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Solution Manual For Fluid Mechanics Fundamentals and Applications 4th Edition by Cengel Cimbala – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Chapter 1 Introduction and Basic Concepts

Solutions Manual for

Fluid Mechanics: Fundamentals and Applications

Fourth Edition

Yunus A. Çengel & John M. Cimbala

McGraw-Hill Education, 2018

Chapter 1

INTRODUCTION AND BASIC CONCEPTS

1-1

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

Introduction, Classification, and System

1-1C

Solution We are to define a fluid and how it differs between a solid and a gas.

Analysis A substance in the liquid or gas phase is referred to as a fluid. A fluid differs from a solid in that a solid

can resist an applied shear stress by deforming, whereas a fluid deforms continuously under the influence of shear

stress, no matter how small. A liquid takes the shape of the container it is in, and a liquid forms a free surface in a larger

container in a gravitational field. A gas, on the other hand, expands until it encounters the walls of the container and fills

the entire available space.

Discussion The subject of fluid mechanics deals with ball fluids, both gases and liquids.

1-2C

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 their classification.

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

1-3C

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.

1-2

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-4C

Solution engine is internal or external.

We are to determine whether the flow of air over the wings of an aircraft and the flow of gases through a jet

Analysis engine.

The flow of air over the wings of an aircraft is external since this is an unbounded fluid flow over a surface.

The flow of gases through a jet engine is internal flow since the fluid is completely bounded by the solid surfaces of the

Discussion body and wings).

If we consider the entire airplane, the flow is both internal (through the jet engines) and external (over the

1-5C

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.

1-6C

Solution We are to define the Mach number of a flow and the meaning for a Mach number of 2.

Analysis The Mach number of a flow is defined as the ratio of the speed of flow to the speed of sound in the

flowing fluid. A Mach number of 2 indicate a flow speed that is twice the speed of sound in that fluid.

Discussion Mach number is an example of a dimensionless (or nondimensional) parameter.

1-3

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-7C

Solution We are to discuss if the Mach number of a constant-speed airplane is constant.

Analysis considerably from point to point in the atmosphere.

No. The speed of sound, and thus the Mach number, changes with temperature which may change

1-8C

Solution incompressible.

We are to determine if the flow of air with a Mach number of 0.12 should be approximated as

Analysis Gas flows can often be approximated as incompressible if the density changes are under about 5 percent,

which is usually the case when Ma < 0.3. Therefore, air flow with a Mach number of 0.12 may be approximated as being

incompressible.

Discussion Air is of course a compressible fluid, but at low Mach numbers, compressibility effects are insignificant.

1-9C

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.

1-10C

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

Analysis The region of flow (usually near a wall) in which the velocity gradients are significant and frictional

effects are important is called the boundary layer. 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 some larger value

sufficiently far from the surface. 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).

1-4

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-11C

Solution We are to define a steady-flow process.

Analysis system boundaries.

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

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

1-12C

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 both shear stresses and additional normal stresses besides pressure, but when a

fluid is at rest, the only normal stress is the pressure, and there are no shear stresses.

1-13C

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.

1-5

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-14C

Solution nozzle.

We are to discuss how to select system when analyzing the acceleration of gases as they flow through a

Analysis When analyzing the acceleration of gases as they flow through a nozzle, a wise choice for the system is the

volume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet cross-sections. This is

a control volume (or open system) since mass crosses the boundary.

Discussion It would be much more difficult to follow a chunk of air as a closed system as it flows through the nozzle.

1-15C

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 selected region in space. Mass may cross

the boundary of a control volume or open system.

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.

1-16C

Solution We are to discuss how to select system for the operation of a reciprocating air compressor.

Analysis We would most likely take the system as the air contained in the piston-cylinder device. This system is a

closed or fixed mass system when it is compressing and no mass enters or leaves it. However, it is an open system

during intake or exhaust.

Discussion In this example, the system boundary is the same for either case – closed or open system.

1-6

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

Mass, Force, and Units

1-17C

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/s2. 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.

1-18C

Solution We are to discuss the difference between pound-mass (lbm) and pound-force (lbf).

Analysis system.

The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English

Discussion You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as

appropriate since the two units have different dimensions.

1-19C

Solution We are to explain why the light-year has the dimension of length.

Analysis In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity

and time. Hence, this product forms a distance dimension and unit.

1-20C

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

Analysis There is no acceleration (car moving at constant velocity), thus the net force is zero in both cases.

Discussion acceleration, there must be zero net force.

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

1-7

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $3.30. The steak that is a better

1-21

Solution buy is to be determined.

Assumptions The steaks are of identical quality.

Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. We choose 1

kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be

⎛ ⎞

⎛ ⎞ ⎛ ⎞ ⎟

$3.15 16 oz 1 lbm Unit Cost =

12 ounce steak:

⎟ ⎟

⎜

⎜ ⎜ ⎟ =

⎟ ⎟

⎜

⎜ ⎜ ⎟

⎟ ⎜ ⎜

⎟

⎜ ⎟

12 oz 1 lbm 0.45359 kg $9.26/kg

⎝ ⎠

⎜

⎝ ⎠ ⎝ ⎠

320 gram steak:

$3.30 ⎛ ⎞ ⎛ ⎞ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ =

1000 g Unit Cost =

⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎝ ⎠ ⎝ ⎠

320 g 1 kg $10.3/kg

Therefore, the steak at the traditional market is a better buy.

Discussion Notice the unity conversion factors in the above equations.

1-22

Solution Analysis The mass of an object is given. Its weight is to be determined.

Applying Newton’s second law, the weight is determined to be

2 (150 kg)(9.6 m/s )

= = =

W mg 1440 N

1-23

Solution The mass of a substance is given. Its weight is to be determined in various units.

Analysis Applying Newton’s second law, the weight is determined in various units to be

⎛ ⎞ ⎟

1 N

⎜ ⎟

⎜ = = = ⎟

W mg 9.81 N

(1 kg)(9.81 m/s ) 1 kg m/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

⎛ ⎞ ⎟

1 kN

⎜ ⎟

⎜ = = = ⎟

W mg 0.00981 kN

(1 kg)(9.81 m/s ) 1000 kg m/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

2 (1 kg)(9.81 m/s )

W mg 2

= = = ⋅

1 kg m/s

1 N 1 kgf

⎛ ⎞⎛ ⎞ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ = = = ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟

W mg 1 kgf

(1 kg)(9.81 m/s ) 1 kg m/s 9.81 N

⋅ ⎟ ⎟ ⎜ ⎜ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎟

2.205 lbm

⎜ ⎟

⎜ = = = ⋅

W mg 2

⎟

⎜ ⎟

(1 kg) (32.2 ft/s ) 1 kg

71 lbm ft/s

⎜ ⎟ ⎟

⎜ ⎝ ⎠

2.205 lbm 1 lbf

⎛ ⎞ ⎛ ⎞ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ = = = ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⋅ ⎟ ⎟ ⎜ ⎜ ⎝ ⎠ ⎝ ⎠

W mg 2.21 lbf

(1 kg) (32.2 ft/s ) 1 kg 2

32.2 lbm ft/s

1-8

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-24

Solution determined.

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

The density of air is constant throughout the room.

The density of air is given to be ρ = 1.16 kg/m3

The mass of the air in the room is

m = ρV = (1.16 kg/m3)(3 × 5 × 7 m3) = 121.8 kg ≅ 122 kg

Room air

3 x 5 x 7 m3

Thus,

⎛ ⎞ ⎟

1 N

⎜ ⎟

⎜ = = = ⎟

W mg 1195 N

(121.8 kg)(9.81 m/s ) 1 kg m/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

Discussion Note that we round our final answers to three or four 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.

1-25

Solution A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and

kJ are to be determined.

Analysis The resistance heater consumes electric energy at a rate of 3 kW or 3 kJ/s. Then the total amount of electric energy

used in 2 hours becomes

Total energy = (Energy per unit time)(Time interval)

= (3 kW)(2 h)

= 6 kWh

Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,

Total energy = (6 kWh)(3600 kJ/kWh)

= 21,600 kJ

Discussion Note kW is a unit for power whereas kWh is a unit for energy.

1-9

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-26E

Solution spring and beam scales on the moon.

Analysis

An astronaut takes his scales with him to the moon. It is to be determined how much he weighs on the

(a) A spring scale measures weight, which is the local gravitational force applied on a body:

⎛ ⎞ ⎟

1 lbf

⎜ ⎟

⎜ = = = ⎟

W mg 33.2 lbf

(195 lbm)(5.48 ft/s ) 32.2 lbm ft/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scale

reads what it reads on earth,

W 195 lbf

Discussion scale would you consider to be more accurate?

The beam scale may be marked in units of weight (lbf), but it really compares mass, not weight. Which

1-27

Solution determined.

Analysis From Newton’s second law, the applied force is

⎛ ⎞ ⎟

1 N

⎜ ⎟

⎜ = = = × = ≅

F ma m 5300 N

(6 g) (90 kg)(6 9.81 m/s ) ⎟

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

5297 N 1 kg m/s

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

The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be

Discussion The man feels like he is six times heavier than normal. You get a similar feeling when riding an elevator to

the top of a tall building, although to a much lesser extent.

1-10

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-28

Solution A rock is thrown upward with a specified force. The acceleration of the rock is to be determined.

