MECHANICS of FLUIDS, 4 TH EDITION, SI

A STUDENT S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS, 4 TH EDITION, SI MERLE C. POTTER DAID C. WIGGERT BASSEM H. RAMADAN

STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS FOURTH EDITION, SI MERLE C. POTTER Michigan State University DAID C. WIGGERT Michigan State University BASSEM RAMADAN Kettering University

Contents Preface iv Chapter Basic Considerations Chapter Fluid Statics 5 Chapter Introduction to Fluids in Motion 9 Chapter 4 The Integral Forms of the Fundamental Laws 7 Chapter 5 The Differential Forms of the Fundamental Laws 59 Chapter 6 Dimensional Analysis and Similitude 77 Chapter 7 Internal Flows 87 Chapter 8 External Flows 9 Chapter 9 Compressible Flow Chapter Flow in Open Channels 4 Chapter Flows in Piping Systems 6 Chapter Turbomachinery 75 Chapter Measurements in Fluid Mechanics 89

Preface for the Student This manual provides the solutions to the problems whose answers are provided at the end of our book, MECHANICS OF FLUIDS, SI. In many cases, the solutions are not as detailed as the examples in the book; they are intended to provide the primary steps in each solution so you, the student, are able to quickly review how a problem is solved. The discussion of a subtle point, should one exist in a particular problem, is left as a task for the instructor. In general, some knowledge of a problem may be needed to fully understand all of the steps presented. This manual is not intended to be a self-paced workbook; your instructor is critically needed to provide explanations, discussions, and illustrations of the myriad of phenomena encountered in the study of fluids, but it should give you considerable help in working through a wide variety of problems. The degree of difficulty and length of solution for each problem varies considerably. Some are relatively easy and others quite difficult. Typically, the easier problems are the first problems for a particular section. We continue to include a number of multiple-choice problems in the earlier chapters, similar to those encountered on the Fundamentals of Engineering Exam (the old EIT Exam) and the GRE/Engineering Exam. These problems will provide a review for the Fluid Mechanics part of those exams. They are all four-part, multiple-choice problems and are located at the beginning of the appropriate chapters. The examples and problems have been carefully solved with the hope that errors have not been introduced. Even though extreme care is taken and problems are reworked, errors creep in. We would appreciate knowing about any errors that you may find so they can be eliminated in future printings. Please send any corrections or comments to MerleCP@att.net.We have class tested most of the chapters with good response from our students, but we are sure that there are improvements to be made. East Lansing, Michigan Flint, Michigan Merle C. Potter David C. Wiggert Bassem Ramadan

CHAPTER Basic Considerations FE-type Exam Review Problems: Problems. to. Chapter / Basic Considerations. (C) m F/a or kg N/m/s N s /m. (B) [μ] [τ/(du/dy)] (F/L )/(L/T)/L F. T/L. (A) 8 9.6 Pa.6 Pa.6 npa.4 (C) The mass is the same on earth and the moon, so we calculate the mass using the weight given on earth as: m W/g 50 N/9.8 m/s 5.484 kg Hence, the weight on the moon is: W mg 5.484.6 40.77 N The shear stress is due to the component of the force acting tangential to the area:.5 (C) Fshear Fsinθ 400sin 0 0 N Fshear 0 N τ 84 Pa or 84 kpa A 4 50 m.6 (B) 5.6 C.7 (D).8 (A) Using Eqn. (.5.): ρ water ( T 4) (80 4) 00 00 968 kg/m 80 80 du The shear stress is given by: τ μ dr We determine du/dr from the given expression for u as: du d ( 500r ) 50, 000r dr dr At the wall r cm 0.0 m. Substituting r in the above equation we get: 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations du dr 50,000r 00 /s The density of water at 0 C is N s/m Now substitute in the equation for shear stress to get du τ μ N s/m 00 /s N/m Pa dr.9 (D) Using Eqn. (.5.6), β 0 (for clean glass tube), and σ 0.076 N/m for water (Table B. in Appendix B) we write: 4σ cosβ 4 0.076 N/m h m or 00 cm ρgd 6 00 kg/m 9.8 m/s m where we used N kg m/s. (C) Density. (C) Assume propane (C H 8 ) behaves as an ideal gas. First, determine the gas R 8.4 kj/kmol K constant for propane using u R 0.885 kj/kg M 44. kg/kmol p then, m RT 800 kn/m 4 m 59.99 kg 60 kg 0.885 kj/(kg K) ( + 7) K Consider water and ice as the system. Hence, the change in energy for the system is zero. That is, the change in energy for water should be equal to the change in energy for the ice. So, we write Δ Eice Δ Ewater. (B) m 0 kj/kg m c Δ T ice water water The mass of ice is calculate using 6 ρ 5 cubes 00 kg/m 40 m /cube 0. kg mice ( ) ( ) Where we assumed the density of ice to be equal to that of water, namely 00 kg/m. