Baixe Thermodynamics by Yunus Cengel 5th Edition [Solution Manual]-1 e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Mecânica, somente na Docsity!
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles.
1-2C 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.
1-3C There is no truth to his claim. It violates the second law of thermodynamics.
Mass, Force, and Units
1-4C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. 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 is 1 lbf.
1-5C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-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 is 1 kg-force.
1-6C There is no acceleration, thus the net force is zero in both cases.
1-7 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/m^3.
Analysis The mass of the water in the tank and the total mass are m tank = 3 kg
V =0.2 m^3
mw = ρV = (1000 kg/m H 2 O
(^3) )(0.2 m (^3) ) = 200 kg
m total = mw + m tank = 200 + 3 = 203 kg
Thus,
1991 N
1 kg m/s
1 N
(203 kg)(9.81m/s^2 ) (^2) =
W = mg =
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/m^3.
ROOM AIR
6X6X8 m^3
Analysis The mass of the air in the room is
m = ρ V=(1.16 kg/m^3 )(6× 6 × 8 m^3 )= 334.1 kg
Thus,
=^3277 N
= =^22
1 kg m/s
1 N
W mg (334.1kg)(9.81m/s )
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 1% 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 99. W (^) s = 0 99. mg (^) s =0 99. ( m )( .9 807 )
Substituting,
- 99 ( 9. 81 )= ( 9. 81 − 3. 32 × 10 −^6 z )→ z = 29,539 m Sea level
1-10E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the spring and beam scales in space.
Analysis ( a ) A spring scale measures weight, which is the local gravitational force applied on a body:
= 25.5^ lbf
= =^22
32.2lbmft/s
1 lbf W mg (150lbm)(5.48ft/s )
( b ) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scale will read what it reads on earth,
W = 150 lbf
1-11 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 the Newton's second law, the force applied is
= 5297 N
= = = ×^22
1 kgm/s
1 N
F ma m ( 6 g) (90kg)(6 9.81m/s )
1-14 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 given to be 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
× = 0.41%
× =
%Reduction in weight %Reductionin 100 g
g g
Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
Systems, Properties, State, and Processes
1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system.
1-16C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system.
1-17C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
1-18C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight.
1-19C A process during which a system remains almost in equilibrium at all times is called a quasi- equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes.
1-20C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric.
1-21C The state of a simple compressible system is completely specified by two independent, intensive properties.
1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated room is a simple compressible system.
1-23C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries.
1-24C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O =
1000 kg/m^3 ). That is, SG = ρ/ ρH2O. When specific gravity is known, density is determined from
ρ = SG× ρ H2O.
1-25 EES The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be 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 sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km.
Properties The density data are given in tabular form as
r , km z , km^ ρ, kg/m^3
0 5 10 15 20 25
0
1
z, km
, kg/m ρ
3
Analysis Using EES, (1) Define a trivial function rho= a+z in 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 + cz^2 = 1.20252 – 0.101674 z + 0.0022375 z^2 for the unit of kg/m^3 , (or, ρ( z ) = (1.20252 – 0.101674 z + 0.0022375 z^2 )× 109 for the unit of kg/km^3 )
where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m^3.
( b ) The mass of atmosphere can be evaluated by integration to be
4 [ ( 2 ) / 2 ( 2 ) / 3 ( 2 ) / 4 / 5 ]
4 5 0
2 3 0 0
2 0 0
2 0
2 0
2 0
2 0
2 0
2 0
ar h r a br h a br cr h b cr h ch
m dV a bz cz r z dz a bz cz r rz z dz
h z
h z V = + + + + + + + +
= =
where r 0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 10^9 for the density unity kg/km^3 , the mass of the atmosphere is determined to be
m = 5.092 × 1018 kg
Discussion Performing the analysis with excel would yield exactly the same results.
