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A STUDENT’S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS, 4TH EDITION, SI MERLE C. POTTER DAVID C. WIGGERT BASSEM H. RAMADAN STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS FOURTH EDITION, SI MERLE C. POTTER Michigan State University DAVID C. WIGGERT Michigan State University BASSEM RAMADAN Kettering University Contents Preface iv Chapter 1 Basic Considerations 1 Chapter 2 Fluid Statics 15 Chapter 3 Introduction to Fluids in Motion 29 Chapter 4 The Integral Forms of the Fundamental Laws 37 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 109 Chapter 9 Compressible Flow 131 Chapter 10 Flow in Open Channels 141 Chapter 11 Flows in Piping Systems 161 Chapter 12 Turbomachinery 175 Chapter 13 Measurements in Fluid Mechanics 189 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 1 / Basic Considerations CHAPTER 1 Basic Considerations FE-type Exam Review Problems: Problems 1.1 to 1.13 1.1 (C) m = F/a or kg = N/m/s2 = N·s2/m 1.2 (B) [μ] = [τ/(du/dy)] = (F/L2)/(L/T)/L = F.T/L2 1.3 (A) 2.36 ×10−8 Pa = 23.6 ×10−9 Pa = 23.6 nPa The mass is the same on earth and the moon, so we calculate the mass using the weight given on earth as: 1.4 (C) m = W/g = 250 N/9.81 m/s2 = 25.484 kg Hence, the weight on the moon is: W = mg = 25.484 × 1.6 = 40.77 N The shear stress is due to the component of the force acting tangential to the area: 1.5 (C) Fshear = F sin θ = 4200sin 30D = 2100 N F 2100 N τ = shear = = 84 ×103 Pa or 84 kPa 2 −4 A 250 ×10 m 1.6 (B) – 53.6°C Using Eqn. (1.5.3): 1.7 (D) ρ water = 1000 − (T − 4) 2 (80 − 4) 2 = 1000 − = 968 kg/m3 180 180 The shear stress is given by: τ = μ 1.8 (A) du dr We determine du/dr from the given expression for u as: du d ⎡ 10 (1 − 2500r2 ) ⎤⎦ = −50, 000r = dr dr ⎣ At the wall r = 2 cm = 0.02 m. Substituting r in the above equation we get: 1 © 2012 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 1 / Basic Considerations du = 50, 000r = 1000 1/s dr The density of water at 20°C is 10−3 N ⋅ s/m2 Now substitute in the equation for shear stress to get τ =μ du = 10−3 N ⋅ s/m 2 × 1000 1/s = 1 N/m 2 = 1 Pa dr Using Eqn. (1.5.16), β = 0 (for clean glass tube), and σ = 0.0736 N/m for water (Table B.1 in Appendix B) we write: 1.9 (D) h= 4σ cosβ 4 × 0.0736 N/m ×1 = = 3 m or 300 cm ρ gD 1000 kg/m3 × 9.81 m/s2 ×10 ×10−6 m where we used N = kg×m/s2 1.10 (C) 1.11 (C) Density Assume propane (C3H8) behaves as an ideal gas. First, determine the gas R 8.314 kJ/kmol ⋅ K = 0.1885 kJ/kg constant for propane using R = u = M 44.1 kg/kmol pV 800 kN/m 2 × 4 m3 then, m = = = 59.99 kg ≈ 60 kg RT 0.1885 kJ/(kg ⋅ K) × (10 + 273) 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 ΔEice = ΔEwater mice × 320 kJ/kg = mwater × cwater ΔT The mass of ice is calculate using mice = ρ V = 5 cubes × (1000 kg/m3 ) × ( 40 × 10−6 m3 /cube ) = 0.2 kg 1.12 (B) Where we assumed the density of ice to be equal to that of water, namely 1000 kg/m3. 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 mwater = ρ V = (1000 kg/m3 ) × 2 liters (10−3 m3 /liter ) = 2 kg Solving for the temperature change for water we get 0.2 kg × 320 kJ/kg = 2 kg × 4.18 kJ/kg ⋅ K × ΔT ⇒ ΔT = 7.66D C 2 © 2012 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 1 / Basic Considerations 1.13 (D) Since a dog’s whistle produces sound waves at a high frequency, the speed of sound is c = RT = 287 J/kg ⋅ K × 323 K = 304 m/s where we used J/kg = m2/s2. Dimensions, Units, and Physical Quantities a) density = 1.16 M FT 2 /L = = FT 2 /L4 3 3 L L c) power = F × velocity = F × L/T = FL/T M/T FT 2 /L = 2 = FT/L3 A LT e) mass flux = 1.18 ∴ [C] = N/kg = (kg⋅ m/s2)/kg = m/s2 b) N = [C] kg m m + c + km = f . Since all terms must have the same dimensions (units) 2 s s we require: kg 1.20 [c] = kg/s, [k] = kg/s2 = N ⋅ s 2 / m ⋅ s 2 = N / m, [f] = kg ⋅ m/s2 = N Note: we could express the units on c as [c] = kg/s = N ⋅ s2 /m ⋅ s = N ⋅ s/m 1.22 a) 1.25 × 108 N a) 20 cm/hr = 20 1.24 c) 6.7 × 108 Pa e) 5.2 × 10−2 m2 cm m hr × × = 5.556 × 10−5 m/s hr 100 cm 3600 s c) 500 hp = 500 hp × 745.7 W = 37, 285 W hp 2 ⎛ 100 cm ⎞ 10 2 × e) 2000 kN/cm = 2000 ⎟ = 2 × 10 N/m 2 ⎜ cm ⎝ m ⎠ 2 1.26 kN The mass is the same on the earth and the moon, so we calculate the mass, then calculate the weight on the moon: m = 27 kg ∴ Wmoon = (27 kg) × (1.63) = 44.01 N 3 © 2012 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 1 / Basic Considerations Pressure and Temperature Use the values from Table B.3 in the Appendix: 1.28 b) At an elevation of 1000 m the atmospheric pressure is 89.85 kPa. Hence, the absolute pressure is 52.3 + 89.85 = 142.2 kPa d) At an elevation of 10,000 m the atmospheric pressure is 26.49 kPa, and the absolute pressure is 52.3 + 26.49 = 78.8 kPa p = po e−gz/RT = 101 e−(9.81 × 4000)/[(287) × (15 + 273)] = 62.8 kPa 1.30 From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is % error = 62.8 − 61.6 × 100 = 1.95 % 61.6 Using Table B.3 and linear interpolation we write: 1.32 10,600 − 10,000 (223.6 − 216.7) = 221.32 K 12,000 − 10,000 5 or (221.32 − 273.15) = −51.83°C 9 T = 223.48 + The normal force due pressure is: Fn = (120, 000 N/m 2 ) × 0.2 × 10−4 m 2 = 2.4 N The tangential force due to shear stress is: Ft = 20 N/m 2 × 0.2 × 10−4 m 2 = 0.0004 N 1.34 The total force is F = Fn2 + Ft2 = 2.40 N ⎛ 0.0004 ⎞ The angle with respect to the normal direction is θ = tan−1 ⎜ ⎟ = 0.0095° ⎝ 2.4 ⎠ Density and Specific Weight Using Eq. 1.5.3 we have ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3 γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3 Using Table B.1 the density and specific weight at 70°C are 1.36 ρ = 977.8 kg/m3 γ = 977.8 × 9.81 = 9592.2 N/m3 % error for ρ = 976 − 978 × 100 = −0.20% 978 % error for γ = 9560 − 9592 × 100 = −0.33% 9592 4 © 2012 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 1 / Basic Considerations 1.38 b) m = γV g = 12,400 N/m3 × 500 × 10−6 m3 9.77 m/s 2 = 0.635 kg Viscosity Assume carbon dioxide is an ideal gas at the given conditions, then p 200 kN/m3 ρ= = = 2.915 kg/m3 RT ( 0.189 kJ/kg ⋅ K )( 90 + 273 K ) γ= 1.40 W mg = = ρ g = 2.915 kg/m3 × 9.81 m/s2 = 28.6 kg/m2 ⋅ s2 = 28.6 N/m3 V V From Fig. B.1 at 90°C, μ ≅ 2 ×10−5 N ⋅ s/m2 , so that the kinematic viscosity is ν= μ 2 ×10−5 N ⋅ s/m2 = = 6.861× 10−6 m2 /s 2.915 kg/m3 ρ The kinematic viscosity cannot be read from Fig. B.2 since the pressure is not at 100 kPa. The shear stress can be calculated using τ = μ du /dy . From the given velocity distribution, u = 120(0.05 y − y2 ) we get ⇒ du = 120(0.05 − 2 y ) dy From Table B.1 at 10°C for water, μ = 1.308 ×10−3 N ⋅ s/m2 1.42 So, at the lower plate where y = 0, we have ( ) du = 120(0.05 − 0) = 6 s−1 ⇒ τ = 1.308 × 10−3 × 6 = 7.848 ×10−3 N/m2 dy At the upper plate where y = 0.05 m, du dy = 120(0.05 − 2 × 0.05) = 6 s−1 ⇒ τ = 7.848 × 10−3 N/m2 y = 0.05 5 © 2012 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 1 / Basic Considerations The shear stress can be calculated using τ = μ du /dr . From the given velocity u = 16(1 − r2 ro2 ) ⇒ we get At the centerline, r = 0, so 1.44 At r = 0.25 cm, ⇒ du du = 16(−2 r ro2 ). Hence, = 32 r ro2 dr dr du = 0, and hence τ = 0. dr du 0.25 100 = 32 r ro2 = 32 = 3200 s−1 , ⇒ 2 dr ( 0.5 100 ) τ = 3200 × μ = 3200 (1 × 10−3 ) = 3.2 N/m2 At the wall, r = 0.5 cm, ⇒ du 0.5 100 = 32 r ro2 = 32 = 6400 s−1 , ⇒ 2 dr ( 0.5 100 ) τ = 6400 × μ = 6400 (1 × 10−3 ) = 6.4 N/m2 Use Eq.1.5.8 to calculate the torque, T = 2π R3ω Lμ h where h = (2.6 − 2.54) 2 = 0.03 cm = 0.03 × 10−2 m The angular velocity ω = 1.46 2π × 2000 rpm = 209.4 rad/s 60 The viscosity of SAE-30 oil at 21°C is μ = 0.2884 Ns/m2 (Figure B.2) T= 2π × (1.27 ×10−2 m)3 × 209.4 rad/s ×1.2 m × 0.2884 Ns/m2 (0.03 ×10−2 )m = 3.10 N ⋅ m power = Tω = 3.1× 209.4 = 650 W = 0.65 kW Assume a linear velocity in the fluid between the rotating disk and solid surface. The velocity of the fluid at the rotating disk is V = rω , and at the solid surface du rω V = 0. So, = , where h is the spacing between the disk and solid surface, dy h and ω = 2π × 400 60 = 41.9 rad/s . The torque needed to rotate the disk is 1.48 T = shear force × moment arm Due to the area element shown, dT = dF × r = τdA × r τ dr r where τ = shear stress in the fluid at the rotating disk and du rω dA = 2π rdr ⇒ dT = μ × 2π rdr × r = μ × 2π r2 dr dy h 6 © 2012 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 1 / Basic Considerations 2πμω 3 πμω 4 r dr = R h 2h 0 R T =∫ The viscosity of water at 16°C is μ = 1.12 ×10−3 Ns/m2 4 ⎛ 0.15 ⎞ π 1.12 × 10 Ns/m ( 41.9 rad/s ) ⎜ ⎟ π μω 4 ⎝ 2 ⎠ = 1.16 × 10−3 N ⋅ m R = ⇒T = 2h 2 × 2 × 10−3 m ( −3 2 ) ( If τ = μ AeCy 1.50 ) du = constant, and μ = AeB/T = AeBy/K = AeCy, then dy du du = constant. ∴ = De−Cy, where D is a constant. dy dy Multiply by dy and integrate to get the velocity profile y u = ∫ De−Cy dy = − 0 ( ) D −Cy y = E eCy − 1 e 0 C where A, B, C, D, E, and K are constants. Compressibility 1.54 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 = 10°C) and using Table B.1 we find B = 211×107 Pa , and ρ = 999.7 kg/m3 ⇒ c = 211×107 N/m2 = 1453 m/s 999.7 kg/m3 The distance across the lake is, L = cΔt = 1453 × 0.62 = 901 m b) Using Table B.2 and T = 37°C we find B = 226 ×107 N/m2 , ρ = 993 kg/m3 1.56 The speed of sound in water is calculated using: c= B ρ ⇒c= ( 226 ×107 ) / 993 = 1508 m/s 7 © 2012 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 1 / Basic Considerations Surface Tension 2σ R For a spherical droplet the pressure is given by p = Using Table B.1 at 15°C the surface tension σ = 7.41×10−2 N/m 1.58 ⇒ p= 2σ 2 × 0.0741 N/m = = 29.6 kPa 5 × 10−6 m R For bubbles: ⇒ p = 4σ 2 × 0.0741 N/m = = 59.3 kPa R 5 × 10−6 m For a spherical droplet the net force due to the pressure difference Δp between the inside and outside of the droplet is balanced by the surface tension force, which is expressed as: 1.60 Δp = 2σ 2 × 0.025 N/m ⇒ pinside − poutside = = 10 kPa 5 × 10−6 m R Hence, pinside = poutside + 10 kPa = 8000 kPa + 10 kPa = 8010 kPa In order to achieve this high pressure in the droplet, diesel fuel is usually pumped to a pressure of about 2000 bar before it is injected into the engine. See Example 1.4: 1.62 4 × ( 0.47 N/m ) cos130D 4σ cosβ = h= ρ gD 13.6 × 1000 kg/m 3 × 9.81 m/s 2 × 2 × 10−2 m ( ) ( ) = −4.53 × 10−4 m = −0.453 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 = 2σ L or ρ gV = 2σ L 1.64 ⎛ π d2 ⎞ L⎟ The volume of the needle is V = ⎜ ⎝ 4 ⎠ σL ⎛ π d2 ⎞ ⇒⎜ L ⎟ ρ g = 2σ L ⎝ 4 ⎠ σL needle W 8σ ∴d = πρg 8 © 2012 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 1 / Basic Considerations There is a surface tension force on the outside and on the inside of the ring. Each surface tension force = σ × π D. 1.66 Neglecting the weight of the ring, the free-body diagram of the ring shows that F = 2σπ D F D Vapor Pressure To determine the temperature, we can determine the absolute pressure and then use property tables for water. The absolute pressure is p = −80 + 92 = 12 kPa. 1.68 From Table B.1, at 50°C water has a vapor pressure of 12.3 kPa; so T = 50°C is a maximum temperature. The water would “boil” above this temperature. 1.70 At 40°C the vapor pressure from Table B.1 is 7.38 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 1.72 atmospheric pressure to be 100 kPa, we have 10,000 kPa + 100 kPa = 600x ∴x = 16.83 km. Ideal Gas Assume air is an ideal gas and calculate the density inside and outside the house using Tin = 15°C, pin = 101.3 kPa, Tout = −25°C, and pout = 85 kPa p 101.3 kN/m3 ρin = = = 1.226 kg/m3 RT 0.287 kJ/kg ⋅ K × (15 + 273 K) 1.74 ρout = 85 = 1.19 kg/m3 0.287 × 248 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 © 2012 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 1 / Basic Considerations The weight can be calculated using, W = γ V = ρ g V 1.76 Assume air in the room is at 20°C and 100 kPa and is an ideal gas: p 100 ⇒ρ= = = 1.189 kg/m3 RT 0.287 × 293 ⇒ W = γ V = ρ g V = 1.189 kg/m3 × 9.81 m/s 2 × (10 × 20 × 4 m3 ) = 9333 N The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have 100,000 N/m2 ×1 m2 = m × 9.81 m/s2 This gives the mass of the air m = 10,194 kg Since air is an ideal gas we can write pV = mRT or 1.78 V = mRT 10,194 kg × 0.287 kJ/kg ⋅ K × (15+273 K ) = = 8426 m3 p 100 kPa Assuming a spherical volume ⇒ V = π d 3 6 , which gives 13 ⎛ 6 × 8426 m3 ⎞ d =⎜ ⎟ = 25.2 m π ⎝ ⎠ 10 © 2012 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 1 / Basic Considerations First Law The fist law of the thermodynamics is applied to the mass: Q1− 2 − W1− 2 = ΔPE + ΔKE + ΔU In this case, since there is no change in potential and internal energy, i.e., ΔPE = 0, and Δ U = 0, and there is no heat transfer to or from the system, Q1− 2 = 0. The above equation simplifies to: −W1−2 = ΔKE 2 where W1− 2 = ∫ Fdl 1 a) F = 200 N ⇒ W1−2 = −200 N ×10 m = −2000 N ⋅ m or J Note that the work is negative in this case since it is done on the system. Substituting in the first law equation we get: 1.80 2000 N ⋅ m = 1 2 × 2000 N ⋅ m m (V22 − V12 ) ⇒ V2 = (10 m/s)2 + = 19.15 m/s 2 15 kg 10 b) F = 20 s ⇒ W1− 2 = − ∫ 20 sds = − 10 s2 10 0 = −1000 N ⋅ m 0 1000 N ⋅ m = 1 2 × 1000 N ⋅ m m (V22 − V12 ) ⇒ V2 = (10 m/s)2 + = 15.28 m/s 2 15 kg 10 c) F = 200 cos (π s 20 ) ⇒ W1− 2 = − ∫ 200 cos (π s 20 ) ds 0 ⇒ W1− 2 = −200 × ( 20 π ) sin (10π 20 ) = − 4000 π N ⋅ m 1 2 × 4000 π N ⋅ m 4000 π N ⋅ m = m (V22 − V12 ) ⇒ V2 = (10 m/s)2 + = 16.42 m/s 2 15 kg Applying the first law to the system (auto + water) we write: Q1− 2 − W1− 2 = ΔPE + ΔKE + ΔU 1.82 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, ΔPE = 0 . The first law equation reduces to: 1 mV12 = ΔU H2O = [ mcΔT ]H O 2 2 where, mH2O = ρ V = 1000 kg/m3 × 2000 cm3 ×10−6 m3 cm3 = 2 kg 11 © 2012 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 1 / Basic Considerations For water c = 4180 J/kg ⋅ K. Substituting in the above equation 2 1 ⎛ 100 ×1000 ⎞ m/s ⎟ = 2 kg × 4180 J/kg ⋅ K × ΔT ×1500 kg × ⎜ 2 ⎝ 3600 ⎠ ⇒ ΔT = 69.2°C 2 For a closed system the work is W1− 2 = ∫ pdV 1 For air (which is an ideal gas), pV = mRT , and p = 2 2 1 1 W1− 2 = ∫ pdV = ∫ mRT V 2 mRT dV V dV = mRT ∫ = mRT ln 2 V V V1 1 Since T = constant, then for this process p1V1 = p2 V2 or 1.84 V2 p1 = V1 p2 Substituting in the expression for work we get: W1−2 = mRT W1−2 = 2 kg × ( 287 J kg ⋅ K ) × 294 K ln p2 p1 1 = −116,980 J = −116.98 kJ 2 The 1st law states that Q1− 2 − W1− 2 = mΔu = mcv ΔT = 0 since ΔT = 0. ∴ Q = W = −116.98 kJ 2 For a closed system the work is W1− 2 = ∫ pdV 1 If p = constant, then W1−2 = p ( V2 − V1 ) 1.86 Since air is an ideal gas, then p1 V1 = mRT1 , and p2 V2 = mRT2 W1− 2 = mR (T2 − T1 ) = mR ( 2T1 − T1 ) = mRT1 = 2 kg × 0.287 kJ/kg ⋅ K × 423 K = 243 kJ 12 © 2012 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 1 / Basic Considerations Isentropic Flow Since the process is adiabatic, we assume an isentropic process to estimate the maximum final pressure: ⎛T ⎞ p2 = p1 ⎜ 2 ⎟ ⎝ T1 ⎠ 1.88 k / k −1 1.4/0.4 ⎛ 423 ⎞ = (150 + 100) ⎜ ⎟ ⎝ 293 ⎠ = 904 kPa abs or 804 kPa gage. Note: We assumed patm = 100 kPa since it was not given. Also, a measured pressure is a gage pressure. Speed of Sound b) c = kRT = 1.4 × 188.9 × 293 = 266.9 m/s 1.90 d) c = kRT = 1.4 × 4124 × 293 = 1301 m/s Note: We must use the units on R to be J/kg.K in the above equations. 1.92 b) The sound will travel at the speed of sound c = kRT = 1.4 × 287 × 293 = 343 m/s The distance is calculated using L = cΔt = 343× 8.32 = 2854 m 13 © 2012 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 1 / Basic Considerations 14 © 2012 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 2 / Fluid Statics CHAPTER 2 Fluid Statics FE-type Exam Review Problems: Problems 2-1 to 2-9 The pressure can be calculated using: p = γ Hg h were h is the height of mercury. 2.1 (C) p = γ Hg h = (13.6 × 9810 N/m3 ) × (28.5 × 0.0254) = 96,600 Pa = 96.6 kPa Since the pressure varies in a vertical direction, then: 2.2 (D) 2.3 (C) p = p0 − ρ gh = 84, 000 Pa − 1.00 kg/m3 × 9.81 m/s 2 × 4000 m = 44.76 kPa pw = patm + γ m hm − γ water hw = 0 + 30,000 × 0.3 − 9810 × 0.1 = 8020 Pa = 8.02 kPa This is the gage pressure since we used patm = 0. Initially, the pressure in the air is p Air ,1 = −γ H = −(13.6 × 9810) × 0.16 = −21,350 Pa. 2.4 (A) After the pressure is increased we have: p Air ,2 = −21, 350 + 10, 000 = −11, 350 = −13.6 × 9810 H 2 . ∴ H 2 = 0.0851 m = 8.51 cm The moment of force P with respect to the hinge, must balance the moment of the hydrostatic force F with respect to the hinge, that is: (2 × 5 ) × P = F × d 3 2.5 (B) 5 F = γ hA = 9.81 kN/m3 × 1 m × (2 × × 3 m 2 )] ∴ F = 98.1 kN 3 The location of F is at 3 3 ( 3.33) 12 I yp = y + = 1.67 + = 2.22 m ⇒ d = 3.33 − 2.22 = 1.11 m yA 1.67(3.33 × 3) 3.33 × P = 98.1× 1.11 ∴ P = 32.7 kN The gate opens when the center of pressure is at the hinge: 2.6 (A) 1.2 + h I 11.2 + h b(1.2 + h)3 /12 y= + 5. y p = y + = + = 5 + 1.2 2 Ay 2 (1.2 + h)b(11.2 + h) / 2 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 = 1.08 m. 15 © 2012 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 2 / Fluid Statics 2.7 (D) The hydrostatic force will pass through the center, and so FH will be balanced by the force in the hinge and the force P will be equal to FV. ∴ P = FV = 9.81× 4 × 1.2 × w + 9.81× (π × 1.22 / 4) × w = 300. ∴ w = 5.16 m. 2.8 (A) The weight is balanced by the buoyancy force which is given by FB = γ V where V is the displaced volume of fluid: 900 × 9.81 = 9810 × 0.01 × 15w. ∴ 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: p plug = pinitial + γ gasoline ΔZ where ΔZ = Δx 2.9 (A) ax g p plug = 20, 000 + 6660 × (1.2 × 5 ) = 24, 070 Pa 9.81 Fplug = p plug A = 24,070 × π × 0.022 = 30.25 N Pressure Since p = γ h, then h = p/γ 2.12 a) h = 250,000/9810 = 25.5 m c) h = 250,000/(13.6 × 9810) = 1.874 m 2.14 2.16 This requires that pwater = pHg ⇒ (γ h )water = (γ h )Hg b) 9810 × h = (13.6 × 9810) × 0.75 ∴h = 10.2 m Δp = −γΔz ⇒ Δp = – 1.27 × 9.81 (3000) = –37,370 Pa or –37.37 kPa From the given information the specific gravity is S = 1.0 + z/100 since S(0) = 1 and S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3. Using dp = −γ dz then, by integration we write: 2.18 10 ⎛ z2 2 ⎞ dp = 1000(1 + z /100) gdz = 1000 g z + ⎜ ⎟ ∫0 ∫0 100 ⎠ ⎝ p ⎛ 102 ⎞ p = 1000 × 9.81⎜10 + ⎟ = 103,000 Pa or 103 kPa 2 ×100 ⎠ ⎝ 16 © 2012 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 2 / Fluid Statics Note: we could have used an average S: Savg = 1.05, so that ρ avg = 1050 kg/m3 and so p = γ h ⇒ p = 1050 × 9.81 × 10 = 103,005 Pa From Eq. (2.4.8): p = patm [(T0 − α z ) / T0 ]g / α R = 100 [(288 − 0.0065 × 300)/288]9.81/0.0065 × 287 = 96.49 kPa Assuming constant density, then 2.20 100 ⎛ ⎞ p = patm − ρ gh = 100 − ⎜ ⎟ × 9.81× 300 /1000 = 96.44 kPa ⎝ 0.287 × 288 ⎠ 96.44 − 96.49 × 100 = −0.052% 96.49 % error = Since the error is small, the density variation can be ignored over heights of 300 m or less. dp dρ Eq. 1.5.11 gives B = ρ ρ gdh = B ρ dρ dρ or ρ 2 But, dp = ρgdh. Therefore T = g dh B Integrate, using ρ0 = 1064 kg/m3, and B = 2.1 × 109 N/m2 ρ ∫ 2 2.22 dρ h g dh = 2 B ρ 0 ∫ This gives ρ = ⎛ ⎛1 1 ⎞ ⎜ 9.81 ∴ −⎜ − ⎟=⎜ ⎝ ρ 1064 ⎠ ⎜ 2.1×109 N/m 2 ⎝ ( ) ⎞ ⎟ h = 4.67 ×10−9 h ⎟⎟ ⎠ 1 9.4 ×10 h −4 − 4.67 ×10−9 h h ∫ Now p = ρ gdh = 0 g ∫ 9.4 × 10−4 − 4.67 ×10−9 h dh 0 = g −4.67 × 10 −9 ln(9.4 × 10 −4 − 4.67 × 10−9 h) If we assume ρ = const: p = ρ gh = 1064 × 9.81× h = 10, 438 h b) For h = 1500 m: paccurate = 15,690 kPa and pestimate = 15,657 kPa. % error = 2.24 15, 657 − 15, 690 × 100 = −0.21% 15, 690 Use Eq. 2.4.8: p= 9.81 ×287 0.0065 101(288 − 0.0065 z / 288) 17 © 2012 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 2 / Fluid Statics a) for z = 3000 ∴p = 69.9 kPa. c) for z = 9000 ∴p = 30.6 kPa. Manometers 2.26 Referring to Fig. 2.7b, the pressure in the pipe is: p = γh = (13.6 × 9810) × 0.25 = 33,350 Pa or 33.35 kPa Referring to Fig. 2.7a, the pressure in the pipe is p = ρgh. If p = 2400 Pa, then 2400 = ρgh = ρ × 9.81 h or ρ = 2.28 a) ρ = 2400 = 680 kg/m3 9.81× 0.36 c) ρ = 2.30 2400 9.81 h ∴The fluid is gasoline 2400 = 999 kg/m3 9.81× 0.245 ∴The fluid is water See Fig. 2.7b: The pressure in the pipe is given by p1 = –γ1h + γ2H p1= –0.86 × 9810 × 0.125 + 13.6 × 9810 × 0.24 = 30,695 Pa or 30,96 kPa pwater − poil = γ oil × 0.3 + γ Hg × H − γ water × 0.2 2.32 40,000 – 16,000 = 920 × 9.81 × 0.3 + 13,600 × 9.81 × H −1000 × 9.81 × 0.2 Solving for H we get: H = 0.1743 m or 17.43 cm pwater = γ Hg ( 0.16 ) − γ water (0.02) − γ Hg (0.04) − γ water (0.02) 2.34 Using γ water = 9.81 kN/m3 and γ Hg = 13.6 × 9.81 kN/m3 pwater = 15.62 kPa 2.36 pwater – 9.81 × 0.12 – 0.68 × 9.81 × 0.1 + 0.86 × 9.81 × 0.1 = poil With pwater = 15 kPa, poil = 14 kPa pgage = pair + γ water × 4 where, pair = patm − γ Hg × H 2.38 ⇒ pgage = −γ Hg × H + γ water × 4 pgage = −13.6 × 9.81 × 0.16 + 9810 × 4 = 17.89 kPa Note: we subtracted atmospheric pressure since we need the gage pressure. 18 © 2012 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 2 / Fluid Statics 2.40 p + 9810 × 0.05 + 1.59 × 9810 × 0.07 – 0.8 × 9810 × 0.1 = 13.6 × 9810 × 0.05 ∴p = 5873 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 10 cm Water Δh A 9 cm 7 cm Δh Mercury 40 2.42 o From the previous problem we have: ( pB )1 = pA + γ water × 0.07 − γ HG × 0.09sin 40 − γ oil × 0.1 = 2.11 kPa (1) For the new condition: ( pB )2 = pA + γ water × ( 0.07 + Δh ) − γ HG × 0.11sin 40 − γ oil × ( 0.1 − Δh sin 40 ) (2) where Δh in this case is calculated from the new manometer reading as: Δh + Δh / sin 40 = 11 − 9 cm ⇒ Δh = 0.783 cm Subtracting Eq.(1) from Eq.(2) yields: ( pB )2 − ( pB )1 = γ water × ( Δh ) − γ HG × 0.02sin 40 − γ oil × ( −Δh sin 40 ) Substituting the value of Δh gives: ( pB )2 = 2.11 + ⎡⎣( 0.00783) − 13.6 × 0.02sin 40 − 0.87 × ( −0.00783sin 40 )⎤⎦ × 9.81 = 0.52 kPa a) Using Eq. (2.4.16): 2.44 p1 = γ1 ( z2 − z1 ) + γ 2 h + ( γ 3 − γ 2 ) H where h = z5 − z2 =17 − 16 = 1 cm 4000 = 9800(0.16 – 0.22) + 15,600(0.01) + (133,400 − 15,600)H ∴H = 0.0376 m or 3.76 cm 19 © 2012 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 2 / Fluid Statics From No. 2.30: poil = 14.0 kPa 2.46 From No. 2.36: poil = pwater – 9.81 × (0.12+Δz) – 0.68 × 9.81 × (0.1−2Δz) + 0.86 × 9.81× (0.1−Δz) ∴Δz = 0.0451 m or 4.51 cm Forces on Plane Areas The hydrostatic force is calculated using: F = γh A where, h = 10 m, and 2.48 A = π R2 = π ( 0.15 m ) hence, F = 9810 ×10 × π ( 0.15) = 6934 N 2 2 For saturated ground, the force on the bottom tending to lift the vault is: F = pc A = 9800 × 1.5 × (2 × 1) = 29,400 N 2.50 The weight of the vault is approximately: W = ρgV walls W = 2400 × 9.81 ⎡⎣ 2 ( 2 × 1.5 × 0.1) + 2 ( 2 × 1× 0.1) + 20 ( 0.8 × 1.3 × 0.1) ⎤⎦ = 28,400 N The vault will tend to rise out of the ground. b) Since the triangle is horizontal the force is due to the uniform pressure at a depth of 10 m. That is, F = pA, where p = γ h = 9.81×10 = 98.1 kN/m2 2.52 The area of the triangle is A = bh 2 = 2.828 × 2 / 2 = 2.828 m2 F = 98.1 × 2.828 = 277.4 kN a) F = γ hA = 9.81× 6 × π 22 = 739.7 kN yp = y + I π × 24 / 4 = 6+ = 6.167 m Ay 4π × 6 ∴(x, y)p = (0, –0.167) m c) F = 9.81 × (4 + 4/3) × 6 = 313.9 kN 2.54 y p = 5.333 + 4/2.5 = 1.5 x 3 × 4 3 / 36 = 5.50 m 5.333 × 6 ∴y = –1.5 y ∴x = 0.9375 ∴(x, y)p = (0.9375, –1.5) m 2.56 F = γh A = 9810 × 6 × 20 = 1.777 × 106 N, or 1177 kN 20 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. x Chapter 2 / Fluid Statics I 4 × 5 3 / 12 yp = y + = 7.5 + = 7.778 m 7.5 × 20 Ay ΣMHinge = 0 ⇒(10 – 7.778) 1177 = 5 P ∴P = 523 kN The vertical height of water is h = 1.22 − 0.42 = 1.1314 m The area of the gate can be split into two areas: A = A1 + A2 or A = 1.2 × 1.1314 + 0.4 × 1.1314 = 1.8102 m2 Use 2 forces: F1 = γ h1A1 = 9810 × 0.5657 × (1.2 ×1.1314) = 7534 N F2 = γ h2 A2 = 9810 × 2.60 1.1314 × (0.4 × 1.1314) = 1674 N 3 2 The location of F1 is at y p1 = ( 1.1314 ) = 0.754 m, and F2 is at 3 yp2 = y + I2 1.1314 0.4 × 1.13143 / 36 = 0.5657 m = + A2 y 3 0.4 × (1.1314 / 2) × (1.1314 / 3) ΣM hinge = 0 : 7534 × 1.1314 + 1674 × (1.1314 − 0.5657 ) − 1.1314 P = 0 3 ∴P = 3346 N The gate is about to open when the center of pressure is at the hinge. 2.62 b) y p = 1.2 + H = (2.0/2 + H ) + b × 23 /12 (1 + H )2b ∴H = 0.6667 m A free-body-diagram of the gate and block is sketched. T Sum forces on the block: ΣFy = 0 2.64 T 0 ∴W = T + FB where FB is the buoyancy force which is given by FB = γ ⎡⎣π R (3 − H ) ⎤⎦ 2 F stop FB yp FH W Take moments about the hinge: T × 3.5 = FH × (3 − yp ) Rx Ry where FH is the hydrostatic force acting on the gate. It is, using 21 © 2012 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 2 / Fluid Statics h = 1.5 m and A = 2 × 3 = 6 m2 FH = γ hA = ( 9.81 kN/m3 )(1.5 m × 6 m2 ) = 88.29 kN From the given information ( ) 2 33 /12 I yp = y + = 1.5 + =2m 1.5 × 6 yA ∴T = 88.29 × ( 3 − 2 ) 3.5 = 25.23 kN FB = W − T = 70 − 25.23 = 44.77 kN. H =3 m− 44.77 kN ( 9.81 kN/m ) π (1 m ) 2 3 ∴ γπ R2 ( 3 − H ) = 44.77 = 1.55 m 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 = W1 + W2 . The force F3 acting on the bottom of the dam can be divided into two forces: Fp1 due to the uniform pressure distribution and Fp2 due to the linear pressure distribution. b) 2.66 W1 = 2.4 × 9810 × 18.9 × 1.8 = 801 kN W2 = 2.4 × 9810 × 18.9 × 7.2 / 2 = 1602 kN W3 = 9810 × (18 × 6.86/2) = 605.7 kN F1 = 9810 × 9 × 18 = 1589 kN assume 1 m deep W3 F1 W F2 = 9810 × 1.5 × 3 = 44.14 kN F2 O F3 Fp1 = 9810 × 3 × 9 = 265 kN Fp2 = 9810 × 15 × 9 / 2 = 662.2 kN ΣM O : (1589)(6) + (265)(4.5) + (662.2)(6) − (801)(0.9) − (44.14)(3/3) − (1602)(4.2) − (605.7)(6.37) = 3392 kN ⋅ m > 0. ∴will tip 22 © 2012 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 2 / Fluid Statics Forces on Curved Surfaces 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: 2.68 ( FH ) water = ( γ hA) (FH )oil = ( γ hA ) oil water = 9.81× 2 × (4 ×10) = 784.8 kN = 0.86 × 9.81×1× 20 = 168.7 kN ΣM: 2 P = 784.8 × 2 − 168.7 × 2. ∴P = 616.1 kN A free-body-diagram of the volume of water in the vicinity of the surface is shown. F1 Force balances in the horizontal and vertical directions give: FH FH = F2 F2 FV = W + F1 . W FV . A xV B where FH and FV are the horizontal and vertical components of the force acting on the water by the surface AB. Hence FH = F2 = ( 9.81 kN/m3 ) ( 8 + 1)( 2 × 4 ) = 706.3 kN The line of action of FH is the same as that of F2. Its distance from the surface is 2.70 ( ) 4 23 12 I yp = y + = 9+ = 9.037 m 9×8 yA To find FV we find W and F1: π ⎡ ⎤ W = γ V = ( 9.81 kN/m3 ) ⎢2 × 2 − ( 22 ) ⎥ × 4 = 33.7 kN 4 ⎣ ⎦ F1 = 9.81 kN/m3 ( 8 × 2 × 4 ) = 628 kN ∴ FV = F1 + W = 33.7 + 628 = 662 kN To find the line of action of FV, we take moments at point A: FV × xV = F1 × d1 + W × d2 where d1 = 1 m, and d2 = 2R 2× 2 = = 1.553 m: 3(4 − π ) 3(4 − π ) 23 © 2012 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 2 / Fluid Statics ∴ xV = F1 × d1 + W × d2 628 ×1 + 33.7 ×1.553 = = 1.028 m 662 FV Finally, the forces FH and FV that act on the surface AB are equal and opposite to those calculated above. So, on the surface, FH acts to the right and FV acts downward. 