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Unformatted text preview: Chapter 7 • Flow Past Immersed Bodies 573 P7.15 Discuss whether fully developed laminar incompressible flow between parallel plates, Eq. (4.143) and Fig. 4.16b, represents an exact solution to the boundarylayer equations (7.19) and the boundary conditions (7.20). In what sense, if any, are duct flows also boundarylayer flows?
Fig. 4.16 Solution: The analysis for flow between parallel plates leads to Eq. (4.143):
2 2 dp dp ⎛ dp ⎞ h ⎛ y ⎞ u=⎜ ⎟ 1 − 2 ⎟ ; v = 0; = constant < 0; = 0, u( ± h) = 0 ⎜ h⎠ ⎝ dx ⎠ 2μ ⎝ dx dy It is indeed a “boundary layer,” with v u and ∂ p/∂ y ≈ 0. The “freestream” is the centerline velocity, umax = (−dp/dx)(h2/2μ). The boundary layer does not grow because it is constrained by the two walls. The entire duct is filled with boundary layer. Ans. P7.16 A thin flat plate 55 by 110 cm is The viscosity is immersed in a 6m/s stream of SAE 10 oil at incorrect. 20°C. Compute the total friction drag if the Please read the stream is parallel to (a) the long side and table to get the (b) the short side. viscosity of SAE 10 oil. The correct answer Solution: For SAE 30 oil at 20°C, take ρ = 891 kg/m3 and μ = 0.29 kg/m⋅s. for (a) 107.1N, 891(6.0)(1.1) 1.328 (b) 151.45N (a) L = 110 cm, Re L = = 20300 (laminar), CD = ≈ 0.00933 0.29 (20300)1/2
⎛ρ⎞ ⎛ 891 ⎞ 2 F = CD ⎜ ⎟ U 2 (2bL) = 0.00933 ⎜ ⎟ (6) [2(0.55)(1.1)] ≈ 181 N ⎝2⎠ ⎝2⎠ Ans. (a) The drag is 41% more if we align the flow with the short side: (b) L = 55 cm, ReL = 10140, CD = 0.0132, F ≈ 256 N (41% more) Ans. (b)
P7.17 Helium at 20°C and low pressure flows past a thin flat plate 1 m long and 2 m wide. It is desired that the total friction drag of the plate be 0.5 N. What is the appropriate absolute pressure of the helium if U = 35 m/s? 604 Solutions Manual • Fluid Mechanics, Fifth Edition P7.60 A fishnet consists of 1mmdiameter strings overlapped and knotted to form 1 by 1cm squares. Estimate the drag of 1 m2 of such a net when towed normal to its plane at 3 m/s in 20°C seawater. What horsepower is required to tow 400 ft2 of this net? Fig. P7.60 Solution: For seawater at 20°C, take ρ =1025 kg/m3 and μ = 0.00107 kg/m⋅s. Neglect the knots at the net’s intersections. Estimate the drag of a single onecentimeter strand:
Re D = Fone strand 1025(3)(0.001) ≈ 2900; Fig. 7.16a or Fig. 5.3a: CD ≈ 1.0 0.00107 ρ ⎛ 1025 ⎞ 2 = C D U 2 DL = (1.0) ⎜ (3) (0.001)(0.01) ≈ 0.046 N/strand ⎝2⎟ ⎠ 2 one m2 contains 20,000 strands: F1 sq m ≈ 20000(0.046) ≈ 920 N Ans. (a) To tow 400 ft 2 = 37.2 m 2 of net, F = 37.2(920) ≈ 34000 N ≈ 7700 lbf If U = 3 m ft = 9.84 , Tow Power = FU = (7700)(9.84) ÷ 550 ≈ 140 hp s s Ans. (b) P7.61 A filter may be idealized as an array of cylindrical fibers normal to the flow, as in Fig. P7.61. Assuming that the fibers are uniformly distributed