SOLUTIONS MANUAL for Fluid Mechanics; Fundamentals and Applications. 4th Edition.

Solu�ons Manual for Fluid Mechanics Fundamentals and Applica�ons 4th Edi�on By Yunus Cengel, John Cimbala (All Chapters 100% Original Verified, A+ Grade) All Chapters Solutions Manual Supplement files download link at the end of this file.

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

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1-1

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

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

their classification.

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. s

1-2

Chapter 1 Introduction and Basic Concepts 1-4C Solution

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

engine is internal or external.

Analysis

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 engine.

Discussion

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

body and wings).

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

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

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

considerably from point to point in the atmosphere.

1-8C Solution

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

incompressible.

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

Chapter 1 Introduction and Basic Concepts 1-11C 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.

1-12C 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

Chapter 1 Introduction and Basic Concepts 1-14C Solution

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

nozzle.

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

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

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

system.

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

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.

1-7

Chapter 1 Introduction and Basic Concepts 1-21 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

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 ⎟⎞ ⎜ ⎟ = $9.26/kg Unit Cost = ⎜⎜ ⎜⎝ 12 oz ⎟⎟⎠ ⎜⎝1 lbm ⎟⎟⎠ ⎜⎜⎝ 0.45359 kg ⎟⎟⎠

12 ounce steak: 320 gram steak:

⎛ $3.30 ⎞⎟ ⎛1000 g ⎞⎟ ⎟ ⎜⎜ ⎟ = $10.3/kg Unit Cost = ⎜⎜ ⎜⎝ 320 g ⎠⎟⎟ ⎝⎜ 1 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

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

Analysis

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

W = mg = (150 kg)(9.6 m/s2 ) = 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 = (1 kg)(9.81 m/s2 ) ⎜⎜ ⎟⎟ = 9.81 N ⎜⎜1 kg ⋅ m/s2 ⎟⎟ ⎝ ⎠ ⎛ ⎞⎟ 1 kN ⎜ W = mg = (1 kg)(9.81 m/s2 ) ⎜⎜ ⎟⎟ = 0.00981 kN 2 ⎜⎜1000 kg ⋅ m/s ⎟⎟ ⎝ ⎠

W = mg = (1 kg)(9.81 m/s2 ) = 1 kg ⋅ m/s2 ⎛ 1 N ⎞⎛ ⎟⎜ 1 kgf ⎞⎟⎟ ⎜ W = mg = (1 kg)(9.81 m/s2 ) ⎜⎜ ⎟⎟⎟⎜⎜ ⎟ = 1 kgf 2 ⎜⎜1 kg ⋅ m/s ⎟⎜⎜ 9.81 N ⎟⎟ ⎝ ⎠⎝ ⎠ ⎛ 2.205 lbm ⎞⎟ ⎜ ⎟⎟ (32.2 ft/s2 ) = 71 lbm ⋅ ft/s2 W = mg = (1 kg) ⎜⎜ ⎜⎜ 1 kg ⎟⎟ ⎝ ⎠ ⎛ 2.205 lbm ⎞⎟ ⎛ ⎞⎟ 1 lbf ⎜ ⎟⎟ (32.2 ft/s2 ) ⎜⎜ ⎟⎟ = 2.21 lbf W = mg = (1 kg) ⎜⎜ ⎜ 2 ⎜⎜ 1 kg ⎟⎟ ⎜⎜ 32.2 lbm ⋅ ft/s ⎟⎟ ⎝ ⎠ ⎝ ⎠

1-8

Chapter 1 Introduction and Basic Concepts 1-24

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.

Properties

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

Analysis

The mass of the air in the room is

Room air 3 x 5 x 7 m3

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

3

Thus, ⎛ 1 N ⎞⎟ ⎜ ⎟ W = mg = (121.8 kg)(9.81 m/s2 ) ⎜⎜ 2⎟ ⎟ = 1195 N ⎜⎜⎝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

Chapter 1 Introduction and Basic Concepts 1-26E

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

Solution

spring and beam scales on the moon.

Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body: ⎛ ⎞⎟ 1 lbf ⎜ W = mg = (195 lbm)(5.48 ft/s2 ) ⎜⎜ ⎟⎟⎟ = 33.2 lbf 2 ⎜⎜ 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

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

scale would you consider to be more accurate?

1-27

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

Solution determined.

Analysis

From Newton's second law, the applied force is ⎛ 1 N ⎞⎟ ⎜ ⎟⎟ = 5297 N ≅ 5300 N F = ma = m(6 g) = (90 kg)(6 × 9.81 m/s2 ) ⎜⎜ ⎜⎜1 kg ⋅ m/s2 ⎟⎟ ⎝ ⎠

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

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

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 ⎞⎟ ⎜ ⎟⎟ = 97.9 N W = mg = (10 kg)(9.79 m/s2 ) ⎜⎜ ⎜⎜1 kg ⋅ m/s2 ⎟⎟ ⎝ ⎠

Then the net force that acts on the rock is

Fnet = Fup − Fdown = 280 − 97.9 = 182.1 N

Rock

From Newton's second law, the acceleration of the rock becomes a=

Discussion

2 F 182.1 N ⎛⎜⎜1 kg ⋅ m/s ⎞⎟⎟ = ⎟ = 18.2 m/s2 ⎜ 10 kg ⎜⎜⎝ 1 N ⎠⎟⎟ m

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

1-29

Solution The previous problem is reconsidered. The entire software solution is to be printed out, including the numerical results with proper units.

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

Chapter 1 Introduction and Basic Concepts 1-30

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

Solution

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 Δg 9.807 − 9.767 % Reduction in weight = % Reduction in g = ×100 = ×100 = 0.41% 9.807 g 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)

z

where a = gs = 9.807 m/s2 is the value of gravitational acceleration at sea level and b = 3.32 × 10−6 s−2.

In our case,

W = m(a − bz) = 0.99Ws = 0.99mgs We cancel out mass from both sides of the equation and solve for z, yielding a − 0.99 gs z= b

0 Sea level

Substituting, z=

9.807 m/s2 − 0.99 (9.807 m/s2 ) 3.32 ×10−6 1/s2

= 29,539 m ≅ 29,500 m

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

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, g = 9.807 − (3.32 ×10−6 s−2 )(354,000 m ) = 8.6317 m/s2

The astronaut’s weight would therefore be ⎛ N ⎞⎟ ⎟ = 690.538 N W = mg = (80.0 kg) (8.6317 m/s2 )⎜⎜ ⎜⎝ kg ⋅ m/s2 ⎠⎟⎟ 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

⎛ ⎝

V = ⎜⎜7.0

⎞⎛ min ⎟⎞⎜⎛ L ⎟⎜ hr ⎟⎞⎛ 1 m ⎟⎞ 3 ⎟⎜⎜ ⎟⎜24 ⎟ = 10.08 m ⎟⎟⎠⎜⎜60 ⎟ ⎜ ⎝ ⎠ min hr ⎝ day ⎟⎠⎟⎝1000 L ⎟⎠

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

→

(101.325 kPa) (10.08 m 3 ) ⎛ kN/m 2 ⎞⎟⎛ kJ ⎞ PV ⎜⎜1 ⎟ ⎟⎜ m= = ⎜⎜⎝ kPa ⎟⎟⎠⎜⎜⎝1 kN ⋅ m ⎟⎟⎠ = 11.936 kg RT ⎛⎜ kJ ⎞⎟ ⎟ (298.15 K) ⎜0.287 ⎜⎝ 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

Chapter 1 Introduction and Basic Concepts 1-34

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

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 = Fthrust V = (1500 N)(70.0 m/s) ⎜⎜ ⎟⎟⎟ = 105 kW ⎜⎜ ⎟⎟⎟ = 141 hp ⎜⎜⎝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

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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:

⎛ ⎞⎟ 1N FL = (1700 lbf) ⎜⎜ ⎟ = 7560 N ⎝ 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 ⎜ ⎟⎟ = 14.393 Btu ≅ 14.4 Btu W = F Δz = (280 lbf)(40.0 ft) ⎜⎜ ⎜⎜ 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,

Δt =

1 hp W 14.393 Btu ⎛⎜ ⎟⎟⎞ = 5.8182 s ≅ 5.82 s = ⎜⎜ ⎟

⎟ 3.50 hp ⎜⎝ 0.7068 Btu/s ⎠ W

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

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

The density of water is constant throughout.

Properties

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

Analysis

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

mtank= 6 kg

mw =ρV =(1000 kg/m3)(0.18 m3) = 180 kg

V = 0.18 m

mtotal = mw + mtank = 180 + 6 = 186 kg

3

H2O

Thus, ⎛ 1 N ⎟⎞ ⎜ ⎟⎟ = 1825 N W = mg = (186 kg)(9.81 m/s2 ) ⎜⎜ ⎜⎜1 kg ⋅ m/s2 ⎟⎟ ⎝ ⎠

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.,

V =

V Δt

=

1.5 L ⎛ 60 s ⎞⎟ ⎜⎜ ⎟ = 31.58 L/min ≅ 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, ⎛1 min ⎟⎞⎛⎜ 1 m 3 ⎞⎟ ⎟ = 0.526 kg/s m = ρV = (999.1 kg/m 3 )(31.58 L/min) ⎜⎜⎜ ⎟⎜ ⎝ 60 s ⎟⎜ ⎠⎝⎜1000 L ⎟⎠⎟ 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.,

m = ρ

V Δt

= (999.1 kg/m 3 )

1.5 L ⎛⎜ 1 m 3 ⎞⎟ ⎟ = 0.526 kg/s ⎜ 2.85 s ⎜⎝⎜1000 L ⎠⎟⎟

1-16

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.

W = F Δz = mgΔz ⎛ 1 N ⎟⎞⎛ 1 kJ ⎞ ⎟⎟ = 1.5976 kJ ≅ 1.60 kJ ⎟⎜ = (90.5 kg)(9.807 m/s2 )(1.80 m )⎜⎜ 2 ⎟⎜ ⎝⎜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,

W 1.5976 kJ ⎛1000 W ⎟⎞ ⎜⎜ W = = ⎟ = 129.88 W ≅ 130 W 12.3 s ⎝ 1 kJ/s ⎟⎠ Δt 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 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 t=

V V

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

1-17

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

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⋅m2/s3 Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m2/s3 for power. Putting the given information into perspective, we have

W [kg⋅m2/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⋅m2/s3” for power is to multiply mass with the square of the velocity and divide by time. Therefore, the desired relation is

W is proportional to mV 2 /t or,

W = CmV 2 /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

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

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

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

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

2*x-y+z=9 3*x^2+2*y=z+2 x*y+2*z=14

Answers: x = 1.556, y = 0.6254, 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

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

x^3-y^2=10.5 3*x*y+y=4.6

Answers:

x = 2.215, y = 0.6018

Discussion

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

1-22

Chapter 1 Introduction and Basic Concepts 1-52

Solution

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

3.5x3 – 10x0.5 – 3x = −4.

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

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

x^2*y-z=1.5 x-3*y^0.5+x*z=-2 x+y-z=4.2

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

Chapter 1 Introduction and Basic Concepts

Review Problems

1-54E

Solution

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

hour.

Assumptions

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

during the hour of operation.

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 ⎝ 1h 3412.14 Btu/h ⎠⎟ Energy efficiency ratio is defined as the amount of heat removed from the cooled space in Btu for 1 Wh (watt(b) 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

Chapter 1 Introduction and Basic Concepts 1-55

Solution

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

different locations are to be determined.

Analysis

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

the relation ⎛ 1 N ⎞⎟ ⎜ ⎟ W = mg = (65 kg)(9.807 − 3.32 ×10−6 z ) ⎜⎜ 2⎟ ⎟ ⎜⎜⎝1 kg ⋅ m/s ⎟⎠

(where z is in units of m/s2 )

Sea level:

(z = 0 m): W = 65 × (9.807 − 3.32 × 10−6 × 0) = 65 × 9.807 = 637.5 N

Denver:

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

Mt. Ev.:

(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

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

expressed in N and kgf.

Analysis

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

thrust developed is expressed in two other units as

Thrust in N:

⎛ 4.448 N ⎞⎟ 5 Thrust = (85,000 lbf) ⎜⎜ ⎟ = 3.78× 10 N ⎝ 1 lbf ⎠⎟

Thrust in kgf:

⎛ 1 kgf ⎞⎟ 4 Thrust = (37.8×105 N) ⎜⎜⎜ ⎟ = 3.85× 10 kgf ⎝ 9.81 N ⎠⎟

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

Chapter 1 Introduction and Basic Concepts 1-57

Solution

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

Table A-7.

Analysis

Using the data from Table A-7, we have

5.857×10−4 = a10b /(293−c ) 4.460×10−4 = a10b /(313−c ) 3.510×10−4 = a10b /(333−c ) which is a nonlinear system of three algebraic equations. Using EES or any other computer code, one finds

a = 8.493×10−6 Pa ⋅ s

b = 534.5 k

c = 2.27 K

Then the viscosity correlation for methanol becomes

μ = (8.493×10−6 )×10534.5 /(T −2.27) For T = 50°C = 323 K the correlation gives μ = 3.941×10−4 Pa ⋅ s, which is nicely agreeing with the data in Table A-7.

1-58

Solution

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

relation is to be determined.

Analysis

We have the dimensions for each term except FL.

[ g ] = ⎡⎢⎣ LT −2 ⎤⎥⎦ [ D] = [ L ]

[VL ] = ⎡⎢⎣ LT −1 ⎤⎥⎦ and −2 ⎤ 1/ 2 1/ 2 ⎡ −1 ⎤ ⎡ ⎡ −1 ⎤ ⎣⎢ LT ⎦⎥ = [ FL ] ⎣⎢ 2 LT ⎦⎥ [ L ] = 2 [ FL ] ⎣⎢ LT ⎦⎥

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

1-26

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

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

m is proportional to ρ D2V or,

m = C ρ D 2V where the constant of proportionality is C = π/4 so that m = ρ(π D2 /4)V Discussion

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

1-27

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

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⋅m3

(b) 1 kN⋅m/kg

(c) 0.001 kJ

(d) 1 N⋅m

(e) 1 m2/s2

(c) kcal

(d) hph

(e) kW

Answer: (e) 1 m2/s2

1-64

Which is a unit for power? (a) Btu

(b) kWh

Answer: (e) kW

1-29

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

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

Chapter 2 Properties of Fluids

Solutions Manual for Fluid Mechanics: Fundamentals and Applications Fourth Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill Education, 2018

Chapter 2 PROPERTIES OF FLUIDS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill Education: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill Education.

2-1

Chapter 2 Properties of Fluids

Density and Specific Gravity

2-1C Solution

We are to discuss the difference between intensive and extensive properties.

Analysis

Intensive properties do not depend on the size (extent) of the system but extensive properties do depend

on the size (extent) of the system.

Discussion

An example of an intensive property is temperature. An example of an extensive property is mass.

2-2C Solution

We are to discuss the difference between mass and molar mass.

Analysis

Mass m is the actual mass in grams or kilograms; molar mass M is the mass per mole in grams/mol or

kg/kmol. These two are related to each other by m = NM, where N is the number of moles.

Discussion

Mass, number of moles, and molar mass are often confused. Molar mass is also called molecular weight.

2-3C Solution

We are to define specific gravity and discuss its relationship to density.

Analysis

The specific gravity, or relative density, is defined as the ratio of the density of a substance to the density

of some standard substance at a specified temperature (the standard is water at 4°C, for which ρH2O = 1000 kg/m3). That is, SG = ρ /ρH2O . When specific gravity is known, density is determined from ρ = SG × ρH2O .

Discussion

Specific gravity is dimensionless and unitless [it is just a number without dimensions or units].

2-2

Chapter 2 Properties of Fluids 2-4C Solution

We are to decide if the specific weight is an extensive or intensive property.

Analysis

The original specific weight is

γ1 =

W

V

If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V /2. The specific weight of one of these halves is

γ=

W /2 = γ1 V /2

which is the same as the original specific weight. Hence, specific weight is an intensive property.

Discussion

If specific weight were an extensive property, its value for half of the system would be halved.

2-5C Solution

We are to discuss the applicability of the ideal gas law.

Analysis

A gas can be treated as an ideal gas when it is at a high temperature and/or a low pressure relative to its

critical temperature and pressure.

Discussion

Air and many other gases at room temperature and pressure can be approximated as ideal gases without any

significant loss of accuracy.

2-6C Solution

We are to discuss the difference between R and Ru.

Analysis

Ru is the universal gas constant that is the same for all gases, whereas R is the specific gas constant that is

different for different gases. These two are related to each other by R = Ru / M , where M is the molar mass (also called the molecular weight) of the gas.

Discussion

Since molar mass has dimensions of mass per mole, R and Ru do not have the same dimensions or units.

2-3

Chapter 2 Properties of Fluids 2-7 Solution

The pressure in a container that is filled with air is to be determined.

Assumptions

At specified conditions, air behaves as an ideal gas.

Properties

The gas constant of air is R = 0.287

Analysis

The definition of the specific volume gives

v=

V m

=

kJ ⎛⎜ kPa ⋅ m3 ⎟⎞ kPa ⋅ m3 ⎟⎟ = 0.287 (see also Table A-1). ⎜⎜ kg ⋅ K ⎜⎝ kJ ⎟⎠ kg ⋅ K

0.075 m 3 = 0.075 m 3 /kg 1 kg

Using the ideal gas equation of state, the pressure is

Pv = RT → P = Discussion

RT

v

=

(0.287 kPa ⋅ m3 /kg ⋅ K)(27 + 273 K) 0.075 m3 /kg

= 1148 kPa

In ideal gas calculations, it saves time to convert the gas constant to appropriate units.

2-8E Solution

The volume of a tank that is filled with argon at a specified state is to be determined.

Assumptions

At specified conditions, argon behaves as an ideal gas.

Properties

The gas constant of argon is obtained from Table A-1E, R = 0.2686 psia⋅ft3/lbm⋅R.

Analysis

According to the ideal gas equation of state,

V= Discussion

mRT (1 lbm)(0.2686 psia ⋅ ft 3 /lbm ⋅ R)(100 + 460 R) = = 0.7521 ft 3 P 200 psia

In ideal gas calculations, it saves time to write the gas constant in appropriate units.

2-9E Solution

The specific volume of oxygen at a specified state is to be determined.

Assumptions

At specified conditions, oxygen behaves as an ideal gas.

Properties

The gas constant of oxygen is obtained from Table A-1E, R = 0.3353 psia⋅ft3/lbm⋅R.

Analysis

According to the ideal gas equation of state,

v= Discussion

RT (0.3353 psia ⋅ ft 3 /lbm ⋅ R)(80 + 460 R) = = 4.53 ft 3 /lbm P 40 psia

In ideal gas calculations, it saves time to write the gas constant in appropriate units.

2-4

Chapter 2 Properties of Fluids 2-10 Solution

The volume and the weight of a fluid are given. Its mass and density are to be determined.

Analysis

Knowing the weight, the mass and the density of the fluid are determined to be

m=

ρ=

W 225 N ⎛⎜1 kg ⋅ m/s3 ⎞⎟ = ⎜ ⎟⎟ = 23.0 kg g 9.80 m/s2 ⎜⎜⎝ 1 N ⎟⎠

m

V

Discussion

=

23.0 kg = 0.957 kg/L 24 L

Note that mass is an intrinsic property, but weight is not.

2-11E

An automobile tire is under-inflated with air. The amount of air that needs to be added to the tire to raise its

Solution

pressure to the recommended value is to be determined.

Assumptions

1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.

Properties

The gas constant of air is Ru = 53.34

Analysis

The initial and final absolute pressures in the tire are

ft ⋅ lbf ⎛⎜ 1 psia ⎞⎟ psia ⋅ ft 3 0.3704 = ⎟ ⎜ lbm ⋅ R ⎜⎝144 lbf/ft 2 ⎠⎟ lbm ⋅ R

P1 = Pg1 + Patm = 22 + 14.6 = 36.6 psia

Tire 2.60 ft3 70°F 22 psia

P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is (36.6 psia)(2.60 ft 3 ) P1V = = 0.4847 lbm m1 = RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(70 + 460 R)

Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes m2 =

(44.6 psia)(2.60 ft 3 ) P2V = = 0.5907 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(70 + 460 R)

Thus the amount of air that needs to be added is

Δm = m2 − m1 = 0.5907 − 0.4847 = 0.106 lbm Discussion

Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.

2-5

Chapter 2 Properties of Fluids 2-12 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined.

Solution

Assumptions

1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.

Properties

The gas constant of air is R = 0.287

Analysis

Initially, the absolute pressure in the tire is

kJ ⎛⎜ kPa ⋅ m3 ⎟⎞ kPa ⋅ m3 ⎟ = 0.287 . ⎜⎜ ⎟ kg ⋅ K ⎝⎜ kJ ⎟⎠ kg ⋅ K

P1 = Pg + Patm = 210 + 100 = 310 kPa

Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire is determined from

P1V1 P2V2 = T1 T2

⎯⎯ → P2 =

323 K T2 P1 = (310 kPa) = 336 kPa T1 298 K

Tire 25°C, 210 kPa

Thus the pressure rise is Δ P = P2 − P1 = 336 − 310 = 26.0 kPa

The amount of air that needs to be bled off to restore pressure to its original value is m1 = m2 =

Discussion

P1V (310 kPa)(0.025 m3 ) = = 0.0906 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(298 K) P2V (310 kPa)(0.025 m3 ) = = 0.0836 kg RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(323 K) Δm = m1 − m2 = 0.0906 − 0.0836 = 0.0070 kg

Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.