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. Similarly, the mass of water is calculated using m ρ 00 kg/m liters m /liter kg water ( ) ( ) Solving for the temperature change for water we get 0. kg 0 kj/kg kg 4.8 kj/kg K ΔT Δ T 7.66 C 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations. (D) Since a dog s whistle produces sound waves at a high frequency, the speed of sound is c RT 87 J/kg K K 04 m/s where we used J/kg m /s. Dimensions, Units, and Physical Quantities.6 a) density M L FT / L 4 FT / L L c) power F velocity F L/T FL/T e) mass flux MT / FT / L A LT FT/ L.8 b) N [C] kg [C] N/kg (kg m/s )/kg m/s.0 m m kg + c + km f. Since all terms must have the same dimensions (units) s s we require: [c] kg/s, [k] kg/s N s / m s N/m, [f] kg m/s N Note: we could express the units on c as [c] kg/s N s /m s N s/m. a).5 8 N c) 6.7 8 Pa e) 5. m.4 a) c) e) cm m hr 5 0 cm/hr 0 5.556 m/s hr 0 cm 600 s 745.7 W 500 hp 500 hp 7, 85 W hp kn 0 cm 000 kn/cm 000 N/m cm m.6 The mass is the same on the earth and the moon, so we calculate the mass, then calculate the weight on the moon: m 7 kg W moon (7 kg) (.6) 44.0 N 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations Pressure and Temperature.8.0. Use the values from Table B. in the Appendix: b) At an elevation of 00 m the atmospheric pressure is 89.85 kpa. Hence, the absolute pressure is 5. + 89.85 4. kpa d) At an elevation of,000 m the atmospheric pressure is 6.49 kpa, and the absolute pressure is 5. + 6.49 78.8 kpa p p o e gz/rt e (9.8 4000)/[(87) (5 + 7)] 6.8 kpa From Table B., at 4000 m: p 6.6 kpa. The percent error is 6.8 6.6 % error 0.95 % 6.6 Using Table B. and linear interpolation we write:,600,000 T.48 + (.6 6.7). K,000,000 or (. 7.5) 5 9 5.8 C.4 4 The normal force due pressure is: F (,000 N/m ) 0. m.4 N n 4 The tangential force due to shear stress is: F 0 N/m 0. m 0.0004 N The total force is F F n Ft +.40 N The angle with respect to the normal direction is θ tan 0.0004.4 0.0095 t Density and Specific Weight.6 Using Eq..5. we have ρ 00 (T 4) /80 00 (70 4) /80 976 kg/m γ 9800 (T 4) /8 9800 (70 4) /80 9560 N/m Using Table B. the density and specific weight at 70 C are ρ 977.8 kg/m γ 977.8 9.8 959. N/m % error for ρ % error for γ 976 978 978 9560 959 959 0 0.0% 0 0.% 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations.8 b) 6 γ,400 N/m 500 m m 0.65 kg g 9.77 m/s iscosity Assume carbon dioxide is an ideal gas at the given conditions, then ρ p 00 kn/m.95 kg/m RT ( 0.89 kj/kg K)( 90 + 7 K).40 W mg γ ρg.95 kg/m 9.8 m/s 8.6 kg/m s 8.6 N/m 5 From Fig. B. at 90 C, μ N s/m, so that the kinematic viscosity is μ ν ρ.95 kg/m 5 N s/m 6 6.86 m /s The kinematic viscosity cannot be read from Fig. B. since the pressure is not at 0 kpa. The shear stress can be calculated using τ μ du/ dy. From the given velocity distribution, u y y du dy (0.05 ) we get (0.05 y).4 From Table B. at C for water, μ.08 N s/m So, at the lower plate where y 0, we have du dy ( ) (0.05 0) 6 s τ.08 6 7.848 N/m At the upper plate where y 0.05 m, du dy y 0.05 (0.05 0.05) 6 s τ 7.848 N/m 5 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations.44.46 The shear stress can be calculated using τ μ du/ dr. From the given velocity du u 6( r ro ) we get 6( rro ). dr Hence, du rro dr du At the centerline, r 0, so 0, and hence τ 0. dr At r 0.5 cm, du 0.5 0 rro 00 s, dr ( ) τ 00 μ 00. N/m At the wall, r 0.5 cm, ( ) τ 6400 μ 6400 6.4 N/m ( 0.5 0) du 0.5 0 rro 6400 s, dr ( 0.5 0) π R ωlμ Use Eq..5.8 to calculate the torque, T h where h (.6.54) 0.0 cm 0.0 m π The angular velocity ω 000 rpm 09.4 rad/s 60 The viscosity of SAE-0 oil at C is μ 0.884 Ns/m (Figure B.) π (.7 m) 09.4 rad/s. m 0.884 Ns/m T. N m (0.0 )m power Tω. 09. 4 650 W 0.65 kw.48 Assume a linear velocity in the fluid between the rotating disk and solid surface. The velocity of the fluid at the rotating disk is rω, and at the solid surface 0. So, du r ω, where h is the spacing between the disk and solid surface, dy h and ω π 400 60 4.9 rad/s. The torque needed to rotate the disk is T shear force moment arm τ dr Due to the area element shown, dt df r τda r r where τ shear stress in the fluid at the rotating disk and du rω da πrdr dt μ πrdr r μ πr dr dy h 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations R πμω πμω T r dr R h h 0 4 The viscosity of water at 6 C is μ. Ns/m T 0.5 π (. Ns/m )( 4.9 rad/s) πμω R h m 4 ( ) 4.6 N m.50 du If τ μ dy constant, and μ AeB/T Ae By/K Ae Cy, then Ae Cy du du constant. dy dy De Cy, where D is a constant. Multiply by dy and integrate to get the velocity profile y Cy D Cy y Cy u De dy e E e C 0 0 ( ) where A, B, C, D, E, and K are constants. Compressibility.54 The sound will travel across the lake at the speed of sound in water. The speed of B sound in water is calculated using c, where B is the bulk modulus of ρ elasticity. Assuming (T C) and using Table B. we find 7 B Pa, and ρ 999.7 kg/m c 999.7 kg/m 7 N/m 45 m/s The distance across the lake is, L cδt 45 0.6 90 m.56 b) Using Table B. and T 7 C we find The speed of sound in water is calculated using: c B 7 c 6 / 99 508 m/s ρ ( ) 7 B 6 N/m, ρ 99 kg/m 7 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations Surface Tension.58.60.6 σ For a spherical droplet the pressure is given by p R Using Table B. at 5 C the surface tension σ 7.4 N/m σ 0.074 N/m p 6 9.6 kpa R 5 m 4σ 0.074 N/m For bubbles: p 6 59. kpa R 5 m For a spherical droplet the net force due to the pressure difference Δ p between the inside and outside of the droplet is balanced by the surface tension force, which is expressed as: σ 0.05 N/m Δ p pinside poutside 6 kpa R 5 m Hence, p p + kpa 8000 kpa + kpa 80 kpa inside outside In order to achieve this high pressure in the droplet, diesel fuel is usually pumped to a pressure of about 000 bar before it is injected into the engine. See Example.4: 4σcosβ h ρ gd ( ) ( ).6 00 kg/m 9.8 m/s ( m) 4 4 0.47 N/m cos 4.5 m 0.45 mm Note that the minus sign indicates a capillary drop rather than a capillary rise in the tube. Draw a free-body diagram of the floating needle as shown in the figure. The weight of the needle and the surface tension force must balance: W σ L or ρg σ L.64 The volume of the needle is ρ 4 d πd L g L 8σ πρg σ πd L 4 σl W σl needle 8 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations.66 There is a surface