Solution manual for fox and mcdonalds introduction to fluid mechanics 9th edition by pritchard

Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.1 Given: [Difficulty: 3] Common Substances Tar Sand “Silly Putty” Jello Modeling clay Toothpaste Wax Shaving cream Some of these substances exhibit characteristics of solids and fluids under different conditions Find: Explain and give examples Solution: Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures At high pressures or over long periods, they exhibit fluid characteristics At higher temperatures, all three liquefy and become viscous fluids Modeling clay and silly putty show fluid behavior when sheared slowly However, they fracture under suddenly applied stress, which is a characteristic of solids Toothpaste behaves as a solid when at rest in the tube When the tube is squeezed hard, toothpaste “flows” out the spout, showing fluid behavior Shaving cream behaves similarly Sand acts solid when in repose (a sand “pile”) However, it “flows” from a spout or down a steep incline Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.2 Given: Five basic conservation laws stated in Section 1-4 Write: A word statement of each, as they apply to a system Solution: Assume that laws are to be written for a system [Difficulty: 2] a Conservation of mass — The mass of a system is constant by definition b Newton's second law of motion — The net force acting on a system is directly proportional to the product of the system mass times its acceleration c First law of thermodynamics — The change in stored energy of a system equals the net energy added to the system as heat and work d Second law of thermodynamics — The entropy of any isolated system cannot decrease during any process between equilibrium states e Principle of angular momentum — The net torque acting on a system is equal to the rate of change of angular momentum of the system Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.3 [Difficulty: 3] Open-Ended Problem Statement: The barrel of a bicycle tire pump becomes quite warm during use Explain the mechanisms responsible for the temperature increase Discussion: Two phenomena are responsible for the temperature increase: (1) friction between the pump piston and barrel and (2) temperature rise of the air as it is compressed in the pump barrel Friction between the pump piston and barrel converts mechanical energy (force on the piston moving through a distance) into thermal energy as a result of friction Lubricating the piston helps to provide a good seal with the pump barrel and reduces friction (and therefore force) between the piston and barrel Temperature of the trapped air rises as it is compressed The compression is not adiabatic because it occurs during