Analysis The weight of the rock is

⎛ ⎞ ⎟

1 N

⎜ ⎟

⎜ = = = ⎟

W mg 97.9 N

(10 kg)(9.79 m/s ) 1 kg m/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

Then the net force that acts on the rock is

net up down F F F

=− =− = 280 97.9 182.1 N

Rock

From Newton’s second law, the acceleration of the rock becomes

182.1 N 1 kg m/s

⎛ ⎞ ⋅ ⎟

⎜ ⎟

⎜ = = = ⎟

⎜ ⎟

18.2 m/s

10 kg 1 N

⎜ ⎟ ⎟

⎜ ⎝ ⎠

Discussion This acceleration is more than twice the acceleration at which it would fall (due to gravity) if dropped.

1-29

Solution numerical results with proper units.

The previous problem is reconsidered. The entire software solution is to be printed out, including the

Analysis The problem is solved using EES (Engineering Equation Solver), and the solution is given below.

m=10 [kg]

F_up=280 [N]

g=9.79 [m/s^2]

W=m*g

F_net = F_up – F_down

F_down=W

F_net=a*m

SOLUTION

a=18.21 [m/s^2]

F_down=97.9 [N]

F_net=182.1 [N]

F_up=280 [N]

g=9.79 [m/s^2]

m=10 [kg]

W=97.9 [N]

The final results are W = 97.9 N and a = 18.2 m/s2, to three significant digits, which agree with the results of the previous

problem.

Discussion Items in quotation marks in the EES Equation window are comments. Units are in square brackets.

1-11

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-30

Solution Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent

reduction in the weight of an airplane cruising at 13,000 m is to be determined.

Properties The gravitational acceleration g is 9.807 m/s2 at sea level and 9.767 m/s2 at an altitude of 13,000 m.

Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is

equivalent to the percent reduction in the gravitational acceleration, which is determined from

Δ −

= = × = × = 0.41%

9.807 9.767 % Reduction in weight % Reduction in 100 100 9.807

Therefore, the airplane and the people in it will weigh 0.41% less at 13,000 m altitude.

Discussion Note that the weight loss at cruising altitudes is negligible. Sorry, but flying in an airplane is not a good way

to lose weight. The best way to lose weight is to carefully control your diet, and to exercise.

1-31

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

= =− ( ) W mg m a bz

where a = gs

In our case,

= 9.807 m/s2 is the value of gravitational acceleration at sea level and b = 3.32 × 10−6

s−2

=− = = ( ) 0.99 0.99 s s W m a bz W mg

We cancel out mass from both sides of the equation and solve for z, yielding

−

0.99 s a g

Substituting,

Sea level

2 2

9.807 m/s 0.99 9.807 m/s

−

( )

z 29,500 m

= = ≅

−

6 2

29,539 m

3.32 10 1/s

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.

1-12

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-32

Solution A relation for the gravitational constant as a function elevation is given. The weight of an astronaut in the

International Space Station is to be determined.

Analysis At the altitude of the Space Station,

( )( ) 6 2 g− −

=− × =

2 9.807 3.32 10 s 354,000 m 8.6317 m/s

The astronaut’s weight would therefore be

⎛ ⎞ ⎟

⎜ ⎟ = = = ⎜ ⎟

W mg

N (80.0 kg) 8.6317 m/s ( )2

⎜ ⎟

⎜

⋅ ⎝ ⎠

690.538 N kg m/s

Or, rounding off, W = 690.5 N. This is the astronauts weight if he were at that elevation, but not in orbit.

However, since the satellite and its occupants are in orbit, the astronaut feels weightless. In other words, if the

person were to step on a bathroom scale, the reading would be zero.

The astronaut feels a gravitational acceleration of 8.93 m/s2. But since the satellite and its occupants are in orbit,

they are in free fall – constantly accelerating (falling) at this rate, but also moving horizontally at a speed such that the

elevation of the satellite remains constant, and its circular orbit is maintained.

The astronaut feels weightless only because he or she is in a steady circular orbit around the earth. The gravitational

acceleration on a satellite is in fact the centripetal acceleration (towards the earth, i.e., “down”) that maintains the satellite’s

circular orbit. The astronaut “feels” weightless while in orbit in the same way that a person jumping off a diving board

“feels” weightless during free fall.

Discussion The astronaut’s actual weight is only about 12% smaller that on the earth’s surface! It is a common

misconception that space has zero gravity. In fact, gravity does indeed decrease away from earth’s surface, but it is still

fairly strong even at altitudes where geosynchronous satellites orbit. In fact, if you think about it, satellites could not

maintain an orbit at all without gravity acting on them. The simple linear equation we used here for gravitational decay with

elevation breaks down at high elevation, and a more appropriate equation should be used.

1-33

Solution The mass of air that a person breathes in per day is to be determined.

