1 1. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) (45 mn), (b) (67 300 ms), (c) 373(10 6 )4 1> mm. a) (45 mn) = 345(10-3 ) N4 = 0.181 N b) (67 300 ms) = 367.3(10 3 )(10-3 ) s4 = 4.53(10 3 ) s c) 373(10 6 ) 4 1> mm = 373(10 6 ) 4 1> (10-3 ) m = 6.9 m a) 0.181 N b) 4.53(10 3 ) s c) 6.9 m 1
1. 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) (0.0763 Ms)>63 mg, (c) (4.78 mm)(63 Mg). a) 749 mm>63 ms = 749(10-6 ) m>63(10-3 ) s = 11.88(10-3 ) m>s = 11.9 mm>s b) (34 mm)(0.0763 Ms)>63 mg = 334(10-3 ) m4 30.0763(10 6 ) s4 > 363(10-6 )(10 3 ) g4 c) (4.78 mm)(63 Mg) = 34.78(10-3 ) m4 363(10 6 ) g4 = 9.86(10 6 ) m # s>kg = 9.86 Mm # s>kg = 1.57(10 6 ) g # m = 1.6 Mg # m a) 11.9 mm>s b) 9.86 Mm # s>kg c) 1.6 Mg # m
1 3. Represent each of the following quantities with combinations of units in the correct SI form, using an appropriate prefix: (a) GN # mm, (b) kg >mm, (c) N>ks, (d) kn>ms. a) GN # mm = (10 9 )N(10-6 )m = 10 3 N # m = kn # m b) kg>mm = (10 3 )g>(10-6 )m = 10 9 g>m = Gg>m c) N>ks = N>(10 3 s) = 10-6 N>s = mn>s d) kn>ms = (10 3 )N>(10-6 )s = 10 9 N>s = GN>s a) kn # m b) Gg>m c) µn>s d) GN>s 3
*1 4. Convert the following temperatures: (a) 0 C to Kelvin, (b) 500 K to degrees Celsius. a) T K = T C + 73 = 0 C + 73 = 93 K b) T K = T C + 73; 500 K = T C + 73 T C = 7 C 4
1 5. Mercury has a specific weight of 133 kn>m 3 when the temperature is 0 C. Determine its density and specific gravity at this temperature. g = rg 133(10 3 ) N>m 3 = r Hg (9.81 m>s ) r Hg = 13 558 kg>m 3 = 13.6 Mg>m 3 S Hg = r Hg r w = 13 558 kg>m3 1000 kg>m 3 = 13.6 r Hg = 13.6 Mg>m 3 S Hg = 13.6 5
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 m 3. From the table in Appendix A, the gas constant for air is R = 86.9 J>kg # K. p = rrt 680(10 3 ) N>m = r(86.9 J>kg # K)(70 + 73) K r = 6.910 kg>m 3 The weight of the air in the tank is W = rg V = (6.910 kg>m 3 )(9.81 m>s )(1.35 m 3 ) = 91.5 N 91.5 N 6
1 7. The bottle tank has a volume of 1.1 m 3 and contains oxygen at an absolute pressure of 1 MPa and a temperature of 30 C. Determine the mass of oxygen in the tank. From the table in Appendix A, the gas constant for oxygen is R = 59.8 J>kg # K. p = rrt 1(10 6 ) N>m = r(59.8 J>kg # K)(30 + 73) K r = 15.44 kg>m 3 The mass of oxygen in the tank is m = rv = (15.44 kg>m 3 )(0.1 m 3 ) = 18.3 kg 18.3 kg 7
*1 8. The bottle tank has a volume of 0.1 m 3 and contains oxygen at an absolute pressure of 8 MPa and temperature of 0 C. Plot the variation of the pressure in the tank (vertical axis) versus the temperature for 0 C T 80 C. Report values in increments of T = 10 C. T C ( C) 0 30 40 50 60 70 80 p(mpa) 8.00 8.7 8.55 8.8 9.09 9.37 9.64 p(mpa) 10 9 From the table in Appendix A, the gas constant for oxygen is R = 59.8 J>(kg # K). For T = (0 C + 73) K = 93 K, p = rrt 8(10 6 ) N>m = r359.8 J>(kg # K)4(93 K) r = 105.10 kg>m 3 Since the mass and volume of the oxygen in the tank remain constant, its density will also be constant. p = rrt p = (105.10 kg>m 3 ) 359.8 J>(kg # K)4(TC + 73) p = (0.0730 T C + 7.4539)(10 6 ) Pa p = (0.0730T C + 7.4539) MPa where T C is in C. 8 7 6 5 4 3 1 0 10 0 30 40 50 (a) 60 70 80 T C (ºC) The plot of p vs T C is shown in Fig. a. 8
1 9. Determine the specific weight of carbon dioxide when the temperature is 100 C and the absolute pressure is 400 kpa. From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>kg # K. p = rrt 400(10 3 ) N>m = r(188.9 J>kg # K)(100 + 73) K r = 5.677 kg>m 3 The specific weight of carbon dioxide is g = rg = (5.677 kg>m 3 )(9.81 m>s ) = 55.7 N>m 3 55.7 N>m 3 9
1 10. Dry air at 5 C has a density of 1.3 kg>m 3. But if it has 100% humidity at the same pressure, its density is 0.65% less. At what temperature would dry air produce this same density? For both cases, the pressures are the same. Applying the ideal gas law with r 1 = 1.3 kg>m 3, r = (1.3 kg>m 3 )(1-0.0065) = 1.005 kg>m 3 and T 1 = (5 C + 73) = 98 K, Then p = r 1 RT 1 = (1.3 kg>m 3 ) R (98 K) = 366.54 R p = r RT ; 366.54 R = (1.005 kg>m 3 )R(T C + 73) T C = 6.9 C 6.9 C 10
1 11. The tank contains air at a temperature of 15 C and an absolute pressure of 10 kpa. If the volume of the tank is 5 m 3 and the temperature rises to 30 C, determine the mass of air that must be removed from the tank to maintain the same pressure. For T 1 = (15 + 73) K = 88 K and R = 86.9 J>kg # K for air, the ideal gas law gives p 1 = r 1 RT 1 ; 10(10 3 ) N>m = r 1 (86.9 J>kg # K)(88 K) Thus, the mass of air at T 1 is r 1 =.5415 kg>m 3 m 1 = r 1 V = (.5415 kg>m 3 )(5 m 3 ) = 1.70768 kg For T = (73 + 30) K = 303 K and R = 86.9 J>kg # K p = r 1 RT ; 10(10 3 ) N>m = r (86.9 J>kg # K)(303 K) r =.4157 kg>m 3 Thus, the mass of air at T is m = r V = (.4157 kg>m 3 )(5 m 3 ) = 1.07886 kg Finally, the mass of air that must be removed is m = m 1 - m = 1.70768 kg - 1.07886 kg = 0.69 kg 0.69 kg 11
*1 1. 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 From Appendix A, at T 1 = 5 C, 1r w 1 = 1000.0 kg>m 3. The volume of the water is V = Ah. Thus, V 1 = (9 m)(4 m)(3.03 m). Then (r w ) 1 = m ; 1000.0 kg>m 3 m = V 1 36 m (3.03 m) m = 109.08(10 3 ) kg At T = 35 C, (r w ) = 994.0 kg>m 3. Then (r w ) = m V ; 994.0 kg>m 3 = 109.08(103 ) (36 m )h h = 3.048 m = 3.05 m 1