EES Solution for final result:
a=1. b=-0. c=0. r= h= m=4pi(ar^2h+r(2a+br)h^2/2+(a+2br+cr^2)h^3/3+(b+2cr)h^4/4+ch^5/5)*1E+
Pressure, Manometer, and Barometer
1-34C The pressure relative to the atmospheric pressure is called the gage pressure , and the pressure relative to an absolute vacuum is called absolute pressure.
1-35C The atmospheric 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.
1-36C 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.
1-37C 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 will be the same.
1-38C Pascal’s principle 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.
1-39C 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.
1-40 The pressure in a vacuum chamber is measured by a vacuum gage. The absolute pressure in the chamber is to be determined.
Analysis The absolute pressure in the chamber is determined from
35 kPa P abs = P atm− P vac= 92 − 35 = 57 kPa Pabs
Patm = 92 kPa
1-41E The pressure in a tank is measured with a manometer by measuring the differential height of the manometer fluid. The absolute pressure in the tank is to be determined for the cases of the manometer arm with the higher and lower fluid level being attached to the tank.
28 in
P atm = 12.7 psia
SG = 1.
Air
Assumptions The fluid in the manometer is incompressible.
Properties The specific gravity of the fluid is given to be SG = 1.25. The density of water at 32°F is 62.4 lbm/ft^3 (Table A-3E)
Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, 3 3
ρ=SG ×ρH 2 O =(1.25)(62. 4 lbm/ft )= 78. 0 lbm/ft
The pressure difference corresponding to a differential height of 28 in between the two arms of the manometer is
- 26 psia 144 in
1 ft 32.174lbmft/s
1 lbf (78 lbm/ft)(32.174ft/s)(28/12ft) 2
2 2
∆ P = ρ gh =
Then the absolute pressures in the tank for the two cases become:
( a ) The fluid level in the arm attached to the tank is higher (vacuum):
P abs = P atm− P vac= 12. 7 − 1. 26 = 11.44 psia
( b ) The fluid level in the arm attached to the tank is lower:
P abs = P gage+ P atm= 12. 7 + 1. 26 = 13.96 psia
Discussion Note that we can determine whether the pressure in a tank is above or below atmospheric pressure by simply observing the side of the manometer arm with the higher fluid level.
1-44 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined.
Assumptions The variation of the density of the liquid with depth is negligible.
Analysis The gage pressure at two different depths of a liquid can be expressed as
P 1 = ρ gh 1 and P 2 = ρ gh 2
h 2
2
h 1
1
Taking their ratio,
1
2 1
2 1
2 h
h gh
gh P
P
Solving for P 2 and substituting gives
= = ( 28 kPa)= 84 kPa 3 m
9 m 1 1
2 (^2) h P
h P
Discussion Note that the gage pressure in a given fluid is proportional to depth.
1-45 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined.
Assumptions The liquid and water are incompressible.
Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m^3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water,
SG (0.85)(100 0 kg/m^3 ) 850 kg/m^3
ρ= × ρ H 2 O = =
Analysis ( a ) Knowing the absolute pressure, the atmospheric pressure can be determined from P atm
h
= 96.0kPa P
2
3 2
atm
1000 N/m
1 kPa (145kPa) (1000kg/m)(9.81m/s )(5m)
P P ρ gh
( b ) The absolute pressure at a depth of 5 m in the other liquid is
= 137.7 kPa
2
3 2
atm
1000 N/m
1 kPa (96.0kPa) (850kg/m )(9.81m/s )(5m)
P P ρ gh
Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected.
1-46E It is to be shown that 1 kgf/cm^2 = 14.223 psi.
Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have
- 20463 lbf 1 N
0.22481lbf 1 kgf 9.80665N (9.80665N) =
and
= = 14.223^ psi
= =^2
2 (^2 2214). 223 lbf/in 1 in
2.54cm 1 kgf/cm 2. 20463 lbf/cm ( 2. 20463 lbf/cm )
1-47E The weight and the foot imprint area of a person are given. The pressures this man exerts on the ground when he stands on one and on both feet are to be determined.
Assumptions The weight of the person is distributed uniformly on foot imprint area.