2.72 Place the resultant FH + FV at the center. FV passes through the hinge. The moment of FH must equal the moment of P with respect to the hinge: 2 × (9.81 × 1 × 10) = 2.8 P ∴P = 70.1 kN The resultant FH + FV of the unknown liquid acts through the center of the circular arc. FV passes through the hinge. Thus, we use only ( FH ) . Assume 1 m wide: FH = γ x hA = γ x ( R 2 )( R ×1) = γ x R2 2 The horizontal force due to the water is Fw = γ w hA = γ w ( R 2 )( R ×1) = γ w R2 2 2.74 The weight of the gate is W = Sγ w V = 0.2γ w (π R2 4) ×1 Summing moments about the hinge: Fw ( R 3) +W ( 4R 3π ) = FH × R ⎛ R2 ⎞ R ⎛ πR 2 ⎞ 4 R ⎛ R 2 ⎞ ⎟⎟ × + ⎜⎜ 0.2 × 9810 ⎟× ⎟× R a) ⎜⎜ 9810 × = ⎜γ x 2 ⎠ 3 ⎝ 4 ⎟⎠ 3π ⎜⎝ 2 ⎟⎠ ⎝ ∴ γ x = 4580 N/m3 The pressure in the dome is: a) p = 60,000 – 9810 × 3 – 0.8 × 9810 × 2 = 14,870 Pa or 14.87 kPa The force is F = pAprojected = (π × 32) × 14.87 = 420.4 kN 2.76 b) From a free-body diagram of the dome filled with oil: W Fweld + W = pA Using the pressure from part (a): Fweld = 14,870 × π × 32 – (0.8 × 9810) × pA Fweld 1⎛4 ⎞ ⎜ π × 3 3 ⎟ = –23,400 or –23.4 kN ⎝ ⎠ 2 3 24 © 2012 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 2 / Fluid Statics Buoyancy Under static conditions the weight of the barge + load = weight of displaced water. 2.78 (a) 20,000 + 250,000 = 9810 × 3 (6d + d 2 /2). ∴d2 + 12d – 18.35 = 0 ∴d = 1.372 m 2.80 2.82 The weight of the cars will be balanced by the weight of displaced water: 15,000 × 60 = 9810(7.5 × 90 × Δd ) ∴Δd = 0.136 m or 13.6 cm T + FB = W (See Fig. 2.11 c) T = 40,000 – 1.59 × 9810 × 2 = 8804 N or 8.804 kN At the limit of lifting: FB = W+pA where p is the pressure acting on the plug. (b) Assume h > 4.5 + R and use the above equation with 2.86 R = 0.4 m and h = 4.92 FB = γwV = γw ×3×(π R2 − Asegment ) = 9810 × 3 × 0.2803 = 8249 N W + pA = 6670 N + 9810 × 4.92 × π ( 0.1) = 8234 N 2 Hence, the plug will lift for h > 4.92 m (a) When the hydrometer is completely submerged in water: ⎡ π × 0.0152 ⎤ π × 0.0052 W = γ w V ⇒ (0.01 + mHg )9.81 = 9810 ⎢ × 0.15 + × 0.12 ⎥ 4 4 ⎢⎣ ⎥⎦ 2.88 ∴mHg = 0.01886 kg When the hydrometer without the stem is submerged in a fluid: W = γ x V ⇒ (0.01 + 0.0189)9.81 = S x × 9810 × π ( 0.015 ) 4 2 × 0.15 ∴Sx = 1.089 25 © 2012 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 2 / Fluid Statics Stability With ends horizontal I o = π d 4 64 The displaced volume is V = W γ water ⇒ V = ( γ xπ d 2 h / 4 ) / 9810 = 8.01× 10−5 γ x d 3 since h = d The depth the cylinder will sink is depth = 2.90 V 8.01× 10−5 γ x d 3 = = 10.20 × 10−5 γ x d A π d2 / 4 The distance CG is CG = h − 10.2 × 10 −5 γ x d / 2 . Then 2 IO π d 4 / 64 d − +10.2 ×10−5γ x d / 2 > 0 GM = − CG = 3 −5 V 8.01×10 γ xd 2 This gives (divide by d and multiply by γx): 612.8 – 0.5 γx + 5.1 × 10−5 γ 2x > 0. Consequently γx > 8368 N/m3 As shown, y = 2.92 G= or 16 × 9 + 16 × 4 = 6.5 cm above the bottom edge. 16 + 16 4γ × 9.5 + 16γ × 8.5 + 16SAγ × 4 = 6.5 cm 0.5γ × 8 + 2γ × 8 + SAγ ×16 ∴130 + 104 SA = 174 + 64 SA ∴ SA = 1.1 The centroid C is 1.5 m below the water surface 2.94 γx < 1436 N/m3 Using Eq. 2.4.47: GM = ∴ CG = 1.5 m A × 83 / 12 − 1.5 = 1.777 − 1.5 = 0.277 > 0 A ×8×3 ∴The barge is stable Linearly Accelerating Containers (a) pmax = −1000 × 20 (0 − 4) – 1000 (9.81) (0 − 2) = 99,620 Pa 2.96 (c) pmax = −1000 × 18 (0 – 3.6) – 1000 (9.81 + 18) (0 – 1.8) = 114,860 Pa or 114.86 kPa 26 © 2012 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 2 / Fluid Statics Use Eq. 2.5.2: ⎛ 8a x ⎞ ⎟ b) 60,000 = –1000 ax (–8) – 1000 (9.81 + 10) ⎜ − 2.5 + ⎜ ⎟ 9 . 81 ⎝ ⎠ 2.98 60 = 8 ax + 49.52 – 19.81 8ax or ax – 1.31 = 1.574 19.81 ax ∴ax = 0.25, 4.8 m/s2 a x2 – 5.1 ax + 1.44 = 0 a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2 p(z) = –1000 × 10 (–7.626) – 1000 × 9.81(z) = 76,260 – 9810 z 2.5 ∴ FAB = ∫ (76,260 − 9810z)4dz = 640,000 N or 640 kN 0 2.100 b) The pressure on the bottom BC is p(x) = –1000 × 10 (x – 7.626) = 76,260 – 10,000 x 7.626 ∴ FBC = ∫ (76,260 − 10,000 x)4dx = 1.163 × 106 N or 1163 kN 0 Use Eq. 2.5.2 with position 1 at the open end: b) pA = –1000 × 10 (0.9 – 0) = –9000 Pa z 1 A pB = –1000 × 10 (0.9) – 1000 × 9.81(−0.6) = –3114 Pa 2.102 pC = –1000 × 9.81 × (–0.6) = 5886 Pa C B x e) pA = 1000 × 18 ( −1.5 × 0.625) = −16,875 Pa pB = 1000 × 18 ( −1.5 × 0.625 ) – 1000 × 9.81 ( −0.625) = –10,740 Pa pC = –1000 × 9.81 ( −0.625) = 6130 Pa Rotating Containers Use Eq. 2.6.4 with position 1 at the open end: 2.104 a) p A = z A 1 1 × 1000 × 102 (0 – 0.92) = –40,500 Pa 2 ω pB = –40,500 + 9810 × 0.6 = –34,600 Pa r pC = 9810 × 0.6 = 5886 Pa C B 27 © 2012 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 2 / Fluid Statics The air volume before and after is equal 1 ∴ π r02 h = π × 0.6 2 × 0.2 2 z 2 ∴ r02 h = 0.144 r0 h (a) Using Eq. 2.6.5: r02 × 5 2 / 2 = 9.81 h ∴h = 0.428 m 1 ∴pA = × 1000 × 52 × 0.62 – 9810 (–0.372) 2 1 A r = 8149 Pa (c) For ω = 10, part of the bottom is bared 1 2 1 2 π × 0.6 2 × 0.2 = π r02 h − π r12 h1 2.106 z Using Eq. 2.6.5: ω 2 r02 = h, 2g ∴ 0.144 = 2g ω ω 2 r12 2 2g h2 − h 2 − h12 = 2g ω2 h12 r0 = h1 h or A 0.144 × 10 2 2 × 9.81 r h1 1 Also, h – h1 = 0.8. 1.6 h – 0.64 = 0.7339. ∴h = 0.859 m, r1 = 0.108 m 1 ∴pA = × 1000 × 102 (0.62 – 0.1082) = 17,400 Pa 2 p (r ) = 1 ρω 2r 2 − ρ g[0 − (0.8 − h)] 2 p(r ) = 500ω 2r 2 + 9810(0.8 − h) p(r ) = 500ω (r − r ) 2 2.108 2 2 1 a) F = ∫ p2π rdr = 2π dA = 2πrdr if h < 0.8 dr if h > 0.8 0.6 ∫ (12,500r 3 + 3650r )dr = 6670 N 0 (We used h = 0.428 m) c) F = ∫ p2π rdr = 2π 0.6 ∫ (50,000(r 3 − 0.1082 r )dr = 9520 N −0.108 (We used r1 = 0.108 m) 28 © 2012 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 3 / Introduction to Fluids in Motion CHAPTER 3 Introduction to Fluids in Motion FE-type Exam Review Problems: Problems 3-1 to 3-9 nˆ ⋅ V = 0 3.1 (D) (nxˆi + ny ˆj) ⋅ (3ˆi − 4ˆj) = 0 or 3nx − 4ny = 0 Also nx2 + ny2 = 1 since n̂ is a unit vector. A simultaneous solution yields nx = 4 / 5 and n y = 3 / 5. (Each with a negative sign would also be OK) a= 3.2 (C) ∂V ∂V ∂V ∂V +u +v +w = 2xy (2 yˆi ) − y 2 (2xˆi − 2 yˆj) = −16ˆi + 8ˆi + 16ˆj ∂t ∂x ∂y ∂z ∴ a = (−8)2 + 162 = 17.89 m/s ax = 3.3 (D) = ∂u ∂u ∂u ∂u ∂u 10 ∂ ⎡ +u +v +w =u = 10(4 − x) −2 ⎤ 2 ⎣ ⎦ ∂t ∂x ∂y ∂z ∂x (4 − x) ∂x 10 10 1 10(−2)(−1)(4 − x)−3 = × 20 × = 6.25 m/s2 2 4 8 (4 − x) 3.4 (C) 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. 3.5 (B) V 2 p γ water h 9810 × 0.800 = = = ρ ρ air 2 1.23 3.6 (C) V12 p V22 + = 2g γ 2g V12 + 0.200 = 0.600 2g ∴V = 113 m/s ∴V = 2 × 9.81× 0.400 = 2.80 m/s The manometer reading h implies: 3.7 (B) V12 p1 V22 p2 2 + = + or V22 = (60 − 10.2) ρ ρ 2 2 1.13 ∴V2 = 9.39 m/s The temperature (the viscosity of the water) and the diameter of the pipe are not needed. 29 © 2012 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 3 / Introduction to Fluids in Motion 3.8 (A) V12 p1 V22 p2 + = + 2g γ 2g γ 3.9 (D) p1 = 800,000 V22 = 9810 2 × 9.81 ∴V2 = 40 m/s ρ 902 V22 − V12 ) = 302 − 152 ) = 304,400 Pa ( ( 2 2 Flow Fields a) u = 3.14 dx = 2t + 2 dt v= dy = 2t dt y x = t 2 + 2t + c1 y = t 2 + c2 (27, 21) = y+2 y ∴parabola x Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian: a) cos α = Several college students would be positioned at each intersection and quantities would be recorded as a function of time. V ⋅ ˆi 1+ 2 = = 0.832 V 32 + 22 V ⋅ nˆ = 0 3.18 (35, 25) 39.8o ∴ x 2 − 2 xy + y 2 = 4 y 3.16 streamlines t=5s ∴ nx = c) cos α = ∴α = 33.69D (3ˆi + 2ˆj) ⋅ (nxˆi + ny ˆj) = 0 2 3 1 , ny = − or nˆ = (2ˆi − 3ˆj) 13 13 13 5 V ⋅ ˆi = = 0.6202 V 52 + (−8)2 V ⋅ nˆ = 0 3 ny = − nx 3nx + 2ny = 0 ⎫ ⎪ 2 ⎬ ∴ 2 2 9 nx + ny = 1 ⎪⎭ nx2 + nx2 = 1 4 (5ˆi − 8ˆj) ⋅ (nx ˆi + ny ˆj) = 0 ∴α = −51.67D 8 nx = n y 5nx − 8ny = 0 ⎫ ⎪ 5 ⎬ ∴ 2 2 64 2 nx + ny = 1 ⎪⎭ ny + ny2 = 1 25 30 © 2012 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 3 / Introduction to Fluids in Motion ∴ny = 5 8 1 ˆ ˆ , nx = or nˆ = (8i + 5j) 89 89 89 ∂V ∂V ∂V ∂V +v +w + = 2 x(2ˆi ) + 2 y (2ˆj) = 4 xˆi + 4 yˆj = 8ˆi − 4ˆj ∂x ∂y ∂z ∂t 3.20 b) u 3.22 The vorticity ω = 2Ω. a) ω = −40ˆi c) ω = 12ˆi − 4kˆ 40 ⎞ 40 ⎞ sin θ ⎛ 40 ⎞ ⎛ ⎛ 80 ⎞ ⎛ a) ar = ⎜10 − 2 ⎟ cos θ ⎜ 3 ⎟ cos θ − ⎜10 + 2 ⎟ ⎜ 1 − ⎟ ( − sin θ ) r ⎠ r ⎠ r ⎝ r2 ⎠ ⎝ ⎝r ⎠ ⎝ 2 1⎛ 40 ⎞ − ⎜ 10 + 2 ⎟ sin 2 θ = (10 − 2.5)( −1)1.25( −1) = 9.375 m/s2 r⎝ r ⎠ 3.24 40 ⎞ 40 ⎞ sin θ ⎛ 40 ⎞ ⎛ ⎛ 80 ⎞ ⎛ aθ = ⎜ 10 − 2 ⎟ cos θ ⎜ 3 ⎟ sin θ + ⎜ 10 + 2 ⎟ ⎜ 10 + 2 ⎟ cos θ ⎝ ⎝r ⎠ ⎝ r ⎠ r ⎠ r ⎝ r ⎠ 1⎛ 1600 ⎞ − ⎜ 100 − 4 ⎟ sin θ cos θ = 0 r⎝ r ⎠ since sin(180°) = 0 aφ = 0 3.26 a= ∂V ∂V ∂V ∂V ∂u ˆ +u +v +w = i ∂t ∂x ∂y ∂z ∂t For steady flow ∂u / ∂t = 0 so that a = 0 3.28 3.30 b) u = 2(1 − 0.52 )(1 − e −t /10 ) = 1.875 m/s at t = ∞ −4 Dρ ∂ρ ∂ρ ∂ρ ∂ρ =u +v +w + = 10(−1.23 ×10−4 e −3000×10 ) Dt ∂x ∂y ∂z ∂t = −9.11× 10 −4 kg/m 3 ⋅ s 3.32 Dρ ∂ρ =u = 4 × (0.01) = 0.04 kg/m3⋅s Dt ∂x 31 © 2012 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 3 / Introduction to Fluids in Motion ∂u ⎫ + V ⋅∇u ⎪ ∂t ⎪ ∂v ∂V ⎪ + V ⋅∇v ⎬ ∴ a = + (V ⋅∇)V ay = ∂t ∂ t ⎪ ∂w ⎪ + V ⋅∇w ⎪ az = ∂t ⎭ ax = 3.34 2π kˆ = 7.272 × 10−5 kˆ rad/s 24 × 60 × 60 V = 5( −0.707ˆi − 0.707kˆ ) = −3.535ˆi − 3.535kˆ m/s Ω= A = 2Ω × V + Ω × ( Ω × r ) = 2 × 7.272 ×10−5 kˆ × (−3.535ˆi − 3.535kˆ ) + 7.272 ×10−5 kˆ × ⎡7.272 ×10−5 kˆ × (6.4 ×106 )(−0.707ˆi + 0.707kˆ ) ⎤ ⎣ ⎦ 3.36 = −52 ×10−5 ˆj + 0.0224ˆi m/s 2 Note: We have neglected the acceleration of the earth relative to the sun since it is quite small (it is d 2S /dt 2 ). The component ( −51.4 × 10 −5 ˆj) is the Coriolis acceleration and causes air motions to move c.w. or c.c.w. in the two hemispheres. Classification of Fluid Flows 3.38 3.42 a, c, e, f, h Steady: a) inviscid Unsteady: b, d, g c) inviscid e) viscous inside the boundary layers and separated regions. VL 3.46 Re = 3.48 a) Re = ν = VD ν 0.18 × 0.75 = 9640 1.4 ×10−5 = 1.2 × 0.01 = 795 1.51× 10−5 g) viscous. ∴Turbulent Always laminar Assume the flow is parallel to the leaf. Then 3 × 105 = VxT /ν 3.50 ∴ xT = 3 × 105ν / V = 3.5 × 105 × 1.4 × 10−4 / 6 = 8.17 m The flow is expected to be laminar 32 © 2012 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 3 / Introduction to Fluids in Motion Dρ ∂ρ ∂ρ ∂ρ ∂ρ =u +v +w + =0 Dt ∂x ∂y ∂z ∂t 3.52 For a steady, plane flow ∂ρ /∂t = 0 and w = 0 ∂ρ ∂ρ +v =0 ∂x ∂y u Then Bernoulli’s Equation 3.54 V2 p = 2 ρ Use ρ = 1.1 kg/m3 a) v = 2 p / ρ = 2 × 2000 /1.1 = 60 m/s 3.56 3.58 V2 p + =0 2 ρ ∴V = V 2 p U∞2 p∞ + = + 2 ρ 2 ρ −2 p ρ = 2 × 2000 = 57.0 m/s 1.23 b) Let r = rc : d) Let θ = 90 D : pT = ρ 2 U ∞2 3 p 90 = − ρU ∞2 2 V 2 p U ∞2 p + = + ∞ 2 ρ 2 ρ a) p = ρ 2 ( ) U ∞2 − u 2 = ρ⎡ 20π ⎞ ⎛ ⎢102 − ⎜10 + ⎟ 2 ⎢⎣ 2π x ⎠ ⎝ 2⎤ ⎡ ⎛ 1 ⎞2 ⎤ ⎥ = 50 ρ ⎢1 − ⎜ 1 + ⎟ ⎥ ⎥⎦ ⎢⎣ ⎝ x ⎠ ⎥⎦ ⎛2 1 ⎞ = −50 ρ ⎜ + 2 ⎟ ⎝x x ⎠ 3.60 c) p = ρ 2 ( ) U ∞2 − u 2 = ρ⎡ 60π ⎞ ⎛ ⎢302 − ⎜ 30 + ⎟ 2 ⎢⎣ 2π x ⎠ ⎝ 2⎤ ⎡ ⎛ 1 ⎞2 ⎤ ⎥ = 450 ρ ⎢1 − ⎜1 + ⎟ ⎥ ⎥⎦ ⎢⎣ ⎝ x ⎠ ⎥⎦ ⎛2 1 ⎞ = −450 ρ ⎜ + 2 ⎟ ⎝x x ⎠ 33 © 2012 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 3 / Introduction to Fluids in Motion Assume the velocity in the plenum is zero. Then 3.62 V12 p1 V22 p2 2 + = + or V22 = (60 − 10.2) 2 2 1.13 ρ ρ ∴V2 = 9.39 m/s We found ρ = 113 . kg / m 3 in Table B.2. Bernoulli from the stream to the pitot probe: V2 +p pT = ρ 2 Manometer: pT + γ H − γ Hg H − γ h = p − γ h 3.64 Then, ρ V2 + p + γ H − γ Hg H = p 2 a) V 2 = (13.6 − 1)9800 (2 × 0.04) 1000 b) V 2 = (13.6 − 1)9800 (2 × 0.1) 1000 ∴V 2 = γ Hg − γ ρ (2 H ) ∴ V = 3.14 m/s ∴ V = 4.97 m/s The pressure at 90° from Problem 3.58 is p90 = −3 ρU ∞2 / 2. The pressure at the 3.66 stagnation point is pT = ρU ∞2 / 2. The manometer provides: pT − γH = p90 1 3 × 1.204U ∞2 − 9800 × 0.04 = − × 1.204U ∞2 . 2 2 3.68 Bernoulli: V22 p V2 p + 2 = 1 + 1 2g γ 2g γ Manometer: p1 + γ z + γ Hg H − γ H − γ z = ∴ U ∞ = 12.76 m/s V22 γ + p2 2g Substitute Bernoulli’s into the manometer equation: V2 p1 + γ Hg − γ H = 1 γ + p1 2g ( ) b) Use H = 0.05 m: V12 × 9800 = (13.6 − 1)9800 × 0.05 2 × 9.81 ∴ V1 = 3.516 m/s Substitute into Bernoulli: p1 = V22 − V12 202 − 3.5162 γ= × 9800 = 193,600 Pa 2g 2 × 9.81 34 © 2012 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 3 / Introduction to Fluids in Motion Write Bernoulli’s equation between points 1 and 2 along the center streamline: p1 + ρV12 2 + γ z1 = p2 + ρV22 2 + γ z2 Since the flow is horizontal, z1 = z2 and Bernoulli’s equation becomes 3.70 p1 + 1000 × 0.52 1.1252 = p2 + 1000 × 2 2 From fluid statics, the pressure at 1 is p1 = γ h = 9810 × 0.25 = 2452 Pa and at 2, using p2 = γH, Bernoulli’s equation predicts 2452 + 1000 × 0.52 1.1252 = 9810H + 1000 × 2 2 ∴ H = 0.1982 m or 19.82 cm Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where p1 = 0 and V 1 = 0. Bernoulli’s equation gives 0= 3.72 3.74 V22 p2 + 2 ρ 1 ∴ p2 = − ρ V22 2 a) ρ = p 90 1 = = 1.239 kg/m3 ∴ p2 = − × 1.239 × 1002 = −6195 Pa RT 0.287 × 253 2 c) ρ = p 92 1 = = 1.094 kg/m3 ∴ p2 = − × 1.094 × 1002 = −5470 Pa 2 RT 0.287 × 293 Bernoulli across nozzle: V12 p1 V22 p2 + = + 2 ρ 2 ρ Bernoulli to max. height: p V12 p1 V22 + + h1 = + 2 + h2 2g γ 2g γ ∴V2 = 2 p1 / ρ ∴ h2 = p1 / γ a) V2 = 2 p1 / ρ = 2 × 700,000 /1000 = 37.42 m/s h2 = p1 / γ = 700,000/9800 = 71.4 m b) V2 = 2 p1 / ρ = 2 ×1, 400,000 /1000 = 52.92 m/s h2 = p1 / γ = 1,400,000/9800 = 142.9 m 3.76 V12 p1 V22 p2 + = + 2 ρ 2 ρ p2 = −100,000 Pa, the lowest possible pressure. 35 © 2012 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 3 / Introduction to Fluids in Motion a) 600,000 V22 100,000 = − 1000 2 1000 ∴ V 2 = 37.4 m/s b) 300,000 V22 100,000 = − 1000 2 1000 ∴ V 2 = 28.3 m/s b) p1 = 3.78 d) p1 = ρ 902 2 V22 − V12 ) = 2 − 102 ) = − 43,300 Pa ( ( 2 2 ρ 1.23 2 V22 − V12 ) = 2 − 102 ) = −59.0 Pa ( ( 2 2 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and a point 1 in between the exit and the center of the tube at a radius r less than R: 3.80 V12 p1 V22 p2 + = + 2 ρ 2 ρ ∴ p1 = ρ V22 − V12 2 Since V 2 < V 1 , we see that p1 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). 3.82 3.84 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. stagnation 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. 36 © 2012 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 4-1 to 4-15 4.1 (B) 4.2 (D) m = ρ AV = p 200 AV = π × 0.042 × 70 = 0.837 kg/s RT 0.287 × 293 Refer to the circle of Problem 4.27: 4.3 (A) 4.4 (D) Q = AV = (π × 0.42 × 75.7 × 2 − 0.10 × 0.40 × sin 75.5 ) × 3 = 0.516 m3/s 360 WP V22 − V12 p2 − p1 = + γQ 2g γ ∴WP = 40 kW 4.5 (A) 0= V22 − V12 2g + p2 − p1 γ WP 1200 − 200 = γ × 0.040 γ and energy required = −1202 p 0= + 2 . ∴ p2 = 7, 200,000 Pa 2 × 9.8 9810 Manometer: γ H + p1 = ρ g 4.6 (C) Energy: K V22 + p2 2g 7.962 100,000 = 2 × 9.81 9810 or 9810 × 0.02 + p1 = ρ g 4.7 (B) K V 2 Δp = 2g γ 7.962 100,000 = 2 × 9.81 9810 V= V22 2g ∴ K = 3.15 Combine the equations: 9810 × 0.02 = 1.2 × hL = K 40 = 47.1 kW 0.85 V12 2 ∴V1 = 18.1 m/s Q 0.040 = = 7.96 m/s A π × 0.042 ∴ K = 3.15 37 © 2012 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) WP V22 − V12 Δp = + γQ 2g γ WP = QΔp = 0.040 × 400 = 16 kW 36.0 + 15 = 4.9 (D) η = 4.582 p 7.162 + B + 3.2 2 × 9.81 9810 2 × 9.81 16 = 18.0 kW 0.89 ∴ pB = 416,000 Pa In the above energy equation we used hL = K V= 4.10 (A) WP V2 Q 0.2 with V = = = 4.42 m/s 2g A π × 0.22 Q 0.1 = = 19.89 m/s A π × 0.042 HP = Energy—surface to entrance: ∴ HP = V22 p2 V2 + + z2 + K 2 2g γ 2g 19.892 180,000 19.892 + + 50 + 5.6 = 201.4 m 2 × 9.81 9810 2 × 9.81 ∴WP = γ QH P /η P = 9810 × 0.1× 201.4 / 0.75 = 263,000 W 4.11 (A) After the pressure is found, that pressure is multiplied by the area of the window. The pressure is relatively constant over the area. V12 p1 V22 p2 . + = + 2g γ 2g γ 4.12 (C) p1 = 9810 × (6.252 − 1) × 12.732 = 3, 085, 000 Pa 2 × 9.81 p1 A1 − F = ρ Q(V2 − V1 ). 3, 085, 000 × π × 0.052 − F = 1000 × 0.1× 12.73(6.25 − 1) ∴ F = 17,500 N 4.13 (D) 4.14 (A) 4.15 (A) − Fx = m(V2 x − V1x ) = 1000 × 0.01 × 0.2 × 50(50cos 60 − 50) = −2500 N − Fx = m (Vr 2 x − Vr1x ) = 1000 × π × 0.02 2 × 60 × (40 cos 45 − 40) = 884 N Power = Fx × VB = 884 × 20 = 17, 700 W Let the vehicle move to the right. The scoop then diverts the water to the right. Then, F = m(V2 x − V1x ) = 1000 × 0.05 × 2 × 60 × [60 − (−60)] = 720,000 N 38 © 2012 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 4.16 b) The energy transferred to or from the system must be zero: Q − W = 0 4.20 b) The conservation of mass. d) The energy equation. System-to-Control-Volume Transformation nˆ 1 = − 4.24 1 ˆ 1 ˆ i− j = −0.707(ˆi + ˆj) 2 2 nˆ 2 = 0.866ˆi − 0.5ˆj . nˆ 3 = −ˆj V1n = V1 ⋅ nˆ1 = 10ˆi ⋅ ⎡⎣−0.707(ˆi + ˆj)⎤⎦ = −7.07 m/s V2n = V2 ⋅ nˆ 2 = 10ˆi ⋅ (0.866ˆi − 0.5ˆj) = 8.66 m/s V = V ⋅ nˆ = 10ˆi ⋅ (−ˆj) = 0 3n 4.26 3 2 (B ⋅ nˆ ) A = 15(0.5ˆi + 0.866ˆj) ⋅ ˆj (10 × 12) = 15 × 0.866 × 120 = 1559 cm3 Volume = 15 sin 60 ×10 ×12 = 1559 cm3 Conservation of Mass Use Eq. 4.4.2 with m V representing the mass in the volume: 0= dmV dmV + ρ A2V2 − ρ A1V1 + ∫ ρ nˆ ⋅ VdA = dt dt c.s. 4.32 = Finally dmV = m − ρQ dt π× A1V1 = A2V2 4.34 dmV + ρQ − m dt m = ρ AV = 1000π Q = AV = π 32 10 4 32 104 32 104 × 18 = π × 62 10 4 V2 ∴V2 = 4.5 m/s ×18 = 51 kg/s × 18 = 0.051 m3 /s 39 © 2012 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 ρ1A1V1 = ρ2 A2V2 4.38 p1 500 = = 4.433 kg/m3 RT 0.287 × 393 1246 ρ2 = = 8.317 kg/m3 0.287 × 522 ρ1 = 4.433 π × 0.052 × 600 = 8.317 π × 0.052 V2 ∴V2 = 319.8 m/s m = ρ 1 A1V1 = 20.89 kg/s Q 1 = A1V1 = 4.712 m3 /s b) (2 × 1.5 + 1.5 × 1.5) 3 = π d 22 2 × 4 2 4.40 Q2 = 2.512 m3/s ∴d2 = 4.478 m a) Since the area is rectangular, V = 5 m/s. m = ρ AV = 1000 × 0.08 × 0.8 × 5 = 320 kg/s 4.42 c) V × 0.08 = 10 × 0.04 + 5 × 0.02 + 5 × 0.02 m = ρ AV = 1000 × 0.08 × 0.8 × 7.5 = 480 kg/s 4.44 m ρ = 0.32 m3/s ∴V = 7.5 m/s Q= m ρ = 0.48 m3 /s If dm/dt = 0, then ρ 1 A 1V 1 = ρ 2 A 2 V 2 + ρ 3 A 3 V 3 . In terms of m 2 and Q 3 this becomes, letting ρ1 = ρ2 = ρ3 1000 × π × 0.02 2 × 12 = m2 + 1000 × 0.01 min = mout + m 4.46 Q= ∴ m2 = 5.08 kg/s ⎡ 0.1 ⎤ ⎢⎣ 0 ⎥⎦ ρ × 0.2 × 2 × 10 = ⎢ ρ ∫ 10(20 y − 100 y 2 )2dy + ρ × 0.1 × 2 × 10 ⎥ + m Note: We see that at y = 0.1 m the velocity u(0.1) = 10 m/s. Thus, we integrate to y = 0.1, and between y = 0.1 and 0.2 the velocity u = 10: ⎡4 ⎤ 4ρ = ⎢ ρ + 2ρ ⎥ + m ⎣3 ⎦ m = ∫ ρ VdA = ∴ m = 0.6667 ρ = 0.82 kg/s 0 .1 ∫ 1165(1 − 1.42 y )(y − 5 y 2 )12 × 1.5dy 0 0 .1 4.48 = 20,970 ∫ (y − 6.42 y 2 + 7.1 y 3 ) dy = 63.7 kg/s 0 2 2 V = umax = × 0.6 = 0.4 m/s (See Prob. 4.43b) 3 3 40 © 2012 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= 1165 + 1000 = 1082 kg/m3 2 ∴ρVA = 1082 × 0.4 × (1.5 × 0.1) = 64.9 kg/s Thus, ρV A ≠ m since ρ = ρ(y) and V = V(y) so that ρV ≠ ρV 4.50 m3 of H 2O 4 m3 of air = 1.5 × (1.5 h) × 9000 × 5 2000 × π × 0.00153 3 3 s m of air ∴h = 0.565 m 4.52 min = m2 + m3 V1 = 20 m/s (see Prob. 4.43c) 20 × 1000π × 0.022 = 10 + 1000π × 0.022 × V3 ∴ V3 = 12.04 m/s The control surface is close to the interface at the instant shown. Ve ∴Vi = interface velocity 4.54 n̂ n̂ Vi ρe AeVe = ρi AV i i 1.5 × π × 0.152 × 300 = 8000 π × 12 × Vi 0.287 × 673 ∴Vi = 0.244 m/s For an incompressible flow (low speed air flow) 4.56 ∫ udA = A V 2 0.2 ∫ 20 y 1/5 2 A1 × 0.8dy = π × 0.152 V2 0 5 20 × 0.8 0.26/5 = π × 0.152 V2 6 ∴ V2 = 27.3 m/s Draw a control volume around the entire set-up: dmtissue + ρV2 A2 − ρV1 A1 dt ⎛ d2 − d2 ⎞ 2 = m tissue + ρπ ⎜⎜ 2 ⎟⎟ h2 − ρπ ( h1 tan φ ) h1 4 ⎝ ⎠ 0= 4.58 or mtissue ⎡ d 2 − d 22 ⎤ = ρπ ⎢ h2 + h12 h1 tan 2φ ⎥ 4 ⎢⎣ ⎥⎦ 41 © 2012 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 dm + ρ A2V2 − ρ AV 1 1 dt 0= 4.60 = m + 1000(π × 0.0032 × 0.02 − 10 × 10−6 / 60) m = 3.99 ×10−4 kg/s 0= 4.62 dm + ρ 3 Q 3 − ρ 1 A 1V 1 − m 2 where m = ρ Ah dt a) 0 = 1000π × 0.62 h + 1000 × 0.6 / 60 − 1000π × 0.022 × 10 − 10 ∴ h = 0.0111 m/s or 11.1 mm/s Choose the control volume to consist of the air volume inside the tank. The conservation of mass equation is 0= d ∫ ρ dV + CS∫ ρ V ⋅ n dA dt CV Since the volume of the tank is constant, and for no flow into the tank, the equation is 0=V 4.64 dρ + ρeVe Ae dt Assuming air behaves as an ideal gas, ρ = p . At the instant of interest, RT dρ 1 dp . Substituting in the conservation of mass equation, we get = dt RT dt ( ) 1.8 kg/m3 ( 200 m/s ) π ( 0.015 m ) ⎡ ⎤ ρeVe Ae dp kJ ( RT ) = − 0.287 =− × 298 K ⎥ ⎢ 3 dt V 1.5 m kg ⋅ K ⎣ ⎦ ∴ 2 dp = −14.5 kPa/s dt Energy Equation W = Tω + pAV + μ 4.66 du Abelt dy = 20 × 500 × 2π /60 + 400 × 0.4 × 0.5 × 10 + 1.81× 10−5 × 100 × 0.5 × 0.8 = 1047 + 800 + 0.000724 = 1847 W 42 © 2012 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 80% of the power is used to increase the pressure while 20% increases the internal energy (Q = 0 because of the insulation). Hence 4.68 mΔu = 0.2 W 1000 × 0.02 × 4.18ΔT = 0.2 × 500 4.70 − WT = −40 × 0.89 mg ∴ΔT = 0.836°C b) WT = 40 × 0.89 × (90,000/60) × 9.81 = 523,900 W V12 p1 V2 p + + z1 = 2 + 2 + z2 2g γ 2g γ 42 42 +2= + h2 2 × 9.81 19.62h22 4.72 2.82 = 0.82 + h2 h22 Continuity: 1 × 4 = h2 V2 This can be solved by trial-and-error: h2 = 2.7 m : 2.82 ? 2.81 h2 = 2.72 m : 2.82 ? 2.83 ∵h2 = 2.71 m h2 = 0.6 m : 2.82 ? 2.87 h2 = 0.62 : 2.82 ? 2.75 ∵h2 = 0.61 m Manometer: Position the datum at the top of the right mercury level. 9810 × 0.4 + 9810 z2 + p2 + 4.74 V22 × 1000 = (9810 × 13.6) × 0.4 + 9810 × 2 + p1 2 Divide by γ = 9810: 0.4 + z 2 + Energy: p2 γ + V22 p = 13.6 × 0.4 + 2 + 1 2g γ V12 p1 V2 p + + z1 = 2 + 2 + z2 2g γ 2g γ Subtract (1) from (2): With z1 = 2 m (2) V12 = 12.6 × 0.4 2g Q = 7 × 10 −3 = π × (0.025) 2V1 4.76 (1) ∴V1 = 9.94 m/s ∴V1 = 3.75 m/s Continuity: π × (0.025) 2V1 = π × (0.0375) 2V2 Energy: V12 p1 V22 p2 V2 + = + + 0.37 1 2g γ 2g γ 2g ∴V2 = 1.58 m/s 43 © 2012 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 ⎡ 3.57 2 1.582 ⎤ ∴ p2 = 414 × 103 + 1000 ⎢ 0.63 − ⎥ = 414.3 kPa 19.62 19.62 ⎥⎦ ⎢⎣ V1 = Q/A1 = 4.78 Energy: 0.08 π × 0.032 ∴V2 = 9V1 = 254.6 m/s = 28.29 m/s V12 p1 V22 p2 V2 + = + + 0.2 1 2g γ 2g γ 2g ⎡ 254.6 2 28.29 2 ⎤ 6 ∴ p1 = 9810 ⎢ − 0.8 ⎥ = 32.1× 10 Pa 2 × 9 . 81 2 × 9 . 81 ⎣ ⎦ a) Energy: V02 p0 V2 p + + z0 = 2 + 2 + z2 2g γ 2g γ ∴ V 2 = 2 gz 0 = 2 × 9.81 × 2.4 = 6.862 m/s Q = AV = 0.8 × 1 × 6.862 = 5.49 m3 /s 4.80 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: V22 ∴ z0 = +h 2g ∴ V2 = 2 g( z 0 − h) = 2 × 9.81 × 2 = 6.264 m/s ∴Q = 0.8 × 6.264 = 5.01 m3/s Note: z0 is measured from the channel bottom in the 2nd geometry. ∴z0 = H + h 4.82 V02 p0 V22 p2 + + z0 = + + z2 2g γ 2g γ 80,000 V22 +4= 9810 2 × 9.81 ∴V2 = 19.04 m/s b) Q = A2V2 = π × 0.092 × 19.04 = 0.485 m3/s Manometer: γ H + γ z + p1 = 13.6γ H + γ z + p2 Energy: p1 γ + Combine energy and manometer: V2 = p1 γ = 12.6 H + p2 γ V 12 p 2 V 22 = + . 2g γ 2g 4.84 Continuity: ∴ 12.6 H = V22 − V12 2g ⎛ d4 ⎞ ∴ V12 = 12.6 H × 2 g ⎜ 14 − 1⎟ ⎜d ⎟ ⎝ 2 ⎠ d12 V1 d 22 44 © 2012 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 1/2 d 2 π ⎛ 12.6 H × 2 g ⎞ ∴ Q = V1π 1 = ⎜ 4 4 ⎟ 4 4 ⎜⎝ d1 / d2 − 1 ⎟⎠ a) Energy from surface to outlet: 4.86 Energy from constriction to outlet: Continuity: V1 = 4V2 d12 ⎛ H = 12.35d12d22 ⎜ 4 ⎜ d − d4 ⎝ 1 2 V22 =H 2g ∴ V22 = 2 gH V12 p2 V22 + = + γ 2g γ 2g p1 With p1 = pv = 2450 Pa and p2 = 100,000 Pa 2450 16 100,000 1 + × 2 gH = + × 2 gH 9810 2 × 9.81 9810 2 × 9.81 Energy: surface to surface: z0 = z2 + hL 4.88 Continuity: V1 = 4V2 ∴ V 12 = 160 g 30 = Energy: surface to constriction: ∴z1 = −40.4 m 1/2 ⎞ ⎟⎟ ⎠ ∴H = 0.663 m ∴30 = 20 + 2 V22 2g ∴ V 22 = 10g 160 g (−94,000) + + z1 2g 9810 ∴H = 40.4 + 20 = 60.4 m Velocity at exit = Ve Velocity in constriction = V1 Velocity in pipe = V2 Ve2 Energy: surface to exit: =H 2g 4.90 Continuity across nozzle: V2 = ∴Ve2 = 2 gH D2 Ve Also, V1 = 4V2 d2 Energy: surface to constriction: H = a) 5 = V12 pv + 2g γ ⎞ −97,550 1 ⎛ D4 16 × × 2 g × 5 + ∴ D = 0.131 m ⎜⎜ ⎟ ⎟ 2g ⎝ 9810 0.24 ⎠ m = pAV = 1000 × π × (0.025) 2 × 36 = 70.65 kg/s 4.92 ⎡ 92 − 362 830 ×1000 ⎤ + WP = 70.65 × 9.81 ⎢ ⎥ / 0.85 =18,500 W or 18.5 kW = 24.8 hp 9810 ⎦ ⎣ 2 × 9.81 45 © 2012 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 ⎡ 0 − 10.2 2 −600,000 ⎤ −WT = 2 × 1000 × 9.81 ⎢ + ⎥ × 0.87 9810 ⎦⎥ ⎣⎢ 2 × 9.81 4.94 ∴WT = 1.304 × 106 W We used V2 = Q 2 = = 10.2 m/s A2 π × 0.252 ∴ ηT = 0.924 ⎡ V 2 − V12 p2 p1 ⎤ c a) Q − WS = mg ⎢ 2 + − + z2 − z1 + v (T2 − T1 ) ⎥ γ 2 γ1 g ⎢⎣ 2 g ⎥⎦ The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13 4.96 γ1 = p1g 85 × 9.81 = = 9.92 N/m3 RT1 0.287 × 293 γ2 = 600 × 9.81 20,500 = T2 0.287 T2 ⎡ 2002 ⎤ 600,000T2 85,000 716.5 (T2 − 293) ⎥ ∴ −(−1,500,000) = 5 × 9.81 ⎢ + − + 20,500 9.92 9.81 ⎣ 2 × 9.81 ⎦ ∴T2 = 572 K or 299°C Be careful of units. p2 = 600,000 Pa cv = 716.5 J/kg ⋅ K ⎡V 2 V2⎤ Energy: surface to exit: −WTηT = mg ⎢ 2 − 20 + 4.5 2 ⎥ 2 g ⎥⎦ ⎢⎣ 2 g 4.98 V2 = 15 = 13.26 m/s π × 0.6 2 mg = Qγ = 15 × 9810 = 147,150 N/s ⎡ 13.262 13.262 ⎤ −WT × 0.8 = 147,150 ⎢ − 20 + 4.5 ⎥ 2 × 9.81 ⎥⎦ ⎢⎣ 2 × 9.81 4.100 ∴WT = 5390 kW Choose a control volume that consists of the entire system and apply conservation of energy: V12 p2 V22 + z = HT + + +z +h HP + + γ 2g 1 γ 2g 2 L p1 46 © 2012 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 0.5 m Section (1) Section (2) Pump We recognize that HT = 0, V1 is negligible and hL = 210 V 2/2g where V = Q/A and V2 = Q / A2 . Rearranging we get: HP = p2 − p1 V22 + + z2 − z1 + hL γ 2g ( ) 6.3 ×10−6 m3 /s (0.321) 2 Q = 0.321 m/s ⇒ = 210 = 1.1 m V= = h L 2 × 9.81 A π 2.5 ×10−3 m 2 V2 = ( ) Q 6.3 × 10−6 m3 /s = = 12.53 m/s A2 π 4 × 10−4 m 2 ( ) Substituting the given values we get: 95 − 100 ) kN/m2 ( (12.53 m/s ) = + 0.5 m + 2 HP ( 6.660 kN/m3 2 9.81 m/s2 ) + 1.1 m = 8.85 m The power input to the pump is: ( )( ) WP = γ QH P /ηP = 6660 N/m3 6.3 ×10−6 m3/s ( 8.85 m ) / 0.75 = 0.5 W Energy: across the nozzle: ∴ 4.102 p1 V12 p2 V22 + = + γ 2g γ 2g 400,000 V12 6.252V12 + = 9810 2 × 9.81 2 × 9.81 V2 = 52 V1 = 6.25V1 22 ∴V1 = 4.58 m/s , VA = 7.16 m/s V2 = 28.6 m/s Energy: surface to exit: HP +15 = 28.62 4.582 7.162 +1.5 + 3.2 2× 9.81 2× 9.81 2× 9.81 ∴ HP = 36.8 m 47 © 2012 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 ∴WP = γ QH P 9810 × (π × 0.012 ) × 28.6 × 36.8 = = 3820 W ηP 0.85 Energy: surface to “A”: 15 = 7.162 p 7.162 + A + 3.2 2 × 9.81 9810 2 × 9.81 ∴ p A = 39,400 Pa Energy: surface to “B”: 4.582 pB 7.162 36.0 + 15 = + + 3.2 2 × 9.81 9810 2 × 9.81 ∴ pB = 416,000 Pa Depth on raised section = y 2 . Continuity: 3 × 3 = V2 y2 32 V2 + 3 = 2 + (0.4 + y2 ) 2g 2g Energy (see Eq. 4.5.21): 92 ∴ 3.059 = + y2 2 g y22 4.104 Trial-and-error: y23 − 3.059 y22 + 4.128 = 0 or y2 = 2.0 : y2 = 1.8 : − 0.11 ? 0 ⎫ ⎬ + 0.05 ? 0 ⎭ y2 = 2.1: y2 = 2.3 : − 0.1 ? 0 ⎫ ⎬ + 0.1 ? 