Solution manual engineering fluid mechanics 9th ED book

Situation: This problem involves the viscosity and density of air and water.... Situation: This problem involves viscosity of SAE 10W-30 oil, kerosene and water.... Situation: Informatio

Notes to instructors

Introduction

The following ideas and information are provided to assist the instructor in the design and implementation

of the course Traditionally this course is taught at Washington State University and the University of Idaho as a three-credit semester course which means 3 hours of lecture per week for 15 weeks Basically the first 11 chapters and Chapter 13 (Flow Measurements) are covered in Mechanical Engineering Chapters 12 (Compressible Flow) and Chapter 14 (Turbomachinery) may be covered depending on the time available and exposure to compressible flow in other courses (Thermodynamics) Open channel flow (Chapter 15) is generally not covered in Mechanical Engineering When the text is used in Civil Engineering, Chapters 1-11 and 13 are nominally covered and Chapters

14 and 15 may be included if time permits and exposure to open channel flow may not be available in other courses The book can be used for 10-week quarter courses by selecting the chapters, or parts of the chapters, most appropriate for the course.

Author Contact

Every effort has been made to insure that the solution manual is error free If errors are found (and they will be!) please contact Professors Crowe or Elger.

Mechanical Engineering Dept School of Mechanical Eng & Matl Science University of Idaho Washington State University

Moscow, ID 83844-0902 Pullman, WA 99164-2920 Phone (208) 885-7889 Phone (509) 335-3214 Fax (208) 885-9031 Fax (509) 335-4662 e-mail: delger@uidaho.edu e-mail: crowe@mme.wsu.edu

Design and Computer Problems

Design problems (marked in the text in blue) are those problems that require engineering practices such

as estimation, making asummptions and considering realistic materials and components These problems provide a platform for student discussion and group activity One approach is to divide the class into small groups of three or four and have these groups work on the design problems together Each group can then report on their design to the rest of the class The role of the professor is to help the student learn the practices of the design review—that is, teach the student to ask in-depth questions and teach them how to develop meaningful and in-depth answers This dialogue stimulates interest and class discussion Solutions to most design problems are included in the solution manual.

Computer-oriented problems (marked in the text is blue) are those problems may best be solved using software such as spreadsheets, TK Solver or MathCad The choice is left to the student The answer book also includes the results for the computer-oriented problems.

Situation: An engineer needs density for an experiment with a glider.

Local pressure = 27.3 in.-Hg = 92.45 kPa

Find: (a) Calculate density using local conditions

(b) Compare calculated density with the value from Table A.2, and make a mendation

recom-Properties: From Table A.2, Rair= 287 kg· KJ , ρ = 1.22 kg/ m3

1 The density difference (local conditions versus table value) is about 12% Most

of this difference is due to the effect of elevation on atmospheric pressure

of elevation are significant

1

Situation: Carbon dioxide is at 300 kPa and 60oC.

Situation: Methane is at 500 kPa and 60oC.

Find: Density and specific weight

kg· K.APPROACH

First, apply the ideal gas law to find density Then, calculate specific weight using

3

Situation: Natural gas (10◦C)is stored in a spherical tank Atmospheric pressure is

Situation: Water and air are at T = 100oC and p = 5 atm.

Find: Ratio of density of water to density of air

ρwater

9584.73

5

Situation: Oxygen (p = 400 psia, T = 70◦F)fills a tank Tank volume = 10 ft3 Tankweight =100 lbf.

Find: Weight (tank plus oxygen)

Properties: From Table A.2, RO 2 = 1555ft·lbf/(slug ·oR)

APPROACH

Apply the ideal gas law to find density of oxygen Then find the weight of the oxygenusing specific weight (γ) and add this to the weight of the tank

ANALYSIS

Ideal gas law

assump-Situation: Air is at an absolute pressure of p = 600 kPa and a temperature of

T = 50oC

Find: (a) Specific weight, and (b) density

Properties: From Table A.2, R = 287 kg· KJ

Situation: Consider a mass of air with a volume of 1 cubic mile.

kg

Properties: From Table A.2, ρair = 0.00237 slugs/ft3

Assumptions: The density of air is the value at sea level for standard conditions.ANALYSIS

Situation: This problem involves the effects of temperature on the properties of air.The application is a bicyclist.