2-6

Chapter 2 Properties of Fluids 2-13 Solution

A balloon is filled with helium gas. The number of moles and the mass of helium are to be determined.

Assumptions

At specified conditions, helium behaves as an ideal gas.

The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ ⎜⎛ kPa ⋅ m3 ⎟⎞ kPa ⋅ m3 ⎟⎟ = 8.31447 Ru = 8.31447 . ⎜⎜ kmol ⋅ K ⎜⎝ kJ ⎠⎟ kmol ⋅ K Properties

Analysis

He D=9m 20°C 200 kPa

The volume of the sphere is

4 3

4 3

V = π r 3 = π(4.5 m)3 = 381.704 m3 Assuming ideal gas behavior, the number of moles of He is determined from (200 kPa)(381.704 m 3 ) PV N= = = 31.321 kmol ≅ 31.3 kmol RuT (8.31447 kPa ⋅ m 3 /kmol ⋅ K)(293.15 K)

Then the mass of He is determined from m = NM = (31.321 kmol)(4.003 kg/kmol) = 125.38 kg ≅ 125 kg

Discussion

Although the helium mass may seem large (about the mass of an adult football player!), it is much smaller than that of the air it displaces, and that is why helium balloons rise in the air.

2-7

Chapter 2 Properties of Fluids 2-14 Solution A balloon is filled with helium gas. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted.

The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ ⎜⎛ kPa ⋅ m3 ⎟⎞ kPa ⋅ m3 ⎟⎟ = 8.31447 Ru = 8.31447 . ⎜⎜ kmol ⋅ K ⎝⎜ kJ ⎠⎟ kmol ⋅ K Properties

Analysis

The EES Equations window is shown below, followed by the two parametric tables and the plot (we overlaid the two cases to get them to appear on the same plot).

P = 200 kPa: P = 100 kPa:

P = 200 kPa

P = 100 kPa

Discussion

Mass increases with diameter as expected, but not linearly since volume is proportional to D3.

2-8

Chapter 2 Properties of Fluids 2-15 A cylindrical tank contains methanol at a specified mass and volume. The methanol’s weight, density, and specific gravity and the force needed to accelerate the tank at a specified rate are to be determined.

Solution

Assumptions

1 The volume of the tank remains constant.

Properties

The density of water is 1000 kg/m3.

Analysis

The methanol’s weight, density, and specific gravity are m W = mg = 60 kg × 9.81 2 = 589 N s m 60 kg ρ= = = 800 kg/m 3 V 1 m3 75 L × 1000 L ρ 800 kg/m3 SG = = = 0.800 ρH2 O 1000 kg/m3

The force needed to accelerate the tank at the given rate is

(

)

m F = ma = (60 kg)× 0.25 2 = 15 N s

2-16 Solution

The cylinder conditions before the heat addition process is specified. The pressure after the heat addition

process is to be determined.

Assumptions

1 The contents of cylinder are approximated by the air properties.

2 Air is an ideal gas.

Analysis

The final pressure may be determined from the ideal gas relation

P2 = Discussion

⎛1500 + 273 K ⎞⎟ T2 (1800 kPa) = 4414 kPa P1 = ⎜⎜ ⎜⎝ 450 + 273 K ⎠⎟⎟ T1

Note that some forms of the ideal gas equation are more

convenient to use than the other forms.

2-9

Combustion chamber 1.8 MPa 450°C

Chapter 2 Properties of Fluids 2-17 A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be

Solution

calculated, and the mass of the atmosphere using the correlation is to be estimated.

Assumptions

1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly spherical with a radius of 6377 km at sea

level, and the thickness of the atmosphere is 25 km.

Properties

The density data are given in

tabular form as a function of radius and elevation, where r = z + 6377 km:

r, km 6377

z, km

ρ, kg/m3

0

1.225

6378

1

1.112

6379

2

1.007

6380

3

0.9093

6381

4

0.8194

6382

5

0.7364

6383

6

0.6601

6385

8

0.5258

6387

10

0.4135

6392

15

0.1948

6397

20

0.08891

6402

25

0.04008

Analysis

Using EES, (1) Define a trivial function “rho= a+z” in the Equation window, (2) select new parametric

table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select Plot and click on curve fit to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2 for the unit of kg/m3, (or, ρ(z) = (1.20252 – 0.101674z + 0.0022375z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation gives ρ = 0.600 kg/m3. (b) The mass of atmosphere is evaluated by integration to be

m=

∫ ρdV = ∫ V

h z =0

(a + bz + cz 2 )4π(r0 + z)2 dz = 4π

∫

h z =0

(a + bz + cz2 )(r02 + 2r0 z + z2 )dz

= 4π ⎡⎢ ar02 h + r0 (2a + br0 )h2 / 2 + (a + 2br0 + cr02 )h3 / 3 + (b + 2cr0 )h 4 / 4 + ch5 / 5⎤⎥ ⎣ ⎦ where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere. Also, a = 1.20252, b = −0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 to convert the density from units of kg/km3 to kg/m3, the mass of the atmosphere is determined to be approximately

m = 5.09×1018 kg 2-10

Chapter 2 Properties of Fluids EES Solution for final result: a = 1.2025166 b = –0.10167 c = 0.0022375 r = 6377 h = 25 m = 4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9 Discussion

At 7 km, the density of the air is approximately half of its value at sea level.

2-11

Chapter 2 Properties of Fluids 2-18 Using the data for the density of R-134a in Table A-4, an expression for the density as a function of

Solution

temperature in a specified form is to be obtained.

Analysis

An Excel sheet gives the following results. Therefore we obtain

ρ (kg/m3 ) = −0.037T 2 +18.016T −855.201, T (K)

Temp

Temp,K

Density

Rel. Error, %

-20

253

1359

-1.801766

-10

263

1327

-0.2446119

0

273

1295

0.8180695

10

283

1261

1.50943695

20

293

1226

1.71892333

30

303

1188

1.57525253

40

313

1147

1.04219704

50

323

1102

0.16279492

60

333

1053

-1.1173789

70

343

996.2

-2.502108

80

353

928.2

-3.693816

90

363

837.7

-3.4076638

100

373

651.7

10.0190272

The relative accuracy is quite reasonable except the last data point.

2-12

Chapter 2 Properties of Fluids 2-19 Solution The difference between specific gravity and specific weight is to be explained and the specific weight of the

substances in Table 2-1 are to be determined. Also, specific volume of a liquid is to be determined.

Analysis (a) Specific gravity is nondimensional, and is the ratio of the density of the fluid to the density of water at 4oC.

Specific weight is dimensional, and is simply the product of the density of the fluid and the gravitational acceleration. (b) Excel was used, and below is a printout from the spreadsheet.

To convert from SG to γs, we multiply by ρwater and by g, where

ρwater = g= Substance Water Blood Seawater Gasoline Ethyl alcohol Mercury Balsa wood Dense oak wood Gold Bones (low) Bones (high) Ice Air (at 1 atm)

1000 9.807 SG 1 1.06 1.025 0.68 0.79 13.6 0.17 0.93 19.3 1.7 2 0.916 0.001204

kg/m3 m/s2 γs (N/m3) 9807 10395.42 10052.175 6668.76 7747.53 133375.2 1667.19 9120.51 189275.1 16671.9 19614 8983.212 11.807628

(c) Specific volume is defined as v = 1/ρ. But ρ = SG x ρwater. Thus, v = 1/(SG x ρwater) = 0.00168919 m3/kg Discussion It is easy to confuse specific weight, specific gravity, and specific volume, so be careful with these terms.

Excel shines in cases where there is a lot of repetition.

2-13

Chapter 2 Properties of Fluids

Vapor Pressure and Cavitation

2-20C Solution

We are to define vapor pressure and discuss its relationship to saturation pressure.

Analysis

The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase

equilibrium with its liquid at a given temperature. In general, the pressure of a vapor or gas, whether it exists alone or in a mixture with other gases, is called the partial pressure. During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent since the vapor is pure.

Discussion

Partial pressure is not necessarily equal to vapor pressure. For example, on a dry day (low relative

humidity), the partial pressure of water vapor in the air is less than the vapor pressure of water. If, however, the relative humidity is 100%, the partial pressure and the vapor pressure are equal.

2-21C Solution

We are to discuss whether the boiling temperature of water increases as pressure increases.

Analysis

Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with pressure.

The higher the pressure, the higher the saturation or boiling temperature.

Discussion

This fact is easily seen by looking at the saturated water property tables. Note that boiling temperature and

saturation pressure at a given pressure are equivalent.

2-22C Solution

We are to determine if temperature increases or remains constant when the pressure of a boiling substance

increases.

Analysis

If the pressure of a substance increases during a boiling process, the temperature also increases since the

boiling (or saturation) temperature of a pure substance depends on pressure and increases with it.

Discussion

We are assuming that the liquid will continue to boil. If the pressure is increased fast enough, boiling may

stop until the temperature has time to reach its new (higher) boiling temperature. A pressure cooker uses this principle.

2-14

Chapter 2 Properties of Fluids 2-23C Solution

We are to define and discuss cavitation.

Analysis

In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure

drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon is a common cause for drop in performance and even the erosion of impeller blades.

Discussion

The word “cavitation” comes from the fact that a vapor bubble or “cavity” appears in the liquid. Not all

cavitation is undesirable. It turns out that some underwater vehicles employ “super cavitation” on purpose to reduce drag.

2-24E Solution

The minimum pressure on the suction side of a water pump is given. The maximum water temperature to

avoid the danger of cavitation is to be determined.

Properties

The saturation temperature of water at 0.70 psia is 90°F (Table A-3E).

Analysis

To avoid cavitation at a specified pressure, the fluid temperature everywhere in the flow should remain

below the saturation temperature at the given pressure, which is

Tmax = Tsat @ 0.70 psia = 90°F Therefore, T must remain below 90°F to avoid the possibility of cavitation.

Discussion

Note that saturation temperature increases with pressure, and thus cavitation may occur at higher pressure at

locations with higher fluid temperatures.

2-25 Solution

The minimum pressure in a pump to avoid cavitation is to be determined.

Properties

The vapor pressure of water at 20°C is 2.339 kPa.

Analysis

To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the vapor (or

saturation) pressure at the given temperature. That is, Pmin = Psat@20°C = 2.339 kPa Therefore, the lowest pressure that can exist in the pump is 2.339 kPa. Discussion

Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater

at higher fluid temperatures.

2-15 . ..

Chapter 2 Properties of Fluids 2-26 Solution

The minimum pressure in a piping system to avoid cavitation is to be determined.

Properties

The vapor pressure of water at 30°C is 4.246 kPa.

Analysis

To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapor (or

saturation) pressure at the given temperature. That is,

Pmin = Psat@30°C = 4.246 kPa Therefore, the pressure should be maintained above 4.246 kPa everywhere in flow.

Discussion

Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater

at higher fluid temperatures.

2-16 . ..

Chapter 2 Properties of Fluids

Energy and Specific Heats

2-27C Solution

We are to define total energy and identify its constituents.

Analysis

The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic,

electrical, and surface tension effects, the total energy of a system consists of the kinetic, potential, and internal

energies.

Discussion

All three constituents of total energy (kinetic, potential, and internal) need to be considered in an analysis of

a general fluid flow.

2-28C Solution

We are to list the forms of energy that contribute to the internal energy of a system.

Analysis

The internal energy of a system is made up of sensible, latent, chemical, and nuclear energies. The

sensible internal energy is due to translational, rotational, and vibrational effects.

Discussion

We deal with the flow of a single phase fluid in most problems in this textbook; therefore, latent, chemical,

and nuclear energies do not need to be considered.

2-29C Solution

We are to discuss the relationship between heat, internal energy, and thermal energy.

Analysis

Thermal energy is the sensible and latent forms of internal energy. It does not include chemical or

nuclear forms of energy. In common terminology, thermal energy is referred to as heat. However, like work, heat is not a property, whereas thermal energy is a property.

Discussion

Technically speaking, “heat” is defined only when there is heat transfer, whereas the energy state of a

substance can always be defined, even if no heat transfer is taking place.

2-17 . ..

Chapter 2 Properties of Fluids 2-30C Solution

We are to define and discuss flow energy.

Analysis

Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids at

rest do not possess any flow energy.

Discussion

Flow energy is not a fundamental quantity, like kinetic or potential energy. However, it is a useful concept

in fluid mechanics since fluids are often forced into and out of control volumes in practice.

2-31C Solution

We are to compare the energies of flowing and non-flowing fluids.

Analysis

A flowing fluid possesses flow energy, which is the energy needed to push a fluid into or out of a

control volume, in addition to the forms of energy possessed by a non-flowing fluid. The total energy of a non-flowing fluid consists of internal and potential energies. If the fluid is moving as a rigid body, but not flowing, it may also have kinetic energy (e.g., gasoline in a tank truck moving down the highway at constant speed with no sloshing). The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies.

Discussion

Flow energy is not to be confused with kinetic energy, even though both are zero when the fluid is at rest.

2-32C Solution

We are to explain how changes in internal energy can be determined.

Analysis

Using specific heat values at the average temperature, the changes in the specific internal energy of ideal

gases can be determined from Δu = cv,avg ΔT . For incompressible substances, cp ≅ cv ≅ c and Δu = cavg ΔT .

Discussion

If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.

2-33C Solution

We are to explain how changes in enthalpy can be determined.

Analysis

Using specific heat values at the average temperature, the changes in specific enthalpy of ideal gases can be

determined from Δh = c p,avg ΔT . For incompressible substances, cp ≅ cv ≅ c and Δh = Δu + vΔP ≅ cavg ΔT + vΔP .

Discussion

If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.

2-18

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Chapter 2 Properties of Fluids 2-34E Solution

We are to estimate the energy required to heat up the water in a hot-water tank.

Assumptions

1 There are no losses. 2 The pressure in the tank remains constant at 1 atm. 3 An approximate analysis is

performed by replacing differential changes in quantities by finite changes.

Properties

The specific heat of water is approximated as a constant, whose value is 0.999 Btu/lbm⋅R at the average

temperature of (60 + 110)/2 = 85oF. In fact, c remains constant at 0.999 Btu/lbm⋅R (to three digits) from 60oF to 110oF. For this same temperature range, the density varies from 62.36 lbm/ft3 at 60oF to 61.86 lbm/ft3 at 110oF. We approximate the density as constant, whose value is 62.17 lbm/ft3 at the average temperature of 85oF. Analysis

For a constant pressure process, Δ u ≅ cavg Δ T . Since this is energy per unit mass, we must multiply by the

total mass of the water in the tank, i.e., Δ U ≅ mcavg Δ T = ρV cavg Δ T . Thus,

⎛ 35.315 ft 3 ⎞⎟ ⎟ = 31,135 Btu ≅ 31,100 Btu ΔU ≅ ρV cavg ΔT = (62.17 lbm/ft 3 )(75 gal)(0.999 Btu/lbm ⋅ R)[(110 − 60)R]⎜⎜⎜ ⎜⎝ 264.17 gal ⎠⎟⎟ where we note temperature differences are identical in oF and R. Discussion

We give the final answer to 3 significant digits. The actual energy required will be greater than this, due to

heat transfer losses and other inefficiencies in the hot-water heating system.

2-35 Solution

The total energy of saturated water vapor flowing in a pipe at a specified velocity and elevation is to be

determined.

Analysis

The total energy of a flowing fluid is given by (Eq. 2-8) e= h+

V2 + gz 2

The enthalpy of the vapor at the specified temperature can be found in any thermo text to be 2745.9 kJ/kg. Then the total energy is determined as m 35 ) ( J m J s e = 2745.9×10 + + (9.81 )×(25 m ) ≅ 2.7468×10 = 2746.8 kJ/kg kg 2 kg s 2

3

6

2

Note that only 0.031% of the total energy comes from the combination of kinetic and potential energies, which explains why we usually neglect kinetic and potential energies in most flow systems.

2-19 . ..

Chapter 2 Properties of Fluids

Compressibility

2-36C Solution

We are to define the coefficient of volume expansion.

Analysis

The coefficient of volume expansion represents the variation of the density of a fluid with temperature at

constant pressure. It differs from the coefficient of compressibility in that the latter represents the variation of pressure of a fluid with density at constant temperature.

Discussion

The coefficient of volume expansion of an ideal gas is equal to the inverse of its absolute temperature.

2-37C Solution

We are to discuss the coefficient of compressibility and the isothermal compressibility.

Analysis

The coefficient of compressibility represents the variation of pressure of a fluid with volume or density

at constant temperature. Isothermal compressibility is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure.

Discussion

The coefficient of compressibility of an ideal gas is equal to its absolute pressure.

2-38C Solution

We are to discuss the sign of the coefficient of compressibility and the coefficient of volume expansion.

Analysis

The coefficient of compressibility of a fluid cannot be negative, but the coefficient of volume expansion can

be negative (e.g., liquid water below 4°C).

Discussion

This is the reason that ice floats on water.

2-20 . ..

Chapter 2 Properties of Fluids 2-39E Solution

We are to estimate the density as water is heated, and we are to compare to the actual density.

Assumptions

1 The coefficient of volume expansion is constant in the given temperature range. 2 The pressure remains

constant at 1 atm throughout the heating process. 3 An approximate analysis is performed by replacing differential changes in quantities by finite changes.

Properties

The density of water at 60°F and 1 atm pressure is ρ1 = 62.36 lbm/ft3. The coefficient of volume expansion

at the average temperature of (60 + 130)/2 = 95°F is β = 0.187 × 10−3 R−1.

The change in density due to the change of temperature from 60°F to 130°F at constant pressure is

Analysis

⎛ 1 ⎞⎛ lbm ⎞ lbm Δρ ≅ −βρΔT = −⎜⎜0.187×10−3 ⎟⎟⎟⎜⎜⎜62.36 3 ⎟⎟⎟ ⎣⎡(130 − 60) R ⎦⎤ = −0.816 3 ⎝ R ⎠⎝ ft ⎠ ft where we note temperature differences are identical in oF and R. Thus, the density at the higher temperature is approximated as ρ2 ≈ ρ1 + Δρ = 62.36 – 0.816 = 61.54 lbm/ft3. From the appendices, we see that the actual density at 130oF is 61.55 lbm/ft3.

Thus, the approximation is extremely good, with an error of less than 0.02%.

Discussion

Note that the density of water decreases while being heated, as expected. This problem can be solved more

accurately using differential analysis when functional forms of properties are available.

2-40 The volume of an ideal gas is reduced by half at constant temperature. The change in pressure is to be

Solution

determined.

Assumptions

The process is isothermal and thus the temperature remains constant.

Analysis

For an ideal gas of fixed mass undergoing an isothermal process, the ideal gas relation reduces to

P2V2 PV = 11 T2 T1

→

P2V2 = P1V1

→

V V2

P2 = 1 P1 =

V1 P1 = 2 P1 0.5V1

Therefore, the change in pressure becomes

ΔP = P2 − P1 = 2 P1 − P1 = P1 Discussion

Note that at constant temperature, pressure and volume of an ideal gas are inversely proportional.

2-21 . ..

Chapter 2 Properties of Fluids 2-41 Water at a given temperature and pressure is compressed to a high pressure isothermally. The increase in the density of water is to be determined.

Solution

Assumptions

1 The isothermal compressibility is constant in the given pressure range. 2 An approximate analysis is

performed by replacing differential changes by finite changes.

Properties

The density of water at 20°C and 1 atm pressure is ρ1 = 998 kg/m3. The isothermal compressibility of water

is given to be α = 4.80 × 10−5 atm−1.

When differential quantities are replaced by differences and the properties α and β are assumed to be

Analysis

constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT The change in density due to a change of pressure from 1 atm to 400 atm at constant temperature is Δ ρ = αρ Δ P = (4.80 ×10− 5 atm -1 )(998 kg/m 3 )(400 − 1)atm = 19.2 kg/m 3

Discussion

Note that the density of water increases from 998 to 1017.2 kg/m3 while being compressed, as expected.

This problem can be solved more accurately using differential analysis when functional forms of properties are available.

2-22 . ..

Chapter 2 Properties of Fluids 2-42 The percent increase in the density of an ideal gas is given for a moderate pressure. The percent increase in density of the gas when compressed at a higher pressure is to be determined.

Solution

Assumptions

The gas behaves an ideal gas.

Analysis

For an ideal gas, P = ρRT and (∂P / ∂ρ )T = RT = P / ρ , and thus κideal gas = P . Therefore, the coefficient

of compressibility of an ideal gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. ΔP ΔP ≅ and rearranging Substituting κ = P into the definition of the coefficient of compressibility κ ≅ − Δv / v Δρ / ρ gives

Δρ ΔP = ρ P Therefore, the percent increase of density of an ideal gas during isothermal compression is equal to the percent

increase in pressure.

At 10 atm: At 1000 atm:

Δρ ΔP 11−10 = = = 0.10 = 10% ρ P 10 Δρ ΔP 101−100 = = = 0.01 = 1% ρ P 100

Therefore, a pressure change of 1 atm causes a density change of 10% at 10 atm and a density change of 1% at 1000 atm.

Discussion

If temperature were also allowed to change, the relationship would not be so simple.

2-23 . ..

Chapter 2 Properties of Fluids 2-43 Saturated refrigerant-134a at a given temperature is cooled at constant pressure. The change in the density of the refrigerant is to be determined.

Solution

Assumptions

1 The coefficient of volume expansion is constant in the given temperature range. 2 An approximate

analysis is performed by replacing differential changes in quantities by finite changes.