Analysis The total volume of air breathed in per day is

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎛ ⎞ ⎟

⎜ ⎟ = = ⎜ ⎟ ⎟

L min hr 1 m 7.0 60 24 10.08 m

⎟⎜ ⎜

⎜ ⎟ ⎟

⎟⎜ ⎜

⎜ ⎟

⎜ ⎟ ⎜

⎟ ⎟

⎜ ⎜ ⎟

⎝ ⎠ ⎜ ⎝ ⎠

min hr day 1000 L

⎝ ⎠⎝ ⎠

We use the ideal gas law to calculate the total mass of air,

3 2

( ) ⎛ ⎞⎛ ⎞

⎟

⎜ ⎟

P mRT m

⎟

⎜

= → = = = ⎜ ⎟

(101.325 kPa) 10.08 m kN/m kJ 1 1 11.936 kg kPa kN m

⎜

⎜ ⎜

⎟ ⎟

⎛ ⎞ ⎜ ⎝ ⎠ ⋅

⎟

⎜ ⎟

kJ 0.287 (298.15 K) kg K

⎟

⎝ ⎠

⎜ ⎟

⎜

⋅ ⎝ ⎠

So, to two significant digits, an average person breathes in about 12 kg of air per day, or about 17% of the average

person’s mass.

Discussion This is a lot more air than you probably thought! We breathe in more air (in terms of mass) than the mass of

food that we eat!

1-13

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-34

Solution cause of the error is to be determined.

During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable

Analysis The two terms on the right-hand side of the equation

E = 16 kJ + 7 kJ/kg

do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by

mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally

homogeneous; that is, every term in the equation will have the same unit.

Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage.

1-35

Solution We are to calculate the useful power delivered by an airplane propeller.

Assumptions 1 The airplane flies at constant altitude and constant speed. 2 Wind is not a factor in the calculations.

Analysis At steady horizontal flight, the airplane’s drag is balanced by the propeller’s thrust. Energy is force times

distance, and power is energy per unit time. Thus, by dimensional reasoning, the power supplied by the propeller must

equal thrust times velocity,

1 kW 1.341 hp

⎛ ⎞ ⎛ ⎞ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ = = = = ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟

W F V 105 kW 141 hp

thrust

(1500 N)(70.0 m/s) 1000 N m/s 1 kW

⋅ ⎟ ⎟ ⎜ ⎜ ⎝ ⎠ ⎝ ⎠

where we give our final answers to 3 significant digits.

Discussion We used two unity conversion ratios in the above calculation. The actual shaft power supplied by the

airplane’s engine will of course be larger than that calculated above due to inefficiencies in the propeller.

1-14

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-36

Solution We are to calculate lift produced by an airplane’s wings.

Assumptions 1 The airplane flies at constant altitude and constant speed. 2 Wind is not a factor in the calculations.

Analysis At steady horizontal flight, the airplane’s weight is balanced by the lift produced by the wings. Thus, the net

lift force must equal the weight, or FL = 1700 lbf. We use unity conversion ratios to convert to newtons:

1 N

⎛ ⎞ ⎟

⎜ = = ⎟

L F 7560 N

(1700 lbf) 0.22481 lbf

⎜ ⎟

⎜ ⎝ ⎠

where we give our final answers to 3 significant digits.

Discussion The answer is valid at any speed, since lift must balance weight in order to sustain straight, horizontal flight.

As the fuel is consumed, the overall weight of the aircraft will decrease, and hence the lift requirement will also decrease. If

the pilot does not adjust, the airplane will climb slowly in altitude.

1-37E

Solution We are to estimate the work required to lift a fireman, and estimate how long it takes.

Assumptions 1 The vertical speed of the fireman is constant.

Analysis

(a) Work W is a form of energy, and is equal to force times distance. Here, the force is the weight of the fireman (and

equipment), and the vertical distance is Δz, where z is the elevation.

⎛ ⎞ ⎟

1 Btu

⎜ ⎟

= Δ= = ≅

W F z 14.4 Btu

(280 lbf)(40.0 ft) ⎜

⎟

⎜ ⎟

⋅ ⎟

14.393 Btu 778.169 ft lbf

⎜ ⎝ ⎠

where we give our final answer to 3 significant digits, but retain 5 digits to avoid round-off error in part (b).

(b) Power is work (energy) per unit time. Assuming a constant speed,

⎛ ⎞ ⎟

1 hp

Δ= = = ≅

14.393 Btu 5.8182 s

⎜ ⎟

⎜ ⎟ ⎟

⎜ ⎝ ⎠

3.50 hp 0.7068 Btu/s

5.82 s

Again we give our final answer to 3 significant digits.

Discussion The actual required power will be greater than calculated here, due to frictional losses and other

inefficiencies in the boom’s lifting system. One unity conversion ratio is used in each of the above calculations.

1-15

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-38

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

Assumptions Properties Analysis The density of water is constant throughout.

