1. 1
Comments on
Thermodynamics: An Engineering Approach, 7th ed. (in SI and
English Units), McGraw-Hill, 2011
Yunus. A. Cengel and Michael. A. Boles
By: Mohsen Hassan vand
Saturday, April 12, 2014

2. 2
Notations
Pg x : Pg x
P x : Paragraph number x
L x : L number x
LB x : Lx from bottom
PB x : Px from bottom
: is suggested to be changed to
----------------------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------------------
1. In PREFACE
2. In PREFACE, Pg xix, L 2
hydroulic hydraulic
3. In PREFACE, Pg xx, P 1, L 7-8,
that numbers without units are meaningless that numbers without units are physically meaningless
Note: Numbers are mathematical entities and therefore, mathematically meaningful.
4. Pg 7, LB 8
….which is defined as the energy …. In the metric system, the amount of energy …
….which is defined as the heat …. In the metric system, the amount of heat …
Because energies other than heat, such as kinetic or potential energy or work do not raise the temperature.
5. Pg 7,
FIGURE 1–9 A body weighing 150 lbf on earth will weigh only 25 lbf on the moon.

3. 3
FIGURE 1–9 A body weighing 150 lbf on earth weighs only 25 lbf on the moon.
6. Pg 8, P 1, last L
… a rated power of 50kW will generate 120000 kWh of electricity per year.
… a rated power of 50kW will generate 120000 kWh of electrical energy per year.
Note : kWh is the unit for electrical energy, not for electricity (coulomb)
7. Pg 20, 0°C (273.15°C) 0°C (273.15°K)
with an uncertainly of _0.005°C with an uncertainty of _0.005°C)
the impurities in a gas absorbed by the walls …. the impurities in a gas adsorbed by the walls ….
- Please note the technical difference between the terms “absorption” and “adsorption”.
8. Pg 28,in EXAMPLE 1–7
1000 kg/m3
) ( 10.1 m) (1000 kg/m3
) ( 10.1 m)
9. Pg 33, on FIGURE 1–61
Solution is not a point ( in the figure, the peak point), it is a process. Therefore on the figure:
EASY WAY Solution I: EASY WAY
HARD WAY Solution II: HARD WAY
SOLUTION Answer
10. In problem 1-93E the following sentence is unnecessary and can be deleted:
Specific heat is defined as the amount of energy needed to increase the temperature of a unit mass of a
substance by one degree.
If you don’t delete it, it should be corrected(energy heat).
11. In problem 1-117E FIGURE P1–117 FIGURE P1–112E
12. The lines alignment or indentation of problem 1-124 should be justified.
13. In problem 1-125 : (CDrag Afront,….) (CDrag , Afront,….)
14. In Pg 53, Section 2-2, L 1
Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electric, …

4. 4
Energy can exist in numerous forms such as thermal, mechanical (pressure, kinetic, potential), electric,…
15. In Pg 54, in FIGURE 2–4,
On the figure: Steam Fluid
Under the figure: … the flow of steam… … the flow of fluid…
16. In Pg 55, in FIGURE 2–6,
It is suggested to replace the following symbol
with something else(e.g, a nuclear power plant), resembling a peaceful application of nuclear energy
17. In Pg 56, FIGURE 2–7,
It seems that the objective of the figure is to show how a disorganized form of energy is converted to an
organized form. But in the hydro-dam shown, it is the gravitational potential energy ( a macroscopic and
organized form of energy) that is converted to kinetic energy(again, a macroscopic and organized form of energy).
Therefore it is suggested the figure to be replaced with something such as the following:
18. In Pg 57, Ls 3-4
complete fission of 1 kg of uranium-235 releases 6.73 × 1010
kJ of heat, which … 3000 tons of coal
complete fission of 1 kg of uranium-235 releases 8.314 × 1010
kJ of heat, which … 3700 tons of coal
Note : 8.314 × 1010
kJ is the energy converted into heat in an operating thermal nuclear reactor.
19. In Pg 57, P 2, L 4,
… hydrogen atoms into a helium atom. … hydrogen nuclei into a helium nucleus.
In last P,
But such high temperatures … in the stars or … atomic bombs …. a small atomic bomb.
But such high temperatures … in the centers of stars or … nuclear bombs …. a small nuclear bomb.
20. In Pg 57, EXAMPLE 2–1, Assumptions 2

5. 5
Nuclear fuel is completely converted to thermal energy.
Nuclear energy is completely converted to thermal energy.
21. Pg 63, LB 5
A small change in volume, for example, is represented by dV,
An infinitesimal change in volume, for example, is represented by dV,
22. Pg 65, in the L after the relation(2-18)
….and I is the number of electrical charges… ….and I is the amount of electrical charges…
23. Pg 67, FIGURE 2–32
Note: The displacement caused by F is x not dx [or, dx is caused by dF not by F].
24. Pg 68, L 2
by replacing pressure P by its counterpart in solids, normal stress /n F A ,
by substituting
.nF A
for solids ( is the )n normal stress
Because in the work expression, relation (2–30), there is no pressure P term to be replaced by its …
25. In Pg 69, the section Nonmechanical Forms of Work is a subsection of the section 2–5 MECHANICAL
FORMS OF WORK.
It seems not logical to put Nonmechanical Forms of Work under the title MECHANICAL FORMS OF
WORK.
26. Pg 71 on the FIGURE 2–42, and also FIGURE 2–43
(Adiabatic) Adiabatic
27. Pg 73, item 3
Mass Flow, m Mass Flow, m
Also, in the first L under item 3,
Noting that energy can be transferred in the forms of heat, work, and mass,
Noting that energy can be transferred in the forms of heat, work, and by mass,
28. Pg 78, in relation (2–41)
Performance Efficiency
Because the title of the section 2-7 is ENERGY CONVERSION EFFICIENCIES, not ENERGY

