Comments on Thermodynamics: An Engineering Approach, 7th ed. (SI Units),

McGraw-Hill, 2011 Yunus. A. Cengel and Michael. A. Boles

By: Mohsen Hassan vand

Saturday, May 03, 2014

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

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1.In FIGURE 1–9 it would be more interesting and relevant if a chubby woman is shown instead of the slime one shown.

2.Page 10, EXAMPLE 1–3, last line 1 lbm 1 kg

3. Page 26, P3, L2 “Since the gravitational effects of gases are negligible”

“Since the gravitational effect on gases is negligible” 4.In Prob. 1-48, it seems that the mentioned pressure (1250 mm Hg!) for a toy balloon is too

high. 5.In EXAMPLE 2–1,

6.Page 59, L3 under the relation (2–13) In the absence of any irreversible losses In the absence of any losses Note: The concept of irreversibility has not been defined in the text yet. It is defined in next sections of the book. In page 82, last line

irreversibilities losses 7.Page 66, P4,

(1) there must be a force acting on the boundary (1) there must be a force acting on the body or its boundary

8.In FIGURE 2–45,

Note: Compare FIGURE 2–45 with FIGURE 2–43.

9.Page 80, relation (2–43), net fuel

10. Under the FIGURE 2–69, …several times its weight in CO2

…several times its mass in CO2

11.Page 114, beginning of P 3, Once boiling starts,

Once vaporization starts, 12.Page 136, P 3, regarding the following phrase:

“…neon, krypton, and even heavier gases such as carbon dioxide…” it is to be noted that krypton(M=83.8) is heavier than carbon dioxide(M=44)(not vice versa).

13.Page 136, EXAMPLE 3–10, LB 6 Therefore, the absolute ….by 6.9% Therefore, the absolute ….by 3.29%

Note: T/T×100=9.8/298×100=3.29% 14.Page 137, last line of P 1:

ideal-gas relations should not be used ideal-gas relation should not be used 15.Page 145, EXAMPLE 3–13, near bottom of the page

28.013 kg/mol

28.013 kg/kmol 16. Prob. 3-34, in kPa

in MPa 17.Prob. 3-49, Prob. 3-53 Prob.3-48 18. In Prob. 4-21,

According to the problem statement T= 180 0C , but according to the related figure T= 120 0C !?

19.Page 217, in relation (5–16), in the first integral, 20. In EXAMPLE 5–1

21.Page 223, P 2, L 3,

… energy transported by mass through an inlet or exit is obtained by integration.

…energy transported by mass through an inlet or exit is obtained by integration.

22.Page 223, bottom of the page,

23.Page 225, in (5-36) and similar relations

24.Page 227, section 5-4, L2-L4, regarding the following sentence: “The components of a steam power plant …. (Fig. 5–24).”

it should be noted that the Fig. 5–24 is gas power plant, not a steam power plant.

Also, in the caption for FIGURE 5–24: … it weighs 12.5 tons … a mass of 12.5 tons

25.Pages 228-229, There are two figures under the same number (two FIGURE 5–26).

26.Page233, section Throttling Valves, L 3 According to the following statement:

“ Unlike turbines,… they produce a pressure drop without involving any work.” the turbines produce work through pressure drop. But we know that turbines are basically designed in such a manner to have a pressure drop as low as possible.

What happens in turbines is primarily an expansion process, not a pressure drop. It is better to differentiate between pressure drop or pressure loss (caused by friction, dissipating the work) and pressure decrease ( caused by expansion, producing work).

27.Page234, L4, with large exposed surface areas with large exposed surface areas per volume. In EXAMPLE 5–8

(h2=h1)

h2=h1

28.In EXAMPLE 5–13 m1=P1V1/RT1

m1=P1 /RT1

m2=P2V2/RT2

m2=P2 /RT2

29.Page246, bottom of the page,

30.In FIGURE 5–51,

31.Page 248, P 2, Regarding the following statement: “Note that the fluid ….no- slip condition …. As a result, …. is zero.”

it should be noted that the pressure work along the portions of the control surface that

coincide with nonmoving solid surfaces is zero in both viscous (satisfying no- slip condition ) and invicid flows (not satisfying no- slip condition, and there may exist a tangential velocity component on the surface), because what matters in flow work is the normal component of the flow velocity, not tangential component.

32.Prob. 5-9 weight fraction

mass fraction

33.Prob. 5-43C Somebody proposes the following system …, is the proposed system sound?

Somebody proposes the following process …, is the proposed process sound?

34.Prob. 5-56 FIGURE P5–56

FIGURE P5–55 On the figure, 0.1MPa

0.5MPa 35.On the FIGURE P5–140 350C

300C 36.Prob. 5-145

According to the problem statement: “Each outlet has the same diameter as the inlet”, but according to the FIGURE P5–145, outlets and inlet have different diameters!