a finite time interval Heat is transferred from the warm compressed air in the pump barrel to the cooler surroundings This raises the temperature of the barrel, making its outside surface warm (or even hot!) to the touch Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.4 (Difficulty: 1) Given: Data on oxygen tank Find: Mass of oxygen Solution: Compute tank volume, and then us e oxygen density to find the mass The given or available data is: D = 16⋅ft For oxygen the critical temperature and pressure are: p = 1000⋅psi T = (77 + 460)⋅R T = 537⋅R Tc = 279⋅R p c = 725.2⋅psi (data from NIST WebBook) so the reduced temperature and pressure are: Using a compressiblity factor chart: Z = 0.948 Since this number is close to 1, we can assume ideal gas behavior Therefore, the governing equation is the ideal gas equation p = ρ⋅R O2⋅T where V is the tank volume V = π⋅D Full file at https://TestbankDirect.eu/ V = π × (16⋅ft) and ρ= V = 2144.7⋅ft M V Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Hence: Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.10 [Difficulty: 4] NOTE: Drag formula is in error: It should be: FD = ⋅ π⋅ V⋅ d Given: Data on sphere and formula for drag Find: Diameter of gasoline droplets that take second to fall 10 in Solution: Use given data and data in Appendices; integrate equation of motion by separating variables FD = 3πVd a = dV/dt Mg The data provided, or available in the Appendices, are: − lbf ⋅ s μ = 4.48 × 10 ⋅ ft ρw = 1.94⋅ slug ft SG gas = 0.72 ρgas = SGgas⋅ ρw M⋅ Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects) dV so g− Integrating twice and using limits ⋅ π⋅ μ⋅ d V( t) = M ρgas = 1.40⋅ dV dt slug ft = M ⋅ g − ⋅ π⋅ μ⋅ V⋅ d = dt ⋅V M⋅ g 3⋅ π⋅ μ ⋅ d Replacing M with an expression involving diameter d − 3⋅ π⋅ μ ⋅ d ⎞ ⎛ ⋅t ⎜ M ⋅⎝1 − e ⎠ π⋅ d M = ρgas ⋅ ⎡ ⎛ − 3⋅ π⋅ μ ⋅ d ⋅ t ⎢ ⎜ M M x( t) = ⋅ ⎢t + ⋅⎝e − 3⋅ π⋅ μ ⋅ d ⎣ 3⋅ π⋅ μ ⋅ d ⎞⎤ ⎥ 1⎠⎥ ⎦ ⎡⎢ ⎛ − 18⋅ μ ⋅ t ⎜ ρgas ⋅ d ⋅ g ⎢ ρgas ⋅ d ⎜ ρgas⋅ d2 x( t) = ⋅ ⎢t + ⋅⎝e − 18⋅ μ 18⋅ μ ⎣ ⎞⎥⎤ ⎥ 1⎠⎥ ⎦ M⋅ g This equation must be solved for d so that x ( ⋅ s) = 10⋅ in The answer can be obtained from manual iteration, or by using Excel's Goal Seek −3 ⋅ in 10 0.75 7.5 x (in) x (in) d = 4.30 × 10 0.5 0.25 2.5 0.025 0.05 0.075 t (s) 0.1 0.25 0.5 0.75 t (s) Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d, with V = 0.25 m/s (allowing for the fact that M is a function of d)! Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.12 [Difficulty: 3] Given: Data on sphere and terminal speed Find: Drag constant k, and time to reach 99% of terminal speed Solution: Use given data; integrate equation of motion by separating variables kVt − 13 M = × 10 The data provided are: mg ft Vt = 0.2⋅ s ⋅ slug M⋅ Newton's 2nd law for the general motion is (ignoring buoyancy effects) dV dt M ⋅ g = k ⋅ Vt Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) − 13 k = × 10 ⋅ slug × 32.2⋅ ft 2 s × 0.2⋅ ft s × = M⋅ g − k⋅ V k = 1.61 × 10 slug⋅ ft ⋅ M k − 13 ⋅ slug × ⎝ k M⋅ g − 11 1.61 × 10 ⋅ lbf ⋅ s × lbf ⋅ s slug⋅ ft t = 0.0286 s Full file at https://TestbankDirect.eu/ g− = dt k M ⋅V ⎠ V = 0.198 ⋅ ft Vt ⋅ V⎞ V = 0.99⋅ Vt We must evaluate this when t = −1 × 10 ⋅ ln⎛⎜ − M⋅ g ft dV t=− k = so − 11 lbf ⋅ s lbf ⋅ s To find the time to reach 99% of Vt, we need V(t) From 1, separating variables Integrating and using limits (1) ⎛ ⋅ ln⎜ − 1.61 × 10 ⎜ ⎝ ft s − 11 lbf ⋅ s ⋅ ft × − 13 × 10 ⋅ slug × s 32.2⋅ ft × 0.198 ⋅ ft s × slug⋅ ft ⎞ lbf ⋅ s ⎠ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.7 (Difficulty: 2) 1.7 A rocket payload with a weight on earth of 2000 𝑙𝑙𝑙 is landed on the moon where the acceleration due to the moon’s gravity 𝑔𝑚 ≈ payload’s moon weight 𝑔𝑒 Find the mass of the payload on the earth and the moon and the Given: Rocket payload weight on earth 𝑊𝑒 = 2000 𝑙𝑙𝑙 The acceleration due to the moon’s gravity 𝑔 𝑔𝑚 ≈ 𝑒 Find: The mass of payload on earth 𝑀𝑒 and on moon 𝑀𝑚 in SI and EE units The payload’s moon weight 𝑊𝑚 Solution: Basic equation: Newton’s law applied to mass and weight Gravity on the moon relative to that on Earth: 𝑀= 𝑊 𝑔 The value of gravity is: 𝑔𝑚 ≈ 𝑔𝑒 𝑔𝑒 = 32.2 The mass on earth is: 𝑓𝑓 𝑠2 𝑊𝑒 2000 𝑙𝑙𝑙 = = 62.1 𝑠𝑠𝑠𝑠 𝑓𝑓 𝑔𝑒 32.2 𝑠 The mass on moon is the same as it on earth: 𝑀𝑒 = The weight on the moon is then 𝑀𝑚 = 62.1 𝑠𝑠𝑠𝑠 𝑊𝑚 = 𝑀𝑚 𝑔𝑚 = 𝑀𝑚 � Full file at https://TestbankDirect.eu/ 𝑔𝑒 𝑔𝑒 𝑊𝑒 � = 𝑀𝑒 � � = = 333 𝑙𝑙𝑙 6 Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.8 (Difficulty: 1) 1.8 A cubic meter of air at 101 𝑘𝑘𝑘 and 15 ℃ weighs 12.0 𝑁 What is its specific volume? What is the specific volume if it is cooled to −10 ℃ at constant pressure? Given: Specific weight 𝛾 = 12.0 𝑁 𝑚3 at 101 𝑘𝑘𝑘 and 15 ℃ Find: The specific volume 𝑣 at 101 𝑘𝑘𝑘 and 15 ℃ Also the specific volume 𝑣 at 101 𝑘𝑘𝑘 and −10 ℃ Assume: Air can be treated as an ideal gas Solution: Basic equation: ideal gas law: 𝑝𝑝 = 𝑅𝑅 The specific volume is equal to the reciprocal of the specific weight divided by gravity 𝑣1 = 𝑔 𝛾 Using the value of gravity in the SI units, the specific volume is The temperature conditions are 𝑚 𝑔 9.81 𝑠2 𝑚3 = 0.818 𝑣1 = = 𝛾 12.0 𝑁 𝑘𝑘 𝑇1 = 15 ℃ = 288 𝐾, 𝑇2 = −10 ℃ = 263𝐾 For 𝑣2 at the same pressure of 101 𝑘𝑘𝑘 and cooled to −10 ℃ we have, because the gas constant is the same at both pressures: 𝑅𝑇1 𝑣1 𝑇1 𝑝 = = 𝑣2 𝑅𝑇2 𝑇2 𝑝 So the specific volume is 𝑣2 = 𝑣1 𝑚3 263 𝐾 𝑚3 𝑇2 = 0.818 × = 0.747 𝑇1 𝑘𝑘 288 𝐾 𝑘𝑘 Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.9 (Difficulty: 2) 1.9 Calculate the specific weight, specific volume and density of air at 40℉ and 50 𝑝𝑝𝑝𝑝 