6. 6
CONVERSION PERFORMANCE
29. Pg 81, in last paragraph, 0.54 g of hydrocarbons 0.54 g of unburned hydrocarbons
30. Pg 83, on the FIGURE 2–60
V1 ×0 P1×Patm P2×Patm V1 =0 P1=Patm P2=Patm
31. Pg 85, in EXAMPLE 2–16 In the following formula :
Elect,out should be corrected to shaft,out
Please see relation (2-46).
32. Pg 86, L 6 1/0.93.2 1/0.932
33. Pg 88, last P , L 1
nitric oxides nitrogen oxides
because nitric oxide is nitrogen monoxide, (NO), but nitrogen oxide can refer to a binary compound of
oxygen and nitrogen, or a mixture of such compounds:
Nitric oxide, also known as nitrogen monoxide, (NO), Nitrogen dioxide (NO2), Nitrous oxide (N2O),
Nitrosylazide (N4O), Nitrate radical (NO3), Dinitrogen trioxide (N2O3), Dinitrogen tetroxide (N2O4),
Dinitrogen pentoxide (N2O5), Trinitramide (N(NO2)3) (source: Wikipedia)
34. In Pg 89, under the title “The Greenhouse Effect:…”, it is written that: “ the glass allows the solar
radiation to enter freely but blocks the infrared radiation emitted by the interior surfaces. This causes a rise
in the interior temperature as a result of the thermal energy buildup in the car”.
I think that the temperature rise inside the car is mainly due to the temperature rise of its roof and its body.
Because even if we use car sunshade, that is if we completely block the solar radiation entrance, the car
interior gets hot again(I know this from experience). The car body is made of a metal that absorbs the solar
radiation efficiently. After getting hot, the body (roof and body) transmits its thermal energy partially toward
car inside.
35. Pg 90, L 1, 360 ppm (or 0.36 percent) 360 ppm (or 0.036 percent)
36. Pg 90, LB 14,
Therefore, a car emits about 12,000 lbm of CO2 ….. weight of a typical car
Therefore, a car emits about 12,000 lbm of CO2 ….. mass of a typical car

7. 7
37. Pg 93, in P 3, Ls 1-2
Convection is called forced convection if the fluid is forced to flow in a tube or over a surface by external
means such as a fan, pump, or the wind.
Convection is called forced convection if the fluid is forced to flow in a tube or over a surface by external
means such as a fan or a pump.
Wind itself is basically caused by natural convection.
38. Pg 97, left column , last L,
The general mass and energy balances… The general energy balance …
39. Pg 98, in problem 2-13
Determine the mechanical energy of air… Determine the mechanical energy of wind …
40. Pg 100, problem 2-39 from stop from rest
41. Pg 103, problem 2-76
The inlet and outlet diameters of the pipe are 8 cm and 12 cm
The inlet and outlet diameters of the pipe are 12 cm and 8 cm
It seems that because the outlet pressure is greater than inlet pressure, outlet diameter should be less than
inlet diameters( due to strength of materials aspects)
The figure should also be corrected.
42. Pg 103, problem 2-88
… and consumes …… 1200 therms of natural gas.
… and consumes …… 1200 therms of natural gas per year.
43. Pg 108, problem 2-133 60 km/s 60 km/h
44. Pg 109, problem 2-144, 7-m long,10-m wide, 10-m long,7-m wide,
45. Pg 109, problem 2-145, … that reviews that… … that reviews the…
46. In Pg 112, inside the FIGURE 3–2 (b), Air Vapor
47. Pg 114, PB 2, L 1
Midway about the vaporization line (state 3, Fig. 3–8)
Midway about the vaporization line (state 3, Fig. 3–11)
48. Pg 116, TABLE 3–1

8. 8
Saturation (boiling) pressure of water at various temperatures
Saturation (sublimation and boiling) pressure of water at various temperatures
Note: Under 0.61 kPa, there is no boiling but sublimation
49. Pg 116, P 3, L 3
Tsat = f (Psat). A plot of Tsat versus Psat, such as the one given for water in Fig. 3–12
Psat= f (Tsat). A plot of Psat versus Tsat, such as the one given for water in Fig. 3–12
50. In Pg123, FIGURE 3–24 and FIGURE 3–25 Volume Specific Volume
51. In Pg 125, FIGURE 3–28 Specific temperature Temperature
52. In Pg 128, FIGURE 3–35 Sat. liquid vg Sat. vapor vg
53. In Pg 130, P 2, L 1
In the region …above the critical point temperature, a substance exists as superheated vapor.
In the region …under the critical point temperature, a substance exists as superheated vapor.
54. In Pg 133, middle parts of the Pg
In our case, the given u value is 1600 In our case, the given u value is 1600 kJ/kg
55. In Pg 135, L 2 before the relation (3-11) molecular weight molecular mass
56. In Pg 136, in EXAMPLE 3–10
220+90=315 kPa 220+95=315 kPa and (25 + 273K) (25 + 273) K
57. In Pg 137, Ls 6, 7
in the vicinity of the critical point and the saturated vapor L (over 100 percent).
in the vicinity of the critical point (over 100 percent)and the saturated vapor L.
58. In Pg 139, item 1,
At very low pressures …temperature (Fig. 3–52), At very low pressures …temperature (Fig. 3–52).
59. Pg 141, section 3-8, L 5
but we shall discuss only three: …. but we shall discuss only four: …. and Virial Equation of State
Note: Virial Equation of State has also been discussed in the text.
60. Pg 142, L 2,

