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Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

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    SOLUTIONS MANUAL Fluid Mechanics for Engineers First Edition David A. Chin January 2018

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    Chapter 1 Properties of Fluids 1.1. (a) Find the dimensions of C L as follows, C L = F L 1 2 ρV 2 A → C L = MLT − 2 (ML − 3 )(LT − 1 ) 2 (L 2 ) = MLT − 2 MLT − 2 = 1 Therefore C L is dimensionless and does not depend on the system of units used. (b) No adjustment factor is necessary when USCS units are used instead of SI units. 1.2. (a) Inserting the dimensions of the variables in the given equation yields ρ d 2 z d t 2 + a d z d t + bz = c ┌ M L 3 L T 2 + a L T + b [L] = c ┐ ┌ ┐ ┌ ┐ ┌ M L 2 T 2 + a L T + b [L] = c ┐ ┌ ┐ Therefore, the required dimensions of the parameters a , b , and c are, a = M L 3 T , b = M L 3 T 2 , c = M L 2 T 2 ┌ ┐ ┌ ┐ ┌ ┐ (b) If ρ ∗ , z ∗ , and t ∗ are the given variables in nonstandard units, then the conversion factors are: ρ ρ ∗ = 10 3 , z z ∗ = 10 − 3 , t t ∗ = 3600 1.3. (a) Inserting the dimensions of the variables in the given equation yields Q = 1 n A 5 3 P 2 3 S 1 2 0 → L 3 T − 1 = 1 ( ) (L 2 ) 5 3 L 2 3 ( − ) 1 2 → L 3 T − 1 = L 8 3 − Since the dimension of the left-had side of the equation is not equal to the dimension on the right-hand side of the equation, the given equation is not dimensionally homogeneous .

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    (b) If the length units are changed from m to ft and 1 m = 3.281 ft, then inserting this conversion factor into the given equations requires that (3 . 281) − 3 Q ' = 1 n (3 . 281) − 2 A ' 5 3 [(3 . 281) − 1 P ' ] 2 3 S 1 2 0 ┌ ┐ where the primed quantities have length units of ft. Simplifying the above equation and removing the primes gives Q = 3 . 281 1 3 n A 5 3 P 2 3 S 1 2 0 → Q = 1 . 486 n A 5 3 P 2 3 S 1 2 0 Therefore, the conversion factor to be added is 1.486 . 1.4. − − − · · − Quantity Dimension Typical SI Unit energy 2 T − 2 FL = ML J force F = MLT − 2 N heat 2 T 2 FL = ML J moment 2 T − 2 FL = ML N m momentum MV = MLT − 1 kg m/ s power 2 FLT − 1 T − 3 = ML W pressure T − 2 FL − 2 = ML − 1 Pa strain LL − 1 = stress T − 2 FL − 2 = ML − 1 Pa work 2 T 2 FL = ML J 1.5. Given With Prefix 5 . 63 × 10 7 N 56 . 3 MN 8 . 27 × 10 5 Pa 827 kPa 2 3 . 86 × 10 − 4 m 2 0 . 0386 km 7 . 88 × 10 5 m 788 km 1.6. Quantity USCS Abbreviation In SI Units 12 gallons per minute 12 gpm 45.4 L/min 55 miles per hour 55 mph 88.5 km/h 5 feet per second 5 ft/s 1.5 m/s 125 cubic feet per minute 125 cfm 3.54 m 3 /min 1000 gallons 1000 gal 3785 L 25 acres 25 ac 10.1 ha 500 horsepower 500 hp 373 kW