37.In EXAMPLE 6–2

38.Page 284, last paragraph, L3 SEER is the ratio the total amount…

SEER is the ratio of total amount…

39.Page 292,LB 4, As dT approaches zero,…

As T approaches zero,…. Note: dT is by definition a T approaching zero.

40.Page 298, under relation (6-15) Several functions ….. satisfy this equation

Any continuous and monotonic function….. satisfies this equation,

41. Page 304, in EXAMPLE 6–6 (20+273K)/( -8+273K)

(20+273) K /( -8+273) K 42.Page 305, in EXAMPLE 6–7

(-5+273K)/( 21+273K)

(-5+273) K /( 21+273) K 43.Page 312, in FIGURE P6–38,

5 hp

3.7 kW 44.Page 314, Prob. 6-55,

COP the refrigerator is 1.2

COP of the refrigerator is 1.2 45.Page 321, Prob. 6-140,

$33 $33/year

46.Page 331,L 4 under relation (7-6), thermal energy reservoirs that can absorb or supply heat indefinitely at a constant temperature.

thermal energy reservoirs that can absorb or supply finite amounts of heat at a constant temperature.

Note: In a physical thermal reservoir, if the heat absorption or heat supply is indefinite, its temperature cannot remain constant. Please see page275, section 6-2, L 3:

“that can supply or absorb finite amounts of heat without undergoing any change in temperature.”

47.Page 342, FIGURE 7–21 T= 0 K , Entropy= 0

T= 0 , Entropy= 0 or T= 0 K , Entropy= 0

T= 0 K, Entropy= 0 kJ/mol.K

48.Page 345, L1 after relation (7-23) or Gibbs, equation

or Gibbs equation

49.Page 356, EXAMPLE 7–11, Solution Helium is … to a specified pressure

Solution Helium is … to a specified temperature

50.Page 359, EXAMPLE 7–12, Analysis We take first the turbine and then the pump as the system.

Analysis We take first the pump and then the compressor as the system.

51.Page 375, under the relation (7–81), if its temperature is constant

if its temperature is uniform

52., EXAMPLE 7–19 , page 379 (12.65 kJ/K)

12.65 kJ/K 53.Page 381,

(100 +273 K)

(100 +273) K

54.Page 384, FIGURE 7–72

55.Page 391, LB 4, other noncondensable gases such as oil vapors other condensable gases such as oil vapors

56.Problem 7-20 100K

1000K

57.Problem 7-239 (e) 1.00 (e) 1 Note: the efficiency is always 1.

58.Page 427, in section Analysis, L10 without wasting a single kilojoule

without wasting a single joule

59.Page 440, L 1 since V2 = V1 for an isolated system since V2 = V1 or P0 =0 for an isolated system

Note: In isolated systems V2 can be different from V1 , if P0 =0.

60.Page 445, last line( and also page 446, line 4) }(kJ/kPa.m3)

(kJ/kPa.m3) }

61.Page 452, bottom of the page 8193-4731=3463

8193-4731=3462 and

Discussion Note that 3463 of…..3463/8193

Discussion Note that 3462 of…..3462/8193

62.Page 458, middle of the page

…+ Qout (since W=0 ke pe 0 ) …+ out (since W=0 and ke pe 0 )

63.Page 465

64.Please recheck the answers data of Prob.8-46. 65.Prob.8-56, 17 °F

17°C 66.Prob.8-119, (a) 4.61 kg (a) 4.61 kg /s 67.Page 499, section 9-7, P 1

In this paragraph, you have deduced that the thermal efficiency of an Otto or Diesel engine is less than that of a Carnot engine because these cycles are externally irreversible. But the thermal efficiency of cycles are defined in terms of internal , not external parameters. I think the thermal efficiency of Otto or Diesel engine is less than that of a Carnot engine because the average temperature at which heat is supplied is less than the maximum temperature, and the average temperature at which heat is rejected is greater than the minimum temperature in the cycle.

68.Page 502, EXAMPLE 9–4, in Analysis section ….and it is rejected again isothermally… ….and a fraction of it is rejected isothermally…

69.Page 522, P2, last line, with less fuel per passenger mile with less fuel per passenger kilometer

70.Page 525, EXAMPLE 9–10 Analysis We take…

Analysis (a)We take….

71.Page 526, bottom of the page, Substituting, the second-law efficiency of this power plant …

Substituting, the second-law efficiency of this cycle 72.Page 529, LB 5,

An extra 50 km… An extra 50 kg…

73.Page 530,P 2,L 6 …due to a rise in ambient temperature or just road friction

…due to a rise in ambient temperature or just tires heat-up(caused by their cyclic inflexion- deflexion). Note: Except during acceleration or deceleration, there is no slippage and therefore no heating due to friction between tire and road.