Analysis The weight of the man is given to be 200 lbf. Noting that pressure is force per unit area, the pressure this man exerts on the ground is
( a ) On both feet: = = 2.78psi ×
= = 2. 78 lbf/in 2 36 in
200 lbf 2
2 A^2
W
P
( b ) On one foot: = = = 5. 56 lbf/in = 5.56psi 36 in
200 lbf 2 A^2
W
P
Discussion Note that the pressure exerted on the ground (and on the feet) is reduced by half when the person stands on both feet.
1-48 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined.
Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible.
Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be
= 1.37m^2
2
1000 N/m
1 kPa 1 kgm/s
1 N
0.5kPa
(70kg)(9.81m/s ) P
mg P
W
A
Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size.
1-52 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined. 780 mbar
h =?
Assumptions The variation of air density and the gravitational acceleration with altitude is negligible.
Properties The density of air is given to be ρ = 1.20 kg/m^3.
Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain 930 mbar
(0.930 0.780) bar 100,000N/m
1 bar 1 kg m/s
1 N
(1.20kg/m)(9.81m/s )()
2 2
3 2
air bottom top
air bottom top
h
gh P P
W A P P
It yields h = 1274 m
which is also the distance climbed.
1-53 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined.
Assumptions The variation of air density with altitude is negligible.
Properties The density of air is given to be ρ = 1.18 kg/m^3. The density of mercury is 13,600 kg/m^3.
730 mmHg
h
755 mmHg
Analysis Atmospheric pressures at the top and at the bottom of the building are
100.70 kPa
1000 N/m
1 kPa 1 kg m/s
1 N
(13,600kg/m)(9.807m/s )(0.755m)
97.36 kPa
1000 N/m
1 kPa 1 kg m/s
1 N
(13,600kg/m)(9.807m/s )(0.730m)
2 2
3 2
bottom bottom
2 2
3 2
top top
P g h
P ρgh
Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain
(100.70 97.36) kPa 1000 N/m
1 kPa 1 kgm/s
1 N
(1.18kg/m)(9.807m/s )()
2 2
3 2
air bottom top
air bottom top
h
gh P P
W A P P
It yields h = 288.6 m
which is also the height of the building.
1-54 EES Problem 1-53 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.
P_bottom=755"[mmHg]" P_top=730"[mmHg]" g=9.807 "[m/s^2]" "local acceleration of gravity at sea level" rho=1.18"[kg/m^3]" DELTAP_abs=(P_bottom-P_top)CONVERT('mmHg','kPa')"[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rhogh/1000 "[kPa]" "Equ. 1-16. Delta P due to the air fluid column height, h, between the top and bottom of the building." "Instead of dividing by 1000 Pa/kPa we could have multiplied rhog*h by the EES function, CONVERT('Pa','kPa')" DELTAP_abs=DELTAP_h
SOLUTION Variables in Main DELTAP_abs=3.333 [kPa] DELTAP_h=3.333 [kPa] g=9.807 [m/s^2] h=288 [m] P_bottom=755 [mmHg] P_top=730 [mmHg] rho=1.18 [kg/m^3]
1-55 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined.
Assumptions The variation of the density of water with depth is negligible.
Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m^3. P atm
Sea h
P
Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kg/m^3 :
SG (1.03)(100 0 kg/m^3 ) 1030 kg/m^3
ρ= × ρ H 2 O = =
The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location:
= 404.0 kPa
2
3 2
atm
1000 N/m
1 kPa (101kPa) (1030kg/m)(9.807m/s )(30m)
P P ρ gh
1-58 EES Problem 1-57 is reconsidered. The effect of the spring force in the range of 0 to 500 N on 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 problem is solved using EES, and the solution is given below.
g=9.807"[m/s^2]" P_atm= 95"[kPa]" m_piston=4"[kg]" {F_spring=60"[N]"} A=35CONVERT('cm^2','m^2')"[m^2]" W_piston=m_pistong"[N]" F_atm=P_atmACONVERT('kPa','N/m^2')"[N]" "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston"[N]" P_gas=F_gas/A*CONVERT('N/m^2','kPa')"[kPa]"
Fspring [N] Pgas [kPa] 0 106. 55.56 122. 111.1 138 166.7 153. 222.2 169. 277.8 185. 333.3 201. 388.9 217. 444.4 233. 500 249.