1 13. The tank initially contains carbon dioxide at an absolute pressure of 00 kpa and temperature of 50 C. As more carbon dioxide is added, the pressure is increasing at 5 kpa>min. Plot the variation of the pressure in the tank (vertical axis) versus the temperature for the first 10 minutes. Report the values in increments of two minutes. Assume the tank expands to keep the density constant. p(kpa) 00 5 50 75 300 35 T C ( C) 50.00 90.38 130.75 171.1 11.50 51.88 From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>(kg # K). For T = (50 C + 73) K = 33 K, p = rrt 00(10 3 ) N>m = r3188.9 J>(kg # K)4(33 K) r = 3.779 kg>m 3 Since the mass and the volume of carbon dioxide in the tank remain constant, its density will also be constant. p = rrt The plot of p vs T C is shown in Fig. a p = (3.779 kg>m 3 )3188.9 J>(kg # K)4(TC + 73) K p = (0.619 T C + 169.04)(10 3 ) Pa p = (0.619 T C + 169.04) kpa where T C is in C p(kpa) 350 300 50 00 150 100 50 0 50 100 150 00 50 300 (a) T C (ºC) p = (0.619 T c + 169) kpa, where T c is in C 13
1 14. A -kg mass of oxygen is held at a constant temperature of 50 and an absolute pressure of 0 kpa. Determine its bulk modulus. E V = - p = rrt dp = drrt dp dv>v = - dpv dv E V = - drrt V dv = - drpv rdv r = m V dr = - mdv V E V = mdv pv V (m>v)dv = P = 0 kpa Note: This illustrates a general point. For an ideal gas, the isothermal (constanttemperature) bulk modulus equals the absolute pressure. 0 kpa 14
1 15. The tank contains kg of air at an absolute pressure of 400 kpa and a temperature of 0 C. If 0.6 kg of air is added to the tank and the temperature rises to 3 C, determine the pressure in the tank. For T 1 = 0 + 73 = 93 K, p 1 = 400 kpa and R = 86.9 J>kg # K for air, the ideal gas law gives p 1 = r 1 RT 1 ; 400(10 3 ) N>m = r 1 (86.9 J>kg # K)(93 K) Since the volume is constant. Then r 1 = 4.7584 kg>m 3 V = m 1 r 1 = m r ; r = m m 1 r 1 Here m 1 = kg and m = ( + 0.6) kg =.6 kg.6 kg r = a kg b (4.7584 kg>m3 ) = 6.1859 kg>m 3 Again applying the ideal gas law with T = (3 + 73) K = 305 K p = r RT = (6.1859 kg>m 3 )(86.9 J>kg # k)(305 K) = 541.30(103 ) Pa = 541 kpa 541 kpa 15
*1 16. The 8-m-diameter spherical balloon is filled with helium that is at a temperature of 8 C and a pressure of 106 kpa. Determine the weight of the helium contained in the balloon. The volume of a sphere is V = 4 3 pr 3. For Helium, the gas constant is R = 077 J>kg # K. Applying the ideal gas law at T = (8 + 73) K = 301 K, Here p = rrt; 106(10 3 ) N>m = r(077 J>kg # K)(301 K) Then, the mass of the helium is Thus, r = 0.1696 kg>m 3 V = 4 3 pr 3 = 4 3 p(4 m)3 = 56 3 p m3 M = r V = (0.1696 kg>m 3 )a 56 3 p m3 b = 45.45 kg W = mg = (45.45 kg)(9.81 m>s ) = 445.90 N = 446 N 16
1 17. What is the increase in the density of helium when the pressure changes from 30 kpa to 450 kpa while the temperature remains constant at 0 C? This is called an isothermal process. Applying the ideal gas law with T 1 = (0 + 73) K = 93 K, p 1 = 30 kpa and R = 077 J>(kg # k), p 1 = r 1 RT 1 ; 30(10 3 ) N>m = r 1 (077 J>(kg # K))(93 K) r 1 = 0.3779 kg>m 3 For p = 450 kpa and T = (0 + 73) K = 93 K, p = r RT ; 450(10 3 ) N>m = r (077 J>(kg # k))(93 K) Thus, the change in density is r = 0.7394 kg>m 3 r = r - r 1 = 0.7394 kg>m 3-0.3779 kg>m 3 = 0.3615 kg>m 3 = 0.36 kg>m 3 0.36 kg>m 3 17
1 18. Sea water has a density of 1030 kg>m 3 at its surface, where the absolute pressure is 101 kpa. Determine its density at a depth of 7 km, where the absolute pressure is 70.4 MPa. The bulk modulus is.33 GPa. Since the pressure at the surface is 101 kpa, then p = 70.4-0.101 = 70.3 MPa. Here, the mass of seawater is constant. To find V 0 >V, use E V = -dp> (dv>v). So, M = r 0 V 0 = rv r = r 0 a V 0 V b L r = r 0 e p>e V dv V = - 1 E V L dp ln a V V 0 b = - 1 E V p V 0 V = e p>e V = (1030 kg>m 3 (70.3 MPa>.33 GPa) )e = 1061.55 kg>m 3 = 1.06(10 3 ) kg>m 3 1.06(10 3 ) kg>m 3 18
1 19. The tank is fabricated from steel that is 0 mm thick. If it contains carbon dioxide at an absolute pressure of 1.35 MPa and a temperature of 0 C, determine the total weight of the tank. The density of steel is 7.85 Mg>m 3, and the inner diameter of the tank is 3 m. Hint: The volume of a sphere is V = 4 3 pr 3. From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>kg # K. p = rrt 1.35(10 6 ) N>m = r co (188.9 J>kg # K)(0 + 73) K r co = 4.39 kg>m 3 Then, the total weight of the tank is W = r st g V st + r co g V co W = 37.85(10 3 ) kg>m 3 4 (9.81 m>s )a 4 3 b(p)c a3.04 3 mb - a 3.00 3 mb d + (4.39 kg>m 3 )(9.81 m>s )a 4 3 3 b(p)a3.00 mb W = 47.5 kn 47.5 kn 19
*1 0. Water at 0 C is subjected to a pressure increase of 44 MPa. Determine the percent increase in its density. Take E V =.0 GPa. r r 1 = m>v - m>v 1 m>v 1 = V 1 V - 1 To find V 1 >V, use E V = -d p > (dv>v). dv V = - dp E V L V 1 V dv V = - 1 E V L ln a V 1 V b = 1 E V p V 1 V = e p>e V So, since the bulk modulus of water at 0 C is E V =.0 GPa, r r 1 = e p>e V - 1 p 1 p dp = e (44 MPa)>.0 GPa) - 1 = 0.00 =.0, 0
1 1. The rain cloud has an approximate volume of 1150 km 3 and an average height, top to bottom, of 105 m. If a cylindrical container m in diameter collects 50 mm of water after the rain falls out of the cloud, estimate the total mass of rain that fell from the cloud. 105 m The volume of the rain water collected is m V r = p(1 m) (0.05 m) = 0.05p m 3 Then the mass of the rain water collected is m r = r w V r = (1000 kg>m 3 )(0.05p m 3 ) = 50p kg The volume of the cloud that contain this same amount of water is Thus, the density of the cloud is V c = p(1 m) (105 m) = 105p m 3 r c = m c V c = Then total mass of the cloud is 50p kg 105p m 3 = 0.476 kg>m 3 m c = r c V c = (0.476 kg/m 3 )c (1150 km 3 ) a 10000 m 1 km b 3 d = 547.6(10 9 ) kg = 548(10 9 ) kg 548(10 9 ) kg 1
1. The container is filled with water at a temperature of 5 C and a depth of.5 m. If the container has a mass of 30 kg, determine the combined weight of the container and the water. 1 m.5 m From Appendix A, r w = 997.1 kg>m 3 at T = 5 C. Here the volume of water is V = pr h = p(0.5 m) (.5 m) = 0.65p m 3 Thus, the mass of water is M w = r w V = 997.1 kg>m 3 (0.65p m 3 ) = 1957.80 kg The total mass is M T = M w + M c = (1957.80 + 30) kg = 1987.80 kg Then the total weight is W T = M T g = (1987.80 kg)(9.81 m>s ) = 19500 N = 19.5 kn 19.5 kn