9. 9
including two of the effects not considered in the ideal-gas model
including two of the effects not existed in the ideal-gas model
61. Pg 154, problem 3-44 ,
…both the initial, intermediate, and final states of the water
…the initial, intermediate, and final states of the water
Note: The number of initial, intermediate, and final states are three states, not two.
62. Pg 154, problem 3-47 , …heat transfer to the pan …heat transfer to the water
63. Pg 155, problem 3-56E , psi psia
64. Pg 155, problem 3-61 , (a) The volume of the tank, (a) The volume of the cylinder,
65. Pg 156, problem 3-78 , 2.21m3 2.21m3
66. Pg 159, problem 3-120E , 0.5 Ibm 0.5 lbm
67. Pg 159, problem 3-125 ,
Determine the final pressure in the tank. Determine the final pressure in the tanks.
68. Pg 166 and 167, in EXAMPLEs 4–2 and 4–3, there are two items under the same title, that is, under the title
Analysis. It is suggested the two items in each case to be combined together.
69. Pg171, EXAMPLE 4–5
Assumptions 1 The tank is stationary…. Assumptions 1 The device is stationary….
70. Pg 174, bottom of the EXAMPLE 4–6, kJkg] kJ/kg]
71. Pg 175, in FIGURE 4–20 Chage change
72. Pg 179, EXAMPLE 4–7,
where a = 28.11, b = 0.1967 × 10-2
, c =0.4802 × 10-5
, and d = -1.966 × 10-9
.
where a = 28.11 kJ/kmol.K , b = 0.1967 × 10-2
kJ/kmol.K2
, c =0.4802 × 10-5
kJ/kmol.K3
, and d = -1.966 ×
10-9
kJ/kmol.K4
.
73. Pg 185, EXAMPLE 4–11, in the last relation 0.001 m3
kg 0.001 m3
/kg
74. Pg 185, EXAMPLE 4–12 , L 1

10. 10
…. an insulated tank …. an insulated rigid tank
Also,
75. Pg 186, EXAMPLE 4–13 , Heating of Aluminum Rods in a Furnace
Considering the title of the example, it is suggested to change all “oven”’s in the text of the example to
“furnace”.
76. Pg 188,
In the text, the sentence “ For an average male (30 years old, 70 kg, 1.8-m2
body surface area), the basal
metabolic rate is 84 W.” refers to a male, therefore, the picture in FIGURE 4–37, should be the picture of
a man ( a male) not a woman.
77. Pg193, Table 4-3,
The upper and lower limits …. body indexes of 19 and 25, respectively.
The upper and lower limits …. body indexes of 25 and, 19 respectively.
Also, in the same Pg,
78. Pg 198, FIGURE P4–25, Fluid Water
79. Pg 199, problem 4.38, 224 224
80. Pg 200, problem 4.43, 0.3 m3 0.3 m3
0.1 m3 0.1 m3
81. Pg 201, problem 4.66E,
Nitrogen gas to …. Nitrogen gas at ….
Nitrogen gas at 20 psia and 100 °F initially occupies ….
Nitrogen gas initially at 20 psia and 100 °F occupies ….
because when we say “initially occupies a volume of 1 ft3
in a rigid container… ” , it means that the gas
may occupy a different volume at final state. But the container is rigid, and the gas volume is the same both
at initial and final states.
82. Pg 202, problem 4.70E, A 3-ft3….. A 3-ft3
….

11. 11
83. Pg 203, problem 4.81,
in a cylinder device fitted with a piston-cylinder. in a cylinder device fitted with a piston.
84. Pg 205, problem 4-106,
Determine … weights of Candy … Answer: 6.5 kg Determine … masses of Candy … Answer: 6.5 kg
85. Pg 206, problem 4-115E, 19 BMI 25 19 BMI 25
86. Pg 209, problem 4-147, Ls 3 and 5, tank cylinder
87. Pg 210, problem 4-153, 682.1 k 682.1 K 680.9 k 680.9 K
88. Pg 212, problem 4-168,
1.5 kg of liquid … is to be heated at 95°C… 1.5 kg of liquid … is to be heated to 95°C…
89. Pg 212, problems 4-171, 4-172 (2.7 × 104
) 2.7 × 104
90. Pg 217, L 3,
dm m
in the same Pg, L 2, it is said that Ac is a point function.
But, in point functions we can attribute a value to the function at each point. We cannot attribute a surface
Ac to a point, Ac is an integrated quantity over a domain of points.
91. Pg 217, FIGURE 5–3,
The average velocity Vavg is defined as the average speed through a cross section.
The average speed Vavg is defined as the average speed through a cross section.
92. Pg 218, first L after relation (5-9),
…aare the total rates… …are the total rates…
93. Pg 219, relation (5-16),
94. Pg 220, L 3 before relation (5-20),
The conservation of mass relations is simplified…
The conservation of mass relation is simplified… or The conservation of mass relation can be simplified…
95. Pg 222, relation (3),
96. Pg 224, FIGURE 5–14 ,

12. 12
nonflowing fluid nonflowing fluid
In a nonflowing fluid the kinetic energy is zero.
97. Pg 233, bottom of the Pg,
98. Pg 249, FIGURE 5–53,
It is suggested:
99. Pg 250,
It is suggested that the relations (5-58) and (5-59) to be broken and written in two Ls.
100. Pg 251,
The expression “for each exit” can be deleted because it doesn’t add any information .
101. Pg 256, problem 5-53, 10 L/S 10 L/s
102. Pg 260, problem 5-94, item (c) adiabetic adiabatic
103. Pg 267, problem 5-152,
hot- water heater water heater See also Pg 227, last P, L 1.
104. In Pg 268, problem 5-155, it is said “ Each outlet has the same diameter as the inlet”, but Fig P5-155
doesn’t show this.
105. Pg 288, last P, L 4, the ratio the total… the ratio of the total…
106. Pg 292, FIGURE 6–28,
It is suggested that the upper right part of the figure to be modified as follows:

13. 13
Also, in FIGURE 6–28,
107. Pg 296, P 3, L 2, as shown in Fig. 6–33. as shown in Fig. 6–33c.
108. Pg 296, last P, L 6, As dT approaches zero, As T approaches zero,
109. Pg 297, last P 6, Ls 2 and 3,
…undergoing a constant-pressure (thus constant-temperature) phase-change process,
…undergoing a constant-pressure phase-change process(thus constant-temperature),
110. Pg 302, L 2 after relation (6–15)
Several functions (T) satisfy this equation… Any monotonic function (T) satisfy this equation…
111. Pg 303, L 1 before relation (6–17)
The temperatures on these two scales differ by a constant 273.15: T(°C)= T (K)- 273.15
The temperatures on these two scales differ by a constant 273.15 K: T(°C)= T (K)- 273.15 K
Note that in the relation T(°C)= T (K)- 273.15 K, there is a hidden dimensional coefficient as follows:
T(°C)=
°
[T (K) - 273.15 K]
112. Pg 308, last part of the EXAMPLE 6–6 (in Discussion part),
Carnot cycle …. Reversed Carnot
113. Pg 311, item 4, L 1,
This can be done by placing a flashlight into the refrigerator,
This can be done by placing an switched-ON flashlight into the refrigerator,

14. 14
114. Pg 314, left column, P 4,
Any device that violates the first or …. Any device that may violate the first or….
115. Pg 314, right column, P 3,
The thermal efficiency of a Carnot heat engine, as well as all other reversible heat engines, is given by
This is the maximum efficiency a heat engine operating between two reservoirs at temperatures TH and TL
can have.
The thermal efficiency of a Carnot heat engine, is given by
No engine operating between two heat reservoirs can be more efficient than a Carnot engine operating
between the same reservoirs.
116. Pg 315, problem 6-24,
The Department of Energy projects that … 1995 and 2010, the United States will need …
The Department of Energy projected that … 1995 and 2010, the United States would need …
The publication date of the book is 2011. The text of the problem should be updated.
117. Pg 317, problem 6-52E, Ibm/h lbm/h and Ibm lbm
118. Pg 318, problem 6-57, L 5,
….the COP the refrigerator is 1.2, …. …. the COP of the refrigerator is 1.2, ….
119. Pg 327, problem 6-159 (hA)H (hA)H
120. Pg 328, problem 6-178, 0.95 kW/m2 0.95 kW/m2
121. Pg 342, in relation (7-13)
Because indicates that entropy is the same at all points, but indicates that entropy is the
same only at end points.
Note: Regarding the content of the second paragraph under the relation (7-13), it should be noted that in a
system which is not in thermodynamic equilibrium, we cannot attribute a single entropy value to the whole
system at any moment. So, when it is said that a process is isentropic, it is implicitly assumed that the system

15. 15
is in equilibrium at all times, and the process is reversible. Consequently, the process is adiabatic, too
(according to q=Tds=0 ).
Conclusion: An isentropic process is by definition a reversible adiabatic process.
122. Pg 347, FIGURE 7–24 should be changed.
123. Pg 350, last sentence,
Note: Mathematically, the expression is wrong.
To be more exact, in solids and liquids , not .
124. Pg 353,
125. Pg 362, relation (7-55)
126. Pg 362, EXAMPLE 7–12, L 2
…assuming that the steam exists as (a) saturated liquid and (b) saturated vapor ...
…assuming that the water exists as (a) saturated liquid and (b) saturated vapor ...
Also, in Pg 363,
(a) In this case, steam is a saturated liquid initially, (a) In this case, water is a saturated liquid initially,
and,
(b) This time, steam is a saturated vapor…. (b) This time, water is a saturated vapor….
127. Pg 364, at the bottom of the Pg, wrev wact wrev wact
128. Pg 367, last L,
In this case, the pressure ratio across each stage is the same, and its value is Px =….
In this case, the pressure ratio across each stage is the same, so Px =….
Note: Px is not the pressure ratio.
129. Pg 369, L 2,
This parameter is the isentropic or adiabatic efficiency,
This parameter is the isentropic or reversible adiabatic efficiency,
130. Pg 369, last P, L 1, T T

16. 16
131. Pg 374, P 2,
This gives an average temperature of 849 K, … by reevaluating … at 749 K and …
This gives an average temperature of 849 K, … by reevaluating …at 849 K and …
132. Pg 377, the sentence before relation (7-75),
When the properties of the mass change during the process, …
When the properties of the mass are not uniform across the inlets and outlets cross sections, …
133. Pg 377, P 3,
Regarding the expression:
“Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature
difference, … always cause the entropy of a system to increase,”
friction, mixing, and chemical reactions always cause the entropy of a system to increase , but heat transfer
not, and its effect depends on its sense of transfer. For example, when there is a finite temperature
difference, but the sense of heat transfer is from the system, the entropy of the system decreases.
Heat transfer through a finite temperature difference always causes entropy generation , not increasing the
entropy.
134. Pg 381, EXAMPLE 7–18, in Analysis section, h2 h1 h2 h1.
135. Pg 385, item (a), (100 + 273 K) (100 + 273) K
136. Pg 392, P 2, Ls 2 and 4,
1/0.8 = 1.25 kW 1 kW /0.8 = 1.25 kW and 1/0.95 = 1.05 kW 1 kW /0.95 =1.05 kW
137. Pg 401, problem 7-48E, …. shown in Fig. P7-8E …. shown in Fig. P7-48E
(a) Pg 404, problem 7-90, 565 K 565 K
138. Pg 406, problem 7-112, L 2,
Steam enters the pump as saturated liquid… Water enters the pump as saturated liquid…
139. Pg 409, problem 7-146,
A well-insulated heat exchanger … Disregarding any heat loss from the heat exchanger, determine …
A well-insulated heat exchanger … determine …
140. Pg 413, problem 7-188,