0⎭ ∴ y2 = 1.85 m ∴ y2 = 2.22 m The depth that actually occurs depends on the downstream conditions. We cannot select a “correct” answer between the two. The average velocity at section 2 is also 8 m/s. The kinetic-energy-correction factor for a parabola is 2 (see Example 4.9). The energy equation is: V12 p1 V22 p2 + = α2 + + hL 2g γ 2g γ 4.106 82 150,000 82 110,000 + =2 + + hL 2 × 9.81 9810 2 × 9.81 9810 a) V = 1 ∫ VdA = A 4.108 α= = 1 AV 3 ∫V π × 0.012 3 dA = ⎛ r2 ⎞ 20 10 ⎜ 1 − 2π rdr = ⎜ 0.012 ⎟⎟ 0.012 ⎝ ⎠ 0.01 1 ∫ 0 π × 0.012 × 53 ∫ 0 3⎛ ⎡ 0.012 0.014 ⎤ − ⎢ ⎥ = 5 m/s 4 × 0.012 ⎦⎥ ⎣⎢ 2 3 r2 ⎞ 10 ⎜1 − 2π rdr ⎜ 0.012 ⎟⎟ ⎝ ⎠ 0.01 1 ∴ hL = 0.815 m 2000 ⎛ 0.012 3 × 0.014 3 × 0.016 0.018 ⎞ − + − ⎜ ⎟ = 2.00 0.012 × 5 3 ⎝ 2 4 × 0.012 6 × 0.014 8 × 0.016 ⎠ 48 © 2012 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 4.110 ⎛ V22 − V12 ⎞ Engine power = FD × V∞ + m ⎜ + u2 − u1 ⎟ ⎜ ⎟ 2 ⎝ ⎠ ⎡ V 2 − V12 ⎤ + cv (T2 − T1 ) ⎥ m f q f = FD V∞ + m ⎢ 2 2 ⎣⎢ ⎦⎥ 0 = α2 4.112 V22 p2 V2 p ν LV + + z2 − 1 − 1 − z1 + 32 2g γ 2g γ gD2 V2 10−6 × 180V 0=2 − 0.35 + 32 × 2 × 9.81 9.81× 0.022 V 2 + 14.4V − 3.434 = 0 ∴ V = 0.235 m/s and Q = 7.37 × 10−5 m3/s Energy from surface to surface: HP = 4.114 V22 p2 V2 p V2 + + z2 − 1 − 1 − z1 + K 2g γ 2g γ 2g a) H P = 40 + 5 Q2 = 40 + 50.7 Q 2 π × 0.04 2 × 2 × 9.81 Try Q = 0.25 : H P = 43.2 (energy) H P = 58 (curve) Try Q = 0.30 : H P = 44.6 (energy) H P = 48 (curve) Solution: Q = 0.32 m3 /s Momentum Equation V12 p1 V22 p2 + = + 2g γ 2g γ 4.116 b) V2 = V12 400,000 16 V12 + = 2 × 9.81 9810 2 × 9.81 d2 V1 = 4 V1 (d / 2)2 ∴ V1 = 7.303 m/s 400,000π × 0.032 − F = 1000π × 0.032 × 7.303(4 × 7.303 − 7.303) ∴ F = 679 N 49 © 2012 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 V12 p1 V22 p2 + = + 2g γ 2g γ V0π × 0.012 = Ve × 0.006 × 0.15 ∴ Ve = 11.1 m/s ΣFx = m(V2 x − V1x ) a) V2 = 102 V1 = 1.562 V1 82 V12 400,000 2.441 V12 + = 2 × 9.81 9810 2 × 9.81 ∴ V1 = 23.56 m/s 4.118 ∴ p1 A1 − F = m(V2 − V1) 400,000π × 0.052 − F = 1000π × 0.052 × 23.56(0.562 × 23.56) c) V2 = 102 V12 400,000 39.06 V12 6 25 V = . V + = 1 1 2g 9810 2g 42 ∴ V1 = 4.585 m/s 400,000π × 0.052 − F = 1000π × 0.052 × 4.585(5.25 × 4.585) V12 p1 V22 p2 + = + 2g γ 2g γ V2 = 4 V1 b) V12 = 4.120 ∴ F = 692 N ∴ F = 2275 N 15 V12 p1 ∴ = 2g γ 2 × 9.81 × 400,000 = 53.33 15 × 9810 ∴V1 = 7.30 m/s, V2 = 29.2 m/s p1 A1 − Fx = m(V2 x − V1x ) ∴ Fx = 400,000π × 0.04 2 + 1000π × 0.04 2 × 7.32 = 2280 N Fy = m(V 2 y − V1y ) = 1000π × 0.042 × 7.3 × ( 29.2 ) = 1071 N π × 0.0252 × 4 = π (0.0252 − 0.022 )V2 AV 1 1 = A2V2 ∴ V2 = 11.11 m/s 4.122 p1 γ + V12 p2 V22 = + 2g γ 2g ⎛ 11.112 − 42 ⎞ p1 = 9810 ⎜ = 53, 700 Pa ⎜ 2 × 9.81 ⎟⎟ ⎝ ⎠ p1A1 F p1 A1 − F = m(V2 − V1 ) ∴ F = 53,700π × 0.0252 − 1000π × 0.0252 × 4(11.11 − 4) = 49.6 N 4.124 Continuity: 6 V1 = 0.2 V2 ∴ V2 = 30 V1 Energy (along bottom streamline): 50 © 2012 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 V12 p1 V2 p + + z1 = 2 + 2 + z 2 2g γ 2g γ V22/900 2 × 9.81 +6= V22 F F1 2 × 9.81 + 0.2 F2 ∴ V2 = 10.67, V1 = 0.36 m/s Momentum: F1 − F2 − F = m(V 2 − V1 ) 9810 × 3(6 × 4) − 9810 × 0.1(0.2 × 4) − F = 1000 × (0.2 × 4) × 10.67(10.67 − 0.36) ∴ F = 618,000 N Continuity: (F acts to the right on the gate) V2 y2 = V1y1 = 4V2 y1 ∴ y2 = 4 y1 1/2 ⎤ Use the result of Example 4.12: 4.126 ⎛ ⎞ 1⎡ 8 y2 = ⎢ − y1 + ⎜ y12 + y1V12 ⎟ 2⎢ g ⎝ ⎠ ⎣ ⎥ ⎥ ⎦ a) y2 = 4 × 0.8 = 3.2 m 1⎡ 8 ⎛ ⎞ 3.2 = ⎢ −0.8 + ⎜ 0.82 + × 0.8 × V12 ⎟ 2 ⎣⎢ 9.81 ⎝ ⎠ 1/ 2 ⎤ ∴ V1 = 8.86 m/s ⎥ ⎦⎥ Refer to Example 4.12: γ 4.128 ⎛ y1 6⎞ y1w − γ × 1 × 2w = ρ × 6w ⎜ 3 − ⎟ 2 y1 ⎠ ⎝ ⎛ y −2⎞ γ ∴ ( y12 − 4) = 18 ρ ⎜ 1 ⎟ 2 ⎝ y1 ⎠ or (y1 + 2)y1 = V1A1 = 2V2 A2 4.130 (V1y1 = 2 × 3) 36 = 3.67 ∴ y1 = 0.77 m, V1 = 7.8 m/s 9.81 V2 = 15 π × 0.052 = 30 m/s 2π × 0.0252 p1 γ + V12 p2 V22 = + 2g γ 2g 302 − 152 ∴ p1 = 9810 = 337,500 Pa 2 × 9.81 ΣFx = m ( V2 x − V1x ) p1A1 − F = m(−V1) ∴ F = p1A1 + mV1 = 337,500π × 0.052 + 1000π × 0.052 × 152 = 4420 N 51 © 2012 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 a) ∑ Fx = m(V2x − V1x ), V1 = 4.132 − F = −mV1 m 300 = = 38.2 m/s ρ A1 1000 × π × 0.052 ∴ F = 300 × 38.2 = 11, 460 N V2 F V1 b) − F = mr (V1 − VB )( cos α − 1) ∴ F = 300 × 28.2 (38.2 − 10) = 6250 N 38.2 b) − F = mr (V1 − VB )( cos α − 1) −700 = 1000π × 0.042 (V1 − 8)2 (0.866 − 1) V1 = 40.24 m/s 4.134 2 ∴ m = ρ AV 1 1 = 1000π × 0.04 × 40.24 = 202 kg/s VB = Rω = 0.5 × 3.0 = 15 m/s 4.136 −Rx = m(V1 − VB )(cosα − 1) = 1000π × 0.0252 × 40 × 25(0.5 − 1) ∴Rx = 982 N ∴W = 10RxVB = 10 × 982 × 15 = 147,300 W −Fx = m(V1 − VB )(cos120° − 1) = 4π × 0.022 × (400 − 180)2 (−0.5 − 1) 4.138 ∴Rx = 365 N VB = 1.2 × 150 = 180 m/s W = 15 × 365 × 180 = 986,000 W The y-component force does no work. b) 100sin 30 = Vr1 sin α1 ⎫⎪ ⎬ ∴α1 = 47 , Vr1 = Vr 2 = 68.35 m/s 100 cos 30 − 40 = Vr1 cos α1 ⎪⎭ ⎫⎪ ⎬ V2 = 38.9 m/s, α 2 = 29.5 V2 cos 60 = 68.35cos α 2 − 40⎪⎭ V2 sin 60 = 68.35sin α 2 4.140 − Rx = m(V2 x − V1x ) = 1000π × 0.0152 × 100(−38.9cos 60 − 100cos 30 ) ∴ Rx = 7500 N ∴ W = 12VB Rx = 12 × 40 × 7500 = 3.60 × 106 W 4.142 To find F, sum forces normal to the plate: ΣFn = m ⎡ (Vout )n − V1n ⎤ ⎣ ⎦ 52 © 2012 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 a) ∴ F = 1000 × 0.02 × 0.4 × 40 ⎡ −( − 40 sin 60 ) ⎤ = 11, 080 N ⎣ ⎦ (We have neglected friction) ΣFt = 0 = m2V2 + m3 (−V3 ) − m1 × 40sin 30 Bernoulli: V1 = V2 = V3 0 = m2 − m3 − 0.5m1 ⎫ ∴ m2 = 0.75m1 = 0.75 × 320 = 240 kg/s ⎬ m3 = 80 kg/s Continuity: m1 = m2 + m3 ⎭ ∴ F = mr (V1r )n = 1000 × 0.02 × 0.4(40 − VB )2 sin 60 4.144 Fx = 8(40 − VB2 ) sin 2 60 W = VB Fx = 8VB (40 − VB ) 2 × 0.75 = 6(1600VB − 80VB2 + VB3 ) dW = 6(1600 − 160VB + 3VB2 ) = 0 dVB ∴ VB = 13.33 m/s F = mr (V1 − VB )(cos α − 1) = 90 × 0.8 × 2.5 × 13.89 × (−13.89)(−1) = 34, 700 N 4.146 50 × 1000 ⎛ ⎞ = 13.89 m/s ⎟ ⎜ VB = 3600 ⎝ ⎠ ∴ W = 34, 700 × 13.89 = 482,000 W or 647 hp To solve this problem, choose a control volume attached to the reverse thruster vanes, as shown below. The momentum equation is applied to a free body diagram. Momentum: − Rx = 0.5m [ (Vr 2 ) x + (Vr 3 ) x ] − m × ( Vr1 ) x Assume the pressure in the gases equals the atmospheric pressure and that Vr1 = Vr2 = Vr3. Hence Vr2 (Vr1 ) x = Vr1 = 800 m/s , and 4.148 (Vr 2 )x = (Vr3 )x = −Vr1 × sin α where α = 20° . Then, momentum is Rx Vr1 − Rx = 0.5m [ −2Vr1 sin α ] − m × Vr1 = − mVr1 ( sin α + 1) Momentum: − Rx = 0.5m [ −2Vr1 sin α ] − m × Vr1 = −mVr1 ( sin α + 1) Vr3 The mass flow rate of the exhaust gases is m = mair + mfuel = mair (1 + 1 40 ) = 100 ×1.025 = 102.5 kg/s 53 © 2012 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 Substituting the given values we calculate the reverse thrust: ( ) Rx = (102.5 kg/s )( 800 m/s ) 1 + sin 20 = 110 kN Note that the thrust acting on the engine is in the opposite direction to Rx, and hence it is referred to as a reverse thrust; its purpose is to decelerate the airplane. For this steady-state flow, we fix the boat and move the upstream air. This provides us with the steady-state flow of Fig. 4.17. This is the same as observing the flow while standing on the boat. W = FV1 4.150 20,000 = F × 50 × 1000 3600 ∴ F = 1440 N V2 + 13.89 (V2 − 13.89) 2 30.6 + 13.89 ∴ Q = A3V3 = π × 12 × = 69.9 m3 /s 2 F = m(V2 − V1 ) ηp = 1440 = 1.23π × 12 × V1 13.89 = = 0.625 V3 22.24 or V2 = 72 × ∴ V2 = 30.6 m/s 62.5% Fix the reference frame to the boat so that V1 = 36 × 4.152 (V1 = 13.89 m/s) 5 = 10 m/s 18 5 = 20 m/s 18 ∴ F = m(V2 − V1 ) = 1000π × ( 0.25 ) × 2 10 + 20 (20 − 10) = 29.44 kN 2 W = F × V1 = (24 kN) × 10 m/s = 294.4 kW or 395 hp m = 1000 × π × ( 0.25) × 2 10 + 20 = 2944 kg/s 2 0.2 = V1A1 = V1 × .2 × 1.0 ∴ V1 = 1 m/s ∴ V1 max = 2 m/s 4.154 0.1 0.1 0 0 ∴ V1 ( y ) = 20(0.1 − y ) flux in = 2 ∫ ρ V 2dy = 2 ∫ 1000 × 202 (0.1 − y) 2 dy = 800,000 The slope at section 1 is −20 Continuity: A1V1 = A2V2 0.13 = 267 N 3 ∴ V2 ( y) = −20 y + A ∴ V2 = 2V1 = 2 m/s 54 © 2012 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 V2 (0) = A ⎫ ⎬ ∴ V2 = A − 1/ 2 V2 (0.05) = A − 1⎭ 2 = A− 0.05 flux out = 2 ∫ 0 1 ∴ A = 2.5 ∴V2 (y) = 2.5 − 20y 2 0.05 ⎡ ( y − 0.125)3 ⎤ 1000(2.5 − 20 y ) dy = 800,000 ⎢ ⎥ 3 ⎣⎢ ⎦⎥ 0 2 = 800,000 [0.00153] 3 = 408.3 N ∴change = 408 − 267 = 141 N From the c.v. shown: ( p1 − p2 )π r02 = τ w 2π r0 L ∴τ w = 4.156 ∴ Δp ro du =μ 2L dr w τw2πroL du 210 × 0.019 = dr w 2 × 9 × 0.001 p1A1 p2A2 = 220 m/s/m mtop = ρ A1V1 − ρ ∫ V2 ( y )dA 4.158 2 ⎡ ⎤ = 1.23 ⎢ 2 ×10 × 32 − ∫ (28 + y 2 )10dy ⎥ = 65.6 kg/s ⎢⎣ ⎥⎦ 0 2 F − = ∫ ρ V 2 dA + mtop V1 − m1V1 = 1.23∫ (28 + y 2 ) 210dy + 65.6 × 32 − 1.23 × 20 × 322 2 0 ∴ F = 3780 N Momentum and Energy a) Energy: 4.160 V12 V2 + z1 = 2 + z2 + hL 2g 2g See Problem 4.125(a) 82 1.9122 + 0 .6 = + 2.51 + hL 2 × 9.81 2 × 9.81 ∴ hL = 1.166 m ∴ losses = γ A1V1hL = 9810 × (0.6 × 1) × 8 × 1.166 = 54,900 W/m of width 55 © 2012 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 Moment of Momentum Ve = m 4 = = 19.89 m/s ρ Ae 1000 × 4 × π × 0.0042 MI = ∫ r × (2Ω × V) ρ d V Velocity in arm = V 0.3 = 4 ∫ rˆi × (−2Ωkˆ × V ˆi ) ρ Adr c.v. 0 = 8 ρ AV Ωkˆ 0.3 ∫ rdr = −0.36ρ AV Ωkˆ 0 ∑M = 0 4.164 and d ∫ (r × V ) ρ d V = 0 dt c.v. ∫ (r × V)V ⋅ nˆ ρ dA = 0.3ˆi × ( 0.707Ve ˆj + 0.707Vekˆ ) Ve ρ Ae c.s. The z-component of ∫ r × V(V ⋅ nˆ ) ρ dA = 0.3 × 0.707Ve Ae ρ 2 c.s. Finally −( M I ) z = 0.36 ρ AV Ω = 4 × 0.3 × 0.707Ve2 Ae ρ 0.36Ω = 4 × 0.3 × 0.707 ×19.89 ∴Ω = 46.9 rad/s m = 10 = ρ AV = 1000π × 0.012 V0 Continuity: Using AV = Ae Ve ∴ V0 = 31.8 m/s V0π × 0.012 = V π × 0.012 + Ve × 0.006(r − 0.05) V0π × 0.012 = Ve × 0.006 × 0.15 ∴ Ve = 11.1 m/s ∴ V = V0 − 19.1(r − 0.05)Ve = 42.4 − 212r 0.05 4.166 MI = ∫ 2rˆi × (+2Ωkˆ × V0ˆi ) ρ Adr + 0 = 4ΩV0 ρ Akˆ 0.2 ∫ 2rˆi × [+2Ωkˆ × (42.4 − 212r )ˆi ]ρ Adr 0.05 0.05 ∫ rdr + 4Ωρ Akˆ 0 0.2 ∫ (42.4r − 212r 2 )dr 0.05 212 ⎡ 42.4 ⎤ = ⎢ (0.22 − 0.052 ) − (0.23 − 0.053 ) ⎥ kˆ 3 ⎣ 2 ⎦ = (0.05Ω + 0.3Ω )kˆ = 0.35Ωkˆ 56 © 2012 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 0.2 ∫ rˆi × (−Ve ˆj)Ve ρ × 0.006dr = −11.12 × 1000 × 0.006 0.05 0.2 ∫ rdr kˆ = −13.86 kˆ 0.05 ∴Ω = 39.6 rad/s ∴ −0.35Ω = −13.86 0.0082 V= × 19.89 = 3.18 m/s 0.022 See Problem 4.165. Ve = 19.89 m/s 0.3 ⎡ ⎛ dΩ ˆ ⎞ ˆ ⎤ M I = 4 ∫ rˆi × ⎢ (−2Ωkˆ × V ˆi ) + ⎜ − k ⎟ × ri ⎥ρ Adr A = π × 0.012 , Ae = π × 0.0042 ⎝ dt ⎠ ⎣ ⎦ 0 = −8 ρ AV Ωkˆ 0.3 ∫ rdr − 4 ρ A 0 dΩ ˆ k dt 0.3 ∫r 2 dr 0 dΩ ˆ k = −360 AV Ωkˆ − 36 A dt 4.168 ∫ (r × V)z (V ⋅ nˆ ) ρ dA = 212Ve Aekˆ 2 c.s. Thus, 360 AV Ω + 36 A dΩ dΩ = 212Ve2 Ae or + 31.8Ω = 373 dt dt The solution is Ω = Ce −31.8t + 11.73. The initial condition is Ω( 0 ) = 0 ∴ C = −1173 . . Finally Ω = 11.73(1 − e −31.8t ) rad/s 57 © 2012 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 58 © 2012 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 5 / The Differential Forms of the Fundamental Laws CHAPTER 5 The Differential Forms of the Fundamental Laws Differential Continuity Equation in − m out = m ∂melement . This is expressed as ∂t ⎡ ⎣ ρ vr (rdθ dz) − ⎢ ρ vr + ∂ ∂ ( ρvr ) dr ⎤⎥ (r + dr)dθ dz + ρ vθ drdz − ⎡⎢ ρ vθ + ( ρ vθ )dθ ⎤⎥ drdz ∂r ∂θ ⎦ ⎣ ⎦ dr ⎞ ∂ dr ⎞ ∂ ⎡ ⎛ dr ⎞ ⎤ ⎛ ⎡ ⎤⎛ + ρ vz ⎜ r + ⎟ dθ dr − ⎢ ρ vz + ( ρ vz )dz ⎥ ⎜ r + ⎟ dθ dr = ⎢ ρ ⎜ r + ⎟ dθ drdz ⎥ 2⎠ ∂z 2⎠ ∂t ⎣ ⎝ 2⎠ ⎝ ⎣ ⎦⎝ ⎦ 5.2 Subtract terms and divide by rdθdrdz : − ρ vr r − r + dr ∂ 1 ∂ ∂ r + dr / 2 ∂ r + dr / 2 ( ρ vr ) − ( ρ vθ ) − ( ρ vz ) = ρ r ∂r r ∂θ ∂z r ∂t r Since dr is an infinitesimal, (r + dr )/r = 1 and (r + dr /2)/r = 1. Hence ∂ρ ∂ 1 ∂ ∂ 1 + ( ρ vr ) + ( ρ vθ ) + ( ρ vz ) + ρ vr = 0 ∂t ∂r r ∂θ ∂z r This can be put in various forms. If ρ = const, it divides out For a steady flow ∂ρ = 0. Then, with v = w = 0, Eq. 5.2.2 yields ∂t ∂ ( ρ u) = 0 ∂x 5.4 or ρ du dρ +u =0 dx dx Partial derivatives are not used since there is only one independent variable. ∂ ∂ρ = 0, ≠ 0. Since water can be considered to be incompressible, we ∂t ∂z ∂ρ ∂ρ demand that D ρ /Dt = 0. Equation (5.2.8) then provides u +w = 0, ∂x ∂z assuming the x-direction to be in the direction of flow. There is no variation with y. Also, we demand that ∇ ⋅ V = 0, or Given: 5.6 59 © 2012 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 5 / The Differential Forms of the Fundamental Laws ∂u ∂w + =0 ∂x ∂z We can use the ideal gas law, ρ = 5.7 p . Then, the continuity equation (5.2.7) RT Dρ = − ρ∇ ⋅ V becomes, assuming RT to be constant Dt p 1 Dp =− ∇ ⋅ V or RT Dt RT 1 Dp = −∇ ⋅ V p Dt a) Use cylindrical co-ordinates with vθ = vz = 0 : 1 ∂ ( rv r ) = 0 r ∂r Integrate: rvr = C 5.8 ∴ vr = C r b) Use spherical coordinates with vθ = vφ = 0 : 1 ∂ 2 (r vr ) = 0 r 2 ∂r Integrate: r 2vr = C ∴ vr = C r2 (a) Since the flow is steady and incompressible then VA = constant, where the constant is determined by using the conditions at the inlet that is, ( VA )inlet = 40 ×1 = 40 m3 /s. And, since the flow is inviscid, the velocity is 5.9 uniform in the channel, so u = V . Hence, at any x position within the channel the velocity u can be calculated using u = V = 40/A. Since the flow area is not constant it is given by A = 2hw, where the vertical distance h is a function of x and can be determined as, h = 0.15x + 0.5H . Substituting, we obtain the following expression for the velocity: u( x) = 40 20 = m/s 2(0.15x + 0.5) 0.15x + 0.5 (b) To determine the acceleration in the x-direction, we use (see Eq. 3.2.9) du where ax = u dx 60 © 2012 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 5 / The Differential Forms of the Fundamental Laws du 3 =− dx (0.15x + 0.5) 2 Hence, the expression for acceleration is ax = 20 −3 −60 × = m/s 2 2 3 0.15x + 0.5 (0.15x + 0.5) (0.15x + 0.5) Note that the minus sign indicates deceleration of the fluid in the x-direction. ∂u ∂v ⎛ ∂u ⎞ + = 0, we write v = − ⎜ ⎟dy + C. ∂x ∂y ⎝ ∂x ⎠ ∂u 3 ∂v With the result from Problem 5.9: =− = − , we integrate to 2 ∂x ∂y (0.15x + 0.5) find ∫ (a) Using the continuity equation v( x, y ) = 3y (0.15x + 0.5)2 and since v = 0 at y = 0, then C = 0. 5.10 (b) To determine the acceleration in the y-direction, we use (see Eq. 3.2.9) ay = u ∂v ∂v +v ∂x ∂y From part (a) we have 0.9 y ∂v =− ∂x (0.15x + 0.5)3 Substituting in the expression for acceleration we get −0.9 y 3y −9 y 20 3 ay = × + × = 3 2 2 0.15x + 0.5 (0.15x + 0.5) (0.15x + 0.5) (0.15x + 0.5) (0.15x + 0.5) 4 5.11 ⎛ ∂u ∂v ⎞ Dρ = − ρ∇ ⋅ V = − ρ ⎜ + ⎟ = −2.3(200 × 1 + 400 × 1) = −1380 kg/m3 ⋅ s Dt ⎝ ∂x ∂y ⎠ 5.12 In a plane flow, u = u( x, y) and v = v( x, y). Continuity demands that ∂u ∂u ∂v ∂v = 0 and hence + = 0. If u = const, then = 0. Thus, we also ∂x ∂x ∂y ∂y have v = const and D ρ /Dt = 0. 5.13 If u = C1 and v = C2 , the continuity equation provides, for an incompressible flow 61 © 2012 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 5 / The Differential Forms of the Fundamental Laws ∂u ∂v ∂w + + =0 ∂x ∂y ∂z ∴ ∂w = 0 so w = C3 ∂z The z-component of velocity w is also constant. We also have Dρ ∂ρ ∂ρ ∂ρ ∂ρ =0= +u +v +w Dt ∂t ∂x ∂y ∂z The density may vary with x, y, z and t. It is not, necessarily, constant. 5.14 ∂u ∂v + =0 ∂x ∂y ∴ A+ ∂v =0 ∂y ∴ v = − Ay But, v( x, 0) = 0 = f ( x) ∂u ∂v + =0 ∂x ∂y 5.15 ∴ v(x, y) = ∴v = ∴ 5y 2 − 5x2 ∫ (x 2 ( x 2 + y 2 )5 − 5 x(2 x) 5x 2 − 5y 2 ∂v ∂u =− =− = − ∂y ∂x (x 2 + y 2 )2 (x 2 + y 2 )2 dy + f (x) = 5y x + y2 2 + f (x) f (x) = 0 5y x + y2 2 From Table 5.1: 5.16 +y ) 2 2 ∴ v(x, y) = − Ay + f (x) 1 ∂ 1 ∂vθ 1⎛ 0.4 ⎞ (rvr ) = − = − ⎜ 10 + 2 ⎟ sin θ r ∂r r ∂θ r⎝ r ⎠ 0.4 ⎞ 0.4 ⎞ ⎛ ⎛ ∴ rvr = ⎜ 10 + 2 ⎟ sin θ dr + f (θ ) = ⎜10r − ⎟ sin θ + f (θ ) r ⎠ r ⎠ ⎝ ⎝ ∫ 0.4 ⎞ ⎛ 0.2vr (0.2,θ ) = ⎜10 × 0.2 − ⎟ sin θ + f (θ ) = 0 0.2 ⎠ ⎝ ∴ f (θ ) = 0 0.4 ⎞ ⎛ ∴ vr = ⎜10 − 2 ⎟ sin θ r ⎠ ⎝ From Table 5.1: 5.17 1 ∂ 1 ∂ vθ −20 ⎛ 1 ⎞ = ( rvr ) = − ⎜1 + 2 ⎟ cosθ r ∂r r ∂θ r ⎝ r ⎠ 1⎞ ⎛ ⎛ 1⎞ ∴ rvr = −20 ⎜1 + 2 ⎟ cos θ dr + f (θ ) = −20 ⎜ r − ⎟ cos θ + f (θ ) ⎝ r ⎠ ⎝ r⎠ ∫ vr (1,θ ) = −20(1 − 1) cos θ + f (θ ) = 0 ∴ f (θ ) = 0 1⎞ ⎛ ∴ vr = −20 ⎜1 − 2 ⎟ cos θ ⎝ r ⎠ 62 © 2012 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 5 / The Differential Forms of the Fundamental Laws From Table 5.1, spherical coordinates: ∴ 5.18 1 ∂ 2 1 ∂ ( r vr ) = − ( vθ sin θ ) 2 ∂r r sin θ ∂θ r 1 ∂ 2 1 ⎛ 40 ⎞ ( r v ) = + 10 ⎜ ⎟ 2sinθ cosθ r r sinθ ⎝ r2 ∂ r r3 ⎠ 40 ⎞ 80 ⎞ ⎛ ⎛ ∴ r 2 vr = r ⎜ 10 + 3 ⎟ 2 cos θ dr + f (θ ) = ⎜ 10r 2 − ⎟ cos θ + f (θ ) r ⎠ r ⎠ ⎝ ⎝ ∫ 80 ⎞ ⎛ 4vr (2,θ ) = ⎜10 × 22 − ⎟ cos θ + f (θ ) = 0 2 ⎠ ⎝ ∴ f (θ ) = 0 80 ⎞ ⎛ ∴ vr = ⎜10 − 3 ⎟ cos θ r ⎠ ⎝ Continuity: 5.19 ρ= ∂ ( ρu) = 0 ∂x ∴ρ du dρ +u =0 dx dx p 124 × 103 = = 1.554 kg/s RT 287 × 278 ∴ du 159 − 137 = = 220 m/s/m dx 0.1 dρ ρ du 1.554 =− =− × 220 = −2.32 kg/m4 dx u dx 147 ∂u ∂v + =0 ∂x ∂y ∂ ⎡ −20( 1 − e − x )⎤ = −20e − x ⎦ ∂x⎣ Hence, in the vicinity of the x-axis: ∂v = 20e − x ∂y 5.20 But v = 0 if y = 0 and v = 20 ye − x + C ∴C = 0 v = 20 ye − x = 20(0.2)e −2 = 0.541 m/s From Table 5.1, 1 ∂ ∂v (rvr ) + z = 0 r ∂r ∂z ∂ ⎡ −20(1 − e − z ) ⎤ = −20e − z ⎦ ∂z ⎣ Hence, in the vicinity of the z-axis: 1 ∂ r2 (rvr ) = 20e − z and rvr = 20e − z + C r ∂r 2 5.21 But vr = 0 if r = 0 ∴C = 0 vr = 10re − z = 10(0.2)e −2 = 0.271 m/s 63 © 2012 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 5 / The Differential Forms of the Fundamental Laws The velocity is zero at the stagnation point. Hence, 0 = 10 − The continuity equation for this plane flow is 5.22 we see that 40 R2 ∴R = 2 m ∂u ∂u ∂v = 80 x −3 , + = 0. Using ∂x ∂x ∂y ∂v = −80 x −3 near the x-axis. Consequently, for small Δy , ∂y Δv = −80 x −3Δy so that v = −80(−3)−3 (0.1) = 0.296 m/s The velocity is zero at the stagnation point. Hence 0 = (40/R2 ) − 10 ∴ R = 2 m Use continuity from Table 5.1: 1 ∂ 2 1 ∂ 20 (40 − 10r 2 ) = − r vr = 2 2 ∂r r r r ∂r ( ) Near the negative x-axis continuity provides us with 5.23 1 ∂ 20 ( vθ sinθ ) = r sinθ ∂θ r Integrate, letting θ = 0 from the y-axis: vθ sin θ = −20 cos θ + C Since vθ = 0 when θ = 90°, C = 0. Then, with α = tan −1 vθ = −20 Continuity: 5.24 cos θ cos88.091 0.0333 = −20 = −20 = 0.667 m/s sin θ sin 88.091 0.999 ∂u ∂v + =0 ∂x ∂y ∴ ∴Δv = v − 0 = −220Δy b) ax = u 0. 1 = 1.909° 3 Δv Δu 13.5 − 11.3 m/s =− =− = −220 Δy Δx 2 × 0.005 m ∴ v = −220 × 0.004 = −0.88 m/s ∂u = 12.6 × (+220) = 2772 m/s 2 ∂x Differential Momentum Equation ΣFy = may . For the fluid particle occupying the volume of Fig. 5.3: 5.25 ∂τ yy dy ⎞ ∂τ zy dz ⎞ ∂τ xy dx ⎞ ⎛ ⎛ ⎛ ⎜⎜τ yy + ⎟⎟ dxdz + ⎜⎜ τ zy + ⎟⎟ dxdy + ⎜⎜ τ xy + ⎟ dydz ∂y 2 ⎠ ∂z 2 ⎠ ∂x 2 ⎟⎠ ⎝ ⎝ ⎝ 64 © 2012 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 5 / The Differential Forms of the Fundamental Laws ∂τ yy dy ⎞ ∂τ zy dz ⎞ ∂τ xy dx ⎞ ⎛ ⎛ ⎛ − ⎜⎜ τ yy − ⎟⎟ dxdz − ⎜⎜ τ zy − ⎟⎟ dxdy − ⎜⎜ τ xy − ⎟⎟ dydz 2 2 2 ∂ ∂ ∂ y z x ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + ρ g y dx dy dz = ρ dx dy dz Dv Dt Dividing by dx dy dz , and adding and subtracting terms: ∂τ xy ∂x + ∂τ yy ∂y + ∂τ zy ∂z + ρ gy = ρ Dv Dt Check continuity: ∂ u ∂ v ∂ w ( x 2 + y 2 )10 − 10 x(2x) ( x 2 + y 2 )10 − 10 y(2 y) + + = + =0 ∂x ∂y ∂z ( x2 + y 2 )2 ( x2 + y 2 )2 Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7, give with g x = g y = 0: ρu 5.26 ∴ ∂p ∂u ∂u + ρv =− ∂x ∂y ∂x ∂p 10 y −20 xy 100( x 2 + y 2 ) y 10 x 10 y 2 − 10 x 2 = −ρ 2 − ρ = ρ ∂x x + y 2 (x2 + y 2 )2 x2 + y 2 (x2 + y 2 )2 ( x 2 + y 2 )3 ∂p ∂v ∂v ρu + ρ v = − ∂x ∂y ∂y ∂p −20 xy 10 y 10 x 2 − 10 y 2 100( x 2 + y 2 ) y 10 x ∴ = −ρ 2 −ρ 2 =ρ ∂y x + y 2 ( x2 + y 2 )2 x + y 2 (x2 + y 2 )2 ( x 2 + y 2 )3 ∴∇p = ∂p ˆ ∂p ˆ 100 y ρ ˆ 100 x ρ ˆ 100 ρ j= 2 i j i+ + = ( xˆi + yˆj) 2 2 2 2 2 2 2 2 ∂x ∂y (x + y ) (x + y ) (x + y ) Check continuity (cylindrical coord. from Table 5.1): 5.27 1 ∂ 1 ∂v θ 10 ⎛ 1⎞ 1⎞ −10 ⎛ ( rv r ) + = ⎜ 1 + 2 ⎟ cos θ + ⎜ 1 + 2 ⎟ cos θ = 0. ∴It is a r ∂r r ∂θ r ⎝ r ⎠ r ⎝ r ⎠ possible flow. For Euler’s Eqs. (let v = 0 in the momentum eqns of Table 5.1) in cylindrical coord: 2 ∂p v2 ∂v v ∂v 100 ρ ⎛ 1⎞ 1⎞ ⎛ ⎛ 20 ⎞ 1 + 2 ⎟ sin 2 θ − 10 ρ ⎜1 − 2 ⎟ cos 2 θ ⎜ 3 ⎟ = ρ θ − ρ vr r − ρ θ r = ⎜ ∂r r ∂r r ∂θ r ⎝ r ⎠ ⎝ r ⎠ ⎝r ⎠ 10ρ ⎛ 1⎞ 2 ⎛ 10 ⎞ − ⎜1 + 2 ⎟ sin θ ⎜10 − 2 ⎟ r ⎝ r ⎠ r ⎠ ⎝ 65 © 2012 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 5 / The Differential Forms of the Fundamental Laws vv ∂v v ∂v 1 ∂p 100 ρ ⎛ 1 ⎞ 1 − 4 ⎟ sin θ cos θ = − ρ r θ − ρ vr θ − ρ θ θ = ⎜ r ∂θ r r ∂θ r ⎝ r ⎠ ∂r 2 1 ⎞ 1 ⎞ ⎛ ⎛ 20 ⎞ 100 ρ ⎛ −10 ρ ⎜1 − 2 ⎟ cos θ sin θ ⎜ 3 ⎟ − ⎜1 + 2 ⎟ sin θ cos θ r ⎝ r ⎠ ⎝ r ⎠ ⎝r ⎠ ∴∇p = ∂p ˆ 1 ∂p ˆ 200 ρ ⎡ 1 200 ρ ⎤ ir + iθ = 3 ⎢ 2 − cos 2θ ⎥ ˆir − 3 sin 2θ ˆiθ ∂r r ∂θ r ⎣r r ⎦ Follow the steps of Problem 5.27. The components of the pressure gradient are vθ2 + vφ2 ∂p ∂v v ∂v =ρ − ρ vr r − ρ θ r ∂r r ∂r r ∂θ 5.28 vv v ∂v ∂v 1 ∂p = − ρ r θ − ρ vr θ − ρ θ θ r ∂θ r r ∂θ ∂r ⎛ 2μ ⎞ ∴p = p −⎜ +λ⎟ ∇⋅V ⎝ 3 ⎠ ⎛ 2μ ⎞ ∴p − p = −⎜ +λ⎟ ∇⋅V ⎝ 3 ⎠ ∂ sˆ Δsˆ Δα nˆ nˆ ≅ =− =− ∂ s Δs RΔα R ∂ sˆ Δsˆ nˆ Δθ ∂θ ≅ = = nˆ ∂t Δt Δt ∂t 5.29 ∴ DV ⎛ ∂V ∂V ⎞ ⎛ ∂θ V 2 ⎞ =⎜ +V − ⎟ nˆ ⎟ sˆ + ⎜ V Dt ⎝ ∂t ∂s ⎠ ⎜⎝ ∂t R ⎟⎠ ⎛ V2 ⎞ ∂V For steady flow, the normal acc. is ⎜ − , the tangential acc. is V ⎟ ⎜ R ⎟ ∂s ⎝ ⎠ For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to Ω. 5.30 Thus, Euler’s equation becomes dΩ ⎞ ⎛ DV + 2Ω × V + Ω × (Ω × r ) + × r ⎟ = −∇p − ρ g dt ⎝ Dt ⎠ ρ⎜ 5.31 τ xx = − p + 2μ ∂u + λ ∇ ⋅ V = −210 kPa ∂x τ yy = τ zz = − p = −210 kPa 66 © 2012 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 5 / The Differential Forms of the Fundamental Laws ⎛ ∂u τ xy = μ ⎜ ⎝ ∂y + ∂v ∂x ⎞ −4 −3 ⎟ = 5 ×10 [32 − 32 ×192 × 0.0025] = 8.32 × 10 Pa ⎠ τ xy τ xz = τ yz = 0 τ xx = 8.32 × 10−3 210 × 10 3 16 y 16 y 2 ∂v ∂u =− = − ∂y ∂x C x9 / 5 C 2 x13/ 5 v( x, o) = 0 ∴ f ( x) = 0 = 3.96 × 10−8 ∴ v( x, y) = 8 = C 10004 / 5 8y 2 Cx9 / 5 − 16 y3 3C 2 x13/ 5 + f ( x) ∴C = 0.0318 ∴ u( x, y ) = 629 yx −4 / 5 − 9890 y 2 x −8 / 5 v( x, y) = 252 y 2 x−9 / 5 − 5270 y3x−13/ 5 5.32 τ xx = − p + 2μ ∂u = −100 + 0 = −100 kPa ∂x τ yy = τ zz = − p = −100 kPa ⎛ ∂u ∂v ⎞ + ⎟ = 2 ×10−5 ⎡629 ×1000−4/5 ⎤ = 5.01×10−5 Pa ⎣ ⎦ ∂ ∂x ⎠ y ⎝ τ xy = μ ⎜ τ xz = τ yz = 0 Du ∂u ⎛ ∂ ∂ ∂ ⎞ = + ⎜ u + v + w ⎟ u = ( V ⋅ ∇ )u ∂y ∂z ⎠ Dt ∂t ⎝ ∂x 5.33 Dv ∂v ⎛ ∂ ∂ ∂ ⎞ = + ⎜ u + v + w ⎟ v = (V ⋅ ∇ )v Dt ∂t ⎝ ∂x ∂y ∂z ⎠ Dw ∂w ⎛ ∂ ∂ ∂ ⎞ = + ⎜ u + v + w ⎟ w = (V ⋅ ∇ )w Dt ∂t ⎝ ∂x ∂y ∂z ⎠ ∴ 5.34 DV Du ˆ Dv ˆ Dw ˆ i+ j+ k = V ⋅ ∇ (uˆi + vˆj + wkˆ ) = (V ⋅ ∇ )V = Dt Dt Dt Dt Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible effects: ρ DV μ ∂ μ ∂ μ ∂ = −∇p + ρ g + μ∇ 2 V + ∇ ⋅ V ˆi + ∇ ⋅ V ˆj + ∇ ⋅ V kˆ Dt 3 ∂x 3 ∂y 3 ∂z 67 © 2012 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 5 / The Differential Forms of the Fundamental Laws = −∇p + ρ g + μ∇ 2 V + ∴ρ μ⎛ ∂ ˆ ∂ ˆ ∂ ˆ⎞ j+ k ⎟∇ ⋅ V ⎜ i+ 3 ⎝ ∂x ∂y DV μ = −∇p + ρ g + μ∇ 2 V + ∇(∇ ⋅ V ) Dt 3 If u = u ( y ), then continuity demands that lower plate), v = 0 ∴ρ 5.35 ∂z ⎠ ∂v = 0 ∴ v = C. But, at y = 0 (the ∂y ∴ C = 0, and v( x, y) = 0 ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂p ∂u ∂u ∂u ⎞ Du = ρ ⎜ + u + v + w ⎟ = − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟ ⎜ Dt ∂x ∂y ∂z ⎠ ∂x ∂y ∂z ⎠⎟ ⎝ ∂t ⎝ ∂x ∴0 = − ∂p ∂ 2u +μ 2 ∂x ay ρ ∂p Dv =0=− Dt ∂y ρ ∂p Dw = 0 = − + ρ (− g ) Dt ∂z ∴ ∂p = −ρ g ∂z The x-component Navier-Stokes equation can be written as ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂p ∂u ∂u ∂u ⎞ + u + v + w ⎟ = − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟ ⎜ ∂x ∂x ∂y ∂z ⎠ ∂x ∂y ∂z ⎟⎠ ⎝ ∂t ⎝ ρ⎜ Based on the given conditions the following assumptions can be made: 5.36 One-dimensional ( v = w = 0 ) ⎛ ∂u ⎞ Steady state ⎜ = 0⎟ ⎝ ∂t ⎠ Incompressible ( ρ = constant ) ⎛ dp ⎞ Zero pressure-gradient ⎜ = 0⎟ ⎝ dx ⎠ ⎛ ∂u ⎞ Fully-developed flow ⎜ = 0⎟ ⎝ ∂x ⎠ ⎛ ∂u ⎞ A wide channel ⎜ = 0⎟ ∂ z ⎝ ⎠ The Navier-Stokes equation takes the simplified form μ ∂ 2u ∂ 2u = 0 or = 0. ∂y 2 ∂y 2 Integrating twice yields, u( y ) = ay + b . To determine a and b we apply the following boundary conditions: u = V1 at y = 0, and u = −V2 at y = h. This gives b = V1 and a = −(V1 + V2 ) / h. The velocity distribution between the plates is 68 © 2012 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 5 / The Differential Forms of the Fundamental Laws then ⎛ V + V2 ⎞ u( y ) = − ⎜ 1 ⎟ y + V1 ⎝ h ⎠ Using the x-component Navier-Stokes equation with x being vertical and the following assumptions: ⎛ ∂u ⎞ Steady state ⎜ = 0⎟ ⎝ ∂t ⎠ One-dimensional ( v = w = 0 ) Incompressible ⎛ ∂u ⎞ Fully-developed flow ⎜ = 0⎟ ⎝ ∂x ⎠ ( ρ = constant ) The x-component Navier-Stokes equation reduces to 0=− 5.37 ∂p ∂2 u − ρg + μ 2 ∂x ∂y Integrate the above differential equation twice (see Problem 5.36): u( y ) = 1 ⎛ dp ⎞ + ρ g ⎟ y 2 + ay + b ⎜ 2 μ ⎝ dx ⎠ Applying the no-slip boundary condition at both plates (see Problem 5.36) we get u( y ) = ( 1 ⎛ dp ⎞ + ρ g ⎟ y 2 − hy ⎜ 2 μ ⎝ dx ⎠ ) Assumptions: One-dimensional ( vr = vθ = 0 ) ⎛ ∂v ⎞ Steady state ⎜ z = 0 ⎟ ⎝ ∂t ⎠ Incompressible ( ρ = constant ) Horizontal ( g z = 0) ⎛ ∂v ⎞ Fully-developed flow ⎜ z = 0 ⎟ ⎝ ∂z ⎠ Dvr 1 ∂p =0=− ρ ∂r Dt 5.38 ρ Dvθ 1 ∂p =0=− ρ r ∂θ Dt ⎛ ∂ 2 v 1 ∂v ⎛ ∂v ∂p Dvz ∂vz vθ ∂vz ∂v ⎞ ∂ 2 vz 1 ∂ 2 vz z + 2 = ρ ⎜ z + vr + + vz z ⎟ = − + μ ⎜ 2z + + 2 ⎜ ⎜ ∂r ∂z Dt ∂r r ∂θ ∂z ⎠⎟ r ∂r r ∂θ ∂z 2 ⎝ ∂t ⎝ ∴0 = − ⎛ ∂2v 1 ∂vz ⎞ ∂p + μ ⎜ 2z + ⎟ ⎜ ∂r r ∂r ⎟⎠ ∂z ⎝ 69 © 2012 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 5 / The Differential Forms of the Fundamental Laws Assumptions: One-dimensional flow ( vθ = vr = 0 ) ⎛ ∂v ⎞ Steady state ⎜ z = 0 ⎟ ⎝ ∂t ⎠ Incompressible ( ρ = constant ) Horizontal ( g z = 0) ⎛ ∂v ⎞ Fully-developed flow ⎜ z = 0 ⎟ ⎝ ∂z ⎠ The Navier-Stokes equation in cylindrical form provides the following equation: 0=− ⎡ ∂ 2 v 1 ∂vz ⎤ ∂p + μ ⎢ 2z + ⎥ ∂z r ∂r ⎥⎦ ⎢⎣ ∂r Rearrange the above equation and integrate: 0=− ∂p μ ∂ ⎛ ∂vz ⎞ + ⎜r ⎟ ∂z r ∂r ⎝ ∂r ⎠ Integrating again yields: vz (r ) = 5.39 ∂vz 1 ∂p ⎛ r ⎞ C1 = ⎜ ⎟+ ∂r μ ∂z ⎝ 2 ⎠ r 1 ∂p ⎛ r 2 ⎞ ⎜ ⎟ + C ln r + C2 μ ∂z ⎜⎝ 4 ⎟⎠ 1 C1 and C2 are determined using the boundary conditions: vz = 0, at r = ro , and vz = Vc at r = ri . Hence Vc = 1 ∂p ⎛ ri2 ⎞ 1 ∂p ⎛ ro2 ⎞ + + C = ln and 0 C r ⎜ ⎟ ⎜ ⎟ + C ln r + C2 2 μ ∂z ⎜⎝ 4 ⎟⎠ 1 i μ ∂z ⎜⎝ 4 ⎟⎠ 1 o Subtracting the second equation from the first yields C1 = Vc − 1 ∂p 2 2 ri − ro 4 μ ∂z ln ( ri ro ) ( ) The drag force on the inner cylinder is zero when the shear stress τ rz on the ⎡ ∂v ∂v ⎤ inner cylinder is zero, i.e., τ rz = μ ⎢ r + z ⎥ = 0 . Since vr = 0 , then ∂r ⎦ r =ri ⎣ ∂z ⎡ ∂v ⎤ τ rz = μ ⎢ z ⎥ = 0 . From the above expression for vz we find ⎣ ∂r ⎦ r =r i ∂vz ∂r = r = ri C 1 ∂p 1 ∂p 2 ri + 1 = 0. Then C1 = − ri . Combining with the above 2μ ∂z ri 2μ ∂z expressions for C1 we solve for Vc . The result is: 70 © 2012 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 5 / The Differential Forms of the Fundamental Laws Vc = Continuity: 5.40 − ( ) 1 ∂p ⎡ 2 2 ri − ro − 2ri2 ln ( ri ro ) ⎤ ⎣ ⎦ 4 μ ∂z 1 ∂ 2 (r vr ) = 0 r 2 ∂r ∴ r 2vr = C ∂p vθ2 ⎛ 2v ⎞ ρ = − + μ ⎜ − 2θ cot θ ⎟ r ∂r ⎝ r ⎠ At r = r1, vr = 0 0=− 0=− ∴C = 0 ∂p ⎡ 1 ∂ ⎛ 2 ∂vθ +μ⎢ ⎜ r ∂r ∂θ ⎣ r ∂r ⎝ vθ ⎤ ⎞ ⎟− ⎥ ⎠ r sin 2 θ ⎦ 1 ∂p r sin θ ∂φ Assumptions: One-dimensional flow ( vz = vr = 0 ) ⎛ ∂v ⎞ Steady state ⎜ z = 0 ⎟ ⎝ ∂t ⎠ Incompressible ( ρ = constant ) Vertical ( gr = gθ = 0) ⎛ ∂v ⎞ Developed flow ⎜ r = 0 ⎟ ⎝ ∂θ ⎠ The simplified differential equation from Table 5.1 is which can be re-written as 5.41 Integrating we get: ∂ 2 vθ ∂r 2 + ∂ ⎛ vθ ∂r ⎜⎝ r ∂ 2 vθ ∂r 2 + 1 ∂vθ vθ − =0 r ∂r r 2 ⎞ ⎟ = 0. ⎠ ∂vθ vθ + = C1 r ∂r The above equation can be re-written as Integrating again yields rvθ = C1 1 ∂ ( rvθ ) = C1 r ∂r r2 r C + C2 ⇒ vθ = C1 + 2 2 2 r Apply the boundary conditions vθ = 0 at r = ri , and vθ = roω at r = ro . We have 0 = C1 ri C2 + 2 ri Solving for C1 and C2 yields C1 = −2 roω = C1 and ω ro2 and C2 = 2 ri2 − ro ro C2 + 2 ro ω ri2ro2 ri2 − ro2 71 © 2012 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 5 / The Differential Forms of the Fundamental Laws ⎛ ω r 2 ⎞ ⎛ ω r 2r 2 ⎞ 1 vθ = ⎜ − 2 o 2 ⎟ r + ⎜ 2 i o2 ⎟ ⎜ r −r ⎟ ⎜r −r ⎟r o ⎠ o ⎠ ⎝ i ⎝ i Finally For an incompressible flow ∇ ⋅ V = 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and 5.3.3: ρ ρ ∂u ⎞ ∂ ⎛ ∂u ∂v ⎞ ∂ ⎛ ∂u ∂w ⎞ Du ∂ ⎛ = ⎜ − p + 2μ ⎟ + μ ⎜ + ⎟ + μ ⎜ + ⎟ + ρ gx ∂x ⎠ ∂y ⎝ ∂y ∂x ⎠ ∂z ⎝ ∂z ∂x ⎠ Dt ∂x ⎝ =− ∂p ∂ 2u ∂ 2u ∂ 2u ∂ ⎛ ∂u ∂v ∂w ⎞ +μ 2 +μ 2 +μ 2 +μ ⎜ + + ⎟ + ρ gx ∂x ∂x ⎝ ∂x ∂y ∂z ⎠ ∂x ∂y ∂z ∴ρ ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ∂p Du = − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gx ∂x ∂z ⎠ Dt ⎝ ∂x ∂y Dv ∂ ⎛ ∂u ∂v ⎞ ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ∂w ⎞ = μ ⎜ + ⎟ + ⎜ − p + 2μ ⎟ + μ ⎜ + ⎟ + ρ gy Dt ∂x ⎝ ∂y ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ∂y ⎠ ∂p ∂ 2v ∂2v ∂ 2v ∂ ⎛ ∂u ∂v ∂w ⎞ =− +μ 2 +μ 2 +μ 2 +μ ⎜ + + ⎟ + ρ gy ∂y ∂y ⎝ ∂x ∂y ∂z ⎠ ∂x ∂y ∂z 5.42 ∴ρ ρ ⎛ ∂2v ∂2v ∂2v ⎞ ∂p Dv = − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gy ⎜ ∂x ∂y ∂z ⎟ Dt ∂y ⎝ ⎠ Dw ∂ ⎛ ∂u ∂w ⎞ ∂ ⎛ ∂v ∂w ⎞ ∂ ⎛ ∂w ⎞ μ⎜ + = ⎟ + ⎜ − p + 2μ ⎟+ μ⎜ + ⎟ + ρ gz Dt ∂x ⎝ ∂z ∂x ⎠ ∂y ⎝ ∂z ∂y ⎠ ∂z ⎝ ∂z ⎠ =− ∴ρ ∂p ∂2w ∂ 2w ∂ 2w ∂ ⎛ ∂u ∂v ∂w ⎞ +μ 2 +μ 2 +μ 2 +μ ⎜ + + ⎟ + ρ gz ∂z ∂z ⎝ ∂x ∂y ∂z ⎠ ∂x ∂y ∂z ⎛ ∂ 2w ∂ 2w ∂ 2w ⎞ ∂p Dw = − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gz ⎜ ∂x Dt ∂z ∂y ∂z ⎟⎠ ⎝ If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with μ = μ( x , y , z ) we arrive at 5.43 ρ ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ∂p Du ∂μ ∂u ∂μ ⎛ ∂u ∂v ⎞ ∂μ ⎛ ∂u ∂w ⎞ = − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟ + 2 + ⎜ + ⎟+ ⎜ + ⎟ ⎜ ⎟ Dt ∂x ∂x ∂x ∂y ⎝ ∂y ∂x ⎠ ∂z ⎝ ∂z ∂x ⎠ ∂y ∂z ⎠ ⎝ ∂x 72 © 2012 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 5 / The Differential Forms of the Fundamental Laws If plane flow is only parallel to the plate, v = w = 0. Continuity then demands that ∂u/∂x = 0. The first equation of (5.3.14) simplifies to ⎛ ∂u ρ⎜ 5.44 ⎝ ∂t +u ⎛ ∂ 2u ∂ 2 u ∂ 2u ⎞ ∂p ∂u ⎞ ∂u ∂u ⎜ ⎟ = − +v + ρ g μ +w + + + ⎟ x ⎜ ∂x 2 ∂y 2 ∂z 2 ⎟ ∂z ⎠ ∂x ∂x ∂y ⎝ ⎠ ρ ∂u ∂ 2u =μ 2 ∂t ∂y We assumed g to be in the y-direction, and since no forcing occurs other than due to the motion of the plate, we let ∂p /∂x = 0. From Eqs. 5.3.10, − 5.45 τ xx + τ yy + τ zz ⎛ 2μ ⎞ ∴p = p −⎜ +λ⎟ ∇⋅V ⎝ 3 ⎠ 3 = p− 2μ ⎛ ∂u ∂v ∂w ⎞ ⎜ + + ⎟− λ ∇⋅V 3 ⎝ ∂x ∂y ∂z ⎠ ⎛ 2μ ⎞ ∴p − p = −⎜ +λ⎟ ∇⋅V ⎝ 3 ⎠ Vorticity ⎛ ∂ ∂ ∂ ⎞ (V ⋅ ∇ )V = ⎜ u + v + w ⎟ (uˆi + vˆj + wkˆ ) ∂y ∂z ⎠ ⎝ ∂x ⎡ ∂ ⎛ ∂w ∂w ∂w ⎞ ∂ ⎛ ∂v ∂v ∂v ⎞ ⎤ ∇ × (V ⋅ ∇ )V = ⎢ ⎜ u +v +w ⎟ − ⎜ u + v + w ⎟ ⎥ ˆi ∂y ∂z ⎠ ∂z ⎝ ∂x ∂y ∂z ⎠ ⎦⎥ ⎣⎢ ∂y ⎝ ∂x ⎡ ∂ ⎛ ∂u ∂u ∂u ⎞ ∂ ⎛ ∂w ∂w ∂w ⎞ ⎤ ˆ + ⎢ ⎜u + v + w ⎟ − ⎜u +v +w ⎟⎥ j ∂y ∂z ⎠ ∂x ⎝ ∂x ∂y ∂z ⎠ ⎦⎥ ⎣⎢ ∂z ⎝ ∂x 5.46 ⎡ ∂ ⎛ ∂v ∂v ∂v ⎞ ∂ ⎛ ∂u ∂u ∂u ⎞ ⎤ + ⎢ ⎜ u + v + w ⎟ − ⎜ u + v + w ⎟ ⎥ kˆ ∂y ∂z ⎠ ∂y ⎝ ∂x ∂y ∂z ⎠ ⎥⎦ ⎢⎣ ∂x ⎝ ∂x Use the definition of vorticity: ω = ( ∂w ∂v ˆ ∂u ∂w ˆ ∂v ∂u ˆ ) j + ( − )k : − )i + ( − ∂y ∂z ∂z ∂x ∂x ∂y ⎡ ∂w ∂v ∂ ∂u ∂w ∂ ∂v ∂x ∂ ⎤ − ) +( − ) + ( − ) ⎥ (uˆi + vˆj + wkˆ ) (ω ⋅ ∇ ) V = ⎢ ( ⎣ ∂y ∂z ∂x ∂z ∂x ∂y ∂x ∂y ∂z ⎦ ⎡ ∂ ∂ ∂ ⎤ ⎡ ∂w ∂v ˆ ∂u ∂w ˆ ∂v ∂u ˆ ⎤ ) j + ( − )k ⎥ − )i + ( − ( V ⋅ ∇ )ω = ⎢ u + v + w ⎥ ⎢ ( ∂y ∂z ⎦ ⎣ ∂y ∂z ∂z ∂x ∂x ∂y ⎦ ⎣ ∂x 73 © 2012 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 5 / The Differential Forms of the Fundamental Laws Expand the above, collect like terms, and compare coefficients of ˆi , ˆj, and kˆ . Studying the vorticity components of Eq. 3.2.21, we see that ωz = −∂u/∂y is the only vorticity component of interest. The third equation of Eq. 5.3.24 then simplifies to Dωz = ν ∇ 2ωz Dt 5.47 =ν ∂ 2ωz ∂y 2 since changes normal to the plate are much larger than changes along the plate, i.e., ∂ωz ∂ω >> z ∂y ∂x If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces to Dω z =0 Dt that is, there is no change in vorticity (along a streamline) between sections 1 and 2. Since (see Eq. 3.2.21), at section 1 ωz = ∂v ∂u − = −10 ∂x ∂y ∂u = 10. This ∂y means the velocity profile at section 2 is a straight line with the same slope of the profile at section 1. Since we are neglecting viscosity, the flow can slip at the wall with a slip velocity u0 ; hence, the velocity distribution at section 2 is u 2 ( y ) = u 0 + 10 y . Continuity then allows us to calculate the profile: we conclude that, for the lower half of the flow at section 2, 5.48 V1 A1 = V2 A2 1 (10 × 0.04)(0.04w) = (u0 + 10 × 0.02 / 2)(0.02w) 2 ∴ u0 = 0.3 m/s Finally u2 ( y) = 0.3 + 10y 5.49 No. The first of Eqs. 5.3.24 shows that Dωx ∂u ∂u ∂u = ωx + ωy + ωz ∂x ∂y ∂z Dt 74 © 2012 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 5 / The Differential Forms of the Fundamental Laws neglecting viscous effects, so that ω y , which is nonzero near the snow surface, creates ω x through the term ωy ∂u/∂y, since there would be a nonzero ∂u/∂y near the tree. Differential Energy Equation ∫ K∇T ⋅ nˆ dA = c.s. c.v. ∫ ∫ ∇ ⋅ (K∇T )d V = c.v. 5.50 ∴ ⎞ ∂ ⎛V2 + gz + u ⎟ρ d V + ⎜⎜ ⎟ ∂t ⎝ 2 ⎠ ∫ c.v. ⎛V2 p⎞ + gz + u + ⎟ ρ V ⋅ nˆ dA ⎜⎜ ρ ⎟⎠ 2 c.s. ⎝ ∫ ⎞ ∂ ⎛ V2 + ρ gz + ρ u ⎟d V + ⎜⎜ ρ ⎟ ∂t ⎝ 2 ⎠ ⎛V2 p⎞ ∇ ⋅ ρV ⎜ + gz + u + ⎟ d V ⎜ 2 ρ ⎟⎠ ⎝ c.v. ∫ ⎡ ⎞ ⎛V2 p ⎞⎤ ∂ ⎛ V2 2 + ρ gz + ρ u ⎟ + ∇ ⋅ ρ V ⎜ + gz + u + ⎟ ⎥ d V = 0 ⎢ −K∇ T + ⎜⎜ ρ ⎟ ⎜ 2 ∂t ⎝ 2 ρ ⎟⎠ ⎥⎦ ⎢ ⎠ ⎝ c.v. ⎣ ∫ ⎛V2 ⎞ p ⎞ V 2 ⎛ ∂ρ ⎡ ∂V ∇p ⎤ ∂ V2 + ∇ ⋅ ρV ⎜ + gz + ⎟ = + ∇ ⋅ ρV ⎟ + ρV ⋅ ⎢ + V ⋅ ∇V + + g∇z ⎥ = 0 ρ ⎜ ⎜ ⎟ t 2 2 2 ρ ∂t ∂ ρ ∂ t ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ continuity momentum ∴−K∇ 2T + ∂ ρ u + ρV ⋅ ∇u = 0 ∂t ρ or Du = K∇ 2T Dt Divide each side by dx dy dz and observe that 5.51 ∂T ∂x − x + dx dx ∂T ∂x x = ∂ T ∂T ∂y 2 ∂x 2 , − y + dy ∂T ∂y y dy = ∂ T 2 ∂x 2 , ∂T ∂z − z + dz ∂T ∂z z dz = ∂ 2T ∂z 2 Eq. 5.4.5 follows. ρ 5.52 D ( h − p /ρ ) Du Dh Dp p D ρ Dh Dp p =ρ =ρ − + =ρ − + [ − ρ∇ ⋅ V ] Dt Dt Dt Dt ρ Dt Dt Dt ρ where we used the continuity equation: D ρ /Dt = − ρ∇ ⋅ V. Then Eq. 5.4. 9 becomes ρ Dh Dp p − + [ − ρ∇ ⋅ V ] = K∇ 2T − p∇ ⋅ V Dt Dt ρ which is simplified to ρ Dp Dh = K∇ 2T + Dt Dt 75 © 2012 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 5 / The Differential Forms of the Fundamental Laws See Eq. 5.4.9: u = cT 5.53 ⎛ ∂T ∂T ∂T ∂T ⎞ 2 ∴ ρc ⎜ +u +v +w ⎟ = K∇ T ∂x ∂y ∂z ⎠ ⎝ ∂t Neglect terms with velocity: ρc ∂T = K∇ 2T ∂t The dissipation function Φ involves viscous effects. For flows with extremely large velocity gradients, it becomes quite large. Then ρ cp 5.54 and DT is large. This leads to very high temperatures on reentry vehicles. Dt u = 10(1 − 10, 000 r 2 ) 5.55 DT =Φ Dt ∴ ∂u = −2r × 105 ∂r (r takes the place of y ) ⎡ 1 ⎛ ∂u ⎞ 2 ⎤ From Eq. 5.4.17, Φ = 2 μ ⎢ ⎜ ⎟ ⎥ = μ × 4r 2 × 1010 ⎢ 2 ⎝ ∂y ⎠ ⎥ ⎣ ⎦ Φ = 1.8 × 10−5 × 4 × 0.012 × 1010 = 72 N/m2 ⋅ s At the wall where r = 0.01 m, At the centerline: ∂u = 0 so Φ = 0 ∂r At a point half-way: Φ = 1.8 × 10−5 × 4 × 0.0052 × 1010 = 18 N/m2 ⋅ s (a) Momentum: 5.56 ∂u ∂ 2u =ν 2 ∂t ∂y ⎛ ∂u ⎞ ∂T ∂ 2T Energy: ρ c = K 2 +μ⎜ ⎟ ∂t ∂y ⎝ ∂y ⎠ 2 ∂u ∂ 2u ∂μ ∂u (b) Momentum: ρ =μ 2 + ∂t ∂y ∂y ∂y ⎛ ∂u ⎞ ∂T ∂ 2T Energy: ρ c = K 2 +μ⎜ ⎟ ∂t ∂y ⎝ ∂y ⎠ 2 76 © 2012 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 6 / Dimensional Analysis and Similitude CHAPTER 6 Dimensional Analysis and Similitude FE-type Exam Review Problems: Problems 6-1 to 6-6 The dimensions on the variables are as follows: L L ML2 ML / T 2 M L [W ] = [ F × V ] = M 2 × = 3 , [ d ] = L, [ Δp ] = = , [V ] = 2 2 T T T T L LT 6.1 (A) First, eliminate T by dividing W by Δp. That leaves T in the denominator so divide by V leaving L2 in the numerator. Then divide by d 2 . That provides π= W ΔpVd 2 V = f ( d , l , g , ω , μ ). The units on the variables on the rhs are as follows: [d ] = L, [l ] = L, [ g ] = 6.2 (A) L T2 , [ω ] = T −1 , [ μ ] = ML T Because mass M occurs in only one term, it cannot enter the relationship. 6.3 (A) Re m = Re p 6.4 (A) Rem = Re p Vm Lm νm Vm Lm νm = = Vp Lp ∴Vm = V p νp Vp Lp νp ∴Vm = V p 6.5 (C) Frm = Fr p Lm ν p ∴Vm = V p From Froude’s number Vm = V p 6.6 (A) Fm* = Fp* or Fm ρmVm2lm2 = Fp ρ pV p2l 2p Lm Lp ν m 2 Vp Vm2 = lm g m l p g p Lp = 12 × 9 = 108 m/s = 4 × 10 1.51×10−5 = 461 m/s 1.31× 10−6 lm 1 = 2 × = 0.5 m/s lp 4 lm . From the dimensionless force we have: lp ∴ Fp = Fm V p2 l 2p Vm2 lm2 = 10 × 25 × 252 = 156,000 N 77 © 2012 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 6 / Dimensional Analysis and Similitude Introduction a) [ m ] = kg/s = N ⋅ s2 /m ⋅ s = N ⋅ s/m 6.8 ∴ c) [ ρ ] = kg/m = N ⋅ s /m ⋅ m = N ⋅ s /m 3 2 e) [W ] = N ⋅ m 3 2 FT L 4 FT 2 ∴ 4 L ∴ FL Dimensional Analysis [V ] = V = f (A, ρ , μ ) 6.10 L M M , [ A ] = L, [ ρ ] = 3 , [ μ ] = T LT L ∴ π1 = f1 (π 20 ) = Const. [V ] = V = f ( H , g , m) 6.12 ρV A μ ∴There is one π − term: π1 = gHm0 ∴ π1 = V2 FD = f (d, A, V , μ , ρ ) ∴ρ L , T VA μ [g] = ∴ π1 = C [ FD ] = =C or Re = Const. L , T2 [ m] = M , [ H ] = L ∴V = gH / C ML L M M d L V , = , = , μ = , ρ = [ ] [ ] [ ] [ ] T LT T2 L3 π1 = FD A a1V b1 ρ c1 , π 2 = dV b2 ρ c2 A a2 , π 3 = μ A a3V b3 ρ c3 ∴ π1 = 6.14 ∴ FD d μ , π2 = , π3 = 2 A ρV A V ρA 2 ⎛d μ ⎞ FD = f1 ⎜ , ⎟ 2 2 ρA V ⎝ A ρ AV ⎠ We could write ⎛ 1 π ⎞ ⎛A μ ⎞ π1 FD = f 2 ⎜ , 3 ⎟ or = f2 ⎜ , ⎟ . This is 2 2 2 ρd V ⎝ d ρdV ⎠ ⎝π2 π2 ⎠ π2 equivalent to the above. Either functional form must be determined by experimentation. 6.16 h = f (σ , d, γ , β , g). [ h ] = L, [σ ] = M M L , d = L, [γ ] = 2 2 , [ β ] = 1, [ g ] = 2 Se 2 [ ] T LT T 78 © 2012 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 6 / Dimensional Analysis and Similitude lect d , γ , g as repeating variables: π 1 = hd a1 γ b1 g c1 , π 2 = σ d a2 γ b2 g c2 , π 3 = β ∴π1 = h/d, π 2 = σ /γ d2 , π 3 = β ∴ h /d = f1 (σ /γ d 2 , β ). Note: gravity does not enter the answer. σ = f ( M , y, I ) [σ ] = 6.18 M ML2 , M = , [ y ] = L, [ I ] = L4 2 [ ] 2 LT T Given that b = −1 , π1 = V = f ( H , g, ρ ) 6.20 [V ] = ∴π1 = VH a g b ρ c = V σI yM = Const. ∴σ = C My I L L M , [ H ] = L, [ g ] = 2 , [ ρ ] = 3 T T L ρ0 g H = Const. ∴ V = Const. gH Density does not enter the expression. Δp = f (V , d ,ν , L, ε , ρ ) [ Δp ] = M L L2 M V d L , = , = , ν = , [ L ] = L, [ e ] = L, [ ρ ] = 3 [ ] 2 [ ] T [ ] T LT L Repeating variables: V , d , ρ 6.22 π1 = ΔpV a1 db1 ρ c1 , π 2 = ν V a2 db2 ρ c2 , π 3 = L V a3 db3 ρ c3 , π 4 = e V a4 db4 ρ c4 By inspection: ∴ π1 = Δp ρV 2 π1 = f1(π 2 , π 3 , π 4 ) Q = f ( R, A, e, S , g ). [Q ] = 6.24 , π2 = ν Vd , π3 = L e , π4 = d d ∴Δp/ρ V 2 = f1(ν /Vd, L/d, e/d) L3 L , [ R ] = L, [ A] = L2 , [ e ] = L, [ s ] = 1, [ g ] = 2 T T There are only two basic dimensions. Choose two repeating variables, R and g. Then π 1 = QR a1 g b1 , π 2 = AR a2 g b2 , π 3 = eR a3 g b3 , π 4 = sR a4 g b4 ∴ π1 = Q A e , π = , π = , π4 = s 2 3 R g R5/ 2 R2 79 © 2012 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 6 / Dimensional Analysis and Similitude ∴π1 = f1 (π 2 , π 3 , π 4 ) ∴ Q/ gR5 = f1 ( A/R2 , e/R, s) FD = f (V , μ , ρ , e, I , d) Repeating variables: V , ρ , d [ FD ] = 6.26 ML L M M , [V ] = , [ μ ] = , [ ρ ] = 3 , [ e ] = L, [ I ] = 1, [ d ] = L 2 T LT T L π 1 = FDV a1 ρ b1 d c1 , π 2 = μV a2 ρ b2 d c2 , π 3 = e V a3 ρ b3 d c3 , π 4 = I V a4 ρ b4 d c4 ∴ π1 = FD μ e , π2 = , π3 = , π 4 = I 2 2 V ρd d ρV d ∴ FD /ρ V 2d 2 = f1 ( μ /ρ Vd, e /d, I ) T = f ( ρ , μ , V , D) = ρ μ V D a b 6.28 c d ML or T2 a b c ⎡M ⎤ ⎡ M ⎤ ⎡ L⎤ =⎢ 3⎥ ⎢ [ L]d ⎥ ⎢ ⎥ ⎣ L ⎦ ⎣ LT ⎦ ⎣ T ⎦ M : 1 = a + b ⇒ a = 1− b T : − 2 = −b − c ⇒ c = 2 − b L: 1 = −3a − b + c + d ⇒ 1 = −3(1 − b) − b + (2 − b) + d ⎛ μ ⎞ 1−b 2−b 2−b ∴ d = 2 − b. T = ρ ( ) μ bV ( ) D( ) = ρ V 2 D 2 ⎜ ⎟ ⎝ ρ VD ⎠ b and T ρ V 2 D2 = φ [ Re] FD = f (V , μ , ρ, d , e, r, c ). Repeating variables: V , ρ , d [ FD ] = ML L M M 1 = μ = ρ = = = = = , [ V ] , [ ] , [ ] , [ d ] L , [ e ] L , [ r ] L , [ c ] T LT T2 L3 L2 π 1 = FDV a ρ b d c , π 2 = μV a ρ b d c , π 3 = eV a ρ b d c , 1 1 1 2 2 2 3 3 3 π 4 = rV a ρ b d c , π 5 = cV a ρ b d c 4 6.30 ∴ π1 = ∴ 4 4 5 5 5 FD μ e r , π2 = , π 3 = , π 4 = , π 5 = cd 2 2 2 ρVd d d ρV d ⎛ μ e r ⎞ FD = f1 ⎜ , , , cd 2 ⎟ 2 2 ρV d ⎝ ρVd d d ⎠ FL = f (V , c , ρ , A c , t , α ). Repeating variables: V , ρ , A c 6.32 [ FL ] = ML L L M , [V ] = , [c ] = , [ ρ ] = 3 , [A c ] = L, [t ] = L, [α ] = 1 2 T T T L π1 = FLV a1 ρ b1 A c c1 , π 2 = cV a2 ρ b2 A c c2 , π 3 = tV a3 ρ b3 A c c3 , π 4 = αV a4 ρ b4 A c c4 80 © 2012 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 6 / Dimensional Analysis and Similitude ∴π1 = ∴ FL c t , π = , π = , π4 = α 2 3 Ac V ρV 2A c2 ⎛c t ⎞ FL = f1 ⎜ , , α ⎟ 2 2 ρV A c ⎝ V Ac ⎠ FD = f (V , ρ , μ , d , L , ρ c , ω ) where d is the cable diameter, L the cable length, ρ c the cable density, and ω the vibration frequency. Repeating variables: V , d , ρ. The π − terms are π1 = 6.34 FD ρV d 2 2 , π2 = Vd ρ μ , π3 = d V ρ , π4 = , π5 = L ρc ωd We then have ⎛ Vdρ d ρ FD V ⎞ = f1 ⎜ , , , ⎟ 2 2 L ρ c ωd ⎠ ρV d ⎝ μ T = g( f , ω , d , H , A, N , h, ρ ). Repeating variables: ω , d , ρ [T ] = 6.36 π1 = ML2 1 1 M , [ f ] = , [ω ] = , [ d ] = L, [ H ] = L, [A] = L, [ N ] = 1, [ h] = L, [ ρ ] = 3 2 T T T L T ρω d 2 5 ∴ , π2 = f ω , π3 = H h A , π4 = , π5 = N, π6 = d d d h⎞ ⎛ f H A = g1 ⎜ , , , N, ⎟ d⎠ ρω d ⎝ω d d T 2 5 d = f (V, Vj , D, σ , ρ, μ, ρ a ). Repeating variables: V j , D, ρ [ d ] = L, [V ] = 6.38 π1 = L L M M M M , [V j ] = , [ D ] = L, [σ ] = 2 , [ ρ ] = 3 , [ μ ] = , [ρa ] = 3 T T LT T L L d V σ μ ρ , π 2 = , π3 = , π4 = , π5 = a 2 ρV j D ρ D Vj ρV j D ∴ ⎛V d σ μ ρa ⎞ ⎟ = f1 ⎜ , , , ⎜ V j ρV j2 D ρV j D ρ ⎟ D ⎝ ⎠ μ = f ( D, H , A, g, ρ, V ) . D = tube dia., H = head above outlet, A = tube length. 6.40 Repeating variables: D, V , ρ π1 = μ ρVD , π2 = H A gD , π3 = , π 4 = 2 D D V 81 © 2012 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 6 / Dimensional Analysis and Similitude ∴ μ ⎛ H A gD ⎞ = f1 ⎜ , , ⎟ ρVD ⎝ D D V2 ⎠ y 2 = f (V1 , y1 , ρ, g). Neglect viscous wall shear. [ y2 ] = L, [V1] = L M L , [ y1] = L, [ ρ ] = 3 , [ g ] = 2 T L T Repeating variables: V1, y1, ρ 6.42 π1 = ∴ y2 gy , π 2 = 21 y1 V1 ( ρ does not enter the problem) ⎛ gy ⎞ y2 = f ⎜ 21 ⎟ ⎜V ⎟ y1 ⎝ 1 ⎠ Similitude Qm Vm A 2m Δpm ρ m Vm2 ( Fp ) m ρ m Vm2 A 2m = , = , = ρ p Vp2 ( Fp ) p ρ p Vp2 A 2p Qp Vp A 2p Δp p 6.44 Q m ρ m Vm3 A 2m = Q p ρ p Vp3 A 2p τ m ρ m Vm2 Tm ρ m Vm2 A 3m , , = = τ p ρ p Vp2 Tp ρ p Vp2 A 3p (Q has same dimensions as W ) a) Rem = Re p 6.46 Vmdm νm = Vpd p ∴ νp Vm d p = =5 V p dm m m ρ mA 2mVm 1 1 = = 2 × 5 ∴ m m = m p = 800 / 5 = 160 kg/s 2 m p ρ p A pV p 5 5 Δpm ρ mVm2 = = 52 ∴Δpm = 25Δp p = 25 × 600 = 15,000 kPa 2 Δp p ρ pV p Rem = Re p 6.48 VmA m νm = V pA p νp ∴ Vm A p ν m ν = = 10 assuming m = 1 Vp A m ν p νp ∴Vm = 10V p = 1000 km/hr This velocity is much too high for a model test; it is in the compressibility region. Thus, small-scale models of autos are not used. Full-scale wind tunnels are common. 82 © 2012 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 6 / Dimensional Analysis and Similitude Properties of the atmosphere at 8 km altitude: T = −37°C +273 = 236 K and pressure = 35.7 kPa, density = 0.526 kg/m3, and viscosity = 1.527 × 10−5 N·s/m2 Properties of air at standard atmosphere: T = 20°C, p = 101.325 kPa, density = 1.204 kg/m3, dynamic viscosity = 1.82 × 10−5 N·s/m2 ⎛ ρVD ⎞ ⎛ ρVD ⎞ Use the Reynolds number to achieve dynamic similarity, ⎜ ⎟ =⎜ ⎟ ⎝ μ ⎠m ⎝ μ ⎠ p 6.50 ⎛ ρD ⎞ ⎛ μ ⎞ Then Vm = Vp ⎜ ⎟ ⎜ ⎟ ⎝ μ ⎠ p ⎝ ρ D ⎠m −5 ⎛ ρ p μ m ⎞ ⎛ Dp ⎞ ⎛ 0.526 ⎞ ⎛ 1.82 ×10 ⎞ ⎛ 10 ⎞ = = 1041 km/hr and Vm = Vp ⎜ 200 ⎟ ⎜⎜ ⎜ ⎟ ⎜ ⎟⎜ −5 ⎟ ⎟ ⎝⎜ 1 ⎠⎟ ⎜ ρ m μ p ⎟ ⎝ Dm ⎟⎠ 1.204 ⎝ ⎠ × 1.527 10 ⎝ ⎠ ⎝ ⎠ ⎛ T ⎞ ⎛ T ⎞ To calculate the thrust apply: ⎜ = ⎟ ⎜ 2 2 2 2⎟ ⎝ ρ V D ⎠ p ⎝ ρ V D ⎠m ⎛ ρ p Vp2 Dp2 ⎞ ⎛ 0.526 ×10412 ×102 ⎞ ⎟ Then Tp = Tm ⎜ 10 = = 1184 N ⎜ 2 2 ⎟ ⎜ ρmVm2 Dm2 ⎟ 1.204 200 1 × × ⎝ ⎠ ⎝ ⎠ Rem = Re p Vmdm νm = Vpd p νp 10−6 ∴ dm = d p = 0.75 × 1× Vm ν p 5 × 10−3 Vp ν m = 0.0015 m = 1.5 mm 6.52 Δpm ρmVm2 1000 = = × 12 = 1.11 2 Δp p ρ pV p 1000 × 0.9 Find ν oil using Fig. B.2. Then Re m = Re p ∴ 6.54 VmA m νm = V pA p νp Vm A p ν m ν 1 = = 30 m = Vp A m ν p νp 30 2 Frm = Frp ∴ ∴ νm 1 = ν p 164 ∴ A Vm = m Ap Vp 2 Frm = Frp Vp Vm2 = A m gm A p g p Vp 1 Vm2 V = ∴ m = 30 Vp A m gm A p g p ∴ν m = 6.1× 10−9 m 2 /s Impossible! 6.56 a) Qm Vm A 2m = Qp V p A 2p ∴ Qm = Q p Vm A 2m 1 1 = 2× × 2 = 0.00632 m 3 /s 2 Vp A p 10 10 83 © 2012 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 6 / Dimensional Analysis and Similitude Fm ρmVm2A2m b) = Fp ρ pV p2A2p ∴ Fp = Fm V p2 A2p Vm2 A2m = 12 ×10 × 102 = 12,000 N Neglect viscous effects, and account for wave (gravity) effects. 2 Frm = Frp 6.58 ∴ Vp Vm2 = A m gm A p g p ∴ ωm = ω p Tm ρmVm2A3m = Tp ρ pV p2A3p ∴ Vm A = m Vp Ap ωm Vm / A m = ω p Vp / A p Vm A p 1 = 600 × × 10 = 1897 rpm Vp A m 10 ∴Tp = Tm V p2 A3p Vm2 A3m = 1.2 × 10 × 103 = 120,000 N ⋅ m Check the Reynolds number: Re p = V pdp /ν p = 15 × 2/10 −6 = 30 × 106 This is a high-Reynolds-number flow Re m = 6.60 2 × 2/30 10 −6 = 1.33 × 105 This may be sufficiently large for similarity. If so Wm ρmVm3A 2m 23 1 = = 3 × 2 = 2.63 × 10−6 3 2 W p ρ pV p A p 15 30 ∴W p = (2 × 2.15) / 2.63 × 10−6 = 1633 kW Re p = 20 × 10 = 13.3 × 10 6 . This is a high-Reynolds-number flow. −5 1.5 × 10 For the wind tunnel, let Rem = 105 = Vm × 0.4/1.5 × 10−5 For the water channel, let Rem = 105 = Vm × 0.1/1× 10−6 6.62 ∴ Vm ≥ 3.75 m/s ∴ Vm ≥ 1.0 m/s Either could be selected. The more convenient facility would be chosen. Fm1 Fm2 = ρ m1 Vm21 A 2m1 ρ m2 Vm22 A 2m2 3.2 = Fm2 W ρ mVm3 A 2m 153 × 0.42 m = ρ V 3A 2 = 203 × 102 W p p p p ∴ Fm2 1000 2.42 0.12 = 3.2 × = 4.16 N 1.23 152 0.42 203 102 ∴ Wp = (15 × 3.2) 3 × = 71,100 W or 95 hp 15 0.42 84 © 2012 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 6 / Dimensional Analysis and Similitude Mach No. is the significant parameter: M m = M p 6.64 b) V p = Vm cp cm Fp = Fm = Vm Tp Tm ρ mVp2A 2p ρ mVm2 A 2m = 10 × 0.601 × Vp2 Vm2 = A m gm A p g p a) Frm = Frp 255.7 = 186 m/s 296 = 200 ∴ 1862 × 202 = 2080 N 2 200 Vm A = m Vp Ap ωm Vm A p 1 = = × 10 ω p Vp A m 10 6.66 VmA m b) Rem = Re p = νm V pA p νp ∴ ∴ωm = 2000 × Vm A p = = 10 Vp A m ωm Vm A p 1 = = 10 × = 1 ω p Vp A m 10 Rem = Re p ∴ VmA m νm = VpA p ∴ωm = 2000 rpm ∴ Vm = Vp νp 10 = 6320 rpm 10 Ap Am = 10 × 10 = 100 m/s This is too large for a water channel. Undoubtedly this is a high-Re flow. Select a speed of 5 m/s. For this speed 6.68 Rem = 5 × 0.1 = 5 × 105 −6 1× 10 where we used A m = 0.1 (A p = 1 m, i.e., the dia. of the porpoise) ωm = ωp (Vm/Vp )(A p/A m ) = 1× (5/10) ×10 = 5 motions/s Normalized Differential Equations 6.70 ρ* = y ρ * u v x , t = tf , u* = , v* = , x* = , y* = . Substitute in: A A ρ0 V V f ρ0 ∂ρ * V ∂( ρ *u* ) V ∂(ρ *v* ) ρ ρ + + =0 0 0 A ∂x* A ∂y* ∂t* Divide by ρ0 V /A : 85 © 2012 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 6 / Dimensional Analysis and Similitude ∴ f A ∂ρ * ∂ ∂ + * ( ρ *u* ) + * ( ρ *v* ) = 0 * V ∂t ∂x ∂y V* = Parameter = fA V V * tU p h , t = , ∇* = A ∇ , p * = , h* = . Euler’s equation is then 2 U A A ρU U 2 DV* U2 * * A ρ = −ρ ∇ p − ρ g ∇*h* * A Dt A A 6.72 Divide by ρU 2 /A : DV* gA = −∇* p* − 2 ∇*h* * Dt U Parameter = gA U2 The only velocity component is u. Continuity then requires that ∂u/∂x = 0 (replace z with x and vz with u in the equations written using cylindrical coordinates). The x-component Navier-Stokes equation is ⎛ ∂ 2u 1 ∂u 1 ∂ 2u ∂ 2u ∂u ∂u vθ ∂u ∂u 1 ∂p =− + gx +ν ⎜ 2 + + vr + +u + + ⎜ ∂r ρ ∂x ∂t ∂r r ∂θ ∂x r ∂r r 2 ∂θ 2 ∂x 2 ⎝ This simplifies to ⎛ ∂ 2u 1 ∂u ⎞ 1 ∂p ∂u =− +ν ⎜ 2 + ⎟ ⎜ ∂r ∂t ρ ∂x r ∂r ⎟⎠ ⎝ 6.74 b) Let u* = u/V , x* = x /d, t* = tν /d 2 , p* = p /ρ V 2 and r* = r /d : ρ V 2 ∂p* ν V ⎛ ∂ 2u* 1 ∂u* ⎞ + + ⎜ ⎟ ρ d ∂x* d2 ⎜⎝ ∂r*2 r* ∂r* ⎟⎠ d2 ∂t* The normalized equation is ν V ∂u* ∂u* ∂t u* = 6.76 u , U * =− = − Re v* = ∂p* ∂x * v , U + ∂ 2u* ∂r T* = *2 + 1 ∂u* r ∂r T , T0 * * x x* = , A where Re = y* = y , A Vd ν ∇*2 = A 2∇ 2 ⎡UT0 ∂T * UT0 ∂T * ⎤ K *2 * ρc p ⎢ + ⎥ = 2 T0∇ T * * A ∂y ⎥⎦ A ⎢⎣ A ∂x Divide by ρ c pUT0 /A : ∂T * ∂T * K + * = ∇*2T * * c U ρ A ∂x ∂y p Parameter = K μ 1 1 = μ c p ρU A Pr Re 86 © 2012 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 7 / Internal Flows CHAPTER 7 Internal Flows FE-type Exam Review Problems: Problems 7-1 to 7-12 7.1 (D) 7.2 (A) 7.3 (D) The flow in a pipe may be laminar at any Reynolds number between 2000 and perhaps 40,000 depending on the character of the flow. 7.4 (D) The friction factor f depends on the velocity (the Reynolds number). 7.5 (A) L V2 Δp = f D 2g 7.6 (B) e 0.15 = = 0.0075 ∴ Moody's diagram gives, assuming Re > 105 D 20 15 V 2 f = 0.034 Then 60,000 = 0.034 × 0.02 2 × 9.81 ∴V = 6.79 m/s and Q = AV = π × 0.012 × 6.79 = 0.00214 m3 /s Check the Reynolds number: Re = 7.7 (D) 6.79 × 0.02 10 −6 e 0.26 = = 0.00325 D 80 Assume Re > 3 × 105 hL 1 V2 = sin θ = f L D 2g sin 30D = 0.026 Check Re: Re = VD ν = 5.49 × 0.08 10 −6 = 1.36 ×105 ∴ OK Then f = 0.026 1 V2 0.08 2 × 9.81 = 4.39 ×105 ∴V = 5.49 m/s ∴ OK e 0.26 4 × 0.06 = = 0.0043 Re = = 3 × 10 4 ∴ f = 0.033 from Moody's diagram −6 D 60 8 × 10 7.8 (B) 20 42 L V2 Δp = −γ f + γ Δh = −9810 × 0.033 × + 9810 × 20 = 108,000 Pa 0.06 2 × 9.81 D 2g 7.9 (A) R= A 4× 4 Q 0.02 = = 1 cm V = = =12.5 m/s P 4× 4 A 0.04 × 0.04 87 © 2012 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 7 / Internal Flows Re = 4RV ν 4 × 0.01×12.5 = 10−6 e 0.046 = = 0.00115 ∴ f = 0.021 4 R 4 ×10 = 5 ×105 L V2 40 12.52 Δp = f = 0.021× × = 167 Pa 4R 2 g 4 × 0.01 2 × 9.81 7.10 (B) Viscous effects (losses) are important. 7.11 (A) A negative pressure must not exist anywhere in a water system for a community since a leak would suck into the system possible impurities. 1 1 0.8 × 2.4 × 0.482 / 3 × 0.0021/ 2 = 4.39 m3 /s AR 2 / 3 S 1/ 2 = 0.012 n Q= 7.12 (C) where we have used R = A Pwetted = 0.8 × 2.4 = 0.48 m 0.8 + 0.8 + 2.4 Laminar or Turbulent Flow 7.14 Re = Vh 7.16 Re = Vh ν ν = = Vh 1×10 b) 1500 = −6 (1/ 2) × 1.4 10 −6 V ×1 ∴V = 0.0015 m/s 1×10−6 ∴Very turbulent = 700,000 Entrance and Developed Flow LE = 0.065 Re D 7.18 a) LE = 0.065 × c) LE = 0.065 × V= 7.20 0.025 π × 0.03 2 Re = VD V= ν 0.1592 × 0.04 1.31×10−6 0.1592 × 0.04 0.661×10−6 = 8.84 Re = π × 0.022 = 0.1592 m/s × 0.04 = 12.6 m × 0.04 = 25.0 m 8.84 × 0.06 1.007 ×10 ∴ LE = 120 × 0.06 = 7.2 m 7.22 0.0002 −6 = 5.3 ×105 ∴Turbulent ∴Developed LE = 0.04 Re × h = 0.04 × 7700 × 0.012 = 3.7 m 88 © 2012 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 7 / Internal Flows ( LE )min = 0.04 ×1500 × 0.012 = 0.72 m Re = 7.24 VD 0.2× 0.04 = = 8000 ν 10−6 a) If laminar, LE = 0.065 × Re × D = 0.065 × 8000 × 0.04 = 20.8 m Li = LE /4 = 20.8/4 = 5.2 m Laminar Flow in a Pipe 7.28 In a developed flow, dh/dx = slope of the pipe and p is a linear function of x so that dp/dx = const. Therefore, d(p + γh)/dx = const and it can be moved outside the integral. Then r d( p + γ h) 0 2 2 1 d( p + γ h) ⎛ r04 r02 2 ⎞ r02 d( p + γ h) ( r r ) rdr − = ⎜⎜ − × r0 ⎟⎟ = 0 ∫ 2 dx dx dx 4 μ r02 2 μ r 0 ⎝ 4 2 ⎠ 8μ 0 2 1500 = VD ν Eq. 7.3.14: 7.30 ∴α = 8ν V r02g = = V × 0.01 ∴ V = 0.0992 m/s 6.61×10−7 Δp Δh 8μV +γ = 2 L L r0 8× 6.61×10−7 × 0.0992 0.005 × 9.81 2 Δh 8μV ∴ = L r02 ρ g α Δh L α = Δh/ L = 0.00214 rad or 0.123D Q = AV = π × 0.0052 × 0.0992 = 7.79 ×10−6 m3 /s 0.01 Eq. 7.3.14: Q= ∫ 0 9810(0.00015) 2 ×10 −3 (0.02 y − y 2 )100dy = 73,600 × (0.01× 0.012 − 7.32 0.013 ) = 0.049 m3/s 3 For a vertical pipe Δh = L. Thus ρ gr02 gr02 9.81× .012 1.226 × 10−4 V= = = = 8μ 8ν 8ν ν b) V = 1.226 × 10−4 = 0.33 m/s ∴ Q = 1.04 × 10−4 m3 /s Re = 17.8 ∴ laminar 0.34/917 89 © 2012 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 7 / Internal Flows Neglect the effects of the entrance region and assume developed flow for the whole length. Also, assume p inlet = γh (neglect V 2 /2 g compared to 4 m). 7.34 ∴Δp = γ h − 0 ∴ ρ gh = 8μ VL ∴V = r02 ghr02 9.81× 4 × 0.00252 = = 0.766 m/s 8ν L 8 ×1×10−6 × 40 ∴ Q = AV = π × 0.00252 × 0.766 = 1.5 × 10−5 m3s LE = 0.065 × Re = 2000 = 7.36 Δp = VD ν Δp = = 8μ VL Re = 40,000 = 7.38 0.766 × 0.005 × 0.005 = 1.2 m 1× 10−6 r02 VD ν V × 0.02 1.5 ×10−5 = = ∴ V = 1.5 m/s 8 ×1.9 × 10−5 × 1.5 × 10 (0.01)2 V × 0.1 = 23 Pa ∴ V = 6.04 m/s ∴ Vmax = 2V = 12.1 m/s 1.51×10−5 8μ VL 8 ×1.81×10−5 × 6.04 ×10 = = 3.5 Pa LE = 0.065 × 40,000 × 0.1 = 260 m r02 0.052 V =− R2 Δp + γΔh R2 γ (− L) γ R2 9800 × 0.0012 =− = = = 1.225 m/s 8μ 8μ L 8μ L 8 ×10−3 Q = AV = π × 0.0012 ×1.225 = 3.85×10−6 m3/s 7.40 Re = VD ν = 122.5 × 00.02 = 2450 10 − 6 It probably is not a laminar flow. Since Re > 2000 it would most likely be turbulent. If the pipe were smooth, disturbance and vibration free, with a wellrounded entrance it could be laminar. ⎛ r2 ⎞ a) Combine Eq. 7.3.12 and 7.3.15: u = umax ⎜ 1 − 2 ⎟ ⎝ r0 ⎠ 7.42 ⎛ r2 ⎞ u(r ) = V = 2V ⎜ 1 − 2 ⎟ ⎜ r ⎟ 0 ⎠ ⎝ ∴r2 = r02 and r = 0.707 r0 2 b) From Eq. 7.3.17: τ = Cr where C is a constant. Then τ w = Cr0 If τ = τ w /2, then τ w /2 = Cr and r = τ w /2C = r0 /2 90 © 2012 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 7 / Internal Flows Re = 20,000 = 7.44 hL = Δp γ = VD ν 8μVL γ r02 = = V × 0.05 1.1×10−6 ∴ V = 0.44 m/s 8 ×1.1×10−3 × 0.44 ×10 9810 × (0.025)2 = 0.0063 m r0 Δp (0.025) × 0.0063 × 9810 = = 0.08 Pa 2L 2 × 10 LE = 0.065 × 20,000 × 0.05 = 65 m τ0 = Q=− See Example 7.2: 7.46 π (−Δp) ⎡ 4 4 (r22 − r12 ) 2 ⎤ ⎢ r2 − r1 − ⎥ 8μ L ⎢⎣ ln(r2 / r1 ) ⎥⎦ ∴Q = 2 2 2 ⎡ −4 4 4 (0.03 − 0.02 ) ⎤ 3 0.03 0.02 − − ⎢ ⎥ = 7.25 × 10 m /s −5 ln(0.03/0.02) ⎦⎥ 8 × 1.81× 10 × 10 ⎣⎢ ∴V = 7.25 ×10−4 Q = = 0.462 m/s A π (0.032 − 0.022 ) π × 10 τ r1 = μ Δp ⎡ r22 − r12 1 ⎤ ∂u =− ⎢ 2r1 − ⎥ ∂r r =r1 4 L ⎢⎣ ln(r2 / r1 ) r1 ⎦⎥ 10 ⎡ 0.032 − 0.022 1 ⎤ =− × ⎢ 2 × 0.02 − ⎥ = 0.0054 Pa 4 × 10 ⎣⎢ ln(0.03/0.02) 0.02 ⎦⎥ From Example 7.2 u(r ) = 1 dp ⎡ 2 2 r22 − r12 r⎤ ln ⎥ ⎢ r − r2 + 4 μ dx ⎣⎢ ln(r1 / r2 ) r2 ⎦⎥ As r1 → 0, ln(r1 / r2 ) → −∞ so that 7.48 ( ) r22 − r12 ln(r / r2 ) → 0 ln(r1 / r2 ) Thus, u(r ) = 1 dp 2 r − r02 4 μ dx As r1 → r2 , r22 − r12 0 = . ∴Differentiate w.r.t r1: ln(r1 / r2 ) 0 Also, ln w here r2 = r0 . See Eq. 7.3.11 −2r1 = −2r12 1/ r1 ⎛ y⎞ y r = ln ⎜ 1 − ⎟ ≅ − , where y = r2 − r r2 r2 ⎝ r2 ⎠ 91 © 2012 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 7 / Internal Flows ∴ u(r ) = 1 dp ⎛ 2 2 2r12 ⎜ r − r2 + 4μ dx ⎜⎝ r2 ⎞ 1 dp ⎡ 2 2r12 2 − + y⎟ = y r y 2 ⎟ 4μ dx ⎢⎢ r2 ⎠ ⎣ ⎤ 1 dp 2 ( y − ay ) y⎥ = ⎥⎦ 4μ dx Laminar Flow Between Parallel Plates There is no pressure gradient ∴Eq. 7.4.13 gives u = V y a The friction balances the weight component. 7.50 τ A = W sin θ a) μ ∂u V 0.2 τ =μ =μ =μ 0.0004 a ∂y 0.2 × 1× 1 = 40 sin 20D 0.0004 dh = sin θ dx Q=− a /2 ∫ 0 = γ sin θ 2 ( y − ay) 2μ y a = 0.012 m γ sin θ 2 50γ sin θ ⎛ a3 a3 ⎞ 25γ a3 ( y − ay)50dy = − sin θ ⎜ − ⎟= 2μ 2μ ⎜⎝ 24 8 ⎟⎠ 12μ 25 12 × 10 ∴V = 7.52 Hence, u( y) = − W ∴ μ = 0.0274 N ⋅ s/m 2 The depth of water is a/2 with the maximum velocity at the surface: − τA −3 × 9810 × sin 20D × 0.0123 = 12.1 m 3 /s 12.1 = 40.3 m/s 0.006 × 50 ∴ Re = Va/2 ν = 40.3 × 0.006 10−6 = 241,000 The assumption of laminar flow was not a good one, but we shall stay with it to answer the remaining parts. umax 9810 × sin 20D ⎛ 0.0122 ⎞ = ⎜ ⎟ = 60.4 m/s 2 × 10−3 ⎜⎝ 4 ⎟⎠ τ0 = μ ∂u γ sinθ 9810sin 20D =− (−a) = × 0.012 = 20.1 Pa ∂y y=0 2 2 Obviously the flow would be turbulent and the above analysis would have to be modified substantially for an actual flow. 92 © 2012 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 7 / Internal Flows Δp = Eq. 7.4.17: 12 μ VL a2 ∴V = 50 × 0.022 12 × 1.81×10−5 × 60 = 1.53 m/s ∴ Q = AV = 0.02 × 0.9 ×1.53 = 0.028 m3/s 7.54 This is maximum since laminar flow is assumed. Check the Reynolds number: Re = Va ν = 1.53 × 0.02 1.51× 10−5 = 2030 This is marginally high. Care should be taken to eliminate vibrations, disturbances, or rough walls. 