: Point (b) is located on the stream line. Determine the flowrate?

Solution

𝑥 = 𝑉𝑥. 𝑡 = 𝑉𝑛 cos 𝜃. 𝑡

𝑉𝑛 =𝑥

𝑐𝑜𝑠 𝜃. 𝑡=

42.43

𝑡 … … … . .1

𝑦 = 𝑉𝑛 𝑠𝑖𝑛 𝜃 . 𝑡 −1

2𝑔𝑡2

𝑉𝑛 =𝑦 +

12

𝑔𝑡2

𝑠𝑖𝑛 𝜃. 𝑡=

21.21

𝑡+ 6.94𝑡 … … … … … 2

42.43

𝑡=

21.21

𝑡+ 6.94𝑡 → 6.94𝑡 =

21.21

𝑡 → 𝑡 = 1.749𝑠𝑒𝑐

Vn=42.43/1.749=24.26m/s

Q=Vn.Anozzle=0.107m3/s

Example 4.22: Water is flowing through a jet with a velocity of 45m/s.

Calculate the value of 𝜃 to be the water reach the hole in the wall.


Fluid Mechanic

105

𝑡2 = √2ℎ

𝑔= √

2×30

9.81= 2.47𝑠𝑒𝑐

𝑋 = 𝑉𝑥. 𝑡2 = 𝑉𝑛 𝑐𝑜𝑠 𝜃. 𝑡2

30 = = 45 × 𝑐𝑜𝑠 𝜃 × 2.47 → 𝜃 = 74.34°

9) Pumps and Turbines

The energy line of liquid flow through turbine drops down directly due to

consumption of energy by turbine which call turbine head (ht) . While the

energy line of liquid flow through pump rises up directly due to adding of

energy to the flow by pump which call pump head (hp). So, the Bernoulli’s

equation will be:

Fig. 4.10


Fluid Mechanic

106

Pumps and turbines power

The head is the energy of unit weight: ℎ =𝐸

𝑊 𝑡ℎ𝑒𝑛, 𝐸 = 𝑊 × ℎ

Power is the energy per unit time: 𝑃𝑜𝑤𝑒𝑟 = 𝑊ℎ

𝑡 𝑤ℎ𝑖𝑙𝑒 𝑄𝑤 =

𝑊

𝑡= 𝑄𝛾

∴ 𝑃𝑜𝑤𝑒𝑟 = 𝑄𝛾ℎ

and 𝑃𝑢𝑚𝑝 𝑝𝑜𝑤𝑒𝑟 = 𝑄𝛾ℎ𝑝

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑝𝑜𝑤𝑒𝑟 = 𝑄𝛾ℎ𝑡

Example 4.23: Draw the E.G.L. and H.G.L. of the pipe system in Fig. and

determine the power of pump. The discharge is 0.15m3/s. neglect the friction of

pipe.


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Example 4.24: The depth of water in tank shown in Fig. is 10m and discharge

required through the system is 0.15m3/s. Determine the velocity and the

pressure in each pipe, the power of the pump. Plot E.G.L. and H.G.L.


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Example 4.25: Calculate the depth of water in tank shown in Fig. which will

produce a discharge of 85 l/s. The input power of the turbine is 15kW. What

flowrate may be expected if the turbine is removed?


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Solved Problems

Problem 4.1: water is pumped from a large lake into an irrigation canal of

rectangular cross section producing the flow situation shown in Fig. calculate

the required pump power.

Problem 4.2: A pump is to be installed to increase the flowrate through the

system by 20%. Calculate the required power.

Problem 4.3: If h1=120mm and h2=180mm.Calculate the power of the pump

shown in Fig.


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Problem 4.4: calculate the pump power shown in Fig.

Problem 4.5: What is the maximum power of the turbine extract from the flow

before cavitation will occur at some point in the system? The atmospheric

pressure 102kPa abs and vapor pressure is 3.5kPa abs.

Problem 4.6: Calculate the pump power that will send the jet over the wall.


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Problem 4.7: If the cavitation anywhere in pipe system shown in Fig. is to be

avoided, what is the diameter of the largest nozzle which may be used?

Atmospheric pressure 100kPa abs and vapor pressure 10kPa abs.

Problem 4.8: What is the smallest nozzle diameter that will produce the

maximum possible flowrate through the pipe system? Atmospheric pressure

98.6kPa abs and vapor pressure 9.8kPa abs.

Problem 4.9: The pressure in the closed tank shown in Fig. is gradually

increased. Calculate the gage reading at which cavitation will appear, in the

25mm contraction. Atmospheric pressure 94kPa abs and vapor pressure 13kPa

abs.


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