Find: a.) Plot air density versus temperature for a range of -10oC to 50oC

b.) Plot tire pressure versus temperature for the same temperature range

Assumptions: For part b, assume that the bike tire was initially inflated to ptire = 450

1.10 1.15 1.20 1.25 1.30 1.35 1.40

Temperature, oC

-20 -10 0 10 20 30 40 50 60 380

Situation: A design team needs to know how much CO2 is needed to inflate a rubberraft.

Raft is shown in the sketch below

Inflation pressure is 3 psi above local atmospheric pressure Thus, inflation pressure

is 17.7 psi = 122 kPa

Find: (a)Estimate the volume of the raft

2.) Assume that the volume of the raft can be approximated by a cylinder of diameter0.45 m and a length of 16 m (8 meters for the length of the sides and 8 meters forthe lengths of the ends plus center tubes)

The final mass (5.66 kg = 12.5 lbm) is large This would require a large and

that is driven by cost

Situation: The application is a helium filled balloon of radius r = 1.3 m.

p = 0.89 bar = 89 kPa

Find: Weight of helium inside balloon

Situation: In the wine and beer industries, fermentation involves glucose (C6H12O6)

from the vat

C6H12O6 → 2(CH3CH2OH) + 2(CO2)The initial specific gravity is 1.08

Specific gravity of alcohol is 0.80

Saturated solution (water + sugar) has a specific gravity of 1.59

Find: (a.) Final specific gravity of the wine

(b.) Percent alcohol content by volume after fermentation

Assumptions: All of the sugar is converted to alcohol

APPROACH

Imagine that the initial mixture is pure water plus saturated sugar solution and thenuse this visualization to find the mass of sugar that is initially present (per unit

produced (per unit of volume) Then, solve for the problem unknowns

ANALYSIS

The initial density of the mixture is

ρmix = ρwVw+ ρsVs

Vo

volume of the mixture is the volume of the pure water plus the volume of saturatedsolution

ms

converts to 0.51 kg of alcohol so the final density of alcohol is

Percent alcohol by volume = 13.7%

15

Situation: This problem involves the viscosity and density of air and water.

Situation: Air at 10oC and 60oC.

Properties: From table A.3, ν60= 1.89× 10−5 m2/s, ν10 = 1.41× 10−5 m2/s.APPROACH

Use properties found in table A.3

ANALYSIS

∆vair,10→60 = (1.89− 1.41) × 10−5 = 4.8×10−6 m2/s

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Situation: This problem involves viscosity of SAE 10W-30 oil, kerosene and water.

Situation: Air and water at 20oC.

Find: (a)Ratio of dynamic viscosity of air to that of water

(b)Ratio of kinematic viscosity of air to that of water

Properties: From Table A.3, µair,20◦ C = 1.81× 10−5 N·s/m2; ν = 1.51× 10−5 m2/sFrom Table A.5, µwater,20◦ C = 1.00× 10−3 N·s/m2; ν = 1.00× 10−6 m2/s

Situation: Sutherland’s equation and the ideal gas law describe behaviors of commongases.

temperature T and pressure p

”o” defines the reference state

Situation: The viscosity of air is µair(15oC) = 1.78× 10−5 N·s/m2.

Properties: From Table A.2, S = 111K

¶3/2

288 + 111

473 + 111

= 1.438Thus

1.78× 10−5N· s/ m2¢

Situation: Kinematic viscosity of methane at 15o

C and 1 atm is 1.59 × 10−5m2/ s

Properties: From Table A.2, S = 198 K

νo

µo

ρoρIdeal-gas law

µT

To

¶5/2

To+ S

T + Sso

¶5/2

288 + 198

473 + 198

= 1.252and

23

Situation: Nitrogen at 59o

F has a dynamic viscosity of 3.59 × 10−7lbf· s/ ft2

ft2

¶

µ = 4.30× 10−7 lbf-s/ft2

Situation: Helium at 59o

F has a kinematic viscosity of 1.22 × 10−3ft2/ s

Situation: Information about propane is provided in the problem statement.Find: Sutherland’s constant.