1 3. If 4 m 3 of helium at 100 kpa of absolute pressure and 0 C is subjected to an absolute pressure of 600 kpa while the temperature remains constant, determine the new density and volume of the helium. From the table in Appendix A, the gas constant for helium is R = 077 J>kg # K, p 1 = r 1 RT 1 100(10 3 ) N>m 3 = r(077 J>kg # K)(0 + 73) K r 1 = 0.1643 kg>m 3 T 1 = T p 1 = r 1RT 1 p r RT p 1 = r 1 p r 100 kpa 0.1643 kg>m3 = 600 kpa r r = 0.9859 kg>m 3 = 0.986 kg>m 3 The mass of the helium is m = r 1 V 1 = (0.1643 kg>m 3 )(4 m 3 ) = 0.6573 kg Since the mass of the helium is constant, regardless of the temperature and pressure, m = r V 0.6573 kg = (0.9859 kg>m 3 )V V = 0.667 m 3 r = 0.986 kg>m 3 V = 0.667 m 3 3
*1 4. A solid has a density of 4500 kg>m 3. When a change in pressure of 5.50 MPa is applied, the density increases to 4750 kg>m 3. Determine the approximate bulk modulus. Differentiate V = m, we obtain r dv = - mdr r Substitute these results into Eq 1 1, E V = - dp dv>v = - dp a - mdr = r b n(m>r) Therefore E V = dp dr>r 5.50110 6 Pa a 4750 kg>m3-4500 kg>m 3 = 99.0110 6 ) Pa = 99.0 MPa 4500 kg>m 3 b The more precise solution can be obtained from Then p E V = L dp p 0 rdr L r 0 r = p - p 0 ln a r b r 0 E V = 5.50110 6 Pa 4750 kg>m3 ln a 4500 kg>m 3 b = 101.7(10 6 ) Pa = 10 MPa 4
1 5. If the bulk modulus for water at 0 C is.0 GPa, determine the change in pressure required to reduce its volume by 0.3%. Here it is regained that dv V = - 0.3 = 0.003. The application of Eq. 1 8 gives 100 E V = - dp dv>v ;.01109 N>m = - dp -0.003 p = dp = 6.60(10 6 ) Pa = 6.60 MPa 6.60 MPa 5
1 6. The density of sea water at its surface is 100 kg>m 3, where the absolute pressure is 101.3 kpa. If at a point deep under the water the density is 1060 kg>m 3, determine the absolute pressure in MPa at this point. Take E V =.33 GPa. Differentiate V = m, we obtain p Substitute these results into Eq. 1 8, Using this result, E V = dv = - mdr r E V = - dp dv>v = - dp a - mdr r b nam r b dp dr>r ;.331109 N>m = = dp dr>r p - 101.3110 3 N>m a 1060 kg>m3-100 kg>m 3 100 kg>m 3 b p = 91.47110 6 Pa = 91.5 MPa 91.5 MPa 6
1 7. Two measurements of shear stress on a surface and the rate of change in shear strain at the surface for a fluid have been determined by experiment to be t 1 = 0.14 N>m, (du>dy) 1 = 13.63 s -1 and t = 0.48 N>m, (du>dy) =153 s -1. Classify the fluid as Newtonian or non-newtonian. Applying Newton s Law of viscosity, t 1 = m 1 a du dy b ; 0.14 N>m = m 1 (13.63 s -1 ) m 1 = 0.0107 N # s>m 1 t = m a du dy b ; 0.48 N>m = m (153 s -1 ) m = 0.003137 N # s>m Since m 1 m then m is not constant. It is an apparent viscosity. The fluid is non-newtonian. non-newtonian 7
*1 8. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = (1y 1>4 ) mm>s, where y is in mm. Determine the shear stress within the fluid at y = 8 mm. Take m = 0.5(10-3 ) N # s>m. 16 mm y 4 mm/s u P Since the velocity distribution is not linear, the velocity gradient varies with y. u = 1y 1>4 du dy = 3y-3>4 At y = 8 mm, t = m du dy = 0.5(10-3 ) N # s>m 33(8 mm) -3>4 s -1 4 t = 0.315 mpa Note: When y = 0, du S, so that t S. Hence the equation can not be used at dy this point. 8
1 9. At a particular temperature the viscosity of an oil is m = 0.354 N # s>m. Determine its kinematic viscosity. The specific gravity is S o = 0.868. The density of the oil can be determined from r o = S o r w = 0.868(1000 kg>m 3 ) = 868 kg>m 3 y o = m o r o = 0.354 N # s>m 868 kg>m 3 = 0.4078(10-3 ) m >s = 0.408(10-3 ) m >s 0.408(10-3 ) m >s 9
1 30. Determine the constants B and C in Andrade s equation for water if it has been experimentally determined that m = 1.00(10-3 ) N # s>m at a temperature of 0 C and that m = 0.554(10-3 ) N # s>m at 50 C. The Andrade s equation is m = Be C>T At T = (0 + 73) K = 93 K, m = 1.00(10-3 ) N # s>m. Thus 1.00(10-3 ) N # s>m = Be C>93 K ln31.00(10-3 ) 4 = ln(be C>93 ) -6.9078 = ln B + ln e C>93-6.9078 = ln B + C>93 ln B = -6.9078 - C>93 (1) At T = (50 + 73) K = 33 K, m = 0.554(10-3 ) N # s>m. Thus, Equating Eqs. (1) and () 0.554(10-3 ) N # s>m = Be C>33 ln30.554(10-3 ) 4 = ln(be C>33 ) -7.4983 = ln B + ln e C>33-7.4983 = ln B + C 33 ln B = -7.4983 - -6.9078 - C 93 = -7.4983 - C 33 0.5906 = 0.31699(10-3 ) C C 33 C = 1863.10 = 1863 K () Substitute this result into Eq. (1) B = 1.7316(10-6 ) N # s>m = 1.73(10-6 ) N # s>m C = 1863.10 = 1863 K B = 1.73(10-6 ) N # s>m 30
1 31. The viscosity of water can be determined using the empirical Andrade s equation with the constants B = 1.73(10-6 ) N # s>m and C = 1863 K. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10 C and T = 80 C. The Andrade s equation for water is m = 1.73(10-6 )e 1863>T At T = (10 + 73) K = 83 K, m = 1.73(10-6 )e 1863 K>83 K = 1.5(10-3 ) N # s>m From the Appendix at T = 10 C, At T = (80 + 73) K = 353 K, m = 1.31(10-3 ) N # s>m m = 1.73(10-6 )e 1863 K>353 K = 0.339(10-3 ) N # s>m From the Appendix at T = 80 C, m = 0.356(10-3 ) N # s>m At T = 83 K, m = 1.5(10-3 ) N # s>m At T = 353 K, m = 0.339(10-3 ) N # s>m 31
*1 3. The constants B = 1.357(10-6 ) N # s> (m # K 1> ) and C = 78.84 K have been used in the empirical Sutherland equation to determine the viscosity of air at standard atmospheric pressure. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10 C and T = 80 C. The Sutherland Equation for air at standard atmospheric pressure is m = 1.357(10-6 )T 3> T + 78.84 At T = (10 + 73) K = 83 K, From Appendix A at T = 10 C, m = 1.357(10-6 )(83 3> ) 83 + 78.84 = 17.9(10-6 ) N # s>m m = 17.6(10-6 ) N # s>m At T = (80 + 73) K = 353 K, From Appendix A at T = 80 C, m = 1.357(10-6 )(353 3> ) 353 + 78.84 = 0.8(10-6 ) N # s>m m = 0.9(10-6 ) N # s>m 3