What are the values if the air is then compressed isentropically to 100 psia? Given: Air temperature: 40℉, Air pressure 50 psia Find: The specific weight, specific volume and density at 40℉ and 50 psia and the values at 100 psia after isentropic compression Assume: Air can be treated as an ideal gas Solution: Basic equation: 𝑝𝑝 = 𝑅𝑅 The absolute temperature is The gas constant is 𝑇1 = 40℉ = 500°𝑅 The specific volume is: 𝑅 = 1715 𝑓𝑓 ∙ 𝑙𝑙𝑙 𝑠𝑠𝑠𝑠 ∙ °𝑅 𝑓𝑓 ∙ 𝑙𝑙𝑙 1715 𝑅𝑇1 𝑓𝑓 𝑠𝑠𝑠𝑠 ∙ °𝑅 𝑣1 = = × 500°𝑅 = 119.1 144𝑖𝑖2 𝑝 𝑠𝑠𝑠𝑠 50𝑝𝑝𝑝𝑝 × 𝑓𝑓 The density is the reciprocal of the specific volume 𝜌1 = 𝑠𝑠𝑠𝑠 = 0.0084 𝑣1 𝑓𝑓 Using Newton’s second law, the specific weight is the density times gravity: 𝛾1 = 𝜌𝜌 = 0.271 𝑙𝑙𝑙 𝑓𝑓 For the isentropic compression of air to 100 psia, we have the relation for entropy change of an ideal gas: 𝑠2 − 𝑠1 = 𝑐𝑝 ln Full file at https://TestbankDirect.eu/ 𝑇2 𝑇1 − 𝑅 ln 𝑝2 𝑝1 Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.33 [Difficulty: 2] Given: Equation for maximum flow rate Find: Whether it is dimensionally correct If not, find units of 2.38 coefficient Write a SI version of the equation Solution: Rearrange equation to check units of 0.04 term Then use conversions from Table G.2 or other sources (e.g., Google) mmax⋅ T0 2.38 = "Solving" the equation for the constant 2.38: At ⋅ p Substituting the units of the terms on the right, the units of the constant are slug s ×R 1 × ft × psi = slug s ×R × ft 2 × in lbf × lbf ⋅ s slug ⋅ ft 2 = R ⋅ in ⋅ s ft 2 c = 2.38⋅ Hence the constant is actually R ⋅ in ⋅ s ft For BG units we could start with the equation and convert each term (e.g., At), and combine the result into a new constant, or simply convert c directly: 1 2 c = 2.38⋅ R ⋅ in ⋅ s ft 2 = 2.38⋅ R ⋅ in ⋅ s ft × ⎛ K ⎞ × ⎛ 1⋅ ft ⎞ × 1⋅ ft ⎜ ⎜ 0.3048m ⋅ ⎝ 1.8⋅ R ⎠ ⎝ 12⋅ in ⎠ c = 0.04⋅ K ⋅s m so mmax = 0.04⋅ Full file at https://TestbankDirect.eu/ At ⋅ p T0 with At in m2, p in Pa, and T0 in K Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.34 [Difficulty: 1] Given: Equation for mean free path of a molecule Find: Dimensions of C for a diemsionally consistent equation Solution: Use the mean free path equation Then "solve" for C and use dimensions The mean free path equation is "Solving" for C, and using dimensions m λ = C⋅ C= ρ⋅ d λ⋅ ρ⋅ d m L× C= M L M Full file at https://TestbankDirect.eu/ ×L =0 The constant C is dimensionless Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.36 Given: Data on a container and added water Find: Weight and volume of water added Solution: Use Appendix A For the empty container Wc = 3.5⋅ lbf For the filled container M total = 2.5⋅ slug The weight of water is then Ww = M total ⋅ g − Wc [Difficulty: 1] The temperature is ft 1⋅ lbf ⋅ s Ww = 2.5⋅ slug × 32.2⋅ × − 