S

To

= 0.964

Situation: Information about ammonia is provided in the problem statement.

Find: Sutherland’s constant

Situation: Information about SAE 10W30 motor oil is provided in the problem ment.

state-Find: The viscosity of motor oil at 60◦C, µ(60oC), using the equation µ = Ceb/T.APPROACH

solve for the unknown value of viscosity

ANALYSIS

µ

·b(1

Situation: Information about grade 100 aviation oil is provided in the problem ment

state-Find: µ(150oF), using the equation µ = Ceb/T

APPROACH

solve for the unknown value of viscosity

ANALYSIS

µ

·b(1

ft2

¶

29

Situation: This problem involves the creation of a computer program to find

Find: Develop a computer program and carry out the activities described in thetextbook

Situation: Oil (SAE 10W30) fills the space between two plates Plate spacing is

∆y = 1/8 = 0.125 in

Lower plate is at rest Upper plate is moving with a speed u = 25 ft/ s

Find: Shear stress

Assumptions: 1.) Assume oil is a Newtonian fluid 2.) Assume Couette flow (linearvelocity profile)

¶

=

µ5.2× 10−4 lbf· s

Situation: Air and water at 40◦C and absolute pressure of 170 kPa

Find: Kinematic and dynamic viscosities of air and water

Water data from Table A.5, µwater = 6.53× 10−4 N·s/m2, ρwater = 992 kg/m3.APPROACH

dynamic and absolute viscosity

Situation: Water flows near a wall The velocity distribution is

·

a³yb

´1/6¸

b1/6

16y5/6

6b

µby

µby

Situation: Information is provided in problem statement.

Find: Shear stress at walls

0.00 0.02 0.04 0.06 0.08 0.10

Velocity

Situation: Information is provided in problem statement.

Find: (a) Maximum and minimum shear stress

(b) Maximum shear stress at wall

Situation: Glycerin is flowing in between two stationary plates The plate spacing is

a.) Velocity and shear stress at12 mm from wall (i.e at y = 12 mm)

b.) Velocity and shear stress at the wall (i.e at y = 0 mm)

APPROACH

shear stress using τ = µ (du/dy), where the rate-of-strain (i.e the derivative du/dy)

is found by differentiating the velocity distribution

du

ddy

µ

2µ

dpdx

¶ddy

¶

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2 As expected, the shear stress at the wall is larger than the shear stress awayfrom the wall This is because shear stress is maximum at the wall and zeroalong the centerline (i.e at y = B/2)

Situation: Laminar flow occurs between two parallel plates–details are provided inthe problem statement.

Find: Is the maximum shear greater at the moving plate or the stationary plate?ANALYSIS

Observation of the velocity gradient lets one conclude that the pressure gradient dp/ds

is negative Also ut is negative Therefore |τh| > |τ0| The maximum shear stressoccurs at y = H

Maximum shear stress occur along the moving plate where y = H

39

Situation: Laminar flow occurs between two parallel plates–details are provided inthe problem statement.

Find: Position (y) of zero shear stress

Situation: Laminar flow occurs between two parallel plates–details are provided inthe problem statement.

Find: Derive an expression for plate speed (ut)to make the shear stress zero at y = 0.ANALYSIS

Situation: A damping device is described in the problem statement.

Find: Torque on shaft

Situation: Oxygen at 50◦F and 100◦F.

Find: Ratio of viscosities: µ100

Situation: This problem involves a cylinder falling inside a pipe that is filled with oil.Find: Speed at which the cylinder slides down the pipe.

C) = 0.35 N·s/m2.ANALYSIS

Situation: This problem involves a cylinder falling inside a pipe–details are provided

in problem statement

Find: Weight of cylinder

Situation: A disk is rotated very close to a solid boundary–details are provided inproblem statement.