tension force on the outside and on the inside of the ring. Each surface tension force σ π D. Neglecting the weight of the ring, the free-body diagram of the ring shows that F σπ D F D apor Pressure.68.70.7 To determine the temperature, we can determine the absolute pressure and then use property tables for water. The absolute pressure is p 80 + 9 kpa. From Table B., at 50 C water has a vapor pressure of. kpa; so T 50 C is a maximum temperature. The water would boil above this temperature. At 40 C the vapor pressure from Table B. is 7.8 kpa. This would be the minimum pressure that could be obtained since the water would vaporize below this pressure. The inlet pressure to a pump cannot be less than 0 kpa absolute. Assuming atmospheric pressure to be 0 kpa, we have,000 kpa + 0 kpa 600x x 6.8 km. Ideal Gas Assume air is an ideal gas and calculate the density inside and outside the house T 5 C, p. kpa, T 5 C, and p 85 kpa using in in out out.74 ρ. kn/m in p.6 kg/m RT 0.87 kj/kg K (5 + 7 K) 85 ρ out.9 kg/m 0.87 48 Yes. The heavier air outside enters at the bottom and the lighter air inside exits at the top. A circulation is set up and the air moves from the outside in and the inside out. This is infiltration, also known as the chimney effect. 9 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations The weight can be calculated using, W γ ρg.76 Assume air in the room is at 0 C and 0 kpa and is an ideal gas: p 0 ρ.89 kg/m RT 0.87 9 W γ ρg.89 kg/m 9.8 m/s ( 0 4 m ) 9 N The pressure holding up the mass is 0 kpa. Hence, using pa W, we have 0,000 N/m m m 9.8 m/s This gives the mass of the air m,94 kg Since air is an ideal gas we can write p mrt or.78 ( ) mrt,94 kg 0.87 kj/kg K 5+7 K 846 m p 0 kpa Assuming a spherical volume πd 6, which gives d 6 846 m 5. m π 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations First Law The fist law of the thermodynamics is applied to the mass: Q W Δ PE+Δ KE+Δ U In this case, since there is no change in potential and internal energy, i.e., Δ PE 0, and Δ U 0, and there is no heat transfer to or from the system, Q 0. The above equation simplifies to: W ΔKE where W Fdl a) F W 00 N 00 N m 000 N m or J.80 Note that the work is negative in this case since it is done on the system. Substituting in the first law equation we get: ( ) 000 N 000 N m m m ( m/s) + 9.5 m/s 5 kg b) 0 F 0s W 0sds s 00 N m 0 ( ) 00 N 00 N m m m ( m/s) + 5.8 m/s 5 kg 0 c) 00cos( π 0) 00cos( π 0) F s W s ds ( ) ( ) W 00 0 π sin π 0 4000 π N m ( ) 4000 π 4000 N m N π m m ( m/s) + 6.4 m/s 5 kg.8 Applying the first law to the system (auto + water) we write: Q W Δ PE+Δ KE+Δ U In this case, the kinetic energy of the automobile is converted into internal energy in the water. There is no heat transfer, Q 0, no work W 0, and no change in potential energy, Δ PE 0. The first law equation reduces to: [ ] m Δ U mcδ T where, m HO HO HO ρ 6 00 kg/m 000 cm m cm kg 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations For water c 480 J/kg K. Substituting in the above equation 0 00 500 kg m/s kg 480 J/kg K ΔT 600 Δ T 69. C For a closed system the work is W pd For air (which is an ideal gas), p mrt, and mrt p.84 W pd mrt d mrt d mrt ln p Since T constant, then for this process p p or p p Substituting in the expression for work we get: W mrt p W kg ( 87 J kg K) 94 K ln 6,980 J 6.98 kj The st law states that Q W mδ u mc Δ T 0 since Δ T 0. Q W 6.98 kj v.86 For a closed system the work is W pd If p constant, then W p( ) Since air is an ideal gas, then p mrt, and p mrt ( ) ( ) W mr T T mr T T mrt kg 0.87 kj/kg K 4 K 4 kj 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations Isentropic Flow Since the process is adiabatic, we assume an isentropic process to estimate the maximum final pressure:.88 k/ k.4/0.4 T 4 p p T 9 (50 + 0) 904 kpa abs or 804 kpa gage. Note: We assumed p atm 0 kpa since it was not given. Also, a measured pressure is a gage pressure. Speed of Sound.90.9 b) c krt.4 88.9 9 66.9 m/s d) c krt.4 44 9 m/s Note: We must use the units on R to be J/kg. K in the above equations. b) The sound will travel at the speed of sound c krt.4 87 9 4 m/s The distance is calculated using L cδ t 4 8. 854 m 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Basic Considerations 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics CHAPTER Fluid Statics FE-type Exam Review Problems: Problems - to -9. (C). (D). (C).4 (A).5 (B) The pressure can be calculated using: p γ h were h is the height of mercury. p γ h (.6 98 N/m ) (8.5 0.054) 96,600 Pa 96.6 kpa Hg Since the pressure varies in a vertical direction, then: Hg p p0 ρgh 84, 000 Pa.00 kg/m 9.8 m/s 4000 m 44.76 kpa p p + γ h γ h 0 + 0,000 0. 98 0. 800 Pa 8.0 kpa w atm m m water w This is the gage pressure since we used p atm 0. Initially, the pressure in the air is pair, γ H (.6 98) 0.6,50 Pa. After the pressure is increased we have: pair,,50 +,000,50.6 98 H. H 0.085 m 8.5 cm The moment of force P with respect to the hinge, must balance the moment of 5 the hydrostatic force F with respect to the hinge, that is: ( ) P F d 5 F γ ha 9.8 kn/m m ( m )] F 98. kn The location of F is at ( ) I. yp y+.67 +. m d... m ya.67(. ). P 98.. P.7 kn The gate opens when the center of pressure is at the hinge:.6 (A). + h I. + h b(. + h) / y + 5. yp y + + 5 +. Ay (. + h) b(. + h) / This can be solved by trial-and-error, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields h.08 m. 5 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics.7 (D).8 (A) The hydrostatic force will pass through the center, and so F H will be balanced by the force in the