and have drag coefficients given by Fig 7.16a, derive an approximate expression for the pressure drop Δp through a filter of thickness L. Solution: Consider a filter section of height H and width b and thickness L. Let N be the number of fibers of diameter D per Chapter 7 • Flow Past Immersed Bodies 607 P7.66 A sphere of density ρs and diameter D is dropped from rest in a fluid of density ρ and viscosity μ. Assuming a constant drag coefficient Cdo , derive a differential equation for the fall velocity V(t) and show that the solution is
⎡ 4 gD(S − 1) ⎤ V=⎢ ⎥ ⎢ 3Cdo ⎥ ⎣ ⎦
1/ 2 tanh Ct
1/ 2 ⎡ 3gCdo (S − 1) ⎤ C=⎢ ⎥ 2 ⎣ 4S D ⎦ where S = ρs /ρ is the specific gravity of the sphere material. Solution: Newton’s law for downward motion gives
∑ Fdown = ma down , or: W − B − CD Fig. P7.66 W dV π , where A = D2 2 g dt 4 π dV = β − α V2 , and W − B = ρ(S − 1)g D3 . Rearrange to 6 dt V2 A = ρ β = g ⎜ 1 − ⎟ and α = ⎝ S⎠ ⎛ 1⎞ ρgCD A
2W Separate the variables and integrate from rest, V = 0 at t = 0: ∫ dt = ∫ dV/(β − α V2), or: V = β tanh t αβ = Vfinal tanh(Ct) Ans. α ( ) where Vfinal ⎡ 4gD(S − 1) ⎤ =⎢ ⎥ ⎣ 3C D ⎦ 1/2 ρs ⎡ 3gC D (S − 1) ⎤ and C = ⎢ 2 ⎥ , S = ρ >1 ⎣ 4S D ⎦
1/2 612 Solutions Manual • Fluid Mechanics, Fifth Edition P7.72 A settling tank for a municipal water supply is 2.5 m deep, and 20°C water flows through continuously at 35 cm/s. Estimate the minimum length of the tank which will ensure that all sediment (SG = 2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and (b) 100 μm.
Fig. P7.72 Solution: For water at 20°C, take ρ = 998 kg/m3 and μ = 0.001 kg/m⋅s. The particles travel with the stream flow U = 35 cm/s (no horizontal drag) and fall at speed Vf with drag equal to their net weight in water:
Wnet = (SG − 1) ρ w g π
6 D3 = Drag = C D ρw
2 Vf2 π
4 D2 , or: Vf2 = 4(SG − 1)gD 3C D where CD = fcn(ReD) from Fig. 7.16b. Then L = Uh/Vf where h = 2.5 m. (a) D = 1 mm: Vf2 = 4(2.55 − 1)(9.81)(0.001) , iterate Fig. 7.16b to CD ≈ 1.0, 3CD
(0.35)(2.5) ≈ 6.3 m 0.14 Ans. (a) Re D ≈ 140, Vf ≈ 0.14 m/s, hence L = Uh/Vf = (b) D = 100 μm: Vf2 = 4(2.55 − 1)(9.81)(0.0001) , iterate Fig. 7.16b to CD ≈ 36, 3CD
0.35(2.5) ≈ 120 m 0.0075 Ans. (b) Re D ≈ 0.75, Vf ≈ 0.0075 m/s, L = 618 Solutions Manual • Fluid Mechanics, Fifth Edition P7.80 A heavy sphere attached to a string should hang at an angle θ when immersed in a stream of velocity U, as in Fig. P7.80. Derive an expression for θ as a function of the sphere and flow properties. What is θ if the sphere is steel (SG = 7.86) of diameter 3 cm and the flow is sealevel standard air at U = 40 m/s? Neglect the string drag. Solution: For sealevel air, take ρ = 1.225 kg/m3 and μ = 1.78E−5 kg/m⋅s. The sphere should hang so that string tension balances the resultant of drag and net weight: Fig. P7.80 3 Wnet −1 ⎡ ( ρs − ρ )g(π /6)D ⎤ tan θ = , or θ = tan ⎢ Symbolic answer. 2 2⎥ Drag ⎣ (π /8)CD ρ U D ⎦ For the given numerical data, first check Re and the drag coefficient, then find the angle: Re D = 1.225(40)(0.03) ≈ 83000, Fig. 7.16b: CD ≈ 0.5 1.78E−5