The density of water is given to be ρ = 1000 kg/m3

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

=ρV =(1000 kg/m3)(0.18 m3) = 180 kg

mtank= 6 kg

V= 0.18 m3

H2O

mtotal = m

+ mtank = 180 + 6 = 186 kg

Thus,

⎛ ⎞ ⎟

1 N

⎜ ⎟

⎜ = = = ⎟

W mg 1825 N

(186 kg)(9.81 m/s ) 1 kg m/s

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

Discussion Note the unity conversion factor in the above equation.

1-39

Solution We are to calculate the volume flow rate and mass flow rate of water.

Assumptions 1 The volume flow rate, temperature, and density of water are constant over the measured time.

Properties The density of water at 15oC is ρ = 999.1 kg/m3

Analysis The volume flow rate is equal to the volume per unit time, i.e.,

⎛ ⎞ ⎟

1.5 L 60 s 31.58 L/min

⎜ = = = ≅

⎟

Δ 31.6 Lpm

⎜ ⎟

2.85 s 1 min

⎜ ⎝ ⎠

where we give our final answer to 3 significant digits, but retain 4 digits to avoid round-off error in the second part of the

problem. Since density is mass per unit volume, mass flow rate is equal to volume flow rate times density. Thus,

⎛ ⎞ ⎛ ⎞ ⎟

3 1 min 1 m

⎜ ⎟ = = = ⎜ ⎟

m ρ

⎜ ⎟

(999.1 kg/m )(31.58 L/min) 60 s 1000 L

⎜ ⎟

⎟⎜

⎜ ⎝ ⎠⎝ ⎠

⎜ ⎟

0.526 kg/s

Discussion We used one unity conversion ratio in the first calculation, and two in the second. If we were interested only

in the mass flow rate, we could have eliminated the intermediate calculation by solving for mass flow rate directly, i.e.,

⎛ ⎞ ⎟

⎜ ⎟ = = = ⎜ ⎟

3 1.5 L 1 m (999.1 kg/m ) m

Δ ⎝ ⎠

⎜ ⎟

⎜

0.526 kg/s 2.85 s 1000 L

1-16

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-40

Solution We are to estimate the work and power required to lift a crate.

Assumptions 1 The vertical speed of the crate is constant.

Properties The gravitational constant is taken as g = 9.807 m/s2

Analysis (a) Work W is a form of energy, and is equal to force times distance. Here, the force is the weight of the

crate, which is F = mg, and the vertical distance is Δz, where z is the elevation.

= Δ= Δ

W F z mg z

( )( )( )

1 N ⎛ ⎞⎛ ⎞

⎟

⎜

⎟ = = ≅

1 kJ 90.5 kg 9.807 m/s 1.80 m ⎜ ⎟

⎜ ⎟

⎜

⎟ ⎟

⎜

1.60 kJ

⎜ ⎝ ⎠ ⋅ ⋅ ⎝ ⎠

1.5976 kJ 1 kg m/s 1000 N m

where we give our final answer to 3 significant digits, but retain 5 digits to avoid round-off error in part (b).

(b) Power is work (energy) per unit time. Assuming a constant speed,

⎛ ⎞ ⎟

1.5976 kJ 1000 W 129.88 W

⎜ = = = ≅

⎟

⎜ ⎟

12.3 s 1 kJ/s

⎜ ⎝ ⎠

Again we give our final answer to 3 significant digits.

130 W

Discussion The actual required power will be greater than calculated here, due to frictional losses and other

inefficiencies in the forklift system. Three unity conversion ratios are used in the above calculations.

1-41

Solution A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to

be obtained for the filling time.

Assumptions Gasoline is an incompressible substance and the flow rate is constant.

Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit

of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds.

Putting the given information into perspective, we have

t [s] ↔ V[L], and V [L/s}

It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate.

Therefore, the desired relation is

Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.

1-17

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-42

Solution A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for the

volume of the pool.

Assumptions Water is an incompressible substance and the average flow velocity is constant.

Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter, and flow

velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be arranged such that

we end up with the unit of seconds. Putting the given information into perspective, we have

V[m3] is a function of t [s], D [m], and V [m/s}

It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square

of D. Therefore, the desired relation is

V = CD2Vt

where the constant of proportionality is obtained for a round hose, namely, C = π/4 so that V= (πD2/4)Vt.

Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach.

1-18

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-43

Solution It is to be shown that the power needed to accelerate a car is proportional to the mass and the square of the

velocity of the car, and inversely proportional to the time interval.

Assumptions The car is initially at rest.