1 33. Determine the constants B and C in the Sutherland equation for air if it has been experimentally determined that at stan dard atmospheric pressure and a temperature of 0 C, m = 18.3(10-6 ) N # s>m, and at 50 C, m = 19.6(10-6 ) N # s>m. The Sutherland equation is m = BT 3> T + C At T = (0 + 73) K = 93 K, m = 18.3(10-6 ) N # s>m. Thus, 18.3(10-6 ) N # s>m = B(933> ) 93 K + C B = 3.6489(10-9 )(93 + C) (1) At T = (50 + 73) K = 33 K, m = 19.6(10-6 ) N # s>m. Thus Solving Eqs. (1) and () yields 19.6(10-6 ) N # s>m = B(333> ) 33 K + C B = 3.3764(10-9 )(33 + C) () B = 1.36(10-6 ) N # s> (m K 1 ) C = 78.8 K B = 1.36 (10-6 ) N # s> (m # K 1, C = 78.8 K 33
1 34. An experimental test using human blood at T = 30 C indicates that it exerts a shear stress of t = 0.15 N>m on surface A, where the measured velocity gradient at the surface is 16.8 s -1. Since blood is a non-newtonian fluid, determine its apparent viscosity at the surface. A Here du dy = 16.8 s-1 and t = 0.15 N>m. Thus t = m a du dy ; 0.15 N>m = m a (16.8 s -1 ) m a = 8.93(10-3 ) N # s>m Realize that blood is a non-newtonian fluid. For this reason, we are calculating the apparent viscosity. 8.93(10-3 ) N # s>m 34
1 35. When the force of 3 mn is applied to the plate the line AB in the liquid remains straight and has an angular rate of rotation of 0. rad>s. If the surface area of the plate in contact with the liquid is 0.6 m, determine the approximate viscosity of the liquid. B B A u 4 mm 3 mn The shear stress acting on the fluid contact surface is t = P A = 3(10-3 ) N 0.6 m = 5(10-3 ) N m Since line AB is a straight line, the velocity distribution will be linear. Here, the velocity gradient is a constant. The velocity of the plate is U = au # = (0.004 m)(0. rad>s) = 0.8(10-3 ) m>s Then, t = m du dy 5(10-3 ) N>m = m a 0.8(10-3 ) m>s 0.004 m b m = 0.05 N # s>m Alternatively, t = m du dt 5(10-3 ) N>m = m(0. rad>s) m = 0.05 N # s>m m = 0.05 N # s>m 35
*1 36. The conical bearing is placed in a lubricating Newtonian fluid having a viscosity m. Determine the torque T required to rotate the bearing with a constant angular velocity of v. Assume the velocity profile along the thickness t of the fluid is linear. v T R Since the velocity distribution is linear, the velocity gradient will be constant. The velocity of the oil in contact with the shaft at an arbitrary point is U = vr. Thus, From the geometry shown in Fig. a, t = m du dy = mvr t u t z = r tan u dz = dr tan u (1) T R ds Also, from the geometry shown in Fig. b, Equating Eqs. (1) and (), dr tan u dz = ds cos u () = ds cos u ds = dr sin u df φ r (a) dz z The area of the surface of the differential element shown shaded in Fig. a is da = prds = p rdr. Thus, the shear force the oil exerts on this area is sin u df = tda = a mvr ba p pmv rdrb = t sin u t sin u r dr dz θ (b) ds Considering the moment equilibrium of the shaft, Fig. a, ΣM z = 0; T - L rdf = 0 T = rdf = pmv r 3 dr L t sin u L0 = pmv R t sin u ar4 4 b ` 0 = pmvr4 t sin u R 36
1 37. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = (1y 1>4 ) mm>s, where y is in mm. Determine the minimum shear stress within the fluid. Take m = 0.5(10-3 ) N # s>m. 16 mm y 4 mm/s u P Since the velocity distribution is not linear, the velocity gradient varies with y. u = 1y 1>4 du dy = 3y-3>4 The velocity gradient is smallest when y = 16 mm. Thus, t min = m du dy = 30.5(10-3 ) N # s>m 4 33(16 mm) -3>4 s -1 4 t min = 0.1875 mpa Note: When y = 0, du S, so, that t S. Hence the equation can not be used at dy this point. 0.1875 mpa 37
1 38. The velocity profile for a thin film of a Newtonian fluid that is confined between a plate and a fixed surface is defined by u = (10y - 0.5y ) mm>s, where y is in mm. Determine the shear stress that the fluid exerts on the plate and on the fixed surface. Take m = 0.53 N # s>m. 4 mm y 36 mm/s u P Since the velocity distribution is not linear, the velocity gradient varies with y. u = (10y - 0.5y ) mm>s du = (10-0.5y) s-1 dy At the plate t p = m du dy = (0.53 N # s>m ) 310-0.5(4 mm) s -1 4 = 4.6 Pa At the fixed surface t fs = m du dy = (0.53 N # s>m ) 3(10-0) s -1 4 = 5.3 Pa t p = 4.6 Pa, t fs = 5.3 Pa 38
1 39. If a force of P = N causes the 30-mm-diameter shaft to slide along the lubricated bearing with a constant speed of 0.5 m>s, determine the viscosity of the lubricant and the constant speed of the shaft when P = 8 N. Assume the lubricant is a Newtonian fluid and the velocity profile between the shaft and the bearing is linear. The gap between the bearing and the shaft is 1 mm. 50 mm 0.5 m/s P Since the velocity distribution is linear, the velocity gradient will be constant. N 3p(0.015 m)4(0.05 m) t = m du dy = m a 0.5 m>s 0.001 m b m = 0.8498 N # s>m Thus, 8 N 3p(0.015 m)4(0.05 m) = (0.8488 N # v s>m ) a 0.001 m b v =.00 m>s Also, by proportion, a N A b a 8 N A b = 0.5 m>s m a b t m a v t b v = 4 m>s =.00 m>s m = 0.849 N # s>m v =.00 m>s 39
*1 40. The velocity profile of a Newtonian fluid flowing over U a fixed surface is approximated by u = U sin a p h yb. Determine the shear stress in the fluid at y = h and at y = h>. The viscosity of the fluid is m. h y u Since the velocity distribution is not linear, the velocity gradient varies with y. u = U sin a p h yb du dy = U a p h b cos a p h yb At y = h, t = m du dy = mu a p h b cos p h (h) t = 0; At y = h>, t = m du dy = mu a p h b cos p h ah b t = 0.354pmU h 40