hinge and the force P will be equal to F. P F 9.8 4. w+ 9.8 ( π. / 4) w 00. w 5.6 m. The weight is balanced by the buoyancy force which is given by F γ where is the displaced volume of fluid: B 900 9.8 98 0.0 5 w. w 6 m The pressure on the plug is due to the initial pressure plus the pressure due to the acceleration of the fluid, that is: pplug pinitial + γ gasolineδ Z where.9 (A) a x Δ Z Δ x g 5 p plug 0, 000 + 6660 (. ) 4, 070 Pa 9.8 F p A 4,070 π 0.0 0.5 N plug plug Pressure..4 Since p γ h, then h p/γ a) h 50,000/98 5.5 m c) h 50,000/(.6 98).874 m This requires that p p ( γh) ( γh) water Hg water Hg b) 98 h (.6 98) 0.75 h. m.6 Δ p γδz Δp.7 9.8 (000) 7,70 Pa or 7.7 kpa.8 From the given information the specific gravity is S.0 + z/0 since S(0) and S().. By definition ρ 00 S, where ρ water 00 kg/m. Using dp γ dz then, by integration we write: p z dp 00( + z /0) gdz 00g z + 0 0 0 p 00 9. 8 +,000 Pa or kpa 0 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics Note: we could have used an average S: S avg.05, so that ρ avg 50 kg/m and so p γ h p 50 9.8,005 Pa.0 / αr From Eq. (.4.8): p patm [( T0 α z) / T0]g 0 [(88 0.0065 00)/88] 9.8/0.0065 87 96.49 kpa Assuming constant density, then 0 p p atm ρgh 0 9.8 00 /00 0.87 88 96.44 kpa 96.44 96.49 % error 0 0.05% 96.49 Since the error is small, the density variation can be ignored over heights of 00 m or less. Eq..5. gives B ρgdh d ρ or ρ dp B ρ But, dp ρgdh. Therefore d ρ dρ ρ T g dh B Integrate, using ρ 0 64 kg/m, and B. 9 N/m. ρ h dρ g dh ρ B 0 9.8 h 4.67 ρ 64 9. ( N/m ) This gives ρ 4 9 9.4 4.67 h Now h h g p ρ gdh dh 4 9 9.4 4.67 h 0 0 g 4 9 ln(9.4 4.67 h) 9 4.67 9 h If we assume ρ const: p ρ gh 64 9.8 h,48 h b) For h 500 m: p accurate 5,690 kpa and p estimate 5,657 kpa. 5, 657 5, 690 % error 0 0.% 5, 690.4 Use Eq..4.8: p (88 0.0065 / 88) 9.8 z 0.0065 87 7 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics a) for z 000 p 69.9 kpa. c) for z 9000 p 0.6 kpa. Manometers.6.8 Referring to Fig..7b, the pressure in the pipe is: p γh (.6 98) 0.5,50 Pa or.5 kpa Referring to Fig..7a, the pressure in the pipe is p ρgh. If p 400 Pa, then 400 400 ρgh ρ 9.8 h or ρ 9.8 h 400 a) ρ 680 kg/m The fluid is gasoline 9.8 0.6 c) 400 ρ 9.8 0.45 999 kg/m The fluid is water.0..4.6.8 See Fig..7b: The pressure in the pipe is given by p γ h + γ H p 0.86 98 0.5 +.6 98 0.4 0,695 Pa or 0,96 kpa p p γ 0.+ γ H γ 0. water oil oil Hg water 40,000 6,000 90 9.8 0. +,600 9.8 H 00 9.8 0. Solving for H we get: H 0.74 m or 7.4 cm ( ) p γ 0.6 γ (0.0) γ (0.04) γ (0.0) water Hg water Hg water Using γ 9.8 kn/m and γ.6 9.8 kn/m water Hg p water 5.6 kpa p water 9.8 0. 0.68 9.8 0. + 0.86 9.8 0. p oil With p water 5 kpa, p oil 4 kpa pgage pair + γ water 4 where, pair patm γ Hg H p γ H + γ 4 gage Hg water p gage.6 9.8 0.6 + 98 4 7.89 kpa Note: we subtracted atmospheric pressure since we need the gage pressure. 8 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics.40 p + 98 0.05 +.59 98 0.07 0.8 98 0..6 98 0.05 p 587 Pa or 5.87 kpa The distance the mercury drops on the left equals the distance along the tube that the mercury rises on the right. This is shown in the sketch. Oil (S 0.87) B Water Δh cm A 9 cm 7 cm Δh 40 o Mercury.4.44 From the previous problem we have: ( p ) p γwater γhg γoil B + 0.07 0.09sin 40 0.. kpa () A For the new condition: ( ) + γ ( 0.07 +Δ ) γ 0.sin 40 γ ( 0. Δ sin 40) p p h h () B A water HG oil where Δh in this case is calculated from the new manometer reading as: Δ h+δ h/ sin 40 9 cm Δ h 0.78 cm Subtracting Eq.() from Eq.() yields: ( p ) ( p ) γ ( Δh) γ 0.0sin 40 γ ( Δhsin 40) B B water HG oil Substituting the value of Δh gives: ( ) ( ) ( ) p B.+ 0.0078.6 0.0sin 40 0.87 0.0078sin 40 9.8 0.5 kpa a) Using Eq. (.4.6): ( ) ( ) p γ z z + γh+ γ γ H where h z5 z7 6 cm 4000 9800(0.6 0.) + 5,600(0.0) + (,400 5,600)H H 0.076 m or.76 cm 9 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics.46 From No..0: p oil 4.0 kpa From No..6: p oil p water 9.8 (0.+Δz) 0.68 9.8 (0. Δz) + 0.86 9.8 (0. Δz) Δz 0.045 m or 4.5 cm Forces on Plane Areas.48.50 The hydrostatic force is calculated using: F A πr π ( 0.5 m) hence, F π ( ) γ ha where, h m, and 98 0.5 694 N For saturated ground, the force on the bottom tending to lift the vault is: F p c A 9800.5 ( ) 9,400 N The weight of the vault is approximately: W ρ g walls ( ) ( ) ( ) W 400 9.8.5 0. + 0. + 0 0.8. 0. 8,400 N The vault will tend to rise out of the ground..5 b) Since the triangle is horizontal the force is due to the uniform pressure at a depth of m. That is, F pa, where p γ h 9.8 98. kn/m The area of the triangle is A bh.88 /.88 m.54 F 98..88 77.4 kn a) F γha 9.8 6 π 79.7 kn y p 4 I π /4 y+ 6 + 6.67 m (x, y) p (0, 0.67) m Ay 4π 6 c) F 9.8 (4 + 4/) 6.9 kn y p 5 + 4 /. 6 5.50 m y.5 5. 6 4/.5.5 x 0.975 x (x, y) p (0.975,.5) m y x.56 F γha 98 6 0.777 6 N, or 77 kn 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics I yp y + + 4 5 / 75. 7.778 m Ay 75. 0 M Hinge Σ 0 ( 7.778) 77 5 P P 5 kn The vertical height of water is h. 0.4.4 m The area of the gate can be split into two areas: A A+ A or A..4 + 0.4.4.8 m Use forces: F γ h A 98 0.5657 (..4) 754 N.60.4 F γ ha 98 (0.4.4) 674 N The location of F is at y p (. 4 ) 0. 754 m, and F is at y p I.4 0.4.4 / 6 y+ + 0.5657 m Ay 0.4 (.4 / ) (.4 / ).4 ΣM hinge 0 : 754 + 674 (.4 0.5657).4P 0 P 46 N The gate is about to open when the center of pressure is at the hinge..6 b) b / yp. + H (.0/ + H) + ( + H)b H 0.6667 m A free-body-diagram of the gate and block is sketched. T Sum forces on the block: Σ F 0 W T + F y B 0 F stop T.64 where F B is the buoyancy force which is given by F B γ πr ( H) F H y p F B W Take moments about the hinge: R x T.5 F ( y ) H p Ry where F H is the hydrostatic force acting on the gate. It is, using 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics h.5 m and A 6 m FH ( )( ) γ ha 9.8 kn/m.5 m 6 m 88.9 kn From the given information y p ( ) I / y+.5 + m ya.5 6 ( ) 88.9 T 5. kn.5 F W T 70 5. 