Analysis The power needed for acceleration depends on the mass, velocity change, and time interval. Also, the unit of

power W is watt, W, which is equivalent to

W = J/s = N⋅m/s = (kg⋅m/s2)m/s = kg⋅m

2/s3

Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m

2/s3 for power. Putting the

given information into perspective, we have

W [kg⋅m

2/s3] is a function of m [kg], V [m/s], and t [s]

It is obvious that the only way to end up with the unit “kg⋅m

2/s3” for power is to multiply mass with the square of the

velocity and divide by time. Therefore, the desired relation is

2 is proportional to / W mV t

or,

2 / W CmV t

where C is the dimensionless constant of proportionality (whose value is ½ in this case).

Discussion Note that this approach cannot determine the numerical value of the dimensionless numbers involved.

1-19

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

Modeling and Solving Engineering Problems

1-44C

Solution We are to discuss the importance of modeling in engineering.

Analysis Modeling makes it possible to predict the course of an event before it actually occurs, or to study

various aspects of an event mathematically without actually running expensive and time-consuming experiments.

When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions

and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and

principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate

approach, and the results are interpreted.

Discussion In most cases of actual engineering design, the results are verified by experiment – usually by building a

prototype. CFD is also being used more and more in the design process.

1-45C

Solution We are to discuss the difference between analytical and experimental approaches.

Analysis The experimental approach (testing and taking measurements) has the advantage of dealing with the

actual physical system, and getting a physical value within the limits of experimental error. However, this approach is

expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage

that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made

in the analysis.

Discussion Most engineering designs require both analytical and experimental components, and both are important.

Nowadays, computational fluid dynamics (CFD) is often used in place of pencil-and-paper analysis and/or experiments.

1-20

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-46C

Solution We are to discuss choosing a model.

Analysis The right choice between a crude and complex model is usually the simplest model that yields adequate

results. Preparing very accurate but complex models is not necessarily a better choice since such models are not much use

to an analyst if they are very difficult and time consuming to solve. At a minimum, the model should reflect the essential

features of the physical problem it represents. After obtaining preliminary results with the simpler model and optimizing the

design, the complex, expensive model may be used for the final prediction.

Discussion Cost is always an issue in engineering design, and “adequate” is often determined by cost.

1-47C

Solution We are to discuss the difference between accuracy and precision.

Analysis Accuracy refers to the closeness of the measured or calculated value to the true value whereas precision

represents the number of significant digits or the closeness of different measurements of the same quantity to each

other. A measurement or calculation can be very precise without being very accurate, and vice-versa. When

measuring the boiling temperature of pure water at standard atmospheric conditions (100.00oC), for example, a temperature

measurement of 97.861oC is very precise, but not as accurate as the less precise measurement of 99.0oC.

Discussion Accuracy and precision are often confused; both are important for quality engineering measurements.

1-48C

Solution We are to discuss how differential equations arise in the study of a physical problem.

Analysis The description of most scientific problems involves equations that relate the changes in some key variables

to each other, and the smaller the increment chosen in the changing variables, the more accurate the description. In the

limiting case of infinitesimal changes in variables, we obtain differential equations, which provide precise mathematical

formulations for the physical principles and laws by representing the rates of changes as derivatives.

Discussion As we shall see in later chapters, the differential equations of fluid mechanics are known, but very difficult

to solve except for very simple geometries. Computers are extremely helpful in this area.

1-21

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-49C

Solution We are to discuss the value of engineering software packages.

Analysis Software packages are of great value in engineering practice, and engineers today rely on software

packages to solve large and complex problems quickly, and to perform optimization studies efficiently. Despite the

convenience and capability that engineering software packages offer, they are still just tools, and they cannot replace

traditional engineering courses. They simply cause a shift in emphasis in the course material from mathematics to physics.

Discussion While software packages save us time by reducing the amount of number-crunching, we must be careful to

understand how they work and what they are doing, or else incorrect results can occur.

1-50

Solution We are to solve a system of 3 equations with 3 unknowns using appropriate software.

Analysis 2*x-y+z=9

3*x^2+2*y=z+2

x*y+2*z=14

Answers: x = 1.556, y = 0.6254, Using EES software, copy the following lines and paste on a blank EES screen to verify the solution:

z = 6.513

Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve.

1-51

Solution We are to solve a system of 2 equations and 2 unknowns using appropriate software.

Analysis x^3-y^2=10.5

3*x*y+y=4.6

Answers: x = 2.215, y = 0.6018

Using EES software, copy the following lines and paste on a blank EES screen to verify the solution:

Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve.

1-22

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-52

Solution 3.5x

3 – 10x

0.5 – 3x = −4.

We are to determine a positive real root of the following equation using appropriate software:

Analysis Using EES software, copy the following lines and paste on a blank EES screen to verify the solution:

3.5*x^3-10*x^0.5-3*x = –4

Answer: x = 1.554

Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve.

1-53

Solution We are to solve a system of 3 equations with 3 unknowns using appropriate software.