1 41. The tank containing gasoline has a long crack on its side that has an average opening of 10 mm. The velocity through the crack is approximated by the equation u = 10(10 9 ) 310(10-6 y - y )4 m>s, where y is in meters, measured upward from the bottom of the crack. Find the shear stress at the bottom, at y = 0 and the location y within the crack where the shear stress in the gasoline is zero. Take m g = 0.317(10-3 ) N # s>m. 10 m y(m) Gasoline is a Newtonian fluid. 10(10 6 ) u 10(10 9 )[10(10 6 )y y ] m s The rate of change of shear strain as a function of y is du dy = 10(109 ) 310(10-6 ) - y4 s -1 At the surface of crack, y = 0 and y = 10(10-6 ) m. Then or du dy ` = 10(10 9 ) 310(10-6 ) - (0)4 = 100(10 3 ) s -1 y = 0 du dy ` = 10(10 9 ) 510(10-6 ) - 310(10-6 )46 = -100(10 3 ) s -1 y = 10 (10-6 ) m Applying Newton s law of viscosity, du t y = 0 = m g dy ` = 30.317(10-3 ) N # s>m 4 3100(10 3 ) s -1 4 = 31.7 N>m y = 0 t = 0 when du dy = 0. Thus 10(10-6 ) - y = 0 du dy = 10(109 ) 310(10-6 ) - y4 = 0 y = 5(10-6 ) m = 5 mm (a) u(m s) t y = 0 = 31.7 N>m t = 0 when y = 5 µm 41
1 4. The tank containing gasoline has a long crack on its side that has an average opening of 10 mm. If the velocity profile through the crack is approximated by the equation u = 10(10 9 )310(10-6 y - y )4 m>s, where y is in meters, plot both the velocity profile and the shear stress distribution for the gasoline as it flows through the crack. Take m g = 0.317(10-3 ) N # s>m. 10 m y(10-6 m) 0 1.5.50 3.75 5.00 u(m>s) 0 0.1094 0.1875 0.344 0.50 6.5 7.50 8.75 10.0 0.344 0.1875 0.1094 0 y(10 6 m) 10.0 7.50 5.00.50 0 0.10 0.0 (a) 0.30 u(m s) Gasoline is a Newtonian fluid. The rate of change of shear strain as a function of y is Applying Newton s law of viscoscity, du dy = 10(109 ) 310(10-6 ) - y4 s -1 t = m du dy = 30.317(10-3 ) N # s>m 4 510(10 9 ) 310(10-6 ) - y4 s -1 6 t = 3.17(10 6 ) 310(10-6 ) - y4 N>m The plots of the velocity profile and the shear stress distribution are shown in Fig. a and b respectively. y(10-6 m) 0 1.5.50 3.75 5.00 t(n>m ) 31.70 3.78 15.85 7.95 0 6.5 7.50 8.75 10.0-7.95-15.85-3.78-31.70 y(10 6 m) 10.0 7.50 5.00.50 40 30 0 10 0 10 0 30 τ(ν m ) 40 (b) y = 1.5 (10-6 ) m, u = 0.109 m>s, t = 3.8 N>m 4
1 43. The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u = (10y - 0.5y ) mm>s, where y is in mm. Determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 5000 mm in contact with the fluid. Take m = 0.53 N # s>m. 4 mm y 36 mm/s u P Since the velocity distribution is not linear, the velocity gradient varies with y. u = (10y - 0.5y ) mm>s du dy = (10-0.5y) s-1 At the plate t p = m du dy = (0.53 N # s>m ) 310-0.5(4 mm)4 s -1 = 4.56 Pa P = t p A = 3(4.56) N>m 4 35000(10-6 ) m 4 = 1.3 mn 1.3 MN 43
*1 44. The 0.15-m-wide plate passes between two layers, A and B, of oil that has a viscosity of m = 0.04 N # s>m. Determine the force P required to move the plate at a constant speed of 6 mm>s. Neglect any friction at the end supports, and assume the velocity profile through each layer is linear. A B 0.0 m 6 mm 4 mm P F A The oil is a Newtonian fluid. P Considering the force equilibrium along the x axis, Fig. a, + S ΣF x = 0; P - F A - F B = 0 P = F A + F B Since the velocity distribution is linear, the velocity gradient will be constant. t A = m du dy = (0.04 N # s>m )a 6 mm>s b = 0.04 Pa 6 mm t B = m du dy = (0.04 N # s>m )a 6 mm>s b = 0.06 Pa 4 mm P = (0.04 N>m )(0. m)(0.15 m) + (0.06 N>m )(0. m)(0.15 m) = 3.00 mn F B (a) 44
1 45. The 0.15-m-wide plate passes between two layers A and B of different oils, having viscosities of m A = 0.03 N # s>m and m B = 0.01 N # s>m. Determine the force P required to move the plate at a constant speed of 6 mm>s. Neglect any friction at the end supports, and assume the velocity profile through each layer is linear. A B 0.0 m 6 mm 4 mm P F A The oil is a Newtonian fluid. P Considering the force equilibrium along the x axis, Fig. a, ΣF x = 0; P - F A - F B = 0 F B (a) P = F A + F B Since the velocity distribution is linear, the velocity gradient will be constant. t A = m du dy = (0.03 N # s>m )a 6 mm>s b = 0.03 Pa 6 mm t B = m du dy = (0.01 N # s>m )a 6 mm>s b = 0.015 Pa 4 mm P = (0.03 N>m )(0. m)(0.15 m) + (0.015 N>m )(0. m)(0.15 m) = 1.35 mn 1.35 mn 45
1 46. The tape is 10 mm wide and is drawn through an applicator, which applies a liquid coating (Newtonian fluid) that has a viscosity of m = 0.830 N # s>m to each side of the tape. If the gap between each side of the tape and the applicator s surface is 0.8 mm, determine the torque T at the instant r = 150 mm that is needed to rotate the wheel at 0.5 rad>s. Assume the velocity profile within the liquid is linear. 30 mm 0.5 rad/s r 150 mm T Considering the moment equilibrium of the wheel, Fig. a, ΣM A = 0; T - P(0.15 m) = 0 Since the velocity distribution is linear, the velocity gradient will be constant. P = t(a) = m(a) du dy P = F = (0.0334 N) 0.15 m W T 0.5 rad>s(0.15 m) P = (0.830 N # s>m )()(0.03 m)(0.01 m) a b 0.0008 m 0 x P = 0.04669 N Thus T = (0.04669 N)(0.15 m) = 7.00 mn # m (a) 0 y 7.00 mn # m 46
1 47. Disks A and B rotate at a constant rate of v A = 50 rad>s and v B = 0 rad>s, respectively. Determine the torque T required to sustain the motion of disk B. The gap, t = 0.1 mm, contains SAE 10 oil for which m = 0.0 N # s>m. Assume the velocity profile is linear. A 100 mm v A t B Oil is a Newtonian fluid. The velocities of the oil on the surfaces of disks A and B are U A = v A r = (50r) m>s and U B = v B r = (0r) m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, y T v B 0 rad/s du dy = U A - U B = t 50r - 0r 0.1(10-3 ) m = 300(103 )r s -1 U A 50r Applying Newton s Law of viscosity, t = m du dy = (0.0 N # s>m ) 3300(10 3 )r4 = (6000r) N>m The shaded differential element shown in Fig. b has an area of da = pr dr. Thus, df = tda = (6000r)(pr dr) = 1(10 3 )pr dr. Moment equilibrium about point O in Fig. b requires t = 0.1(10 3 ) m (a) U B 0r u a+ ΣM O = 0; T - L r df = 0 0.1 m T - r31(10 3 )pr dr4 = 0 L 0 0.1 m T = 1(10 3 )pr 3 dr L 0 = 1(10 3 )p a r 4 0.1 m 4 b ` 0 = 0.94 N # m 0.1 m T 0 r dr df da (b) 0.94 N # m 47