Properties

The density of saturated liquid R-134a at 10°C is ρ1 =1261 kg/m3. The coefficient of volume expansion at

the average temperature of (10+0)/2 = 5°C is β = 0.00269 K–1.

When differential quantities are replaced by differences and the properties α and β are assumed to be

Analysis

constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT The change in density due to the change of temperature from 10°C to 0°C at constant pressure is Δ ρ = −βρΔ T = −(0.00269 K − 1 )(1261 kg/m 3 )(0 − 10)K = 33.9 kg/m 3

Discussion

Noting that Δρ = ρ2 − ρ1 , the density of R-134a at 0°C is

ρ 2 = ρ1 + Δ ρ = 1261 + 33.9 = 1294.9 kg/m 3

which is almost identical to the listed value of 1295 kg/m3 at 0°C in R-134a table in the Appendix. This is mostly due to β varying with temperature almost linearly. Note that the density increases during cooling, as expected.

2-24 . ..

Chapter 2 Properties of Fluids 2-44 A water tank completely filled with water can withstand tension caused by a volume expansion of 0.8%.

The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined.

Solution

Assumptions

1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing

differential changes in quantities by finite changes. 3 The effect of pressure is disregarded.

Properties

The average volume expansion coefficient is given to be β = 0.377 × 10–3 K–1 (Table A-3 at 40°C).

Analysis

When differential quantities are replaced by differences and the properties α and β are assumed to be

constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT A volume increase of 0.8% corresponds to a density decrease of 0.8%, which can be expressed as Δρ = −0.008ρ . Then the decrease in density due to a temperature rise of ΔT at constant pressure is −0.008ρ = −βρΔT Solving for ΔT and substituting, the maximum temperature rise is determined to be ΔT =

Discussion

0.008 0.008 = = 21.2 K = 21.2°C β 0.377×10−3 K -1

This result is conservative since in reality the increasing pressure will tend to compress the water and

increase its density.

2-25 . ..

Chapter 2 Properties of Fluids 2-45 A water tank completely filled with water can withstand tension caused by a volume expansion of 0.4%.

The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined.

Solution

Assumptions

1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing

differential changes in quantities by finite changes. 3 The effect of pressure is disregarded.

Properties

The average volume expansion coefficient is given to be β = 0.377 × 10–3 K–1 (Table A-3 at 40°C).

Analysis

When differential quantities are replaced by differences and the properties α and β are assumed to be

constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT A volume increase of 0.4% corresponds to a density decrease of 0.4%, which can be expressed as Δρ = −0.004ρ . Then the decrease in density due to a temperature rise of ΔT at constant pressure is −0.004 ρ = −βρΔT Solving for ΔT and substituting, the maximum temperature rise is determined to be ΔT =

Discussion

0.004 0.004 = = 10.6 K = 10.6°C β 0.377×10−3 K -1

This result is conservative since in reality the increasing pressure will tend to compress the water and

increase its density. The change in temperature is exactly half of that of the previous problem, as expected.

2-26 . ..

Chapter 2 Properties of Fluids 2-46 Solution The density of seawater at the free surface and the bulk modulus of elasticity are given. The density and pressure at a depth of 2500 m are to be determined.

Assumptions 1 The temperature and the bulk modulus of elasticity of seawater is constant. 2 The gravitational acceleration remains constant.

Properties The density of seawater at free surface where the pressure is given to be 1030 kg/m3, and the bulk modulus of elasticity of seawater is given to be 2.34 × 109 N/m2.

Analysis

The coefficient of compressibility or the bulk modulus of elasticity of fluids is expressed as

⎛ ∂P ⎞ κ = ρ ⎜⎜⎜ ⎟⎟⎟ ⎝ ∂ρ ⎠⎟

or

κ=ρ

T

dP dρ

(at constant T )

The differential pressure change across a differential fluid height of dz is given as

z=0

dP = ρ gdz

z

Combining the two relations above and rearranging,

κ=ρ

ρ gdz dz = gρ 2 dρ dρ

→

dρ gdz = κ ρ2

2500 m

Integrating from z = 0 where ρ = ρ 0 = 1030 kg/m 3 to z = z where ρ = ρ gives

∫

ρ

ρ0

dρ g = κ ρ2

z

∫ dz

→

0

1 1 gz − = ρ0 ρ κ

Solving for ρ gives the variation of density with depth as

ρ=

1 (1/ ρ0 ) −( gz / κ)

Substituting into the pressure change relation dP = ρ gdz and integrating from z = 0 where P = P0 = 98 kPa to z = z where P = P gives P

∫ dP = ∫ P0

0

z

gdz (1/ ρ0 ) − ( gz / κ)

→

⎛ 1 ⎟⎟⎞ P = P0 + κln ⎜⎜⎜ ⎟ ⎜⎝1 − (ρ0 gz / κ) ⎟⎟⎠

which is the desired relation for the variation of pressure in seawater with depth. At z = 2500 m, the values of density and pressure are determined by substitution to be

1 = 1041 kg/m 3 1/(1030 kg/m3 ) − (9.81 m/s2 )(2500 m) /(2.34 ×109 N/m 2 ) ⎛ ⎞⎟ 1 P = (98,000 Pa) + (2.34 ×109 N/m 2 ) ln ⎜⎜⎜ ⎟⎟ 3 2 9 2 ⎝⎜1 − (1030 kg/m )(9.81 m/s )(2500 m) /(2.34 ×10 N/m ) ⎠⎟

ρ=

= 2.550 ×10 7 Pa = 25.50 MPa

since 1 Pa = 1 N/m2 = 1 kg/m⋅s2 and 1 kPa = 1000 Pa. Discussion

Note that if we assumed ρ = ρo = constant at 1030 kg/m3, the pressure at 2500 m would be P = P0 + ρ gz =

0.098 + 25.26 = 25.36 MPa. Then the density at 2500 m is estimated to be Δ ρ = ρα Δ P = (1030)(2340 MPa)− 1 (25.26 MPa) = 11.1 kg/m 3 and thus ρ = 1041 kg/m

3

2-27 . ..

Chapter 2 Properties of Fluids 2-47E Solution The coefficient of compressibility of water is given. The pressure increases required to reduce the volume of water by 1 percent and then by 2 percent are to be determined.

Assumptions

1 The coefficient of compressibility is constant. 2 The temperature remains constant.

Properties

The coefficient of compressibility of water is given to be 7×105 psia.

Analysis

(a) A volume decrease of 1 percent can mathematically be expressed as

Δv

v

=

ΔV

V

= −0.01

The coefficient of compressibility is expressed as ⎛ ∂P ⎞ ΔP κ = −v ⎜⎜ ⎟⎟⎟ ≅ − ⎜⎝ ∂v ⎠ Δv /v T Rearranging and substituting, the required pressure increase is determined to be

⎛ Δv ⎟⎞ 5 ΔP = −κ ⎜⎜⎜ ⎟ = −(7×10 psia)(−0.01) = 7,000 psia ⎝ v ⎟⎠ (b) Similarly, the required pressure increase for a volume reduction of 2 percent becomes

⎛ Δv ⎟⎞ 5 ΔP = −κ ⎜⎜⎜ ⎟ = −(7×10 psia)(−0.02) = 14,000 psia ⎝ v ⎟⎠ Discussion

Note that at extremely high pressures are required to compress water to an appreciable amount.

2-28 . ..

Chapter 2 Properties of Fluids 2-48 Solution determined.

Assumptions

The water contained in a piston-cylinder device is compressed isothermally. The energy needed is to be 1 The coefficient of compressibility of water remains unchanged during the compression.

Analysis We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as 2

∫ PdV

W =−

(1)

1

From the definition of coefficient of compressibility we have

κ=−

dP dV/V

Rearranging we obtain

dV dP =− V κ which can be integrated from the initial state to any state as follows: V dV P dP P − P0 V =− → ln = − V0 κ V0 V P0 κ

∫

∫

from which we obtain

P = P0 − κ ln

V V0

Substituting in Eq. 1 we have

∫

W =−

V1

V0

⎡ ⎤ V1 V⎞ V ⎜⎜ P0 − κ ln ⎟⎟ dV = ⎢κV ln − ( P0 + κ )V ⎥ ⎢⎣ ⎥⎦ V V0 ⎟⎠ V0 V0 ⎜ ⎝ 0

∫

PdV = −

V1 ⎛

or

W = ( P0 + κ )(V0 − V1 ) + κV0 ln

V1 V0

In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as

V1 − V0 ≅ −α ( P1 − P0 ) V0 or

V1 ≅ V0 (1− α ( P1 − P0 ))

where α is the isothermal compressibility of water, which is at 4.80 × 10−5 atm−1 at 20°C. Realizing that 10 kg water occupies initially a volume of V0 = 10 × 10−3 m3, the final volume of water is determined to be

V1 ≅ (0.01 m3 )×⎡⎣1−(4.80×10−5 atm−1 )×(100 atm −1 atm)⎤⎦ = 9.952×10−3 m3 Then the work done on the water is

W = (1 atm + 2100 atm)×(10×10−3 m3 − 9.952×10−3 m3 ) +2100 atm ×(10 ×10−3 m3 ) ln from which we obtain

W = 2.903×10−4 atm ⋅ m3 ≅ 29.4 J since 1 atm = 101325 Pa.

2-29 . ..

9.952 ×10−3 m3 10 ×10−3 m3

Chapter 2 Properties of Fluids 2-49 The water contained in a piston-cylinder device is compressed isothermally and the pressure increases linearly. The energy needed is to be determined.

Solution

Assumptions

1 The pressure increases linearly.

Analysis

We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as 2

∫ PdV

W =−

(1)

1

For a linear pressure increase we take

P + P 100 atm +1 atm P = Pave = 1 2 = = 50.5 atm 2 2 In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as (Eq. 3−23)

V1 − V0 ≅ −α ( P1 − P0 ) V0 or

V1 ≅ V0 (1− α ( P1 − P0 )) where α is the isothermal compressibility of water, which is at 4.80 × 10−5 atm−1 at 20°C. Realizing that 10 kg water occupies initially a volume of V0 = 10 × 10−3 m3, the final volume of water is determined to be

V1 ≅ (0.01 m3 )×⎡⎣1−(4.80×10−5 atm−1 )×(100 atm −1 atm)⎤⎦ = 9.952×10−3 m3 Therefore the work expression becomes V1

∫ PdV = −P (V −V ) = −(50.5 atm)×(9.952×10 m −10×10 m )

W =−

V0

−3

ave

1

3

−3

3

0

or

W = 2.424×10−3 atm ⋅ m3 = 246 J Thus, we conclude that linear pressure increase approximation does not work well since it gives almost ten times larger work.

2-30 . ..

Chapter 2 Properties of Fluids 2-50 We are to estimate the density of air at two temperatures using the Boussinesq approximation, and we are to compare to the actual density at those temperatures.

Solution

Assumptions

1 The coefficient of volume expansion is constant in the given temperature range. 2 The pressure remains

constant at 95.0 kPa throughout the flow field. 3 Air is taken as an ideal gas.

Properties

The reference density of air at T0 = 40°C (313.15 K) and P = 95.0 kPa is

ρ0 =

⎛ kJ ⎞⎟⎛⎜ kN/m2 ⎞⎟ P 95.0 kPa 3 ⎜⎜ = ⎟⎟⎜⎜ ⎟⎟ = 1.05703 kg/m RT0 (0.2870 kJ/kg ⋅ K )(313.15K ) ⎜⎝ kN ⋅ m ⎠⎜ ⎝ kPa ⎠⎟

where we have kept 6 significant digits to avoid round-off error in subsequent calculations. The coefficient of volume expansion at at T0 = 40°C (313.15 K) is β = 1/T0 = 3.1934 × 10–3 K–1.

Analysis

Using the Boussinesq approximation at T = 20oC , we calculate

Boussinesq density at 20oC: ρ20 = ρ0 ⎡⎣⎢1 − β (T − T0 )⎤⎦⎥ = (1.05703 kg/m 3 ) ⎡⎢1 − (3.1934 ×10−3 K -1 )(20 − 40) K ⎤⎥ = 1.1245 kg/m3 ≅ 1.12 kg/m 3 ⎣ ⎦

where we note temperature differences are identical in oC and K. Similarly, at T = 60oC, Boussinesq density at 60oC: ρ60 = ρ0 ⎡⎣⎢1 − β (T − T0 )⎤⎦⎥ = (1.05703 kg/m 3 ) ⎢⎡1 − (3.1934 ×10−3 K -1 )(60 − 40) K ⎥⎤ = 0.98953 kg/m 3 ≅ 0.990 kg/m 3 ⎣ ⎦ The actual values of density are obtained from the ideal gas law, yielding ρ20, actual = 1.1292 kg/m3 ≈ 1.13 kg/m3; the Boussinesq approximation has a percentage error of –0.41%. Similarly, ρ20, actual = 0.99358 kg/m3 ≈ 0.994 kg/m3; the Boussinesq approximation has a percentage error of –0.41%. Thus, the Boussinesq approximation is extremely good for this range of temperature.

Discussion

In the calculation of percentage error, we used five digits of precision to avoid round-off error. Note that the

density of air decreases when heated, as expected (hot air rises). The Boussinesq approximation is often used in computational fluid dynamics (CFD).

2-31 . ..

Chapter 2 Properties of Fluids 2-51 Using the definition of the coefficient of volume expansion and the expression βideal gas = 1/ T , it is to be

Solution

shown that the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature.

Assumptions

The gas behaves as an ideal gas.

Analysis The coefficient of volume expansion β can be expressed as β =

Δv / v 1 ⎛⎜ ∂v ⎞⎟ . ⎜ ⎟ ≈ v ⎜⎝ ∂T ⎟⎠P ΔT

Noting that βideal gas = 1/ T for an ideal gas and rearranging give

Δv

v

=

ΔT T

Therefore, the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the

percent increase in absolute temperature.

Discussion

We must be careful to use absolute temperature (K or R), not relative temperature (oC or oF).

2-32 . ..

Chapter 2 Properties of Fluids

Speed of Sound

2-52C Solution

We are to define and discuss sound and how it is generated and how it travels.

Analysis

Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium.

It travels by wave propagation. Sound waves cannot travel in a vacuum.

Discussion

Electromagnetic waves, like light and radio waves, can travel in a vacuum, but sound cannot.

2-53C Solution

We are to discuss whether sound travels faster in warm or cool air.

Analysis

Sound travels faster in warm (higher temperature) air since c = kRT .

Discussion

On the microscopic scale, we can imagine the air molecules moving around at higher speed in warmer air,

leading to higher propagation of disturbances.

2-54C Solution

We are to compare the speed of sound in air, helium, and argon.

Analysis

Sound travels fastest in helium, since c = kRT and helium has the highest kR value. It is about 0.40 for

air, 0.35 for argon, and 3.46 for helium.

Discussion

We are assuming, of course, that these gases behave as ideal gases – a good approximation at room

temperature.

2-33 . ..

Chapter 2 Properties of Fluids 2-55C Solution

We are to compare the speed of sound in air at two different pressures, but the same temperature.

Analysis

Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on

temperature only. Therefore, the speed of sound is the same in both mediums.

Discussion

If the temperature were different, however, the speed of sound would be different.

2-56C Solution

We are to examine whether the Mach number remains constant in constant-velocity flow.

Analysis

In general, no, because the Mach number also depends on the speed of sound in gas, which depends on the

temperature of the gas. The Mach number remains constant only if the temperature and the velocity are constant.

Discussion

It turns out that the speed of sound is not a strong function of pressure. In fact, it is not a function of

pressure at all for an ideal gas.

2-57C Solution

We are to state whether the propagation of sound waves is an isentropic process.

Analysis

Yes, the propagation of sound waves is nearly isentropic. Because the amplitude of an ordinary sound

wave is very small, and it does not cause any significant change in temperature and pressure.

Discussion

No process is truly isentropic, but the increase of entropy due to sound propagation is negligibly small.

2-58C Solution

We are to discuss sonic velocity – specifically, whether it is constant or it changes.

Analysis

The sonic speed in a medium depends on the properties of the medium, and it changes as the properties

of the medium change.

Discussion

The most common example is the change in speed of sound due to temperature change.

2-34 . ..

Chapter 2 Properties of Fluids 2-59 The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound.

Solution

The isentropic relation Pvk = A where A is a constant can also be expressed as

Analysis

⎛ 1 ⎞k P = A ⎜⎜ ⎟⎟⎟ = Aρ k ⎝v⎠

Substituting it into the relation for the speed of sound,

⎛ ∂ P ⎞ ⎛ ∂ ( Aρ)k ⎟⎞ ⎟⎟ = kAρ k−1 = k ( Aρ k ) / ρ = k ( P / ρ) = kRT c2 = ⎜⎜⎜ ⎟⎟⎟ = ⎜⎜⎜ ⎝ ∂ρ ⎟⎠ ⎝⎜ ∂ρ ⎟⎠ s

s

since for an ideal gas P = ρRT or RT = P/ρ. Therefore, c = kRT , which is the desired relation.

Discussion

Notice that pressure has dropped out; the speed of sound in an ideal gas is not a function of pressure.

2-35 . ..

Chapter 2 Properties of Fluids 2-60 Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle.

Solution

Assumptions

1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process.

Properties

The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Its constant pressure specific heat and specific

heat ratio at room temperature are cp = 0.8439 kJ/kg⋅K and k = 1.288.

Analysis

(a) At the inlet

⎛1000 m2 / s2 ⎟⎞ ⎟ = 540.3 m/s c1 = k1 RT1 = (1.288)(0.1889 kJ/kg ⋅ K)(1200 K)⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎟⎟⎠ Thus,

Ma1 =

1200 K 50 m/s

V1 50 m/s = = 0.0925 c1 540.3 m/s

Carbon dioxide

400 K

(b) At the exit,

⎛1000 m2 / s2 ⎟⎞ ⎟ = 312.0 m/s c2 = k2 RT2 = (1.288)(0.1889 kJ/kg ⋅ K)(400 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎟⎟⎠ The nozzle exit velocity is determined from the steady-flow energy balance relation, 0 = h2 − h1 +

V2 2 − V12 2

→

0 = c p (T2 − T1 ) +

0 = (0.8439 kJ/kg ⋅ K)(400 −1200 K) +

V2 2 − V12 2

V2 2 − (50 m/s)2 ⎛⎜ 1 kJ/kg ⎞⎟ ⎜⎜⎝1000 m 2 / s2 ⎟⎟⎠ 2

⎯⎯ →

V2 = 1163 m/s

Thus,

Ma 2 =

V2 1163 m/s = = 3.73 c2 312 m/s

Discussion

The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database): → c1 = 516 m/s,

V1 = 50 m/s, Ma1 = 0.0969

At 400 K: cp = 0.9383 kJ/kg⋅K, k = 1.252 → c2 = 308 m/s,

V2 = 1356 m/s, Ma2 = 4.41

At 1200 K: cp = 1.278 kJ/kg⋅K, k = 1.173

Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are significant.

2-36 . ..

Chapter 2 Properties of Fluids 2-61 Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger.

Solution

Assumptions

1 N2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible.

Properties

The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at

room temperature are cp = 1.040 kJ/kg⋅K and k = 1.4.

Analysis

The speed of sound at the inlet is c1 = k1 RT1 ⎛1000 m 2 / s2 ⎟⎞ ⎟ = (1.400)(0.2968 kJ/kg ⋅ K)(283 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎟⎟⎠ = 342.9 m/s

120 kJ/kg

Thus,

Ma1 =

150 kPa 10°C 100 m/s

V1 100 m/s = = 0.292 c1 342.9 m/s

Nitrogen

100 kPa 200 m/s

From the energy balance on the heat exchanger, qin = c p (T2 − T1 ) +

V2 2 − V12 2

120 kJ/kg = (1.040 kJ/kg.°C)(T2 −10°C) +

(200 m/s)2 − (100 m/s)2 ⎛⎜ 1 kJ/kg ⎞⎟ ⎟ ⎜⎜⎝ 2 1000 m 2 / s2 ⎟⎠

It yields T2 = 111°C = 384 K

⎛1000 m2 / s2 ⎞⎟ ⎟ = 399 m/s c2 = k2 RT2 = (1.4)(0.2968 kJ/kg ⋅ K)(384 K)⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎠⎟⎟ Thus,

Ma 2 =

V2 200 m/s = = 0.501 c2 399 m/s

Discussion

The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database): → c1 = 343 m/s,

V1 = 100 m/s, Ma1 = 0.292

At 111°C cp = 1.041 kJ/kg⋅K, k = 1.399 → c2 = 399 m/s,

V2 = 200 m/s, Ma2 = 0.501

At 10°C : cp = 1.038 kJ/kg⋅K, k = 1.400

Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are almost identical to the values obtained assuming constant specific heats.

2-37 . ..

Chapter 2 Properties of Fluids 2-62 Solution

The speed of sound in refrigerant-134a at a specified state is to be determined.

Assumptions

R-134a is an ideal gas with constant specific heats at room temperature.

Properties

The gas constant of R-134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108.

Analysis

From the ideal-gas speed of sound relation,

⎛1000 m2 / s2 ⎟⎞ ⎟ = 176 m/s c = kRT = (1.108)(0.08149 kJ/kg ⋅ K)(70 + 273 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎟⎟⎠ Discussion

Note that the speed of sound is independent of pressure for ideal gases.

2-63 The Mach number of an aircraft and the speed of sound in air are to be determined at two specified

Solution

temperatures.

Assumptions

Air is an ideal gas with constant specific heats at room temperature.

Properties

The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4.