7.56 Assume laminar: −3 12 μ VL 6 12 × 9.4 × 10 V × 0.1 Δp = ∴ 4.2 × 10 = ∴ V = 93 m/s (0.5 × 10−3 ) 2 a2 ∴Q = AV = (0.0005 × 0.1) × 93 = 4.6 ×10−3 m3/s = 4.6 L/s u= 7.58 1 dp 2 U du 1 dp U ( y − ay ) + y τ = μ = (2 y − a) + μ 2μ dx a dy 2 dx a a) τ y =0.006 = 0.006 c) Q = ∫ 0 1 U (−20)(0.006) + 1.95 × 10−5 =0 2 0.006 ∴U = 18.5 m/s 1 U ⎡ (−20)( y 2 − 0.006 y ) + ⎢ −5 0.006 ⎣ 2 × 1.95 × 10 ⎤ y ⎥dy ∴U = −6.15 m/s ⎦ a) The solution for laminar flow between two parallel plates is given in Eq. (7.4.13) as u( y) = ( ) 1 d U ( p + γ h ) y2 − ay + y 2 μ dx a For a horizontal flow dh/dx = 0, and hence the velocity profile is given by u( y) = 7.60 ( ) 1 dp 2 U y − ay + y 2 μ dx a b) To determine the pressure gradient we set u = 0 at y = a/2, that is 0= 1 dp ⎛ a2 a2 ⎞ U ⎜ − ⎟+ 2 μ dx ⎝ 4 2 ⎠ a ⎛a⎞ ⎜ ⎟ ⎝2⎠ Simplifying and solving for the pressure gradient we get ( dp = 4 μU a2 = 4 0.4 N ⋅ s/m2 dx ) ( 2 m/s ) ( 0.01m ) 2 = 32 kPa/m 93 © 2012 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 7 / Internal Flows Note that since the pressure gradient is positive in the flow direction, it is considered to be an adverse pressure gradient. U y a vθ = 7.62 ∴τ = μ dvθ 0.2 × 30 = 0.1× = 750 Pa dy 0.0008 T = F × R = τπ D × L × R = 750π × 0.4 × 0.8 × 0.2 = 151 N ⋅ m Neglect the shear on the cylinder bottom; assume a linear velocity profile: vθ = 7.64 U 0.1× 30 y= y = 3000 y a 0.001 ∴τ = μ dvθ = 0.42 × 3000 = 1260 Pa dy ∴T = F × R = π DL ×τ × R = π × 0.2 × 0.1×1260 × 0.1 = 7.9 N ⋅ m Assume that all losses occur in the 8-m-long channel. The velocity through the straws and screens is so low that the associated losses will be neglected. Assume a developed flow in the channel: V= 7.66 Re×ν 7000 ×1.5 × 10−5 = = 8.75 m/s 0.012 h Δp = 12 × 1.8 × 10 −5 × 8.75 × 8 = 105 Pa 0.012 2 Energy: Δp ΔpAV = Δp m = W ( ρ AV ) = fan ρη = ρη η 105 × 1.2 × 0.012 × 8.75 = 18.9 W 0.7 Laminar Flow Between Rotating Cylinders Use Eq. 7.5.19: T= 4πμ r12r22 Lω1 r22 − r12 = 4π × 0.035 × 0.022 × 0.032 × 0.4 × (3000 × 2π /60) ∴ T = 0.040 N ⋅ m 7.68 Re = 0.032 − 0.022 = Tω = 0.04 × (3000 × 2π /60) = 12.6 W ∴W ω r1 (r2 − r1 ) ρ (3000 × 2π /60) × 0.02 × (0.01) × 917 = = 1650 μ 0.035 ∴Eq. 7.5.15 is OK 94 © 2012 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 7 / Internal Flows r22ω2 ⎛ r12 ⎞ d ⎛ vθ ⎞ 2 μ r12ω2 With ω1 = 0, Eq. 7.5.15 is vθ = 2 2 ⎜ r − ⎟ τ 2 = μ r ⎜ ⎟ = 2 2 r2 − r1 ⎝ r ⎠ dr ⎝ r ⎠ r2 − r1 7.70 2μr12ω2 4πμr12r22 Lω2 ∴T2 = τ 2 A2r2 = 2 2 2π r2 Lr2 = r2 − r1 r22 − r12 Turbulent Flow Let u = u + u′, v = v + v′, w = w + w′. The continuity equation becomes ∂ ∂ ∂ ∂u ∂v ∂w ∂u′ ∂v′ ∂w′ (u + u′) + ( v + v′) + (w + w′) = + + + + + ∂x ∂y ∂z ∂x ∂y ∂z ∂x ∂y ∂z Now, time-average the above equation recognizing that 7.72 ∂u ∂u = and ∂x ∂x ∂u′ ∂ ∂u ∂v ∂w = u′ = 0. Then, + + = 0. Substitute this back into the continuity ∂x ∂x ∂x ∂y ∂z equation, so that ∂u′ ∂v′ ∂w′ + + =0 ∂x ∂y ∂z Use the fact that 7.74 ∂ ∂ u′v′ = u′v′. See Eq. 7.6.2. This is equivalent to ∂y ∂y T ⎞ 1T ∂ ∂ ⎛1 ⎜ ∫ u′v′dt ⎟ = ∫ (u′v′)dt ⎟ T ∂y ∂y ⎜⎝ T 0 0 ⎠ which is obviously correct. Also ⎛ ∂u′ ∂v′ ∂w′ ∂ ∂ ∂ u′u′ + u′v′ + u′w′ = u′ ⎜ + + ⎜ ∂x ∂y ∂z ∂x ∂y ∂z ⎝ ⎞ ∂u′ ∂u′ ∂u′ + v′ + w′ ⎟ + u′ ⎟ ∂x ∂y ∂z ⎠ Time average both sides and obtain the result. 95 © 2012 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 7 / Internal Flows u = Σui /11 = 16.2 m/s 7.76 t 0 u′ −0.1 v = Σvi /11 = −1.6 m/s 0.01 0.02 0.03 9.5 −5.6 1.1 u′2 0.01 90.2 31.4 1.2 v′ 3.2 −3.8 −7 5.1 v′ 2 10.2 14.4 49 u′v′ −0.32 −36.1 39.2 u′2 = 51.2 m 2 /s 2 η =− 26 5.6 0.05 0.06 −11 −6 0.9 121 36 0.81 5.7 −4.4 0.2 32.5 19.4 −62.7 26.4 0.07 v′ = v − v 0.08 0.09 0.1 12.4 −9.5 3 5.4 153.8 90.2 9 29.2 8.3 −3.6 −6.6 3.1 0.04 68.9 13.0 43.6 9.6 0.2 102.9 34.2 −19.8 16.7 u′v′ = 9.7 m2 /s2 v ′2 = 26.1 m 2 /s 2 u′v′ −2.52 =− = 0.016 m2 /s (23 18.2) / 0.03 − du / dy u′v′ Kuv = 7.78 0.04 u′ = u − u = u′2 v′2 −2.52 = −0.125 29 14 A m = η / ∂u / ∂y = 0.016/(23 −18.2) / 0.03 = 0.01 m or 1 cm b) Re = 7.80 0.2 × 0.2 10 −6 = 40,000, e = 0.0013 D From the Moody diagram, this is in the transition zone where δ ν may be near, in magnitude, to e. The pipe is rough. b) With Re = 40,000, e /D = 0.0013 the Moody diagram gives f = 0.026. Then 0.026 uτ = 0.130/1000 = 0.01140 m/s 1000 × 0.2 2 = 0.130 Pa τ0 = 8 7.82 Eq. 7.6.17: V= 7.84 r ⎛ ⎞ umax = uτ ⎜ 2.44 ln o + 8.5 ⎟ e ⎝ ⎠ 0.1 ⎛ ⎞ = 0.0114 ⎜ 2.44 ln + 8.5 ⎟ = 0.262 m/s 0.00026 ⎝ ⎠ Q 70 ×10−3 = = 6.2 m/s A π × (0.6)2 Re = VD ν = 6.2 × 0.12 = 7.4 × 105 −6 1.007 ×10 From Table 7.1, n ≅ 8.5. From Eq. 7.6.20 umax = (n + 1)(2n + 1) 9.5 ×18 ×V = × 6.2 = 7.34 m/s 2 2n 2 × 8.52 96 © 2012 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 7 / Internal Flows With n = 7, Eq. 7.6.21 gives f = 1 n 2 = 1 = 0.0204 49 y 1 1 ∴τ 0 = ρ V 2 f = × 1000 × 102 × 0.0204 = 1020 Pa 8 2 Since τ varies linearly with r and is zero at r = 0 τ = τ0 7.86 τturb C (r = 0) τlam τ r 1020 r = 20,400 r = r0 0.05 τ lam = μ ⎡ ∂u 1 1 −6 / 7 ⎤ 10−3 × 12.24 −6 / 7 = 10−3 ⎢umax = 0.00268 y −6 / 7 y y ⎥= 1/ 7 1/ 7 ∂y 7 r0 ⎣ ⎦ 7 × 0.05 τturb =τ −τlam = 20,400r − 0.00268y−6/7 where y + r = 0.05 τ lam ( y ) is good away from y = 0 (the wall) τ lam (0.00625) = 0.21, τ lam (0.003125) = 0.38 τ lam (0.00156) = 0.68, τ lam (0.00078) = 1.24 dp 2τ 2 ×1020 =− 0 =− = −40,800 Pa/m 0.05 dx r0 We must find uτ : V= Q 1.2 = = 19.63 m/s A π × 0.42 Re = 19.63 × 0.8 = 71,400 2.2 × 10−4 e 0.26 = = 0.000325 ∴ Moody diagram ⇒ f = 0.021 D 800 7.88 τ0 = 1 1 fρV 2 = × 0.021 × 917 × 19.63 2 = 928 Pa 8 8 uτ = τ 0 /ρ = 928/917 = 1.006 m/s 1.006 × 0.4 ⎡ ⎤ ∴ umax = 1.006 ⎢ 2.44 ln + 5.7 ⎥ = 24.2 m/s −4 2.2 × 10 ⎣ ⎦ a) From a control volume of the 10-m section of pipe 7.90 Δp π D2 4 = τ 0π DL ∴τ 0 = 0.12 × 5000 = 15 Pa 4 × 10 Assume n = 7. Then Eq. 7.6.21 gives f = 1 1 = = 0.0204 2 n 49 97 © 2012 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 7 / Internal Flows From Eq. 7.3.19, 8τ 0 8 × 15 = = 6.41 or V = 2.53 m/s ρ f 917 × 0.0204 2.53 × 0.12 = 1381. This suggests laminar flow. Use Eq. 7.3.14: 2.2 × 10 −4 Check: Re = V= V2 = r02 Δp 0.062 × 5000 = = 1.12 m/s 8μ L 8 × (917 × 2.2 ×10−4 ) ×10 Check: Re = 1.12 × 0.12 = 611 2.2 × 10−4 OK Finally, Q = AV = π × 0.06 2 × 1.12 = 0.0127 m 3 /s Turbulent Flow in Pipes and Conduits a) Re = VD ν = (0.020 / π × 0.042 ) × 0.08 = 3.18 × 105 −6 10 e = 0 ∴ f = 0.0143 D b) Eq. 7.6.26 provides f by trial-and-error. Try f = 0.0143 from Moody’s diagram: ( 7.92 ) 1 = 0.86 ln 3.18 ×105 0.0143 − 0.8 ∴ f = 0.0146 f Another iteration may be recommended but this is quite close. The value for f is essentially the same using either method. The equations could be programmed on a computer. a) Re = 0.025 × 0.04 64ν 64 × 10−6 = 1000 ∴ laminar ∴ f = = = 0.064 VD 0.025 × 0.04 10−6 c) Re = 2.5× 0.04 = 100,000 10−6 7.94 V= Q 1.65 × 10−3 VD 1.45 × 0.038 = = 1.45 m/s Re = = = 4.8 × 104 2 −6 ν A π (0.019) 1.14 × 10 hL = f 7.96 e 0.26 = = 0.0065 ∴ f = 0.034 D 40 L V2 180 1.452 = f = 508 f D 2g 0.038 19.62 a) e 0.00026 = = 0.0068 D 0.038 c) e 0.000046 = = 0.0012 D 0.038 ∴ f = 0.035 ∴ f = 0.0245 ∴ hL = 508 × 0.035 = 17.8 m ∴ hL = 508 × 0.0245 = 12.4 m 98 © 2012 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 7 / Internal Flows V= Q 0.08 = = 4.73 m/s A π × 0.0752 Δp = γ f a) Re = 7.98 e 0.15 = = 0.001 D 150 L V2 10 4.732 = (917 × 9.81) f × = 6.84 ×105 f D 2g 0.15 2 × 9.81 4.73 × 0.15 64 = 322 ∴ laminar and Δp = 6.84 × 105 × = 1.36 × 105 Pa 0.0022 Re Eq. 7.6.29 is not applicable to a laminar flow c) Re = 4.73 × 0.15 = 16,300 0.000044 ∴ f = 0.029 Δp = 6.84 × 105 × 0.029 = 20,000 Pa 0.9 ⎧ ⎛ 4.4 × 10−5 × 0.15 ⎞ ⎤ ⎫⎪ 0.08 × 10 ⎪ ⎡ 0.15 ⎢ Δp = 9.8 × 917 ×1.07 + 4.62 ⎜ ⎨ln ⎟ ⎥⎬ 0.08 9.8 × 0.155 ⎪ ⎢ 3.7 × 50 ⎝ ⎠ ⎥⎦ ⎭⎪ ⎩ ⎣ = 20,700 Pa −2 2 V= 7.100 Q 8.3 × 10−3 = = 2.94 m/s A π (0.03) 2 L V2 ∴ f = 0.0167 ∴Δp − γΔh = γ f D 2g ∴ Δp = 9810 × 0.0167 V= 7.102 2.94 × 0.06 = 1.3 × 105 1.31× 10−6 e =0 D (Recall: Δp = p1 − p 2 , Δh = h2 − h1 ) 90 2.942 × + 9810 × 90sin 30D = 550 × 103 Pa. = 550 kPa 0.06 19.62 5 Q = = 9.95 m/s A π × 0.42 Re = 9.95 × 0.8 = 8 × 106 −6 10 e 1.6 = = 0.002 (using an average “e” value) ∴ f = 0.0237 D 800 ∴Δp = γ f L V2 100 9.952 = 9810 × 0.0237 × × = 147,000 Pa D 2g 0.8 2 × 9.81 Use Eq. 7.6.30: hL = 7.104 Re = Δp γ = 200,000 = 20.4 m ν = 10−6 m2/s 9810 ⎡ 9.81× 0.045 × 20.4 ⎤ b) Q = −0.965 ⎢ ⎥ 100 ⎣ ⎦ 0.5 ⎡ 0.046 ⎛ 3.17 ×10−12 × 100 ⎞0.5 ⎤ ln ⎢ +⎜ ⎟ ⎥ ⎢ 3.7 × 40 ⎝ 9.81× 0.043 × 20.4 ⎠ ⎥ ⎣ ⎦ = 0.0033 m3 /s 99 © 2012 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 7 / Internal Flows Use Eq. 7.6.30: p 200 = = 3.38 kg/m3 RT 0.189 × 313 b) ρ = 7.106 e 0.046 = = 0.000104 3.7 D 3.7 ×120 = ρQ m ν= ∴ hL = 400 = 12.1 m 9.81× 3.38 μ 1.3 ×10−5 = = 3.8 ×10−6 m2 /s ρ 3.38 ⎡ 9.81 × 0.125 × 12.1 ⎤ Q = − 0.965 ⎢ ⎥ 400 ⎣ ⎦ 0.5 0.5 ⎡ ⎛ 3.17 × 3.82 × 10−12 × 400 ⎞ ⎤ ln ⎢ 0.000104 + ⎜ ⎟ ⎥ 9.81 × 0.123 × 12.1 ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ = 0.0205 = 0.069 kg/s ∴m 5.2 ⎤ 2 4.75 ⎡ 1.25 ⎛ LQ ⎞ 9.4 ⎛ L ⎞ ⎢ Use Eq. 7.6.31: e = 0 D = 0.66 e ⎜ ⎟ +ν Q ⎜ ⎟ ⎥ gh gh ⎢ ⎝ L ⎠ ⎥⎦ ⎝ L ⎠ ⎣ a) hL = Δp γ = 200,000 = 20.4 m 9810 ν = 10−6 m2 /s 5.2 ⎡ −6 100 ⎞ ⎤ 9.4 ⎛ D = 0.66 ⎢10 × 0.002 ⎜ ⎟ ⎥ ⎝ 9.81× 20.4 ⎠ ⎦⎥ ⎣⎢ 7.108 c) hL = Δp γ = 200,000 = 25.2 m 7930 0.04 = 0.032 m ν = 2.1× 10−6 m 2 /s 5.2 ⎡ ⎛ 100 ⎞ ⎤ D = 0.66 ⎢2.1×10−6 × 0.0029.4 ⎜ ⎟ ⎥ ⎝ 9.81× 25.2 ⎠ ⎦⎥ ⎣⎢ a) V = 0.04 0.4/60 L V2 2 0.00849/ D , h f = = L D 2g π D 2 /4 0.04 or 3 = f = 0.031 m 1200 (0.00849/D 2 ) 2 or D 2 × 9.8 f = 680 D5 7.110 Guess: f = 0.02. Then the above equation gives D = 0.124 m Check: V = 0.551 m/s, Re = 0.551× 0.25 e 0.0015 = 1.05 × 105 , = = 1.2 × 10−5 −6 124 D 1.31× 10 ∴ f = 0.0175 . Use f = 0.0175 and D = 0.121 m b) Use Eq. 7.6.31: 100 © 2012 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 7 / Internal Flows 9.4 5.2 ⎤ 2 4.75 ⎡ ⎛ 1200 ⎞ ⎥ −5 1.25 ⎛ 1200 × (0.4/60) ⎞ −6 ⎛ 0.4 ⎞ D = 0.66 ⎢ (1.5 × 10 ) ⎜ 1.31 10 + × × ⎟ ⎜ ⎟ ⎜ ⎟ 9.8 × 3 ⎢ ⎝ 60 ⎠ ⎝ 9.8 × 3 ⎠ ⎥ ⎝ ⎠ ⎣ ⎦ 0.04 = 0.127 m Exiting K.E. = V 2 0.8922 = = 0.041 m 2 g 2 × 9.8 ∴ negligible Use Eq. 7.6.30: 80 0.02 × 0.04 , ν = 10−6 m 2 /s, use D = 4 R = 4 = 0.027 m 9810 2(0.02 + 0.04) 0.5 ⎡ 0.0015 ⎛ 3.17 × 10−12 × 2 × 9810 ⎞0.5 ⎤ ⎡ 9.81× 0.0275 80 ⎤ × +⎜ Q = −0.965 ⎢ ⎟ ⎥ ⎥ ln ⎢ 3 × 2 9810 3.7 27 × × 9.81 0.027 80 ⎢ ⎣ ⎦ ⎝ ⎠ ⎥⎦ ⎣ hL = 7.112 = 0.000143 m3 /s b) R = 7.114 A 0.6 × 1.2 = = 0.3 m, hL = 10,000 × 0.0015 = 15 m (from energy eq.) Pwet 2.4 ⎡ 9.8 × 1.25 × 15 ⎤ Q = −0.965 ⎢ ⎥ ⎣ 10,000 ⎦ 0.5 0.5 ⎡ 1.5 ⎛ 3.17 × 10 −12 × 10,000 ⎞ ⎤ 3 +⎜ ln ⎢ ⎟ ⎥ = 0.257 m /s 9.8 × 1.23 × 15 ⎢ 3.7 × 1200 ⎝ ⎥ ⎠ ⎦ ⎣ Minor Losses 7.116 V2 Referring to the equation Δp = − ρ Δn (n is in the direction R of the center of curvature), we observe that Δp is negative all along the line CB from C to B in Fig. P7.115. Hence, the pressure decreases from C to B with pC > pB. Using Bernoulli’s p B equation we see that VB > VC. Fluid moves from the high pressure region at the outside of the bend toward the low pressure region at the outside of the bend creating a secondary flow. a) V1 = 63.7, V2 = 15.9, 7.118 7.120 Energy: 0 = V= pC ρ1 = 1.78 kg/m3 15.92 − 63.7 2 p2 − 50,000 (63.7 − 15.9)2 + + 0.4 2 × 9.81 1.78 × 9.81 2 × 9.81 ∴ p2 = 52,600 Pa Q 3.3 ×10−3 L V2 Neglect pipe friction, i.e., f ≅0 = = 1.6 m/s D 2g A π (0.025)2 101 © 2012 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 7 / Internal Flows Energy: 0 = 1.682 − 02 1.682 + 0 − 1.8 + (K + 0.03) 2 × 9.81 2 × 9.81 ∴ K = 11.48 Simple Piping Systems Assume completely turbulent regime: e 0.046 = = 0.0012 D 40 Energy: 7.122 0= ∴ f = 0.0205, Re > 106 2 V2 ⎡ L ⎤V − 40 + ⎢ f + 2Kelbow + Kentrance ⎥ 2g ⎣ D ⎦ 2g 2 110 ⎡ ⎤ V 40 = ⎢1 + 0.0205 × + 2 × 1.0 + 0.5⎥ 0.04 ⎣ ⎦ 2 × 9.81 ∴ V = 3.62 m/s Re = ∴ V = 3.50 m/s (Assume a screwed elbow) 3.62 × 0.04 = 1.4 ×105 −6 10 ∴ Try f = 0.022 ∴Q = AV = π × 0.022 × 3.5 = 0.0044 m3/s The EGL and HGL have sudden drops at the elbows, and a gradual slope over the pipe length. Assume screwed elbows: 0 = V2 L ⎞V2 ⎛ − 2 + ⎜ Kentrance + 2K elbows + f ⎟ 2g D ⎠ 2g ⎝ b) Assume 7.124 26 ⎞ V 2 ⎛ f = 0.018 : 2 = ⎜ 0.8 + 1 + 2 × 0.8 + 0.018 ⎟ 0.08 ⎠ 2 × 9.81 ⎝ ∴ Re = 2.06 × 0.08 1.14 × 10 −6 = 1.5 × 105 ∴ V = 2.06 m/s f = 0.0164 ∴V = 2.12 Q = π × 0.042 × 2.12 = 0.011 m3/s Assume constant water level, neglect the velocity head at the pipe exit and let p exit = 0. Then the energy equation says hL = 0.8 m: 0.8 = f 7.126 V2 V2 10 + 2 × 1.5 0.025 2 × 9.8 2 × 9.8 Try f = 0.02. Then V = 0.319 m/s Re = or 0.8 = 400 fV 2 + 0.153V 2 0.313 × 0.025 = 1.1× 104 ∴ f = 0.029 −6 0.73 ×10 Try f = 0.029 Then V = 0.262 m/s ∴ Q = AV = π × 0.01252 × 0.262 = 1.28 × 10−4 m3 /s 102 © 2012 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 7 / Internal Flows Finally 7.128 Δt = − V 10 = = 78.1 sec or 1.3 min Q 0.128 V32 p0 V2 1.2 V12 1.2 V22 − − 1.2 + 0.8 1 + f1 + f2 2g γ 2g 0.008 2 g 0.005 2 g Energy: 0= Continuity: V3 = Assume f1 = f 2 = 0.02 Then energy says V1 = 2.09 m/s Check: Re1 = Finally V3 = 16V1 = 46.1 m/s Energy: 0 = −3 + f Assume turbulent: Re = Check: 25 64 V2 , V2 = V1 4 25 2.09 × 0.008 10−6 = 16,700 ∴ f = 0.026 Then V1 = 2.88 m/s 300 V 2 0.0094 2 × 9.8 f = 0.04 Then V = 0.215 m/s 0.215 × 0.0094 10−6 = 2020 ∴ marginally laminar Assume laminar flow with f = 64/ Re. Then 7.130 3= 64 ×10−6 300 V2 × × V × 0.0094 0.0094 2 × 9.8 Check: Re = 0.271× 0.0094 10−6 ∴ V = 0.271 m/s = 2540 ∴turbulent The flow is neither laminar nor turbulent but may oscillate between the two. The wall friction is too low in the laminar state so it speeds up and becomes turbulent. The wall friction is too high in the turbulent state so it slows down and becomes laminar, etc., etc. Q 0.01 = = 7.96 m/s A π × 0.022 e 0.0015 = = 3.8 × 10−5 D 40 V= 7.132 Re = 7.96 × 0.04 1.14 × 10 −6 = 2.8 × 105 ∴ f = 0.0145 800 ⎞ 7.962 ⎛ H P = 80 − 10 + ⎜ 0.5 + 1.0 + 0.0145 = 1010 m. ⎟ 0.04 ⎠ 2 × 9.81 ⎝ = γ QH P = 9810 × 0.01×1010 = 117,000 W ∴W P ηP 0.85 103 © 2012 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 7 / Internal Flows 0= 7.962 1670 − 100,000 L ⎞ 7.962 ⎛ + + 0 − 10 + ⎜ 0.5 + 0.0145 ∴ L = 13.0 m ⎟ 2 × 9.81 9810 0.04 ⎠ 2 × 9.81 ⎝ V12 690×103 V22 (4V1)2 + = = Energy across nozzle (neglect losses): 2g 9810 2g 2g ∴ V1 = 9.6 m/s Re = 9.6 × 0.05 1.007 × 10 −6 = 4.8 × 105 e 0.000045 = = 0.0009 D 0.05 ∴ f = 0.02 7.134 HP = (4 × 9.62 )2 360 ⎞ 9.62 ⎛ − 18 + ⎜ 0.5 + 0.02 = 677.2 m ⎟ 19.62 0.05 ⎠ 19.62 ⎝ = γ QH P = 9810 × (π × ( 0.025 ) × 9.6) × 677.2 =166.97 ×103 W or 166.97 kW ∴W P ηP 0.75 2 0= 9.62 (2.3 − 100)103 L ⎞ 9.62 ⎛ + − 18 + ⎜ 0.5 + 0.02 ∴ L = 11.1 m ⎟ 19.62 9810 0.05 ⎠ 19.62 ⎝ e = 0.0013 ∴ f = 0.021 Q = 0.0314V D 300 ⎞ V 2 ⎛ H P = −20 + ⎜ 0.5 + 1.0 + 0.021 = −20 + 1700Q 2 ⎟ 0.2 ⎠ 2 g ⎝ 7.136 Try Q = 0.23 m3 /s: ( H P )E = 70 m ( H P )C = 70 m = 9810 × 0.23 × 70 = 190,000 W ∴W P 0.83 EGL HGL Re = 1.5 × 106 ∴OK e 0.046 = = 0.00029 ∴ f = 0.015 Q = π × 0.082 V = 0.0201V D 160 50 ⎞ Q2 ⎛ H P = 25 + ⎜ 0.5 + 2 × 0.4 + 1 + 0.015 = 25 + 882Q 2 ⎟ 2 0.16 ⎝ ⎠ 2 g × 0.0201 Try Q = 0.23 m3/s (H P )E = 72 m ( H P )C. = 70 m 7.138 = 9810 × 0.23 × 72 = 195,000 W a) ∴ W P 0.83 b) 0 = p 11.442 10 ⎞ 11.442 ⎛ + 2 − 8 + ⎜ 0.5 + 0.015 ∴ pin = −81,000 Pa ⎟ 2 × 9.81 9810 0.16 ⎠ 2 × 9.81 ⎝ c) 72 = p 11.442 10 ⎞ 11.442 ⎛ + 3 − 8 + ⎜ 0.5 + 0.015 ∴ pout = 625,000 Pa ⎟ 2 × 9.81 9810 0.16 ⎠ 2 × 9.81 ⎝ 104 © 2012 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 7 / Internal Flows d) EGL HGL e 1.65 = = 0.0014 ∴ f = 0.021 Q = π × 0.62 V = 1.13V D 1200 1000 ⎞ Q2 ⎛ − HT = −60 + ⎜ 0.8 + 1 + 0.021 ∴ HT = 60 − 0.77Q 2 ⎟ 2 1.2 ⎝ ⎠ 1.13 × 2 × 9.81 7.140 = γ QH η = 9810 × (60Q − 0.77Q3 ) × 0.9 = 530,000Q − 6800Q3 W T T T = 500,000 Q (power in watts) But, the characteristic curve is W T Hence, 500,000Q = 530,000Q − 6800Q3 ∴Q = 2.1 m3/s = 1.05 MW ∴W T Open-Channel Flow τ w Aw = W sin θ 7.142 sin θ = 0.001 L τwAw or τ w × w ×1 = ( w × d ×1) ρ w g sin θ τ w = ρ w dg sin θ = 1000 ×1.8 × 9.81× 0.001 W θ = 17.6 Pa a) Q = 1.0 1.0 ⎛ 2 × 0.6 ⎞ AR2 / 3S1/ 2 = × (2 × 0.6) ⎜ ⎟ n 0.012 ⎝ 2 + 2 × 0.6 ⎠ 2/3 × 0.0011/ 2 = 1.64 m3/s b) Planed wood and finished concrete have the same “n” (i.e., roughness) 7.144 ∴Use a value for “e” which is relatively small, say e = 0.6 mm. The solution is not too sensitive to this choice. Then, with R = 0.375 m e e 0.6 = = = 0.0004 D 4 R 4 × 375 ∴ f = 0.016 105 © 2012 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 7 / Internal Flows hL = f L V2 = LS D 2g ∴V 2 = 2 × 9.81× 0.001× (4 × 0.375) 0.016 ∴V = 1.36 m/s 1.36 × (4 × 0.375) ⎛ ⎞ ∴ Q = 1.36 × 2 × 0.6 = 1.64 m 3 /s ⎜ Check: Re = = 2.0 × 106 ∴ OK ⎟ −6 10 ⎝ ⎠ a) R = 1.2 × 0.5 + 0.5 × 0.5 = 0.325 m ∴ Q = 1.0 × 0.85 × 0.3252 / 3 × 0.0011/ 2 = 0.794 m3 /s 1.2 + 2 × 0.5/0.707 V= 7.146 0.016 Q 0.794 = = 0.934 m/s A 0.85 b) Use an “e” for rough concrete, e = 3 mm e 3 ∴ = = 0.0023 ∴ f = 0.025 D 4 × 325 hL = LS = f L V2 4R 2g ∴V 2 = 4 × 0.325 × 2 × 9.81× 0.001 and V = 1.01 m/s 0.025 ∴Q = 1.01× 0.85 = 0.86 m3/s Assume y > 3. R = 30 + 20( y − 3) 20 y − 30 10 y − 15 = = 26 + 2( y − 3) 2 y + 20 y + 10 ⎛ 10 y − 15 ⎞ 1.0 100 = (20 y − 30) ⎜ ⎟ 0.022 ⎝ y + 10 ⎠ 7.148 Or ⎛ 10 y − 15 ⎞ 6.96 = (2 y − 3) ⎜ ⎟ ⎝ y + 10 ⎠ Try y = 5: 6.96 =? 12.3 2/3 × 0.0011/2 2/3 y y = 4: 6.96 =? 7.36 ⎫ y = 3.8: 6.96 =? 6.47 ⎪ ⎪ ⎬ ∴ y = 3.91 m ? y = 3.9: 6.96 = 6.91⎪ ⎪⎭ 106 © 2012 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 7 / Internal Flows 2/3 2⎞ ⎛π 1 π 2 2 × 0.6 ⎟ × 0.0011/2 Try y = 0.6 m : Q = × × (0.6) ⎜ ⎜ 0.6π ⎟ 0.013 2 ⎝ ⎠ y − 2 y − 0.6 = 0.62 m3/s cosα = = 2 0.6 ∴ y > 0.6 m. R = 7.150 A = 2π (0.6) Q= A 2π (0.6) × α /180 180 − α + ( y − 0.6)0.6 sin α 180 α 0.6 1 AR2 / 3 × 0.0011/ 2 = 0.66 ∴ AR2 / 3 = 0.27 0.013 ⎫ Try y = 0.63: α = 87.13D , A = 0.6012 m 2 , R = 0.33 m. 0.287 =? 0.27 ⎪ ⎪ ⎬ y = 0.612 m D 2 ? Try y = 0.612: α = 88.85 , A = 0.58 m , R = 0.312 m. 0.267 = 0.27 ⎪ ⎪⎭ 107 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. y Chapter 7 / Internal Flows 108 © 2012 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 8 / External Flows CHAPTER 8 External Flows FE-type Exam Review Problems: Problems 8-1 to 8-8 8.1 (C) 8.2 (C) 8.3 (B) Re = VD ν = 0.8 × 0.008 1.31× 10−6 = 4880 Assume a large Reynolds number so that CD = 0.2. Then 8.4 (B) 2 F= 1 1 ⎛ 80 ×1000 ⎞ 2 ρV 2 ACD = ×1.23 × ⎜ ⎟ × π × 5 × 0.2 = 4770 N 2 2 ⎝ 3600 ⎠ Assume a Reynolds number of 105. Then CD = 1.2 8.5 (D) 8.6 (C) F= 1 1 ρV 2 ACD ∴ 60 = × 1.23 × 402 × 4 × D × 1.2 ∴ D = 0.0041 m 2 2 Re = VD Re = VD ν ν = = 40 × 0.0041 10 −6 4 × 0.02 1.6 × 10 −5 = 1.64 ×105 ∴ CD = 1.2 The assumption was OK = 5000 ∴ St = 0.21 = ∴ f = 42 Hz (cycles/second) distance = 8.7 (C) fD f × 0.02 = V 4 V 4 m/s = = 0.095 m/cycle f 42 cycles/s By reducing the separated flow area, the pressure in that area increases thereby reducing that part of the drag due to pressure. From Fig. 8.12a, C L = 1.1 C L = 8.8 (B) ∴V 2 = 1 2 FL ρ V 2cL 2W 2 ×1200 × 9.81 = = 1088 and V = 33.0 m/s ρ cLCL 1.23 ×16 ×1.1 109 © 2012 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 8 / External Flows Separated Flows Re = 5 = 8.10 VD 5 × 1.51× 10−5 = 3.78 × 10−5 m 20 ∴D = ν inviscid flow inviscid flow no separation separated region boundary layer near surface viscous flow near sphere boundary layer separated region 8.12 building wake inviscid flow Re = VD 8.14 ν = 20 × D 1.51×10 −5 = 13.25 ×105 D b) Re = 13.25 ×105 × 0.06 = 7.9 ×104 ∴Separated flow Ftotal = Fbottom + Ftop = 20,000 × 0.3 × 0.3+10,000 × 0.3 × 0.3 = 2700 N Flift = 2700 cos 10° = 2659 N Fdrag = 2700 sin 10° = 469 N 8.16 CL = CD = FL 1 2 ρV A 2 FD 1 2 ρV A 2 = = 2659 1 2 × 1000 × 5 × 0.3 × 0.3 2 469 1 2 × 1000 × 5 × 0.3 × 0.3 2 = 2.36 = 0.417 If C D = 1.0 for a sphere, Re = 100 (see Fig. 8.8). ∴ 8.18 V × 0.1 ν = 100, V =1000ν b) V = 1000 × 1.46 ×10−5 = 0.798 m/s 0.015 × 1.22 1 ∴ FD = × (0.015 ×1.22) × 0.7982 π × 0.052 × 1.0 2 110 © 2012 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 8 / External Flows = 4.58 ×10−5 N The velocities associated with the two Reynolds numbers are V1 = 8.20 Re1ν 3 × 105 ×1.5 × 10−5 = = 101 m/s 0.0445 D Re 2 ν 6 × 104 ×1.5 ×10−5 = = 20 m/s V2 = 0.0445 D The drag, between these two velocities, is reduced by a factor of 2.5, i.e., (CD )high = 0.5 and (CD )low = 0.2. Thus, between 20 m/s and 100 m/s, the drag is reduced by a factor of 2.5. This would significantly lengthen the flight of the ball. 8.22 1 V × 0.2 4.2 = × 1000V 2π × 0.12 C D ∴ V 2C D = 0.267 Re = = 2 × 105 V − 6 2 10 Try CD = 0.5 : a) Re1 = 25 × 0.05 1.08 ×10 −5 ∴ V = 0.73 m/s Re = 1.46 ×105 ∴ OK = 1.2 ×105 Re2 = 1.8 × 105 Re3 = 2.4 × 105 Assume a rough cylinder (the air is highly turbulent). ∴ ( CD )1 = 0.7, ( CD )2 = 0.8, ( CD )3 = 0.9 8.24 1 ∴ FD = × 1.45 × 252 (0.05 × 10 × 0.7 + 0.075 × 15 × 0.8 + 0.1× 20 × 0.9) = 1380 N 2 1 M = ×1.45 × 252 (0.05 ×10 × 0.7 × 40 + 0.075 ×15 × 0.8 × 27.5 + 0.1× 20 × 0.9 × 10) 2 = 25,700 N ⋅ m Since the air cannot flow around the bottom, we imagine the structure to be mirrored as shown. Then L / D = 40/5 = 8 8.26 Remin = VDmin ∴CD = 0.66CD∞ = 30 × 2 1.5 × 10−5 ∴ C D = 1.0 × 0.66 = 0.66 ν = 4 × 106 1 ⎛ 2+8 ⎞ ∴ FD = × 1.22 × 302 × ⎜ × 20 ⎟ × 0.66 = 36,000 N 2 ⎝ 2 ⎠ 111 © 2012 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 8 / External Flows 8.28 4 1 4 FB + FD = FW 1.22 × π (0.25)3 + ×1.22V 2π (0.25)2 CD = 9810S π (0.25)3 3 2 3 V × 0.5 Re = = 3.4 ×104 V ∴1 + 0.153V 2C D = 820S −5 1.47 ×10 a) S = 0.005 V 2C D = 20.26 Assume atmospheric turbulence, i.e., rough Try C D = 0.4 : V = 7.1 m/s Re = 2.4 ×105 ∴ C D = 0.3 and V = 8.22 m/s From Table 8.2 CD = 0.35 8.30 1 FD = ×1.22V 2 × 3.2 × 0.35 = 0.683V 2 2 2 80 ×1000 ⎛ 80 ×1000 ⎞ a) FD = 0.683 × ⎜ = 7500 W or 10 hp ⎟ = 337 N ∴ W = 337 3600 ⎝ 3600 ⎠ Re = 8.32 VD ν a) FD = = (40,000/3600)0.6 = 4.42 × 105 ∴ C D = 0.35 from Fig. 8.8. −5 1.51× 10 1 1 ρ V 2 AC D = ×1.204 × (40, 000/3600) 2 × 0.6 × 6 × 0.35 = 93.6 N 2 2 With the deflector the drag coefficient is 0.76 rather than 0.96. The required power, directly related to fuel consumed, is reduced by the ratio of 0.76/0.96. The cost per year without the deflector is Cost = (200,000/1.2) × 0.25 = $41,667 8.34 With the deflector it is Cost = 41,667 × 0.76/0.96 = $32,986 The savings is $41,667 − 32,986 = $8,800 8.36 FD = 1 1 ρ V 2 ACD = ×1.22 × (27.8 ×1.6)2 × π × 0.052 ×1.1 = 10.43 N 2 2 = F × V × 2 = 10.43 × (27.8 ×1.6) × 2 = 226 W or 1.24 hp W D The net force acting up is (use absolute pressure) 4 4 120 Fup = π × 0.43 × 1.21 × 9.8 − 0.5 − π × 0.43 × 9.8 = 2.16 N 3 3 2.077 × 293 8.38 From a force triangle (2.16 N up and FD to the right), we see that tan α = Fup /FD a) FD = 2.16 / tan 80° = 0.381. Assume CD = 0.2: 1 0.381 = × 1.21V 2π × 0.42 × 0.2 2 ∴ V = 2.50 m/s 112 © 2012 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 8 / External Flows Check Re: Re = 2.5 × 0.8 = 1.33×105 −5 1.51×10 Too low. Use C D = 0.5: 1 0.381 = × 1.21V 2π × 0.42 × 0.5 2 c) FD = 2.16 / tan 60° = 1.25. Assume CD = 0.5: 1 1.25 = × 1.21V 2π × 0.42 × 0.5 2 Check Re: Re = ∴ V = 1.58 m/s ∴ V = 2.86 m/s 2.86 × 0.8 = 1.5 ×105 1.51×10−5 ∴OK Power to move the sign: FDV = 1 ρ V 2 ACD × V 2 1 = × 1.21× 11.112 × 0.72 × 1.1× 11.11 = 657 J/s. 2 8.40 This power comes from the engine: 657 = (12,000 × 1000)m × 0.3 ∴ m = 1.825 × 10−4 kg/s Assuming the density of gas to be 900 kg/m3 1825 . × 10 − 4 × 10 × 3600 × 6 × 52 × 8.42 1000 × 0.30 = $683 900 = 40 × 746η = F × V = 1 ρ V 2 AC × V = 1 ρ AC V 3 W D D D 2 2 1 ∴ 40 × 746 × 0.9 = × 1.22 × 3 × 0.35V 3 ∴ V = 34.7 m/s or 125 km/hr 2 Vortex Shedding 40 > 8.44 VD ν = 1.8D 1.141×10 10,000 < VD ν −6 = ∴ D < 2.5 ×10−5 m 1.8 D 1.141×10−6 ∴ D > 0.0063 m or 0.63 m 113 © 2012 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 8 / External Flows St = 8.46 fD 0.002 × 2 = V V Re = VD ν = Try St = 0.21: V = 0.0191 m/s V ×2 10−6 Use Fig. 8.9 Re = 38 ×103 ∴ OK Streamlining Re = 8.48 25 × 0.15 1.5 × 10 −5 1 = 2.5 × 105 FD = × 1.225 × 252 × 1.0 × 0.8 × (1.8 × 0.5 ) = 82.7 N 2 The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1. We have = F × V = 82.7 × 25 = 2067.5 W or W D (C D )streamlined = 0.035 Re = VD ν = 2 × 0.8 10 −6 L 4 = =5 D 0.8 8.50 ∴ FD = 2.9 N = 1.6 × 106 2.77 kW = 72.5 W or 0.72 kW. = 0.1 hp W ∴ C D = 0.45 from Fig. 8.8. ∴ C D = 0.62 × 0.45 = 0.28 Because only one end is free, we double the length. FD = 1 1 ρ V 2 AC D = ×1000 × 22 × 0.8 × 2 × 0.28 = 900 N 2 2 If streamlined, CD = 0.03 × 0.62 = 0.0186 1 ∴ FD = ×1000 × 22 × 0.8 × 2 × 0.0186 = 60 N 2 V = 50 ×1000/3600 = 13.9 m/s Re = 13.9 × 0.3 = 2.8 × 105 ∴ C D = 0.4 −5 1.5 × 10 We assumed a head diameter of 0.3 m and used the rough sphere curve. 