44.77 kn. γπ R H 44.77 B 44.77 kn H m.55 m ( 9.8 kn/m ) π ( m) ( ).66 The dam will topple if there is a net clockwise moment about O The weight of the dam consists of the weight of the rectangular area + a triangular area, that is: W W+ W. The force F acting on the bottom of the dam can be divided into two forces: Fp due to the uniform pressure distribution and Fp due to the linear pressure distribution. W.4 98 8.9.8 80 kn assume m deep b) W.4 98 8.9 7. / 60 kn W 98 (8 6.86/) 605.7 kn F 98 9 8 589 kn F 98. 5 44. 4 kn F p 98 9 65 kn W F W F O F F p 98 5 9 / 66. kn Σ M O : (589)(6) + (65)(4.5) + (66.)(6) (80)(0.9) (44.4)(/) (60)(4.) (605.7)(6.7) 9 kn m > 0. will tip 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics Forces on Curved Surfaces.68.70 Since all infinitesimal pressure forces pass through the center, we can place the resultant forces at the center. Since the vertical components pass through the bottom point, they produce no moment about that point. Hence, consider only horizontal forces: ( γ ha) ( γ ha) ( F ) 9.8 (4 ) 784.8 kn H water water ( F ) 0.86 9.8 0 68.7 kn H oil oil ΣM: P 784. 8 68. 7. P 66. kn A free-body-diagram of the volume of water in the vicinity of the surface is shown. Force balances in the horizontal and vertical directions give: FH F F W + F where FH and F are the horizontal and vertical components of the force acting on the water by the surface AB. Hence F F 9.8 kn/m 8 + 4 706. kn H ( )( )( ) The line of action of F H is the same as that of F. Its distance from the surface is y p ( ) I 4 y+ 9 + 9.07 m ya 9 8 F B. W F F F H x.a To find F we find W and F : π W γ ( 9.8 kn/m ) ( ) 4.7 kn 4 F ( ) 9.8 kn/m 8 4 68 kn F F + W.7 + 68 66 kn To find the line of action of F, we take moments at point A: F x F d + W d where R d m, and d.55 m: 4 4 ( π) ( π) 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics F d+ W d 68 +.7.55 x.08 m F 66 Finally, the forces F H and F that act on the surface AB are equal and opposite to those calculated above. So, on the surface, F H acts to the right and F acts downward..7 Place the resultant FH + F at the center. F passes through the hinge. The moment of F H must equal the moment of P with respect to the hinge: (9.8 ).8 P P 70. kn The resultant F + F H of the unknown liquid acts through the center of the circular arc. F passes through the hinge. Thus, we use only ( F H ). Assume m wide: ( )( ) F γ ha γ R R γ R H x x x.74 F γ ha γ R R γ R The horizontal force due to the water is ( )( ) The weight of the gate is W Sγw γw( πr ) 0. 4 w w w w Summing moments about the hinge: ( ) ( 4 π ) F R + W R F R w H R R R R R a) π 4 98 + 0. 98 γ x R 4 π γ x 4580 N/m.76 The pressure in the dome is: a) p 60,000 98 0.8 98 4,870 Pa or 4.87 kpa The force is F pa projected (π ) 4.87 40.4 kn b) From a free-body diagram of the dome filled with oil: W F weld + W pa pa Using the pressure from part (a): F weld F weld 4,870 π (0.8 98) 4 π,400 or.4 kn 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics Buoyancy.78 Under static conditions the weight of the barge + load weight of displaced water. (a) 0,000 + 50,000 98 (6 d + d /). d + d 8.5 0 d.7 m.80.8.86 The weight of the cars will be balanced by the weight of displaced water: 5,000 60 98(7.5 90 Δd ) Δd 0.6 m or.6 cm T + F B W (See Fig.. c) T 40,000.59 98 8804 N or 8.804 kn At the limit of lifting: F B W+pA where p is the pressure acting on the plug. (b) Assume h> 4.5 + Rand use the above equation with R 0.4 m and h 4.9 ( ) F γ γ πr A 98 0.80 849 N B w w segment ( ) W + pa 6670 N + 98 4.9 π 0. 84 N Hence, the plug will lift for h > 4.9 m (a) When the hydrometer is completely submerged in water:.88 W γ w π 0.05 π 0.005 (0.0 + m Hg )9.8 98 0.5 + 0. 4 4 m Hg 0.0886 kg When the hydrometer without the stem is submerged in a fluid: W γ x S x.089 ( 0.05) π (0.0+ 0.089)9.8 Sx 98 0.5 4 5 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics Stability.90 With ends horizontal ( γ xπ ) 4 Io π d 64 The displaced volume is W γ water d h/ 4 / 98 8.0 γ d since h d 5 x The depth the cylinder will sink is depth A 5 8.0 γ xd.0 π d /4 5 γ d The distance CG is CG h 5. γ x d /. Then 4 IO π d / 64 d 5 80. γ xd GM CG +. d / > x 5 γ x 0 This gives (divide by d and multiply by γ x ): 6.8 0.5 γ x + 5. 5 γ x > 0. Consequently γ x > 868 N/m or γ x < 46 N/m.9.94 6 9 + 6 4 As shown, y 6.5 cm above the bottom edge. 6 + 6 4γ 9.5 + 6γ 8.5 + 6SAγ 4 G 6.5 cm 0.5γ 8 + γ 8 + S γ 6 A + 4 S A 74 + 64 S A S A. The centroid C is.5 m below the water surface CG.5 m Using Eq..4.47: 8 / GM.5.777.5 0.77 0 8 > The barge is stable Linearly Accelerating Containers.96 (a) p max 00 0 (0 4) 00 (9.8) (0 ) 99,60 Pa (c) p max 00 8 (0.6) 00 (9.8 + 8) (0.8) 4,860 Pa or 4.86 kpa 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics Use Eq..5.:.98 b) 60,000 00 a x ( 8) 00 (9.8 + ).5 + 8a x 9.8 8a 60 8 a x + 49.5 9.8 x or a x..574 9.8 a x 5. a x +.44 0 a x 0.5, 4.8 m/s a x a) The pressure on the end AB (z is zero at B) is, using Eq..5. p(z) 00 ( 7.66) 00 9.8(z) 76,60 98 z.0 5. FAB (76,60 98 z)4dz 640,000 N or 640 kn 0 b) The pressure on the bottom BC is p(x) 00 (x 7.66) 76,60,000 x 7. 66 FBC (76,60,000 x)4dx.6 6 N or 6 kn 0 Use Eq..5. with position at the open end: b) p A 00 (0.9 0) 9000 Pa z A p B 00 (0.9) 00 9.8( 0.6) 4 Pa. p C 00 9.8 ( 0.6) 5886 Pa e) p A 00 8 ( 5. 