17. 17
141. Pg 415, problem 7-203,
…by an adiabatic 1.3-kW compressor to …. this adiabatic 0.7-kW compressor …
…by an adiabatic 1.3-kW compressor to …. this adiabatic 1.3-kW compressor …
142. Pg 416, problem 7-207,
It is better to change the text as follows:
… a 0.35-L canned drink that explodes … how many kg of TNT…
… a 0.35-L canned drink that explodes … how many grams of TNT…
Note that the answer is just 6 g.
143. Pg 417, problem 7-213,
… outer surfaces of the bottom of the tank… … outer surfaces of the bottom of the pan…
144. Pg 420, problem 7-241,
An apple with an average mass of 0.15 kg… An apple with a mass of 0.15 kg…
145. Pg 421, problem 7-258,
The highest … that condenses at the turbine exit The highest … that condenses up to the turbine exit
Note: condensation does not occur at the turbine exit, it starts at a point inside the turbine and continues
throughout the turbine until the turbine exit.
146. Pg 422,
In problem 7-259, practically the magnitude of 0.2°C for water temperature rise due to friction seems too
big.
147. Pg 425, P 1, last sentence,
Regarding the sentence,
“However, the atmosphere is in the dead state, and the energy it contains has no work potential.”,
it should be noted that the atmosphere is out of (far from) equilibrium and therefore it is not in a dead
state, and the energy it contains has great work potentials. Actually, there is a great potential of energy
(e.g., in the form of wind kinetic energy near ground surface and jet streams in upper layers of atmosphere)

18. 18
that can be used for producing electrical power.
The following sentence should also be corrected:
FIGURE 8–4 The atmosphere contains a tremendous amount of energy, but no exergy.
because there are tidal, solar, geothermal, and wind forms of exergies in atmosphere( which can be called
atmospheric exergies).
148. Pg 426, LB2 and LB3,
Betz’s law …. is slowed to one-third of its initial velocity. Betz’s law …. is slowed to 16/27 of its
initial velocity.
149. Pg 430, P 3,
The reversible work …. by considering a series of imaginary reversible heat engines ….
The reversible work …. by considering a reversible heat engine ….
Here, the source temperature is a function of time, but uniform throughout at each moment. That is, at
all times we have the same heat engine.
Here, imagination of a series of reversible heat engines is meaningless. How do those imaginary heat
engines are connected together at any instant t ? What is the difference between their source temperatures ?
150. Pg 434, last sentence,
it would supply the house with 26.7 units of heat (extracted from the cold outside air) ….
it would supply the house with 26.7 units of heat (25.7 of which extracted from the cold outside air) ...
151. Pg 435, last sentence of EXAMPLE 8–6 86.3 96.3
152. Pg 443, L 1, 2 before relation (8-28)
when the fluid properties are variable … when the fluid properties are nonuniform …
153. Pg 454, bottom of the Pg
154. Pg 459, last P, L 3, … exergy expanded is the… … exergy expended is the…
155. Pg 462, middle of the Pg,
156. Pg 467, FIGURE 8–49

19. 19
In figures (a) and (b), the expression “Variation of mental alertness with time” is superfluous and can be
deleted.
157. Pg 469,
158. Pg 469,
159. Pg 472, problem 8-36,
This phrase is not correct: …and electrical work is done on the water…
No work is done on the water. Electrical work is first converted to internal energy in the resistance, and
then transferred as heat to water.
160. Pg 483, problem 8-128, 4.61 kg 4.61 kg/s
161. Pg 497, P 1 L 7,8
during the combustion process during the compression process
162. Pg 505, last P, Ls 2 and 3,
Here the isothermal expansion and compression processes are executed in a compressor and a turbine,
respectively
Here the isothermal compression and expansion processes are executed in a compressor and a turbine,
respectively
163. Pg 507, P 3, Ls 5-7,

20. 20
… and thus a working fluid that … can be utilized as the working fluid.
… and thus a working fluid that … can be utilized.
164. Pg 526, FIGURE 9–53 … it weighs 6800 kg… … its mass is 6800 kg…
165. Pg 527, P 4, L 4, gases are expended… gases are expanded….
166. Pg 529, EXAMPLE 9–10
This example has no title.
EXAMPLE 9–10 EXAMPLE 9–10 Second-Law Analysis of an Otto Cycle
Also,
Processes 2-3 and 4-1 are constant-volume…. ( a )Processes 2-3 and 4-1 are constant-volume….
167. Pg 534, last P, L 3
….is used to overcome aerodynamic drag (i.e., to push the air out of the way)
….is used to overcome aerodynamic drag (i.e., to overcome the air flow friction)
Note: In a frictionless flow too, the car pushes the air out of the way, but it needs no power to do that. The
air is pushed away in front of the car, and instead the air is pulled toward the rear part of the car. This is
just a kinematic effect caused by the law of mass conservation. The root of aerodynamic drag (a
dynamic effect) is friction, which causes form or pressure drag. Therefore, the car needs power to
overcome this frictional effect not to push the air out of the way.
168. Pg 538,left column, P 3,
… regeneration ,a process during which … to a thermal energy storage device (called a regenerator)
… regeneration ,a process during which … to a thermal energy exchange device (called a regenerator)
Note: In a steady state process, a regenerator does not store energy; it is a heat exchanger and exchanges
thermal energy between two flow streams.
169. Pg 539, problem 9-13, 1.2 1-2
170. Pg 540, problem 9-22,
If the entropy increase during the isothermal heat rejection process is 0.25 kJ/kg .K
If the entropy increase during the isothermal heat addition process is 0.25 kJ/kg .K

21. 21
171. Pg 542, problem 9-52E, L 2, The slate of The state of
172. Pg 546, problem 9-114, L 7,
determine the isentropic efficiency of the turbine and the compressor
determine the isentropic efficiencies of the turbine and the compressor
173. Pg 549, problem 9-160E, L 3, I800 1800
174. Pg 550, problem 9-175, L 6, turbine efficiencies turbine isentropic efficiencies
175. Pg 551, problem 9-184,
176. Pg 552, problem 9-191, … and turbine efficiencies … and turbine isentropic efficiencies
177. Pg 554, problem 9-212, last sentence, Are there… inputs. Are there… inputs?
178. Pg 556, item1 under section 10-1,
which is 374°C for water which is about 374°C for water
179. Pg 560, EXAMPLE 10–1, last P,
The following reasoning seems to be wrong:
“The difference between the two efficiencies is due to the large external irreversibility in Rankine cycle
caused by the large temperature difference between steam and combustion gases in the furnace.”
because according to efficiency definition, the external irreversibility has no effect on efficiency in an ideal
Rankine cycle.
Here in this example, the difference between the two efficiencies is mainly due to the lower
average temperature during heat addition process.
180. Pg 571, FIGURE 10–14
181. Pg 572, P 2, L 7,
182. Pg 579, P 2, L 7, EXAMPLE 10–6, last sentence,