8.52 FD = 1 1 ρ V 2 AC D = × 1.2 × 13.92 (π × 0.32 /4) × 0.4 = 3.3 N 2 2 FD = 1 1 ρ V 2 ACD = ×1.2 ×13.92 (π × 0.32 /4) × 0.035 = 0.29 N 2 2 114 © 2012 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 8 / External Flows Cavitation CL = 8.54 FL 1 2 ρV A 2 = 200,000 1 × 1000 × 122 × 0.4 × 10 2 CD = 0.0165 = ? σ crit = 0.75 > ∴α ≅ 3D = 0.69 FD 1 ×1000 ×122 × 0.4 ×10 2 ∴ FD = 4800 N (9810 × 0.4 + 101,000) − 1670 1 ×1000 ×122 2 = 1.43 p∞ = 9810 × 5 + 101,000 = 150,000 Pa pv = 1670 Pa Re = σ= 8.56 150,000 − 1670 1 ×1000 × 202 2 ∴ FD = = 0.74 ∴ no cavitation 20 × 0.8 10 −6 = 16 × 106 ∴ C D = C D (0)(1 + σ ) = 0.3(1 + 0.74) = 0.52 1 1 ρ V 2 AC D = ×1000 × 202 × π × 0.42 × 0.52 = 52,000 N 2 2 Note: We retain 2 sig. figures since C D is known to only 2 sig. figures. Added Mass ΣF = ma 8.58 4 400 a) 400 − 9810 × π × 0.23 = a 3 9.81 ∴ a = 1.75 m/s 2 4 4 ⎛ 400 1 ⎞ b) 400 − 9810 × π × 0.23 = ⎜ + × 1000 × π × 0.23 ⎟ a 3 3 ⎝ 9.81 2 ⎠ ∴ a = 1.24 m/s 2 Lift and Drag on Airfoils The total aerodynamic drag consists of both lift and drag that is: FTotal = FL + FD ⇒ FTotal = FL2 + FD2 = 18 kN ⇒ FL2 + FD2 = (18 kN ) 8.60 2 which can be combined with FL = 3FD to yield 9 FD2 + FD2 = 324 ⇒ FD = 324 10 = 5.69 kN and hence, FL = 3 × 5.69 = 17.1 kN so 115 © 2012 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 8 / External Flows CL = CL = 8.62 FL 1 2 ρ V 2cL FL 1 2 ρV A 2 = = 17.1×103 1 × 1.2 × 61.12 × 1.3 ×10 2 1000 × 9.81 1 × 0.412 × 802 × 15 2 = 0.496 = 0.587 ∴α = 3.2D C D = 0.0065 = F V = ⎛ 1 × 0.412 × 802 × 15 × 0.0065 ⎞ × 80 = 10,300 W or 13.8 hp W D ⎜ ⎟ ⎝2 ⎠ 8.64 C L = 1.22 = 8.66 C L = 1.22 = 1500 × 9.81 + 3000 1 ×1.007 × V 2 × 20 2 1500 × 9.81 + 9000 1 × 1.22 × V 2 × 20 2 a) C L = 1.72 = 8.68 ∴ V = 38.0 m/s ∴ V = 39.9 m/s 250,000 × 9.81 ∴ V = 75.2 m/s 1 ×1.05 × V 2 × 60 × 8 2 % change = 75.2 − 69.8 × 100 = 7.77% increase 69.8 Vorticity, Velocity Potential, and Stream Function ∇p ⎡∂ V ⎤ ∇× ⎢ + (V ⋅∇)V + − ν∇ 2 V ⎥ = 0 ρ ⎣ ∂t ⎦ ∇× ∂V ∂ ∂ω = (∇ × V ) = ∂t ∂t ∂t ∇× ∇p ρ = 1 ρ ∇ × ∇p = 0 ∇ × (∇ 2 V ) = ∇ 2 (∇ × V ) = ∇ 2ω (we have interchanged derivatives) 8.70 ⎡1 ⎤ 1 ∇ × [ (V ⋅∇ )V ] = ∇ × ⎢ ∇V 2 − V × (∇ × V ) ⎥ = (∇ × ∇V 2 ) − ∇ × (V × ω) ⎣2 ⎦ 2 = V (∇ ⋅ ω) − ω (∇ ⋅ V ) + (V ⋅∇)ω − (ω ⋅∇)V = ( V ⋅ ∇ )ω − (ω ⋅ ∇ ) V since ∇ ⋅ ω = ∇ ⋅ (∇ × V ) = 0 and ∇ ⋅ V = 0 There results: ∂ω + (V ⋅∇)ω − (ω ⋅∇)V −ν∇ 2ω = 0 ∂t This is written as Dω = (ω ⋅∇)V + ν∇ 2ω Dt 116 © 2012 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 8 / External Flows x-comp: 8.72 y-comp: z-comp: ∂ωx ∂ω ∂ω ∂ω ∂u ∂u ∂u + u x + v x + w x = ωx + ωy + ωz + ν∇ 2ωx ∂t ∂x ∂y ∂z ∂x ∂y ∂z ∂ω y ∂t +u ∂ω y ∂x +v ∂ω y ∂y +w ∂ω y ∂z = ωx ∂v ∂v ∂v + ωy + ωz + ν∇ 2ω y ∂x ∂y ∂z ∂ωz ∂ω ∂ω ∂ω ∂w ∂w ∂w + u z + v z + w z = ωx + ωy + ωz + ν∇ 2ωz ∂t ∂x ∂y ∂z ∂x ∂y ∂z ⎛ ∂w ∂v ⎞ ˆ ⎛ ∂u ∂w ⎞ ˆ ⎛ ∂v ∂u ⎞ ˆ a) ∇× V = ⎜ − ⎟i +⎜ − ⎟ j+ ⎜ − ⎟k = 0 ⎝ ∂y ∂z ⎠ ⎝ ∂z ∂x ⎠ ⎝ ∂x ∂y ⎠ ∴irrotational ∂φ = 10x ∴φ = 5x2 + f ( y) ∂x ∂φ ∂f = = 20y ∂y ∂y ∴ f = 10y2 + C Let C = 0 ∴ φ = 5 x 2 + 10 y 2 8.74 ⎛ − y 1 ( x 2 + y 2 ) −1/ 2 2 x − x 1 ( x2 + y 2 )−1/ 2 2 y ⎞ 2 2 ˆ ˆ ⎟ kˆ = 0 c) ∇ × V = 0i + 0 j + ⎜ − 2 2 2 2 ⎜ ⎟ x +y x +y ⎝ ⎠ ∴irrotational ∂φ x = ∂x x2 + y 2 ∴φ = x 2 + y 2 + f ( y ) ∂f y ∂φ 1 2 = ( x + y 2 )−1/ 2 2 y + = ∂y 2 ∂y x2 + y 2 ∴ ∂f = 0 ∴ f = C. Let C = 0 ∂y ∴φ = x 2 + y 2 u= ∂ψ = 100 ∴ψ = 100 y + f ( x) ∂y v=− df ∂ψ =− = 50 ∂x dx ∴ f = −50 x + C ∴ψ ( x, y ) = 100 y − 50 x 8.76 u= (We usually let C = 0) ∂φ = 100 ∴φ = 100x + f ( y) ∂x v= ∂φ df = = 50 ∂y dy ∴ f = 50 y + C ∴φ ( x, y ) = 100 x + 50 y 117 © 2012 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 8 / External Flows 8.78 u= 2y ∂ψ ∂φ = 20 2 = 2 ∂y ∂x x +y v= ∂f ∂f ∂φ 40 / x 40 x 2x =− + =− 2 + = −20 2 ∴ f = C. Let C = 0 2 2 2 ∂y ∂y ∂y x +y x + y2 1+ y / x φ = −40 tan −1 ∴φ = −40 tan −1 y + f ( y) x y x b) Polar coord: φ = 10r cos θ + 5ln r 2 ∂φ 10r 1 ∂ψ = 10 cos θ + 2 = ∂r r ∂θ r (See Eq. 8.5.14.) ∴ψ = 10r sin θ + 10θ + f (r ) 1 ∂φ ∂ψ df = −10sin θ = − = −10sin θ − ∴ f = C ∴ψ = 10r sin θ + 10θ ∂r r ∂θ dr ∴ψ ( x, y ) = 10 y + 10 tan −1 c) v = 8.80 u= 10y ∂φ = 2 ∂y x + y 2 ∂φ 10 x = 10 + 2 ∂x x + y2 Bernoulli: y x Along x-axis (y = 0) v = 0 Along x-axis u = 10 + V2 p V2 p + + gz = ∞ + ∞ + gz∞ 2 ρ 2 ρ (10 + 10/x)2 p 102 100,000 + = + ρ ρ 2 2 e) (assume z = z∞ ) ⎛2 1 ⎞ ∴ p = 100 − 50 ⎜ + 2 ⎟ kPa ⎝x x ⎠ ay = v ∂v/∂y + u ∂v/∂x = 0 on x-axis ( 10 x ax = u∂u/∂x + v ∂u/∂y = (10 + 10/x ) −10/x2 ) ∴ ax ( −2, 0) = (10 − 5) ( −10/4 ) = −12.5 m/s 2 118 © 2012 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 8 / External Flows Superposition of Simple Flows ψ = 9y + a) vr = 0.45π θ = 9r sin θ + 0.225θ 2π 1 ∂ψ 0.225 = 9cosθ + =0 r ∂θ r At θ = π , 8.82 Stag. pt: 0.225 = 9 ∴ rs = 0.025 m rs (−0.025, 0) c) q = U × H = Δψ ∴9H = 0.225π Thickness = 2 H = π 20 d) vr (1, π ) = 9 cos π + φ= ∴H = π 40 m or 0.157 m 0.225 = −8.25 ∴ u = 8.25 m/s 3 1/2 −2π 1/2 2π ⎡ ln ( x + 1) 2 + y 2 ⎤ + ln ⎡ ( x − 1) 2 + y 2 ⎤ + 2 x ⎦ ⎣ ⎦ 2π ⎣ 2π 1 1 = ln ⎡(x +1)2 + y2 ⎤ − ln ⎡(x −1)2 + y2 ⎤ + 2x ⎣ ⎦ ⎦ 2 2 ⎣ 1 1 2( x + 1) 2( x − 1) ∂φ 2 2 = − +2 u= ∂x ( x + 1) 2 + y 2 ( x − 1) 2 + y3 8.84 Along the x-axis (y = 0), v = 0 and u = Set u = 0 : Stag. pts.: ( x − 1) + y 2 2 − y ( x − 1) 2 + y 2 1 1 − +2 x +1 x −1 ( 2, 0), (− 2, 0) u(0, 4) = φ= y 1 1 − = 2, or x 2 = 2 ∴ x = ± 2 x −1 x +1 u(4, 0) = 8.86 v= 1 1 − + 2 = 1.867 m/s −4 + 1 −4 − 1 1 1 + 42 − −1 1 + 42 + 2 = 2.118 m/s v(−4, 0) = 0 v(0, 4) = 4 1 + 42 − 4 1 + 42 =0 1/ 2 2π 1/ 2 2π ⎡ ln ( y − 1) 2 + x 2 ⎤ + ln ⎡ ( y + 1) 2 + x 2 ⎤ + U∞ x ⎦ ⎦ 2π ⎣ 2π ⎣ 119 © 2012 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 8 / External Flows 1 1 = ln ⎡( y − 1)2 + x2 ⎤ + ln ⎡( y + 1)2 + x2 ⎤ + U∞ x ⎦ 2 ⎣ ⎦ 2 ⎣ y a) Stag. pts. May occur on x-axis, y = 0 ∂φ u= ∂x = y =0 x 1 + x2 ∴ x 2 + 0.2 x + 1 = 0 + x x 1 + x2 + 10 ∴no stagnation points exist on the x-axis (They do exist away from the x-axis) h 1 Along the y-axis: u( y) = 10 q = ∫ udy = (2π ) = π m2 /s 2 0 h ∴π = ∫ 10dy = 10 h ∴ h = 0.314 m 0 ψ= 4π 20π ln r = 2θ + 10 ln r at ( x, y) = (0,1) which is (r,θ ) = (1, π /2). θ+ 2π 2π b) vr = 2 10 and vθ = − . From Table 5.1 (use the l.h.s. of momentum): r r Dvr vθ2 ∂vr vθ2 2 ⎛ 2 ⎞ 100 ar = + = vr − = ⎜ − 2 ⎟ − 3 = −104 m/s 2 Dt r ∂r r r⎝ r ⎠ r aθ = 8.88 Dvθ vr vθ ∂v vv 2 ⎛ 10 ⎞ 2(−10) + = vr θ + r θ = ⎜ 2 ⎟ + =0 Dt r ∂r r r⎝r ⎠ r3 ∴ a(0,1) = −104i r or (ax , ay ) = (0, −104) m/s2 c) vr (14.14, π /4) = 2 = 0.1414, 14.14 vr (0.1, π /2) = 2 = 20, 0.1 vθ (14.14, π /4) = − vθ (0.1, π /2) = 10 = −0.707 m/s 14.14 −10 = −100 m/s −0.1 20,000 0.14142 + 0.707 2 p 202 + 1002 Bernoulli: + = + ∴ p = 13,760 Pa 1.2 2 1.2 2 We used ρair = 1.2 kg/m3 at standard conditions. 8.90 b) vθ = − ∂ψ μ sin θ ⎛ 4⎞ = −U∞ sin θ − = ⎜ −4 − 2 ⎟ sin θ = −8sin θ 2 ∂r rc ⎝ 1 ⎠ 120 © 2012 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 8 / External Flows c) pc = p∞ + ρ v2 V∞2 42 82 sin 2 θ − ρ θ = 50,000 + 1000 × − 1000 2 2 2 2 −x ∴ pc = 58 − 32sin 2 θ kPa π /2 d) Drag = 2 ∫ (58 − 32sin 2 α ) cos α ×1×1dα − 26 × 2 ×1 x = −1 0 −vr ⎡ ⎛ 1 ⎞⎤ = 2 ⎢58 − 32 ⎜ ⎟⎥ − 52 = 42.7 kN ⎝ 3 ⎠⎦ ⎣ vθ = −2 × 20sin θ − Γ 2π × 0.4 0 = −2 × 20sin 270D − Γ = 2π rc2ω ∴ω = 8.92 For one stag. pt.: vθ = 0 at θ = 270D : Γ 2π × 0.4 Γ ∴Γ = 2 × 20 × 2π × 0.4 = 100.5 m 2 /s 100.5 = 2π rc2 (See the figure in Problem 8.89c) 2π × 0.42 = 100 rad/s (See Example 8.12) Min. pressure occurs where vθ is max, i.e., θ = π / 2. There vθ = −2 × 20 ×1 − ∴ pmin = p∞ + 8.94 100.5 = 80 m/s 2π × 0.4 v2 V∞2 202 802 ρ − θ ρ = 0+ × 1.22 − × 1.22 = −3660 Pa 2 2 2 2 At 9000 m, ρ = 0.47 kg/m3 Lift = ρU ∞ ΓL = 0.47 × (105) × (1400) × 18 = 1244 × 103 N = 1244 kN Place four sources as shown. Then, with q = 2π for each: u( x, y ) = 8.96 + v ( x, y ) = x−2 ( x − 2) + ( y − 2) x+2 2 2 ( x + 2) 2 + ( y + 2) 2 y−2 ( x − 2) + ( y − 2) 2 2 + x+2 + + y ( x + 2) 2 + ( y − 2) 2 x−2 x ( x − 2) 2 + ( y + 2) 2 y+2 ( x − 2) + ( y + 2) 2 2 + y−2 ( x + 2) + ( y − 2) 2 2 + y+2 ( x + 2) 2 + ( y + 2) 2 ∴v(4, 3) = (0.729, 0.481) m/s 121 © 2012 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 8 / External Flows Boundary Layers Recrit = 8.98 b) ν = 8.100 U ∞ xT ∴ xT = ν 6 × 105ν = 2700ν 222.2 μ = 3.22 ×10−5 m2 /s ∴ xT = 2700 × 3.22 ×10−5 = 8.7 ×10−2 m or 8.7 cm ρ a) Use Recrit = 3 × 105 = 10 xT ∴ xT = 0.03 m or 3 cm c) Use Recrit = 3 × 105 = 10 xT ∴ xT = 0.03 m or 3 cm e) Re = 6 × 104 = 10−6 10−6 10 xgrowth ∴ xgrowth = 0.006 m 10−6 or 6 mm The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19. The pressure distribution (see solution 8.89) on the surface is p = p 0 − 2 ρU ∞2 sin 2 α where rc α = x (α is zero at the stagnation point). Then 8.102 p( x) = 20,000 − 2 ×1000 ×102 sin 2 ( x/2) = 20 − 200 sin 2 ( x /2) kPa The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall: vθ ( r = 2) = −10 sin θ − 10 sin θ = −20 sin(π − α ) = 20 sin α ∴ U ( x) = 20 sin( x /2) The height h above the plate is h( x) = mx + 0.4 0.1 = m × 2 + 0.4 ∴ m = −0.15 ∴h(x) = 0.4 − 0.15x Continuity: 6× 0.4 = U(x)h ∴U(x) = 8.104 U ( x) = Euler’s Eqn: 2.4 or 0.4 − 0.15x 16 2.67 − x ρu ∂p ∂u =− ∂x ∂x ∴ dp 16 16 256 =ρ × = 2 dx 2.67 − x (2.67 − x) (2.67 − x)3 122 © 2012 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 8 / External Flows Von Kármán Integral Equation δ τ 0 = −δ = −δ δ dp d d + U ( x ) ∫ ρudy − ∫ ρu 2 dy dx dx 0 dx 0 δ ⎞ d δ dp d δ dU ⎛ uUdy udy + ρ − ρ ρu 2 dy ⎜ ⎟− ∫ ∫ ∫ dx dx 0 dx ⎝ 0 ⎠ dx 0 8.106 where we have used g δ ⎞ df dfg dg ⎛ = −f ⎜⎜ Here U = g , f = ∫ ρudy ⎟⎟ dx dx dx ⎝ 0 ⎠ δ δ dp d dU ∴τ 0 = −δ + ∫ ρu(U − u)dy − ρ udy dx dx 0 dx ∫0 ( ρ = const.) δ dU d a) If dp/dx = 0 then = 0 and τ 0 = ρ ∫ u(U∞ − u)dy dx dx 0 τ0 = ρ δ δ πy⎛ πy⎞ πy y⎤ d d ⎡ 2δ d ⎛ 2δ δ ⎞ U∞2 sin 1 − sin dy = ρU ∞2 cos − − = ρU∞2 − ⎟ ⎜ ⎜ ⎟ ∫ ⎢ dx 0 2δ ⎝ 2δ ⎠ dx ⎣ π 2δ 2 ⎦⎥ 0 dx ⎝ π 2 ⎠ Also, we have τ 0 = μ ∴ μU ∞ 8.108 b) τ 0 = μU∞ ∂u π = μU∞ cos 0 ∂y y =0 2δ π dδ = 0.137 ρU ∞2 2δ dx ∫ ∴ v = U∞ 0 8.110 U∞ × νx 2 4.79 ⎝ ∴ δ = 4.79 νx U∞ ⎛ ayx −3/ 2 ⎞ ⎞ ay ∂ ∂v ⎛ ay ⎞ = = − =− U U sin ⎟⎟ cos ⎟⎟ ∞ ∞⎜ ⎜ ⎟ ⎜ ∂ ∂ x y 2 x ⎝ x⎠ ⎠ ⎝ ⎠ ⎛ U∞ ⎞ U∞3 cos ⎜⎜ 0.328y ⎟⎟ dy = 0.0316 ν νx ⎠ ν ⎝ 0.164 y U∞ x U∞ dx U∞ U∞ = 0.328μU∞ 2 4.79 ν x νx ∂x δ ν π 1 ⎛ c) ∂u = U∞ ∂ sin ⎜ π y ⎜ ∂x ∴ δ dδ = 11.5 3/ 2 δ ⎡⎛ ∫ y cos ⎢⎣⎢⎜⎜⎝ 0.189 0 U∞ ⎞ ⎤ ⎟ y ⎥ dy ν ⎟⎠ ⎦⎥ δ /2 ⎧δ /6 y⎞ d ⎪ ⎛ y 1 ⎞⎛ y 1 ⎞ 2 y⎛ τ 0 = ⎨ ρ 3U∞ ⎜1 − 3 ⎟ dy + ρU∞2 ⎜ + ⎟⎜1 − − ⎟ dy δ⎝ δ⎠ dx ⎪ ⎝ δ 3 ⎠⎝ δ 3 ⎠ δ /6 ⎩0 ∫ ∫ 123 © 2012 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 8 / External Flows δ + ∫ δ /2 = 3U d ρU∞2 (0.1358δ ) = μ ∞ dx δ ∴δ ⎫ y 2 ⎞⎛ y 2⎞ ⎪ + ⎟ ⎜1 − − ⎟ dy ⎬ ⎝ 3δ 3 ⎠ ⎝ 3δ 3 ⎠ ⎪ ⎭ ⎛ ρU∞2 ⎜ dδ μ = 22.08 dx ρU∞ Thus ⎛ 6.65 ν ⎞ 2 −1/ 2 , τ 0 (x) = 0.1358ρU∞2 ⎜⎜ ⎟⎟ = 0.451ρU∞ Rex U∞ 2 U x ∞ ⎠ ⎝ νx δ (x) = 6.65 % error for δ = 6.65 − 5 ×100 = 33% 5 % error for τ 0 = 0.451 − 0.332 × 100 = 36% 0.332 The given velocity profile is that used in Example 8.13. There we found δ = 5.48 ν x/U∞ = 5.48 10−6 x/10 = 0.00173 x = 0.00173 3 = 0.003 m Assume the streamline is outside the b.l. Continuity is then 0.003 10 × 0.02 = ∫ 0 8.112 ⎛ 2y y2 ⎞ 10 ⎜ − dy + (h − 0.003)10 ⎜ 0.003 0.0032 ⎟⎟ ⎝ ⎠ = 0.02 + 10h − 0.03 1 δd = 10 0.003 ∫ 0 ∴ h = 0.021 m or 2.1 cm ⎛ 20 y 10 y 2 ⎞ 1 + ⎜⎜10 − ⎟⎟ dy = ( 0.03 − 0.03 + 0.01) = 0.001 m 2 0.003 0.003 ⎠ 10 ⎝ h − 2 = 2.1 − 2 = 0.1 cm or 0.001 m The streamline moves away from the wall a distance δ d δ 3⎞ 3⎞ ⎛ ⎛ a) u = U ∞ ⎜ 3 y − 1 y ⎟ δ d = 1 U ∞ ⎜ 1 − 3 y + 1 y ⎟ dy = δ − 3 δ + 1 δ = 0.375δ ⎜ 2 δ 2 δ3 ⎟ ⎜ 2 δ 2 δ3 ⎟ 4 8 U∞ ⎝ ⎠ ⎝ ⎠ ∫ 0 8.114 From Eq. 8.6.16, δ d = 0.375 × 4.65 θ= νx U∞ = 1.74 νx U∞ % error = 1.2% δ 3 ⎞⎛ ⎛ 3 y 1 y3 ⎞ 2 3 y 1 y U − 1 − + dy = 0.139δ ⎜ ⎟⎜ ∞⎜ 3 ⎟⎜ 3⎟ ⎟ 2 δ 2 2 δ 2 U∞2 δ δ ⎝ ⎠⎝ ⎠ 0 1 ∫ 124 © 2012 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 8 / External Flows ∴θ = 0.139 × 4.65 a) δ = 4.65 b) νx U∞ νx U∞ = 0.648 νx % error = U∞ 1/ 2 ⎛ 1.5 ×10−5 × 6 ⎞ = 4.65 ⎜ ⎟⎟ ⎜ 4 ⎝ ⎠ τ 0 = 0.323ρU∞2 0.648 − 0.644 × 100 = 0.62% 0.644 ν xU∞ = 0.0221 m 2 ⎛ 1.5 × 10 = 0.323 × 1.22 × 4 ⎜ ⎜ ⎝ 6× 4 −5 ⎞1/ 2 ⎟⎟ ⎠ = 0.00498 Pa 1/ 2 −5 ⎞ ⎛ c) Drag = 1 ρU∞2 Lw ×1.29 ν = 1 ×1.22 × 42 × 6 × 5 ×1.29 ⎜ 1.5 ×10 ⎟ ⎜ 6× 4 ⎟ 2 LU∞ 2 ⎝ ⎠ d) 8.116 = 0.299 N ⎡ 3y 3 y3 ⎤ dδ ∂u = U∞ ⎢ − 2 + ⎥ ∂x 2 δ 4 ⎥⎦ dx ⎢⎣ 2δ ⎡ ⎤ dδ 42 y 3 3× 4 3 = 4 ⎢− y + × ⎥ 2 −5 2 4.654 × (1.5 ×10−5 × 3)2 ⎦⎥ dx ⎣⎢ 2 × 4.65 ×1.5 ×10 × 3 ∴ ∂u 4.65 1.5 × 10−5 1 = 4 ⎡ −6166 y + 2.53 × 107 y3 ⎤ = −64.1y + 2.63 × 105 y3 ⎣ ⎦ ∂x 2 4 3 δ ∫ ∴v = − 0 ∂u 64.1 2.63 × 105 × 0.01562 − × 0.01564 = 0.00391 m/s dy = ∂x 2 4 where δ x =3 = 4.65 1.5 ×10−5 × 3 = 0.01560 m 4 Laminar and Turbulent Boundary Layers 8.118 −5 ⎞ ⎛ a) δ = 0.38 × 6 ⎜ 1.5 × 10 ⎟ ⎜ 20 × 6 ⎟ ⎝ ⎠ 0.2 ⎛ 1.5 × 10−5 ⎞ 1 = 0.0949 m τ 0 = × 1.22 × 202 × 0.059 ⎜ ⎜ 20 × 6 ⎟⎟ 2 ⎝ ⎠ = 0.6 Pa 8.120 0.2 ⎡ ⎛ 1.47 ×10−5 ⎞ ⎛ 1.47 × 10−5 ⎞ ⎤ 1 2 ⎢ ⎥ a) Drag = ×1.225 × 6 × (3.6 × 15) 0.074 ⎜ − 1060 ⎜ ⎟ ⎜ 6 × 3.6 ⎟ ⎜ 6 × 3.6 ⎟⎟ ⎥ ⎢ 2 ⎝ ⎠ ⎝ ⎠⎦ ⎣ = 1191(4.32 ×10−3 − 0.72 ×10−3 ) = 4.3 N 125 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 0.2 Chapter 8 / External Flows ⎛ 1.5 ×10−5 ⎞ 60 × 1000 a) U∞ = = 16.67 m/s δ = 0.38 × 100,000 ⎜ ⎜ 16.67 × 105 ⎟⎟ 3600 ⎝ ⎠ 0.2 = 235 m 0.2 ⎡ ⎛ 1.5 ×10−5 ⎞ ⎤ 1 1 2 2 ⎢ ⎥ = 0.0618 Pa τ 0 = ρU∞ c f = ×1.22 ×16.67 × 0.059 ⎜⎜ 5⎟ ⎟ ⎥ ⎢ 2 2 16.67 10 × ⎝ ⎠ ⎦ ⎣ 8.122 b) τ 0 = ∴ uτ = 1 1 0.455 ρU∞2 c f = ×1.22 ×16.672 = 0.151 Pa 2 2 2 5 ⎡ ⎛ 16.67 ×10 ⎞ ⎤ ⎢ln ⎜⎜ 0.06 ⎟⎥ 1.5 ×10−5 ⎠⎟ ⎥⎦ ⎢⎣ ⎝ 0.151 16.67 0.351δ = 0.351 m/s ∴ = 2.44 ln + 7.4 ∴ δ = 585 m 1.22 0.351 1.5 ×10−5 Both (a) and (b) are in error; however, (b) is more accurate. a) Use Eq. 8.6.40: c f = b) τ 0 = 8.124 uτ = c) δν = d) 0.455 ⎡ ⎛ 90 × 6 ⎞ ⎤ ⎟⎥ ⎢ ln ⎜ 0.06 1.47 ×10−5 ⎠ ⎦ ⎣ ⎝ 2 = 0.00213 1 1 ρU ∞2 c f = × 1.225 × 902 × 0.00213 = 10.52 Pa 2 2 10.52 = 2.93 m/s 1.225 5ν 5 ×1.47 × 10−5 = = 2.51× 10−5 m uτ 2.93 90 2.93δ = 2.44 ln + 7.4 2.93 1.47 ×10−5 ∴δ = 0.071 m Assume flat plates with dp/dx = 0. C f = 8.126 0.523 ⎡ ⎛ 10 ×100 ⎞ ⎤ ⎟⎥ ⎢ln ⎜ 0.06 10−6 ⎠ ⎦ ⎣ ⎝ 2 = 0.00163 1 ∴Drag = 2 × × 1000 × 102 × 10 × 100 × 0.00163 = 163,000 N 2 To find δ max we need uτ : 126 © 2012 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 8 / External Flows 1 2 τ 0 = × 1000 × 102 0.455 ⎡ ⎛ 10 × 100 ⎞ ⎤ ⎟⎥ ⎢ ln ⎜ 0.06 10−6 ⎠ ⎦ ⎣ ⎝ 2 = 70.9 Pa ∴ uτ = 10 0.266δ = 2.44 ln + 7.4 0.266 10−6 70.9 = 0.266 m/s 1000 ∴δ max = 0.89 m Laminar Boundary-Layer Equations ∂ψ ∂u ∂ 2ψ ∂ψ u= , = , v=− , ∂y ∂x ∂x∂y ∂x 8.128 ∂u ∂ 2ψ = , ∂y ∂y 2 ∂ 2u ∂y 2 = ∂ 3ψ ∂y 3 Substitute into Eq. 8.6.45 (with dp /dx = 0) : ∂ψ ∂ 2ψ ∂ψ ∂ 2ψ ∂3ψ − =ν 3 ∂y ∂x∂y ∂x ∂y 2 ∂y u= U∞ dF ∂η ∂ψ = U∞ν x = U∞ν x F '(η ) = U∞ F '(η ) dη ∂y νx ∂y We used Eq. 8.6.50 and Eqs. 8.6.48. 8.130 v=− ∂ψ ∂ =− ∂x ∂x ( 1 U∞ν ∂F ∂η F − U∞ν x 2 x ∂η ∂x =− U∞ ⎛ 1 −3/2 ⎞ 1 U∞ν F − U∞ν x F ' y − x ⎟ 2 x ν ⎜⎝ 2 ⎠ =− y U ∞ U ∞ν 1 U ∞ν 1 U ∞ν F+ F'= (η F '− F ) 2 x 2 νx x 2 x a) τ 0 = 0.332 × 1.22 × 52 b) δ = 5 8.132 c) vmax = δ ) U∞ν x F = − 1.5 × 10−5 = 0.0124 Pa 2×5 1.5 × 10−5 × 2 = 0.0122 m 5 1.5 ×10−5 × 5 ⎤ η ( F ' − F ) = × 0.8605 = 0.00527 m/s ⎥⎦ x ⎢⎣ 2 2 max ν U∞ ⎡ 1 δ δ dF dF vx d) Q = ∫ u(1× dy) = ∫ U∞ dy = ∫ U∞ dη dη dη U∞ 0 0 0 127 © 2012 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 8 / External Flows = U∞ νx U∞ [ F (δ ) − F (0)] = 5 1.5 ×10−5 × 2 × 3.28 = 0.04 m3 /s/m 5 At x = 2 m, Re = 5 × 2/10−6 = 107 ∴Assume turbulent from the leading edge. 1 0.455 1 0.455 ρU∞2 = ×1000 × 52 = 32.1 Pa 2 2 2 ⎡(ln 0.06 ×107 ) ⎤ [ln(0.06 Re x )] 2 ⎣ ⎦ a) τ 0 = b) uτ = 8.134 τ0 32.1 = = 0.1792 m/s ρ 1000 5 0.1792δ = 2.44 ln + 7.4 0.1792 10−6 ∴δ = 0.0248 m c) Use the 1/7 the power-law equation: 0.0248 Q= ∫ 0 1/7 ⎛ y ⎞ 5⎜ ⎟ ⎝ 0.0248 ⎠ dy = 0.109 m3 /s/m From Table 8.5 we interpolate for F' = 0.5 to be η= 0.5 − 0.3298 ( 2 − 1) + 1 = 1.57 0.6298 − 0.3298 1.57 = 8.136 v= U∞ 5 =y νx 1.5 ×10−5 × 2 νU ∞ ⎛ 1 ⎞ ⎜ ⎟ (ηF '− F ) x ⎝ 2⎠ = ∴ y = 0.00385 m or 3.85 mm 1.5 ×10−5 × 5 (0.207) = 0.00127 m/s 2 −5 ⎛ ∂u ∂v ⎞ ν 2 1.5 × 10 2 = 0.291(1.2)5 = 0.011 Pa τ = μ⎜ + ⎟ = F " ρU ∞ xU∞ 2×5 ⎝ ∂y ∂x ⎠ ⎛ 3 y 1 y3 ⎞ u = U∞ ⎜ − ⎜ 2 δ 2 δ 3 ⎟⎟ ⎝ ⎠ 8.138 y U cubic For the Blasius profile: see Table 8.5. (This is Blasius only a sketch. The student is encouraged to draw the profiles to scale.) 8.140 A: ∂p < 0. (favorable) ∂x 128 © 2012 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 8 / External Flows B: ∂p ≅0 ∂x y C ∂p C: > 0 (unfavorable) ∂x D: ∂p >0 ∂x E: ∂p <0 ∂x D δD δ C E δΒ δ δ E Α A 129 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. B Chapter 8 / External Flows 130 © 2012 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 9 / Compressible Flow CHAPTER 9 Compressible Flow Introduction c p = cv + R c p = kcv ∴ cp = 9.2 ∴ cp = cp ⎛ 1⎞ + R or c p ⎜1 − ⎟ = R k ⎝ k⎠ Rk k −1 Speed of Sound Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change: Q − W V 2 − V12 p2 p1 S = 2 + − + u − u m 2 ρ 2 ρ1 2 1 ~ + pv = u ~ + p / ρ. Therefore Enthalpy is defined in Thermodynamics as h = u Q − W V 2 − V12 S = 2 + h2 − h1 2 m 9.4 Assume the fluid is an ideal gas with constant specific heat so that Δh = c p ΔT . Then 2 2 Q − W S = V2 − V1 + c (T − T ) 1 p 2 m 2 Next, let c p = cv + R and k = c p /c v so that c p / R = k /( k − 1). Then, with the ideal gas law p = ρ RT , the first law takes the form Q − W V22 − V12 k ⎛ p2 p1 ⎞ S = + − ⎟ ⎜ m k − 1 ⎝ ρ 2 ρ1 ⎠ 2 The speed of sound is given by c = dp/dρ 9.6 For an isothermal process TR = p /ρ = K, where K is a constant. This can be differentiated: dp = Kdρ = RTdρ. Hence, the speed of sound is c = RT . 131 © 2012 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 9 / Compressible Flow For water Bulk modulus = ρ dp = 2110 × 106 Pa dρ 9.8 Since ρ = 1000 kg/m 3 , we see that c= dp 2110 ×106 = = 1453 m/s 1000 dρ Since c = 1450 m/s for the small wave, the time increment is 9.10 9.12 Δt = d 10 = = 0.0069 s c 1450 c = kRT = 1.4 × 287 × 263 = 325 m/s ∴ d = ct = 256 × 1.21 = 393 m c = 1.4 × 287 × 263 = 256 m/s sin α = sin α = 0.256 ∴ tan α = 0.2648 = 9.14 Δt = 1 c = M V 1000 ∴ L = 3776 m L 3776 = 3.776 s 1000 V 1000 m L Eq. 9.2.4: ΔV = − Δp Δp 0.3 =− =− = −0.35 m/s ρc ρ kRT 1.225 1.4 × 287 × 288 Energy Eq: 9.16 V2 (V + ΔV )2 + c pT = + c p (T + ΔT ) 2 2 ∴ΔT = (ΔV )2 ∴ 0 = V ΔV + + c p ΔT 2 −cΔV − 1.4 × 287 × 288 m/s (−0.35 m/s) = = 0.012 K cp 1005 (N-m)/(kg ⋅ K) Use N = kg ⋅ m ⋅ s −2 (Units can be a pain!) 132 © 2012 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 9 / Compressible Flow Isentropic Flow a) ps = patm + 10 = 69.9 + 10 = 79.9 kPa abs V p1 = 69.9 kPa abs 1 s Vs=0 From 1 → s : 9.18 V12 p1 ps + = 2 ρ1 ρs 1/ k ⎛p ⎞ ρs = ρ1 ⎜ s ⎟ ⎝ p1 ⎠ V12 69,900 79,900 ∴ + = 2 0.906 0.997 Is pr < 0.5283p0 ? 1/1.4 ⎛ 79.9 ⎞ = 0.906 ⎜ ⎟ ⎝ 69.9 ⎠ ∴ V1 = 77.3 m/s 0.5283 × 200 = 105.7 kPa a) pr < 0.5283p0 ∴ choked flow ∴ Me = 1 1000 × 298 = 9.20 ρe = = 0.997 kg/m3 ∴ Ve2 = kRTe pe = 105.7 kPa 1.4 × 287Te + 1000Te . ∴ Te = 248.1 K, Ve = 315.8 m/s. 2 105.7 = 1.484 × π × 0.012 × 315.8 = 0.1473 kg/s = 1.484 kg/m3 ∴ m 0.287 × 248.1 b) pr > 0.5283p0 ρ0 = ∴ M e < 1 1000 × 298 = Ve2 1.4 130,000 + × 2 0.4 ρe 1.4 130 ⎛ ρe ⎞ = 200 ⎜⎝ 2.338 ⎟⎠ 200 = 2.338 ∴ ρ e = 1.7187 kg/m3 ∴ Ve = 257.9 m/s 0.287 × 298 = 1.7187 × π × 0.012 × 257.9 = 0.1393 kg/s ∴m a) pr < 0.5283 p0 ∴ Me = 1 ∴ pe = 0.5283 × 200 = 105.7 kPa Te = 0.8333 × 298 = 248.3 K ρe = 105.7 = 1.483 kg/m3 0.287 × 248.3 Ve = 1.4 × 287 × 248.3 = 315.9 m/s = 1.483 × π × 0.012 × 315.9 = 0.1472 kg/s ∴m 9.22 b) pr > 0.5283 p0 ∴ pe = 130 kPa, ρe = pe = 0.65 ∴ M e = 0.81, Te = 0.884T0 p0 130 = 1.719 kg/m3 , Ve = 0.81 1.4 × 287 × 263.4 = 263.5 m/s 0.287 × 263.4 = 1.719 × π × 0.012 × 263.5 = 0.1423 kg/s ∴m 133 © 2012 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 9 / Compressible Flow pe = 0.5283 × 400 = 211.3 kPa abs 9.24 Te = 0.8333 × 303 = 252.5 K = Ve = 1.4 × 287 × 252.5 = 318.5 m/s ∴ m 211.3 π × 0.052 × 318.5 = 7.29 kg/s 0.287 × 252.5 pe = 0.5283 p0 = 101.3 kPa ∴ p0 = 192 kPa Te = 0.8333 × 280 = 233.3 K Ve = 1.4 × 287 × 233.3 = 306.2 m/s 9.26 = [101.3 ×103 /(287 × 233.3)] × π × (0.3)2 × 306.2 = 1.31 kg/s m p0 = 2 ×192 pe = 0.5283 p0 = 203 kPa, Te = 233.3 K, Ve = 306.2 m/s = [203×103 /(287 × 233.3)]× π × (0.03)2 × 306.2 = 2.62 kg/s m 1.667 / 0.667 5193 × 300 = 1.667 × 2077 Te ⎛ 225 ⎞ + 5193 Te ∴ Te = 225 K ∴ pe = 200 ⎜ ⎟ 2 ⎝ 300 ⎠ = 97.45 kPa abs Next, Tt = 225 K, pt = 97.45 kPa; ∴ Vt = 1.667 × 2077 × 225 = 882.6 m/s ρt = 9.28 97.45 = 0.2085 kg/m3 2.077 × 225 V 2 1.667 pe 5193 × 300 = e + 2 0.667 ρ e = 0.2085×π ×0 .032 ×882.6 = ρ eπ × 0.0752 Ve 1.667 ρe ⎛ ⎞ pe = 200 ⎜ ⎟ ⎝ 200 / 2.077 × 300 ⎠ = 1330 ρ e1.667 kPa Ve2 + 3324 × 103 × 9.54Ve−0.667 2 or 3.116 ×106 = Ve2 + 63 420 ×103 Ve−0.667 Trial-and-error: Ve = 91.8 m/s ∴ ρe = 0.3203 kg/m3 and pe = 199.4 kPa abs We need to determine the Mach number at the exit. Since the M = 1 at the throat, then A* = Athroat = 9.7 cm 2 . Hence, the area ratio at the exit is 9.30 Ae A* = 13 9.7 = 1.34 . Using the air tables, we find two possible solutions, one for subsonic flow, and the other for supersonic flow in the diverging section of the nozzle. At the exit: Subsonic Flow: ⇒ M e = 0.5, Te T0 = 0.9524, and pe p0 = 0.8430 Hence, Ve = Me ce = Me kRTe = 0.5 1.4 × 287 ( 0.9524 × 295) = 168 m/s Supersonic Flow: ⇒ M e = 1.76, Te T0 = 0.6175, and pe p0 = 0.1850. 134 © 2012 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 9 / Compressible Flow Hence, Ve = M e ce = M e kRTe = 1.76 1.4 × 287 ( 0.6175 × 295 ) = 476 m/s ρ1 = p1 / RT1 = (310 + 101.3)103 /(287 × 289) = 4.96 kg/m3 1/1.4 ⎛ 351.3 ⎞ ρ 2 = 4.96 ⎜ ⎟ ⎝ 411.3 ⎠ 9.32 = 4.43 kg/m3 V1 × 4.96 × (0.1)2 = V2 × 4.43 × (0.05)2 ∴ V2 = 4.48 V1 V12 1.4 411.3 × 103 4.482 V12 1.4 351.3 × 103 + × = + × 2 0.4 4.96 2 0.4 4.43 ∴ V1 = 36.5 m/s = 4.96π × (0.05) 2 × 36.5 = 1.42 kg/s ∴m Vt2 = kRT 1000 × 293 = 1.4 / 0.4 ⎛ 244 ⎞ ∴ pt = 500 ⎜ ⎟ ⎝ 293 ⎠ 9.34 1000 × 293 = pe ρ e1.4 = 1.4 × 287 Tt + 1000 Tt ∴ Tt = 244.0 K Vt = 313.1 m/s 2 = 263.5 kPa abs ∴ ρt = Ve2 1.4 pe + 2 0.4 ρ e 263.5 = 3.763 kg/m3 0.287 × 244 3.763 × π × 0.0252 × 313.1 = ρ eπ × 0.0752Ve 263,500 3.7631.4 ∴ 293,000 = Ve2 + 1.014 ×106 Ve−0.4 . Trial-and-error: Ve = 22.2 m/s, 659 m/s 2 ∴ ρe = 5.897, 0.1987 kg/m3 ∴ pe = 494.2 kPa, 4.29 kPa abs M t = 1 ∴ pt = 0.5283 × 830 = 438.5 kPa, Tt = 0.8333 × 289 = 241 K ∴ ρt = 6.34 kg/m 3 = 14.6 = 6.34 m 9.36 π dt2 4 1.4 × 287 × 241 ∴ dt = 0.097 m p 103 = = 0.125 ∴ M e = 2.014, Te = 0.552 × 289 = 160 K, Ve = 2.014 1.4 × 287 × 160 p0 830 = 253 m/s A A* = 1.708 ∴ π de2 4 = 1.708 π × 0.0972 4 ∴ de = 0.127 m 135 © 2012 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 9 / Compressible Flow Using compressible flow tables for air, we determine the pressure ratio and temperature ratio for M = 2.8 to be: p T = 0.03685, and = 0.3894 p0 T0 9.38 ∴ p = 0.03685 × p0 = 129 kPa abs and T = 0.3894 × T0 = 125 K ∴ V = Mc = 2.8 kRT = 2.8 1.4 × 287 × 320 = 1004 m/s Let Mt = 1. Neglect viscous effects. M1 = 9.40 ∴ A A* = 1.5007 ∴ At = 150 = 0.430 1.4 × 287 × 303 A1 π × 0.052 π dt2 = = ∴ dt = 0.0816 m or 8.16 cm 1.5007 1.5007 4 Isentropic flow. Since k = 1.4 for nitrogen, the isentropic flow table may be used. At M = 3, A A* i = 4.235 M>1 Vi = 3 1.4 × 297 × 373 = 1181 m/s 9.42 ∴ Ai = t ρi = Ve ~= 0 M<1 e Mt = 1 100 = 0.9027 kg/m3 0.297 × 373 m 10 0.00938 = = 0.00938 m2 ∴ At = = 0.00221 m2 ρi Vi 0.9027 ×1181 4.235 At M = 3, T = 0.3571 T0 , p = 0.02722 p0 ∴T0 = Te = 373 100 = 1044 K or 772DC p0 = = pe = 3670 kPa abs 0.3571 0.02722 Assume pe = 101 kPa. Then ρ e = 9.44 = ρ AV 2 F = mV Mt = 1 9.46 101 = 0.4198 kg/m3 0.189 × 1273 80,000 × 9.81 = 0.4198π × 0.252 V 2 6 ∴ V = 1260 m/s Ae = 4; ∴ M e = 2.94, pe = 0.02980 p0 A* Te = 0.3665 T0 = 0.3665 × 300 = 110.0 K, pe = 100 = 0.0298 p0 ∴ p0 = 3356 kPa abs F p0A0 ∴Ve = 2.94 1.4 × 287 ×109.95 = 618 m/s 136 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Ve Chapter 9 / Compressible Flow ∴ FB = −100 π × 0.052 × 6182 + 3,356,000π × 0.22 = 412,000 N 0.287 ×109.95 Normal Shock a) 0.9850 ×1000 = ρ 2V2 80,000 − p2 = 0.985 × 1000(V2 − 1000) ⎞ V22 − 10002 1.4 ⎛ p2 80 ⎛ ⎞ + − 287 × 283 ⎟ = 0 ⎜ ρ1 = = 0.9850 kg/m3 ⎟ ⎜ 2 0.4 ⎝ ρ 2 0.287 × 283 ⎝ ⎠ ⎠ V22 10002 1.4 V2 − + × (−985V2 + 1,065,000) − 284,300 = 0 2 2 0.4 985 9.48 ∴ 3V22 − 3784V2 + 784,300 = 0 ∴ V2 = 261 m/s ρ 2 = 3.774 kg/m3 Substitute in and find p2 = 808 kPa abs M1 = 1000 808 = 2.966 T2 = = 746 K or 473DC 0.287 × 3.774 1.4 × 287 × 283 M2 = 261 = 0.477 1.4 × 287 × 746 ρ 2 p2 T1 2 k M12 − k + 1 = = ρ1 p1 T2 k +1 ⎡ ( k + 1) 2 M12 ( k + 1)M12 = k −1 2 ⎤ 2 2 + ( k − 1)M12 ⎢⎣1 + 2 M1 ⎥⎦ [4 k M1 − 2 k + 2] M12 = 9.50 k + 1 p2 k − 1 (This is Eq. 9.4.12). Substitute into above: + 2k p1 2k ⎡ ⎤ ⎡ ⎤ p p ( k + 1) ⎢ ( k + 1) 2 + ( k − 1) ⎥ ( k + 1) ⎢ ( k + 1) 2 + k − 1⎥ p1 p1 ρ2 ⎣ ⎦ ⎣ ⎦ = = ⎧ ⎫ ( k + 1) 2 + ( k − 1)( k + 1) p 2 ρ1 p2 + ( k − 1) ⎬ 4 k + ( k − 1) ⎨ ( k + 1) p1 p1 ⎩ ⎭ = k − 1 + ( k + 1) p2 / p1 k + 1 + ( k − 1) p2 / p1 p2 ρ k +1 >> 1, 2 = p1 ρ1 k − 1 For a strong schock in which If M 2 = 0.5, then M1 = 2.645 ∴ V1 = 2.645 1.4 × 287 × 293 = 908 m/s 9.52 p2 = 8.00 × 200 = 1600 kPa abs ρ2 = 1600 = 8.33 kg/m3 0.287 × (2.285 × 293) 137 © 2012 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 9 / Compressible Flow p1 = 0.2615 × 101 = 26.4 kPa 9.54 T1 = 223.3 K M1 = 1000 / 1.4 × 287 × 223.3 = 3.34 ∴ M 2 = 0.4578 p2 = 12.85 × 26.4 = 339 kPa T2 = 3.101× 223.3 = 692.5 K For isentropic flow from 2 → 0: For M = 0.458, p = 0.866p0 and T = 0.960 T0 A * A T0 = 692.5 / 0.96 = 721 K or 448D C ∴ p0 = 339 / 0.866 = 391 kPa abs = 4 ∴ M e = 0.147 pe = 0.985 p0 ∴ p0 = 101 = 102.5 kPa abs 0.985 M t = 1 pt = 0.5283 ×102.5 = 54.15 kPa Tt = 0.8333 × 298 = 248.3 K 9.56 ∴ ρt = 54.15 = 0.7599 kg/m3 0.287 × 248.3 Vt = 1.4 × 287 × 248.3 = 315.9 m/s = 0.7599 × π × 0.0252 × 315.9 = 0.471 kg/s. If throat area is reduced, M t ∴m = 0.7599 × π × 0.022 × 315.9 = 0.302 kg/s remains at 1, ρt = 0.7599 kg/m3 and m pe = 100.3 kPa = p2 ∴ p1 = A A* = 4 ∴ M1 = 2.94, and p2 / p1 = 9.918 100.3 10.11 = 10.11 kPa At M1 = 2.94, p / p0 = 0.0298 ∴ p0 = = 339.3 kPa 9.918 0.0298 M t = 1, pt = 0.5283 × 393.3 = 179.2 kPa Tt = 0.8333 × 290 = 242 K 9.58 ∴Vt = 1.4 × 287 × 242 = 312 m/s M1 = 2.94, p1 = 10.11 kPa T1 = 0.3665 × 290 = 106.3 K ∴V1 = 2.94 1.4 × 287 × 106.3 = 607.6 m/s M 2 = 0.4788, pe = 100.3 kPa Te = T2 = 2.609 ×106.3 = 277.3 ∴ V2 = 0.4788 1.4 × 287 × 277.3 = 160 m/s Vapor Flow ⎛ 655 ⎞ pt = 0.546 p0 = 0.546 ×1200 = 655 kPa Tt = 673 ⎜ ⎟ ⎝ 1200 ⎠ 9.60 ∴ ρt = 655 = 2.42 kg/m3 0.462 × 585 = ρt At Vt ∴ 4 = 2.42 × m π × dt2 4 0.3/1.3 = 585 K Vt = 1.3 × 462 × 585 = 593 m/s (M t = 1) × 593 ∴ dt = 0.060 m or 6 cm 138 © 2012 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 9 / Compressible Flow ⎛ 101 ⎞ Te = 673 ⎜ ⎟ ⎝ 1200 ⎠ 0.3/1.3 = 380.2 K ∴ ρe = 101 = 0.575 kg/m3 0.462 × 380.2 Ve2 + 1872 × 380.2 = 1872 × 673 (Energy from 0 → e) (c p = 1872 J/kg ⋅ K) 2 ∴ Ve = 1050 m/s ∴ 4 = 0.575(π de2 /4) × 1050 ∴ de = 0.092 m or 9.2 cm ⎛ 546 ⎞ M e = 1, pe = 0.546 ×1000 = 546 kPa Te = 643 ⎜ ⎟ ⎝ 1000 ⎠ 9.62 ∴ ρe = 0.3/1.3 = 559 K 546 ×103 = 2.11 kg/m3 Ve = 1.3 × 462 × 559 = 579 m/s 462 × 559 3.65 = 2.11 π de2 4 × 579 ∴ de = 0.0617 m or 6.17 cm Oblique Shock Wave M1 = 800 = 2.29 1.4 × 287 × 303 V2 V1 From Fig. 9.15, β = 46°, 79° a) β = 46° ∴ M1n = 2.29 sin 46° = 1.65 ∴ M 2n = 0.654 = M 2 sin(46° − 20°) ∴ M 2 = 1.49 9.64 p2 = 3.01× 40 = 120.4 kPa abs T2 = 1.423 × 303 = 431 K V2 = 1.4 × 287 × 431 × 1.49 = 620 m/s a detached shock c) V1 = 35o M1n = 3.5sin 35° = 2.01 ∴ M 2n = 0.576 T2 = 1.696 × 303 = 514 K M 2 = 0.576/ sin(35° − 20°) = 2.26 9.66 θ1 = 20° = θ 2 ∴ β 2 = 47° M2n = 2.26sin 47° = 1.65 ∴ M3n = 0.654 = M3 sin(47° − 20°) ∴ M3 = 1.44 T3 = 1.423 × 514 = 731 K V3 = M3 kRT3 = 1.44 1.4 × 287 × 731 = 780 m/s 139 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 20o Chapter 9 / Compressible Flow M1 = 3, θ = 10° ∴ β1 = 28° M1n = 3sin 28° = 1.41 ∴ M 2n = 0.736 ∴ p2 = 2.153 × 40 = 86.1 kPa abs 9.68 M2 = 0.736 = 2.38 sin(28° − 10°) ∴ p3 = 6.442 × 86.1 = 555 kPa abs ( p3 )normal = 10.33 × 40 = 413 kPa abs Expansion Waves θ1 = 26.4° For M = 4, θ = 65.8° (See Fig. 9.18.) ∴θ = 65.8 − 26.4 = 39.4° 9.70 T2 = T1 T0 T2 1 = 273 × 0.2381 = 117 K T1 T0 0.5556 T2 = −156°C ∴ V2 = 4 1.4 × 287 × 117 = 867 m/s a) θ1 = 39.1° θ 2 = 39.1 + 5 = 44.1° ∴ M u = 2.72 p2u = (20/0.0585) × 0.04165 = 14.24 kPa abs 9.72 For θ = 5° and M = 2.5, β = 27° M1n = 2.5sin 27° = 1.13 ∴ M 2n = 0.889 ∴ p2A = 1.32 × 20 = 26.4 kPa abs MA = M2 = 0.889/ sin(27° − 5°) = 2.37 If θ = 5° with M1 = 4, then Fig. 9.15 → β =18° M1n = 4sin18° = 1.24 ∴ M 2n = 0.818 M1 M2u ∴ p2A = 1.627 × 20 = 32.5 kPa ∴ M 2A shock 0.818 = = 3.64 sin(18° − 5°) At M1 = 4, θ1 = 65.8° At 75.8°, M2u = 4.88 p2u = p1 shock M2l p0 p2 0.002177 = 20 0.006586 p p0 = 6.61 kPa CL = CD = Lift 32.5 A cos 5° − 20 × A/2 − 6.61× ( A/2) × cos10° = = 0.0854 1 1 2 2 ρ V1 A ×1.4 × 4 × 20 A 2 2 Drag 1 ρV12 A 2 = 32.5A sin5° − 6.61× ( A/2) × sin10° = 0.010 1 2 ×1.4 × 4 × 20A 2 140 © 2012 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 10 / Flow in Open Channels CHAPTER 10 Flow in Open Channels Introduction α = cos −1 (1 − 2 × 0.3) = 1.159 rad or 66.4° Q2B gA 3 = 10.2 = Q 2d sin α g ⎡ d 2/ 4 (α − sin α cos α ) ⎤ ⎣ ⎦ 3 42 sin(1.159) =1 9.81× (d5/ 64) × 1.159 − sin1.159cos1.159)3 This equation reduces to: 192.1 = d 5 ∴ d = 2.86 m Uniform Flow 10.4 141 © 2012 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 10 / Flow in Open Channels c1 A 5/ 3 S 0 . Substitute in n P 2/ 3 appropriate expressions for A and P, and solve for y 0 by trial and error. Use Chezy-Manning equation in the form Q = (a) 35 = 1 2 ⎡ ⎤ ⎢⎣ 4.5 y 0 + 2 y 0 (2.5 + 3.5) ⎥⎦ 5/3 0.015 ⎡ 4.5 + y 0 ( 1 + 2.52 + 1 + 3.52 ) ⎤ ⎣⎢ ⎦⎥ 0.00035 2/3 (4.5 y0 + 3y02 )5/3 This reduces to 28.06 = (4.5 + 6.33y0 ) 2/3 Solving, y0 = 2.15 m, A = 4.5 × 2.15 + 3 × 2.152 = 23.5 m 2 , 10.6 P = 4.5 + 6.33 × 2.15 = 18.1 m (c) ⎡1.82 / 4 (α − sin α cos α ) ⎤ ⎦ 3.3 = ⎣ 2/3 0.012(1.8α ) This reduces to 2.63 = 0.001 (α − sin α cos α )5 / 3 α 2/3 Solving, α = 1.94 rad, y0 = A= 5/3 1.8 (1 − cos1.94) = 1.255 m 2 1.82 (1.94 − sin1.94 cos1.94) = 1.845 m2 4 P = 1.8 × 1.94 = 3.5 m b = 0, m1 = 8, m2 = 0 1 1 8 A = by + y 2 (m1 + m2 ) = m1 y 2 = y 2 = 4 y 2 2 2 2 P = b+ y 10.8 Q = AR ( 2/3 ) ( 1 + m12 + 1 + m22 = y 2 ⎞ S0 2 ⎛ 4y = 4y ⎜ ⎟ n ⎝ 9.062 y ⎠ 2/3 ) ( 1 + m12 + 1 = y 0.0005 = 3.456 y8/3 0.015 (a) y = 0.12 m, Q = 3.956(0.12)8 / 3 = 0.0121 m 3 /s (b) ⎛ 0.08 ⎞ Q = 0.08 m /s, y = ⎜ ⎟ ⎝ 5.456 ⎠ 3 ) 65 + 1 = 9.062 y 3/8 = 0.244 m 142 © 2012 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 10 / Flow in Open Channels Energy Concepts q = 2gy 2 (E − y ) , E = constant, dq 1 4 gy ( E − y ) − 2 gy 2 2 gy ( E − y ) − gy 2 = = dy 2 q 2 gy 2 ( E − y ) 10.10 Setting dq / dy = 0 and noting that q ≠ 0, so that the numerator is zero, one can solve for y at the condition q = qmax (note: y = 0 is a trivial solution): gy[ 2(E − y ) − y ] = 0, 