Analysis

From the definitions of the speed of sound and the Mach number,

(a) At 300 K,

⎛1000 m 2 / s2 ⎞⎟ ⎟ = 347 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(300 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎠⎟⎟ and

Ma =

V 330 m/s = = 0.951 c 347 m/s

(b) At 800 K,

⎛1000 m2 / s2 ⎟⎞ ⎟ = 567 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(800 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎟⎟⎠ and

Ma = Discussion

V 330 m/s = = 0.582 c 567 m/s Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a

rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature.

2-38 . ..

Chapter 2 Properties of Fluids 2-64E Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior.

Solution

Assumptions

Steam is an ideal gas with constant specific heats.

Properties

The gas constant of steam is R = 0.1102 Btu/lbm·R. Its specific heat ratio is given to be k = 1.3.

Analysis

From the ideal-gas speed of sound relation,

⎛ 25,037 ft 2 / s2 ⎞⎟ ⎟ = 2040 ft/s c = kRT = (1.3)(0.1102 Btu/lbm ⋅ R)(1160 R)⎜⎜⎜ ⎜⎝ 1 Btu/lbm ⎠⎟⎟ Thus,

Ma =

V 900 ft/s = = 0.441 c 2040 ft/s

Discussion

Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case.

2-39 . ..

Chapter 2 Properties of Fluids 2-65E Solution

Problem 2-64E is reconsidered. The variation of Mach number with temperature as the temperature changes

between 350° and 700°F is to be investigated, and the results are to be plotted.

Analysis

The EES Equations window is printed below, along with the tabulated and plotted results.

T=Temperature+460 R=0.1102 V=900 k=1.3 c=SQRT(k*R*T*25037) Ma=V/c

Temperature,

Mach number

T, °F

Ma

350 375

0.528 0.520

400 425

0.512 0.505

450 475

0.498 0.491

500 525

0.485 0.479

550 575

0.473 0.467

600 625

0.462 0.456

650 675

0.451 0.446

700

0.441

Discussion

Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.

2-40 . ..

Chapter 2 Properties of Fluids 2-66 The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined.

Solution

Assumptions

Air is an ideal gas with constant specific heats at room temperature.

Properties

The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded.

Analysis

The final temperature of air is determined from the isentropic relation of ideal gases, ⎛ P ⎞( k −1) / k ⎛ 0.4 MPa ⎟⎞(1.4−1) /1.4 T2 = T1 ⎜⎜⎜ 2 ⎟⎟⎟ = (350.2 K) ⎜⎜ = 215.2 K ⎟ ⎝ 2.2 MPa ⎟⎠ ⎜⎝ P1 ⎟⎠

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as

Ratio = Discussion

k RT T c2 350.2 = 1 1 = 1 = = 1.28 c1 T2 k2 RT2 215.2

Note that the speed of sound is proportional to the square root of thermodynamic temperature.

2-67 The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined.

Solution

Assumptions

Helium is an ideal gas with constant specific heats at room temperature.

Properties

The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667.

Analysis

The final temperature of helium is determined from the isentropic relation of ideal gases, ⎛ P ⎞( k −1) / k ⎛ 0.4 ⎞(1.667−1) /1.667 T2 = T1 ⎜⎜⎜ 2 ⎟⎟⎟ = (350.2 K) ⎜⎜ ⎟⎟⎟ = 177.0 K ⎝ 2.2 ⎠ ⎜⎝ P1 ⎠⎟

The ratio of the initial to the final speed of sound can be expressed as

Ratio = Discussion

k RT T c2 350.2 = 1 1 = 1 = = 1.41 c1 T2 k2 RT2 177.0

Note that the speed of sound is proportional to the square root of thermodynamic temperature.

2-41 . ..

Chapter 2 Properties of Fluids 2-68 Solution

The Mach number of a passenger plane for specified limiting operating conditions is to be determined.

Assumptions

Air is an ideal gas with constant specific heats at room temperature.

Properties

The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4.

Analysis

From the speed of sound relation

⎛1000 m2 / s2 ⎞⎟ ⎟ = 293 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(-60 + 273 K) ⎜⎜⎜ ⎜⎝ 1 kJ/kg ⎠⎟⎟ Thus, the Mach number corresponding to the maximum cruising speed of the plane is

Ma = Discussion

Vmax (945 / 3.6) m/s = = 0.897 c 293 m/s Note that this is a subsonic flight since Ma < 1. Also, using a k value at −60°C would give practically the

same result.

2-42 . ..

Chapter 2 Properties of Fluids

Viscosity

2-69C Solution

We are to discuss Newtonian fluids.

Analysis

Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called

Newtonian fluids. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids.

Discussion

In the differential analysis of fluid flow, only Newtonian fluids are considered in this textbook.

2-70C Solution

We are to define and discuss viscosity.

Analysis

Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the

internal frictional force that develops between different layers of fluids as they are forced to move relative to each other.

Viscosity is caused by the cohesive forces between the molecules in liquids, and by the molecular collisions in gases. In general, liquids have higher dynamic viscosities than gases.

Discussion

The ratio of viscosity μ to density ρ often appears in the equations of fluid mechanics, and is defined as the

kinematic viscosity, ν = μ /ρ.

2-71C Solution

We are to discuss how kinematic viscosity varies with temperature in liquids and gases.

Analysis

(a) For liquids, the kinematic viscosity decreases with temperature. (b) For gases, the kinematic

viscosity increases with temperature.

Discussion

You can easily verify this by looking at the appendices.

2-43 . ..

Chapter 2 Properties of Fluids 2-72C We are to compare the settling speed of balls dropped in water and oil; namely, we are to determine which will reach the bottom of the container first.

Solution

Analysis

When two identical small glass balls are dropped into two identical containers, one filled with water and the

other with oil, the ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.

Discussion Oil is very viscous, with typical values of viscosity approximately 800 times greater than that of water at room temperature.

2-73E The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be

Solution

determined.

R

l = 0.035 in fluid

Assumptions

1 The inner cylinder is completely submerged in the fluid. 2 The viscous effects on the two ends of the

inner cylinder are negligible. 3 The fluid is Newtonian.

Analysis Substituting the given values, the viscosity of the fluid is determined to be μ=

Discussion

TA (1.2 lbf ⋅ ft)(0.035/12 ft) = = 2.72 ×10−4 lbf ⋅ s/ft 2 3 2

4 π R nL 4 π (3 /12 ft)3 (250 / 60 s-1 )(5 ft) 2

This is the viscosity value at temperature that existed during the experiment. Viscosity is a strong

function of temperature, and the values can be significantly different at different temperatures.

2-44 . ..

Chapter 2 Properties of Fluids 2-74 The viscosities of carbon dioxide at two temperatures are given. The constants of Sutherland correlation for carbon dioxide are to be determined and the viscosity of carbon dioxide at a specified temperature is to be predicted and

Solution

compared to the value in table A-10.

Analysis

Sutherland correlation is given by Eq. 2−32 as

μ=

a T 1 + b/T

where T is the absolute temperature. Substituting the given values we have μ1 =

μ2 =

a T1 1 + b/T1 a T2

1 + b/T2

a 50 + 273.15 a 323.15 → 1.612 ×10−5 = b b 1+ 1+ 50 + 273.15 323.15

=

=

a 200 + 273.15 a 473.15 → 2.276 ×10−5 = b b 1+ 1+ 200 + 273.15 473.15

which is a nonlinear system of two algebraic equations. Using EES or any other computer code, one finds the following result:

a = 1.633×10−6 kg/ (m ⋅ s ⋅ K1/2 )

b = 265.5 K

Using these values the Sutherland correlation becomes

μ=

1.633×10−6 T 1 + 265.5/T

Therefore the viscosity at 100°C is found to be

μ=

1.633×10−6 373.15 = 1.843×10−5 Pa ⋅ s 1 + 265.5/373.15

The agreement is perfect and within approximately 0.1%.

2-45 . ..

Chapter 2 Properties of Fluids 2-75 Solution The velocity profile of a fluid flowing though a circular pipe is given. The friction drag force exerted on the pipe by the fluid in the flow direction per unit length of the pipe is to be determined.

Assumptions

The viscosity of the fluid is constant.

Analysis

The wall shear stress is determined from its definition to be

τw = −μ

nμumax −nr n−1 du d ⎛ rn ⎞ = −μumax ⎜⎜1− n ⎟⎟⎟ = −μumax = n ⎜ dr r=R dr ⎝ R ⎠r =R R R r=R

Note that the quantity du /dr is negative in pipe flow, and the negative sign

u(r) = umax(1-rn/Rn)

is added to the τw relation for pipes to make shear stress in the positive (flow) direction a positive quantity. (Or, du /dr = – du /dy since y = R – r).

Then the friction drag force exerted by the fluid on the inner surface of the

R r

pipe becomes nμumax (2π R ) L = 2 nπμumax L R Therefore, the drag force per unit length of the pipe is F = τ w Aw =

0 umax

F / L = 2nπμumax .

Discussion

Note that the drag force acting on the pipe in this case is independent of the pipe diameter.

2-76 Solution determined.

The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be

R

l = 1 mm fluid Assumptions

1 The inner cylinder is completely submerged in oil. 2 The viscous effects on the two ends of the inner

cylinder are negligible. 3 The fluid is Newtonian.

Analysis

Substituting the given values, the viscosity of the fluid is determined to be TA (0.8 N ⋅ m)(0.001 m) = = 0.0128 N ⋅ s/m 2 μ= 2 3 2

4 π R nL 4 π (0.075 m)3 (300 / 60 s-1 )(0.75 m)

Discussion

This is the viscosity value at the temperature that existed during the experiment. Viscosity is a strong

function of temperature, and the values can be significantly different at different temperatures.

2-46 . ..

Chapter 2 Properties of Fluids 2-77 A thin flat plate is pulled horizontally through an oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity. The location in oil where the velocity is zero and the force that needs to be

Solution

applied on the plate are to be determined.

Assumptions

1 The thickness of the plate is negligible. 2 The velocity profile in each oil layer is linear.

Properties

The absolute viscosity of oil is given to be μ = 0.027 Pa⋅s = 0.027 N⋅s/m2.

Analysis

(a) The velocity profile in each oil layer relative to the fixed wall is as shown in the figure below. The point of zero velocity is indicated by point A, and its distance from the lower plate is determined from geometric considerations

(the similarity of the two triangles in the lower oil layer) to be

2.6 − y A 3 = 0.3 yA

→ yA = 0.23636 mm Fixed wall

h1=1 mm

V = 3 m/s

h2=2.6 mm y

F

A yA

Vw= 0.3 m/s Moving wall

(b) The magnitudes of shear forces acting on the upper and lower surfaces of the plate are

Fshear, upper = τ w, upper As = μ As

du V −0 3 m/s = μ As = (0.027 N ⋅ s/m 2 )(0.3× 0.3 m 2 ) = 7.29 N dy h1 1.0 ×10-3 m

Fshear, lower = τ w, lower As = μ As

V − Vw [3 − (−0.3)] m/s du = μ As = (0.027 N ⋅ s/m 2 )(0.3× 0.3 m 2 ) = 3.08 N dy h2 2.6 ×10-3 m

Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be F = Fshear, upper + Fshear, lower = 7.29 + 3.08 = 10.4 N

Discussion

Note that wall shear is a friction force between a solid and a liquid, and it acts in the opposite direction of

motion.

2-47 . ..

Chapter 2 Properties of Fluids 2-78 We are to determine the torque required to rotate the inner cylinder of two concentric cylinders, with the inner cylinder rotating and the outer cylinder stationary. We are also to explain what happens when the gap gets bigger.

Solution

Assumptions

1 The fluid is incompressible and Newtonian. 2 End effects (top and bottom) are negligible. 3 The gap is

very small so that wall curvature effects are negligible. 4 The gap is so small that the velocity profile in the gap is linear.

Inner cylinder V h

u

y

Outer cylinder Analysis

(a) We assume a linear velocity profile between the two walls as sketched – the inner wall is moving at

speed V = ωiRi and the outer wall is stationary. The thickness of the gap is h, and we let y be the distance from the outer wall into the fluid (towards the inner wall). Thus, u=V

y du V and τ = μ =μ h dy h

where

h = Ro - Ri and V = ωi Ri Since shear stress τ has dimensions of force/area, the clockwise (mathematically negative) tangential force acting along the surface of the inner cylinder by the fluid is

F = −τ A = −μ

μωi Ri V 2π Ri L = − 2π Ri L h Ro − Ri

But the torque is the tangential force times the moment arm Ri. Also, we are asked for the torque required to turn the inner cylinder. This applied torque is counterclockwise (mathematically positive). Thus, T = −FRi =

2π Lμωi Ri 3 2π Lμωi Ri 3 = Ro − Ri h

(b) The above is only an approximation because we assumed a linear velocity profile. As long as the gap is very small, and therefore the wall curvature effects are negligible, this approximation should be very good. Another way to think about this is that when the gap is very small compared to the cylinder radii, a magnified view of the flow in the gap appears similar to flow between two infinite walls (Couette flow). However, as the gap increases, the curvature effects are no longer negligible, and the linear velocity profile is not expected to be a valid approximation. We do not expect the velocity to

remain linear as the gap increases.

Discussion

It is possible to solve for the exact velocity profile for this problem, and therefore the torque can be found analytically, but this has to wait until the differential analysis chapter.

2-48 . ..

Chapter 2 Properties of Fluids 2-79 A clutch system is used to transmit torque through an oil film between two identical disks. For specified rotational speeds, the transmitted torque is to be determined.

Solution

Assumptions

1 The thickness of the oil film is uniform. 2 The rotational speeds of the disks remain constant.

Properties

The absolute viscosity of oil is given to be μ = 0.38 N⋅s/m2.

Driving shaft

Driven shaft

30 cm

2 mm

SAE 30W oil

The disks are rotting in the same direction at different angular speeds of ω1 and of ω2 . Therefore, we can Analysis assume one of the disks to be stationary and the other to be rotating at an angular speed of ω1 − ω 2 . The velocity gradient anywhere in the oil of film thickness h is V/h where V = (ω1 − ω2 )r is the tangential velocity. Then the wall shear stress anywhere on the surface of the faster disk at a distance r from the axis of rotation can be expressed as (ω − ω2 )r du V =μ =μ 1 τw = μ dr h h h Then the shear force acting on a differential area dA on the surface and the torque generation associated with it can be expressed as (ω − ω2 )r dF = τ w dA = μ 1 (2πr )dr h (ω − ω2 )r 2 2πμ(ω1 − ω2 ) 3 dT = rdF = μ 1 (2πr )dr = r dr h h Integrating,

T=

2πμ(ω1 − ω2 ) h

∫

D/2

r =0

ω 2r ω1r

D/2

r 3dr =

2πμ(ω1 − ω2 ) r 4 πμ(ω1 − ω2 ) D4 = h 4 r =0 32h

Noting that ω = 2π n , the relative angular speed is

⎛1 min ⎞⎟ ω1 − ω2 = 2π(n 1 − n 2 ) = (2π rad/s)[(1200 −1125) rev/min]⎜⎜⎜ ⎟ = 7.854 rad/s ⎝ 60 s ⎠⎟ Substituting, the torque transmitted is determined to be T=

π(0.38 N ⋅ s/m 2 )(7.854 /s)(0.30 m)4 = 1.19 N ⋅ m 32(0.002 m)

Discussion Note that the torque transmitted is proportional to the fourth power of disk diameter, and is inversely

proportional to the thickness of the oil film.

2-49 . ..

Chapter 2 Properties of Fluids 2-80 Solution

We are to investigate the effect of oil film thickness on the transmitted torque.

Analysis

The previous problem is reconsidered. Using EES software, the effect of oil film thickness on the

torque transmitted is investigated. Film thickness varied from 0.1 mm to 10 mm, and the results are tabulated and plotted. The relation used is T =

πμ(ω1 − ω2 ) D 4 . The EES Equations window is printed below, followed by the 32h

tabulated and plotted results. mu=0.38 [N-s/m^2] n1=1200 [rpm] n2=1125 [rpm] D=0.3 [m] h=0.002 [m] w1=2*pi*n1/60 w2=2*pi*n2/60 Tq=pi*mu*(w1-w2)*(D^4)/(32*h)

Film

Torque

thickness

transmitted

h, mm

T, N⋅m

0.1 0.2

23.73 11.87

0.4 0.6

5.933 3.956

0.8 1

2.967 2.373

2 4

1.187 0.5933

6 8

0.3956 0.2967

10

0.2373

Conclusion Torque transmitted is inversely proportional to oil film thickness, and the film thickness should be as small as

possible to maximize the transmitted torque.

Discussion

To obtain the solution in EES, we set up a parametric table, specify h, and let EES calculate T for each

value of h.

2-50 . ..

Chapter 2 Properties of Fluids 2-81 Solution A block is moved at constant velocity on an inclined surface. The force that needs to be applied in the horizontal direction when the block is dry, and the percent reduction in the required force when an oil film is applied on the surface are to be determined.

Assumptions 1 The inclined surface is plane (perfectly flat, although tilted). 2 The friction coefficient and the oil film thickness are uniform. 3 The weight of the oil layer is negligible.

The absolute viscosity of oil is given to be μ = 0.012 Pa⋅s = 0.012 N⋅s/m2.

Properties

Analysis (a) The velocity of the block is constant, and thus its acceleration and the net force acting on it are zero. A free body diagram of the block is given. Then the force balance gives

∑F =0: ∑F =0: x

F1 − Ff cos 20°− FN 1 sin 20° = 0

(1)

y

FN 1 cos 20°− Ff sin 20°− W = 0

(2)

Friction force: Ff = fFN 1

V= 1.10 m/s Ff 200

F1

(3)

FN1 200 W = 150 N

20

Substituting Eq. (3) into Eq. (2) and solving for FN1 gives FN 1 =

y

0

x

W 150 N = = 177.0 N cos 20°− f sin 20° cos 20°− 0.27 sin 20°

Then from Eq. (1): F1 = Ff cos 20° + FN 1 sin 20° = (0.27×177 N) cos 20° + (177 N)sin 20° = 105.5 N

(b) In this case, the friction force is replaced by the shear force applied on the bottom surface of the block due to the oil. Because of the no-slip condition, the oil film sticks to the inclined surface at the bottom and the lower surface of the block at the top. Then the shear force is expressed as V Fshear = τ w As = μ As h 1.10 m/s = (0.012 N ⋅ s/m 2 )(0.5× 0.2 m 2 ) 4 ×10−4 m = 3.3 N

V= 1.10 m/s 50 cm

0.4 mm

F2 0

20

Fshear = τwAs FN2

W = 150 N

Replacing the friction force by the shear force in part (a),

∑F =0: ∑F =0: x

F2 − Fshear cos 20°− FN 2 sin 20° = 0

(4)

y

FN 2 cos 20°− Fshear sin 20°− W = 0

(5)

Eq. (5) gives FN 2 = (Fshear sin 20° + W ) / cos 20° = [(3.3 N)sin 20° + (150 N)] / cos 20° = 159.7 N Substituting into Eq. (4), the required horizontal force is determined to be F2 = Fshear cos 20° + FN 2 sin 20° = (3.3 N) cos 20° + (159.7 N)sin 20° = 57.7 N Then, our final result is expressed as

Percentage reduced in required force = Discussion surface.

F1 − F2 105.5 − 57.7 ×100% = ×100% = 45.3% F2 105.5

Note that the force required to push the block on the inclined surface reduces significantly by oiling the

2-51 . ..

Chapter 2 Properties of Fluids 2-82 For flow over a plate, the variation of velocity with distance is given. A relation for the wall shear stress is

Solution

to be obtained.

Assumptions

The fluid is Newtonian.

Analysis

Noting that u(y) = ay – by2, wall shear stress is determined from its definition to be τw = μ

Discussion

du d (ay − by 2 ) =μ = μ(a − 2by) y=0 = aμ dy dy y=0 y=0

Note that shear stress varies with vertical distance in this case.

2-83 The velocity profile for laminar one-dimensional flow through a circular pipe is given. A relation for

Solution

friction drag force exerted on the pipe and its numerical value for water are to be determined.

Assumptions

1 The flow through the circular pipe is one-dimensional. 2 The fluid is Newtonian.

Properties

The viscosity of water at 20°C is given to be 0.0010 kg/m⋅s.

Analysis

⎛ r2 ⎞ (a) The velocity profile is given by u(r ) = umax ⎜⎜⎜1 − 2 ⎟⎟⎟ ⎜⎝ R ⎠⎟

where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center, r = 0. The shear stress at the pipe surface is expressed as 2μumax −2r du d ⎛ r2 ⎞ = −μumax ⎜⎜⎜1− 2 ⎟⎟⎟ = −μumax 2 = τw = −μ dr r=R dr ⎝⎜ R ⎟⎠r =R R R r =R

u(r) = umax(1-r2/R2)

R r 0 umax

Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the τw relation for pipes to make shear stress in the positive (flow) direction a positive quantity. (Or, du/dr = −du/dy since y = R – r). Then the friction drag force exerted by the fluid on the inner surface of the pipe becomes

FD = τw As =

2μumax (2πRL) = 4πμLumax R

(b) Substituting the values we get

⎛ 1 N ⎞⎟ ⎟ = 1.13 N FD = 4πμ Lumax = 4π(0.0010 kg/m ⋅ s)(30 m)(3 m/s) ⎜⎜⎜ 2⎟ ⎝⎜1 kg ⋅ m/s ⎠⎟ In the entrance region and during turbulent flow, the velocity gradient is greater near the wall, and thus the Discussion drag force in such cases will be greater.

2-52 . ..

