Fluid Mechanics Answers, Exercises of Fluid Dynamics

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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

Solutions Manual for

Fluid Mechanics: Fundamentals and Applications

Third Edition

Yunus A. Çengel & John M. Cimbala

McGraw-Hill, 2013

CHAPTER 1

INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL

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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

1-5C

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.

1-6C

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.

1-7C

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.

1-8C

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.

1-9C

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.

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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).

1-11C

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

Analysis The classical approach is a macroscopic approach , based on experiments or analysis of the gross behavior of a fluid, without knowledge of individual molecules, whereas the statistical approach is a microscopic approach based on the average behavior of large groups of individual molecules.

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

1-12C

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.

1-13C

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.

1-14C

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.

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Mass, Force, and Units

1-18C

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.

1-19C

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

Analysis The unit kilogram (kg) is the mass unit in the SI system , and it is sometimes called kg-mass , whereas kg- force (kgf) is a force unit. One kg-force is the force required to accelerate a 1-kg mass by 9.807 m/s^2. In other words, the weight of 1-kg mass at sea level on earth is 1 kg-force.

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

1-20C

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

Analysis Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit in the English system. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s^2. In other words, the weight of a 1-lbm mass at sea level on earth is 1 lbf.

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

1-21C

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.

1-22C

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.

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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  = 1000 kg/m^3.

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

mw = V = (1000 kg/m^3 )(0.18 m^3 ) = 180 kg

m total = mw + mtank = 180 + 6 = 186 kg Thus,

^1825 N

 ^22

1 kg m/s

1 N

W mg (186kg)(9.81m/s )

Discussion Note the unity conversion factor in the above equation.

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 ( 200 kg)( 9. 6 m/s^2 ) 1920 N

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

 9.81N

 ^22

1 kg m/s

1 N

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

 0.00981kN 

 ^22

1000 kg m/s

1 kN W mg (1kg)(9.81m/s)

W  mg (1 kg)(9.81m/s^2 ) 1 kg  m/s^2

^1 kgf 

9.81N

1 kgf 1 kg m/s

1 N

W mg (1 kg)(9.81m/s^2 ) 2

 71 lbm  ft/s^2  

  (32.2ft/s ) 1 kg

2.205lbm W mg (1 kg)^2

 2.21lbf 

 ^22

32.2lbm ft/s

1 lbf (32.2ft/s ) 1 kg

2.205lbm W mg (1kg)

m tank= 6 kg V = 0.18 m^3

H 2 O

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

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

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

2 2

1 N

6 g (90 kg)(6 9.81 m/s ) 5297 N

1 kg m/s

F ma m( )

5300 N

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.

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

2 2

1 N

(5 kg)(9.79 m/s ) 48 95 N

1 kg m/s

W mg.

49.0 N

Then the net force that acts on the rock is

Fnet  Fup  Fdown  150 48.95 101. 05 N

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

 20.2 m/s^2 

1 N

1 kgm/s 5 kg

101.05 N^2

m

F

a

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

Rock

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

Analysis The EES Equations window is printed below, followed by the Solution window.

W=mg "[N]" m=5 [kg] g=9.79 [m/s^2] "The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down "[N]" F_up=150 [N] F_down=W "[N]" "The acceleration of the rock is determined from Newton's second law." F_net=am "To Run the program, press F2 or click on the calculator icon from the Calculate menu"

SOLUTION Variables in Main a=20.21 [m/s^2] F_down=48.95 [N] F_net=101.1 [N] F_up=150 [N] g=9.79 [m/s^2] m=5 [kg] W=48.95 [N]

The final results are W = 49.0 N and a = 20.2 m/s^2 , 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.

Solution Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.

Properties The gravitational acceleration g is 9.807 m/s^2 at sea level and 9.767 m/s^2 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

% Reduction in weight % Reduction in 100 100 9 807

g.. g 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.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

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

V

t

V

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

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 m^3. 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 [m^3 ] 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 “m^3 ” for volume is to multiply the quantities t and V with the square of D. Therefore, the desired relation is

V = CD^2 Vt

where the constant of proportionality is obtained for a round hose, namely, C = π /4 so that V = (  D^2 /4) Vt.

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

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

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/s^2 )m/s = kgm^2 /s^3

Therefore, the independent quantities should be arranged such that we end up with the unit kgm^2 /s^3 for power. Putting the given information into perspective, we have

W  [ kgm^2 /s^3 ] is a function of m [kg], V [m/s], and t [s]

It is obvious that the only way to end up with the unit “kgm^2 /s^3 ” 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.

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,

105 kW  141 hp 

1 kW

1.341hp 1000 N m/s

1 kW W  F thrust V (1500N)(70.0m/s)

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.

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Solution A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better 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

12 ounce steak :  $9.26/kg 

0.45359 kg

1 lbm 1 lbm

16 oz 12 oz

UnitCost=

320 gram steak :

 $10.3/kg 

1 kg

1000 g 320 g

UnitCost=

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

Discussion Notice the unity conversion factors in the above equations.

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 20oC is  = 998.0 kg/m^3.

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

2.0 L 60 s 42.105 L/min t 2.85 s 1 min

  ^  

 ^ 

 (^) 42.1 Lpm

V

V

where we give our final answer to 3 significant digits, but retain 5 digits to avoid round-off error in the second part of the problem. Since density is mass per unit volume, mass flow rate is equal to volume flow rate times density. Thus,

  

3 998.0 kg/m^3 42.105 L/min 1 min^ 1 m 60 s 1000 L

m 

 ^ 

 (^) V  0.700 kg/s

Discussion We used one unity conversion ratio in the first calculation, and two in the second. If we were interested only in the mass flow rate, we could have eliminated the intermediate calculation by solving for mass flow rate directly, i.e.,

 

3 998.0 kg/m^3 2.0 L^ 1 m 0.700 kg/s 2.85 s 1000 L

m t

V

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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/s^2.

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  z , where z is the elevation.

  2   2

1 N 1 kJ 90.5 kg 9.807 m/s 1.80 m 1.5976 kJ 1 kg m/s 1000 N m

W  F  z  mg  z     (^)      ^ ^  

1.60 kJ

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

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

1.5976 kJ 1000 W 129.88 W 12.3 s 1 kJ/s

W

W

t

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

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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.

1-49C

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.

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

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

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

Answers : x = 1.556, y = 0.6254, z = 6.

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

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

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

x^3-y^2=10. 3xy+y=4.

Answers : x = 2.215 , y = 0.

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

Solution We are to determine a positive real root of the following equation using EES: 3.5 x^3 – 10 x 0.5^ – 3 x = 4.

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

3.5x^3-10x^0.5-3*x = -

Answer : x = 1.

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

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

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

x^2y-z=1. x-3y^0.5+x*z=- x+y-z=4.

Answers : x = 0.9149, y = 10.95, z = 7.

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