065. ) 6,875 Pa C B x p B 00 8 ( 5. 065. ) 00 9.8 ( 0. 65),740 Pa p C 00 9.8 ( 0. 65) 6 Pa Rotating Containers Use Eq..6.4 with position at the open end: z A.4 a) p A 00 (0 0.9 ) 40,500 Pa p B 40,500 + 98 0.6 4,600 Pa p C 98 0.6 5886 Pa r C B ω 7 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Fluid Statics The air volume before and after is equal 0 06 0 πrh π.. r h 0 0.44 (a) Using Eq..6.5: r 0 5 / 9.8 h h 0.48 m p A 00 5 0.6 98 ( 0.7) A h z r 0 r.6 849 Pa (c) For ω, part of the bottom is bared π Using Eq..6.5: ω 0 g 0.6 0. π r0 h π r h r ω r h, h g g g 044. h h ω ω h h 0.44 9.8 or h A Also, h h 0.8..6 h 0.64 0.79. h 0.859 m, r 0.8 m p A 00 (0.6 0.8 ) 7,400 Pa h z r 0 r pr ( ) [0 (0.8 )] ρω r ρg h pr ( ) 500ω r + 98(0.8 h) if h < 0.8 pr () ω ( r r ) if h > 0.8 500 d r d A πr d r.8 a) 0.6 F p π rdr π (,500 r + 650 r ) dr 6670 N (We used h 0.48 m) c) 0 0.6 0.8 F p π rdr π (50,000( r 0.8 r ) dr 950 N (We used r 0.8 m) 8 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion CHAPTER Introduction to Fluids in Motion FE-type Exam Review Problems: Problems - to -9 nˆ 0 ( n ˆi+ n ˆj) (ˆi 4 ˆj) 0 or n 4n 0 x y x y. (D). (C). (D).4 (C).5 (B).6 (C) Also n x n x ny + since ˆn is a unit vector. A simultaneous solution yields 4 / 5 and n / 5. (Each with a negative sign would also be OK) y ˆ a + u + v + w xy( yi) y (xˆi yˆj) 6ˆi+ 8ˆi+ 6ˆj t x y z a ( 8) + 6 7.89 m/s u u u u u (4 ) ax + u + v + w u x t x y z x (4 x) x ( )( )(4 x) 0 6.5 m/s (4 x) 4 8 The only velocity component is u(x). We have neglected v(x) since it is quite small. If v(x) were not negligible, the flow would be two-dimensional. p γ waterh 98 0.800 m/s ρ ρ. air p + + 0.00 0.600 9.8 0.400.80 m/s g γ g g The manometer reading h implies:.7 (B) or (60.) 9.9 m/s + p p ρ + ρ. The temperature (the viscosity of the water) and the diameter of the pipe are not needed. 9 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion.8 (A) g p p + + γ g γ 800,000 40 m/s 98 9.8 ρ.9 (D) p ( ) ( ) 90 0 5 04,400 Pa Flow Fields dx dy a) u t+ v t dt dt y streamlines t 5 s.4 x t + t + c y t + c (7, ) (5, 5) y + y 9.8 o x xy + y 4 y parabola x.6 Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian: Several college students would be positioned at each intersection and quantities would be recorded as a function of time. a) ˆi + cosα 0.8 α.69 + nx + n 0 n y ˆ 0 (ˆ ˆ) ( ˆ ˆ n i+ j nxi+ nyj) 0 n n 9 x + ny nx + nx y 4 x.8 n, or ˆ (ˆ ˆ x n y n i j) c) ˆi 5 cosα 0.60 α 5.67 5 + ( 8) 5nx 8n 0 n y ˆ 0 (5ˆ 8 ˆ) ( ˆ ˆ n i j nxi+ nyj) 0 n 5 8 n 5 64 x + ny ny + ny x y 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion 5 8 n, or ˆ (8ˆ 5 ˆ y n x n i+ j) 89 89 89.0 b) u + v + w + x( ˆi) + y( ˆj) 4xˆi+ 4yˆj 8ˆi 4ˆj x y z t. The vorticity ω Ω. a) ω 40ˆi c) ω ˆi 4k ˆ a) a r 40 80 40 sinθ 40 cosθ cosθ ( sin θ) + r r r r r 40 + sin θ (.5)( ). 5( ) 9.75 m/s r r.4 40 80 θ aθ θ θ θ r r + + 40 r sin r + 40 cos sin cos r 600 0 sin θcosθ 0 since sin(80 ) 0 4 r r a φ 0.6 a + u t x + v u + w ˆi y z t For steady flow u/ t 0 so that a 0.8 b) t / u ( 0.5 )( e ).875 m/s at t.0 Dρ ρ ρ ρ ρ 4 4 000 u + v + w + (. e ) Dt x y z t 4 9. kg/m s Dρ ρ. u 4 (0.0) 0.04 kg/m s Dt x 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion u ax + u t v.4 ay + v a + ( ) t t w az + w t π ˆ 5 ˆ Ω k 7.7 krad/s 4 60 60 5( 0.707ˆi 0.707 kˆ).55ˆi.55km/s ˆ A Ω + Ω ( Ω r ) 5ˆ ˆ ˆ 5 7.7 k (.55i.55 k) + 7.7 k ˆ.6 7.7 kˆ (6.4 )( 0.707i+ 0.707 k ˆ) 5 ˆj+ 0.04 ˆi m/s 5 6 ˆ 5 Note: We have neglected the acceleration of the earth relative to the sun since it is quite small (it is / d S dt ). The component ( 5.4 5ˆ j ) is the Coriolis acceleration and causes air motions to move c.w. or c.c.w. in the two hemispheres. Classification of Fluid Flows.8 Steady: a, c, e, f, h Unsteady: b, d, g.4 a) inviscid c) inviscid e) viscous inside the boundary layers and separated regions. g) viscous. L.46 Re 0.8 0.75.4 ν 5 9640 Turbulent.48 a) D. 0.0 Re 795 ν 5.5 Always laminar Assume the flow is parallel to the leaf. Then 5 x T/ν.50 5 5 4 x ν /.5.4 / 6 8.7 m T The flow is expected to be laminar 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion.5 Dρ ρ ρ ρ ρ u + v + w + 0 Dt x y z t For a steady, plane flow ρ / t 0 and w 0 ρ ρ Then u + v 0 x y Bernoulli s Equation.54.56.58 p Use ρ. kg/m ρ a) v p / ρ 000 /. 60 m/s p + 0 p 000 57.0 m/s ρ ρ. p U p ρ + + b) Let r rc: pt U ρ ρ d) Let θ 90 : p90 ρu p U p + + ρ ρ.60 a) p ( U u ) ρ ρ 0π + 50ρ + π x x 50ρ + x x c) p ( U u ) ρ ρ 60π 0 0 + 450ρ + π x x 450ρ + x x 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion Assume the velocity in the plenum is zero. Then.6 or (60.) 9.9 m/s + p p ρ + ρ. We found ρ. kg/ m in Table B.. Bernoulli from the stream to the pitot probe: Manometer: pt + γ H γhgh γh p γh p T ρ + p.64 Then, ρ + p+ γh γhg H p γ Hg γ ( H) ρ (.6 )9800 a) ( 0.04) 00.4 m/s (.6 )9800 b) ( 0.) 00 4.97 m/s.66 The pressure at 90 from Problem.58 is p90 ρu /. The pressure at the stagnation point is pt ρu /. The manometer provides: p γ T H p90.04u 9800 0.04.04 U. U.76 m/s Bernoulli: p + g γ p + g γ.68 Manometer: + γ + γhg γ γ γ + p z H H z p g Substitute Bernoulli s into the manometer equation: p + ( γhg γ) H γ + p g b) Use H 0.05 m: Substitute into Bernoulli: 9800 (.6 )9800 0.05.56 m/s 9.8 0 56 p γ. 9800 9,600 Pa g 98. 