22. 22
… that the thermal efficiency would be 50.6. … that the thermal efficiency would be 50.6 percent.
183. Pg 584, P 1,
… dates to the beginning of this century… … dates to the beginning of last century…
184. Pg 591, FIGURE 10–27, Steam pump Water pump
185. Pg 593-594,
186. Pg 598, problem 10-53,
….which discharges into the condenser after being throttled to condenser pressure. Calculate ….per unit
mass of boiler flow rate.
….which discharges into the mixing chamber after being pumped to mixing chamber pressure.
Calculate ….per unit mass of boiler flow rate.
187. Pg 608, problem 10-128,
… of steam that condenses at turbine exit is … of steam that condenses up to turbine exit is
188. Pg 617, section 11-4, P 1, L 3
fluid friction flow friction
Note: “fluid friction” is meaningless.
Also, in P 2, L 8, fluid friction flow friction
189. Pg 620, under relation (11-17),
4 1 4 1
XXQ QL L
X X X X
190. Pg 629, P 1, L 6,

23. 23
for one of the cycles would be different for each cycle would be different
191. Pg 641, FIGURE 11–29
192. Pg 641, FIGURE 11–30
193. Pg 642,
Also,
194. Pg 646, problem 11.33E,
The temperature ….are at 10F and 80F,… The temperatures ….are at 10F and 80F,…
195. Pg 651, problem 11.81, L 2, P11- 61 P11-81
196. Pg 651, problem 11.93,
If the COP of an actual absorption chiller at the same temperature limits has a COP of 0.8
If the actual absorption chiller at the same temperature limits has a COP of 0.8
197. Pg 653, problem 11.114,
The item( c ) should be deleted, because in the problem it has already been mentioned that COP=6.
198. Pg 669, FIGURE 12–9
This sentence is unclear:
The slope of the saturation curve on a P-T diagram is constant at a constant T or P.
That sentence could be corrected as follows:
During a phase-change process, the saturation pressure depends only on the temperature, and it is
independent of the specific volume. Therefore, ( P/ T) vf = ( P/ T ) vg = (dP/dT )sat.

24. 24
199. Pg 670, 646.18 - 504.58 kPa (646.18 - 504.58) kPa
200. Pg 678, FIGURE 12–13
… an h= constant L on a P-T diagram. … an h= constant L on a T- P diagram.
201. Pg 682, last L,
202. Pg 684,
(0.22)(200 )
ln
(0.88)(800 )
kPa
kPa
(0.22)(200 )
ln
(0.88)(800 )
psia
psia
203. Pg 687, in problems 12-23, 12-24 and 12-25,
the enthalpy of vaporization of steam the enthalpy of vaporization of water
204. Pg 687, problem 12-36,
Comparing with problems 12-38,39,40, and 41, it seems that: a = 0.01 m3
/kg (not 1m3
/kg).
205. Pg 687, problem 12-38,
…the entropy of air, in kJ/kg… …the entropy of air, in kJ/K.kg…
206. Pg 687, problem 12-41,
…the entropy of helium, in kJ/kg… …the entropy of helium, in kJ/K.kg…
207. Pg 698, P 1 under relation (13-10), L 5-7,
Regarding the sentence:
“ Dalton’s law disregards the influence of dissimilar molecules in a mixture on each other. As a result, it
tends to underpredict the pressure of a gas mixture for a given Vm and Tm.”
I think if the force between dissimilar molecules is attractive, disregarding of their influence causes the
pressure to be over-predicted. Otherwise, if the force between dissimilar molecules is repulsive, disregarding
of their influence causes the pressure to be under-predicted.

25. 25
208. Pg 703, EXAMPLE 13–3, L 6 her unit mass per unit mass
209. Pg 704, L 5, (600K) 600K
210. Pg 712,
211. Pg 713, L 2 after relation (13-45),
…depends on the mole fraction of the components….
…depends on the mole fraction of the component, yi, ….
212. Pg 713, relations (13-47) , (13-48) and in FIGURE 13–20
, ,
213. Pg 714, relation (13-51b) ,
214. Pg 721, problem 13-10E,
molecular weight molecular mass ( to be affected throughout the text)
215. Pg 723, problem 13-46, Amagad Amagat
216. Pg 727, problem 13-91,
the law of additive volumes the law of Amagat’s additive volumes
217. Pg 728, problem 13-101, A steady steam … A steady stream …
218. Pg 734, bottom of the Pg, Tsat =3.1698kPa Psat =3.1698kPa
219. Pg 739,
220. Pg 742, P 3,
Regarding the following sentence:
We can also reduce the heat loss from the body either by … or by increasing the rate of heat generation

26. 26
within the body by exercising.
I think increasing the rate of heat generation within the body by exercising causes the rate of heat loss of
from the body to be increased, not to be reduced. The rate of heat loss from the body is proportional to
the temperature difference between the body and its surroundings . Exercising increases the skin
temperature , or in other words, increases the temperature difference between the body and its surroundings.
221. Pg 744, bottom of the Pg,
222. Pg 749, P 3 of EXAMPLE 14–7 , in Analysis section
Regarding the sentence:
….and the psychrometric chart of the process are shown in Fig. 14-28.
Where is the psychrometric chart?
223. Pg 759, problem 14-84E ,
… and the rate at which the pipe is cooled. … and the rate at which the air is cooled.
Note : The net heat transfer to the pipe is zero.
224. Pg 762,
For solving the problem14-121E we need Henry’s constant. But the Henry’s law appears first in chapter 16.
225. Pg 770, P2, Ls 1-3,
… and the components that exist after the reaction are called products
… and the components that produced due to the reaction are called products
Note : In all reactions, there remains some reactants after the reaction (which cannot be called products).
226. Pg 774, 22H HN N
22O ON N 22N NN N
227. Pg 779, L 4,
nuclear energy (associated with the atomic structure)
nuclear energy (associated with the nuclear structure)
228. Pg 781, P 4,
…..the heating value of the fuel, which is defined as …. in a steady-flow process and ….
…..the heating value of the fuel, which is defined as …. and ….
Note: The heating value is independent of the path or steadiness or unsteadiness of the process.
229. Pg 782, EXAMPLE 15–5,