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    1.7. (a) 1 hp = 550 ft · lb s × 0 . 3048 m ft × 4 . 448 N lb = 745 . 7 W (b) 1 lb/in 2 = 1 lb in 2 × 4 . 448 N lb × 1 0 . 0254 2 in 2 m 2 = 6 . 894 × 10 3 Pa = 6 . 894 kPa 1.8. From the given data: ρ 0 = 1000 kg/m 3 , and the density deviates most from 1000 kg/m 3 at T = 100 ◦ C, where ρ = 958.4 kg/m 3 . Hence, the maximum error in assuming a density of 1000 kg/m 3 is error = 1000 − 958 . 4 958 . 4 × 100 = 4 . 3% 1.9. From the given data: V 1 = 3 L , ρ 1 = 1030 kg/m 3 , V 2 = 5 L , and ρ = 920 kg/m 3 . The density of the mixture, ρ m , is given by ρ m = ρ 1 V 1 + ρ 2 V 2 V 1 + V 2 = (1030)(3) + (920)(5) 3 + 5 = 961 kg/m 3 Note: The volumes do not need to be converted from L to m 3 since the conversion factor would cancel out. 1.10. (a) The specific weight, γ , is derived from the density, ρ , using the relation: γ = ρg = 9 . 807 ρ . Obtaining ρ from Appendix B.1 gives: Temperature ρ ( ◦ C) (kg/m 3 ) (N/m 3 ) γ 0 999.8 9805 20 998.2 9789 100 958.4 9399 (b) The specific gravity, SG, is derived from the density, ρ , using the relation: SG = ρ/ρ 4 ◦ C . Obtaining ρ from Appendix B.1 gives: Temperature ρ SG ( ◦ C) (kg/m 3 ) (–) 0 999.8 1.000 20 998.2 0.998 100 958.4 0.958 V 1 = 400 L, T 1 = 15 ◦ C , and T 2 = 90 ◦ C. 1.11. From the given data: The densities of water corresponding to T 1 and T 2 (from Appendix B.1) are: ρ 1 = 999 . 1 kg/m 3 and ρ 2 = 965 . 3 kg/m 3 . (a) The initial mass, m 1 , in the tank is given by m 1 = ρ 1 V 1 = (999 . 1)(0 . 4) = 399 . 6 kg

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    The volume of water after heating to 90 ◦ C is given by V 2 = m 1 ρ 2 = 399 . 6 965 . 3 = 0 . 4140 m 3 = 414 . 0 L Therefore, the spilled volume, Δ V , is given by Δ V = V 2 V 1 = 414 400 = 14 L − − (b) The spilled mass, Δ m is given by Δ m = ρ 2 Δ V = (965 . 3)(14 10 − 3 ) = 13 . 5 kg × The percent change in the mass (= percent change in weight) is 13 . 5 / 399 . 6 × 100 = 3.4% 1.12. From the given data: γ = 12 kN/m 3 = 12 000 N/m 3 . For water at 4 ◦ C: ρ w = 1000 kg/m 3 . According to the definitions of density and specific gravity, ρ = γ g = (12000)(9 . 807) = 1224 kg/m 3 SG = ρ ρ w = 1224 1000 = 1 . 224 1.13. From the given data: SG = 1.5. For water at 4 C: ◦ ρ w = 1000 kg/m 3 . According to the definitions of density and specific weight, ρ = SG · ρ w = (1 . 5)(1000) = 1500 kg/m 3 γ = ρg = (1500)(9 . 807) = 14 710 N/m 3 14 . 7 kN m 3 ≃ · 1.14. For any given volume, V , containing a mixture, let C m = mass ratio, ρ f = density of the pure fluid, ρ m = density of the mixture, m f = mass of pure fluid, m m = mass of mixture , m s = mass of solids in the mixture, SG f = specific gravity of pure fluid, and SG m = specific gravity of mixture. Therefore, m f = ρ f V, m m = ρ m V, m s = m m m f = ( ρ m ρ f ) V − − Using these relationships yields, C m = m s m m = ( ρ m − ρ f ) / / V ρ m / / V = 1 − ρ f ρ m → C m = 1 − SG f SG m 1.35. From the given data: ρ = 800 kg/m 3 . For water at 4 ◦ C: ρ w = 1000 kg/m 3 . According to the definitions given in Equations 1.9 and 1.10, γ = ρg = (800)(9 . 807) = 7846 N/m 3 7 . 85 kN/m 3 ≃ SG = ρ ρ w = 800 1000 = 0 . 80