Analysis x^2*y-z=1.5

x-3*y^0.5+x*z=-2

x+y-z=4.2

Using EES software, copy the following lines and paste on a blank EES screen to verify the solution:

Answers: x = 0.9149, y = 10.95, z = 7.665

Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve.

1-23

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

Review Problems

1-54E

Solution hour.

We are to estimate the rate of heat transfer into a room and the cost of running an air conditioner for one

Assumptions during the hour of operation.

1 The rate of heat transfer is constant. 2 The indoor and outdoor temperatures do not change significantly

Analysis

(a) In one hour, the air conditioner supplies 5,000 Btu of cooling, but runs only 60% of the time. Since the indoor and

outdoor temperatures remain constant during the hour of operation, the average rate of heat transfer into the room is the

same as the average rate of cooling supplied by the air conditioner. Thus,

0.60 5000 Btu 1 kW

( ) ⎛ ⎞ ⎟

⎜ = = = ⎟

Q 3,000 Btu/h 0.879 kW

⎜ ⎟

⎜ ⎝ ⎠

1 h 3412.14 Btu/h

(b) Energy efficiency ratio is defined as the amount of heat removed from the cooled space in Btu for 1 Wh (watt-

hour) of electricity consumed. Thus, for every Wh of electricity, this particular air conditioner removes 9.0 Btu from the

room. To remove 3,000 Btu in one hour, the air conditioner therefore consumes 3,000/9.0 = 333.33 Wh = 0.33333 kWh of

electricity. At a cost of 7.5 cents per kWh, it costs only 2.50 cents to run the air conditioner for one hour.

Discussion Notice the unity conversion ratio in the above calculation. We also needed to use some common sense and

dimensional reasoning to come up with the appropriate calculations. While this may seem very cheap, if this air conditioner

is run at these conditions continuously for one month, the electricity will cost ($0.025/h) (24 h/day) (30 day/mo) = $18/mo.

1-24

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-55

Solution different locations are to be determined.

Analysis the relation

The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at

The weight of an 65-kg man at various locations is obtained by substituting the altitude z (values in m) into

⎛ ⎞ ⎟

1 N

−

⎜ ⎟

(65 kg)(9.807 3.32 10 ) 6 2

⎜

= =− × ⎟

⎜ ⎟

(where is in units of m/s ) 1 kg m/s W mg z z

⎜ ⎟

⋅ ⎟

⎜ ⎝ ⎠

Sea level: Denver: Mt. Ev.: (z = 0 m): W = 65 × (9.807 − 3.32 × 10−6 × 0) = 65 × 9.807 = 637.5 N

(z = 1610 m): W = 65 × (9.807 − 3.32 × 10−6 × 1610) = 65 × 9.802 = 637.1 N

(z = 8848 m): W = 65 × (9.807 − 3.32 × 10−6 × 8848) = 65 × 9.778 = 635.5 N

Discussion We report 4 significant digits since the values are so close to each other. The percentage difference in

weight from sea level to Mt. Everest is only about –0.3%, which is negligible for most engineering calculations.

1-56E

Solution expressed in N and kgf.

The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be

Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the

thrust developed is expressed in two other units as

4.448 N

⎛ ⎞ ⎟

Thrust in N:

⎜ = = × ⎟

Thrust (85,000 lbf) 1 lbf

⎜ ⎟

5 3.78 10 N

⎜ ⎝ ⎠

5 1 kgf

⎛ ⎞ ⎟

Thrust in kgf:

⎜ = × = × ⎟

Thrust (37.8 10 N) 9.81 N

⎜ ⎟

4 3.85 10 kgf

⎜ ⎝ ⎠

Discussion Because the gravitational acceleration on earth is close

to 10 m/s2, it turns out that the two force units N and kgf differ by

nearly a factor of 10. This can lead to confusion, and we recommend

that you do not use the unit kgf.

1-25

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-57

Solution Table A-7.

The constants appearing dynamic viscosity relation for methanol are to be determined using the data in

Analysis Using the data from Table A-7, we have

4 /(293 ) 5.857 10 10b c

× =

a− −

4 /(313 ) 4.460 10 10b c

× =

a− −

4 /(333 ) 3.510 10 10b c

× =

a− −

which is a nonlinear system of three algebraic equations. Using EES or any other computer code, one finds

6 8.493 10 Pa s 534.5 k 2.27 K a b c−

= ×⋅ = =

Then the viscosity correlation for methanol becomes

6 534.5/( 2.27) (8.493 10 ) 10 T

= × ×

μ− −

For T μ−

= ° = 4 50 C 323 K the correlation gives 3.941 10 Pa s,

= ×⋅ which is nicely agreeing with the data in Table A-7.

1-58

Solution relation is to be determined.