2E − 3y = 0, ∴ y = 2 E/3 = yc q = V1y1 = 3 × 2.5 = 7.5 m 2 /s E1 = y1 + q2 2 gy12 = 2.5 + 32 = 2.96 m 2 × 9.81 yc = 3 q 2/ g = 7.52/ 9.81 = 1.79 m, Ec = 3 3 yc = × 1.79 = 2.68 m 2 2 (a) E2 = E1 − h = 2.96 − 0.2 = 2.76 m, E2 > Ec ∴ y2 is subcritical 10.12 q2 7.5 2 2.87 ∴ E2 = y 2 + , or 2.76 = y 2 + = y2 + 2 2 2 2gy 2 2 × 9.81y 2 y2 Solving, y2 = 2.13 m, and y2 + h − y1 = 2.13 + 0.2 − 2.5 = −0.17 m (c) Set E2 = Ec = 2.68 m and ∴ maximum h is hmax = E1 − E2 = E1 − Ec = 2.96 − 2.68 = 0.28 m ∴ yc + hmax − y1 = 1.79 + 0.28 − 2.5 = −0.43 m (a) q1 = V1y1 = 3 × 3 = 9 m 2 /s, Q = b1q1 = 3 × 9 = 27 m 3 /s E1 = y1 + 10.14 q12 2 gy12 = 3+ 32 3 = 3.46 m, yc1 = 92 /9.81 = 2.02 m 2 × 9.81 3 Ec1 = × 2.02 = 3.03 m 2 Without change in width at loc. 2, E2 = E1 − h = 3.46 − 0.7 = 2.76 m Since E2 < Ec1 , width must change to prevent choking. Set Ec2 = 2.76 m 2 ∴ yc2 = × 2.76 = 1.84 m, q2 = gyc32 = 9.81×1.843 = 7.81 m2 /s 3 ∴ b2 = Q / q2 = 27 / 7.81 = 3.46 m 143 © 2012 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 10 / Flow in Open Channels q2 5.52 E1 = y1 + = 2.15 + = 2.484 m 2 gy12 2 × 9.81× 2.152 Fr1 = Ec = q gy13 = 5.5 9.81× 2.153 = 0.557 yc = 3 q 2 3 5.52 = = 1.456 ≈ 1.46 m 9.81 g 3 3 yc = × 1.456 = 2.183 ≈ 2.18 m 2 2 (a) The maximum height of the raised bottom at location 2 will be one for which the energy is a minimum: E1 = Ec + h , 2.484 = 2.183 + h , ∴ h = 2.484 − 2.183 = 0.30 m 10.16 (b) (c) (d) Since Fr1 < 1, if h > 0.30 m, subcritical nonuniform flow will occur upstream of the transition. At the canal entrance, the condition of critical flow is Fr 2 = Q 2 B / gA 3 = 1, to be solved for the unknown width b. 10.18 (a) Rectangular channel: B = b, A = by, y = yc ∴ 10.20 Q 2b 3 3 gb y = Q2 2 3 gb y =1 ∴b = Q gy 3 = 18 = 5.75 m 9.81 × 1 Objective is to determine the zero of the function f ( y ) = Q 2 B( y ) / g[ A ( y )]3 − 1, in which eqns. for B(y) and A(y) representing either a circular or trapezoidal section area can be substituted. The false position algorithm is explained in Example 10.10; this was used to determine the roots. The solutions are: (a) yc = 1.00 m, (c) yc = 1.81 m 144 © 2012 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 10 / Flow in Open Channels In the lower section ( y ≤ 1 m), the area is y y A = ∫ bdη = ∫ 3 η dη = 2 y 3/ 2 0 Assume y < 1 m; then Fr 2 = 0 If yc = 1 m, then Q2B gA3 = Q23 y g 8 y 9/ 2 = 3Q 2 8 gy 4 3Q 2 = 1, or Q = 8 × 9.81/ 3 = 5.1 m 3 /s 4 8 × 9.81× 1 ∴ when Q > 5.1 m 3 /s, critical depth will be > 1 m. Cross-sectional area including 10.22 upper region (y > 1 m) is A = 2( 1) 3 /2 + 23( y c − 1) = 23 y c − 21 , and B = 23 m (a) Q 2B 55 2 × 23 = = 1, ( 23y c − 21) 3 = 7092, 3 3 gA 9.81( 23y c − 21) (7092)1/ 3 + 21 ∴ yc = = 1.75 m 23 (b) Q2 B 3 × 3.52 = = 1, yc4 = 0.468, ∴ yc = 0.83 m gA3 8 × 9.81yc4 22.5 1.52 2 3 = 1.5 m /s, yc = (a) q = = 0.612 m 15 9.81 3 1.52 Ec = × 0.612 = 0.919 m, E0 = 1.2 + = 1.28 m = E3 2 2 × 9.81× 1.2 2 E2 = E0 − h = 1.28 − 0.2 = 1.08 m > E3 ∴ no choking. Since losses are neglected throughout the transition, y1 = y0, and y2 can be computed by writing the energy equation between locations 1 and 2: 10.26 1.28 = y2 + 1.52 2 × 9.81× 1.2 2 + h = y2 + 0.115 y22 + 0.2 , ∴ y2 ≈ 0.95 m (b) 145 © 2012 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 10 / Flow in Open Channels (a) B′ = 1 m, H′ = 0.36 m 10.28 Q = 0.11BH (1.522 B) (b) H′ (m) Q m3/s 0.026 ⇒B= B′ H′ = 3.281 and H = = 1.18 0.3048 0.3048 = 0.11× (3.281)(1.18)(1.522×3.218) 0.026 = 0.42 m3 /s 0.15 0.2 0.25 0.30 0.35 0.4 0.45 0.17 0.23 0.29 0.35 0.41 0.47 0.53 ⎛2 2 ⎞ ⎛2 2 ⎞ Q1 = b ⎜⎜ g ⎟⎟ ( y1 − h)3 / 2 , Q2 = b ⎜⎜ g ⎟⎟ ( y2 − h)3 / 2 ⎝3 3 ⎠ ⎝3 3 ⎠ given Q1 , y1 , Q2 , y2 , solve for b and h y1 − h ⎛ Q1 ⎞ =⎜ ⎟ y2 − h ⎝ Q2 ⎠ 10.30 (a) h= 2/3 or h = y1 − y2 (Q1 / Q2 ) 2 / 3 1 − (Q1 / Q2 ) 2 / 3 1.05 − 1.75(0.15 / 30) 2 / 3 = 1.03 m 1 − (0.15 / 30) 2 / 3 ∴b = Q2 30 = = 28.8 m ⎞ 2⎛ 2 ⎞ 2⎛ 2 3/ 2 3/ 2 g ⎟ ( y 2 − h) × 9.81 ⎟ (1.75 − 1.03) ⎜ ⎜ 3⎝ 3 ⎠ 3⎝ 3 ⎠ Momentum Concepts ⎛ y2 q2 ⎞ M = b⎜ + ⎟ ⎝ 2 gy ⎠ 2 q2 M 1⎛ y ⎞ = ⎜ ⎟ + by c2 2 ⎝ y c ⎠ gy c2 y 10.32 2 q2 1⎛ y ⎞ = ⎜ ⎟ + 3 2 ⎝ yc ⎠ gyc ( y / yc ) But q 2 /gyc3 = 1 2 M 1⎛ y ⎞ 1 ∴ 2 = ⎜ ⎟ + (y / yc ) by c 2 ⎝ y c ⎠ 146 © 2012 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 10 / Flow in Open Channels (a) Conditions at location 1: E1 = y1 + q12 2 gy12 = 1.8 + d/2 1.5 2 2 × 9.81 × 1.8 2 = 1.835 m q1 b2 b1 q2 d/2 3 yc = 3 q12 /g = 1.52 /9.81 = 0.612 m. The smallest constriction at location 2 is one that establishes critical flow, i.e., where minimum energy exists: ∴ E 2 = E c , or since E 1 = E 2 , 2 2 2 yc2 = Ec2 = E1 = × 1.835 = 1.22 m 3 3 3 ∴ q2 = g ( yc2 )3 = 9.81×1.223 = 4.22 m 2 /s 10.34 b2 = q1b1 1.5 × 6 = = 2.13 m 4.22 q2 Hence the maximum diameter is d = b 1 − b 2 = 6 − 2.13 = 3.87 m The momentum eqn. cannot be used to determine the drag since no information is given with regard to location downstream of cofferdam. But we do have available the drag relation, which will provide the required drag force: Frontal area: A = y1d = 1.8 × 3.87 = 6.97 m 2 Approach velocity: V1 = q1 / y1 = 1.5 /1.8 = 0.833 m/s ∴ F = C D A ρ V12 / 2 = 0.15 × 6.97 × 1000 × 0.8332 / 2 = 363 N , a rather insignificant drag force! Asill = = 1 wh 2 1 × 2 × 3 × 0.3 2 = 0.27 m 2 10.36 y= 1 y 3 A = my 2 = 3 y 2 147 © 2012 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 10 / Flow in Open Channels M1 = A1y1 + Q2 Q2 Q2 1 = 3 × 0.52 × × 0.5 + = + 0.125 , gA1 3 7.358 9.81× 3 × 0.52 Q2 1 Q2 Q2 2 M 2 = A2 y 2 + = 3 × 1.8 × × 1.8 + = 5.832 + , gA2 3 95.35 9.81 × 3 × 1.8 2 F γ = CD AsillQ2 2 gA12 = 0.4 × 0.27Q2 2 × 9.81× (3 × 0.52 )2 = Q2 102.2 Substitute into momentum eqn: M1 = M 2 + F /γ ∴ 0.125 + Q2 Q2 Q2 = 5.832 + + 7.358 95.35 102.2 1 1 ⎞ ⎛ 1 − − Q2 = (5.832 − 0.125) ⎜ ⎟ ⎝ 7.58 95.35 102.2 ⎠ −1 = 49.37 ∴ Q = 7.03 m3 /s 10.38 Given condition: V2 y2 = 0.4V1y1 Continuity eqn: y1 (V1 + w) = y2 (V2 + w) Momentum eqn: y2 1 ⎡ (V + w ) 2 ⎤ = ⎢ 1+ 8 1 − 1⎥ y1 2 ⎢ gy1 ⎥⎦ ⎣ Unknowns are V2 , y2 and w. Eliminate V2 and w, and solve for y2 . (a) V2 y2 = 0.4 × 1 × 1.5 = 0.6 1.5( 1 + w ) = y 2V2 + y 2 w = 0.6 + y 2 w, or w = 0.9 /( y2 − 1.5) 2 ⎡ ⎤ ⎛ y2 1 ⎢ 0.9 ⎞ ⎥ ∴ = 1 + 8 ⎜1 + ⎟ /(9.81 × 1.5) − 1⎥ or ⎢ 1.5 2 ⎝ y2 − 1.5 ⎠ ⎣⎢ ⎦⎥ 2 ⎛ 0.9 ⎞ 1 − 1 + 0.5437 ⎜1 + ⎟ + 1.333y2 = 0 ⎝ y2 − 1.5 ⎠ 148 © 2012 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 10 / Flow in Open Channels ∴Solving, y2 = 1.77 m ∴ V2 = ∴w = 0.6 0.6 = = 0.339 m/s y 2 1.77 0.9 0.9 = = 3.33 m/s y2 − 1.5 1.77 − 1.5 (a) To compute the discharge, write the energy equation between locations 1 and 2, and solve for q: 2 g ( y2 − y1 ) q= ( y1−2 − y2−2 ) = 2 × 9.81(0.10 − 2.5) ( 2.5 −2 − 0.10 −2 ) = 0.687 m 2 /s ∴ Q = bq = 5 × 0.687 = 3.43 m 3 /s (b) To compute depth downstream, first compute the Froude number at 2: 10.40 Fr2 = ∴ y3 = q gy y2 2 3 2 ( = 0.687 9.81 × 0.10 3 ) 1 + 8 Fr22 − 1 = = 6.934 0.10 2 ( ) 1 + 8 × 6.934 2 − 1 = 0.932 m (c) To calculate power lost, first compute the head loss in the jump: ( y 3 − y 2 )3 (0.932 − 0.10 ) 3 hj = = = 1.544 m 4y 3 y 2 4 × 0.932 × 0.10 = γ Qh = 9810 × 3. 43 × 1.544 = 5.20 × 10 4 watt, or 52.0 kW ∴W j Use Eqn. 10.5.16 with F = 0: M1 = = 10.42 y 12 Q2 ( 2my 1 + 3b) + 6 g(by 1 + my 12 ) 1.12 602 (2 × 3 × 1.1 + 3 × 5) + = 44.55 m3 2 6 9.81(5 × 1.1 + 3 × 1.1 ) ∴ M 2 = M1 or y22 Q2 (2my2 + 3b) + = 44.55 6 g (by2 + my22 ) y22 602 (2 × 3y2 + 3 × 5) + = 44.55 6 9.81(5y2 + 3y22 ) 149 © 2012 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 10 / Flow in Open Channels y 23 + 2.5 y 22 + 367 = 44.55 5 y 2 + 3 y 22 ∴Solving, y2 = 2.55 m Energy loss across jump is h j : h j = E1 − E2 = y1 + Q2 Q2 y − − 2 2 gA12 2 gA22 A1 = 5 × 1.1 + 3 × 1.12 = 9.13 m 2 , A2 = 5 × 2.55 + 3 × 2.552 = 32.26 m 2 , ∴ h j = 1.1 + 60 2 60 2 2.55 − + = 0.575 m 2 × 9.81 × 9.132 2 × 9.81 × 32.26 2 : ∴Power dissipated is W j = γ Qh = 9810 × 60 × 0.575 = 3.38 ×105 W W j j First find Q, then compute y 1 : V2 = Fr2 gy2 = 0.4 9.81× 1.5 = 1.534 m/s ∴ Q = V 2 A2 = 1.534 × 5 × 1.5 = 11.51 m 3 /s q = Q / b = 11.51/ 5 = 2.30 m 2 /s yc = 3 q 2 / g = 3 2.302 / 9.81 = 0.814 m To compute y2 , use momentum eqn, M1 − F / γ = M2 or 10.44 ⎛ y12 q 2 ⎞ C D Aρ V12 / 2 ⎛ y22 q 2 ⎞ =⎜ + ⎜⎜ + ⎟⎟ − ⎜ 2 gy ⎟⎟ bγ 2 ⎠ ⎝ 2 gy1 ⎠ ⎝ Substitute in known data, plus the relation V1 = q / y 1 : y12 2.302 0.35 × 5 × 0.17 × 1000 × 2.302 1.52 2.302 + − = + 2 9.81y1 2 5 × 9810 y12 9.81 × 1.52 which reduces to y12 + 1.078 0.0321 − = 2.97 y1 y12 ∴Solving, y1 = 0.346 m 150 © 2012 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 10 / Flow in Open Channels Nonuniform Gradually Varied Flow 3 yc = 3 q 2 /g = 0.1942 /9.81 = 0.157 m (a) q = Q / b = 0.35 /1.8 = 0.194 m 2 /s Use Chezy-Manning eqn. to compute y 0 : (1.8 y0 )5/3 0.35 × 0.012 Qn = = 0.0329 = (1.8 + 2 y0 ) 2/3 c1 S0 1× 0.0163 ∴Solving, y0 = 0.094 m Ec + h = 3 3 yc + h = × 0.157 + 0.1 = 0.335 m 2 2 E0 = y0 + q2 2 gy02 10.46 0.1942 = 0.311 m 2 × 9.81× 0.0942 = 0.094 + Since E 0 < E c + h , normal conditions cannot exist at loc. 1, and choking will occur at loc. 2. Compute alternate depths at locs. 1 and 3: E 1 = E 3 = E c + h. y+ 0.1942 2 × 9.81y 2 = y+ 0.00192 y2 = 0.335 ∴Solving, y 1 = 0.316 m, y 3 = 0.088 m. (Note: loc. 2 is a critical control, with subcritical flow upstream and supercritical flow downstream.) A jump is located upstream of loc. 1. Find the depth conjugate to y0: Fr03 = q2/gyo3 = 0.1942 /(9.81× 0.0943 ) = 4.62 ∴ycj = y0 2 ( 1+ 8× 4.62 −1) = 0.243 m ( 1+ 8Fr −1) = 0.094 2 2 0 3 yc = 3 q 2 /g = 8.252 /9.81 = 1.91 m q = Q / b = 33 / 4 = 8.25 m 2 /s Use Chezy-Manning eqn. to compute y 0 : 10.48 (4 y0 )5 / 3 Qn 33 × 0.012 = = 13.43 = c1 S0 1 × 0.00087 (4 + 2 y0 ) 2 / 3 ∴Solving, y0 = 2.98 m Since y 0 > y c , mild slope conditions prevail. With yentrance < yc , an M3 profile exists downstream of the entrance. For free outfall conditions, yexit ≅ yc , and an M 2 profile exists upstream of the exit. The M 3 and M 2 profiles are separated by a hydraulic jump located approximately 260 m downstream of the entrance (determined by numerical analysis). 151 © 2012 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 10 / Flow in Open Channels A spreadsheet solution is shown below. An M3 profile is situated upstream, with a hydraulic jump at approximately 150 m followed by an M1 profile. Q= 20 n= 0.014 S0 = 0.001 L= 300 g = 9.81 b= 0 m1 = 3.5 yu = 0.50 yd = 2.50 c1 = 1.00 m2 = 2.5 Depth Residual 10.50 Station 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 yc = 1.554 y0 = 1.809 -1.8E-09 y 0.500 0.600 0.700 0.800 0.900 1.000 1.100 1.200 1.300 1.400 2.500 2.450 2.400 2.350 2.300 2.250 2.200 A 0.750 1.080 1.470 1.920 2.430 3.000 3.630 4.320 5.070 5.880 18.750 18.008 17.280 16.568 15.870 15.188 14.520 0.00 V E ym S(ym) ycj FM Residual Δx x 26.667 36.744 0 3.769 54.49 2.77E-04 18.519 18.079 0.550 5.720E-01 33 33 3.324 37.97 -5.67E-05 13.605 10.135 0.650 2.347E-01 34 67 2.837 28.08 3.64E-06 10.417 6.330 0.750 1.094E-01 35 102 2.578 21.75 -5.48E-04 8.230 4.353 0.850 5.612E-02 36 138 2.365 17.51 -1.19E-04 6.667 3.265 0.950 3.101E-02 36 174 2.184 14.59 -2.52E-05 5.510 2.647 1.050 1.818E-02 36 210 2.028 12.56 -4.53E-06 4.630 2.292 1.150 1.119E-02 35 245 1.893 11.17 -1.48E-04 3.945 2.093 1.250 7.175E-03 32 277 1.778 10.24 -6.60E-05 3.401 1.990 1.350 4.760E-03 28 304 1.684 9.68 -3.90E-05 1.067 2.558 300 1.111 2.513 2.475 1.878E-04 -56 244 1.157 2.468 2.425 2.094E-04 -56 188 1.207 2.424 2.375 2.340E-04 -57 131 1.260 2.381 2.325 2.621E-04 -59 72 1.317 2.338 2.275 2.943E-04 -60 12 1.377 2.297 2.225 3.313E-04 -62 -51 q = Q /b = 15/4 = 3.75 m 2 /s 3 yc = 3 q 2 /g = 3.752 /9.81 = 1.13 m y01 = 0.93 m, yc > y01 ∴upstream reach has a steep slope. y02 = 1.42 m, yc < y02 10.52 ∴downstream reach has a mild slope. Compute depth conjugate to y 0 1 : Fr012 = q 2 /( gy013 ) = 3.752 /(9.81 × 0.933 ) = 1.782 ∴ ycj = y01 ⎡ 0.93 ⎡ 1 + 8Fr012 −1⎤ = 1 + 8 ×1.782 −1⎤⎦ = 1.35 m ⎢ ⎥ ⎦ 2 ⎣ 2 ⎣ 152 © 2012 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 10 / Flow in Open Channels ∴Hydraulic jump occurs upstream, followed by an S 1 curve up to the transition. 10.54 q 2 3 5.752 = = 1.50 m. Since y01 < yc and y02 > yc, there will be a g 9.81 hydraulic jump between A and C. (a) yc = 3 (a) Compute y c using Fr 2 = 1: Q2B gA 3 = Q 2 d sin α g ⎡ (d 2 /4)(α − sin α cos α ) ⎤ ⎣ ⎦ = 0.4175 3 = 2.52 × 2.5sin α 9.81 ⎡ (2.52 /4)(α − sin α cos α ) ⎤ ⎣ ⎦ 3 sin α =1 (α − sin α cos α )3 ∴Solving d 2 α = 1.115 rad and ∴ yc = (1 − cosα ) = 2.5 (1 − cos1.115) = 0.70 m 2 Compute y 0 using 10.56 ⎡ (d 2 /4)(α − sin α cos α ) ⎤ Qn ⎦ =⎣ c1 S0 (α d ) 2/3 which reduces to 1.039 = 5/3 2 ⎡ ⎤ 2.5 × 0.015 ⎣(2.5 /4)(α − sin α cos α ) ⎦ = , 1× 0.001 (α × 2.5)2/3 5/3 (α − sin α cos α )5 / 3 α 2/3 ∴Solving, α = 1.361 rad, and d 2.5 ∴ y0 = (1 − cosα ) = (1 − cos1.361) = 0.99 m 2 2 Since y 0 > y c , a mild slope condition exists. The water surface consists of an M 3 curve beginning at the inlet, followed by a hydraulic jump to an M 2 curve, which terminates at critical depth at the outlet. The numerically-predicted water surface and energy grade line are provided in the following table. 153 © 2012 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 10 / Flow in Open Channels x (m) 0 7 14 20* 20* 230 350 (a) q = E (m) 1.64 1.44 1.29 1.18* 1.07* 1.065 1.05 30 = 6 m2 /s 5 62 E1 = 2 + 2 × 9.81× 22 E2 = 1.8 + 10.58 y (m) 0.40 0.43 0.46 0.49* 0.97* 0.96 0.93 x y E (m) (m) (m) 406 0.90 1.03 442 0.87 1.01 465 0.84 1.00 480 0.82 0.98 490 0.79 0.97 500 0.70 0.95 * hydraulic jump yc = 3 62 =1.54 m 9.81 = 2.459 m 62 2 × 9.81× 1.82 = 2.366 m ym = 1.9 m, Am = 5 × 1.9 = 9.5 m 2 , Pm = 5 + 2 × 1.9 = 8.8 m, Rm = 9.5 / 8.8 = 1.08 m Q 2 n2 1 ⎛ 30 × 0.012 ⎞ =⎜ = 0.0036 ⎟ 2 4/3 4/3 Am Rm 9.5 ⎝ ⎠ 1.08 2 S( y m ) = ∴Δx = E2 − E1 2.366 − 2.459 = = 20.2 m S0 − S( ym )1 −0.001 − 0.0036 (b) An A2 profile is contained within the reach (which has an adverse slope with y2 < y1) Evaluate Q using Δx = 10.60 E2 − E1 S0 − S( ym ) E1 = y1 + Q2 Q2 Q2 = 1.05 + = 1.05 + 135.2 2 gA12 2 × 9.81 × (2.5 × 1.05) 2 E2 = y2 + Q2 Q2 Q2 = 1.2 + = 1.2 + 176.6 2 gA22 2 × 9.81× (2.5 × 1.2)2 ym = 1 1 ( y1 + y2 ) = (1.05 + 1.2) = 1.125 m 2 2 2 ⎛ Qn ⎞ (b + 2 ym ) 4 / 3 S ( ym ) = ⎜ ⎟ 10 / 3 ⎝ c1 ⎠ (bym ) 154 © 2012 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 10 / Flow in Open Channels = Q2 × 0.0132 × (2.5 + 2 ×1.125)4/ 3 = 4.296 × 10−5 Q 2 12 × (2.5 ×1.125)10/ 3 ∴Δ x[S0 − S( ym )] = E2 − E1 50(0.005 − 4.296 × 10−5 Q 2 ) = 1.2 + Q2 Q2 − 1.05 − 176.6 135.2 which reduces to 4.141 × 10 −4 Q 2 = 0.10 ∴ Solving, Q = 15.5 m3 /s ⎛Q⎞ yc = 3 ⎜ ⎟ ⎝b⎠ 2 ⎛ 15.5 ⎞ g =3⎜ ⎟ ⎝ 2.5 ⎠ 2 9.81 = 1.58 m ∴The profile is an S3 curve Write energy eqn. across the gate to compute the discharge: E1 = E2 or y1 + q2 2 gy12 = y2 + q2 2 gy22 ⎛ 1 1⎞ ∴ q = 2g( y 1 − y 2 )⎜ 2 − 2 ⎟ − 1 ⎝ y2 y1 ⎠ 1 ⎞ ⎛ 1 = 2 × 9.81× (1.85 − 0.35) ⎜ − − 1 = 1.93 m2 /s 2 2⎟ ⎝ 0.35 1.85 ⎠ ∴Q = bq = 4 ×1.93 = 7.72 m3/s yc = 3 q 2 / g = 1.932 / 9.81 = 0.72 m 10.62 Compute y 0 using Chezy-Manning relation: (4 y0 )5/ 3 Qn 7.72 × 0.014 = = 3.821 = (4 + 2 y0 )2 / 3 c1 S0 1× 0.0008 ∴ Solving, y 0 = 1.17 m. Since y 0 > y c , mild channel conditions exist. Upstream of the gate there will be an M 1 profile, with an M 3 downstream of the gate, terminating in a hydraulic jump to normal flow conditions. Compute the depth upstream of the jump, conjugate to y 0 : Fr02 = q 2 /gy03 = 1.932 /(9.81× 1.173 ) = 0.237 ∴ ycj = y0 2 ( 1 + 8Fr − 1) = 1.172 ( 1 + 8 × 0.237 − 1) = 0.41 m 2 0 155 © 2012 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 10 / Flow in Open Channels Use the step method to compute water surface and energy grade line: q2 1.93 2 0.190 E=y+ = y + =y+ 2 2 y2 2gy 2 × 9.81y S( y ) = Q 2 n 2 ( 4 + 2y ) 4 /3 7.72 2 × 0.014 2 ( 4 + 2y ) 4 /3 = 4 10 /3 y 10 /3 ( 4 y ) 10 /3 = 1.15 × 10 −4 ( 4 + 2y ) 4 /3 y −10/3 M 1 curve upstream of gate y E ym (m) 1.85 (m) 1.906 (m) 1.7 1.5 1.3 1.2 S( y m ) Δx x (m) (m) 0 1.775 2.514×10−4 −255 1.6 3.336×10−4 −390 1.4 4.824×10−4 −542 1.25 6.627×10−4 −583 1.766 1.584 1.412 1.332 −255 −645 −1187 −1770 M 3 curve downstream of gate y E ym (m) 0.35 (m) 1.901 (m) 0.37 1.758 0.39 0.41 S( y m ) Δx x (m) (m) 0 0.36 0.02741 5 0.38 0.02315 5 0.40 0.01973 5 5 1.639 10 1.540 15 Hydraulic jump occurs ~5 m from the gate. (a) 10.64 156 © 2012 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 10 / Flow in Open Channels Compute normal depth using Chezy-Manning eqn: (3.66 y02 )5/3 Qn 15.38 × 0.017 = = 5.476 = (3.66 + 2 y02 ) 2/3 c1 S0 1× 0.00228 ∴ Solving, y02 = 1.65 m Compute critical depth: q = Q /b = 15.38/3.66 = 4.20 m 2 /s 3 ∴ yc = 3 q 2 /g = 4.202 /9.81 = 1.22 m ∴Upstream of slope change the channel is steep, and downstream, the channel is mild. The hydraulic jump will terminate at normal depth. (a) Find depth conjugate to y 02 : Fr022 = q 2 /gy023 = 4.20 2 /(9.81 × 1.653 ) = 0.400 1.65 1 + 8Fr − 1) = ( 1 + 8 × 0.4 − 1) = 0.87 m ( 2 2 y0 2 ∴ ycj = 10.66 2 02 (b) Step method: E= y+ q2 2 gy 2 = y+ 4.20 2 0.900 = y+ 2 2 2 × 9.81y y Q 2 n 2 (b + 2y ) 4 /3 15.38 2 × 0.017 2 ( 3.66 + 2y ) 4 /3 S( y ) = = (by ) 10 /3 ( 3.66) 10 /3 y 10 /3 = 9.052 ×10−4 (3.66 + 2y)4/3 y−10/3 E ym (m) 0.6 (m) 3.100 (m) 0.7 2.537 y 0.8 0.87 S( y m ) Δx x (m) 0.65 0.03216 18.8 0.75 0.02104 17.6 0.835 0.01536 11.2 (m) 0 18.8 2.206 36.4 2.059 47.6 ≅ L j (c) An M 3 curve exists downstream of the change in channel slope, terminating in a hydraulic jump, approximately 50 m from the slope change. 157 © 2012 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 10 / Flow in Open Channels This is a design analysis problem. Some extensive calculations are required. (a) Find the depth of flow immediately upstream of the jump (call it location 1): V2 = Fr2 = Q 19 19 = = 1.267 m/s, q = =0.95 m 2 /s. A2 0.75 × 20 20 1.267 = 0.467, 9.81× 0.75 y1 = 0.75 2 ( 1 + 8 × 0.467 − 1) = 0.246 m 2 Compute the loss across the jump, and subsequently the dissipated power: 0.952 0.952 − 0.75 − = 3.87 m 2 × 9.81× 0.102 2 × 9.81× 0.752 h j = E1 − E2 = 0.10 + = γ Qh = 9810 × 19 × 3.87 = 7.21× 105 watt, or 721 kW ∴W j 10.68 (b) The length of the apron is the distance from the toe to downstream of the jump. First, compute the distance from ytoe to y1 using a single increment of length: Etoe = 0.10 + ym = 0.95 2 2 × 9.81 × 0.10 2 = 4.700 m, E1 = 0.246 + 0.95 2 2 × 9.81 × 0.246 2 = 1.006 m, 1 (0.10 + 0.246) = 0.173 m, A m = 20 × 0.173 = 3.46 m 2 , 2 ⎛ 19 × 0.014 ⎞ Pm = 20 + 2 × 0.173 = 20.35 m, Sm = ⎜ ⎟ ⎝ 3.46 ⎠ ∴ x1 − xtoe = 2 1 ( 3.46 / 20.35 )4/3 = 0.06275, E1 − Etoe 1.006 − 4.700 = = 59.3 m S0 − Sm 0.0005 − 0.06275 The length of the jump is six times the downstream depth, or 6 × 0.75 = 4.5 m. Hence the required length of the apron is L = 59.3 + 4.5 = 63.8 m, or approximately 65 m. Normal design would require additional length as a safety factor. Use the varied flow function to compute the water surface profile. ⎛ q 2 n2 ⎞ y0 = ⎜ ⎜ S ⎟⎟ ⎝ 0 ⎠ 10.70 3/10 1/ 3 ⎛ q2 ⎞ yc = ⎜ ⎟ ⎜ g⎟ ⎝ ⎠ ⎛ 1.52 × 0.0152 ⎞ =⎜ = 1.627 m ⎜ 0.0001 ⎟⎟ ⎝ ⎠ ⎛ 1.52 ⎞ =⎜ = 0.612 m ⎜ 9.81 ⎟⎟ ⎝ ⎠ Therefore the estuary has a mild slope and an M1 curve exists upstream of the outlet. For a wide rectangular channel (Ex. 10.16), J = 2.5, N = 3.33, M = 3, and N/J = 1.33. Use Eq. 10.7.13 to compute x: 158 © 2012 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 10 / Flow in Open Channels 3 ⎤ 1.627 ⎡ ⎛ 0.612 ⎞ 2.5 x= F (v, 2.5) ⎥ ⎢u − F (u,3.33) + ⎜ ⎟ 0.0001 ⎣⎢ ⎝ 1.627 ⎠ 3.33 ⎦⎥ = 16, 270 [u − F (u,3.33) − 0.04 F (v, 2.5) ] The results of the calculations are as follows: y (m) 7 6 5 4 2 u 4.30 3.67 3.07 2.46 1.23 F(u,3.33) 0.015 0.021 0.032 0.052 0.348 v 6.96 5.67 4.45 3.31 1.32 F(v,2.5) 0.038 0.053 0.074 0.117 0.572 x (m) 69,800 59,700 49,500 39,500 14,700 x – 69,800 (m) 0 −10,100 −20,300 −30,300 −55,100 This is a design analysis problem, and requires extensive calculations that would best be performed on a spreadsheet. Let location A be 400 m upstream of location B. Conditions at B are known. The energy equation between A and B is predicted using total energy H = y + z + α V 2 / 2 g : L H*A ≈ HB + hL ≈ HB + (SA + SB ) 2 In the equation, HB and SB are known, and HA and SA are unknown. A trial and error solution is required. With the table of given data, location A is associated with x = 0, and location B with x = 400 m. The following parameters are computed at location B, where yB = 3.0 m: Location 10.72 Side channel Main channel AB (m2) 60 273 PB (m) 149 97 KB (Eq. 10.7.8) RB (m) 0.40 2.83 11 18,210 Now αB, VB, HB, and SB can be computed: (60 + 273)2 ⎛ 113 18,2103 ⎞ + ⎜ ⎟ = 1.48 (11 + 18,210)3 ⎝ 602 2732 ⎠ Q 280 VB = = = 0.84 m/s AB 60 + 273 αB = SB = 280 2 = 0.000236 (11 + 18, 210)3 H B = 3.0 + 15.1 + 1.48 × (Eq. 10.7.9) (Eq. 10.7.7) 0.84 2 = 18.153 m 2 × 9.81 159 © 2012 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 10 / Flow in Open Channels With L = 400 m, the energy equation between A and B is predicted: H *A = 18.153 + 400 (SA + 0.000236) = 18.20 + 200SA 2 The total energy at location A is H A = yA + zA + α A VA2 V2 = y A + 15.0 + α A A 2g 2g The trial and error solution proceeds by assuming values of yA, computing the corresponding αA, VA, SA, HA, and H *A until the latter two are in close agreement. For yA = 3.2 m, the hydraulic parameters at location A are provided in this table: AA (m2) 172 342 Location Side channel Main channel PA (m) 116 112 RA (m) 1.48 3.05 KA (Eq. 10.7.8) 4470 24,000 The corresponding values of HA and H A* are αA = ⎛ 44703 24, 0003 ⎞ + ⎜ ⎟ = 1.39 (4470 + 24, 000)3 ⎜⎝ 1722 3422 ⎟⎠ VA = Q 280 = = 0.54 m/s AA 172 + 342 SA = (172 + 342) 2 2802 (4470 + 24, 000) 3 = 9.67 × 10−5 H A = 3.2 + 15.0 + 1.39 × (Eq. 10.7.9) (Eq. 10.7.7) 0.542 = 18.22 m 2 × 9.81 H *A = 18.200 + 200 × 0.0000967 = 18.22 m Hence the depth at the upstream location is yA = 3.2 m 160 © 2012 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 11 / Flows in Piping Systems CHAPTER 11 Flows in Piping Systems Steady Flows W f Q Substitute V = and H P = into the energy equation: A γQ p1 γ + W f p2 Q2 Q2 + = + + hL γ 2 gA22 2 gA12 γ Q (a) Note that the kinetic energy terms can be neglected: 11.2 π π A1 = × 0.052 = 0.00196 m2 , A2 = × 0.082 = 0.00503 m2 , 4 4 Q= 0.095 = 0.00158 m3 /s 60 W f 350 × 103 0 .0 0 1 5 8 2 760 × 103 0 .0 0 1 5 8 2 + + = + + 6 .6 9810 9 8 1 0 × 0 .0 0 1 5 8 9810 2 × 9 .8 1 × 0 .0 0 1 9 6 2 2 × 9 .8 1 × 0 .0 0 5 0 3 2 = 750 watt ∴W f = 77.5 + 0.005 + 6.6 35.7 + 0.03 + 0.0645W f Write energy eqn. between locs. 1 and 2: H P + z1 = z2 + f ( L + Le ) D Q2 ⎛π ⎞ 2g ⎜ D2 ⎟ 4 ⎝ ⎠ 2 1/ 5 ⎡ ⎤ Q2 ( L + Le ) f Solving for D 1 , D = ⎢ ⎥ 2 ⎣ 2g(π / 4) ( HP − (z2 − z1 )) ⎦ 11.4 Substitute the relations ⎡ ⎛ e ⎞⎤ f = 1.325 ⎢ ln ⎜ 0.27 + 5.74 Re −0.9 ⎟ ⎥ D ⎠⎦ ⎣ ⎝ −2 and Re = 4Q ΣkD , Le = , f π Dν and solve for D by successive substitution. (a) Compute H P from pump data using linear interpolation: H P = 136 − (136 − 127) (0.757 − 0.95) = 129.1 m (0.757 − 1.01) 161 © 2012 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 11 / Flows in Piping Systems Q = 0.95 m3 /s 1/ 5 ⎡ ⎤ 0.952 (500 + Le ) f ∴D = ⎢ ⎥ 2 ⎣ 2 × 9.81(π / 4) (129.1 − 36) ⎦ Re = D (m) e/D 0.3 0.003 0.4 0.0023 0.39 0.0023 0.39 = 0.24[(500 + Le ) f ]1/ 5 1.401×106 D Le = f 4.67 × 106 3.5 × 106 3.6 × 106 0.026 0.024 0.024 2.50 (m) f 29 41.6 40.6 ∴ D = 1.39 m. (a) Energy eqn. from A to B: ⎛ ΣK1 ΣK2 ⎞ 2 ⎛p ⎞ ⎛p ⎞ 1.85 + Q ⎜ + z ⎟ − ⎜ + z ⎟ = ( R1 + R2 + R3 )Q + ⎜⎜ 2 2⎟ ⎟ ⎝γ ⎠A ⎝ γ ⎠B ⎝ 2 gA1 2 gA2 ⎠ Use Hazen-Williams resistance formula for R: R= 10.59 L C1.85D4.87 ∴ R1 = ΣK1 2 gA12 11.6 ΣK 2 2 gA22 10.59 × 200 = 1071 1001.85 × 0.24.87 = 10.59 ×150 2 = 193.4, = 103.3, R2 = 2 4 1201.85 × 0.254.87 2 × 9.81 × (π / 4) × (0.2) = 3 = 63.5, 2 × 9.81× (π / 4)2 × (0.3)4 R3 = 10.59 × 300 = 271.1 901.85 × 0.34.87 Substitute known data into energy eqn: 250 − 107 = (1071 + 193 + 271)Q1.85 + (103.3 + 63.5)Q 2 143 = 1535Q1.85 + 167Q2 F (Q) = 1535Q1.85 + 167Q2 − 143, Iter. 1 2 3 Q(m3 /s) 0.277* 0.265 0.265 F ′(Q) = 2840Q0.85 + 334Q F′ F 12.60 0.284 Δ Q = − F /F ′ 1046 1007 − 0.01205 − 0.00028 162 © 2012 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 11 / Flows in Piping Systems ∴Q = 0.265 m3 /s * 1/1.85 ⎛ 143 ⎞ Q≅⎜ ⎟ ⎝ 1535 ⎠ (1st iteration) Write the energy equation from the liquid surface (location 1) to the valve exit (location 2): p1 γ + z1 + W f 2 L ⎛ ⎞ Q = ⎜ ΣK + f + 1⎟ + z2 D γQ ⎝ ⎠ 2 gA 2 (a) Rearrange and substitute in known data: P1 11.8 γ + z1 − z2 + 1 W fL ⎞ Q 2 ⎛ f − ⎜ ΣK + + 1⎟ =0 D ⎠ 2 gA2 γ Q ⎝ 110 ×103 1×104 1 + 24 − 18 + 0.68 × 9810 0.68 × 9810 Q (0.5 + 3 × 0.26 + 2 + 0.015 × 450 / 0.30 + 1) 2 Q =0 − 2 × 9.81× (0.7854 × 0.302 )2 1.50 22.4 + − 273.2Q 2 = 0 Q By trial and error, Q ≈ 0.32 m3/s Initially work between locs. B and C: 11.10 ∴W = Le = 8 f ( L + Le ) DΣk , R= f gπ 2 D5 Le1 = 1.2 × 2 8 × 0.015 × 260 = 160, R1 = = 0.1295, 0.015 9.81× π 2 × 1.25 Le 2 = 1× 3 8 × 0.02 × 1150 = 150, R2 = = 1.900, 0.02 9.81× π 2 × 15 Le 3 = 0.5 × 2 8 × 0.018 × 1556 = 56, R3 = = 74.05, 0.018 9.81× π 2 × 0.55 Le 4 = 0.75 × 4 8 × 0.021 × 943 = 143, R4 = = 6.895 0.021 9.81 × π 2 × 0.755 Q2 ⎛ 4 ⎞ ⎜Σ 1 ⎟ ⎜ i =2 Ri ⎟ ⎝ ⎠ 2 = 32 1 1 ⎞ ⎛ 1 + + ⎜ ⎟ 74.05 6.895 ⎠ ⎝ 1.9 2 = 6.022 m, 163 © 2012 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 11 / Flows in Piping Systems ∴Q2 = W / R2 = 6.022 /1.9 = 1.780 m3 /s, Q3 = W / R3 = 6.022 / 74.05 = 0.285 m3 /s, Q4 = W / R4 = 6.022 / 6.895 = 0.935 m3 /s. Check continuity: Q2 + Q3 + Q4 = 1.78 + 0.285 + 0.935 = 3 = Q1, ∴ OK , write energy equation from loc. A to B: To find W P ⎛p ⎞ H P = R1Q12 + ⎜ + z ⎟ = R1Q12 + W + 20 ⎝γ ⎠B ∴ H P = 0.1295 × 32 + 6.022 + 20 = 27.2 m = 1.07 MW = γ Q1HP = 9810 × 3 × 27.2 = 1.07 ×106 W, or W ∴W P P 0.75 η Use SI data from Pbm 11.11 (a): ΣQ = Q1 − Q2 − Q3 = 0, or w( H ) = 30 − H H − 20 H − 18 − − 142.1 232.4 1773 Use false position method of solution: Hr = 11.12 Iter 1 2 3 HA 25 24.57 24.46 Hu 20 20 20 w( H A ) −0.02193 −0.00562 −0.00144 H A w ( Hu ) − Hu w ( H A ) w ( Hu ) − w( H A ) Sign w(H A )w( H r ) ε 0.2317 24.57 −0.00562 + ⎯ 0.2317 24.46 −0.00144 + 0.0045 0.2317 24.43 −0.00030 + 0.0012 w( H u ) Hr w( H r ) ∴ Q1 = (30 − 24.43) /142.1 = 0.198 m3/s Q2 = (24.43 − 20) / 232.4 = 0.138 m3/s Q3 = (24.43 − 18) /1773 = 0.060 m3/s Continuity at junction: Q1 − (Q 2 + Q3 + Q 4 ) = 0. 