Chapter 2 Properties of Fluids 2-84 The velocity profile for laminar one-dimensional flow through a circular pipe is given. A relation for friction drag force exerted on the pipe and its numerical value for water are to be determined.

Solution

Assumptions

1 The flow through the circular pipe is one-dimensional. 2 The fluid is Newtonian.

Properties

The viscosity of water at 20°C is given to be 0.0010 kg/m⋅s.

Analysis

⎛ r2 ⎞ (a) The velocity profile is given by u(r ) = umax ⎜⎜⎜1 − 2 ⎟⎟⎟ ⎜⎝ R ⎠⎟

where R is the radius of the pipe, r is the radial distance from the center of

u(r) = umax(1-r2/R2)

R r

the pipe, and umax is the maximum flow velocity, which occurs at the center, r = 0. The shear stress at the pipe surface can be expressed as

τw = −μ

2μumax −2r du d ⎛ r2 ⎞ = −μumax ⎜⎜⎜1− 2 ⎟⎟⎟ = −μumax 2 = dr r=R dr ⎜⎝ R ⎟⎠r =R R R r =R

0 umax

Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the τw relation for pipes to make shear stress in the positive (flow) direction a positive quantity. (Or, du/dr = −du/dy since y = R – r). Then the friction drag force exerted by the fluid on the inner surface of the pipe becomes

FD = τw As =

2μumax (2πRL) = 4πμLumax R

(b) Substituting, we get

⎛ 1 N ⎞⎟ ⎟ = 2.26 N FD = 4πμ Lumax = 4π(0.0010 kg/m ⋅ s)(30 m)(6 m/s) ⎜⎜⎜ 2⎟ ⎝⎜1 kg ⋅ m/s ⎠⎟ Discussion

In the entrance region and during turbulent flow, the velocity gradient is greater near the wall, and thus the

drag force in such cases will be larger.

2-53 . ..

Chapter 2 Properties of Fluids 2-85 Solution A frustum shaped body is rotating at a constant angular speed in an oil container. The power required to maintain this motion and the reduction in the required power input when the oil temperature rises are to be determined.

Assumptions The thickness of the oil layer remains constant.

Properties

Case

The absolute viscosity of oil is given to be

μ = 0.1 Pa⋅s = 0.1 N⋅s/m2 at 20°C and 0.0078 Pa⋅s at 80°C.

D = 12

Analysis The velocity gradient anywhere in the oil of film thickness h is V/h where V = ωr is the tangential velocity.

Then the wall shear stress anywhere on the surface of the frustum at a distance r from the axis of rotation is du V ωr τw = μ =μ =μ dr h h The shear force acting on differential area dA on the surface, the torque it generates, and the shaft power associated with it

L = 12 d = 4 cm z

SAE 10W oil of film thickness h

r

are expressed as dF = τ w dA = μ

μω h

ωr dA h

dT = rdF = μ

ωr 2 dA h

μω W sh = ω T = r 2 dA h A A Top surface: For the top surface, dA = 2πrdr. Substituting and integrating,

T=

∫

2

∫

∫

D/2

r 2 dA

μω W sh, top = h

2

∫

D/2

r =0

2πμω 2 r (2πr )dr = h 2

r =0

D/2

2πμω 2 r 4 πμω 2 D4 r dr = = h 4 r =0 32h 3

πμω 2 d 4 32h Side surface: The differential area for the side surface can be expressed as dA = 2πrdz. From geometric considerations, Bottom surface: A relation for the bottom surface is obtained by replacing D by d, W sh, bottom =

the variation of radius with axial distance is expressed as r = Differentiating gives dr =

μω W sh, top = h

2

∫

D/2

r =0

d D−d + z.

2 2L

D−d 2L 4π L dz or dz = dr . Therefore, dA = 2πdz = rdr. Substituting and integrating, 2L D−d D−d r2

4π L 4πμω 2 L rdr = D−d h( D − d )

∫

D/2

D/2

r =d / 2

r 3 dr =

4πμω 2 L r 4 πμω 2 L( D2 − d 2 ) = h( D − d ) 4 r = d / 2 16h( D − d )

Then the total power required becomes

πμω 2 D 4 ⎡⎢ 2 L[1 − (d / D)4 )] ⎤⎥ 4 W sh, total = W sh, top + W sh, bottom + W sh, side = 1 ( d / D ) + + ⎥, 32h ⎢⎣ D−d ⎦ where d/D = 4/12 = 1/3. Substituting, π(0.1 N ⋅ s/m )(200/s) (0.12 m) W sh, total = 32(0.0012 m) 2

2

4 ⎡

4 ⎤ ⎢1 + (1/ 3)4 + 2(0.12 m)[1− (1/ 3) )] ⎥ ⎛⎜ 1 W ⎟⎟⎞ = 270 W ⎢ (0.12 − 0.04) m ⎦⎥ ⎜⎝1 Nm/s ⎟⎠ ⎣

Noting that power is proportional to viscosity, the power required at 80°C is μ 0.0078 N ⋅ s/m W sh, total, 80 °C = 80°C W sh, total, 20°C = (270 W) = 21.1 W μ20°C 0.1 N ⋅ s/m 2 2

2-54 . ..

Chapter 2 Properties of Fluids Therefore, the reduction in the required power input at 80°C is Reduction = W sh, total, 20 ° C − W sh, total, 80 ° C = 270 − 21.1 = = 249 W , which is about 92%.

Discussion

Note that the power required to overcome shear forces in a viscous fluid greatly depends on temperature.

-

2-86 We are to determine the torque required to rotate the outer cylinder of two concentric cylinders, with the outer cylinder rotating and the inner cylinder stationary.

Solution

Assumptions

1 The fluid is incompressible and Newtonian. 2 End effects (top and bottom) are negligible. 3 The gap is

very small so that wall curvature effects are negligible. 4 The gap is so small that the velocity profile in the gap is linear.

Inner cylinder h

y

u

V Outer cylinder Analysis

We assume a linear velocity profile between the two walls – the outer wall is moving at speed V = ωoRo and

the inner wall is stationary. The thickness of the gap is h, and we let y be the distance from the outer wall into the fluid (towards the inner wall) as sketched. Thus, u=V

h−y du V and τ = μ = −μ h dy h

where

h = Ro - Ri and V = ωo Ro Since shear stress τ has dimensions of force/area, the clockwise (mathematically negative) tangential force acting along the surface of the outer cylinder by the fluid is

F = −τ A = −μ

μωo Ro V 2π Ro L = − 2π Ro L h Ro − Ri

But the torque is the tangential force times the moment arm Ro. Also, we are asked for the torque required to turn the inner cylinder. This applied torque is counterclockwise (mathematically positive). Thus, T = −FRo =

2π L μωo Ro3 2π L μωo Ro3 = Ro − Ri h

Discussion The above is only an approximation because we assumed a linear velocity profile. As long as the gap is very small, and therefore the wall curvature effects are negligible, this approximation should be very good. It is possible to solve for the exact velocity profile for this problem, and therefore the torque can be found analytically, but this has to wait until the differential analysis chapter.

2-55 . ..

Chapter 2 Properties of Fluids 2-87 A thin flat plate is pulled horizontally through the mid plane of an oil layer sandwiched between two stationary plates. The force that needs to be applied on the plate to maintain this motion is to be determined for this case

Solution

and for the case when the plate.

Assumptions

1 The thickness of the plate is negligible. 2 The velocity profile in each oil layer is linear.

Properties

The absolute viscosity of oil is given to be μ = 0.9 N⋅s/m2.

Analysis

The velocity profile in each oil layer relative to the fixed wall is as shown in the figure.

Stationary surface

h1=2 cm

V = 5 m/s

F

h2=2 cm y Stationary surface

The magnitudes of shear forces acting on the upper and lower surfaces of the moving thin plate are Fshear, upper = τ w, upper As = μ As

Fshear, lower = τ w, lower As = μ As

du V −0 5 m/s = μ As = (0.9 N ⋅ s/m 2 )(0.5× 2 m 2 ) = 225 N dy h1 0.02 m

V − Vw 5 m/s du = μ As = (0.9 N ⋅ s/m 2 )(0.5× 2 m 2 ) = 225 N 0.02 m dy h2

Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be F = Fshear, upper + Fshear, lower = 225 + 225 = 450 N When the plate is 1 cm from the bottom surface and 3 cm from the top surface, the force F becomes Fshear, upper = τ w, upper As = μ As

du V −0 5 m/s = μ As = (0.9 N ⋅ s/m 2 )(0.5× 2 m 2 ) = 150 N dy h1 0.03 m

Fshear, lower = τ w, lower As = μ As

du V −0 5 m/s = μ As = (0.9 N ⋅ s/m 2 )(0.5× 2 m 2 ) = 450 N dy h2 0.01 m

Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be F = Fshear, upper + Fshear, lower = 150 + 450 = 600 N Discussion

Note that the relative location of the thin plate affects the required force significantly.

2-56 . ..

Chapter 2 Properties of Fluids 2-88 A thin flat plate is pulled horizontally through the mid plane of an oil layer sandwiched between two stationary plates. The force that needs to be applied on the plate to maintain this motion is to be determined for this case

Solution

and for the case when the plate .

Assumptions

1 The thickness of the plate is negligible. 2 The velocity profile in each oil layer is linear.

Properties

The absolute viscosity of oil is μ = 0.9 N⋅s/m2 in the lower part, and 4 times that in the upper part.

Analysis We measure vertical distance y from the lower plate. The total distance between the stationary plates is h = h1 + h2 = 4 cm , which is constant. Then the distance of the moving plate is y from the lower plate and h – y from the upper plate, where y is variable.

Stationary surface

h1=h - y

V = 5 m/s

F

h2=y y Stationary surface The shear forces acting on the upper and lower surfaces of the moving thin plate are Fshear, upper = τ w, upper As = μupper As

du V = μupper As dy h−y

Fshear, lower = τ w, lower As = μlower As

du V = μlower As dy y

Then the total shear force acting on the plate becomes ⎛ μupper μlower ⎞⎟ V V ⎟ + μlower As = AsV ⎜⎜⎜ + h−y h−y y ⎠⎟⎟ ⎝ h−y dF The value of y that will minimize the force F is determined by setting =0: dy μupper μ μlower y − lower =0 → = μupper h−y (h − y)2 y2 F = Fshear, upper + Fshear, lower == μupper As

Solving for y and substituting, the value of y that minimizes the shear force is determined to be y= Discussion

μlower / μupper

1 − μlower / μupper

By showing that

h=

1/ 4 (4 cm) = 1 cm 1 − 1/ 4

d2F > 0 at y = 1 cm, it can be verified that F is indeed a minimum at that location and dy2

not a maximum.

2-57 . ..

Chapter 2 Properties of Fluids 2-89 Solution A cylinder slides down from rest in a vertical tube whose inner surface is covered by oil. An expression for the velocity of the cylinder as a function of time is to be derived.

Assumptions

1 Velocity profile in the oil film is linear.

D

L

Cylinder

Oil film, h

Analysis Assuming a linear velocity profile in the oil film the drag force due to wall shear stress can be expressed as FD = μ

dV V A = μ π DL = kV dy h

where V is the instantaneous velocity of the cylinder and k=μ

π DL h

Applying Newton’s second law of motion for the cylinder, we write

mg − kV = m

dV dt

where t is the time. This is a first-order linear equation and can be expressed in standard form as follows:

dV k + V =g dt m

with V (0) = 0

whose solution is obtained to be V (t ) =

mg ( ( ) 1 − e− k/m t ) k

As t → ∞ the second term will vanish leaving us with V (t ) =

mg k

which is constant. This constant is referred to as “limit velocity VL″ . Rearranging for viscosity, we have mgh μ= π DLVL 2-58 . ..

Chapter 2 Properties of Fluids Therefore this equation enables us to estimate dynamic viscosity of oil provided that the limit velocity of the cylinder is precisely measured.

2-59 . ..

Chapter 2 Properties of Fluids

Surface Tension and Capillary Effect

2-90C Solution

We are to define and discuss surface tension.

Analysis

The magnitude of the pulling force at the surface of a liquid per unit length is called surface tension σs.

It is caused by the attractive forces between the molecules. The surface tension is also surface energy (per unit area) since it represents the stretching work that needs to be done to increase the surface area of the liquid by a unit amount.

Discussion Surface tension is the cause of some very interesting phenomena such as capillary rise and insects that can walk on water.

2-91C Solution

We are to define and discuss the capillary effect.

Analysis

The capillary effect is the rise or fall of a liquid in a small-diameter tube inserted into the liquid. It is

caused by the net effect of the cohesive forces (the forces between like molecules, like water) and adhesive forces (the forces between unlike molecules, like water and glass). The capillary effect is proportional to the cosine of the contact angle, which is the angle that the tangent to the liquid surface makes with the solid surface at the point of contact.

Discussion

The contact angle determines whether the meniscus at the top of the column is concave or convex.

2-92C Solution

We are to determine whether the level of liquid in a tube will rise or fall due to the capillary effect.

Analysis

The liquid level in the tube will drop since the contact angle is greater than 90°, and cos(110°) < 0.

Discussion

This liquid must be a non-wetting liquid when in contact with the tube material. Mercury is an example of a

non-wetting liquid with a contact angle (with glass) that is greater than 90o.

2-60 . ..

Chapter 2 Properties of Fluids 2-93C Solution

We are to analyze the pressure difference between inside and outside of a soap bubble.

Analysis

The pressure inside a soap bubble is greater than the pressure outside, as evidenced by the stretch of

the soap film.

Discussion

You can make an analogy between the soap film and the skin of a balloon.

2-94C Solution

We are to compare the capillary rise in small and large diameter tubes.

Analysis

The capillary rise is inversely proportional to the diameter of the tube, and thus capillary rise is greater in

the smaller-diameter tube.

Discussion Note however, that if the tube diameter is large enough, there is no capillary rise (or fall) at all. Rather, the upward (or downward) rise of the liquid occurs only near the tube walls; the elevation of the middle portion of the liquid in the tube does not change for large diameter tubes.

2-95 Solution

The diameter of a soap bubble is given. The gage pressure inside the bubble is to be determined.

Assumptions

The soap bubble is in atmospheric air.

Properties

The surface tension of soap water at 20°C is σs = 0.025 N/m.

The pressure difference between the inside and the outside of a Analysis bubble is given by

4σ ΔPbubble = Pi − P0 = s R

P0

Soap bubble P

In the open atmosphere P0 = Patm, and thus ΔPbubble is equivalent to the gage pressure. Substituting,

4(0.025 N/m) = 100 N/m 2 = 100 Pa (0.00200/2) m 4(0.025 N/m) = 4 N/m 2 = 4 Pa D = 5.00 cm: Pi, gage = ΔPbubble = (0.0500/2) m D = 0.200 cm: Pi, gage = ΔPbubble =

Discussion Note that the gage pressure in a soap bubble is inversely proportional to the radius (or diameter). Therefore, the excess pressure is larger in smaller bubbles.

2-61 . ..

Chapter 2 Properties of Fluids 2-96E Solution

A soap bubble is enlarged by blowing air into it. The required work input is to be determined.

Properties

The surface tension of solution is given to be σs = 0.0027 lbf/ft.

Analysis The work associated with the stretching of a film is the surface tension work, and is expressed in differential form as δWs = σs dAs . Noting that surface tension is constant, the surface tension work is simply surface tension multiplied by the change in surface area,

Air

Ws = σ s ( A2 − A1 ) = 2 πσ s ( D22 − D12 )

The factor 2 is due to having two surfaces in contact with air. Substituting, the required work input is determined to be

Soap bubble P

⎛ ⎞⎟ 1 Btu -7 Ws = 2π(0.0027 lbf/ft) ((2.7 /12 ft)2 − (2.4 /12 ft)2 )⎜⎜ ⎟ = 2.32 ×10 Btu ⎝ 778.169 lbf ⋅ ft ⎠⎟ Discussion Note that when a bubble explodes, an equivalent amount of energy is released to the environment.

2-97 A glass tube is inserted into a liquid, and the capillary rise is

Solution

measured. The surface tension of the liquid is to be determined.

Assumptions

1 There are no impurities in the liquid, and no contamination on

the surfaces of the glass tube. 2 The liquid is open to the atmospheric air.

Air

h

3

Properties

The density of the liquid is given to be 960 kg/m . The contact Liquid

angle is given to be 15°.

Analysis

Substituting the numerical values, the surface tension is

determined from the capillary rise relation to be ρ gRh σs = 2 cos φ =

(960 kg/m 3 )(9.81 m/s2 )(0.0016 / 2 m)(0.005 m) ⎛⎜ 1N ⎟⎞ ⎟ ⎜⎜ 2⎟ 2( cos15°) ⎝⎜ 1 kg ⋅ m/s ⎠⎟

= 0.0195 N/m

Discussion

Since surface tension depends on temperature, the value determined is valid at the liquid’s temperature.

2-62 . ..

Chapter 2 Properties of Fluids 2-98 An air bubble in a liquid is considered. The pressure difference between the inside and outside the bubble is to be determined.

Solution

Properties

The surface tension σs is given for two cases to be 0.08 and 0.12 N/m.

Analysis

Considering that an air bubble in a liquid has only one interface, the pressure difference between the inside

and the outside of the bubble is determined from

ΔPbubble = Pi − P0 =

2σs R

Liquid

Substituting, the pressure difference is determined to be: 2(0.08 N/m) ΔPbubble = = 2133 N/m2 = 2.13 kPa (a) σs = 0.08 N/m: 0.00015/2 m 2(0.12 N/m) ΔPbubble = = 3200 N/m2 = 3.20 kPa (b) σs = 0.12 N/m: 0.00015/2 m Note that a small gas bubble in a liquid is highly pressurized.

Discussion

Air bubble P

The smaller the bubble diameter, the larger the pressure inside the bubble.

2-99 The force acting on the movable wire of a liquid film suspended on a U-shaped wire frame is measured. The surface tension of the liquid in the air is to be determined.

Solution

Assumptions

1 There are no impurities in the liquid, and no contamination on the surfaces of the wire frame. 2 The liquid

is open to the atmospheric air.

Analysis

Substituting the numerical values, the surface tension is determined from the surface tension force relation

to be σs =

Discussion

F 0.030 N = = 0.19 N/m 2b 2(0.08 m)

b

The surface tension depends on temperature. Therefore, the value

determined is valid at the temperature of the liquid.

2-63 . ..

Liquid film

F

Chapter 2 Properties of Fluids 2-100 Solution

A capillary tube is immersed vertically in water. The height of water rise in the tube is to be determined.

Assumptions

1 There are no impurities in water, and no contamination on the surfaces of the tube.. 2 Water is open to the

atmospheric air.

Analysis

The capillary rise is determined from Eq. 2-38 to be h=

2σ S 2 ×(1 N/m )× cos 6 ο cos ∅ = = 0.338 m ρ gR (1000 kg/m 3 )×(9.81 m/s2 )×(0.6 ×10−3 m )

2-101E Solution

A glass tube is inserted into mercury. The capillary drop of mercury in the tube is to be determined.

Assumptions

1 There are no impurities in mercury, and no contamination on the surfaces of the glass tube. 2 The mercury

is open to the atmospheric air.

Properties

The surface tension of mercury-glass in atmospheric air at 68°F (20°C) is obtained from Table 2-4:

σs = (0.440 N/m)(0.22482 lbf/N)(0.3048 m/ft) = 0.030151 lbf/ft. To obtain the density of mercury, we interpolate from

Table A-8E at 68°F, yielding ρ = 845.65 lbm/ft3. The contact angle is given to be 140°.

Analysis Substituting the numerical values, the capillary drop is determined to be

h=

⎛ 32.174 lbm ⋅ ft/s2 ⎞⎟ 2σ s cos φ 2(0.030151 lbf/ft)( cos140°) ⎜⎜ ⎟⎟ = ⎟⎠ 1 lbf ρ gR (845.65 lbm/ft 3 )(32.174 ft/s2 )(0.009 /12) ft ⎜⎜⎝

= −0.072834 ft = −0.87401 in

Thus, the capillary drop is 0.874 in to three significant digits.

Discussion The negative sign indicates capillary drop instead of rise. The drop is very small in this case because of the large diameter of the tube.

2-64 . ..

Air Mercury

h

Chapter 2 Properties of Fluids 2-102 A capillary tube is immersed vertically in water. The maximum capillary rise and tube diameter for the maximum rise case are to be determined.

Solution

Assumptions

1 There are no impurities in water, and no contamination on the surfaces of the tube. 2 Water is open to the

atmospheric air.

Properties

The surface tension is given to be σs = 1 N/m.

Analysis

At the liquid side of the meniscus P = 2 kPa. Therefore the capillary rise would be

h=

(101325 − 2000)×103 Pa Patm − P = = 10.12 m ρg (1000 kg/m3 )×(9.81 m/s2 )

Then the tube diameter needed for this capillary rise is R=

2σ S 2 ×(1 N/m )× cos 6 ο cos ∅ = ≅ 2.0 ×10−5 m = 20 μm ρ gh (1000 kg/m 3 )×(9.81 m/s2 )×(10.12 m )

2-65 . ..

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Chapter 2 Properties of Fluids 2-103 A steel ball floats on water due to the surface tension effect. The maximum diameter of the ball is to be determined, and the calculations are to be repeated for aluminum.

Solution

Assumptions

1 The water is pure, and its temperature is constant. 2 The ball is dropped on water slowly so that the

inertial effects are negligible. 3 The contact angle is taken to be 0° for maximum diameter.