*1 48. If disk A is stationary, v A = 0 and disk B rotates at v B = 0 rad>s, determine the torque T required to sustain the motion. Plot your results of torque (vertical axis) versus the gap thickness for 0 t 0.1 m. The gap contains SAE10 oil for which m = 0.0 N # s>m. Assume the velocity profile is linear. A 100 mm v A t B t(10-3 ) m 0 0.0 0.04 0.06 0.08 0.10 T(N # m) 3.14 1.57 1.05 0.785 0.68 T v B 0 rad/s T(N.m) y 3.5 3.0.5.0 t 1.5 1.0 u 0.5 0 0.0 0.04 0.06 0.08 0.10 t(10 3 m) (a) U B 0r (C) Oil is a Newtonian fluid. The velocities of the oil on the surfaces of disks A and B are U A = v A r = 0 and U B = v B r = (0r) m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, du dy = U A - U B = 0-0r t t = a - 0r b s -1 t 0.1 m 0 r dr Applying Newton s law of viscosity, du t = m ` dy ` = (0.0 N # s>m )a 0r t b = a 0.4r b N>m t T df da (b) 48
*1 48. (continued) The shaded differential element shown in Fig. b has an area of da = pr dr. Thus, df = tda = a 0.4r b(pr dr) = a 0.8p br dr. Moment equilibrium about point O t t in Fig. b requires a+ ΣM O = 0; T - L r df = 0 0.1 m T - r c a 0.8p br dr d = 0 L 0 t 0.1 m T = a 0.8p br 3 dr L t 0 T = a 0.8p ba r 4 0.1 m t 4 b ` 0 T = c 0(10-6 )p t d N # m where t is in m The plot of T vs t is shown Fig. c. 49
1 49. The very thin tube A of mean radius r and length L is placed within the fixed circular cavity as shown. If the cavity has a small gap of thickness t on each side of the tube, and is filled with a Newtonian liquid having a viscosity m, determine the torque T required to overcome the fluid resistance and rotate the tube with a constant angular velocity of v. Assume the velocity profile within the liquid is linear. t t r T A L Since the velocity distribution is assumed to be linear, the velocity gradient will be constant. t = m du dy = m (vr) t F = µ r L t Considering the moment equilibrium of the tube, Fig. a, T ΣM = 0; T - tar = 0 T = (m) (vr) (prl)r t T = 4pmvr 3 L t r O (a) T = 4pmvr 3 L t 50
1 50. The shaft rests on a -mm-thin film of oil having a viscosity of m = 0.0657 N # s>m. If the shaft is rotating at a constant angular velocity of v = rad>s, determine the shear stress in the oil at r = 50 mm and r = 100 mm. Assume the velocity profile within the oil is linear. v rad/s T 100 mm Oil is a Newtonian fluid. Since the velocity distribution is linear, the velocity gradient will be constant. At r = 50 mm, t = m du dy ( rad>s)(50 mm) t = (0.0657 N # s>m )a b mm t = 3.8 Pa At r = 100 mm, ( rad>s)(100 mm) t = (0.0657 N # s>m )a b mm t = 6.57 Pa At r = 50 mm, t = 3.8 Pa At r = 100 mm, t = 6.57 Pa 51
1 51. The shaft rests on a -mm-thin film of oil having a viscosity of m = 0.0657 N # s>m. If the shaft is rotating at a constant angular velocity of v = rad>s, determine the torque T that must be applied to the shaft to maintain the motion. Assume the velocity profile within the oil is linear. v rad/s T 100 mm Oil is a Newtonian fluid. Since the velocity distribution is linear, the velocity gradient will be constant. The velocity of the oil in contact with the shaft at an arbitrary point is U = vr. Thus, df t = m du dy = mvr t Thus, the shear force the oil exerts on the differential element of area da = pr dr shown shaded in Fig. a is df = tda = a mvr t b(pr dr) = pmv r dr t T O r dr Considering the moment equilibrium of the shaft, Fig. a, (a) a+ ΣM O = 0; df - T = 0 Lr T = Lr df = pmv t R r 3 dr L0 = pmv t a r 4 4 b ` R 0 = pmvr4 t Substituting, T = p a0.0657 N # s b( rad>s)(0.1 m)4 m (0.00 m) = 10.3(10-3 ) N # m = 10.3 mn # m 10.3 mn # m 5
*1 5. Water at A has a temperature of 15 C and flows along the top surface of the plate C. The velocity profile is approximated as m A = 10 sin (.5py) m>s, where y is in meters. Below the plate the water at B has a temperature of 60 C and a velocity profile of u B = 4(10 3 )(0.1y - y ), where y is in meters. Determine the resultant force per unit length of plate C the flow exerts on the plate due to viscous friction. The plate is 3 m wide. C 100 mm 100 mm y A B Water is a Newtonian fluid. Water at A, T = 15 C. From Appendix A m = 1.15(10-3 ) N # s>m. Here du A dy = 10 a5p b cos a5p yb = a5p cos 5p yb s-1 At surface of plate C, y = 0. Then du A dy ` = 5p cos c 5p y = 0 (0) d = 5p s-1 Applying Newton s law of viscosity t A y = 0 = m du A dy ` = 31.15(10-3 ) N # s>m 4 (5p s -1 ) = 0.0875p N>m y = 0 Water at B, T = 60 C. From Appendix A m = 0.470(10-3 ) N # s>m. Here du B dy = 34(103 )(0.1 - y)4 s -1 At the surface of plate C, y = 0.1 m. Then du B dy ` = 4(10 3 ) 30.1 - (0.1)4 = -400 s -1 y = 0.1 m Applying Newton s law of viscosity, t B y = 0.1 m = m du B dy ` = 30.470(10-3 ) N # s>m 4(400 s -1 ) = 0.188 N>m y = 0.1 m Here, the area per unit length of plate is A = 3 m. Thus F = (t A + t B )A = (0.0875p N>m + 0.188 N>m )(3 m) = 0.835 N>m 53