3.5⋅ lbf 1⋅ slug ⋅ ft s Ww = 77.0 lbf 90°F = 32.2°C ρ = 1.93⋅ and from Table A.7 slug ft Hence Vw = Mw Ww Vw = g⋅ ρ or ρ s ft 1⋅ slug ⋅ ft Vw = 77.0⋅ lbf × ⋅ × ⋅ × 32.2 ft 1.93 slug 1⋅ lbf ⋅ s Full file at https://TestbankDirect.eu/ Vw = 1.24ft Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.38 Given: Specific speed in customary units Find: Units; Specific speed in SI units [Difficulty: 1] Solution: The units are rpm⋅ gpm or ft ft 4 s Using data from tables (e.g Table G.2) NScu = 2000⋅ rpm⋅ gpm ft 1 ⎛ ⋅ ft ⎞ ⎜ 12 rpm⋅ gpm ⋅ π⋅ rad ⋅ ⎛ × 0.000946⋅ m ⋅ ⎞ NScu = 2000 × × × ×⎜ ⋅ ×⎜ ⋅ rev 60⋅ s ⋅ gal 60⋅ s ⎠ ⎝ ⎝ 0.0254⋅ m ⎠ ft ⎛ m3 ⎞ ⋅⎜ s ⎝ s ⎠ NScu = 4.06⋅ rad m Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.40 Given: [Difficulty: 2] Air at standard conditions – p = 29.9 in Hg, T = 59°F Uncertainty in p is ± 0.1 in Hg, in T is ± 0.5°F Note that 29.9 in Hg corresponds to 14.7 psia Find: Air density using ideal gas equation of state; Estimate of uncertainty in calculated value Solution: ρ= p lbf lb ⋅o R in = 14.7 × × × 144 RT in 53.3 ft ⋅ lbf 519o R ft The uncertainty in density is given by ⎡⎛ p ∂ρ ⎞ ⎛ T ∂ρ ⎞ ⎤ u ρ = ⎢⎜⎜ u p ⎟⎟ + ⎜⎜ uT ⎟⎟ ⎥ ⎢⎣⎝ ρ ∂p ⎠ ⎝ ρ ∂T ⎠ ⎥⎦ p ∂ρ RT = RT = = 1; ρ ∂p RT RT T ∂ρ T p p = ⋅− =− = −1; ρ ∂T ρ RT ρRT ± 0.1 = ± 0.334% 29.9 ± 0.5 uT = = ± 0.0963% 460 + 59 up = Then [ ] 2 u ρ = u 2p + (− uT ) [ u ρ = ± 0.348% = ± 2.66 × 10 − Full file at https://TestbankDirect.eu/ ] 2 = ± 0.334% + (− 0.0963% ) lbm ft Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.42 [Difficulty: 2] m = 1.62 ± 0.01oz (20 to 1) D = 1.68 ± 0.01in (20 to 1) Given: Standard American golf ball: Find: Density and specific gravity; Estimate uncertainties in calculated values Solution: Density is mass per unit volume, so ρ= m m m m =4 = = V πR 4π (D 2) π D ρ= π ×1.62 oz × 0.4536 kg in.3 × × = 1130 kg/m 3 3 16 oz (1.68) in (0.0254) m SG = and ρ ρH O = 1130 kg m3 × = 1.13 m 1000 kg 2 ⎡⎛ m ∂ρ ⎞ ⎛ D ∂ρ ⎞ ⎤2 u ρ = ⎢⎜⎜ u m ⎟⎟ + ⎜⎜ u D ⎟⎟ ⎥ ⎣⎢⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎥⎦ The uncertainty in density is given by m ∂ρ m ∀ = = = 1; ρ ∂m ρ ∀ ∀ um = D ∂ρ D ⎛ 6m ⎞ m = ⋅ ⎜ − ⎟ = −3 = −3; ρ ∂D ρ ⎝ πD ⎠ π ρD ± 0.01 = ± 0.617% 1.62 uD = ± 0.1 = ± 0.595% 1.68 Thus [ ] 2 u ρ = ± u + (− 3u D ) m [ ] 2 = ± 0.617% + (− × 0.595% ) u ρ = ±1.89% = ± 21.4 u SG = u ρ = ±1.89% = ± 0.0214 Finally, ρ = 1130 ± 21.4 kg/m SG = 1.13 ± 0.0214 Full file at https://TestbankDirect.eu/ (20 to 1) (20 to 1) kg m3 Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.43 Given: [Difficulty: 2] Pet food can H = 102 ± mm (20 to 1) D = 73 ± mm (20 to 1) m = 397 ± g (20 to 1) Find: Magnitude and estimated uncertainty of pet food density Solution: Density is ρ= m m m = = or ρ = ρ ( m, D, H ) ∀ πR H π D H From uncertainty analysis: ⎡⎛ m ∂ρ ⎞ ⎛ D ∂ρ ⎞ ⎛ H ∂ρ ⎞ ⎤ u ρ = ± ⎢⎜⎜ u m ⎟⎟ + ⎜⎜ u D ⎟⎟ + ⎜⎜ u H ⎟⎟ ⎥ ⎢⎣⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎝ ρ ∂H ⎠ ⎥⎦ Evaluating: m ∂ρ m ±1 4m = = = 1; um = = ±0.252% ρ ∂m ρ π D H ρ πD H 397 D ∂ρ D ±1 4m 4m = ( −2) = ( −2 ) = −2; u D = = ±137% ρ ∂D ρ ρ 73 πD H πD H H ∂ρ H ±1 4m 4m = ( −1) = ( −1) = −1; u H = = ±0.980% 2 ρ ∂H ρ ρ πD H 102 πD H Substituting: [ 2 2 u ρ = ±2.92% ∀= π D2 H = π × (73) mm × 102 mm × m3 10 mm 397 g m kg × = 930 kg m ρ= = −4 ∀ 4.27 × 10 m 1000 g Thus: ] 2 u ρ = ± (1 × 0.252 ) + (− × 1.37 ) + (− × 0.980) ρ = 930 ± 27.2 kg m (20 to 1) Full file at https://TestbankDirect.eu/ = 4.27 × 10 −4 m Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.44 Given: [Difficulty: 2] Mass flow rate of water determine by collecting discharge over a timed interval is 0.2 kg/s Scales can be read to nearest 0.05 kg Stopwatch can be read to nearest 0.2 s Find: Estimate precision of flow rate calculation for time intervals of (a) 10 s, and (b) Solution: Apply methodology of uncertainty analysis, Appendix F: m& = ∆m ∆t Computing equations: 2 ⎡⎛ ∆m ∂m& ⎞ ⎛ ∆t ∂m& ⎞ ⎤2 u m& = ± ⎢⎜ u ∆m ⎟ + ⎜ u ∆t ⎟ ⎥ ⎠ ⎝ m& ∂∆t ⎠ ⎦⎥ ⎣⎢⎝ m& ∂∆m ∆m ∂m& = ∆t = and m& ∂∆m ∆t Thus ∆t ∂m& ∆t ∆m = ⋅ − = −1 m& ∂∆t ∆m ∆t The uncertainties are expected to be ± half the least counts of the measuring instruments Tabulating results: Water Time Interval, ∆t Uncertainty Error in ∆t in ∆t (s) (s) Collected, Error in ∆m ∆m (kg) (%) Uncertainty Uncertainty in ∆m & in m (%) (%) (kg) 10 ± 0.10 ± 1.0 2.0 ± 0.025 ± 1.25 ± 1.60 60 ± 0.10 ± 0.167 12.0 ± 0.025 ± 0.208 ± 0.267 A time interval of about 15 seconds should be chosen to reduce the uncertainty in results to ± percent Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.45 [Difficulty: 3] Given: Nominal mass flow rate of water determined by collecting discharge (in a beaker) over a timed & = 100 g s ; Scales have capacity of kg, with least count of g; Timer has least interval is m count of 0.1 s; Beakers with volume of 100, 500, 1000 mL are available – tare mass of 1000 mL beaker is 500 g Find: Estimate (a) time intervals, and (b) uncertainties, in measuring mass flow rate from using each of the three beakers Solution: To estimate time intervals assume beaker is filled to maximum volume in case of 100 and 500 mL beakers and to maximum allowable mass of water (500 g) in case of 1000 mL beaker & = m Then Tabulating results ∆m ∆t ∆t = and ∆m ρ∆∀ = & & m m ∆∀ = 100 mL 500 mL 1000 mL ∆t = 1s 5s s Apply the methodology of uncertainty analysis, Appendix E Computing equation: 2 ⎡⎛ ∆m ∂m& ⎞ ⎛ ∆t ∂m& ⎞ ⎤2 u m& = ± ⎢⎜ u ∆m ⎟ + ⎜ u ∆t ⎟ ⎥ ⎠ ⎝ m& ∂∆t ⎠ ⎥⎦ ⎣⎢⎝ m& ∂∆m The uncertainties are ± half the least counts of the measuring instruments: δ∆m = ±0.5 g ∆m ∂m& = ∆t = and