Find: (a) Ratio of shear stress at r = 2 cm to shear stress at r = 3 cm

(b) Speed of oil at contact with disk surface

(c) Shear stress at disk surface

Assumptions: Linear velocity distribution: dV /dy = V /y = ωr/y

Situation: A disk is rotated very close to a solid boundary–details are provided inproblem statement.

Find: Torque to rotate disk

Assumptions: Linear velocity distribution: dV /dy = V /y = ωr/y

Situation: In order to provide damping for an instrument, a disk is rotated in acontainer of oil.

Find: Derive an equation for damping torque as a function of D, S, ω and µ

dTone side = r[τ (2πrdr)]

= r[(µrω/s)(2πrdr)]

dTboth sides = 4(rπµω/s)r3drIntegrate

Situation: One type of viscometer involves the use of a rotating cylinder inside a fixed

Find: (a) Design a viscometer that can be used to measure the viscosity of motor oil.Assumptions:

2 Let I.D of fixed cylinder = 9.00 in = 0.7500 ft

3 Let O.D of rotating cylinder = 8.900 in = 0.7417 ft

Let the applied torque, which drives the rotating cylinder, be produced by a force

force is produced by a weight and pulley system shown in the sketch below

h rc

∆r WPulley

where T = applied torque

rc= outer radius of rotating cylinder

49

τ As where As =area in shear = 2πrch

W rs = rc3µω(2π)h/∆rOr

The weight W will be arbitrarily chosen (say 2 or 3 oz.) and ω will be determined by

ω = Vfall/rs Equation (3) then becomes

1 Specify dimensions of all parts of the instrument

2 Neglect friction in bearings of pulley and on shaft of cylinder

3 Neglect weight of thread or monofilament line

4 Consider degree of accuracy

Situation: Water is subjected to an increase in pressure.Find: Pressure increase needed to reduce volume by 1%.

Situation: Very small spherical droplet of water.

Find: Pressure inside

Situation: A spherical soap bubble has an inside radius R, a wall-thickness t, andsurface tension σ.

Find: (a) Derive a formula for the pressure difference across the bubble

(b) Pressure difference for a bubble with a radius of 4 mm

Assumptions: The effect of thickness is negligible, and the surface tension is that ofpure water

Situation: A water bug with 6 legs, each with a contact length of 5 mm, is balanced

on the surface of a water pond

Find: Maximum mass of bug to avoid sinking

Properties: Surface tension of water, from Table A.5, σ = 0.073 N/m

Cross section

of bug leg

Assume is small Then cos =1; F cos = F θ

0.00438 N9.81 m2/ s

55

Situation: A water column in a glass tube is used to measure pressure.

Part of the water column height is due to pressure in a pipe, and part is due tocapillary rise

Additional details are provided in the problem statement

Find: Height of water column due to surface tension effects

Properties: From Table A.5: surface tension of water is 0.005 lbf/ft

Situation: Two vertical glass plates are spaced 1 mm apart.

Find: Capillary rise (h) between the plates

APPROACH

Apply equilibrium, then the surface tension force equation

ANALYSIS

θ σ

Situation: A spherical water drop has a diameter of 1-mm.

Find: Pressure inside the droplet

APPROACH

Apply equilibrium, then the surface tension force equation

ANALYSIS

Equilibrium (half the water droplet)

Situation: A tube employing capillary rise is used to measure temperature of water.Find: Size the tube (this means specify diameter and length).

Spe-59

Situation: A glass tube is immersed in a pool of mercury–details are provided in theproblem statement.

Find: Depression distance of mercury: d

γdSubstitute in values

= 0.0118 m

∆h = 11.8 mm

Situation: A soap bubble and a droplet of water both with a diameter of 2mm, falling

in air The value of surface tension is equal

Find: Which has the greater pressure inside

ANALYSIS

The soap bubble will have the greatest pressure because there are two surfaces (twosurface tension forces) creating the pressure within the bubble The correct choice isa)

61