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    1.16. From the given data: M = 200 kg, and SG = 1.5. At 4 C the density of water is ◦ ρ w = 1000 kg/m 3 . The volume, V , of the reservoir is given by V = M ρ w SG = 200 (1000)(1 . 5) = 0 . 133 m 3 · 1.17 . From the given data: W c = 10 N, and W t = 50 N. For kerosene at 20 ◦ C , ρ = 808 kg/m 3 and γ = 7924 N/m 3 (Appendix B.4). Using these data gives the following, weight of kerosene, W k = W t W c = 50 10 = 40 N − − volume of kerosene, V k = W k γ = 40 7924 = 5 . 048 × 10 − 3 m 3 = 5 . 05 L mass of kerosene = W k g = 40 9 . 807 = 4 . 08 kg 1.18. The bulk modulus, E v , is defined by Equation 1.12 as E v = − d p d V /V (1) where the density of the fluid, ρ , is defined by ρ = M V (2) where M is the (constant) mass of fluid and V is the volume of fluid that is compressed by the fluid pressure. Differentiating Equation 2 with respect to V gives d ρ d V = − M V 2 (3) Combining Equations 2 and 3 to eliminate M yields d ρ d V = − ρ V or d V V = − d ρ ρ (4) Finally, combining Equations 1 and 4 gives E v = d p d ρ/ρ 1.19. Equation 1.13 can be approximated by Δ ρ ρ = Δ p E v (1) and at 20 ◦ C, E v = 2 . 18 10 6 kPa (Table B.1). For × Δ ρ/ρ = 0.01, Equation 1 becomes 0 . 01 = Δ p 2 . 18 10 6 → Δ p = 2 . 18 × 10 4 kPa = 21 . 8 MPa ×

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    1.20. From the given data: T = 20 C, D = 3 m, R = D/ 2 = 1 . 5 m, and ◦ Δ p = 9 MPa . For wate r at 20 C, ◦ ρ 0 = 998 . 2 kg/m 3 and E v = 2 . 18 10 6 kPa × (from Appendix B.1). Using these data, the volume, V , of the tank, and the initial mass, m 0 is the tank are calculated as V = 4 3 πR 3 = 4 3 π (1 . 5) 3 = 14 . 14 m 3 , m 0 = ρ 0 V = (998 . 2)(14 . 14) = 1 . 411 10 4 kg × From the definition of the bulk modulus, E v , E v ≈ Δ p Δ ρ/ρ 0 → Δ m m 0 = Δ p E v → Δ m = m 0 Δ p E v Substituting the given and derived data into this relationship yields, Δ m = (1 . 411 × 10 4 ) 9 × 10 3 2 . 18 10 6 = 58 . 3 kg × 1.21. From the given data: p 1 = 100 kPa, p 2 = 20 000 kPa, V 1 = 1.700 m 3 , and V 2 = 1.650 m 3 . Using the definition of the bulk modulus given by Equation 1.12, E v ≈ − Δ p Δ V /V = − 20000 − 100 1 . 650 − 1 . 700 1 . 700 = 6 . 766 × 10 5 kPa = 677 MPa 1.22. From the given data: V 1 = 10 m 3 , and Δ p = 10 MPa. For benzene, E v = 1030 MPa (Appendix B.4). Using the definition of the bulk modulus given by Equation 1.12, E v ≈ − Δ p Δ V /V 1 → 1030 ≈ − 10 Δ V 10 → Δ V ≈ 0 . 0971 m 3 1.23. From the given data: T 1 = 10 ◦ C, and T 2 = 100 ◦ C. The average coefficient of volume expansion, β , between T 1 and T 2 is derived from Appendix B.1 as ¯ β = 0 . 418 10 − 3 K − 1 . × (a) Applying Equation 1.19 (with Δ p = 0 ) gives Δ ρ ρ ≈ − ¯ β Δ T = − 0 . 418 × 10 − 3 (100 − 10) = − 0 . 0376 = − 3 . 76% (b) Let A be the surface area of the water in the pot (assumed to be constant), h 1 is the depth of water at T 1 , an d h 1 + Δ h be the depth of water at T 2 . Therefore, since the mass of water is constant, ρ 1 / / Ah 1 = ρ 2 / / A ( h 1 + Δ h ) → ρ 1 h 1 = ρ 2 h 1 + ρ 2 Δ h → Δ h h 1 = ρ 1 − ρ 2 ρ 2 (1) The density of water at T 1 = 10 ◦ C an d T 2 = 100 ◦ C are obtained from Appendix B.1 as ρ 1 = 999 . 7 kg/m 3 and ρ 2 = 958 . 4 kg/m 3 , respectively. Using Equation 1 gives Δ h h 1 = 999 . 7 − 958 . 4 958 . 4 = 0 . 0431 = 4 . 31%