1 53. The read write head for a hand-held music player has a surface area of 0.04 mm. The head is held 0.04 µm above the disk, which is rotating at a constant rate of 1800 rpm. Determine the torque T that must be applied to the disk to overcome the frictional shear resistance of the air between the head and the disk. The surrounding air is at standard atmospheric pressure and a temperature of 0 C. Assume the velocity profile is linear. 8 mm T Here Air is a Newtonian fluid. v = a1800 rev min bap rad 1 rev min ba1 b = 60p rad>s. 60 s Thus, the velocity of the air on the disk is U = vr = (60p)(0.008) = 0.48p m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, du dy = U t = 0.48p m>s 0.04(10-6 )m = 1(106 )p s -1 For air at T = 0 C and standard atmospheric pressure, m = 18.1(10-6 ) N # s>m (Appendix A). Applying Newton s law of viscosity, t = m du dy = 318.1(10-6 ) N # s>m 4 31(10 6 )p s -1 4 = 17.p N>m Then, the drag force produced is F D = ta = (17.p N>m )a 0.04 1000 m b = 8.688(10-6 )p N The moment equilibrium about point O requires a+ ΣM O = 0; T - 38.688(10-6 )p N4(0.008 m) = 0 T = 0.18(10-6 ) N # m = 0.18 mn # m u 0.008 m t = 0.04(10 6 ) m U 0.48 m/s T 0 y (a) F D 8.688(10 6 ) N (b) T = 0.18 mn # m 54
1 54. The city of Denver, Colorado, is at an elevation of 1610 m above sea level. Determine how hot one can prepare water to make a cup of coffee. At the elevation of 1610 meters, the atmospheric pressure can be obtained by interpolating the data given in Appendix A. p atm = 89.88 kpa - a 89.88 kpa - 79.50 kpa b(610 m) = 83.55 kpa 1000 m Since water boils if the vapor pressure is equal to the atmospheric pressure, then the boiling temperature at Denver can be obtained by interpolating the data given in Appendix A. 83.55-70.1 T boil = 90 C + a b(5 C) = 94.6 C 84.6-70.1 Note: Compare this with T boil = 100 C at 1 atm. 94.6 C 55
1 55. As water at 40 C flows through the transition, its pressure will begin to decrease. Determine the lowest pressure it can have without causing cavitation. From Appendix A, the vapor pressure of water at T = 40 C is p y = 7.38 kpa Cavitation (or boiling of water) will occur when the water pressure is equal to or less than p y. Thus, p min = 7.38 kpa 7.38 kpa 56
*1 56. Steel particles are ejected from a grinder and fall gently into a tank of water. Determine the largest average diameter of a particle that will float on the water. Take r = 7850 kg>m 3 and s = 0.076 N>m. Assume that each particle has the shape of a sphere, where V = 4 3 pr 3. The weight of a steel particle is W = r st gv = (7850 kg>m 3 )(9.81 m>s ) c 4 3 p ad b 3 d = 40.31(10 3 ) d 3 The maximum weight the surface tension can support is when it acts vertically upward (In this case, the water surface is in contact with lower hemisphere) as shown in the FBD of steel particle, Fig. a. Consider the force equilibrium along vertical, W (a) r = d Water surface + c ΣF y = 0; (0.076 N>m)c p a d b d - 40.31(103 ) d 3 = 0 d =.3783(10-3 ) m =.38 mm 57
1 57. The blades of a turbine are rotating in water that has a temperature of 30 C. What is the lowest water pressure that can be developed at the blades so that cavitation will not occur? From Appendix A, the vapor pressure of water at T = 30 C is p y = 4.5 kpa Cavitation (boiling of water) will occur if the water pressure is equal or less than p y. Thus p min = p y = 4.5 kpa 4.5 kpa 58
1 58. Water at 15 C is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location in the hose? At a particular temperature, the cavitation will occur when the pressure drops below or equals the water vapor pressure at this temperature. From the table in Appendix A, p v = 1.71 kpa at T = 15 C. Since the cavitation has already occurred, the absolute pressure at the bend cannot be greater than p v. Thus p max = p v = 1.71 kpa 1.71 kpa 59
1 59. Water at 5 C is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location in the hose? From Appendix A, the vapor pressure of water at T = 5 C is p y = 3.17 kpa Cavitation (boiling of water) will occur if the water pressure is equal or less than p y. p max = p y = 3.17 kpa 3.17 kpa 60
*1 60. Determine the distance h that the column of mercury in the tube will be depressed as a function tube s diameter D when the tube is inserted into the mercury at a room temperature of 0 C. Plot this relationship of h (vertical axis) versus D for 1 mm D 4 mm. Give values for increments of D = 0.05 in. Discuss this result. D h 50 D (mm) 1 1.5.0.5 3.0 3.5 4.0 h (mm) 9.01 6.01 4.51 3.61 3.00.58.5 h (mm) 10 8 6 4 σ θ P (a) θ σ 0 1 3 4 (b) D (mm) The pressure on the meniscus shown in its FBD, Fig. a, is p = p mg gh. Consider the force equilibrium along vertical, + c ΣF y = 0; (r Hg gh)a p 4 D b - s(pd) cos u = 0 h = 4s cos u r Hg gd From the table in Appendix A, r Hg = 13550 kg>m 3 and s = 0.466 N>m at T = 0 C. Here u = 50 and D is in mm. Then h = 4(0.466 N>m) cos 50 (13550 kg>m 3 )(9.81 m>s )(D>1000) h = e c 9.0137(10-3 ) 1000 mm d m fa D 1 m b h = a 9.0137 b mm where D is in mm D The values of h for various D are tabulated and the plot of h vs D is shown in Fig. b. 61
1 61. Water in the glass tube is at a temperature of 40 C. Plot the height h of the water as a function of the tube s inner diameter D for 0.5 mm D 3 mm. Use increments of 0.5 mm. Take s = 69.6 mn>m. D h When water contacts the glass wall, u = 0. The weight of the rising column of water is W = g w V = r w g a p 4 D hb = 1 4 pr wgd h σ σ The vertical force equilibrium, Fig. a, requires w + c ΣF y = 0; s(pd) - 1 4 pr wgd h = 0 h = 4s r w gd h From Appendix A, r w = 99.3 kg>m 3 at T = 40 C. Then h = For 0.5 mm D 3 mm 4(0.0696 N>m) (99.3 kg>m 3 )(9.81 m>s )D = 8.6(10-6 ) m D D(mm) 0.5 1.0 1.5.0.5 3.0 h(mm) 57. 8.6 19.07 14.3 11.44 9.53 The plot of h vs D is shown in Fig. b. h(mm) 60 50 40 30 0 10 D (a) 0 0.5 1.0 1.5.0.5 (b) 3.0 D(mm) D = 1.0 mm, h = 8.6 mm 6