0 100 200 300 400 500
100
120
140
160
180
200
220
240
260
F
spring
[N]
P
gas
[kPa]
1-59 [ Also solved by EES on enclosed CD ] Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water.
Properties The densities of water and mercury are given to be ρwater = 1000 kg/m^3 and be ρHg = 13,600 kg/m^3.
Analysis The gage pressure is related to the vertical distance h between the two fluid levels by
g
P
P gh h
gage gage = → =
( a ) For mercury,
0. 60 m 1 kN
1000 kg/ms 1 kPa
1 kN/m (13,600kg/m )(9.81m/s )
80 kPa^22 3 2
gage =
g
P
h
ρ Hg
80 kPa
h
AIR
( b ) For water,
8.16 m 1 kN
1000 kg/ms 1 kPa
1 kN/m (1000kg/m)(9.81m/s )
80 kPa^22 3 2 HO
gage 2
g
P
h
1-61 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined.
0.60 m
P atm = 98 kPa
AIR
Properties The density of oil is given to be ρ = 850 kg/m^3.
Analysis The absolute pressure in the tank is determined from
= 103 kPa
2
3 2
atm
1000 N/m
1 kPa (98kPa) (850kg/m )(9.81m/s )(0.60m)
P P ρ gh
1-62 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined.
Properties The density of mercury is given to be ρ = 13,600 kg/m^3.
AIR
P
15 mm
Analysis ( a ) The pressure in the duct is above atmospheric pressure
since the fluid column on the duct side is at a lower level.
( b ) The absolute pressure in the duct is determined from
= 102 kPa
2 2
3 2
atm
1000 N/m
1 kPa 1 kgm/s
1 N
(100kPa) (13,600kg/m )(9.81m/s )(0.015m)
P P ρ gh
1-63 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined.
AIR
P
45 mm
Properties The density of mercury is given to be ρ = 13,600 kg/m^3.
Analysis ( a ) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.
( b ) The absolute pressure in the duct is determined from
= 106 kPa
2 2
3 2
atm
1000 N/m
1 kPa 1 kgm/s
1 N
(100kPa) (13,600kg/m)(9.81m/s )(0.045m)
P P ρ gh
1-64 The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column.
Assumptions Both mercury and water are incompressible substances.
Properties We take the densities of water and mercury to be 1000 kg/m^3 and 13,600 kg/m^3 , respectively.
Analysis Using the relation P = ρ gh for gage pressure, the high and low pressures are expressed as
10.7 kPa
16.0kPa
1000 N/m
1 kPa 1 kg m/s
1 N
(13,600kg/m)(9.81m/s)(0.08m)
1000 N/m
1 kPa 1 kg m/s
1 N
(13,600kg/m)(9.81m/s)(0.12m)
2 2
3 2 low low
2 2
3 2 high high
^ =
P gh
P gh
Noting that 1 psi = 6.895 kPa,
2.32 psi 6.895 kPa
1 psi high (16.^0 Pa) =
P = and 1.55psi 6.895 kPa
1 psi low (10.7Pa) =
P =
For a given pressure, the relation P = ρ gh
water
can be expressed for
mercury and water as P = ρwater gh and.
Setting these two relations equal to each other and solving for water height gives
P = ρmercury gh mercury
water mercury^ h
mercury
P water gh water mercury gh mercury h water h
ρ
ρ =ρ =ρ → =
Therefore,
1.09 m
1.63m
( 0. 08 m) 1000 kg/m
13 , 600 kg/m
( 0. 12 m) 1000 kg/m
13 , 600 kg/m
3
3 mercury,low water
mercury water,low
3
3 mercury,high water
mercury water,high
h h
h h
Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices.
PROPRIETARY MATERIAL. © 2006 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.