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    1.24. from the given data: β = 5 . 7 10 − 4 K − 1 , × T 1 = 10 ◦ C, and T 2 = 90 ◦ C. Applying Equation 1.19 (with Δ p = 0 ) gives Δ ρ ρ ≈ − β Δ T = − 5 . 7 × 10 − 4 (90 − 10) = − 0 . 0432 = − 4 . 32% 1.25. From the given data: T 1 = 15 ◦ C, Δ V /V 1 = 0 . 01, and β = 9 . 5 10 − 4 K − 1 × . Using the given data, V 2 − V 1 V = 0 . 01 → V 2 V − 1 = 0 . 01 → V 2 V = 1 . 01 (1) 1 1 1 Δ ρ ρ 1 = / / m/V 2 − / / m/V 1 / / m/V 1 → Δ ρ ρ 1 = V 1 V 2 − 1 (2) Δ ρ ρ 1 = − 5 . 7 × 10 − 4 Δ T (3) where Δ T is the maximum allowable temperature rise. Combining Equations 1 to 3 yields Δ T = 10 . 4 ◦ C . 1.26. From the given data: T = 20 ◦ C = 293 K. Noting that the speed of sound, c , is given by c = √ v E /ρ , the calculation of the speed of sound in water and mercury are summarized in the following table: Medium E v ( × 10 6 Pa) ρ (kg/m 3 ) c (m/s) Water 2171 998 1475 Mercury 26200 13550 1390 Therefore, the speed of sound in water at 20 ◦ C is 1475 m/s , and the speed of sound in mercury at 20 ◦ C is 1390 m/s . 1.27. From the given data: c = 1700 m/s, and SG = 1.8. For water at 4 ◦ C, ρ w = 1000 kg/m 3 . Using Equation 1.15, the bulk modulus, E v , is calculated as follows: c = E v SG ρ w → 1700 = E v (1 . 8)(1000) → E v = 3 . 06 × 10 6 Pa = 3 . 06 MPa / · / 1.28. From the given data: n 1 = 10 10 molecules per mm 3 , and T = 15 ◦ C = 288 K. For an ideal gas, R u = 8 . 314 J/(kg K). · (a) Using the given data with the ideal-gas law, n = 10 10 × 6 . 023 10 23 = 1 . 660 × 10 − 14 moles V = 1 mm 3 = 10 − 9 m 3 p = n V R u T = 1 . 660 × 10 − 14 10 − 9 (8 . 314)(288) = 0 . 0397 Pa

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    (b) The term used to describe a gas in which the continuum approximation is not valid is a rarefied gas . 1.29. From the given data: p = 101 kPa, and T = 25 ◦ C = 298 K. For He, R He = 8314 / 4 . 003 = 2077 J/(kg K), and for air, · R air = 8314 / 28 . 96 = 287 . 1 J/(kg K) (from Appendix B.5). Using · the ideal gas law, ρ He = p R He T = 101 × 10 3 (2077)(298) = 0 . 1632 kg/m 3 ρ air = p R air T = 101 × 10 3 (287 . 1)(298) = 1 . 181 kg/m 3 The specific volume is defined by Equation 1.11 as the volume per unit mass, hence v He = 1 ρ He = 1 0 . 1632 = 6 . 13 m 3 /kg v air = 1 ρ air = 1 1 . 181 = 0 . 847 m 3 /kg 1.30. For air at standard atmospheric pressure, p = 101 kPa and R = 287 . 1 J/(kg K). Taking · ρ 1 = density from Appendix B.2, and ρ 2 = density from ideal gas law, gives: T ( ◦ C) ρ 1 (kg/m 3 ) T ( K) ρ 2 (kg/m 3 ) Δ (%) − 40 1 . 514 233 1 . 5098 − 0 . 27 − 20 1 . 394 253 1 . 3905 − 0 . 25 0 1 . 292 273 1 . 2886 − 0 . 26 5 1 . 269 278 1 . 2654 − 0 . 28 10 1 . 246 283 1 . 2431 − 0 . 23 15 1 . 225 288 1 . 2215 − 0 . 29 20 1 . 204 293 1 . 2007 − 0 . 28 25 1 . 184 298 1 . 1805 − 0 . 29 30 1 . 164 303 1 . 1610 − 0 . 25 40 1 . 127 313 1 . 1239 − 0 . 27 50 1 . 092 323 1 . 0891 − 0 . 26 60 1 . 059 333 1 . 0564 − 0 . 24 70 1 . 028 343 1 . 0256 − 0 . 23 80 0 . 9994 353 0 . 9966 − 0 . 28 90 0 . 9718 363 0 . 9691 − 0 . 27 100 0 . 9458 373 0 . 9431 − 0 . 28 200 0 . 7459 473 0 . 7438 − 0 . 29 300 0 . 6158 573 0 . 6140 − 0 . 30 400 0 . 5243 673 0 . 5227 − 0 . 30 500 0 . 4565 773 0 . 4551 − 0 . 31 1000 0 . 2772 1273 0 . 2764 0 . 31 −

