ν δ - 1 -
δ ν ν δ ν ν - 2 -
ρ δ ρ θ θ θ δ τ ρ θ δ δ θ δ δ δ δ τ μ δ μ δ ν δ δ δ - 3 -
τ ρ δ ρ δ ρ δ δ δ δ δ δ δ δ δ δ δ - 4 -
ρ μ ρ μ ρ ρ μ μ ρ - 5 -
ρ τ μ τ μ ρ δ δ δ - 6 -
τ ρ μ τ ρ μ ρ δ θ θ δ θ - 7 -
ν δ δ ρ ν - 8 -
δ δ ρ τ ρ θ θ κ ν κ τ ρ κ δ ν δ ν δ δ δ δ 10 4 10 5 10 6 10 7 0.00493 0.00315 0.00217 0.00158 δ - 9 -
δ θ δ δ δ δ δ δ δ δ δ δ δ δ τ μ ρ δ δ δ δ - 10 -
δ θ ε ε - 11 -
ρ ν ε ν δ δ - 12 -
ρ ρ ε ρ - 13 -
7.6 Experimental External Flows Boundary layer theory is useful but, because of flow separation causes troubles. 1. Drag of two- and three-dimensional bodies: a. Blunt bodies. b. Streamlined shapes. 2. Performance of lifting bodies: a. Airfoils and aircraft. b. Projectiles and finned bodies. c. Birds and insects. Drag of Immersed Bodies Drag : The force on the body along free stream line. Drag and rolling moment. Lifting and yawing moment. Side force and pitching moment When the body has symmetry about the lift-drag axis (airplanes, ships, and cars) : the side force, yaw, and roll vanish, and the problem reduces to a 2-D case: two forces, lift - 14 -
and drag, and one moment, pitch. If the free stream is parallel to the intersection of these two planes : drag only, with no lift, side force, or moments. free-stream velocity and a characteristic length of the body ν (7.61) Characteristic Area ρ (7.62) Drag force = Lift force = ρ (Important) ρ (Important) 1. Frontal area. the body as seen from the stream; suitable for thick, stubby bodies, such as spheres, cylinders, cars, missiles, - 15 -
projectiles, and torpedoes. 2. Planform area, the body area as seen from above; suitable for wide, flat bodies such as wings and hydrofoils. 3. Wetted area, customary for surface ships and barges. Friction Drag and Pressure Drag (7.63) - 16 -
ρ θ - 17 -
Two-dimensional Bodies The drag of some representative wide-span (nearly two-dimensional) bodies All bodies have high at very low (creeping flow), while they spread apart at high Reynolds numbers according to their degree of streamlining. Creeping Flow if the Reynolds number is very small, ; the acceleration terms in the Navier-Stokes equations (7.14b, c) are negligible. The flow is termed creeping flow, of Stokes flow. Continuity and momentum reduce to two linear equations for velocity and pressure: - 18 -
μ Stokes, the sphere drag formula: πμ ρ π ρμ (7.64) EXAMPLE 7.6 A square 6-in piling is acted on by a water flow of 5 ft/s that is 20 ft deep, as shown in Fig. E 7.6. Estimate the maximum bending exerted by the flow on the bottom of the piling. Solution Assume seawater with ρ and kinematic - 19 -
viscosity ν. With a piling width of, we have This is the range where Table 7.2 applies. The worst case - 20 -
occurs when the flow strikes the flat side of the piling.. the frontal area is. The drag is estimated by ρ If the flow is uniform, the center of this force should be at approximately middepth. Therefore the bottom bending moment - 21 -
is Ans. According to the flexure formula from strength of materials, the bending stress at the bottom would be in to be multiplies, of course, by the stress concentration factor due to the built-in end conditions. EXAMPLE 7.8 A high-speed car with, and deploys a 2-m parachute to slow down from an initial velocity of (Fig. E 7.8). Assuming constant, brakes free, and no rolling resistance, calculate the distance and velocity of the car after 1, 10, 100 and 1000 s. For air assume ρ, and neglect interference between the wake of the car and the parachute. Solution Newton's law applied in the direction of motion gives ρ where subscript denotes the car and subscript the parachute. This is of the form - 22 -
Κ Separate the variables and integrate: Κ ρ Κ or Κ Rearrange and solve for the velocity : Κ Κ ρ ⑴ We can integrate this to find the distance traveled: α α α Κ ⑵ Now work out some numbers. From Table 7.3. ; hence π T h e n Κ α Now make a table of the results for and from Eqs. ⑴ and ⑵: ---> Look at p.490 Biological Drag Reduction Most such effort concentrates on rigid-body shapes. A different process occurs in nature. Tree root systems have evolved in several ways to resist wind-induced bending moments. And trunk cross sections have become resistant to bending but relatively easy to twist and reconfigure. - 23 -
Forces on Lifting Bodies Lifting bodies (airfoils, hydrofoils, or vanes) are intended to provide a large force normal to the free stream and as little drag as possible. Lift coefficient: (7.66a) Drag coefficient: (7.66b) α or α ρ ρ - 24 -
The rounded leading edge prevents flow separation there, but the sharp trailing edge cause a tangential wake motion that generates the lift. - 25 -
Figure 7.25 : lift and drag on a symmetric airfoil, NACA 0009 foil, the last digit indicating the thickness of 9 percent. πα The effect of increasing Reynolds number in Fig. 7.25 is to increase the maximum lift and stall angle (without changing the slope appreciably) and to reduce the drag coefficient. For takeoff and landing, the lift is greatly increased by deflecting a split flap, as shown in Fig. 7.25. This makes the airfoil unsymmetric (or effectively cambered) and changed the zero-lift point to α. A lifting craft cruises at low angle of attack, where the lift is much larger than the drag. - 26 -
Finite span can be correlated with slenderness, or aspect ratio, denoted (AR): (7.68) The theory of finite-span airfoils [16] predicts that the effective angle of attack increases, as in Fig.7.27, by the amount α π (7.69) - 27 -
When applied to Eq. (7.67), the finite-span lift becomes π α The associated drag increase is α α, or π (7.70) (7.71) The existence of a maximum lift coefficient implies the existence of a minimum speed, or stall speed, for a craft whose lift supports its weight: ρ or ρ (7.72) - 28 -
- 29 -
EXAMPLE 7.9 An aircraft weighs, has a planform area of, and can deliver a constant thrust of. It has an aspect ratio of, and. Neglecting rolling resistance, estimate the takeoff distance at sea level if takeoff speed equals times stall speed. Take. Solution The stall speed from Eq. (7.72), with sea-level density ρ, is ρ Hence takeoff speed. The drag is estimated from Eq. (7.71) for as π A force balance in the direction of takeoff gives ρ ⑴ Since we are looking for distance, not time, we introduce into Eq. ⑴, separate variables, and integrate: or ⑵ - 30 -
여기서 m = 75000/32.2 = 2329 slug - 31 -