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion Write Bernoulli s equation between points and along the center streamline: ρ ρ p z p z + + γ + + γ Since the flow is horizontal, z z and Bernoulli s equation becomes.70 0.5.5 p+ 00 p + 00 From fluid statics, the pressure at is p γ h 98 0.5 45 Pa and at, using p γh, Bernoulli s equation predicts 0.5.5 45 + 00 98H+ 00 H 0.98 m or 9.8 cm Assume incompressible flow ( < 0 m/s) with point outside the wind tunnel where p 0 and 0. Bernoulli s equation gives.7 p 0 + p ρ ρ p 90 a) ρ.9 kg/m p.9 0 695 Pa RT 0.87 5 p 9 c) ρ.094 kg/m p.094 0 5470 Pa RT 0.87 9 Bernoulli across nozzle: + p p ρ + ρ p / ρ.74 Bernoulli to max. height: a) p g p + + h γ p g / ρ 700,000 /00 7.4 m/s + γ + h h p / γ h p /γ 700,000/9800 7.4 m b) p / ρ, 400,000 /00 5.9 m/s h p /γ,400,000/9800 4.9 m.76 + p p ρ + ρ p 0,000 Pa, the lowest possible pressure. 5 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter / Introduction to Fluids in Motion a) 600,000 0,000 7.4 m/s 00 00.78 b) 00, 000 0, 000 8. m/s 00 00 ρ b) p ( ) ( ) 90 4,00 Pa ρ d) p ( ) ( ). 59.0 Pa Apply Bernoulli s equation between the exit (point ) where the radius is R and a point in between the exit and the center of the tube at a radius r less than R:.80 Since + + p ρ p p ρ ρ <, we see that p is negative (a vacuum) so that the envelope would tend to rise due to the negative pressure over most of its area (except for a small area near the end of the tube)..8.84 A burr downstream of the opening will create a region that acts similar to a stagnation region thereby creating a high pressure since the velocity will be relatively low in that region. The higher pressure at B will force the fluid toward the lower pressure at A, especially in the wall region of slow moving fluid, thereby causing a secondary flow normal to the pipe s axis. This results in a relatively high loss for an elbow. stagnation region 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws CHAPTER 4 The Integral Forms of the Fundamental Laws FE-type Exam Review Problems: Problems 4- to 4-5 4. (B) 4. (D) 4. (A) 4.4 (D) p 00 m ρa A π 0.04 70 0.87 kg/s RT 0.87 9 Refer to the circle of Problem 4.7: 75.7 Q A ( π 0.4 0. 0.40 sin 75.5 ) 0.56 m /s 60 W P γ Q g p p W P 0 00 + γ γ 0.040 γ 40 W P 40 kw and energy required 47. kw 0.85 4.5 (A) 0 p p + g γ p 0 +. p 7, 00,000 Pa 9.8 98 Manometer: ρ γh + p g + p or g 98 0.0 + p ρ g g 4.6 (C) 4.7 (B) 7.96 0,000 Energy: K K.5 9.8 98 Combine the equations: 98 0.0. 8. m/s Δp Q 0.040 hl K 7.96 m/s g γ A π 0.04 7.96 0,000 K K.5 9.8 98 7 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 4.8 (C) 4.9 (D) W P γ Q g W P Δp + γ Δ 6 0.040 400 W Q p 6 kw P 8.0 kw η 0.89 pb 4.58 7.6 6.0 + 5 + +. pb 46,000 Pa 9.8 98 9.8 In the above energy equation we used Q 0. hl K with 4.4 m/s g A π 0. Q 0. 9. 89 m/s A π 004. 4. (A) 4. (A) 4. (C) Energy surface to entrance: p HP z K g + γ + + g 9. 89 80,000 9. 89 H P + + 50 + 5. 6 0. 4 m 9. 8 98 9. 8 W γqh / η 98 0. 0. 4 / 0. 75 6,000 W P P P After the pressure is found, that pressure is multiplied by the area of the window. The pressure is relatively constant over the area. p p (6.5 ).7 + +. p 98,085,000 Pa g γ g γ 9.8 pa F ρq ( )., 085, 000 π 0.05 F 00 0..7(6.5 ) F 7,500 N 4. (D) Fx m ( x x) 00 0.0 0. 50(50cos 60 50) 500 N 4.4 (A) 4.5 (A) F m ( ) 00 π 0.0 60 (40cos 45 40) 884 N x rx rx Power F 884 0 7, 700 W x B Let the vehicle move to the right. The scoop then diverts the water to the right. Then, F m ( x x) 00 0.05 60 [60 ( 60)] 70,000 N 8 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws Basic Laws 4.6 b) The energy transferred to or from the system must be zero: Q W 0 4.0 b) The conservation of mass. d) The energy equation. System-to-Control-olume Transformation 4.4 4.6 ˆ ˆ ˆ 0.707( ˆ ˆ) n i j i+ j nˆ ˆi 0.707( ˆ+ ˆ) i j 7.07 m/s n ˆ ˆ ˆ n nˆ i (0.866i 0.5 j) 8.66 m/s nˆ ˆi ( ˆj ) 0 n nˆ 0.866ˆi 0.5ˆj. nˆ ˆj ( Bn ˆ ) A 5(0.5ˆi + 0.866 ˆj) ˆj( ) 5 0. 866 559 cm olume 5 sin 60 559 cm Conservation of Mass Use Eq. 4.4. with m representing the mass in the volume: 4. dm dm + ρ ˆ da dt n + ρa ρa dt 0 cs.. dm + ρq m dt Finally dm m ρ Q dt 4.4 A A π 4 8 π 6 4 4.5 m/s m ρa 00π 8 5 kg/s 4 Q A π 4 8 0.05 m /s 9 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 4.8 4.40 p 500 ρa ρa ρ 4.4 kg/m RT 0.87 9 46 ρ 8.7 kg/m 0.87 5 4.4 π 0.05 600 8.7 π 0.05 9.8 m/s m ρ A 0.89 kg/s Q A 4.7 m /s Q.5 m /s d b) (.5 +.5.5) π d 4.478 m 4 a) Since the area is rectangular, 5 m/s. 4.4 m ρ A 00 0.08 0.8 5 0 kg/s Q m ρ 0. m /s c) 0.08 0.04 + 5 0.0 + 5 0.0 7.5 m/s m ρ A 00 0.08 0.8 7.5 480 kg/s Q m 0.48 m /s ρ 4.44 4.46 4.48 If dm/ dt 0, then ρ A ρ A + ρ A. In terms of m and Q becomes, letting ρ ρ ρ 00 π 0. 0 m + 00 0. 0 m 5.08 kg/s 0. m in m out + m ρ 0. ρ (0y 0 y )dy+ ρ 0. + m 0 Note: We see that at y 0. m the velocity u(0.) m/s. Thus, we integrate to y 0., and between y 0. and 0. the velocity u : 4 4ρ ρ + ρ + m m 0. 6667ρ 0.8 kg/s 0. 0 0. m ρda 65(. 4 y)( y 5 y ).5dy 0 0,970 ( y 6. 4y + 7. y ) dy 6. 7 kg/s this u max 06. 04m/s. (See Prob. 4.4b) 40 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws ρ 65 + 00 8 kg/m ρ 8 0 4 ( 5 0 ) A... 64.9 kg/s Thus, ρa m since ρ ρ(y) and (y) so that ρ ρ 4.50 4.5 m of HO 4 m of air 000 π 0.005 9000 5.5 (.5 h) m of air s h 0.565 m m m + m 0 m/s (see Prob. 4.4c) in 0 00π 0.0 + 00π 0.0.04 m/s The control surface is close to the interface at the instant shown. i interface velocity e ˆn i ˆn 4.54 ρ A ρ A e e e i i i 8000.5 π 0.5 00 π 0.87 67 i i 0.44 m/s For an incompressible flow (low speed air flow) 4.56 4.58 uda A A 0. 0 /5 0y 0.8dy π 0.5 5 6/5 0 0.8 0. π 0.5 7. m/s 6 Draw a control volume around the entire set-up: or dmtissue 0 + ρa ρa dt d d m tissue + ρπ h 4 ρπ h φ h ( tan ) d d m tissue ρπ h + h h 4 tan φ 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 4.60 dm 0 + ρa ρa dt 6 m + 00( π 0.00 0.0 / 60) m 4.99 kg/s 4.6 4.64 dm dt 0 + Q A m a) ρ ρ where m ρ Ah 0 00π 0. 