27. 27
Also, in this example,
230. Pg 784, L 2 before relation (15–15)
231. Pg 786, left column of the table,
H2O(g) H2O(v)
It is suggested to do the change throughout the text. For example, in Pg 787, L 3:
water exists in the gas phase water exists in the vapor phase
Note: A gas is a state of materials that its temperature is greater than its critical temperature.
232. Pg 786, L 3 under the table, 298 298 298 298h h h h
233. Pg 786, LB 3,
2 The fuel, the air, and the…. 2 The fuel, the oxygen, and the….
234. Pg 791, P 2 under relation (15-20), Ls 7-8,
… third law of thermodynamics in the early part of this century.
… third law of thermodynamics in the early part of last century.
235. Pg 793, P 2,
For the very special … and the partial pressure Pi =1 atm for each component of the reactants and…
For the very special … and the pressure P =1 atm for each component of the reactants and …
Note : in the definition of reversible work or Gibbs function of formation, each component of the reactants
and the products are assumed to be in their pure form( not in a mixture of reactants or products). Therefore
the combined word “partial pressure” is a misnomer or misleading term.

28. 28
236. Pg 794, last P,
…and Ofh for O2 and N2. Assuming …, the fh and h values…
…and 0fh for O2 and N2. Assuming …, the fh and h values…
237. Pg 795, last P, L 1, 4CH 4 CHs s
238. Pg 796, P 2,
Because of its importance, it is suggested that the following sentence :
“ This example shows that even complete combustion processes are highly irreversible.”
to be italicized.
239. Pg 805, problem 15-67, left column, H2O / (g) H2O (v)
240. Pg 806, problem 15-78,
…fuels butane, ethane, methane, and propane …fuels methane, ethane, propane and butane
241. Pg 806, problem 15-79,
The adiabatic flame temperature for a stoichiometric acetylene -oxygen mixture seems to be around
3480 °C, not 8850 °C.
242. Pg 806, problem 15-82, FIGURE 15–82 FIGURE P15–82
243. Pg 808, problem 15-104, left row of the table, kg/kmol kJ/kmol
244. Pg 809, left column, P 1,
methyl alcohol vapor (CH3OH(g)) methyl alcohol vapor (CH3OH(v))
245. Pg 814, FIGURE 16–2
FIGURE 16–2

29. 29
Criteria for chemical equilibrium for a fixed ... Criterion for chemical equilibrium for a fixed ...
246. Pg 814, relation(16-2),
247. Pg 815, , FIGURE 16–4
FIGURE 16–4
Criteria for chemical equilibrium for a fixed ... Criterion for chemical equilibrium for a fixed ...
248. Pg 817, L 1, ….rezpresents the Gibbs function …. represents the Gibbs function
249. Pg 818, EXAMPLE 16–1
To be more accurate, it is better to do the following change:
Note : Please see Pg 821, L 3.
250. Pg 819, bottom of the Pg, Using an equation solves … Using an equation solver …
251. Pg 821, item 6,
When the stoichiometric coefficients are doubled, the value of KP is squared.
When the stoichiometric coefficients are multiplied by n, the value of KP is raised to power n.
252. Pg 821, item 7, Ls 3-4,
…and at even higher temperatures atoms start to lose electrons and ionize…
…and at higher temperatures, atoms even start to lose electrons and ionize…
253. Pg 822, , FIGURE 16–11
254. Pg 823,
1.906CO2 + 0.094CO + 2.074O2 1.906CO2 + 0.094CO + 2.047O2
Also,

30. 30
255. Pg 825, L 4 after relation (2),
….that part of the H2O in the products is dissociated into H2 and OH…
….that part of the H2O in the reactants is dissociated into H2, O2 and OH…
256. Pg 826, section 16-5, L 2,
It was shown … KP of an ideal gas depends… It was shown … KP of an ideal gas mixture depends…
257. Pg 827, EXAMPLE 16–6
258. Pg 830, L 5,
…a single-component two-phase system has one independent property…
…a single-component two-phase system has one independent intensive property…
259. Pg 831, bottom of the Pg,
calcium bicarbonate [Ca(HO3)2] calcium bicarbonate [Ca(HCO3)2]
260. Pg 833, last P, L 4,
…is simply taken to be 1.0, …is simply taken to be 1,
261. Pg 838, problems 16-10 and 16-11,
Determine the Gibbs function… Determine the Gibbs function value…
262. Pg 838, problems 16-12 At temperature… At what temperature…
263. Pg 839, problems 16-36, Compare… Calculate…
264. Pg 839, problems 16-40,
265. Pg 840, problem 16-42,
- In the text, the extent of reaction is shown by not by .
- The problem should be revised as follows:
Show that the extent of the reaction, , for the dissociation reaction is given by…
Note : The is always less than one for the specified reaction( furthermore, for deriving the specified
relation , there is no need for to be less than one).