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    Based on these results, the ideal gas law gives quite accurate estimates with errors less than 0.31% . 1.31. From the given data: ρ = 5 kg/m 3 , and p = 450 kPa. Properties of O 2 from Appendix B.5: c p = 909 J/kg K, c v = 649 J/kg K, and R = · · c p c v − = 909 649 = 260 J/kg K. The ideal − · gas law, Equation 1.24, gives ρ = p RT → 5 = 450 × 10 3 (260) T → T = 346 K = 73 ◦ C 1.32. From the given data: V = 2 m 3 , T = 15 ◦ C = 288 K, and p = 500 kPa. The molar mass of helium is 4.003 g/mol, and hence the gas constant for helium can be taken as R = 8314/4.003 = 2077 J/kg K. The density, mass, and weight of helium in the tank are given by · ρ = p RT = 500 × 10 3 (2077)(288) = 0 . 08610 kg/m 3 M = ρV = (0 . 08610)(2) = 0 . 1722 kg W = M g = (0 . 1722)(9 . 807) = 1 . 689 N 1.33. From the given data: m = 10 kg, T = 15 C = 288 K, p = 10 MPa, and L = 3 D . For pure ◦ oxygen, R = R u /M = 8314 / 32 = 259 . 8 J/(kg · K). Using the given data and the ideal gas law, V = π 4 D 2 L = π 4 D 2 (3 D ) = 3 π 4 D 3 , m = pV RT → V = mRT p Combining these equations gives 3 4 πD 3 = mRT p → D = 4 mRT 3 πp 1 3 = 4(10)(259 . 8)(288) 3 π (10 10 6 ) 1 3 = 0 . 317 m ┌ ┐ ┌ × ┐ Since the length must be three times the diameter, L = 3(0 . 317) = 0 . 950 m. The required dimensions of the tank are a diameter of 317 mm and a length of 0.950 m . 1.34. From the given data: M = 10 kg, T = 60 ◦ C = 333 K, and p = 200 kPa. For air, R = 287.1 J/kg · K. The volume, V , can be derived from the ideal gas law, Equation 1.24, as follows M V = p RT → 10 V = 200 × 10 3 (287 . 1)(333) → V = 4 . 78 m 3 1.35. From the given data: V = 200 L, m = 3 kg, and T = 15 ◦ C = 288 K. This assumes that the temperature of the air in the tank is the same as in the room. For standard air, R = 287 . 1 J/(kg · K). Using the ideal-gas law gives ρ = p RT → p = m V RT = 3 0 . 2 (287 . 1)(288) = 1 . 24 × 10 6 Pa = 1 . 24 MPa

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    1.36 . From the given data: V = 0.1 m 3 , T = 20 C = 293 K, and p = 400 kPa. The gas constant ◦ for air can be taken as R = 287.1 J/kg · K. The density of air in the tank can be calculated using Equation 1.24, which gives ρ = p RT = 400 × 10 3 (287 . 1)(293) = 0 . 4898 kg/m 3 Hence the weight of air in the tank , W , is given by W = ρV g = (0 . 4898)(0 . 1)(9 . 807) = 0 . 4803 N 0 . 48 N ≃ The weight (0.48 N) has been rounded to two significant digits to be consistent with the accuracy of the given data. 1.37. From the given data: V = 10 m 12 m 4 m = 480 m 3 , × × p = 101 . 3 kPa, T 1 = 20 ◦ C = 293 . 15 K, and T = 10 C = 283 . 15 K. 2 ◦ For air, R = 287 . 1 J/(kg K). · (a) Using the given data at a temperature of 20 ◦ C gives ρ 1 = p RT 1 = 101 . 3 × 10 3 (287 . 1)(293 . 15) = 1 . 204 kg/m 3 , m 1 = ρ 1 V = (1 . 204)(480) = 577 . 7 kg W 1 = m 1 g = (577 . 7)(9 . 807) = 5 . 666 10 3 N = 1274 lb × (b) When the temperature in the room is reduced to 10 ◦ C, then ρ 2 = p RT 2 = 101 . 3 × 10 3 (287 . 1)(283 . 15) = 1 . 246 kg/m 3 , m 2 = ρ 2 V = (1 . 246)(480) = 598 . 1 kg change = m 2 − m 1 m 1 × 100 = 598 . 1 − 577 . 7 577 . 7 × 100 = 3 . 53% 1.xx From the given data: V = 600 mL = 6 10 − 4 m 3 , × p 1 = 101 . 3 kPa, T = 30 C = 303 . 15 K, 1 ◦ T 2 = 1 . 5 C = 274 . 65 K. ◦ Applying the ideal gas law, assuming that the number of moles of air above the water remains constant: pV = nRT → n = pV RT → p 1 T 1 = p 2 T 2 → p 2 = p 1 T 2 T 1 = (101 . 3) 274 . 65 303 . 15 = 91 . 78 kPa Therefore, the pressure differential is 101 . 3 91 . 78 = 9.5 kPa − 1.38. From the given data: p 1 = 600 kPa, T = 20 C = 293 K, 1 ◦ and T 2 = 30 ◦ C = 303 K. Taking ρ and R as constants, the ideal gas law gives p 1 T 1 = p 2 T 2 → 600 293 = p 2 303 → p 2 = 620 kPa Therefore the change in pressure is 620 kPa − 600 kPa = 20 kPa . This result would be the same for any gas that obeys the ideal gas law.