A relation for the terminal settling velocity of a solid particle is given. The dimension of a parameter in the

Analysis We have the dimensions for each term except FL.

[ ]

g LT

−

⎡ ⎤

= ⎢ ⎥ ⎣ ⎦

D L

[ ] [ ]

[ ]

V LT

−

⎡ ⎤

= ⎢ ⎥ ⎣ ⎦

and

1 2 1/ 2 1/ 2 1 2 2 L L LT F LT L F LT

[ ] [ ] [ ]− − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ = = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Therefore FL is a dimensionless coefficient that is this equation is dimensionally homogeneous, and should hold for any unit

system.

1-26

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-59

Solution The flow of air through a wind turbine is considered. Based on unit considerations, a proportionality relation is to

be obtained for the mass flow rate of air through the blades.

Assumptions Wind approaches the turbine blades with a uniform velocity.

Analysis The mass flow rate depends on the air density, average wind velocity, and the cross-sectional area which depends

on hose diameter. Also, the unit of mass flow rate m is kg/s. Therefore, the independent quantities should be arranged such

that we end up with the proper unit. Putting the given information into perspective, we have

m [kg/s] is a function of ρ [kg/m3], D [m], and V [m/s}

It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantities ρ and V with

the square of D. Therefore, the desired proportionality relation is

is proportional to m ρ

or,

D V

2 m C D V

ρ=

where the constant of proportionality is C = π/4 so that =

2 ( /4) m D V ρ π

Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach.

1-27

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-60

Solution The volume of an oil tank is given. The mass of oil is to be determined.

Assumptions Oil is a nearly incompressible substance and thus its density is constant.

Analysis A sketch of the system is given below.

Suppose we forgot the formula that relates mass to density and volume. However, we know that mass has the unit of

kilograms. That is, whatever calculations we do, we should end up with the unit of kilograms. Putting the given information

into perspective, we have

ρ = 850 kg/m3 and V = 2 m3

It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities. Therefore, the formula we

are looking for should be

m = ρV

Thus,

m = (850 kg/m3)(2 m3) = 1700 kg

Discussion Note that this approach may not work for more complicated formulas. Nondimensional constants also may be

present in the formulas, and these cannot be derived from unit considerations alone.

1-28

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

Fundamentals of Engineering (FE) Exam Problems

1-61

If mass, heat, and work are not allowed to cross the boundaries of a system, the system is called

(a) Isolated (b) Isothermal (c) Adiabatic (d) Control mass (e) Control volume

Answer (a) Isolated

1-62

The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of aircraft is

(a) Sonic (b) Subsonic (c) Supersonic (d) Hypersonic

Answer (b) Subsonic

1-63

One J/kg is equal to

(a) 1 kPa⋅m

3 (b) 1 kN⋅m/kg (c) 0.001 kJ (d) 1 N⋅m (e) 1 m2/s2

Answer: (e) 1 m2/s2

1-64

Which is a unit for power?

(a) Btu (b) kWh (c) kcal (d) hph (e) kW

Answer: (e) kW

1-29

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-65

The speed of an aircraft is given to be 950 km/h. If the speed of sound at that location is 315 m/s, the Mach number is

(a) 0.63 (b) 0.84 (c) 1.0 (d) 1.07 (e) 1.20

Answer (b) 0.84

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

Vel=950 [km/h]*Convert(km/h, m/s)

c=315 [m/s]

Ma=Vel/c

1-66

The weight of a 10-kg mass at sea level is

(a) 9.81 N (b) 32.2 kgf (c) 98.1 N (d) 10 N (e) 100 N

Answer (c) 98.1 N

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

m=10 [kg]

g=9.81 [m/s^2]

W=m*g

1-30

course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1 Introduction and Basic Concepts

1-67

The weight of a 1-lbm mass is

(a) 1 lbm⋅ft/s2 (b) 9.81 lbf (c) 9.81 N (d) 32.2 lbf (e) 1 lbf

Answer (e) 1 lbf

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

m=1 [lbm]

g=32.2 [ft/s^2]

W=m*g*Convert(lbm-ft/s^2, lbf)

1-68

A hydroelectric power plant operates at its rated power of 12 MW. If the plant has produced 26 million kWh of electricity

in a specified year, the number of hours the plant has operated that year is

(a) 2167 h (b) 2508 h (c) 3086 h (d) 3710 h (e) 8760 h

Answer (a) 2167 h

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES

screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

RatedPower=12000 [kW]

ElectricityProduced=26E6 [kWh]

Hours=ElectricityProduced/RatedPower

Design and Essay Problems

1-69 to 1-72

Solution Students’ essays and designs should be unique and will differ from each other.

1-31

course preparation. If you are a student using this Manual, you are using it without permission.

Solution Manual For Fluid Mechanics Fundamentals and Applications 4th Edition by Cengel Cimbala

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