m& ∂∆m ∆t ∆t ∂m& ∆t ∆m = ⋅− = −1 m& ∂∆t ∆m ∆t δ∆t = 0.05 s [ ] 2 ∴ u m& = ± u ∆m + (− u ∆t ) Tabulating results: Beaker Volume ∆∀ (mL) 100 500 1000 Water Collected ∆m(g) 100 500 500 Error in ∆m (g) Uncertainty in ∆m (%) ± 0.50 ± 0.50 ± 0.50 ± 0.50 ± 0.10 ± 0.10 Time Interval ∆t (s) 1.0 5.0 5.0 Error in ∆t (s) Uncertainty in ∆t (%) Uncertainty & (%) in m ± 0.05 ± 0.05 ± 0.05 ± 5.0 ± 1.0 ± 1.0 ± 5.03 ± 1.0 ± 1.0 Since the scales have a capacity of kg and the tare mass of the 1000 mL beaker is 500 g, there is no advantage in & could be reduced to ± 0.50 percent by using the large beaker if a scale using the larger beaker The uncertainty in m with greater capacity the same least count were available Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.46 Given: [Difficulty: 2] Standard British golf ball: m = 45.9 ± 0.3 g (20 to 1) D = 411 ± 0.3 mm (20 to 1) Find: Density and specific gravity; Estimate of uncertainties in calculated values Solution: Density is mass per unit volume, so ρ= m = ∀ ρ= π m πR 3 = m m = π D3 4π ( D 2) × 0.0459 kg × m = 1260 kg m (0.0411) and ρ SG = ρH O = 1260 kg m3 × m3 = 126 1000 kg The uncertainty in density is given by ⎡⎛ m uρ = ± ⎢⎜⎜ ⎢⎣⎝ ρ m ∂ρ m = ρ ∂m ρ ∂ρ ⎞ ⎛ D ∂ρ ⎞ um ⎟ + ⎜ uD ⎟ ∂m ⎟⎠ ⎜⎝ ρ ∂D ⎟⎠ ⎤2 ⎥ ⎥⎦ ∀ = = 1; ∀ ∀ um = ± D ∂D D ⎛ m⎞ ⎛ 6m ⎞ = −3⎜ ⎟ = −3; = ⎜− ⎟ ρ ∂m ρ ⎝ π D ⎠ ⎝ πD ⎠ Thus [ u ρ = ± u m + (− 3u D ) ] 2 uD = ± [ u SG = u ρ = ± 2.29% = ± 0.0289 ρ = 1260 ± 28.9 kg m (20 to 1) SG = 126 ± 0.0289 (20 to 1) Full file at https://TestbankDirect.eu/ 0.3 = ±0.730% 41.1 = ± 0.654 + (− × 0.730 ) u ρ = ± 2.29% = ± 28.9 kg m Summarizing 0.3 = ±0.654% 45.9 ] 2 Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.48 Given: Data on water Find: Viscosity; Uncertainty in viscosity [Difficulty: 3] Solution: The data is: − N⋅s A = 2.414 × 10 ⋅ B = 247.8⋅ K C = 140 ⋅ K T = 303 ⋅ K m 0.5⋅ K uT = The uncertainty in temperature is u T = 0.171⋅ % 293⋅ K B Also μ( T) = A⋅ 10 ( T− C) − N⋅s μ ( 293⋅ K ) = 1.005 × 10 Evaluating ⋅ m A ⋅ B⋅ ln( 10) d μ ( T) = − dT For the uncertainty B 10 Hence u μ( T) = T d μ( T) ⋅ u T μ( T) dT ⋅ = Full file at https://TestbankDirect.eu/ C −T ln( 10) ⋅ B⋅ T⋅ u T ( C−T ) ⋅ ( C − T) Evaluating u μ( T) = 1.11⋅ % Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.50 [Difficulty: 3] Given: Lateral acceleration, a = 0.70 g, measured on 150-ft diameter skid pad; Uncertainties in Path deviation ±2 ft; vehicle speed ±0.5 mph Find: Estimate uncertainty in lateral acceleration; ow could experimental procedure be improved? Solution: Lateral acceleration is given by a = V2/R From Appendix F, u a = ±[(2 u v ) + ( u R ) ]1/ From the given data, V = aR; V = aR = 0.70 × 32.2 Then uv = ± and uR = ± δV V δR R = ±0.5 ft ft × 75 ft = 41.1 s s mi s ft hr × × 5280 × = ±0.0178 