F= π
8 (0.5)(1.225)(40)2 (0.03)2 ≈ 0.346 N; W = [7.86(998) − 1.225](9.81) (0.03)3 ≈ 1.09 N ∴ θ = tan −1 (1.09/0.346) ≈ 72° 6 π Ans. P7.81 A typical U.S. Army parachute has a projected diameter of 28 ft. For a payload mass of 80 kg, (a) what terminal velocity will result at 1000m standard altitude? For the same velocity and payload, what size dragproducing “chute” is required if one uses a square flat plate held (b) vertically; and (c) horizontally? (Neglect the fact that flat shapes are not dynamically stable in free fall.) Neglect plate weight. Solution: For air at 1000 meters, from Table A3, ρ ≈ 1.112 kg/m3. Convert D = 28 ft = 8.53 m. Convert W = mg = 80(9.81) = 785 N. From Table 73 for a parachute, read CD ≈ 1.2, Then, for part (a),
m ⎛ 1.112 ⎞ 2 π W = 785 N = Drag = 1.2 ⎜ (8.53)2 , solve for U ≈ 4.53 ⎟U ⎝2⎠ s 4
Ans. (a) 646 Solutions Manual • Fluid Mechanics, Fifth Edition For our particular data, evaluate Vf = C= 2(9810 sin 5.71° − 70) m = 46.0 ; 1.225(0.7) s (9810 sin 5.71° − 70)(0.7)(1.225/2) = 0.0197 s−1 1000 We need to know when the car reaches Δx = (20 m)/sin(5.71°) ≈ 201 m. The above expression for V(t) may be readily integrated: Δx = ∫ V dt =
0 t Vf 46.0 ln[ cosh(Ct)] = ln[ cosh(0.0197t)] = 201 m if t ≈ 21.4 s C 0.0197 m km = 65.9 Ans. s h Then, at Δx = 201 m, Δz = 20 m, V = 46.0 tanh[0.0197(21.4)] ≈ 18.3 P7.115 The Cessna Citation executive jet weighs 67 kN and has a wing area of 32 m2. It cruises at 10 km standard altitude with a lift coefficient of 0.21 and a drag coefficient of 0.015. Estimate (a) the cruise speed in mi/h; and (b) the horsepower required to maintain cruise velocity. Solution: At 10 km standard altitude (Table A6) the air density is 0.4125 kg/m3. (a) The cruise speed is found by setting lift equal to weight: Lift = 67000 N = CL ρ ⎛ 0.4125 kg/m 3 ⎞ 2 2 V 2 Awing = 0.21⎜ ⎟ V (32 m ), 2 2 ⎝ ⎠ mi m = 492 h s Ans. (a) Solve V = 220 (b) With speed known, the power is found from the drag: ρ ⎧ ⎫ ⎛ ⎞ ⎛ 0.4125 ⎞ 2 Power = FdragV = ⎜ CD V 2 A⎟ V = ⎨0.015 ⎜ ⎟ (220) (32) ⎬ (220) ⎝ ⎠ ⎝2⎠ 2 ⎩ ⎭
= 1.05 MW = 1410 hp Ans. (b)
P7.116 An airplane weighs 180 kN and has a wing area of 160 m2 and a mean chord of 4 m. The airfoil properties are given by Fig. 7.25. If the plane is designed to land at Vo = 1.2Vstall, using a split flap set at 60°, (a) What is the proper landing speed in mi/h? (b) What power is required for takeoff at the same speed? ...
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This note was uploaded on 05/18/2011 for the course MEEG 332 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
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