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Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 1 INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) 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: 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.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-2 Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.

1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.

1-3C There is no truth to his claim. It violates the second law of thermodynamics.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system. 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.

1-5C 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.

1-6C There is no acceleration, thus the net force is zero in both cases.

1-7E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives

W = mg ⎯ ⎯→ m =

210 lbf W = g 32.10 ft/s 2

⎛ 32.174 lbm ⋅ ft/s 2 ⎜ ⎜ 1 lbf ⎝

⎞ ⎟ = 210.5 lbm ⎟ ⎠

Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be 1 lbf ⎞ ⎛ W = mg = (210.5 lbm)(5.47 ft/s 2 )⎜ ⎟ = 35.8 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠

1-8 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 3

3

m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg

ROOM AIR 6X6X8 m3

Thus, ⎛ 1N W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3277 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 0.5% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z )

In our case, W = 0.995W s = 0.995mg s = 0.995(m)(9.81)

Substituting, 0.995(9.81) = (9.81 − 3.32 × 10

0 −6

z) ⎯ ⎯→ z = 14,774 m ≅ 14,770 m

Sea level

1-10 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

1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying Newton's second law, the weight is determined in various units to be ⎛ 1 kJ/kg ⋅ K ⎞ ⎟⎟ = 1.005 kJ/kg ⋅ K c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 1 kJ/kg ⋅ °C ⎠ ⎛ 1000 J ⎞⎛ 1 kg ⎞ ⎟⎟ = 1.005 J/g ⋅ °C c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟⎜⎜ ⎝ 1 kJ ⎠⎝ 1000 g ⎠ ⎛ 1 kcal ⎞ c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟ = 0.240 kcal/kg ⋅ °C ⎝ 4.1868 kJ ⎠ ⎛ 1 Btu/lbm ⋅ °F ⎞ ⎟⎟ = 0.240 Btu/lbm ⋅ °F c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 4.1868 kJ/kg ⋅ °C ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-5 1-12

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 ⎛ 1N W = mg = (3 kg)(9.79 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 29.37 N ⎟ ⎠

Then the net force that acts on the rock is

Fnet = Fup − Fdown = 200 − 29.37 = 170.6 N

Stone

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

a=

F 170.6 N ⎛⎜ 1 kg ⋅ m/s 2 = m 3 kg ⎜⎝ 1 N

⎞ ⎟ = 56.9 m/s 2 ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-6 1-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] 200

160

2

a [m/s2] 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21

a [m/s ]

m [kg] 1 2 3 4 5 6 7 8 9 10

120

80

40

0 1

2

3

4

5

6

m [kg]

7

8

9

10

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-7 1-14 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 = 25 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.

1-15 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 4 kW or 4 kJ/s. Then the total amount of electric energy used in 2 hours becomes

Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = 8 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy.

1-16 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.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-8 1-17 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...

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 1 INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) 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: 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.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-2 Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.

1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.

1-3C There is no truth to his claim. It violates the second law of thermodynamics.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system. 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.

1-5C 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.

1-6C There is no acceleration, thus the net force is zero in both cases.

1-7E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives

W = mg ⎯ ⎯→ m =

210 lbf W = g 32.10 ft/s 2

⎛ 32.174 lbm ⋅ ft/s 2 ⎜ ⎜ 1 lbf ⎝

⎞ ⎟ = 210.5 lbm ⎟ ⎠

Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be 1 lbf ⎞ ⎛ W = mg = (210.5 lbm)(5.47 ft/s 2 )⎜ ⎟ = 35.8 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠

1-8 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 3

3

m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg

ROOM AIR 6X6X8 m3

Thus, ⎛ 1N W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3277 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 0.5% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z )

In our case, W = 0.995W s = 0.995mg s = 0.995(m)(9.81)

Substituting, 0.995(9.81) = (9.81 − 3.32 × 10

0 −6

z) ⎯ ⎯→ z = 14,774 m ≅ 14,770 m

Sea level

1-10 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

1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying Newton's second law, the weight is determined in various units to be ⎛ 1 kJ/kg ⋅ K ⎞ ⎟⎟ = 1.005 kJ/kg ⋅ K c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 1 kJ/kg ⋅ °C ⎠ ⎛ 1000 J ⎞⎛ 1 kg ⎞ ⎟⎟ = 1.005 J/g ⋅ °C c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟⎜⎜ ⎝ 1 kJ ⎠⎝ 1000 g ⎠ ⎛ 1 kcal ⎞ c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟ = 0.240 kcal/kg ⋅ °C ⎝ 4.1868 kJ ⎠ ⎛ 1 Btu/lbm ⋅ °F ⎞ ⎟⎟ = 0.240 Btu/lbm ⋅ °F c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 4.1868 kJ/kg ⋅ °C ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-5 1-12

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 ⎛ 1N W = mg = (3 kg)(9.79 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 29.37 N ⎟ ⎠

Then the net force that acts on the rock is

Fnet = Fup − Fdown = 200 − 29.37 = 170.6 N

Stone

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

a=

F 170.6 N ⎛⎜ 1 kg ⋅ m/s 2 = m 3 kg ⎜⎝ 1 N

⎞ ⎟ = 56.9 m/s 2 ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-6 1-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] 200

160

2

a [m/s2] 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21

a [m/s ]

m [kg] 1 2 3 4 5 6 7 8 9 10

120

80

40

0 1

2

3

4

5

6

m [kg]

7

8

9

10

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-7 1-14 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 = 25 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.

1-15 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 4 kW or 4 kJ/s. Then the total amount of electric energy used in 2 hours becomes

Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = 8 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy.

1-16 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.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-8 1-17 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...

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