hr 41.1 ft mi 3600 s = ±2 ft × = ±0.0267 75 ft so u a = ± (2 × 0.0178) + (0.0267) 1/ = ±0.0445 u a = ±4.45 percent Experimental procedure could be improved by using a larger circle, assuming the absolute errors in measurement are constant For D = 400 ft; R = 200 ft V = aR; V = aR = 0.70 × 32.2 ft ft × 200 ft = 67.1 = 45.8 mph s s 0.5 = ±0.0109; u R = ± = ± 0.0100 45.8 200 u a = ± (2 × 0.0109) + 0.0100 = ± 0.0240 = ± 2.4% uV = ± [ Full file at https://TestbankDirect.eu/ ] Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Given data: H= δL = δθ = 57.7 0.5 0.2 ft ft deg For this building height, we are to vary θ (and therefore L ) to minimize the uncertainty u H Full file at https://TestbankDirect.eu/ Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar FullPlotting file uat vshttps://TestbankDirect.eu/ θ H 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 uH 4.02% 2.05% 1.42% 1.13% 1.00% 0.95% 0.96% 1.02% 1.11% 1.25% 1.44% 1.70% 2.07% 2.62% 3.52% 5.32% 10.69% Uncertainty in Height (H = 57.7 ft) vs θ 12% 10% 8% uH θ (deg) 6% 4% 2% 0% 10 20 30 40 50 60 70 80 90 θ (o) Optimizing using Solver θ (deg) 31.4 uH 0.947% To find the optimum θ as a function of building height H we need a more complex Solver θ (deg) 50 75 100 125 175 200 250 300 400 500 600 700 800 900 1000 29.9 34.3 37.1 39.0 41.3 42.0 43.0 43.5 44.1 44.4 44.6 44.7 44.8 44.8 44.9 uH 0.992% 0.877% 0.818% 0.784% 0.747% 0.737% 0.724% 0.717% 0.709% 0.705% 0.703% 0.702% 0.701% 0.700% 0.700% Use Solver to vary ALL θ's to minimize the total u H! Total u H's: 11.3% Full file at https://TestbankDirect.eu/ Optimum Angle vs Building Height 50 40 θ (deg) H (ft) 30 20 10 0 100 200 300 400 500 H (ft) 600 700 800 900 1000 Solution Manual for Fox and McDonalds Introduction to Fluid Mechanics 9th Edition by Pritchar Full file at https://TestbankDirect.eu/ Problem 1.52 [Difficulty: 4] Given: American golf ball, m = 1.62 ± 0.01 oz, D = 1.68 in Find: Precision to which D must be measured to estimate density within uncertainty of ± 1percent Solution: Apply uncertainty concepts Definition: Density, ρ≡ m ∀ ∀ = 34 π R = π D6 ⎡⎛ x ∂R ⎤2 ⎞ u R = ± ⎢⎜ u x1 ⎟ + L⎥ ⎢⎣⎝ R ∂x1 ⎥⎦ ⎠ Computing equation: From the definition, ρ = π Dm = π6Dm = ρ (m, D) 3/6 Thus m ∂ρ ρ ∂m = and D ∂ρ ρ ∂D = , so u ρ = ±[(1 u m ) + (3 u D ) ] u 2ρ = u m + u 2D Solving, u D = ± 13 [u ρ − u m2 ] From the data given, u ρ = ±0.0100 um = ±0.01 oz = ±0.00617 1.62 oz 1 u D = ± [(0.0100) − (0.00617) ] = ±0.00262 or ± 0.262% Since u D = ± δDD , then δ D = ± D u D = ±1.68 in.x 0.00262 = ± 0.00441 in The ball diameter must be measured to a precision of ± 0.00441 in.( ± 0.112 mm) or better to estimate density within ± 1percent A micrometer or caliper could be used Full file at https://TestbankDirect.eu/