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    1.39. From the given data: p 1 = 130 kPa, p 2 = 210 kPa, V = 15 L = 0 . 015 m 3 , and T = 30 ◦ C. Required constants: R = 8 . 314 kJ/kmol K, molar mass of air, M = 28.97 kg/kmol, · p atm = 101 kPa. Using these data with the ideal gas law, n 1 = p 1 V 1 RT 1 = (101 + 130)(0 . 015) (8 . 314)(273 . 15 + 30) = 0 . 00137 kmol n 2 = p 2 V 2 RT = (101 + 210)(0 . 015) (8 . 314)(273 . 15 + 30) = 0 . 00185 kmol 2 mass added = ( n 2 − n 1 ) M = (0 . 00185 − 0 . 00137)28 . 97 = 0 . 0139 kg = 13 . 9 g 1.40. Consider three states of the tire: State 1 is the initial state, State 2 is the heated-up state, and State 3 is the cooled-down state. From the given data: p atm = 101 . 3 kPa, V = 15 L , p 1 (gauge) = 207 kPa, p 1 (absolute) = 207 kPa + 101.3 kPa = 308.3 kPa, T 1 = 20 ◦ C = 293 K, T 2 = 60 ◦ C = 333 K, and T 3 = T 1 = 293 K. (a) Between the initial state and the heated-up state, the air density in the tire remains constant, so the ideal-gas law gives ρR = p T = constant → p 1 T 1 = p 2 T 2 → p 2 = T 2 T 1 p 1 = 333 293 (207) = 350 kPa Hence, the resulting gauge pressure is 350 kPa 101.3 kPa = 249 kPa = − 36 lb/in 2 . (b) Between the heated-up state and the cooled-down state, the air density in the tire remains constant and the air pressure at the heated-up state is equal to 308.3 kPa. The ideal-gas law gives ρR = p T = constant → p 3 T 3 = p 2 T 2 → p 3 = T 3 T 2 p 2 = 293 333 (308 . 3) = 271 kPa Hence, the resulting gauge pressure is 271 kPa − 101.3 kPa = 170 kPa = 25 lb/in 2 . 1.41. From the given data: T 1 = 20 ◦ C = 293 K, T 2 = 10 ◦ C = 283 K, p 1g = 12 . 5 lb/in 2 , and p 1 = p 1g + p atm = 12 . 5 + 14 . 7 = 27 . 2 lb/in 2 . (a) In this case the expected pressure, p 2 in a football is given by p 2 = T 2 T 1 p 1 = 283 293 (27 . 2) = 26 . 3 lb/in 2 → p 2g = p 2 − p atm = 26 . 2 − 14 . 7 = 11 . 6 lb/in 2 (b) In this case the required temperature, T 2 on the field is given by p 2 = T 2 T 1 p 1 → (10 . 5 + 14 . 7) = T 2 293 (27 . 2) = 272 K = − 2 ◦ C (c) Theory: The New England Patriots did not inflate the footballs to the required pressure. 1.42. From the given data: p 1 = 207 kPa and p 2 = 241 kPa.