6 h + 00 0. 6 / 60 00π 0. 0 h 0.0 m/s or. mm/s Choose the control volume to consist of the air volume inside the tank. The conservation of mass equation is d 0 ρd + ρ da dt n C Since the volume of the tank is constant, and for no flow into the tank, the equation is dρ dt CS A 0 + ρ e e e p Assuming air behaves as an ideal gas, ρ. At the instant of interest, RT dρ dp. Substituting in the conservation of mass equation, we get dt RT dt ( )( ) π ( ) dp ρ.8 kg/m 00 m/s 0.05 m ea e e kj ( RT ) 0.87 98 K dt.5 m kg K dp 4.5 kpa/s dt Energy Equation 4.66 du W Tω+ pa + μ A dy belt 0 500 π /60 + 400 0.4 0.5 +.8 0 0.5 0.8 47 + 800 + 0. 00074 847 W 5 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 4.68 80% of the power is used to increase the pressure while 0% increases the internal energy ( Q 0 because of the insulation). Hence m Δ u 0. W 00 0. 0 4. 8Δ T 0. 500 Δ T 0.86 C W 4.70 T 40 0. 89 mg T b) W 40 0.89 (90,000/60) 9.8 5,900 W + p + + p z + z g γ g γ 4.7 4 4 + + h 9.8 9.6h 0.8.8 +h h Continuity: 4 h This can be solved by trial-and-error: h.7 m :.8?.8 h.7 m :.8?.8 h 7. m h 0.6 m :.8?.87 h 0.6 :.8?.75 h 06. m Manometer: Position the datum at the top of the right mercury level. 4.74 98 0.4 + 98z + p + 00 (98.6) 0.4 + 98 + p Divide by γ 98: Energy: 0.4 + p p z +.6 0.4 γ + g + + γ () z z p p g + γ + g + γ + () Subtract () from (): With z m.6 0.4 g 9.94 m/s π Q 7 (0.05).75 m/s 4.76 Continuity: Energy: π π (0.05) (0.075).58 m/s p p 0.7 g + γ g + γ + g 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 57. 58. p 44 + 00 0. 6 44. kpa 9. 6 9. 6 4.78 008. Q/ A 8.9 m/s 9 54.6 m/s π 00. Energy: p p 0. g + γ g + γ + g p 98 54. 6 8. 9 08. 98. 98. 6. Pa a) Energy: 0 + 0 + z 0 + + z p p g γ g γ gz 98. 4. 6.86 m/s 0 4.80 4.8 4.84 Q A 0.8 6.86 5.49 m /s For the second geometry the pressure on the surface is zero but it increases with depth. The elevation of the surface is 0.8 m: z0 + h g( z0 h) 98. 6.64 m/s g Q 0.8 6.64 5.0 m /s Note: z 0 is measured from the channel bottom in the nd geometry. z 0 H + h 0 + 0 + z 0 + + z p p g γ g γ b) Q A π 0.09 9.04 80,000 + 4 9.04 m/s 98 9. 8 0.485 m /s p Manometer: γ H+ γz+ p.6γh+ γz+ p p. 6H + γ γ Energy: p p γ + g γ + g. Combine energy and manometer:. 6H g Continuity: d d 4 d.6h g 4 d 44 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws / / d π.6h g H π 4 4.5 4 4 4 4 d / d d d Q d d d a) Energy from surface to outlet: H g gh 4.86 Energy from constriction to outlet: p p + + γ g γ g Continuity: 4 With p p v 450 Pa and p 0,000 Pa 450 6 0,000 + 98 9 8 gh + gh. 98 9. 8 H 0.66 m Energy: surface to surface: z0 z + hl 0 0 + g 4.88 Continuity: 4 60 g g 60 g ( 94,000) Energy: surface to constriction: 0 + + z g 98 z 40.4 m H 40.4 + 0 60.4 m elocity at exit e elocity in constriction elocity in pipe Energy: surface to exit: e H g e gh 4.90 Continuity across nozzle: D Also, 4 d e Energy: surface to constriction: p H + v g γ a) 4 D 4 97,550 5 6 g 5 + D 0. m g 0. 98 4.9 m pa 00 π (0. 05) 6 70. 65 kg/s W P 9 6 80 00 70.65 9.8 + / 0.85 8,500 W or 8.5 kw 4.8 hp 9.8 98 45 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws 4.94 0. 600,000 W T 00 9. 8 + 0. 87 9. 8 98. 04 W W T 6 Q We used. m/s A π 0.5 η T 0. 94 4.96 a) p p c v Q W S mg + + z z+ ( T T) g γ γ g The above is Eq. 4.5.7 with Eq. 4.5.8 and Eq..7. γ pg 85 9.8 600 9.8 0,500 9.9 N/m γ RT 0.87 9 0.87 T T 00 600,000T 85,000 76.5 (,500,000) 5 9.8 + + ( T 9) 9.8 0,500 9.9 9.8 T 57 K or 99 C Be careful of units. p 600,000 Pa c 76.5 J/kg K v 4.98 4.0 Energy: surface to exit: W 0 4 5 TηT mg +. g g 5. 6 m/s mg Qγ 5 98 47,50 N/s π 06..6. 6 W T 0. 8 47,50 0 + 4. 5 W T 590 kw 9. 8 9. 8 Choose a control volume that consists of the entire system and apply conservation of energy: H p z H p z h γ g γ g + + + + + + + P T L 46 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws Carburetor Section () 0.5 m Pump Section () We recognize that H T 0, is negligible and h L /g where Q/A and Q/ A. Rearranging we get: p p HP + + z z+ h γ g 6 ( ) π (.5 m) L 6. m /s Q (0.) 0. m/s hl. m A 9.8 6 Q 6. m /s.5 m/s π 4 ( 4 m) A Substituting the given values we get: H P ( 95 0 ) kn/m (.5 m/s) + 0.5 m + +. m 8.85 m 6.660 kn/m 9.8 m/s The power input to the pump is: P P P ( ) 6 ( )( )( ) W γqh / η 6660 N/m 6. m /s 8.85 m / 0.75 0.5 W Energy: across the nozzle: 5 + + 6.5 p p γ g γ g 4. 400,000 6.5 + 4.58 m/s 98 9.8 9.8 8.6 m/s Energy: surface to exit: 8.6 4.58 7.6 H P + 5 +.5 +. 9.8 9.8 9.8, A 7.6 m/s H P 6.8 m 47 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 / The Integral Forms of the Fundamental Laws QH 98 ( 0.0 ) 8.6 6.8 W γ P π P 80 W ηp 0.85 Energy: surface to A : 7.6 p 7.6 5 + A +. pa 9,400 Pa 9.8 98 9.8 Energy: surface to B : 4.58 p 7.6 6.0 + 5 + B +. pb 46,000 Pa 9.8 98 9.8 Depth on raised section y. Continuity: y Energy (see Eq. 4.5.): + + + y (0.4 ) g g 4.4 9. 059 + y or y. 059 y + 4. 8 0 gy Trial-and-error: y y.0 : 0.? 0.8: + 0.05? 0 y.85 m y.: 0.? 0 y. m y.: + 0.? 0 The depth that actually occurs depends on the downstream conditions. We cannot select a correct answer between the two. 4.6 The average velocity at section is also 8 m/s. The kinetic-energy-correction factor for a parabola is (see Example 4.9). The energy equation is: α p p h g + γ g + γ + L 8 50,000 8,000 + + + 9. 8 98 9. 8 98 h L h 0. 85 m L a) 0.0 4 r 0 0.0 0.0 da π rdr 5 m/s A π 0.0 0.0 0.0 4 0.0 0 4.8 0.0 r da π 0.0 5 0 0.0 α πrdr A 000 00. 5 00 4. 00. 4 00. + 00. 6 00. 6 4 8 00. 8 00. 6 00. 48 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.