74.Page 557,P 2,L 3 ….cavitation, the rapid vaporization and condensation of the fluid at the low-pressure side of the pump impeller

….cavitation, the rapid vaporization of the fluid at low-pressure side and rapid collapsing of the vapor bubbles at high-pressure side of the pump impeller

Note: Bubbles are collapsed, not condensed. Condensation is caused by temperature decrease, and collapsing is caused by pressure increase (in high pressure sides of impellers).

75.Page 585,L 3 That is, 1 kg of exhaust gases… That is, each kg of exhaust gases…

76.Page 587, FIGURE 10–27 Steam pump Water pump

77.Prob. 10-22 …with steam as the working fluid… …with water as the working fluid…

78.Page 608, line 1 after relation (11–3) COPHP > 1 COPHP

1

79.Page 615, L3 after relation (11–9), …proportional to the temperature difference TH -TL ….

…proportional to the temperature ratio TH /TL ….

80.Page 632,

81.Prob. 11-101

FIGURE P11–101 should be shifted to under the problem 11–101. 82.Page 660,

and, on FIGURE 12–5,

83.Page 662, relation (12-9)

84.Page 671, FIGURE 12–10, Regarding the shaded areas on the figure,

Note: In the figure, T= 1°C.

85.Page 672, L1, The difference between cp and cv approaches zero…

The cp and cv approach zero… 86.Page 678 , relation (12-61),

According to the notation used in FIGURE 12–17

87.Pages 698-699 , relation (13-24) and (13-25),

88.Page 700, L8 (2.209kJ/kg.K)(600-298.15) (2.209kJ/kg.K)(600-298.15) K

89.Page 704,

and in page 705

90.Page 706, L1 after relation (13-30), …is the chemical potential,

…is the chemical potential of the i-th species,

91.Page 707, L2 after relation (13-34), The enthalpy of mixing is positive for exothermic mixing processes, negative for endothermic mixing processes, …

The enthalpy of mixing is negative for exothermic mixing processes, positive for endothermic mixing processes, …

92.Page 707, relations (13-36) and (13-37),

93.Page 708, P1 after relation (13-41) , last sentence The sentence “…water and oil won’t even mix at all to form a solution” should be modified. Because we know oil mixes with water, but its solubility under ambient conditions is weak(and at higher temperatures, the solubility of oil in water is considerable).

94.Page 709,

In relation (13-48):

In relation (13-49):

95.Page 710, relation (13-51b),

96.Prob. 13-38 What is the specific weight, in kg/m3 What is the specific weight, in N/m3

97.Page 745, L 1 after relation(14-23)

Eliminating from…. Eliminating from…. 98.Page 748, FIGURE 14–32

Two natural draft… Three natural draft…. 99.Page 750,

Note: In air-conditioning processes no work is produced. 100.Page 765, P 1, L 2

a realistic measure is the cost per unit energy rather than cost per unit volume.

a realistic measure is the cost per unit exergy rather than cost per unit volume. Note: This comment is just a suggestion.

Also, in relation (15–1):

Note: Apples (here, O2 ) and oranges (N2) do not add!

101.Page 780, L 2 before relation (15–15):

102. Page 782,

103.Page 785, L1 before the table,

104.Page 788, L 3 before relation (15-25)

Note: Please note the in relation (15-26). 105.Page 789, relation (15-27)

106.Page 791, bottom of the page, first row of the table

107.Page 793, bottom of the page

108. Prob. 15-65, table H2O/(g) H2O(g)

109.Prob. 15-92, in kJ/kg in MJ/kg

Note: Please see the answer. 110.Prob. 15-100, first column of the table

kg/kmol

kJ/kmol 111.Page 814, EXAMPLE 16–1

112.EXAMPLE 16–2 , middle of the page 815

113.Page 817,

114. Page 855, FIGURE 17–18

Note: o 0

115.Page 864, item 4, In first line: PE >Pb > 0 PE >Pb

0 In sixth line: P b >P F

P F <P b <P E

116.Page 873, L 1 after relation (17–44) From the point of view shown in Fig. 17–42,

From the point of view shown in Fig. 17–41,

117.Page 874, FIGURE 17–42

118.Page 886, FIGURE 17–58

Note: Please see relation (17–67).

119.Prob. 17-62, Answers: 128 kPa, 172 K…

Answers: 172 K, 128 kPa …

120.Prob. 17-77, Answers: 448 kPa, 284 K…

Answers: 284 K, 448 kPa …

121.Prob. 17-139, in the FIGURE P17–139

122.Page 907,TABLE A-2 (c), second column, S2 S

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