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    (a) From the ideal-gas law, ρ = p RT → p 1 T 1 = p 2 T 2 → T 2 T 1 = p 2 p 1 = 241 207 = 1 . 164 Using this relationship, the percentage change in temperature is calculated as follows, % change in T = T 2 − T 1 T 1 × 100 = T 2 T 1 − 1 × 100 = [1 . 164 − 1] × 100 = 16 . 4% ┌ ┐ Note that this percentage change applies to a temperature given in degrees kelvin. (b) From the given data: T 1 = 25 ◦ C = 298 K. 1 . 164(298) = 347 K = 74 ◦ C . Hence, from the result in Part (a), T 2 = Note that thhe percentage change in ◦ C is much higher than the percentage change in K. 1.43. (a) From the given data: D = 6 m, T = 20 ◦ C = 293 K, p = 200 kPa, R u = 8312 J/(kmol K), · and V = πD 3 / 6 = 113 . 1 m 3 . For an ideal gas: pV = nRT → (200 × 10 3 )(113 . 1) = n (8312)(293) → n = 9 . 29 kmol Since the relative molecular mass of He is 4.003, the mass of 9.29 kmol of He is (9.29)(4.003)kg = 37.2 kg . (b) From the given data: T 1 = 25 ◦ C = 298 K, p 1 = 210 kPa + 101 kPa = 311 kPa (assuming atmospheric pressure is 101 kPa), V = 0 . 025 m 3 , an d T 2 = 50 ◦ C = 323 K. From the ideal gas law, V = nRT 1 p 1 = nRT 2 p 2 → p 2 = p 1 T 2 T 1 = 311 323 298 = 337 kPa ( ) ( ) This corresponds to a pressure increase of 337 kPa 311 kPa = 26 kPa . − 1.44. From given data: W rat = 1.5 N, W bal = 0.5 N, ρ air = 1.17 kg/m 3 , p = 100 kPa, T = 25 ◦ C = 298 K. Take R u = 8.314 J/mol K · and molar mass of He, m He , is 4.003 g/mol. To lift the rat: W rat + W bal + ρ He gV bal ' '' ' Weight lifted = ρ air gV bal ' '' ' buoyant force → W rat + W bal = ( ρ air − ρ He ) gV bal (1) For helium in the balloon, R He = R u m He = 8 . 314 4 . 003 = 2 . 077 J/g · K = 2077 J/kg · K ρ He = p R He T = 100 × 10 3 (2077)(298) = 0 . 162 kg/m 3 V bal = nR u T p Substituting into Equation (1) gives 1 . 5 + 0 . 5 = (1 . 17 − 0 . 162)(9 . 81) n (8 . 314)(298) 100 10 3 → n = 8 . 16 mol ┌ × ┐ Hence, the mass of helium required is 8.16 × 4.003 = 32.7 g

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    1.45. Form the given data: z = 10 m, D 1 = 5.0 mm, T 1 = 20 ◦ C, and p atm = 101.3 kPa . At the bubble release location, p 1 = p atm + γz = 101 . 3 + 9 . 79(10) = 199 . 2 kPa , T 1 = 20 ◦ C At the surface of the lake, p 2 = 101.3 kPa and for an isothermal process, p 1 V 1 = p 2 V 2 → V 2 = p 1 p 2 V 1 → 4 3 πr 3 2 = p 1 p 2 4 3 πr 3 1 → r 2 r 1 = D 2 D 1 = p 1 p 2 1 3 → D 2 = ( p 1 p 2 ) 1 3 D 1 = ( 199 . 2 101 . 3 ) 1 3 (5) = 6 . 3 mm ( ) ( ) ( ) Therefore, the diameter of the bubble when it reaches the surface of the lake is 6.3 mm . 1.46. From the given data: V 1 = 1.0 m 3 , V 2 = 0.4 m 3 , an d p 1 = 101 kPa. For air, the ratio of specific heats is given by k = 1.40. (a) Under isentropic conditions, the pressure of the compressed volume, p 2 , is g i ven b y Equations 1.33 and 1.34 as p 1 v k 1 = p 2 v k 2 → (101)(1 . 0) 1 . 40 = p 2 (0 . 4) 1 . 40 → p 2 = 364 kPa (b) Under isothermal conditions, the pressure of the compressed volume, p 2 , is g iven by Equations 1.33 and 1.34 as p 1 v 1 = p 2 v 2 → (101)(1 . 0) = p 2 (0 . 4) → p 2 = 253 kPa 1.xx From the given data: V 0 = 1 L = 10 − 3 m 3 , P 0 = 101 . 3 kPa, p vap = 2 . 337 kPa, T 0 = 20 ◦ C = 293 K, and R = 8 . 3142 J/(mol K). Since, under initial conditions the gas is 50% CO 2 , the · total number of moles, n T , of CO 2 os given by n T = (0 . 5 P 0 ) V 0 RT 0 = (0 . 5 × 101 . 3 × 10 3 )(10 − 3 ) (8 . 3142)(293) = 0 . 02079 mol (1) When filled with 0.75 L of water, the number of moles of CO 2 in the water, n w , is given b y n w = C CO2 · V w = P CO2 H · V w = P CO2 2 . 937 10 6 · (0 . 75) → n w = 2 . 554 × 10 − 7 P CO2 (2) × In the space above the water, the ideal gas law is obeyed such that the number of moles of CO 2 in the gaseous phase is given by n g = P CO2 V g RT 0 = P CO2 (0 . 25 × 10 − 3 ) (8 . 3142)(293) → n g = 1 . 026 × 10 − 7 P CO2 (3) Since the total amount of CO 2 does not change, then Equations1 to 3 can be combined to give n T = n w + n g → 0 . 02079 = (2 . 554+1 . 026) × 10 − 7 P CO2 → P CO2 = 58078 Pa = 58 . 1 kPa

