ÇENGEL fluidos - solution.pdf - Fluid Mechanics - morggi

Chapter 1 Introduction and Basic Concepts

Solutions Manual for

Fluid Mechanics: Fundamentals and Applications

by Çengel & Cimbala

CHAPTER 1

INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL

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Chapter 1 Introduction and Basic Concepts

Introduction, Classification, and System

1-1C

Solution

We are to define internal, external, and open-channel flows.

Analysis External flow is the flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe. The flow

in a pipe or duct is internal flow if the fluid is completely bounded by solid surfaces . The flow of liquids in a pipe is

called open-channel flow if the pipe is partially filled with the liquid and there is a free surface , such as the flow of

water in rivers and irrigation ditches.

Discussion As we shall see in later chapters, there different approximations are used in the analysis of fluid flows based

on their classification.

1-2C

Solution

We are to define incompressible and compressible flow, and discuss fluid compressibility.

Analysis A fluid flow during which the density of the fluid remains nearly constant is called incompressible flow .

A flow in which density varies significantly is called compressible flow . A fluid whose density is practically independent

of pressure (such as a liquid) is commonly referred to as an “incompressible fluid,” although it is more proper to refer to

incompressible flow . The flow of compressible fluid (such as air) does not necessarily need to be treated as compressible

since the density of a compressible fluid may still remain nearly constant during flow – especially flow at low speeds.

Discussion It turns out that the Mach number is the critical parameter to determine whether the flow of a gas can be

approximated as an incompressible flow. If Ma is less than about 0.3, the incompressible approximation yields results that

are in error by less than a couple percent.

1-3C

Solution

We are to define the no-slip condition and its cause.

Analysis A fluid in direct contact with a solid surface sticks to the surface and there is no slip . This is known as

the no-slip condition , and it is due to the viscosity of the fluid.

Discussion

There is no such thing as an inviscid fluid, since all fluids have viscosity.

1-4C

Solution We are to define forced flow and discuss the difference between forced and natural flow. We are also to

discuss whether wind-driven flows are forced or natural.

Analysis In forced flow , the fluid is forced to flow over a surface or in a tube by external means such as a pump or a

fan. In natural flow , any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise

of the warmer fluid and the fall of the cooler fluid. The flow caused by winds is natural flow for the earth, but it is

forced flow for bodies subjected to the winds since for the body it makes no difference whether the air motion is caused

by a fan or by the winds.

Discussion

As seen here, the classification of forced vs. natural flow may depend on your frame of reference.

1-2

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PROPRIETARY MATERIAL

Chapter 1 Introduction and Basic Concepts

1-5C

Solution

We are to define a boundary layer, and discuss its cause.

Analysis When a fluid stream encounters a solid surface that is at rest, the fluid velocity assumes a value of zero at

that surface. The velocity then varies from zero at the surface to the freestream value sufficiently far from the surface. The

region of flow in which the velocity gradients are significant and frictional effects are important is called the

boundary layer . The development of a boundary layer is caused by the no-slip condition .

Discussion As we shall see later, flow within a boundary layer is rotational (individual fluid particles rotate), while that

outside the boundary layer is typically irrotational (individual fluid particles move, but do not rotate).

1-6C

Solution

We are to discuss the differences between classical and statistical approaches.

Analysis The classical approach is a macroscopic approach , based on experiments or analysis of the gross behavior

of a fluid, without knowledge of individual molecules, whereas the statistical approach is a microscopic approach based

on the average behavior of large groups of individual molecules.

Discussion

The classical approach is easier and much more common in fluid flow analysis.

1-7C

Solution

We are to define a steady-flow process.

Analysis A process is said to be steady if it involves no changes with time anywhere within the system or at the

system boundaries.

Discussion

The opposite of steady flow is unsteady flow , which involves changes with time.

1-8C

Solution

We are to define stress, normal stress, shear stress, and pressure.

Analysis Stress is defined as force per unit area , and is determined by dividing the force by the area upon which it

acts. The normal component of a force acting on a surface per unit area is called the normal stress , and the tangential

component of a force acting on a surface per unit area is called shear stress . In a fluid at rest, the normal stress is called

pressure .

Discussion

Fluids in motion may have additional normal stresses, but when a fluid is at rest, the only normal stress is

the pressure.

1-9C

Solution

We are to define system, surroundings, and boundary.