Properties

The surface tension of water at 10°C is σs = 0.0745 N/m (Table 2-4 by interpolation). The contact angle is

taken to be 0°. The densities of steel and aluminum are given to be ρsteel = 7800 kg/m3 and ρAl = 2700 kg/m3.

σ

W = mg

Analysis

The surface tension force and the weight of the ball can be expressed as

Fs = π Dσs and W = mg = ρ gV = ρ gπ D 3 / 6 When the ball floats, the net force acting on the ball in the vertical direction is zero. Therefore, setting Fs = W and solving for diameter D gives D =

6σ s . Substititing the known quantities, the maximum diameters for the steel and aluminum ρg

balls become

Dsteel =

6σs = ρg

⎛ 1 kg ⋅ m/s2 ⎞⎟ 6(0.0745 N/m) ⎜ ⎟⎟ = 2.42×10−3 m = 2.42 mm 3 2 ⎜ ⎟⎠ 1N (7800 kg/m )(9.81 m/s ) ⎜⎜⎝

DAl =

6σs = ρg

⎛ 1 kg ⋅ m/s2 ⎞⎟ 6(0.0745 N/m) ⎜ ⎟⎟ = 4.11×10−3 m = 4.11 mm 3 2 ⎜ ⎜ ⎟⎠ ⎜ 1N (2700 kg/m )(9.81 m/s ) ⎝

Discussion Note that the ball diameter is inversely proportional to the square root of density, and thus for a given material, the smaller balls are more likely to float.

2-66 . ..

Chapter 2 Properties of Fluids 2-104 Nutrients dissolved in water are carried to upper parts of plants.

The height to which the water solution rises in a tree as a result of the capillary

Solution

effect is to be determined.

Assumptions

1 The solution can be treated as water with a contact angle of

15°. 2 The diameter of the tube is constant. 3 The temperature of the water solution is 20°C. Properties

The surface tension of water at 20°C is σs = 0.073 N/m. The

density of water solution can be taken to be 1000 kg/m3. The contact angle is given to be 15°.

Analysis Substituting the numerical values, the capillary rise is determined to be

h=

⎛1 kg ⋅ m/s2 ⎞⎟ 2σs cos φ 2(0.073 N/m)( cos15°) = ⎜⎜ ⎟⎟ = 11.1 m ρ gR (1000 kg/m3 )(9.81 m/s2 )(1.3×10−6 m) ⎜⎝⎜ 1 N ⎟⎠

Other effects such as the chemical potential difference also cause Discussion the fluid to rise in trees.

2-67 . ..

Chapter 2 Properties of Fluids

Review Problems

2-105 A journal bearing is lubricated with oil whose viscosity is known. The torques needed to overcome the bearing friction during start-up and steady operation are to be determined.

Solution

Assumptions

1 The gap is uniform, and is completely filled with oil. 2 The end effects on the sides of the bearing are

negligible. 3 The fluid is Newtonian.

Properties

The viscosity of oil is given to be 0.1 kg/m⋅s at 20°C, and 0.008 kg/m⋅s at 80°C.

R

l = 0.08 cm fluid

Analysis

The radius of the shaft is R = 0.04 m. Substituting the given values, the torque is determined to be

At start up at 20°C:

T=μ

4π 2 R3 nL 4π 2 (0.04 m)3 (1500 / 60 s-1 )(0.55 m) = (0.1 kg/m ⋅ s) = 4.34 N ⋅ m 0.0008 m A

During steady operation at 80°C:

T=μ Discussion

4π 2 R3 nL 4π 2 (0.04 m)3 (1500 / 60 s-1 )(0.55 m) = (0.008 kg/m ⋅ s) = 0.347 N ⋅ m 0.0008 m A Note that the torque needed to overcome friction reduces considerably due to the decrease in the viscosity

of oil at higher temperature.

2-68 . ..

Chapter 2 Properties of Fluids 2-106 A U-tube with a large diameter arm contains water. The difference between the water levels of the two arms is to be determined.

Solution

Assumptions

1 Both arms of the U-tube are open to the atmosphere. 2 Water is at room temperature. 3 The contact angle

of water is zero, φ = 0.

The surface tension and density of water at 20°C are σs = 0.073 N/m and ρ = 1000 kg/m3.

Properties

h

Analysis Any difference in water levels between the two arms is due to surface tension effects and thus capillary rise.

Noting that capillary rise in a tube is inversely proportional to tube diameter there will be no capillary rise in the arm with a large diameter. Then the water level difference between the two arms is simply the capillary rise in the smaller diameter arm,

h=

⎛1 kg ⋅ m/s2 ⎟⎞⎛1000 mm ⎞ 2σs cos φ 2(0.073 N/m)( cos 0°) ⎜⎜ ⎟⎟ = 5.95 mm ⎟⎟⎜⎜ = 3 2 ρ gR (1000 kg/m )(9.81 m/s )(0.0025 m) ⎜⎜⎝ 1 N ⎟⎠⎝⎜ 1 m ⎠⎟

Discussion

Note that this is a significant difference, and shows the importance of using a U-tube made of a uniform

diameter tube.

2-107E Solution A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined.

Assumptions

1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant.

Properties

The gas constant of air is Ru = 53.34

ft ⋅ lbf ⎛⎜ 1 psia ⎟⎞ psia ⋅ ft 3 0.3794 = . The air temperature is 70oF = ⎟ ⎜ lbm ⋅ R ⎜⎝144 lbf/ft 2 ⎟⎠ lbm ⋅ R

70 + 459.67 = 529.67 R Analysis

Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be

V=

m1 RT1 (40 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(529.67 R) = = 392.380 ft 3 P1 20 psia

m2 =

P2V (35 psia)(392.380 ft 3 ) = = 67.413 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(550 R)

Thus the amount of air added is

Air, 40 lbm 20 psia 70°F

Δ m = m2 − m1 = 67.413 − 40.0 = 27.413 lbm ≅ 27.4 lbm

Discussion

As the temperature slowly decreases due to heat transfer, the pressure will also decrease.

2-69 . ..

Chapter 2 Properties of Fluids 2-108 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be Solution

determined.

Assumptions

The tank is insulated so that no heat is transferred.

Analysis

Treating N2 as an ideal gas, the initial and the final masses in the

tank are determined to be m1 =

P1V (800 kPa)(10 m3 ) = = 90.45 kg RT1 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(298 K)

m2 =

P2V (600 kPa)(10 m3 ) = = 69.00 kg RT2 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(293 K)

N2 800 kPa 25°C 10 m3

Thus the amount of N2 that escaped is Δ m = m1 − m2 = 90.45 − 69.00 = 21.5 kg Discussion

Gas expansion generally causes the temperature to drop. This principle is used in some types of

refrigeration.

2-109 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent

Solution

increase in the absolute temperature of the air in the tire is to be determined.

Assumptions

1 The volume of the tire remains constant. 2 Air is an ideal gas.

Analysis

Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before

the trip are

P1V1 P2V2 = T1 T2

→

T2 P 335 kPa = 2= =1.047 T1 P1 320 kPa

Therefore, the absolute temperature of air in the tire will increase by 4.7% during this trip.

Discussion This may not seem like a large temperature increase, but if the tire is originally at 20oC (293.15 K), the temperature increases to 1.047(293.15 K) = 306.92 K or about 33.8oC.

2-70 . ..

Chapter 2 Properties of Fluids 2-110E Solution

The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.

Properties

The vapor pressure of water at 60°F is 0.2563 psia.

Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is Pv = Psat@60°F = 0.2563 psia The minimum pressure in the pump is 0.1 psia, which is less than the vapor pressure. Therefore, there is danger of cavitation in the pump.

Discussion

Note that the vapor pressure increases with increasing temperature, and the danger of cavitation increases at

higher fluid temperatures.

2-111 Air in a partially filled closed water tank is evacuated. The absolute pressure in the evacuated space is to be

Solution

determined.

Properties

The saturation pressure of water at 70°C is 31.19 kPa.

Analysis

When air is completely evacuated, the vacated space is filled with water vapor, and the tank contains a

saturated water-vapor mixture at the given pressure. Since we have a two-phase mixture of a pure substance at a specified temperature, the vapor pressure must be the saturation pressure at this temperature. That is,

Pv = Psat@70°C = 31.19 kPa ≅ 31.2 kPa Discussion

If there is any air left in the container, the vapor pressure will be less. In that case the sum of the component

pressures of vapor and air would equal 31.19 kPa.

2-71 . ..

Chapter 2 Properties of Fluids 2-112 Solution The specific gravities of solid particles and carrier fluids of a slurry are given. The relation for the specific gravity of the slurry is to be obtained in terms of the mass fraction Cs, mass and the specific gravity SGs of solid particles.

Assumptions 1 The solid particles are distributed uniformly in water so that the solution is homogeneous. 2 The effect of dissimilar molecules on each other is negligible.

Consider solid particles of mass ms and volume Vs dissolved in a fluid of mass mf and volume Vm . The total

Analysis

volume of the suspension (or mixture) is Vm = Vs +V f .

Dividing by Vm gives 1=

Vs V f + → Vm Vm

Vf m /ρ m ρ SG m V = 1 − s = 1 − s s = 1 − s m = 1 − Cs,mass mm / ρm mm ρ s SG s Vm Vm

(1)

since ratio of densities is equal two the ratio of specific gravities, and ms / mm = Cs, mass . The total mass of the suspension (or mixture) is mm = ms + m f . Dividing by mm and using the definition Cs, mass = ms / mm give 1 = Cs,mass +

mf mm

= Cs,mass +

ρ fV f ρmVm

→

Vf ρm = ρf (1 − Cs,mass )Vm

(2)

Taking the fluid to be water so that ρm / ρ f = SG m and combining equations 1 and 2 give SG m =

1 − Cs,mass SG m / SG s 1 − Cs,mass

Solving for SGm and rearranging gives SG m =

1 1 + Cs, mass (1 SG s −1)

which is the desired result.

Discussion

As a quick check, if there were no particles at all, SGm = 0, as expected.

2-113 Solution A rigid tank contains an ideal gas at a specified state. The final temperature when half the mass is withdrawn and final pressure when no mass is withdrawn are to be determined.

Analysis

(a) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final

temperature may be determined from the ideal gas relation as T2 =

⎛ 100 kPa ⎟⎞ m1 P2 T1 = (2)⎜⎜⎜ ⎟ (600 K) = 400 K ⎝ 300 kPa ⎟⎠ m2 P1

Ideal gas 300 kPa

(b) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields P2 = Discussion

⎛ 400 K ⎞⎟ T2 P1 = ⎜⎜ ⎟ (300 kPa) = 200 kPa ⎝ T1 600 K ⎠⎟ Note that some forms of the ideal gas equation are more convenient to use than the other forms.

2-72

. ..

600 K

Chapter 2 Properties of Fluids 2-114 Suspended solid particles in water are considered. A relation is to be developed for the specific gravity of the suspension in terms of the mass fraction Cs, mass and volume fraction C s , vol of the particles.

Solution

Assumptions

1 The solid particles are distributed uniformly in water so that the solution is homogeneous. 2 The effect of

dissimilar molecules on each other is negligible.

Consider solid particles of mass ms and volume V s dissolved in a fluid of mass mf and volume Vm . The total

Analysis

volume of the suspension (or mixture) is

Vm = Vs +V f Dividing by Vm and using the definition Cs, vol = Vs /Vm give 1 = Cs,vol +

Vf Vm

→

Vf = 1 − Cs,vol Vm

(1)

The total mass of the suspension (or mixture) is

mm = ms + m f Dividing by mm and using the definition Cs, mass = ms / mm give 1 = Cs,mass +

mf mm

= Cs,mass +

ρ fV f ρmVm

→

ρf ρm

V Vf

= (1 − Cs,mass ) m

(2)

Combining equations 1 and 2 gives ρf ρm

=

1 − Cs,mass 1 − Cs,vol

When the fluid is water, the ratio ρ f / ρm is the inverse of the definition of specific gravity. Therefore, the desired relation for the specific gravity of the mixture is SG m =

1 − Cs,vol ρm = ρf 1 − Cs,mass

which is the desired result.

Discussion

As a quick check, if there were no particles at all, SGm = 0, as expected.

2-73 . ..

Chapter 2 Properties of Fluids 2-115 The variation of the dynamic viscosity of water with absolute temperature is given. Using tabular data, a relation is to be obtained for viscosity as a 4th-order polynomial. The result is to be compared to Andrade’s equation in the

Solution

form of μ = D ⋅ e B / T .

Properties

The viscosity data are given in tabular form as T (K)

μ (Pa⋅s)

273.15

1.787×10-3

278.15

1.519×10-3

283.15

1.307×10-3

293.15

1.002×10-3

303.15

7.975×10-4

313.15

6.529×10-4

333.15

4.665×10-4

353.15

3.547×10-4

373.15

2.828×10-4

Analysis

Using EES, (1) Define a trivial

function “a=mu+T” in the equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify polynomial and enter/edit equation.

The equations and plot are shown here. μ = 0.489291758 − 0.00568904387T + 0.0000249152104T2 − 4.86155745×10–8T3 + 3.56198079×10–11T4 μ = 0.000001475*EXP(1926.5/T) [used initial guess of a0=1.8×10–6 and a1=1800 in mu=a0*exp(a1/T)]

At T = 323.15 K, the polynomial and exponential curve fits give Polynomial: μ(323.15 K) = 0.0005529 Pa⋅s

(1.1% error, relative to 0.0005468 Pa⋅s)

Exponential: μ(323.15 K) = 0.0005726 Pa⋅s

(4.7% error, relative to 0.0005468 Pa⋅s)

Discussion

This problem can also be solved using an Excel worksheet, with the following results:

Polynomial:

A = 0.4893, B = −0.005689, C = 0.00002492, D = −0.000000048612, and E = 0.00000000003562

Andrade’s equation:

1864.06 T

μ = 1.807952 E − 6 * e

2-74 . ..

Chapter 2 Properties of Fluids 2-116 Solution

A newly produced pipe is tested using pressurized water. The additional water that needs to be pumped to

reach a specified pressure is to be determined.

Assumptions

1 There is no deformation in the pipe.

Properties

The coefficient of compressibility is given to be 2.10 × 109 Pa.

Analysis

From Eq. 2-13, we have κ≅

ρ − ρ1 ΔP ΔP Δρ ΔP → = = or 2 Δρ ρ κ ρ1 κ ρ

from which we write ⎛ ⎛ ΔP ⎞⎟ 10 ×106 Pa ⎞⎟ 3 ⎟ = 1004.76 kg/m 3 ρ2 = ρ1 ⎜⎜⎜1 + ⎟ = (1000 kg/m )×⎜⎜⎜1 + ⎟ ⎝ ⎝ κ ⎠ 2.10 ×109 Pa ⎟⎠

Then the amount of additional water is m = Vcyl Δρ =

⎛ ⎞ π D2 π (3 m )2 kg LΔρ = ×(15)×⎜⎜⎜1004.76 3 −1000 kg/m 3 ⎟⎟⎟ = 505 kg ⎝ ⎠ 4 4 m

2-117 Solution

We are to prove that the coefficient of volume expansion for an ideal gas is equal to 1/T.

Assumptions

1 Temperature and pressure are in the range that the gas can be approximated as an ideal gas.

Analysis

The ideal gas law is P = ρ RT , which we re-write as ρ =

1 ⎛ ∂ρ ⎞ P . By definition, β = − ⎜⎜⎜ ⎟⎟⎟ . Thus, ρ ⎝ ∂T ⎠P RT

substitution and differentiation yields ⎛ ⎛ P ⎞ ⎞⎟ ⎜∂⎜ ⎟⎟ 1 ⎜⎜⎜ ⎝⎜ RT ⎠⎟⎟ ⎟⎟ 1⎛ P ⎟⎞ ρ 1 ⎟⎟ = − ⎜⎜− βideal gas = − ⎜ = 1/T ⎟= ⎜ ⎜ ⎟ ρ ⎜ ∂T ⎟ ρ ⎝ RT 2 ⎟⎠ ρ T ⎟⎟ ⎜⎜ ⎝ ⎠P

where both pressure and the gas constant R are treated as constants in the differentiation.

Discussion

The coefficient of volume expansion of an ideal gas is not constant, but rather decreases with temperature.

However, for small temperature differences, β is often approximated as a constant with little loss of accuracy.

2-75 . ..

Chapter 2 Properties of Fluids 2-118

The pressure is given at a certain depth of the ocean. An analytical relation between density and pressure is

Solution

to be obtained and the density at a specified pressure is to be determined. The density is to be compared with that from Eq. 2-13.

Properties The coefficient of compressibility is given to be 2350 MPa. The liquid density at the free surface is given to 3 be 1030 kg/m .

Analysis

(a) From the definition, we have

κ=

dP d ρ dP → = d ρ /ρ ρ κ

Integrating ρ dρ

∫ ρ ∫ ρ0

=

0

P dP

κ

→ ln

ρ P = → ρ = ρ0 e P/κ ρ0 κ

With the given data we obtain

ρ = (1030 kg/m3 )×e100/2350 = 1074 kg/m3 (b) Eq. 2-13 can be rearranged to give Δρ ≅ ρ

ΔP κ

or

ρ − ρ0 ≅ ρ0

P − P0 P − P0 kg kg 100 MPa → ρ ≅ ρ0 + ρ0 = 1030 3 +1030 3 × ≈ 1074 kg/m3 κ κ m m 2350 MPa

which is identical with (a). Therefore we conclude that linear approximation (Eq. 2-13) is quite reasonable.

2-119E Solution The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined.

Assumptions

Air is an ideal gas with constant specific heats at room temperature.

Properties The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded.

Analysis

The final temperature of air is determined from the isentropic relation of ideal gases, ⎛ P ⎞( k −1) / k ⎛ 60 ⎞⎟(1.4−1) /1.4 T2 = T1 ⎜⎜⎜ 2 ⎟⎟⎟ = (240 + 460 R) ⎜⎜ = 496.3 R ⎟ ⎝ 200 ⎠⎟ ⎜⎝ P1 ⎠⎟

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as

Ratio = Discussion

k RT T c2 240 + 460 = 1 1 = 1 = = 1.19 c1 T2 k2 RT2 496.3

Note that the speed of sound is proportional to the square root of thermodynamic temperature.

2-76 . ..

Chapter 2 Properties of Fluids 2-120

A shaft is pulled with a constant velocity through a bearing. The space between the shaft and bearing is filled with a fluid. The force required to maintain the axial movement of the shaft is to be determined.

Solution

Assumptions

1 The fluid is Newtonian.

Properties

The viscosity of the fluid is given to be 0.1 Pa⋅s.

Analysis

The varying clearance h can be expressed as a function of axial coordinate x (see figure). According to this sketch we obtain x h = h1 − (h1 − h2 ) L Assuming a linear velocity distribution in the gap, the viscous force acting on the differential strip element is dF = τ dA = μ

U μU π D × π Ddx = x dx h h1 − (h1 − h2 ) L

Integrating F = μU π D

∫

x=L x =0

dx

x = −μU π D h1 − (h1 − h2 ) L

(

ln h1 − (h1 − h2 )

(h1 − h2 ) L

x ⎤ x=L ⎥ μU π DL h1 L ⎥ ln = ⎥ h1 − h2 h2 ⎥ ⎥ ⎦ x =0

)

For the given data, we obtain F=

(0.1 Pa ⋅ s)(5 m/s) π (80 ×10−3 mm )(400 ×10−3 mm ) (1.2 − 0.4 )×10−3 mm

2-77 . ..

ln

1.2 ≅ 69 N 0.4

Chapter 2 Properties of Fluids 2-121

A shaft rotates with a constant angual speed in a bearing. The space between the shaft and bearing is filled with a fluid. The torque required to maintain the motion is to be determined.

Solution

Assumptions

1 The fluid is Newtonian.

Properties

The viscosity of the fluid is given to be 0.1 Pa⋅s.

Analysis

The varying clearance h can be expressed as a function of axial coordinate x (see figure below).

According to this sketch we obtain x L Assuming a linear velocity distribution in the gap, the viscous force acting on the differential strip element is h = h1 − (h1 − h2 )

dF = τ dA = μ

U μU π D × π Ddx = x dx h h1 − (h1 − h2 ) L

where U = 2nπ / 60 in this case. Then the viscous torque developed on the shaft D dT = dF × = 2

⎛ 2 nπ D ⎞⎟ D μ ⎜⎜ × ⎟⎟ π D × ⎝ 60 dx μnπ 2 D 3 2⎠ 2 dx = x x 120 h1 − (h1 − h2 ) h1 − (h1 − h2 ) L L

Integrating T=

μnπ 2 D 120

3 x= L

∫

dx

x =− − − h h h ( ) 1 2 x =0 1 L

μnπ 2 D 120

(

3 ln h1 − ( h1 − h2 )

( h1 − h2 ) L

x ⎤ x= L ⎥ 1 μnπ 2 D3 L h1 L ⎥ ln = ⎥ 120 h1 − h2 h2 ⎥ ⎥ ⎦ x =0

)

For the given data, we obtain T=

2 −3 −3 1 (0.1 Pa ⋅ s)(1450 rpm ) π (80 ×10 mm )( 400 ×10 mm ) 1.2 ≅ 3.35 N ⋅ m ln (1.2 − 0.4 )×10−3 mm 120 0.4

2-78 . ..

Chapter 2 Properties of Fluids 2-122 Solution

A relation is to be derived for the capillary rise of a liquid between two large parallel plates a distance t

apart inserted into a liquid vertically. The contact angle is given to be φ.

Assumptions

There are no impurities in the liquid, and no contamination on the surfaces of the plates.