1 6. When a can of soda water is opened, small gas bubbles are produced within it. Determine the difference in pressure between the inside and outside of a bubble having a diameter of 0.5 mm. The surrounding temperature is 0 C. Take s = 0.0735 N>m. p 0 The horizontal component of the resultant force due to the pressure on the bubble is equal to the product of the pressure and the vertical projected area of the bubble consider the force equilibrium along horizontal by refering to the FBD of the bubble, Fig. a, + S ΣF x = 0; p i a p 4 d b - p 0 a p 4 d b - s(pd) = 0 p i (a) p i - p 0 = 4s d Here, s = 0.0735 N>m and d = 0.5(10-3 ) m. Then 4(0.0735 N>m) p = p i - p 0 = 0.5(10-3 ) m = 588 Pa 588 Pa 63
1 63. Determine the greatest distance h that the column of mercury in the tube will be depressed when the tube is inserted into the mercury at a room temperature of 0 C. Set D = 3 mm. D h 50 The greatest depressed distance h in Fig. a achieve when s acts vertically downward as shown in the FBD of the mercury meniscus in Fig. b. Here the pressure on the meniscus is p = r Hg gh. Consider the force equilibrium along vertical, Fig. b, + c ΣF y = 0 (r Hg gh)a p 4 D b - s(pd) = 0 h h = 4s r Hg gd From the table in Appendix A, r Hg = 13550 kg>m 3 and s = 0.466 N>m at T = 0 C. Then (a) h = 4(0.466 N>m) (13550 kg>m 3 )(9.81 m>s )(0.003 m) p = 4.674(10-3 ) m σ (b) σ = 4.67 mm 4.67 mm 64
*1 64. The tube has an inner diameter d and is immersed in water at an angle u from the horizontal. Determine the average length L to which water will rise along the tube due to capillary action. The surface tension of the water is s and its density is r. L u d The free-body diagram of the water column is shown in Fig. a. The weight of this column is W = rg V = rgc p a d b L d = prgd L. 4 x For water, its surface will be almost parallel to the surface of the tube (contact angle 0 ). Thus, s acts along the tube. Considering equilibrium along the x axis, ΣF x = 0; s(pd) - prgd L 4 sin u = 0 L = 4s rgd sin u d L (a) W = Pgd L 4 N 65
1 65. The tube has an inner diameter of d = mm and is immersed in water. Determine the average length L to which the water will rise along the tube due to capillary action as a function of the angle of tilt, u. Plot this relationship of L (vertical axis) versus u for 10 u 30. Give values for increments of u = 5. The surface tension of the water is s = 75.4 mn>m, and its density is r = 1000 kg>m 3. L u d u(deg.) 10 15 0 5 30 L(mm) 88.5 59.4 44.9 36.4 30.7 x L(mm) 100 80 60 40 0 L 0.00m = 0.0754 Nm N W = [9.81(10 3 ) h] N (a) 0 5 10 15 0 5 30 The FBD of the water column is shown in Fig. a. The weight of this column is W = rg V = (1000 kg>m 3 )(9.81 m>s ) c p 4 (0.00 m)l d = 39.81(10-3 )pl4 N. For water, its surface will be almost parallel to the surface of the tube (u 0 ) at the point of contact. Thus, s acts along the tube. Considering equilibrium along x axis, ΣF x = 0; (0.0754 N>m)3p(0.00 m)4-39.81(10-3 )pl4 sin u = 0 L = a 0.0154 b m where u is in deg. sin u The plot of L versus u is shown in Fig. a. L = (0.0154>sin u) m 66
1 66. The marine water strider, Halobates, has a mass of 0.36 g. If it has six slender legs, determine the minimum contact length of all of its legs combined to support itself in water having a temperature of T = 0 C. Take s = 7.7 mn>m, and assume the legs are thin cylinders that are water repellent. The force supported by the legs is P = 30.36(10-3 ) kg4 39.81 m>s 4 = 3.5316(10-3 ) N Here, s is most effective in supporting the weight if it acts vertically upward. This requirement is indicated on the FBD of each leg in Fig. a. The force equilibrium along vertical requires P = 3.5316(10 3 ) N l l + c ΣF y = 0; 3.5316(10-3 ) N - (0.077 N>m)l = 0 l = 4.3(10-3 ) m = 4.3 mm Note: Because of surface microstructure, a water strider s legs are highly hydrophobic. That is why the water surface curves downward with u 0, instead of upward as it does when water meets glass. (a) 4.3 mm 67
1 67. Many camera phones now use liquid lenses as a means of providing a quick auto-focus. These lenses work by electrically controlling the internal pressure within a liquid droplet, thereby affecting the angle of the meniscus of the droplet, and so creating a variable focal length. To analyze this effect, consider, for example, a segment of a spherical droplet that has a base diameter of 3 mm. The pressure in the droplet is 105 Pa and is controlled through a tiny hole at the center. If the tangent at the surface is 30, determine the surface tension at the surface that holds the droplet in place. 30 3 mm Writing the force equation of equilibrium along the vertical by referring to the FBD of the droplet in Fig. a +c ΣF z = 0; a105 N m b 3p(0.0015 m) 4 - (s sin 30 )3p(0.0015 m)4 = 0 30º r = 0.0015m 30º s = 0.158 N>m 105 N m (a) 0.158 N>m 68
*1 68. The ring has a weight of 0. N and is suspended on the surface of the water, for which s = 73.6 mn>m. Determine the vertical force P needed to pull the ring free from the surface. Note: This method is often used to measure surface tension. P The free-body diagram of the ring is shown in Fig. a. For water, its surface will be almost parallel to the surface of the wire (u 0 ) at the point of contact, Fig. a. 50 mm + c ΣF y = 0; P - W - T = 0 P - 0. N - (0.0736 N>m)3p(0.05 m)4 = 0 P = 0.46 N P T = 7.36(10 ) N T = 7.36(10 ) N W = 0. N (a) 69
1 69. The ring has a weight of 0. N and is suspended on the surface of the water. If it takes a force of P = 0.45 N to lift the ring free from the surface, determine the surface tension of the water. P The free-body diagram of the ring is shown in Fig. a. For water, its surface will be almost parallel to the surface of the wire (u 0 ) at the point of contact, Fig. a. 50 mm + c ΣF y = 0; 0.45 N - 0. N - 3s(p(0.05 m))4 = 0 s = 0.0716 N>m = 0.0716 N>m P = 0.45 N T = 0.1 T = 0.1 W = 0. N (a) 0.0716 N>m 70