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    The (partial) pressure, P O , of the other gas in the space above the water surface is given by P O = (0 . 5 P 0 ) V 0 V g = (0 . 5 × 101 . 3) 1 0 . 25 = 202 . 6 kPa Therefore, the total pressure, P , in the space above the liquid surface is equal to the sum of the partial pressures of the two gasses plus the vapor pressure of water (2.337 kPa), which gives P = P CO2 + P O + p vap = 58 . 1 + 202 . 6 + 2 . 337 = 263 . 0 kPa 1.47. From the given data: T = 20 ◦ C = 293 K, p = 101 . 3 Pa, f 1 = 0 . 20, and f 2 = 0 . 8. For O 2 , R 1 = 259 . 8 J/(kg K), · and for N 2 , R 2 = 296 . 7 J/(kg K). · Under the given conditions, the densities of O 2 and N 2 , represented by ρ 1 and ρ 2 , are given by ρ 1 = p R 1 T = 101 . 3 × 10 3 (259 . 8)(293) = 1 . 331 kg/m 3 , ρ 2 = p R 2 T = 101 . 3 × 10 3 (296 . 7)(293) = 1 . 166 kg/m 3 (a) For each 1 m 3 of air there is f 1 m 3 of O 2 and f 2 m 3 of N 2 . Therefore, the partial pressures of O 2 and N 2 , denoted by p 1 and p 2 , are given b y p 1 = f 1 p = (0 . 2)(293) = 20 . 3 kPa , p 2 = f 2 p = (0 . 8)(293) = 81 . 0 kPa (b) The density, ρ m of the mixture is given by ρ m = f 1 ρ 1 + f 2 ρ 2 = (0 . 2)(1 . 331) + (0 . 8)(1 . 166) = 1 . 199 kg/m 3 1.48. From the given data: V 1 = 2.0 m 3 , V 2 = 4.0 m 3 , T = 20 C = 293 K, 1 ◦ and p 1 = 100 kPa. For oxygen, c p = 909 J/kg K, · c v = 649 J/kg K, · and k = c p /c v = 909 / 649 = 1 . 40. The gas constant for oxygen, R , is given by R = c p c v = 909 649 = 260 J/kg K − − · (a) Under isentropic conditions, the pressure of the expanded volume, p 2 , is given by Equa- tions 1.33 and 1.34 as p 1 v k 1 = p 2 v k 2 → (100)(2 . 0) 1 . 40 = p 2 (4 . 0) 1 . 40 → p 2 = 37 . 9 kPa The initial density, ρ 1 is given by the ideal gas law as follows, ρ 1 = p 1 RT 1 = 100 × 10 3 (260)(293) = 1 . 131 kg/m 3 The ratio of the final density to the initial density is given by Equation 1.37 as follows: ρ 2 ρ 1 = ( p 2 p 1 ) 1 k → ρ 2 1 . 131 = ( 37 . 9 100 ) 1 1 . 40 → ρ 2 = 0 . 566 kg/m 3

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