Analysis A system is defined as a quantity of matter or a region in space chosen for study . The mass or region

outside the system is called the surroundings . The real or imaginary surface that separates the system from its

surroundings is called the boundary .

Discussion Some authors like to define closed systems and open systems , while others use the notation “system” to

mean a closed system and “control volume” to mean an open system. This has been a source of confusion for students for

many years. [See the next question for further discussion about this.]

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Chapter 1 Introduction and Basic Concepts

1-10C

Solution

We are to discuss when a system is considered closed or open.

Analysis Systems may be considered to be closed or open, depending on whether a fixed mass or a volume in space

is chosen for study. A closed system (also known as a control mass or simply a system ) consists of a fixed amount of

mass, and no mass can cross its boundary . An open system , or a control volume , is a properly selected region in space .

Discussion In thermodynamics, it is more common to use the terms open system and closed system , but in fluid

mechanics, it is more common to use the terms system and control volume to mean the same things, respectively.

Mass, Force, and Units

1-11C

Solution

We are to discuss the difference between pound-mass and pound-force.

Analysis Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit in the

English system . One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s 2 . In other words, the

weight of a 1-lbm mass at sea level on earth is 1 lbf.

Discussion

It is not proper to say that one lbm is equal to one lbf since the two units have different dimensions.

1-12C

Solution

We are to discuss the difference between kg-mass and kg-force.

Analysis The unit kilogram (kg) is the mass unit in the SI system , and it is sometimes called kg-mass , whereas kg-

force (kgf) is a force unit . One kg-force is the force required to accelerate a 1-kg mass by 9.807 m/s 2 . In other words, the

weight of 1-kg mass at sea level on earth is 1 kg-force.

Discussion

It is not proper to say that one kg-mass is equal to one kg-force since the two units have different

dimensions.

1-13C

Solution

We are to calculate the net force on a car cruising at constant velocity.

Analysis

There is no acceleration, thus the net force is zero in both cases .

Discussion By Newton’s second law, the force on an object is directly proportional to its acceleration. If there is zero

acceleration, there must be zero net force.

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PROPRIETARY MATERIAL

Chapter 1 Introduction and Basic Concepts

1-14

Solution

A plastic tank is filled with water. The weight of the combined system is to be determined.

Assumptions

The density of water is constant throughout.

The density of water is given to be ρ = 1000 kg/m 3 .

Properties

Analysis

The mass of the water in the tank and the total mass are

m tank =3 kg

V = 0.2

m 3

m w = ρ V = (1000 kg/m 3 )(0.2 m 3 ) = 200 kg

m total = m w + m tank = 200 + 3 = 203 kg

Thus,

⎛

1 N

⎞

Wmg

(203 kg)(9.81 m/s )

1991 N

≅

1990 N

⎜

⎟

1 kg m/s

⋅

⎝

⎠

where we give the final answer to three significant digits.

Discussion

Note the unity conversion factor in the above equation.

1-15

Solution

The interior dimensions of a room are given. The mass and weight of the air in the room are to be

determined.

Assumptions

The density of air is constant throughout the room.

The density of air is given to be ρ = 1.16 kg/m 3 .

Properties

ROOM

AIR

6X6X8 m 3

Analysis

The mass of the air in the room is

= =

(1.16 kg/m )(6

××

8 m )

334 1 kg

≅ 334 kg

Thus,

⎛

1 N

⎞

Wmg

(334.1 kg)(9.81 m/s )

3277 N

≅

3280 N

⎜

⎟

1 kg m/s

⋅

⎝

⎠

Discussion Note that we round our final answers to three significant digits, but use extra digit(s) in intermediate

calculations. Considering that the mass of an average man is about 70 to 90 kg, the mass of air in the room is probably

larger than you might have expected.

1-16

Solution The variation of gravitational acceleration above sea level is given as a function of altitude. The height at

which the weight of a body decreases by 1% is to be determined.

Analysis

The weight of a body at the elevation z can be expressed as

Wmgm

9 807

− × −

3 32

)

In our case,

099

(

)( .

9807

)

Substituting,

) (

)

(

−

0 99 9 807

9 807

−

3 32

→=

29 540 m

≅ 29,500 m

Sea level

where we have rounded off the final answer to three significant digits.

Discussion This is more than three times higher than the altitude at which a typical commercial jet flies, which is about

30,000 ft (9140 m). So, flying in a jet is not a good way to lose weight – diet and exercise are always the best bet.

1-5

teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

PROPRIETARY MATERIAL

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