Analysis

The magnitude of the capillary rise between two large parallel plates can be determined from a force

balance on the rectangular liquid column of height h and width w between the plates. The bottom of the liquid column is at the same level as the free surface of the liquid reservoir, and thus the pressure there must be atmospheric pressure. This will balance the atmospheric pressure acting from the top surface, and thus these two effects will cancel each other. The weight of the liquid column is t

W = mg = ρ gV = ρ g(w × t × h ) Equating the vertical component of the surface tension force to the weight gives W = Fsurface

→

ρ g(w × t × h ) = 2 wσs cos φ

h

Air Canceling w and solving for h gives the capillary rise to be

Capillary rise:

h=

Liquid W

2σs cos φ ρ gt

Discussion The relation above is also valid for non-wetting liquids (such as mercury in glass), and gives a capillary drop instead of a capillary rise.

2-79 . ..

Chapter 2 Properties of Fluids 2-123

A cylindrical shaft rotates inside an oil bearing at a specified speed. The power required to overcome friction is to be determined.

Solution

Assumptions

1 The gap is uniform, and is completely filled with oil. 2 The end effects on the sides of the bearing are

negligible. 3 The fluid is Newtonian.

Properties

The viscosity of oil is given to be 0.300 N⋅s/m2.

R

l

Analysis

(a) The radius of the shaft is R = 0.05 m, and thickness of the oil layer is A = (10.3 – 10)/2 = 0.15 cm. The

power-torque relationship is

4 π R nL W = ωT = 2πn T where, from Chap. 2, T = μ A Substituting, the required power to overcome friction is determined to be 2

3

6π 3 R3 n 2 L 6π 3 (0.05 m)3 (600 / 60 s-1 )2 (0.50 m) ⎛ 1 W ⎟⎞ ⎜ W = μ = (0.3N ⋅ s/m 2 ) ⎜⎝1 N ⋅ m/s ⎟⎟⎠ = 233 W 0.0015 m A (b) For the case of n = 1200 rpm : 6π 3 R3 n 2 L 6π 3 (0.05 m)3 (1200 / 60 s-1 )2 (0.50 m) ⎛ 1 W ⎟⎞ W = μ = (0.3N ⋅ s/m 2 ) ⎟ = 930 W ⎜⎝⎜ 0.0015 m 1 N ⋅ m/s ⎟⎠ A Note the power dissipated in journal bearing is proportional to the cube of the shaft radius and to the square Discussion of the shaft speed, and is inversely proportional to the oil layer thickness.

2-80 . ..

Chapter 2 Properties of Fluids 2-124

A large plate is pulled at a constant speed over a fixed plate. The space between the plates is filled with engine oil. The shear stress developed on the upper plate and its direction are to be determined for parabolic and linear

Solution

velocity profile cases.

Assumptions

1 The thickness of the plate is not important in this problem.

Properties

The viscosity of oil is μ = 0.8374 Pa⋅s (Table A-7).

Analysis Considering a parabolic profile we would have V 2 = ky, where k is a constant. Since V = U = 4 m/s when y = h = 5 mm = 5 × 10−3m, write

(4 ms ) = k ×(5×10 m) → k = 3200 m /s 2

−3

2

Then the velocity profile becomes

V 2 = 3200 y → V = 56.568 y Assuming Newtonian behavior, the shear stress on the upper wall is τ =μ

⎛ 1 ⎞⎟ dV d ⎟⎟ = μ (56.568 y ) = μ(56.568) ⎜⎜⎜ ⎜⎝ 2 y ⎠⎟ dy dy y= h

⎛ ⎞⎟ ⎛ 1 ⎟⎞ 1 = μ(56.568) ⎜⎜ ⎟ ⎟ = (0.8374 Pa ⋅ s)(56.568) ⎜⎜ ⎟ ⎝⎜ 2 h ⎠ ⎝⎜ 2 0.005 m ⎠⎟ = 325 N/m 2

Since dynamic viscosity of oil is 0.8374 Pa⋅s (see Table A-7). If we assume a linear profile we will have dV U 4 m/s = = = 800 s−1 dy h 5×10−3 m

Then the shear stress in this case would be τ=μ

dV U = μ = (0.8374 Pa ⋅ s)×(800) = 670 N/m 2 dy h

Therefore we conclude that the linear assumption is not realistic since it gives over prediction.

2-81 . ..

Chapter 2 Properties of Fluids 2-125 Solution Air spaces in certain bricks form air columns of a specified diameter. The height that water can rise in those

tubes is to be determined.

Assumptions 1 The interconnected air pockets form a cylindrical air column. 2 The air columns are open to the atmospheric air. 3 The contact angle of water is zero, φ = 0.

Properties The surface tension is given to be 0.085 N/m, and we take the water density to be 1000 kg/m3.

Air Brick h Mercury

Analysis Substituting the numerical values, the capillary rise is determined to be ⎛1 kg ⋅ m/s2 ⎞⎟ 2σ cos φ 2(0.085 N/m)( cos 0°) ⎜⎜ ⎟⎟ = 5.78 m h= s = 3 2 −6 ρ gR (1000 kg/m )(9.81 m/s )(3×10 m) ⎜⎜⎝ 1 N ⎠⎟ Discussion The surface tension depends on temperature. Therefore, the value determined may change with temperature.

2-82 . ..

Chapter 2 Properties of Fluids 2-126 Solution A fluid between two long parallel plates is heated as the upper plate is moving. A relation for the fluid velocity is to be obtained and velocity profile is to be plotted. Also the shear stress is to be calculated and its direction is to be shown.

Assumptions 1 Flow is parallel to plates. 2 Flow is one-dimensional. 3 Pressure is constant. 4 Fluid is Newtonian and incompressible. 5 Gravitational effect is neglected.

Analysis (a) Taking an infinitesimal fluid element and applying force balance (assuming one dimensional flow),

The force equilibrium in x direction yields dτ =0 dy

→ τ = constant

(1)

Then, Newton’s law of viscosity yields τ=μ

du = constant dy

(2)

The viscosity is changing linearly with respect to y. Thus, it can be expressed in the form μ =Ay + B where A, B are constants.

Using (i) at y = 0, μ = 0.90 Pa⋅s and, (ii) at y = 0.0004 m, μ = 0.50 Pa⋅s, A, B can be determined and the viscosity function is found μ = 0.90 − 1000y Substituting the viscosity function in equation (3) yields τ=μ

du du du C = (0.9 −1000 y) = constant = C → = dy dy dy 0.9 −1000 y

2-83 . ..

(3)

Chapter 2 Properties of Fluids u=C

dy

∫ 0.9 −1000 y + D

= E ln (0.9 −1000 y) + D

(4)

where E, D area integration constants to be determined. Using the no-slip boundary conditions: (i) at y = 0, u = 0 and, (ii) at y = 0.0004 m, u = 10 m/s, E, D can be determined and the velocity function is found as u ( y) =

⎞⎟ 10 ⎛ 0.9 ⎜ 9 ⎜⎜⎝ 0.9 −1000 y ⎠⎟⎟ ln 5

(5)

The velocity function is plotted below. The velocity vectors are also shown. For comparison, a linear profile is also plotted.

As seen below, the rate at which velocity increases towards the moving plate is not constant − this rate increases as one approaches the moving plate side. This is expected. The viscosity decreases towards the moving plate. To keep the shear stress constant (as was founded earlier), the velocity should increase more and more (not a constant rate) as one approaches the moving plate.

(b) Using Newton’s law of viscosity τ=μ

du 10 (1000)(0.9) 0.9 −1000 y = (0.9 −1000 y) = 17, 000 Pa 2 9 dy 0.9 ln (0.9 −1000 y) 5

(6)

As found earlier, the shear stress is constant throughout the flow.

The shear stress directions on the top surface of the fluid element adjacent to the moving plate, and on the moving plate are:

2-84 . ..

Chapter 2 Properties of Fluids 2-127 Solution A thrust bearing is operated with a thin film of oil. The ratio of lost power in the thrust bearing to the produced power is to be determined.

Assumptions

1 Fluid is Newtonian and incompressible. 2 Linear velocity assumption is correct in the bearing.

3 Gravitational effect is neglected.

Analysis Due to the linear velocity assumption is correct in the bearing, shear stress on the position r = r;

τ =μ

du Δu U −0 U ωr =μ =μ =μ =μ dr Δr e e e

(1)

Angular velocity, ω = 2πn/60 τ=

2μπnr μπ nr = 60 e 30 e

(2)

Friction force on the rotating differential surface, dA = r⋅dθ⋅dr,

dFS = τ ⋅ dA = τ ⋅ r ⋅ dθ ⋅ dr

(3)

Torque on the rotating axis of this force,

dTS = r ⋅ dFS

(4) 2-85

. ..

Chapter 2 Properties of Fluids

Integrating equation (4), Total Torque, TS =

μπ n 30e

∫

2π

r2

dθ

0

∫ r1

2μπ 2 n (r2 − r1 ) μπ 2 n 4 = (r2 − r14 ) 30e 4 60e 4

r 3 dr =

4

(5)

Total Friction Power (loss power), 3 2 ( ) 3( ) [ 4 1.2 4 ] 2πn μπ 2 n 4 (r2 − r14 ) = μπ n (r24 − r14 ) = 0.038 π 550 1.6 −− (1800)(0.22)10 3 60 60e 1800e 2

PS = ω ⋅ TS =

PS = 4032194 W = 4.032194 MW PS 4.032194 = = 0.08064 ≅ 8.06% P 50

2-86 . ..

(6)

Chapter 2 Properties of Fluids 2-128 Solution A multi-disk Electro-rheological “ER” clutch is considered. The ER fluid has a shear stress that is expressed as τ = τ y + μ(du dy) . A relationship for the torque transmitted by the clutch is to be obtained, and the numerical value of the torque is to be calculated.

Assumptions

1 The thickness of the oil layer between the disks is constant. 2 The Bingham plastic model for shear stress

expressed as τ = τ y + μ(du dy) is valid.

The constants in shear stress relation are given to be μ = 0.1 Pa⋅s and τy = 2.5 kPa.

Properties

h = 1.2 mm Shell R2

R1

Output shaft Input shaft Plates mounted on shell Plates mounted on input shaft Variable magnetic field Analysis (a) The velocity gradient anywhere in the oil of film thickness h is V/h where V = ωr is the tangential velocity relative to plates mounted on the shell. Then the wall shear stress anywhere on the surface of a plate mounted on the input shaft at a distance r from the axis of rotation is expressed as

du V ωr = τy + μ = τy + μ dr h h Then the shear force acting on a differential area dA on the surface of a disk and the torque generation associated with it are expressed as

τw = τ y + μ

⎛ ωr ⎞ dF = τ w dA = ⎜⎜τ y + μ ⎟⎟⎟ (2πr )dr ⎝ h⎠ ⎛ ⎛ ωr ⎞ ωr 3 ⎞⎟ dT = rdF = r ⎜⎜τ y + μ ⎟⎟⎟ (2πr )dr = 2π ⎜⎜⎜τ y r 2 + μ ⎟ dr ⎝ ⎝ h⎠ h ⎠⎟ Integrating, T = 2π

∫

⎡ r 3 μω r 4 ⎤ R2 ⎛ 2 ω r 3 ⎞⎟ ⎥ + ⎟⎟ dr = 2 π ⎢⎢ τ y ⎜⎜⎝⎜τ y r + μ ⎥ h ⎠ h 3 4 r = R1 ⎣ ⎦ R2

⎡ τy ⎤ μω 4 = 2 π ⎢ ( R23 − R13 ) + ( R2 − R14 )⎥ 4h ⎣⎢ 3 ⎦⎥ r=R 1

This is the torque transmitted by one surface of a plate mounted on the input shaft. Then the torque transmitted by both surfaces of N plates attached to input shaft in the clutch becomes ⎡ τy ⎤ μω 4 T = 4π N ⎢ ( R23 − R13 ) + ( R2 − R14 )⎥ 4h ⎣⎢ 3 ⎦⎥ (b) Noting that ω = 2 π n = 2 π (2400 rev/min) = 15, 080 rad/min = 251.3 rad/s and substituting,

⎡ 2500 N/m2 ⎤ (0.1 N ⋅ s/m 2 )(251.3 /s) T = (4π)(11) ⎢⎢ [(0.20 m)3 − (0.05 m)3 ] + [(0.20 m)4 − (0.05 m)4 ]⎥⎥ = 2060 N ⋅ m 3 4(0.0012 m) ⎣ ⎦ Discussion

Can you think of some other potential applications for this kind of fluid?

2-87

. ..

Chapter 2 Properties of Fluids 2-129

A multi-disk magnetorheological “MR” clutch is considered The MR fluid has a shear stress that is

Solution

expressed as τ = τ y + K (du dy)m . A relationship for the torque transmitted by the clutch is to be obtained, and the numerical value of the torque is to be calculated.

Assumptions 1 The thickness of the oil layer between the disks is constant. 2 The Herschel-Bulkley model for shear stress expressed as τ = τ y + K (du dy)m is valid.

The constants in shear stress relation are given to be τy = 900 Pa, K = 58 Pa⋅sm , and m = 0.82.

Properties

h = 1.5 mm Shell R2

R1

Output shaft Input shaft Plates mounted on shell Plates mounted on input shaft Variable magnetic field Analysis (a) The velocity gradient anywhere in the oil of film thickness h is V/h where V = ωr is the tangential velocity relative to plates mounted on the shell. Then the wall shear stress anywhere on the surface of a plate mounted on the input shaft at a distance r from the axis of rotation is expressed as ⎛ du ⎞m ⎛ V ⎞m ⎛ ωr ⎞m τ w = τ y + K ⎜⎜ ⎟⎟⎟ = τ y + K ⎜⎜ ⎟⎟⎟ = τ y + K ⎜⎜ ⎟⎟⎟ ⎝h⎠ ⎝h⎠ ⎝ dr ⎠ Then the shear force acting on a differential area dA on the surface of a disk and the torque generation associated with it are expressed as m +2 ⎞ ⎛ ⎛ ⎛ ⎛ ωr ⎞m ⎞ ⎛ ωr ⎞m ⎞ ω mr ⎟⎟⎟ dr dF = τ w dA = ⎜⎜⎜τ y + K ⎜⎜ ⎟⎟⎟ ⎟⎟⎟ (2πr )dr and dT = rdF = r ⎜⎜⎜τ y + K ⎜⎜ ⎟⎟⎟ ⎟⎟⎟ (2πr )dr = 2π ⎜⎜⎜τ y r 2 + K m ⎟⎟⎠ ⎝ h ⎠ ⎟⎠ ⎝ h ⎠ ⎠⎟ ⎜⎝ h ⎝ ⎝

Integrating, T = 2π

∫

R2 R1

m m +2 ⎞ m ⎛ ⎡ r 3 K ω m r m +3 ⎤ R2 ⎡τ ⎤ ⎟⎟ ⎜⎜τ r 2 + K ω r m +3 ⎥ = 2π ⎢ y ( R23 − R13 ) + K ω + − R1m +3 )⎥ ⎟⎟ dr = 2 π ⎢⎢ τ y m m⎥ m ( R2 ⎜⎜ y ⎢ ⎥⎦ ⎟ h m h m h + + 3 ( 3) 3 ( 3) ⎝ ⎠ ⎣ ⎣ ⎦ R1

This is the torque transmitted by one surface of a plate mounted on the input shaft. Then the torque transmitted by both surfaces of N plates attached to input shaft in the clutch becomes ⎡ τy ⎤ Kωm T = 4π N ⎢ ( R23 − R13 ) + ( R2m +3 − R1m +3 )⎥ m ⎢⎣ 3 ⎥⎦ (m + 3)h (b) Noting that ω = 2 π n = 2 π (3000 rev/min) = 18,850 rad/min = 314.2 rad/s and substituting, ⎡ 900 N/m 2 ⎤ (58 N ⋅ s0.82 /m 2 )(314.2 /s)0.82 T = (4 π )(11) ⎢⎢ [(0.20 m)3 − (0.05 m)3 ] + [(0.20 m)3.82 − (0.05 m)3.82 ]⎥⎥ 0.82 3 (0.82 + 3)(0.0015 m) ⎣ ⎦ = 103.4 kN ⋅ m Discussion

Can you think of some other potential applications for this . fluid?

2-88

. ..

Chapter 2 Properties of Fluids 2-130

The laminar flow of a Bingham plastic fluid in a horizontal pipe of radius R is considered. The shear stress at the pipe wall and the friction drag force acting on a pipe section of length L are to be determined.

Solution

Assumptions

1 The fluid is a Bingham plastic with τ = τ y + μ(du dr ) where τy is the yield stress. 2 The flow through the

pipe is one-dimensional.

R r u(r) 0 The velocity profile is given by u(r ) =

Analysis

τy ΔP 2 (r − R 2 ) + (r − R) where ΔP/L is the pressure drop along 4μ L μ

the pipe per unit length, μ is the dynamic viscosity, r is the radial distance from the centerline. Its gradient at the pipe wall (r = R) is ⎞ ⎛ ΔP τ y ⎟⎞ τy ⎞ du d ⎛ ΔP 2 1 ⎛ ΔP (r − R 2 ) + (r − R)⎟⎟⎟ R + τ y ⎟⎟⎟ = ⎜⎜⎜ = ⎜⎜⎜2r + ⎟⎟ = ⎜⎜ ⎜ ⎟⎠ ⎠ dr r = R dr ⎝ 4 μ L μ ⎝ 4μ L μ ⎟⎠r = R μ ⎝ 2 L r=R Substituing into τ = τ y + μ(du dr ) , the wall shear stress at the pipe surface becomes

τw = τ y + μ

du ΔP ΔP R + τ y = 2τ y + R = τy + dr r = R 2L 2L

Then the friction drag force exerted by the fluid on the inner surface of the pipe becomes

⎛ ⎛ ΔP ⎟⎞ ΔP ⎟⎞ FD = τ w As = ⎜⎜⎜2τ y + R⎟⎟ (2π RL ) = 2π RL ⎜⎜⎜2τ y + R⎟ = 4π RLτ y + π R 2 ΔP ⎝ ⎝ 2L ⎠ 2 L ⎟⎠ Discussion

Note that the total friction drag is proportional to yield shear stress and the pressure drop.

2-89 .. ..

Chapter 2 Properties of Fluids 2-131

A circular disk immersed in oil is used as a damper, as shown in the figure. It is to be shown that the

Solution

damping torque is Tdamping = Cω where C = 0.5πμ (1 a + 1 b ) R 4 .

Assumptions

1 The thickness of the oil layer on each

side remains constant. 2 The velocity profiles are linear on both sides of the disk. 3 The tip effects are negligible.

4 The effect of the shaft is negligible.

R Analysis

The velocity gradient anywhere in the oil y

a

of film thickness a is V/a where V = ωr is the tangential velocity. Then the wall shear stress anywhere on the upper surface of the disk at a distance r from the axis of

b

rotation can be expressed as

Disk

du V ωr =μ =μ dy a a Then the shear force acting on a differential area dA on τw = μ

Damping oil

the surface and the torque it generates can be expressed as ωr dF = τ w dA = μ dA a ωr 2 dT = rdF = μ dA a Noting that dA = 2πrdr and integrating, the torque on the top surface is determined to be

Ttop =

μω a

∫

r 2 dA = A

μω a

∫

R

r =0

r 2 (2πr )dr =

2πμω a

∫

R

r =0

R

r 3dr =

2πμω r 4 πμω R 4 = a 4 r =0 2a

The torque on the bottom surface is obtained by replaying a by b,

πμω R 4 2b The total torque acting on the disk is the sum of the torques acting on the top and bottom surfaces, Tbottom =

Tdamping, total = Tbottom + Ttop =

πμω R 4 ⎛ 1 1 ⎞⎟ ⎜⎜ + ⎟ 2 ⎝ a b ⎠⎟

or, Tdamping, total = Cω

where C =

πμ R 4 ⎛ 1 1 ⎞⎟ ⎜ + ⎟ 2 ⎜⎝ a b ⎠⎟

This completes the proof.

Discussion Note that the damping torque (and thus damping power) is inversely proportional to the thickness of oil films on either side, and it is proportional to the 4th power of the radius of the damper disk.

2-90 .. ..

Chapter 2 Properties of Fluids 2-132 Solution A thin oil film is sandwiched between two large parallel plates with top plate stationary and bottom plate

moving. A third plate is dragged through the oil. The velocity profile is to be sketched and the vertical distance where the velocity is zero is to be determined. Also, the force required to keep the middle plate at constant speed is to be determined.

Assumptions 1 Flow is parallel to plates. 2 Flow is one-dimensional. 3 Pressure is constant. 4 Fluid is Newtonian and incompressible. 5 Gravitational effect is neglected.

Analysis (a) The velocity profiles are expected to be linear with respect to y, and independent of x. At some x location, we sketch the velocity profile in both the upper and lower fluid regions, keeping in mind the no-slip boundary conditions: u = 0 at the top wall, u = 1 m/s at the plate (both sides), and u = −0.3 m/s at the bottom wall.