11.14 Energy eqn. across pipe 1: p0 γ = R1Q12 + H , or Q1 = ( p0 /γ − H ) / R1 in which H = piezometric head at junction. For each branch (i= 2, 3, 4): 164 © 2012 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 11 / Flows in Piping Systems H = RiQ12 or Qi = H / Ri 1/ 2 Substitute into continuity eqn: ⎡ p0 / γ − H ⎤ ⎢ ⎥ ⎣ R1 ⎦ 1/ 2 ⎡H⎤ = ∑⎢ ⎥ i = 2 ⎣ Ri ⎦ 4 4 = H∑ i=2 1 Ri ∴Solving for H, H= = p 0γ ⎡ 1 + R1 ⎢ ⎢⎣ Σ 1 ⎤ ⎥ R i ⎥⎦ 2 300 × 103 / 9810 ⎡ ⎤ 1 1 1 1 + 1.6 × 10 ⎢ + + ⎥ 5 1× 106 1.8 × 106 ⎥⎦ ⎣⎢ 5.3 × 10 2 4 = 30.58 = 26.46 m 1.556 ∴Q1 = (30.58 − 26.46) /1.6 × 104 = 0.01605 m3/s Q2 = 26.46 / 5.3 × 105 = 0.00707 m3/s Q3 = 26.46 /1×106 = 0.00514 m3/s Q4 = 26.46 /1.8 ×106 = 0.00383 m3/s (c) Substitute known data into energy equation, and solve for Q 1 : −2 ⎫ ⎧ ⎡ 1 ⎤ ⎪ 2 ⎪ a0 = ⎨a1 + R1 + ⎢ ∑ ⎥ ⎬ Q1 R ⎢ ⎪ i ⎥ ⎣ ⎦ ⎪⎭ ⎩ 11.16 −2 ⎧⎪ ⎡ ⎤ ⎫⎪ 2 1 1 1 4 2 ∴ 45 = ⎨10 + 34,650 + ⎢ + + ⎥ ⎬ Q1 = 56, 410Q1 82,500 127,900 115,500 ⎣ ⎦ ⎪⎭ ⎪⎩ ∴ Q1 = 45 / 56,410 = 0.0282 m3/s ∴ H = 45 − (104 + 34,650) × 0.02822 = 9.49 m Q2 = 9.49 / 82,500 = 0.0107 m3/s Q3 = 9.49 /127,900 = 0.0086 m3/s Q4 = 9.49 /115,500 = 0.0091 m3/s 165 © 2012 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 11 / Flows in Piping Systems (a) For each pipe compute equivalent length and resistance coefficient: Pipe 1: Le = DΣK / f = 0.05 × 3/ 0.02 = 7.5 R1 = 8 f ( L + Le ) 8 × 0.02 × 37.5 = = 1.983 × 105 2 5 2 5 gπ D 9.81 × π × 0.05 Pipe 2: Le = 0.075 × 5 / 0.025 = 15, R2 = Pipe 3: Le = 0.06 × 1/ 0.022 = 2.7 , 11.18 W= Q2 ⎛ 3 ⎞ ⎜ Σ( Ri ) −1/2 ⎟ ⎜ i =1 ⎟ ⎝ ⎠ 2 = 8 × 0.025 × 55 = 4.788 × 10 4 9.81 × π 2 × 0.0755 R3 = 8 × 0.022 × 62.7 = 1.466 × 105 2 5 9.81× π × 0.06 (0.6 / 60) 2 = 1.125 m [(1.983 × 105 ) −1/ 2 + (4.788 × 104 ) −1/ 2 + (1.466 × 105 )−1/ 2 ]2 ∴ Q1 = W / R1 = 1.125 /1.983 ×105 = 0.00238 m3 /s, or 143 L/min Q2 = W / R2 = 1.125 / 4.788 ×104 = 0.00485 m3 /s, or 290 L/min Q3 = W / R3 = 1.125 /1.466 ×105 = 0.00277 m3 /s, or 166 L/min Write energy eqn. from junction at A to suction side of pump at B: ⎛ ΣK ⎞⎟ Q2 + pB + z H PA = ⎜ R1 + B ⎜ γ 2 gA12 ⎟⎠ ⎝ where H PA = pump head and pB = pressure. The only unknown in the relation is Q, since H PA 11.20 η W PA 1×106 × 0.76 95.6 = = = γQ Q 0.81× 9810 × Q Evaluate the constants: R1 = 8 f1L1 8 × 0.23 × 5000 = = 40.0 2 5 gπ D1 9.81× π 2 × 0.755 pB γ + zB = ΣK = 2 = 0.52 2 gA12 2 × 9.81× (π / 4)2 × 0.754 150 × 103 + 27 = 45.9 m. 0.81× 9810 Substituting into the energy eqn., 95.6 = 40.5Q2 + 45.9 Q Solve using Newton’s method: Q = 1.053 m3/s and HPA = 95.6/1.053 = 90.8 m 166 © 2012 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 11 / Flows in Piping Systems The second reach is now analyzed. Write energy eqn. from suction side of pump at B to the downstream reservoir: ⎛ ΣK ⎟⎞ Q2 + 50 + z B + H PB = ⎜ R2 + ⎜ γ 2 gA22 ⎟⎠ ⎝ pB Evaluate the constants: R2 = 8 f2 L2 8 × 0.023 × 7500 = = 60.1 gπ 2D25 9.81× π 2 × 0.755 ΣK = 10 = 2.61 2 gA12 2 × 9.81 × (π / 4) 2 × 0.75 4 ∴ 45.9 + H PB = (60.1 + 2.6) ×1.0532 + 50 or H PB = 73.6 m ∴W PB = γ QH PB 0.81× 9810 ×1.053 × 73.6 = = 8.10 ×105 W or 0.76 η = 810 kW W PB Compute resistance coefficients: R1 = 10.59 L 10.59 × 200 = = 7.6 1.85 4.87 C D 1301.85 × 0.5 4.87 R2 = 10.59× 600 = 274.6 1301.85 × 0.34.87 R3 = 10.59×1500 = 686.5 1301.85 × 0.34.87 R4 = 10.59×1500 =169.1 1301.85 ×0.44.87 (a) Let H J = piezometric head at junction J. Assume Q1 = 2 m3 /s (out of J). Then H J = 30 + R1Q12 = 30 + 7.6 × 22 = 60.4 m Q2 = (250 − 60.4) / 274.6 = 0.831 m3/s (into J ) 11.22 Q3 = (300 − 60.4) / 686.5 = 0.591 m3/s (into J ) Q4 = (200 − 60.4) /169.1 = 0.909 m3/s (into J ) ∴ ΔQ = −2 + 0.831 + 0.591 + 0.909 = +0.331 Assume Q1 = 2.5 m3/s. Then H J = 30 + 7.6 × 2.52 = 77.5 m Q2 = (250 − 77.5) / 274.6 = 0.793 m3/s Q3 = (300 − 77.5) / 686.5 = 0.569 m3/s Q4 = (200 − 77.5) /169.1 = 0.851 m3/s ∴ΔQ = −2.5 + 0.793 + 0.569 + 0.851 = −0.287 Linear interpolation to find the next estimate of Q 1 yields 167 © 2012 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 11 / Flows in Piping Systems −0.287 − 0.331 −0.287 = 2.5 − 2 2.5 − Q1 ∴Q1 = 2.27 m3 / s Then H J = 30 + 7.6 × 2.27 2 = 69.2 m Q2 = (250 − 69.2)/ 274.6 = 0.811 m3 /s Q3 = (300 − 69.2)/ 686.5 = 0.580 m3 /s Q4 = (200 − 69.2) /169.1 = 0.880 m3 / s ∴ ΔQ = −2.27 + 0.811 + 0.580 + 0.880 = −0.001 ∴ OK Compute the R-values for each pipe: Pipe 1: Le = 1 × 0.1 8 × 0.04 × 5.5 = 2.5 m, R1 = = 1819.6 0.04 9.87 × π 2 × 0.15 Pipe 2: Le = 2 × 0.1 8 × 0.04 × 155 = 5 m, R2 = = 5.31 × 104 2 5 0.04 9.81× π × 0.1 Pipe 3: Le = 2 × 0.1 8 × 0.04 × 605 = 5 m, R3 = = 2 × 105 2 5 0.04 9.81 × π × 0.1 Pipe 4: Le = 4 × 0.1 8 × 0.04 ×1,817 = 10 m, R4 = = 77,750 0.04 9.81× π 2 × 0.15 (a) Compute the discharge and head loss in pipe 2: 11.24 Q2 = 5.2 ×10−3 m3/s ∴(hL )2 = R2Q22 = 5.31×104 × (5.2 ×10−3 )2 = 1.387 m (c) Q2 = 11.6 ×10−3 m3/s = 696 L/min, and from the pump curve, HP = 136 m Write the energy equation from A to C: H C = H P − ( R1 + R2 )Q22 = 136 − 53,120 × (11.6 × 10−3 )2 = 129 m The energy equation from C to D is HC = zD + R4Q42 , where HC is the hydraulic grade line at location C. Therefore the discharge in pipe 4 is Q4 = H C − zD 129 − 127 = = 5.07 L/ s, or 300 L/min 77,750 R4 168 © 2012 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 11 / Flows in Piping Systems (a) From the pump curve, for HP = 138 m, Q = 0.01 m3/s. Write the energy equation from A to C and evaluate the hydraulic grade line HC = H P − (hL )1 − (hL )2 = 138 − (53,120) × (0.01)2 = 132.7 m 11.26 Thus the discharge in pipe 3, between location C and location B is Q3 = ΔQI = HC − HB 132.7 −130 = = 3.67 ×10−3 m3/s or 220 L/min 5 R3 2 ×10 −(±W1 ± W2 ) − 10 −(±W2 ± W3 ) − 15 , ΔQII = G1 + G2 G2 + G3 ± Wi = RQi Qi , Gi = 2 R Qi Loop Pipe I Q 1 −3.5 2 0 W G −24.50 0 0 Σ 24.50 ΔQI = 11.28 II +24.50 − 10 = +1.04 14.00 2 0 3 − 3.5 ΔQII = ΔQI = W G Q −2.46 −12.10 9.84 −2.30 +0.36 +0.26 1.44 +0.43 Σ −11.84 11.28 14.00 +11.84 − 10 = +0.16 11.28 − 0.36 − 0.26 1.44 − 0.43 − 24.5 14.00 − 2.82 − 15.90 11.28 + 2.73 −24.50 14.00 Σ −16.16 12.72 0 Σ 14.00 Q +24.50 − 15 = +0.68 14.00 0 ΔQII = +16.16 − 15 = + 0.09 12.72 HJunction = 50 − R1Q12 = 50 − 2 × 2.32 = 39.4 ∴Q1 = 2.30 (into J ), Q2 = 0.43 (into J ), Q3 = 2.73 (out of J ) 169 © 2012 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 11 / Flows in Piping Systems First, continuity is satisfied as shown in the figure: (a) The hydraulic grade lines at the nodes (i.e., the nodal piezometric heads can now be computed: 11.30 H A = 100 − R1Q12 = 100 − 20 × 0.452 = 95.95 m H B = H A − R2Q22 = 95.95 − 51× 0.282 = 91.95 m HC = H A − R3Q32 = 95.95 − 280 × 0.172 = 87.86 m HD = HB − R5Q52 = 91.95 − 310 × 0.182 = 81.91 m (b) p A = γ ( H A − z A ) = 9810(95.95 − 10) = 8.43 × 105 Pa p B = γ ( H B − z B ) = 9810(91.95 − 20) = 7.06 × 105 Pa pC = γ ( HC − zC ) = 9810(87.86 − 5) = 8.13 ×105 Pa pD = γ ( H D − z D ) = 9810(81.91 − 0) = 8.04 × 105 Pa Assume flow directions as shown. Then ΔQ = −( R1Q12 + R2Q22 − R3Q32 ) 2( R1Q1 + R2Q2 + R3Q3 ) Initial flow assumption: Q1 = 25, Q 2 = 10, Q3 = 25 11.32 1st iteration: ΔQ = −(3 × 252 + 5 × 102 − 2 × 252 ) = −3.21 2(3 × 25 + 5 × 10 + 2 × 25) ∴ Q1 = 25 − 3.21 = 21.79, Q2 = 10 − 3.21 = 6.79 , Q3 = 25 + 3.21 = 28.21 2nd iteration: ΔQ = −(3 × 21.792 + 5 × 6.792 − 2 × 28.212 ) = −0.20 2(3 × 21.79 + 5 × 6.79 + 2 × 28.21) ∴ Q1 = 21.79 − 0.20 = 21.59 , Q2 = 6.79 − 0.20 = 6.59 , Q3 = 28.21 + 0.20 = 28.41 170 © 2012 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 11 / Flows in Piping Systems 3rd iteration: ΔQ = −(3 × 21.592 + 5 × 6.592 − 2 × 28.412 ) = − 0.0041 2(3 × 21.59 + 5 × 6.59 + 2 × 28.41) ∴ Q1 ≅ 21.6, Q2 ≅ 6.6, Q3 ≅ 28.4. Units are L/s A Mathcad solution is provided below. The solution converges after 3 iterations. ORIGIN := 1 Given: L := ⎛ 200 ⎞ ⎜ 300 ⎟ ⋅ m ⎜ ⎟ ⎝ 120 ⎠ D := ΔZ := 50⋅ m ⎛ 1500 ⎞ ⎜ 1000 ⎟ ⋅ mm ⎜ ⎟ ⎝ 1200 ⎠ e := 1⋅ mm f := 1.325⋅ ⎜⎛ ln ⎛⎜ 0.27⋅ 11.34 ⎝ ⎝ γ := 9810⋅ newton 3 m W P := 1920⋅ kW ⎞⎞ D ⎟⎟ i ⎠⎠ e −2 η := 0.82 ⎛ 0.018 ⎞ f = ⎜ 0.02 ⎟ ⎜ ⎟ ⎝ 0.019 ⎠ D ⋅ K ⎞⎤ ⎡⎢ ⎛⎜ i i ⎟⎥ 8⋅ f ⋅ L + ⎢ i⎜ i f ⎟⎥ i ⎠⎦ ⎣ ⎝ R := i 2 ⎛ 2 ⎞ ⎜ 0 ⎟ ⎜ ⎟ ⎝ 10 ⎠ i := 1 .. 3 Resistance coefficients: i K := R = ( i)5 g⋅π ⋅ D ⎛ 0.071 ⎞ 2 ⎜ 0.487 ⎟ s ⎜ ⎟ 5 ⎝ 0.473 ⎠ m 3 Q := 2⋅ Initial flow estimate (note that the discharge is common to all three lines): 1 Hardy Cross iteration: ( 3) − R + R + R ⋅ Q⋅ Q + ΔQ( Q) := 1 2 ( W P⋅ η 3) 2⋅ R + R + R ⋅ Q + N := 4 Solution: j := 1 .. N 1 Q j+ 1 2 γ⋅Q − ΔZ W P⋅ η γ⋅Q 2 ( j) := Q + ΔQ Q j ( j) = ΔQ Q ⎛ 2 ⎞ ⎜ 2.59 ⎟ ⎜ ⎟ m3 Q = ⎜ 2.762 ⎟ ⎜ 2.771 ⎟ s ⎜ ⎟ ⎝ 2.771 ⎠ 0.59 Residual: 0.172 8.456·10 -3 3 m s 1.762·10 -5 171 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. m s Chapter 11 / Flows in Piping Systems The solution was obtained using EPANET, Version 2.0. Link - Node Table: ----------------------------------------------------------------Link Start End Length Diameter ID Node Node m mm ---------------------------------------------------------------1 1 5 200 100 2 5 3 150 50 3 5 6 500 100 4 6 4 35 50 5 6 2 120 100 11.36 Node Results: ---------------------------------------------------------------Node Demand Head Pressure Quality ID LPS m m ---------------------------------------------------------------5 0.00 121.72 121.72 0.00 6 0.00 116.23 116.23 0.00 1 -9.28 125.00 0.00 0.00 Reservoir 2 7.15 115.00 0.00 0.00 Reservoir 3 1.58 118.00 0.00 0.00 Reservoir 4 0.55 116.00 0.00 0.00 Reservoir Link Results: ---------------------------------------------------------------Link Flow Velocity Headloss Status ID LPS m/s m/km ---------------------------------------------------------------1 9.28 1.18 16.40 Open 2 1.58 0.81 24.80 Open 3 7.69 0.98 10.98 Open 4 0.55 0.28 6.55 Open 5 7.15 0.91 10.24 Open (c) ΔQ = 11.38 (d) R1 = −(R1 + R2 + R3) ) Q Q + (a0 + a1Q + a2Q2 + a3Q3 ) + zA − zB 2(R1 + R2 + R3) ) Q − (a1 + 2a2Q + 3a3Q2 ) 8 f L1 8 f L2 8 f L3 = 0.00516, R2 = = 2.538, R3 = = 0.0103 2 5 2 2 5 gπ D1 gπ N D2 gπ 2 D35 Using the equation in part (c), Q = 2.061 m3/s after 5 iterations, beginning with an initial estimate Q = 3 m3/s. 172 © 2012 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 11 / Flows in Piping Systems Iteration Q ΔQ 1 3 − 0.611 2 2.389 − 0.277 3 2.113 − 0.050 4 2.062 − 0.0014 5 2.061 − 5.1 × 10−13 Unsteady Flow Vss = 11.44 2 g ( H1 − H3 ) 2 × 9.81× 3 = = 0.307 m/s fL / D + K 0.025 × 2500 / 0.1 + 0.15 V = 0.99Vss = 0.99 × 0.307 = 0.304 m/s ⎡ (V ss + V V ss L ln ⎢ 2 g(H1 − H 3 ) ⎣ (V ss − V ⎡ ( 0 .3 0 7 + 0 .3 0 7 × 2 5 0 0 = ln ⎢ 2 × 9 .8 1 × 3 ⎣ ( 0 .3 0 7 − t = )( V s s − V 0 ) ⎤ ⎥ )( V s s + V 0 ) ⎦ 0 .3 0 4 )( 0 .3 0 7 − 0 ) ⎤ = 6 9 .0 s 0 .3 0 4 )( 0 .3 0 7 + 0 ) ⎥⎦ First compute the initial velocity V0, with H1-H3 = 8 m, and K0 = 275: V0 = 2 × 9.81 × 8 = 0.552 m/s 0.015 × 800 + 275 0.05 The final steady-state velocity is, with Kss = 5: VSS = 11.46 2 × 9.81× 8 = 0.800 m/s 0.015 × 800 +5 0.05 The steady-state discharge is Q = 0.7854 × 0.052 × 0.800 = 0.00157 m3 / s , and the time to reach 95% of that value is V = 0.95Vss = 0.95 × 0.800 = 0.760 m/s ∴t = 0.800 × 800 ⎡ (0.800 + 0.760)(0.800 − 0.552) ⎤ ln = 8.03 s 2 × 9.81× 8 ⎢⎣ (0.800 − 0.760)(0.800 + 0.552) ⎥⎦ 173 © 2012 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 11 / Flows in Piping Systems 174 © 2012 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 12 / Turbomachinery CHAPTER 12 Turbomachinery Elementary Theory ω = 800 × π / 30 = 83.8 rad/s u1 = ωr1 = 83.8 × 0.04 = 3.35 m/s u 2 = ω r2 = 83.8 × 0.125 = 10.48 m/s Q = 2π r1b1Vn1 , but Vn1 = u1 since β1 = 45D 12.2 ∴Q = 2π × 0.04 × 0.05 × 3.35 = 0.0421 m3 / s Vn2 = Q 0.0421 = = 2.14 m/s 2π r2b2 2π × 0.125 × 0.025 Vt2 = u2 − Vt1 = 0 Vn2 tan β 2 = 10.48 − 2.14 = 6.77 m/s tan 30D (α1 = 90D under ideal conditions) ( ) ∴T = ρQ r2Vt2 − rV 1 t1 = 1000 × 0.0421(0.125 × 6.77 − 0) = 35.6 N ⋅ m W P = ωT = 83.8 × 35.6 = 2980 W H t = ωT / γ Q = 2980 /(9810 × 0.0421) = 7.22 m Q = 7.6 ×10−3 m3/s 12.4 ω = 2000 × π 30 = 209 rad/s 175 © 2012 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 12 / Turbomachinery Q 7.6 ×10−3 = = 1.94 m/s 2π r2b2 2π × 0.0625 × 0.01 Vn2 = u2 = ωr2 = 209 × 0.0625 = 13.06 m/s ∴ Ht = 13.06 u2 (13.06 − 1.94cot 60°) = 15.9 m (u2 − Vn2 cot β 2 ) = 9.81 g = γ QH = 9810 × 0.8 × 7.6 ×10−3 ×15.9 = 948 W W t = 948/ 746 = 1.21 hp W Compute loss in suction pipe: 2 ⎛ L ⎞ Q hL = ⎜ f + Σ K ⎟ ⎝ D ⎠ 2 gA 2 11 ⎛ ⎞ = ⎜ 0.015 × + 2 × 0.19 + 0.8⎟ ⎝ ⎠ 0.1 12.6 0.05 2 2 ⎛ π⎞ 2 × 9.81 × ⎜ ⎟ × 0.1 4 ⎝ 4⎠ = 5.85 m Water at 20°C: γ = 9792 N/m3 , pv = 2340 Pa. Substitute known data into NPSH relation, solving for Δz: ∴Δ z = patm − pv γ 101×103 − 2340 − 5.85 − 3 = 1.23 m − hL − NPSH = 9792 Dimensional Analysis and Similitude Q1 = N ω1 970 Q = 1Q = × 0.8 = 0.65 m 3 / s ω 2 2 N 2 2 1200 ∴from Fig. 12.9, H1 ≅ 11.5 m and W 1 ≅ 91 kW. 2 ⎛ω ⎞ ⎛ 1200 ⎞ ∴ H 2 = ⎜ 2 ⎟ H1 = ⎜ ⎟ ×11.5 = 17.6 m ⎝ 970 ⎠ ⎝ ω1 ⎠ 12.8 2 3 ⎛ 1200 ⎞ = ⎛ ω2 ⎞ W ∴W ⎜ ⎟ 1=⎜ 2 ⎟ × 91 = 172 kW ⎝ 970 ⎠ ⎝ ω1 ⎠ 12.10 CQ = 3 Q Q / 3600 = = 1.061× 10 −4 Q 3 ω D 304 × 0.2053 (Q in m 3 /h) 176 © 2012 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 12 / Turbomachinery CH = gH 9.81H = = 2.526 × 10−3 H (H in m) 2 2 2 2 ω D 304 × 0.205 Tabulate CQ and CH using selected values of Q and H from Fig. 12.6: CQ×10 −3 0 0.53 1.06 1.59 2.12 2.65 3.18 Q (m3/h) 0 50 100 150 200 250 300 H (m) 54 53 52 50 47 41 33 CH ×10−1 1.36 1.34 1.31 1.26 1.19 1.04 0.83 The dimensionless curve shown in Fig. 12.12 is for the 240-mm impeller. Since the impellers are not the same (240 mm versus 205 mm), dynamic similitude does not exist, and thus the curves are not the same. Compute the specific speed: Ω P = 12.12 ω Q ( gH P )3 / 4 = 1800 × π × 0.15 30 = 1.30 , (9.81 × 22)3 / 4 hence use a mixed flow pump. As an alternate, since Ω P is close to unity, a radial flow pump could be employed. Fig. 12.13: At η = 0.75 (best eff. ), CQ ≅ 0.048 , C H ≅ 0.018, C W ≅ 0.0011, C NPSH ≅ 0.023. ω = 750 × π 30 = 78.5 rad/s 1/3 (a) 12.14 ⎛ 1240 ×10−3 ⎞ D = ⎜⎜ ⎟⎟ ⎝ 0.049 × 78.5 ⎠ H= = 0.685 m 0.018 × 78.52 × 0.6852 = 5.3 m 32.2 H NPSH = 0.023 × 78.52 × 0.6852 = 6.78 m 9.81 = 0.0011×1000 × 78.53 × 0.6855 = 80,250 W = 80.25 kW W or 12.16 = 82.25/0.746 = 107.6 hp W ω = 600 × π 30 = 62.8 rad /s, Q = 22.7 / 60 = 0.378 m 3 / s, 177 © 2012 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 12 / Turbomachinery ΩP = ω Q ( gH P ) = 3/ 4 2 62.8 0.378 = 0.751 (9.81×19.5)3/ 4 2 H 2 ⎛ ω 2 ⎞ ⎛ D2 ⎞ =⎜ ⎟ ⎜ ⎟ =2 H1 ⎝ ω1 ⎠ ⎝ D1 ⎠ 2 3 Q2 ⎛ ω 2 ⎞⎛ D2 ⎞ =⎜ ⎟⎜ ⎟ =2 Q1 ⎝ ω1 ⎠⎝ D1 ⎠ 2 ⎛ ω ⎞ ⎛ D ⎞ ⎛ ω ⎞⎛ D ⎞ ∴ ⎜ 2 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟⎜ 2 ⎟ ⎝ ω1 ⎠ ⎝ D1 ⎠ ⎝ ω1 ⎠⎝ D1 ⎠ 12.18 ∴Use a radial flow pump. 3 or ω 2 D2 = ω1 D1 4 ⎛ω ⎞ ∴ ⎜ 2 ⎟ = 2, ω2 = 4 2 ω1 = 1.19 ω1 and D2 = 4 2 D1 = 1.19 D1 ⎝ ω1 ⎠ Assume a pump speed N = 2000 rpm or ω = 2000 × π 30 = 209 rad/s /(γ Q) = 200 ×103 /(8830 × 0.66) = 34.3 m HP = W f ∴Ω P = ω Q ( gH P ) 3/ 4 = 209 0.66 = 2.16 (9.81× 34.3)3/ 4 The specific speed suggests a mixed-flow pump. However, if N = 1000 rpm, a radial-flow pump may be appropriate. Consider both possibilities. Mixed flow: from Fig. 12.14, at best η : CW ≅ 0.0117, CQ ≅ 0.148, CH ≅ 0.067 Use 12.20 CQ = gH Q and CH = 2 P2 3 ωD ωD Combining and solving for D and ω Q / CQ D= gHP / CH ρ= Q CQD3 0.66 / 0.148 = 0.251 m 9.81× 34.3/ 0.067 ∴D = ω= and ω = 0.66 = 282 rad/s or N = 282 × 30 / π = 2674 rpm 0.148 × 0.2513 γ g = 8830 = 900 kg/m3 9.81 178 © 2012 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 12 / Turbomachinery = C ρω 3D5 = 0.0117 × 900 × 2823 × 0.2515 ∴W P W = 2.35 × 105 W or 235 kW Radial flow: from Fig. 12.12, at best η : CW ≅ 0.027, CQ ≅ 0.0165, C H ≅ 0.125 ∴D = ω= 0.66 / 0.0165 = 0.878 m 9.81× 34.3/ 0.125 0.66 59.1× 30 = 59.1 rad/s or N = = 564 rpm 3 π 0.0165 × 0.878 = 0.0027 × 900 × 59.13 × 0.8785 = 2.62 × 105 W W P or 262 kW Hence, a mixed-flow pump is preferred. Use of Turbopumps 12.22 The intersection of the system demand curve with the head-discharge curve yields Q = 2.75 m3 / min, H P = 12.6 m, W P = 7.2 kW. N 1 = 1350 rpm, Q 1 = 2.75 m 3 / min, H 1 = 12.6 m, 12.24 W 1 = 7.2 kW, N 2 = 1200 rpm, D 2 = D1 179 © 2012 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 12 / Turbomachinery 3 ⎛ N ⎞⎛ D ⎞ ⎛ 1200 ⎞ 3 ∴Q2 = Q1 ⎜ 2 ⎟⎜ 2 ⎟ = 2.75 ⎜ ⎟ = 2.44 m /min 1350 N D ⎝ ⎠ ⎝ 1 ⎠⎝ 1 ⎠ 2 2 3 5 ⎛N ⎞ ⎛D ⎞ ⎛ 1200 ⎞ H 2 = H1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 12.6 ⎜ ⎟ = 9.96 m ⎝ 1350 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ 2 ⎛N ⎞ ⎛D ⎞ ⎛ 1200 ⎞ W 2 = W 1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 7.2 ⎜ ⎟ = 5.06 kW ⎝ 1350 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ 3 Efficiency will remain approximately the same. Compute system demand: 2 2 14 ⎛ L ⎞V ⎛ ⎞ 3 = ⎜ 0.01 × + 4 × 0.1⎟ = 0.40 m H P = ⎜ f + ΣK ⎟ ⎝ D ⎠ 2g ⎝ 0.3 ⎠ 2 × 9.81 12.26 Q = VA = 3 × ∴Ω P = π 4 × 0.32 = 0.212 m2 ω Q ( gH P ) 3/ 4 = 31.4 0.212 = 5.19 (9.81 × 0.4)3/ 4 ω = 300 × π / 30 = 31.4 rad/s ∴Axial pump is appropriate. (a) 12.28 The intersection of the pump curve with the demand curve yields H P ≅ 64 m and Q ≅ 280 m3 / h. ∴ From Fig. 12.6, W P ≅ 64 kW and NPSH ≅ 8.3 m 180 © 2012 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 12 / Turbomachinery Use energy eqn. to establish system demand: e⎤ ⎡ f ≅ 1.325 ⎢ln 0.27 ⎥ D⎦ ⎣ H P = Δz + f −2 0.000255 ⎤ ⎡ = 1.325 ⎢ln 0.27 × 0.45 ⎥⎦ ⎣ −2 = 0.017 0.017 × 5000 L Q2 = 192 + Q2 = 192 + 381Q2 2 5 2 D 2 gA 2 × 9.81× (π / 4) × (0.45) From Fig. 12.6, at best ηP , Q ≅ 12.30 240 m3 = 0.067 m3/s, and HP ≅ 65 m 3600 s Assume three pumps in series, so that HP = 3× 65 = 195 m. Then the demand head is H P = 192 + 381× (0.067) 2 = 193.7 m Hence three pumps in series are appropriate. The required power is WP = γ QHP/ηP = 9810× 0.86× 0.067 ×195/0.75 = 146,965 W W P = 146,965/746 = 197 hp or (a) For water at 80°C, pv = 46.4 × 103 Pa, and ρ = 9553 kg/m3. Write the energy equation from the inlet (section i) to the location of cavitation in the pump: pi γ 12.32 + Vi2 pv = + NPSH 2g γ ∴ NPSH = pi − pv γ Vi2 (83 − 46.4) ×103 62 + = + = 5.67 m 2g 9533 2 × 9.81 (b) NPSH1 = 5.67 m, N1 = 2400 rpm, N2 = 1000 rpm, D2/D1 = 4 2 2 ⎛N ⎞ ⎛D ⎞ ⎛ 1000 ⎞ ∴ NPSH 2 = NPSH1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 5.67 ⎜ ⎟ ⎝ 2400 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ 2 ( 4 )2 = 15.8 m Given: 12.34 L = 200 m , D = 0.05 m , Δz = z 2 − z1 = −3 m , V = 3 m/s , ν = 6 × 10 −7 m 2 /s , ρ = 1593 kg/m 3 , pv = 86.2 × 103 Pa , pa = 101 × 103 Pa Compute the pump head: 181 © 2012 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 12 / Turbomachinery Re = 3 × 0.05 6 × 10 −7 = 2.5 × 105 ( ) f = 1.325 ⎡ ln 5.74Re-0.9 ⎤ ⎣ ⎦ −2 = 1.325 ⎡ ln 5.74 × (2.5 × 105 ) −0.9 ⎤ ⎣ ⎦ −2 = 0.015 2 fL ⎞ V 2 ⎛ ⎛ 0.015 × 200 ⎞ 3 = −3 + ⎜ 1 + = 25.0 m H P = Δz + ⎜ 1 + ⎟ ⎟ 0.05 D ⎠ 2g ⎝ ⎠ 2 × 9.81 ⎝ (a) Choose a radial-flow pump. Use Fig. P12.35 to select the size and speed: C H = 0.124, CQ = 0.0165, η = 0.75 Q = 3× ∴D = ∴ω = π 4 4 × 0.052 = 0.00589 m 2 CHQ 2 CQ2 gH P Q CQ D 3 = = 4 0.124 × 0.00589 2 0.01652 × 9.81 × 25.0 0.00589 0.0165 × 0.090 3 = 0.090 m = 490 rad/s, or N = 490 × 30 π = 4680 rpm (b) Available net positive suction head: NPSH = pa − pv 101×103 − 86.2 ×103 − Δz = + 3 = 3.95 m 1593 × 9.81 ρg (c) 182 © 2012 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 12 / Turbomachinery Turbines ω = 120 × π / 30 = 12.6 rad/s α1 = cot −1 (2π r12b1ω / Q + cot β1 ) = cot −1 (2π × 4.52 × 0.85 × 12.6 /150 + cot 75D ) = 6.1D Vt1 = u1 + Vn1 cot β1 = ω r1 + = 12.6 × 4.5 + 12.36 Q cot β1 2π r1b1 150 cot 75D = 58.37 m/s 2π × 4.5 × 0.85 Vt2 = u2 + Vn2 cot β 2 = ω r2 + = 12.6 × 2.5 + Q cot β 2 2π r2b2 150 cot100D = 29.52 m/s 2π × 2.5 × 0.85 ∴ T = ρQ(rV 1 t1 − r2Vt2 ) = 1000 ×150(4.5 × 58.37 − 2.5 × 29.52) = 2.83 × 107 N ⋅ m = ωT = 12.6 × 2.83 × 107 = 3.57 × 108 W or 357 MW W T =W , hence Under ideal conditions ηT = 1, and W T f HT = W T / γ Q = 3.57 × 108 /(9810 × 150) = 243 m N 2 = 240 rpm , W 2 = 2200 kW , D 2 = 0.9 m, W 1 = 9kW , H1 = 7.5 m 3 5 2 2 W 2 ⎛ N 2 ⎞ ⎛ D2 ⎞ H 2 ⎛ N 2 ⎞ ⎛ D2 ⎞ From the similarity rules = ⎜ =⎜ ⎟ ⎜ ⎟ and ⎟ ⎜ ⎟ H1 ⎝ N1 ⎠ ⎝ D1 ⎠ W1 ⎝ N1 ⎠ ⎝ D1 ⎠ 12.38 Substitute second eqn. into the first to eliminate ( N 2 / N 1 ), and solve for D 1 : 1/ 2 3/ 4 ⎞1/ 2 ⎛ H ⎞3/ 4 ⎛W ⎛ 9 ⎞ ⎛ 45 ⎞ 1 2 ∴ D1 = D2 ⎜ ⎟ ⎜ ⎟ = 0.9 ⎜ ⎟ ⎜ ⎟ = 0.221 m ⎝ 2200 ⎠ ⎝ 7.5 ⎠ ⎝ W2 ⎠ ⎝ H1 ⎠ 1/ 2 ⎛H ⎞ N 2 = N1 ⎜ 1 ⎟ ⎝ H2 ⎠ ⎛ D2 ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ = 240 ⎜ ⎟ D ⎝ 45 ⎠ ⎝ 1⎠ 1/ 2 ⎛ 0.9 ⎞ ⎜ ⎟ = 399 rpm ⎝ 0.221 ⎠ 183 © 2012 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 12 / Turbomachinery Write energy eqn. from upper reservoir (loc. 1) to surge tank (loc. 2) and solve for Q: 1/ 2 ⎡ ⎤ D Q = ⎢ 2 gA2 ( z1 − z2 ) ⎥ fL ⎣ ⎦ 1/ 2 ⎡ 2 × 9.81 × (π / 4) 2 × 0.855 ⎤ =⎢ (650 − 648.5) ⎥ 0.025 × 2000 ⎣ ⎦ = 0.401 m 3 / s Apply energy eqn. from loc. 2 to lower reservoir (loc. 3) and determine H T : 2 ⎛ L ⎞ Q HT = z2 − z3 − ⎜ f + K v ⎟ ⎝ D ⎠ 2 gA2 12.40 ⎛ 0.02 × 100 ⎞ = 648.5 − 595 − ⎜ + 1⎟ ⎝ 0.85 ⎠ 0.4012 ⎛π ⎞ 2 × 9.81× ⎜ ⎟ × 0.855 ⎝4⎠ 2 = 53.4 m ∴ W T = γ QHTηT = 9810 × 0.401 × 53.4 × 0.9 = 1.89 × 105 W or 189 kW ∴ From Fig. 12.32, use a Francis turbine. A representative value of the specific speed is 2 (Fig. 12.20): ΩT ( gHT )5 / 4 2(9.81× 53.3)5 / 4 ∴ω = = = 306 rad/s / ρ )1/ 2 (W (2.67 × 105 /1000)1/ 2 T or N = 306 × 30 / π = 2920 rpm ω = 200 × π / 30 = 20.94 rad/s and from Fig. P12.42 at best efficiency (ηT = 0.8), φ ≈ 0.42, cv = 0.94 V1 = cv 2 gHT = 0.94 2 × 9.81× 120 = 45.6 m/s ∴Q = 12.42 WT 4.5 × 10 6 = = 4.78 m 3 / s γ H T η T 9810 × 120 × 0.8 This is the discharge from all of the jets. Determine the wheel radius r : r = φ 2 gHT 0.42 2 × 9.81× 120 = = 0.973 m 20.94 ω Hence, the diameter of the wheel is 2r = 2 × 0.973 = 1.95 m Compute diameter of one jet: D j = 2r / 8 = 1.95 / 8 = 0.244 m, or 244 mm Let N j = no. of jets. Then each jet has a discharge of Q / N j and an area 184 © 2012 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 12 / Turbomachinery Q / Nj V1 Nj = = π 4 D2j ∴Solving for N j : Q 1 4.78 = = 2.24 V1 ⎛ π 2 ⎞ ⎛π ⎞ 2 D 45.6 0.244 × × ⎜ ⎜ ⎟ j ⎟ ⎝4 ⎠ ⎝4⎠ ∴Use three jets ΩT = ω (W T / ρ )1/ 2 ( gHT )5 / 4 = 20.94(4.5 × 106 /1000)1/ 2 = 0.204 (9.81 × 120)5 / 4 Q = 2082 / 6 = 347 m3 / sec (one unit) HT = WT 427,300 × 746 = = 110.1 m γ QηT 9810 × 347 × 0.85 Write energy eqn. from upper reservoir (loc. 1) to lake (loc. 2): 2 ⎛ L ⎞ Q + HT + z2 z1 = ⎜ f + ΣK ⎟ 2 ⎝ D ⎠ 2gA 12.44 ⎛ 0.01 × 390 ⎞ ∴ 309 = ⎜ + 0.5 ⎟ D ⎝ ⎠ which reduces to 2.4 − 347 2 ⎛π ⎞ 2 × 9.81 × ⎜ ⎟ D 4 ⎝4⎠ 2 + 110.1 + 196.5 38,840 4980 − 4 =0 D5 D ∴ Solving, D ≅ 8 m ∴From Fig. 12.32, a Francis or pump/turbine unit is indicated. (a) Let H be the total head and Q the discharge delivered to the turbine; then HT = 0.95H = 0.95 × 305 = 289.8 m and 12.46 Q= W 10.4 × 10 6 T = = 4.30 m 3 / s. γHT ηT 9810 × 289.8 × 0.85 Write energy eqn. from reservoir to turbine outlet: 2 ⎛ L ⎞ Q H = HT +⎜ f +∑K⎟ 2 ⎝ D ⎠ 2gA 4.32 ⎛ 0.02 × 3000 ⎞ ∴ 305 = 289.8 + ⎜ + 2⎟ D ⎝ ⎠ 2 × 9.81× (π / 4)2 D 4 185 © 2012 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 12 / Turbomachinery which reduces to 15.2 − 91.67 3.06 − 4 =0 D5 D ∴Solving D = 1.45 m (b) Compute jet velocity: V1 = cv 2 gHT = 0.98 2 × 9.81× 289.8 = 73.9 m/s The flow through one nozzle is Q/4 and the jet area is πD 2j / 4. Hence π D 2j 4 = Q / 4 4.3/ 4 = = 0.0146 m 2 V1 73.9 4 × 0.0146 ∴ Dj = π = 0.136 m 2 12.48 ⎡ 4.15 × (9.81 × 3.7)5 / 4 ⎤ 6 WT = 1000 ⎢ ⎥ = 4.99 × 10 W (one unit). × 50 / 30 π ⎣ ⎦ Total power developed is 9.21 × 10 6 W. Hence, required number of units is 9.21/ 4.99 = 1.8 ∴ Use two turbines. (a) H T = z1 − z 2 − fL Q 2 D 2 gA 2 0.015 × 350 × 0.252 = 915 − 892 − = 11.8 m 0.3 × 2 × 9.81× (0.7854 × 0.32 ) 2 = γ QH η = 9810 × 0.25 ×11.8 × 0.85 = 2.46 ×104 W or 24.6 kW W T T T (b) Compute the specific speed of the turbine: 12.50 ω=N π 30 ∴ΩT = = 1200 × /ρ ω W T ( gHT ) 5/4 π 30 = = 126 rad/s 126 2.46 × 104 /1000 ( 9.81×11.8) 5/4 = 1.65 Hence, from Fig. 12.20, a Francis turbine is appropriate. (c) From Fig. 12.24, the turbine with ΩT = 1.063 is chosen: CH = 0.23, CQ = 0.13, and ηT = 0.91. 186 © 2012 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 12 / Turbomachinery D= ω= 4 CHQ2 = CQ2 gHT 4 0.23 × 0.252 = 0.293 m or approximately 0.30 m 0.132 × 9.81 × 11.8 Q 0.25 30 = = 71.2 rad/s or N = 71.2 × = 680 rpm 3 3 π CQ D 0.13 × 0.30 W T = γ QHTηT = 9810 × 0.25 × 11.8 × 0.91 = 2.63 × 10 4 W, or 26.3 kW Calculate a new specific speed based on the final design data: ΩT = 71.2 × 2.63 ×104 /1000 (9.81×11.8)5/4 = 0.96 An acceptable value according to Fig. 12.20. 187 © 2012 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 12 / Turbomachinery 188 © 2012 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 13 / Measurements in Fluid Mechanics CHAPTER 13 Measurements in Fluid Mechanics Introduction a) γ Hg h V2 V2 h ×ρ = p = ∴ × 1.22 = ( 9810 × 13.6 ) 2 100 2 100 ∴V = 46.8 h and C = 46.8 13.2 b) V2 h × (0.8217 ×1.22) = (9810 × 13.6) 2 100 Q= ∴ V = 51.6 h and C = 51.6 V 1.9 ×10−3 = = 3.2 ×10−6 m3/s Δt 10 × 60 = ρQ = 997 × 3.2 ×10−6 = 3.19 ×10−3 kg/s m 13.4 Q 3.2 ×10−6 0.65 × 0.0025 V= = = 0.65 m/s ∴ Re = = 1805 2 A (π × (0.0025) /4) 9 ×10−7 The flow is laminar. Q = ⎡π ×12 ×10 + π (22 − 12 ) × 9.9 + π (32 − 22 ) × 9.5 + π (42 − 32 ) × 8.65 + ⎣ π (4.52 − 42 ) × 6.65 + π (52 − 4.52 ) × 2.6⎤⎦ ×10−4 = 0.0592 m3/s 13.6 V= Q 0.0592 = = 7.54 m/s A π × 0.052 See the sketch above: 13.8 p1 − p2 = H. ∴ p1 − p2 = 9810(13.6 − 1) × 0.12 = 14,830 Pa. γ Hg − γ ∴ h1 − h2 = p1 − p2 14,830 = = 1.512 m 9810 9810 D0 15 = = 0.625 D 24 189 © 2012 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 13 / Measurements in Fluid Mechanics a) Assume: Re = 105 ∴ K = 0.68 ∴Q = 0.68π × 0.0752 2 × 9.81×1.512 = 0.0654 m3/s Check: V = Q = 1.45 m/s A Re = 1.45 × 0.24 10 −6 = 3.5 × 105 ∴ K = 0.67 and Q = 0.064 m3/s Q = KA RΔp 6.5 where K = 1 − Dρ Re Re = VD ν = Q4 π Dν a) Assume 13.10 Re = 105 Then K = 0.98 Q = 0.98π × 0.052 Check V = 0.2 × 80,000 = 0.0974 m3/s 0.1×1000 0.0974 12.4 × 0.1 Q = = ∴ = 12.4 m/s Re =1.2 ×106 ∴ K = 0.99 − 2 6 A π × 0.05 10 ∴OK A0 = K= π ⎛ 33.3 ⎞ 2 −4 2 ×⎜ ⎟ = 8.709 × 10 m 4 ⎝ 1000 ⎠ A0 Q 2 g (S − 1)Δh 13.12 = 8.709 × 10 −4 β = 33.3/ 54 = 0.617 (Δh in meters of mercury) Q Q = 73.03 = K(Q , Δh) Δh 2 × 9.81 × ( 13.6 − 1)Δh δK = [ K(Q , Δh) − K(Q + δQ , Δh)] 2 + [ K(Q , Δh) − K(Q , Δh + δ ( Δh))] 2 190 © 2012 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 13 / Measurements in Fluid Mechanics No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 K (Q + δ Q, Δh) 1.147 1.093 1.082 1.095 1.100 1.112 1.096 1.084 1.063 1.083 1.099 1.085 1.103 1.109 1.050 K (Q, Δh) 1.076 1.024 1.017 1.026 1.031 1.041 1.027 1.025 1.005 1.025 1.041 1.027 1.043 1.040 0.9935 K (Q, Δh + δ (Δh)) 1.002 0.9805 0.9862 1.001 1.012 1.025 1.012 1.013 0.9938 1.015 1.032 1.018 1.036 1.033 0.9864 δK 0.1026 0.0816 0.0719 0.0734 0.0716 0.0728 0.0706 0.0602 0.0591 0.0589 0.0587 0.0587 0.0604 0.0694 0.0569 Compare the above with Fig. 13.10, curve labeled “Venturi meters and nozzles” β = 0.6. 13.14 i (S − 1)Δh 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (m water) 0.164 0.277 0.403 0.504 0.680 0.781 0.882 1.033 1.147 1.260 1.424 1.550 1.688 1.764 1.764 x i( 1) y i( 2 ) -1.8079 -1.2837 -0.9088 -0.6852 -0.3857 -0.2472 -0.1256 0.03247 0.1371 0.2311 0.3535 0.4383 0.5235 0.5676 0.5676 Σ -2.5929 -6.3890 -6.1754 -5.9955 -5.8746 -5.7199 -5.6408 -5.5940 -5.5165 -5.4846 -5.4171 -5.3412 -5.3124 -5.2533 -5.2344 -5.2805 -84.2292 xi y i 11.5507 7.9274 5.4487 4.0253 2.2062 1.3944 0.7026 -0.1791 -0.7519 -1.2519 -1.8881 -2.3284 -2.7501 -2.9710 -2.9972 +18.1376 x i2 3.2685 1.6479 0.8259 0.4695 0.1488 0.06111 0.01578 0.001054 0.01880 0.05341 0.1250 0.1921 0.2741 0.3222 0.3222 +7.7464 191 © 2012 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 13 / Measurements in Fluid Mechanics (1) xi = ln[(S − 1)Δh] S = 13.6 (Δh in meters mercury) ∴m = b= (2) yi = ln Q (Q in m3 /s) 18.1376 − ( −2.5929)( −84.2292) / 15 = 0.4902 , 7.7464 − ( 2.5929) 2 / 15 −84.2292 − 0.4902( −2.5929) = −5.5305 15 ∴ C = exp(−5.5305) = 0.00396 . From Problem 13.12, A0 = 8.709 ×10−4 m2 and Kavg = 1.029 . Hence in Eq. 13.3.8 Kavg A 0 2 g = 1.029 × (8.709 ×10−4 ) 2 × 9.81 = 0.00397, and the exponent is 0.500. 192 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.