fluid mechanics 1st & 2nd edition Hibbeler solutions manual pdf | solution

1. https://gioumeh.com/product/fluid-mechanics-solutions/

2. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–1. Represent each of the following quantities with combinations of units in the correct SI form, using an (d) kN>(mm)2. appropriate prefix: (a) mm#MN, (b) Mg>mm, (c) km>ms, SOLUTION a) mm#MN = (10-3 m)(106 N) = 103 N#m = kN#m b) Mg>mm = (106 g2>(10-3 m2 = 109 g>m = Gg>m d) kN>1mm22= (103 N2>(10-3 m)2= 109 N>m2= GN>m2 Ans. c) km>ms = (103 m)>110-3 s2 = 106 m>s = Mm>s Ans. Ans. Ans. Ans: a) kN#m c) Mm>s b) Gg>m d) GN>m2 1 M01_HIBB9290_01_SE_C01_ANS.indd 1 22/02/17 4:50 PM

3. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–2. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) 34.86(106)42 mm, (b) (348mm)3, (c) (83700 mN)2. a) 34.86(106)42 mm = 34.861106242110-3 m2 = 23.6211092 m = 23.6 Gm Ans. SOLUTION b) (348 mm)3= 3348(10-3) m43= 42.14(10-3) m3= 42.1110-32 m3 c) 183,700 mN22= 383,700(10-3) N42= 7.006(103) N2= 7.01(1032 N2 Ans. Ans. Ans: a) 23.6 Gm c) 7.01(1032 N2 b) 42.1110-32 m3 2 M01_HIBB9290_01_SE_C01_ANS.indd 2 22/02/17 4:50 PM

4. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–3. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) 749 mm>63 ms, (b) (34 mm)(0.0763Ms)>263mg, (c) (4.78mm)(263Mg). SOLUTION a) 749 mm>63 ms = 749(10-6) m>63(10-3) s = 11.88(10-3) m>s = 9.86(106) m#s>kg = 9.86 Mm#s>kg = 1.257(106) g#m = 1.26 Mg#m = 11.9 mm>s b) (34 mm)(0.0763 Ms)>263 mg = 334110-32 m430.076311062s4>3263110-6211032 g4 Ans. c) (4.78 mm)(263 Mg) = 34.78(10-3) m43263(106) g4 Ans. Ans. a) 11.9 mm>s Ans: b) 9.86 Mm#s>kg c) 1.26 Mg#m 3 M01_HIBB9290_01_SE_C01_ANS.indd 3 22/02/17 4:50 PM

5. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *1–4. Convert the following temperatures: (a) 250 K to degrees Celsius, (b) 322°F to degrees Rankine, (c) 230°F to degrees Celsius, (d) 40°C to degrees Fahrenheit. SOLUTION a) TK= TC+ 273; 250 K = TC+ 273 TC= -23.0°C Ans. b) TR= TF+ 460 = 322°F + 460 = 782°R 91TF- 322 =5 d) TC=5 Ans. 91230°F - 322 = 110°C c) TC=5 Ans. 91TF- 322; 40°C =5 91TF- 322TF= 104°F Ans. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. Ans: a) -23.0°C b) 782°R c) 110°C d) 104°F 4 M01_HIBB9290_01_SE_C01_ANS.indd 4 22/02/17 4:50 PM

6. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.22 slug>ft3. Determine the weight of the liquid when it is 1–5. The tank contains a liquid having a density of 4 ft 2 ft 1 ft at the level shown. 3 ft SOLUTION The specific weight of the liquid and the volume of the liquid are g = rg = (1.22 slug>ft3)(32.2 ft>s2) = 39.284 lb>ft3 V = (4 ft)(2 ft)(2 ft) = 16 ft3 Then the weight of the liquid is W = gV = (39.284 lb>ft3)(16 ft3) = 628.54 lb = 629 lb Ans. Ans: W = 629 lb 5 M01_HIBB9290_01_SE_C01_ANS.indd 5 22/02/17 4:50 PM

7. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–6. If air within the tank is at an absolute pressure of 680 kPa and a temperature of 70°C, determine the weight of the air inside the tank. The tank has an interior volume of 1.35 m3. SOLUTION From the table in Appendix A, the gas constant for air is R = 286.9 J>kg#K. p = rRT r = 6.910 kg>m3 The weight of the air in the tank is W = rgV = (6.910 kg>m3)(9.81 m>s2)(1.35 m3) 680(103) N>m2= r(286.9 J>kg#K)(70° + 273) K Ans. = 91.5 N This work and Ans: W = 91.5 N 6 M01_HIBB9290_01_SE_C01_ANS.indd 6 22/02/17 4:50 PM

8. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–7. The bottle tank has a volume of 0.35 m3 and contains 40 kg of nitrogen at a temperature of 40°C. Determine the absolute pressure in the tank. SOLUTION The density of nitrogen in the tank is 0.35 m3= 114.29 kg>m3 40 kg r =m V= From the table in Appendix A, the gas constant for nitrogen is R = 296.8 J>kg#K. p = rRT = 10.6211062Pa or sale Applying the ideal gas law, p = (114.29 kg>m3)(296.8 J>kg#K)(40°C + 273) K = 10.6 MPa This work and Ans. Ans: p = 10.6 MPa 7 M01_HIBB9290_01_SE_C01_ANS.indd 7 22/02/17 4:50 PM

9. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *1–8. The bottle tank contains nitrogen having a temperature of 60°C. Plot the variation of the pressure in the tank (vertical axis) versus the density for 0 … r … 5 kg>m3. Report values in increments of ∆p = 50kPa. SOLUTION From the table in Appendix A, the gas constant for nitrogen is R = 296.8 J>kg#K. gas law, p = r1296.8 J>kg#K)(333 K) The constant temperature is T = (60°C + 273) K = 333 K. Applying the ideal p = rRT p = 198,834r2 Pa = 198.8r2 kPa 250 2.53 Ans. r (kg>m3) The plot of p vs r is shown in Fig. a. p(kPa) p(kPa) 150 1.52 200 2.02 300 3.04 350 3.54 their courses and assessing student learning. Dissemination 400 4.05 This work is protected by United States copyright laws 400 and is provided solely for the use of instructors in teaching or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. 300 200 100 r(kgym3) 0 1 2 3 4 5 (a) p = 198.8r2 kPa Ans: 8 M01_HIBB9290_01_SE_C01_ANS.indd 8 22/02/17 4:50 PM

10. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–9. Determine the specific weight of hydrogen when the temperature is 85°C and the absolute pressure is 4 MPa. SOLUTION From the table in Appendix A, the gas constant for hydrogen is R = 4124 J>kg#K. p = rRT r = 2.7093 kg>m3 Then the specific weight of hydrogen is g = rg = (2.7093 kg>m3)(9.81 m>s2) their courses and assessing student learning. Dissemination Applying the ideal gas law, 4(106) N>m2= r(4124 J>kg#K)(85°C + 273) K = 26.58 N>m3 = 26.6 N>m3 This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching Ans. or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. g = 26.6 N>m3 Ans: 9 M01_HIBB9290_01_SE_C01_ANS.indd 9 22/02/17 4:50 PM

11. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–10. Dry air at 25°C has a density of 1.23 kg>m3. But if it 0.65% less. At what temperature would dry air produce this same smaller density? has 100% humidity at the same pressure, its density is SOLUTION r1= 1.23 kg>m3, r2= (1.23 kg>m3)(1 - 0.0065) = 1.222005 kg>m3 For both cases, the pressures are the same. Applying the ideal gas law with T1= (25°C + 273) = 298 K, p = r1RT1= (1.23 kg>m3)R(298 K) = 366.54 R Then p = r2RT2; 366.54 R = (1.222005 kg>m3)R(TC+ 273) and TC= 26.9°C Ans. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. Ans: TC= 26.9°C 10 M01_HIBB9290_01_SE_C01_ANS.indd 10 22/02/17 4:50 PM

12. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–11. The tanker carries 900(103) barrels of crude oil in its hold. Determine the weight of the oil if its specific gravity is 0.940. Each barrel contains 42 gallons, and there are 7.48gal>ft3. SOLUTION The specific weight of the crude oil is go= Sogw= 0.940(62.4 lb>ft3) = 58.656 lb>ft3 The volume of the crude oil is Vo= 3900(103) bl4a42 gal Then, the weight of the crude oil is Wo= goVo= 158.656 lb>ft3235.053511062 ft34 = 296(106) lb ba 7.48 galb = 5.0535(106) ft3 1 ft3 1 bl This work is protected by United States copyright laws = 296.41(106) lb and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination Ans. or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. Ans: Wo= 296(106) lb 11 M01_HIBB9290_01_SE_C01_ANS.indd 11 22/02/17 4:50 PM

13. https://gioumeh.com/product/fluid-mechanics-solutions/ © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *1–12. Water in the swimming pool has a measured depth of 3.03 m when the temperature is 5°C. Determine its approximate depth when the temperature becomes 35°C. Neglect losses due to evaporation. 9 m 4 m SOLUTION From Appendix A,at T1= 5°C, 1rw21= 1000.0 kg>m3. The volume of the water is m V1 V = Ah. Thus, V1= (9 m)(4 m)(3.03 m). Then ; 1000.0 kg>m3= m = 109.08(103) kg m (rw)1= 36 m2(3.03 m) At T2= 35°C, (rw)2= 994.0 kg>m3. Then (rw)2= 994.0 kg>m3= h = 3.048 m = 3.05 m will destroy the integrity of the work and is not permitted. 109.08(103) (36 m2)h m V2 This work is protected by United States copyright laws ; and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) Ans. Ans: h = 3.05 m 12 M01_HIBB9290_01_SE_C01_ANS.indd 12 22/02/17 4:50 PM