31. 31
266. Pg 840, problem 16-52,
267. Pg 840, problem 16-55,
In this problem, and some other problems of this chapter, it is recommended that:
combustion process combustion reaction
268. Pg 842, problem 16-81,
…(c) strong liquid mixture….(d) weak liquid solution…
…(c) strong liquid solution….(d) weak liquid solution…
269. Pg 842, problem 16-88,
…the natural logarithms of the equilibrium constant for the reaction … and…
…the natural logarithms of the equilibrium constants for the reactions … and…
270. Pg 843, problem 16-108,
271. Pg 845, problem 16-125, Ls 4-5,
The combustion …. Consider the combustion ….
272. Pg 850, Where is the FIGURE 17–6?
273. Pg 851, L 6,
This sentence is superfluous and it is better to be deleted: Thus the work supplied to the compressor is
233.9 kJ/kg.
274. Pg 852, L 3,
…second-order term dV 2
…second-order term (dV) 2
Note : dV 2
=2VdV is a first-order term!
275. Pg 852, first P,
I think the following reasoning is wrong:
The amplitude of the ordinary sonic wave is very small and does not cause any appreciable change in the
pressure and temperature of the fluid. Therefore, the propagation of a sonic wave is not only adiabatic but
also very nearly isentropic.
We know that a sonic wave, by definition, is caused by a differential change of pressure in the medium.
Therefore, it is obvious that any change in the pressure, temperature or any other properties of the medium,
including its entropy or its heat exchange with the environment, is also infinitesimal and not appreciable. So,

32. 32
we cannot deduce that because the change in pressure, temperature , entropy or … is infinitesimal, therefore
the propagation of sonic waves is isobar, isotherm, isentropic or…
The propagation of sonic waves is adiabatic because the speed is high and there isn’t enough time for heat
transfer. The propagation of sonic waves is frictionless because the contact surface between the fluid(in the
region near the wave) and pipe is negligible, comparing with the surface through which it propagates.
Therefore, the process of sound propagation is both adiabatic and frictionless, or simply isentropic.
276. Pg 853, L 6, hypersonic when aM 1 hypersonic when aM 5
277. Pg 854,
278. Pg 855, TABLE 17–1
279. Pg 857, P 3, L 1,
Thus the proper …. depends on the highest velocity desired...
Thus the proper …. depends on the highest Mach number desired...
280. Pg 859, FIGURE 17–18
281. Pg 860, in Converging Nozzles section, Ls 3-4,
The reservoir is sufficiently large so that the nozzle inlet velocity is negligible.
The nozzle’s inlet cross section is sufficiently large so that the nozzle inlet velocity is negligible.
Note : When a reservoir is large enough, it can be assumed that its pressure and temperature remain
constant during the process. When a nozzle’s inlet cross section is sufficiently large (relative to nozzle’s
throat), it can be assumed that its inlet velocity is negligible.
282. Pg 862, FIGURE 17–22, the title of horizontal axis,
283. Pg 862, P 2, Ls 3-4,
…is plotted against the static-to-stagnation pressure ratio at the throat Pt /P0.
…is plotted against the back-to-stagnation pressure ratio at the throat Pb /P0.

33. 33
284. Pg 867, item 3, Ls 3-8,
This acceleration comes to a sudden stop, however, as a normal shock develops at a section between the
throat and the exit plane, which causes a sudden drop in velocity to subsonic levels and a sudden increase
in pressure. The fluid then continues to decelerate further in the remaining part of the converging–
diverging nozzle.
Then a sudden drop in velocity to subsonic levels and a sudden increase in pressure occurs, as a normal
shock develops at a section between the throat and the exit plane. The fluid then continues to decelerate
further in the remaining part of the diverging nozzle.
Note: the sentence: “The acceleration comes to a sudden stop” is not meaningful. The acceleration changes
its sign during the shock (before the shock the acceleration is positive, across the shock it is highly negative,
and after the shock it is negative).
285. Pg 869, last P, L 4, Pierre Lapace Pierre Laplace
286. Pg 873, first P, L 2,
Since the flow across the shock is adiabatic and irreversible,
Since the flow across the shock is adiabatic but irreversible,
287. Pg 873, EXAMPLE 17–8, L 1,
…maximum entropy on the Fanno L (point b of Fig.17–31)…
…maximum entropy on the Fanno L (point a of Fig.17–31)…
288. Pg 876, P 2, L 9-11,
Since … the boundary layer growing along the wedge is very thin, and we ignore its effects.
Since … the boundary layer growing along the wedge is very thin, and we ignore its thickness.
Note: The dynamical effects of boundary layer(e.g., drag) cannot be ignored, particularly at high Reynolds
numbers.
289. Pg 877, Ls 1-2 under relation(17-44),
From the point of view shown in Fig. 17–42, From the point of view shown in Fig. 17–41,
290. Pg 887, item 4,
The following reasoning is plausible if T0 remains constant:

34. 34
The cooling effect in this region is due to the large increase in the fluid velocity and the accompanying drop
in temperature in accordance with the relation T0 =T+ V2
/2cp.
Note: But in Rayleigh flow q = cp (T02 - T01), and heating increases the stagnation temperature T0 (therefore
T0 does not remain onstant).
291. Pg 890, FIGURE 17–58, second row, k/(k+1) k/(k-1)
292. Pg 891, section Choked Rayleigh Flow, Ls 8-10,
Regarding the following sentence:
If we keep heating the fluid, we will simply move the critical state further downstream and reduce the flow
rate since fluid density at the critical state will now be lower.
It should be noted that if the pipe length is fixed, the critical state occurs always at pipe exit (therefore the
phrase “move the critical state further downstream” is meaningless), and further heating only reduces the
mass flow rate.
293. Pg 898, problem 17-2C,
How and why is the stagnation enthalpy h0 defined? Why and how is the stagnation enthalpy h0
defined?
294. Pg 898, problem 17-23,
takeoff weight of about 260,000 kg takeoff mass of about 260,000 kg
295. Pg 899, problems 17-48C and 17-49C, a supersonic fluid a supersonic flow
296. Pg 903, problems 17-125,
297. Pg 905, problems 17-157,
298. Pg 905, problems 17-162, item (b), diversion section divering section

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