V = 1.00 m/s Oil

μ, ρ

Oil

μ, ρ

h1

Area = A

F

Area = A h2

yA Vw = 0.300 m/s

Equating triangles, we know that u = 0 at y = yA, where yA /Vw = h2/(Vplate + Vw) or yA = Vw h2/(Vplate + Vw). Plugging in the numerical values, we get yA = (0.300 m/s)(1.65 mm)/(1.00 + 0.300) m/s = 0.381 mm. yA = 0.381 mm (b) To calculate the force needed to pull the plate, we draw a free body diagram of the plate,

F1

F

F2

For constant plate speed (no acceleration), F = F1 + F2. But we can calculate these two forces independently based on the shear stress at the walls of the plate, F1 = τ1 A = μ

⎛ du du ⎟⎟⎞ du du ⎜ + A and F2 = τ 2 A = μ A . Thus, F = F1 + F2 = μ A ⎜⎜ ⎟. dy 1 dy 2 ⎜⎝ dy 1 dy 2 ⎟⎟⎠

Plugging in the numbers, we get

⎡ (1 − 0) m/s (1 − (−0.30)) m/s ⎤ F = (0.0357 N ⋅ s/m 2 )(0.20 ×0.20 m 2 ) ⎢ + ⎥ = 2.55 N ⎢⎣ 0.001 m 0.00165 m ⎥⎦ F = 2.55 N Discussion There are forces acting on the bottom and top walls also. You should be able to calculate these yourself.

2-91 .. ..

Chapter 2 Properties of Fluids

Fundamentals of Engineering (FE) Exam Problems

2-133

The specific gravity of a fluid is specified to be 0.82. The specific volume of this fluid is (a) 0.001 m3/kg

(b) 0.00122 m3/kg

(c) 0.0082 m3/kg

(d) 82 m3/kg

(e) 820 m3/kg Answer (b) 0.00122 m3/kg 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).

SG=0.82 rho_water=1000 [kg/m^3] rho_fluid=SG*rho_water v=1/rho_fluid

2-134

The specific gravity of mercury is 13.6. The specific weight of mercury is (a) 1.36 kN/m3

(b) 9.81 kN/m3

(c) 106 kN/m3

(d) 133 kN/m3

(e) 13,600 kN/m3

Answer (d) 133 kN/m3 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).

SG=13.6 rho_water=1000 [kg/m^3] rho=SG*rho_water g=9.81 [m/s^2] SW=rho*g

2-92 .. ..

Chapter 2 Properties of Fluids 2-135

A 0.08-m3 rigid tank contains air at 3 bar and 127°C. The mass of the air in the tank is (a) 0.209 kg

(b) 0.659 kg

(c) 0.8 kg

(d) 0.002 kg

(e) 0.066 kg

Answer (a) 0.209 kg 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).

V=0.08 [m^3] P=300 [kPa] T=(127+273) [K] R=0.287 [kJ/kg-K] m=(P*V)/(R*T)

2-136

The pressure of water is increased from 100 kPa to 700 kPa by a pump. The density of water is 1 kg/L. If the water temperature does not change during this process, the enthalpy change of the water is (a) 400 kJ/kg

(b) 0.4 kJ/kg

(c) 600 kJ/kg

(d) 800 kJ/kg

(e) 0.6 kJ/kg

Answer (e) 0.6 kJ/kg 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).

P1=100 [kPa] P2=700 [kPa] rho=1000 [kg/m^3] DELTAh=(P2-P1)/rho

2-137

An ideal gas flows in a pipe at 37°C. The density of the gas is 1.9 kg/m3 and its molar mass is 44 kg/kmol. The pressure of the gas is (a) 13 kPa

(b) 79 kPa

(c) 111 kPa

(d) 490 kPa

(e) 4900 kPa

Answer (c) 111 kPa 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).

T=(37+273) [K] rho=1.9 [kg/m^3] MM=44 [kg/kmol] R_u=8.314 [kJ/kmol-K] R=R_u/MM P=rho*R*T

2-93 .. ..

Chapter 2 Properties of Fluids 2-138

Liquid water vaporizes into water vapor as it flows in the piping of a boiler. If the temperature of water in the pipe is 180°C, the vapor pressure of water in the pipe is (a) 1002 kPa

(b) 180 kPa

(c) 101.3 kPa

(d) 18 kPa

(e) 100 kPa

Answer (a) 1002 kPa 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).

T=180 [C] P_vapor=pressure(steam, T=T, x=1)

2-139

In a water distribution system, the pressure of water can be as low as 1.4 psia. The maximum temperature of water allowed in the piping to avoid cavitation is (a) 50°F

(b) 77°F

(c) 100°F

(d) 113°F

(e) 140°F

Answer (d) 113°F 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).

P=1.4 [psia] T_max=temperature(steam, P=P, x=1)

2-140

The pressure of water is increased from 100 kPa to 900 kPa by a pump. The temperature of water also increases by 0.15°C. The density of water is 1 kg/L and its specific heat is cp = 4.18 kJ/kg⋅°C. The enthalpy change of the water during this process is (a) 900 kJ/kg

(b) 1.43 kJ/kg

(c) 4.18 kJ/kg

(d) 0.63 kJ/kg

(e) 0.80 kJ/kg

Answer (b) 1.43 kJ/kg 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).

P1=100 [kPa] P2=900 [kPa] DELTAT=0.15 [C] rho=1000 [kg/m^3] c_p=4.18 [kJ/kg-C] DELTAh=c_p*DELTAT+(P2-P1)/rho

2-94 .. ..

Chapter 2 Properties of Fluids 2-141

An ideal gas is compressed isothermally from 100 kPa to 170 kPa. The percent increase in the density of this gas during this process is (a) 70%

(b) 35%

(c) 17%

(d) 59%

(e) 170%

Answer (a) 70% 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).

P1=100 [kPa] P2=170 [kPa] DELTAP=P2-P1 DELTAP\P=DELTAP/P1*Convert(,%) DELTArho\rho=DELTAP\P

2-142

The variation of the density of a fluid with temperature at constant pressure is represented by (a) Bulk modulus of elasticity

(b) Coefficient of compressibility

(c) Isothermal compressibility

(d) Coefficient of volume expansion

(e) None of these Answer (d) Coefficient of volume expansion

2-143

Water is heated from 2°C to 78°C at a constant pressure of 100 kPa. The initial density of water is 1000 kg/m3 and the volume expansion coefficient of water is β = 0.377×10−3 K−1. The final density of the water is (a) 28.7 kg/m3

(b) 539 kg/m3

(c) 997 kg/m3

(d) 984 kg/m3

(e) 971 kg/m3

Answer (e) 971 kg/m3 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).

T1=2 [C] T2=78 [C] P=100 [kPa] rho_1=1000 [kg/m^3] beta=0.377E-3 [1/K] DELTAT=T2-T1 DELTArho=-beta*rho_1*DELTAT DELTArho=rho_2-rho_1

2-95 .. ..

Chapter 2 Properties of Fluids 2-144

The viscosity of liquids ___________ and the viscosity of gases ___________ with temperature. (a) Increases, increases

(b) Increases, decreases

(d) Decreases, decreases

(e) Decreases, remains the same

(c) Decreases, increases

Answer (c) Decreases, increases

2-145

The pressure of water at atmospheric pressure must be raised to 210 atm to compress it by 1 percent. Then, the coefficient of compressibility value of water is (a) 209 atm

(b) 20,900 atm

(c) 21 atm

(d) 0.21 atm

(e) 210,000 atm

Answer (b) 20,900 atm 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).

P1=1 [atm] P2=210 [atm] DELTArho\rho=0.01 DELTAP=P2-P1 CoeffComp=DELTAP/DELTArho\rho

2-146

The density of a fluid decreases by 3 percent at constant pressure when its temperature increases by 10°C. The coefficient of volume expansion of this fluid is (a) 0.03 K−1

(b) 0.003 K−1

(c) 0.1 K−1

(d) 0.5 K−1

(e) 3 K−1

Answer (b) 0.003 K−1 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).

DELTArho\rho=-0.03 DELTAT=10 [K] beta=-DELTArho\rho/DELTAT

2-96 .. ..

Chapter 2 Properties of Fluids 2-147

The speed of a spacecraft is given to be 1250 km/h in atmospheric air at −40°C. The Mach number of this flow is (a) 35.9

(b) 0.85

(c) 1.0

(d) 1.13

(e) 2.74

Answer (d) 1.13 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=1250 [km/h]*Convert(km/h, m/s) T=(-40+273.15) [K] R=0.287 [kJ/kg-K] k=1.4 c=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Ma=Vel/c

2-148

The dynamic viscosity of air at 20°C and 200 kPa is 1.83×10−5 kg/m⋅s. The kinematic viscosity of air at this state is (a) 0.525×10−5 m2/s

(b) 0.77×10−5 m2/s

(c) 1.47×10−5 m2/s

(d) 1.83×10−5 m2/s

(e) 0.380×10−5 m2/s Answer (b) 0.77×10−5 m2/s 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).

T=(20+273.15) [K] P=200 [kPa] mu=1.83E-5 [kg/m-s] R=0.287 [kJ/kg-K] rho=P/(R*T) nu=mu/rho

2-97 .. ..

Chapter 2 Properties of Fluids 2-149

A viscometer constructed of two 30-cm-long concentric cylinders is used to measure the viscosity of a fluid. The outer diameter of the inner cylinder is 9 cm, and the gap between the two cylinders is 0.18 cm.

The inner cylinder is rotated at 250 rpm, and the torque is measured to be 1.4 N⋅m. The viscosity of the fluid is (a) 0.0084 N⋅s/m2

(b) 0.017 N⋅s/m2

(c) 0.062 N⋅s/m2

(d) 0.0049 N⋅s/m2

(e) 0.56 N⋅s/m2 Answer (e) 0.56 N⋅s/m2 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).

L=0.3 [m] R=0.045 [m] gap=0.0018 [m] n_dot=(250/60) [1/s] T=1.4 [N-m] mu=(T*gap)/(4*pi^2*R^3*n_dot*L)

2-150

A 0.6-mm-diameter glass tube is inserted into water at 20°C in a cup. The surface tension of water at 20°C is σs = 0.073 N/m. The contact angle can be taken as zero degrees. The capillary rise of water in the tube is (a) 2.6 cm

(b) 7.1 cm

(c) 5.0 cm

(d) 9.7 cm

(e) 12.0 cm

Answer (c) 5.0 cm 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).

D=0.0006 [m] R=D/2 sigma_s=0.073 [N/m] phi=0 [degrees] rho=1000 [kg/m^3] g=9.81 [m/s^2] h=(2*sigma_s*cos(phi))/(rho*g*R)

2-98 .. ..

Chapter 2 Properties of Fluids 2-151

A liquid film suspended on a U-shaped wire frame with a 6-cm-long movable side is used to measure the surface tension of a liquid. If the force needed to move the wire is 0.028 N, the surface tension of this liquid in air is (a) 0.00762 N/m (b) 0.096 N/m

(c) 0.168 N/m

(d) 0.233 N/m

(e) 0.466 N/m

Answer (d) 0.233 N/m 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).

F=0.028 [N] b=0.06 [m] sigma_s=F/(2*b)

2-152

It is observed that water at 20°C solution rises up to 20 m height in a tree due to capillary effect. The surface tension of water at 20°C is σs = 0.073 N/m and the contact angle is 20°. The maximum diameter of the tube in which water rises is (a) 0.035 mm

(b) 0.016 mm

(c) 0.02 mm

(d) 0.002 mm

(e) 0.0014 mm

Answer (e) 0.0014 mm 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). h=20 [m] sigma_s=0.073 [N/m] phi=20 [degrees] rho=1000 [kg/m^3] g=9.81 [m/s^2] h=(2*sigma_s*cos(phi))/(rho*g*R) D=2*R

2-99 .. ..

Chapter 2 Properties of Fluids

Design and Essay Problems

2-153 to 2-158 Solution

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

2-100 .. ..

Chapter 2 Properties of Fluids 2-157

We are to determine the inlet water speed at which cavitation is likely to occur in the throat of a convergingdiverging tube or duct, and repeat for a higher temperature.

Solution

Assumptions

1 The fluid is incompressible and Newtonian. 2 Gravitational effects are negligible. 3 Irreversibilities are

negligible. 4 The equations provided are valid for this flow.

Properties

For water at 20oC, ρ = 998.0 kg/m3 and Psat = 2.339 kPa.

Analysis

(a) Two equations are given for velocity, pressure, and cross-sectional area, namely,

V1 A1 = V2 A2

and

P1 + ρ

V12 V2 = P2 + ρ 2 2 2

Solving the first equation for V2 gives

V2 = V1

A1 A2

(1)

Substituting the above into the equation for pressure and solving for V1 yields, after some algebra, V1 =

2 ( P1 − P2 ) 2 ⎛ ⎟⎞ ⎜⎛ A ⎞ ρ ⎜⎜⎜⎜⎜ 1 ⎟⎟⎟ −1⎟⎟⎟ ⎜⎝⎜⎝⎜ A2 ⎠⎟ ⎠⎟⎟

But the pressure at which cavitation is likely to occur is the vapor (saturation) pressure of the water. We also know that throat diameter D2 is 1/20 times the inlet diameter D1, and since A = πD2/4, A1/A2 = (20)2 = 400. Thus, V1 =

2 2 (20.803 − 2.339) kPa ⎛⎜1000 N/m 2 ⎞⎛ ⎟⎟⎜⎜1 kg ⋅ m/s ⎞⎟⎟ = 0.015207 m ⎜⎜ ⎟ kg ⎟⎟⎜⎜ ⎜ kPa N s ⎠⎝ ⎠⎟ 998.0 3 (4002 −1) ⎝ m

So, the minimum inlet velocity at which cavitation is likely to occur is 0.0152 m/s (to three significant digits). The velocity at the throat is much faster than this, of course. Using Eq. (1), Vt = V1

2 ⎛ D ⎟⎞2 A1 π D12 ⎜ 1 ⎟ = 0.015207 ⎛⎜ 20 ⎞⎟ = 6.0828 m/s V = V1 = ⎜ ⎟ 1⎜ ⎜⎝ 1 ⎠⎟ ⎜⎝ Dt ⎟⎟⎠ At π Dt 2

(b) If the water is warmer (50oC), the density reduces to 988.1 kg/m3, and the vapor pressure increases to 12.35 kPa. At these conditions, V1 = 0.0103 m/s. As might be expected, at higher temperature, a lower inlet velocity is required to generate cavitation, since the water is warmer and already closer to its boiling point.

Discussion Cavitation is usually undesirable since it leads to noise, and the collapse of the bubbles can be destructive. It is therefore often wise to design piping systems and turbomachinery to avoid cavitation.

2-101 .. ..

Chapter 2 Properties of Fluids 2-158

We are to explain how objects like razor blades and paper clips can float on water, even though they are much denser than water.

Solution

Analysis

Just as some insects like water striders can be supported on water by surface tension, surface tension is the

key to explaining this phenomenon. If we think of surface tension like a skin on top of the water, somewhat like a

stretched piece of balloon, we can understand how something heavier than water pushes down on the surface, but the surface tension forces counteract the weight (to within limits) by providing an upward force. Since soap decreases

surface tension, we expect that it would be harder to float objects like this on a soapy surface; with a high enough soap concentration, in fact, we would expect that the razor blade or paper clip could not float at all.

Discussion

If the razor blade or paper clip is fully submerged (breaking through the surface tension), it sinks.

2-102 . ..

Chapter 3 Pressure and Fluid Statics

Solutions Manual for Fluid Mechanics: Fundamentals and Applications Fourth Edition Yunus A. <;engel & John M. Cimbala McGraw-Hill Education, 2018

Chapter 3 PRESSURE AND FLUID STATICS

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Chapter 3 Pressure and Fluid Statics Pressure, Manometer, and Barometer

3-lC Solution

We are to discuss the difference between gage pressure and absolute pressure.

Analysis

The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative

to an absolute vacuum is called absolute pressure.

Discussion

Most pressure gages (like your bicycle tire gage) read relative to atmospheric pressure, and therefore read

the gage pressure.

3-2C Solution

We are to compare the pressure on the surfaces of a cube.

Analysis

Since pressure increases with depth, the pressure on the bottom face of the cube is higher than that on

the top. The pressure varies linearly along the side faces. However, if the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube are nearly the same.

Discussion

In the limit of an "infinitesimal cube", we have a fluid particle, with pressure P defined at a "point".

3-3C Solution

We are to explain nose bleeding and shortness of breath at high elevation.

Analysis

Atmospheric air pressure which is the external pressure exerted on the skin decreases with increasing

elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations, and thus lower amount of oxygen per unit volume.

Discussion

People who climb high mountains like Mt. Everest suffer other physical problems due to the low pressure.

Chapter 3 Pressure and Fluid Statics 3-4C Solution

We are to compare the volume and mass flow rates of two fans at different elevations.

Analysis

The density of air at sea level is higher than the density of air on top of a high mountain. Therefore, the

volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher.

Discussion

In reality, the fan blades on the high mountain would experience less frictional drag, and hence the fan

motor would not have as much resistance - the rotational speed of the fan on the mountain may be slightly higher than that at sea level.

3-SC Solution

We are to examine a claim about absolute pressure.

Analysis

No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It

is the gage pressure that doubles when the depth is doubled.

Discussion

This is analogous to temperature scales - when performing analysis using something like the ideal gas law,

you must use absolute temperature (K), not relative temperature (°C), or you will run into the same kind of problem.

3-6C Solution

We are to define Pascal's law and give an example.

Analysis

Pascal's law states that the pressure applied to a confined fluid increases the pressure throughout by

the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal's principle is the operation of the hydraulic car jack.

Discussion

Students may have various answers to the last part of the question. The above discussion applies to fluids at

rest (hydrostatics). When fluids are in motion, Pascal's principle does not necessarily apply. However, as we shall see in later chapters, the differential equations of incompressible fluid flow contain only pressure gradients, and thus an increase in pressure in the whole system does not affect fluid motion.

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Chapter 3 Pressure and Fluid Statics 3-30 Solution

The previous problem is reconsidered. The effect of the spring force in the range of Oto 500 Non the

pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed.

Analysis

The EES Equations window is printed below, followed by the tabulated and plotted results.

g=9.81 [m/s"2]

P_atm= 95 [kPa] m_piston=5 [kg] F_spring=75 [NJ A=35*CONVERT('cm"2','m"2') W_piston=m_piston*g F_atm=P_atm*A*CONVERT('kPa','N/m"2') "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston P_gas=F_gas/A*CONVERT('N/m"2','kPa') Fspring [N]

P9 as [kPa]

0 25 50 75 100 125 150 175 200 225 250 275 300 325 350 375 400 425 450 475 500

109 116.2 123.3 130.4 137.6 144.7 151.9 159 166.2 173.3 180.4 187.6 194.7 201.9 209 216.2 223.3 230.4 237.6 244.7 251.9

Discussion

250

150

100-----����---------------� 300 0 100 200 400 500

The relationship is linear, as expected.

Fspring

[N]

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Chapter 3 Pressure and Fluid Statics

240.

22ft

•

.tot 190

L fBO

..

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80 40

.... � ..., Fltdd HflilaJlt v:,

·••om••,•••• Q•Jttt'Y

11,00.---.-------...-------.----..--------------.-....----.----

E

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o .__......._...____.__....____.__,.._____.__........___..___........___.,__........___..__......

O. �.H 4fft. •ftt MOO t•Oot 191.1 1,.._ .·, (k gtm •jJ Discussion

Many comments are provided in the Equation window above to help you learn some of the features of EES.

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Chapter 3 Pressure and Fluid Statics 3-62C Solution

We are to consider the effect of plate rotation on the hydrostatic force on the plate surface.

Analysis

There will be no change on the hydrostatic force acting on the top surface of this submerged horizontal flat

plate as a result of this rotation since the magnitude of the resultant force acting on a plane surface of a completely submerged body in a homogeneous fluid is equal to the product of the pressure Pc at the centroid of the surface and the area A of the surface.

Discussion

If the rotation were not around the centroid, there would be a change in the force.

3-63C Solution

We are to explain how to determine the horizontal component of hydrostatic force on a curved surface.

Analysis

The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude

and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface.

Discussion

We could also integrate pressure along the surface, but the method discussed here is much simpler and

yields the same answer.

3-64C Solution

We are to explain how to determine the vertical component of hydrostatic force on a curved surface.

Analysis

The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force

acting on the horizontal projection of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block.

We could also integrate pressure along the surface, but the method discussed here is much simpler and Discussion yields the same answer.

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Chapter 3 Pressure and Fluid Statics 3-75 The previous problem is reconsidered. The effect of water depth on the force exerted on the plate by the

Solution

ridge as the water depth varies from Oto 5 m in increments of 0.5 m is to be investigated.

The EES Equations window is printed below, followed by the tabulated and plotted results.

Analysis

g = 9.81 "m/s2" rho=1000 "kg/m3" s=1"m" w = 5"m" A=w*h P_ave = rho*g*h/2000 "kPa" F_R=P_ave*A "kN" y_p=2*h/3 F_ridge =(s+y_p)*F_R/(s+h)

Dept

Pave,

h,m

kPa

FR

Yp

kN

m

F,idge

kN

0.0

0

0.0

0.00

0

0.5

2.453

6.1

0.33

5

1.0

4.905

24.5

0.67

20

1.5 2.0

7.358 9.81

55.2 98.1

1.00 1.33

44

2.5

12.26

153.3

1.67

117

3.0

14.72

220.7

2.00

166

3.5

17.17

300.4

2.33

223

4.0

19.62

392.4

2.67

288

4.5

22.07

496.6

3.00

361

5.0

24.53

613.1

3.33

443

76

§0 l!OQ J iO //

'300 2 rO

:,/

� ::200

,/" /''

"' -� 1 w0_

LIL. 1"00 !50 0 0 Discussion

......--- _...,,.--

/

/

,../ .?

4

The force on the ridge does not increase linearly, as we may have suspected.

3-56

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